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Class 12th Chapter -7 Integrals | NCERT Maths Solution | NCERT Solution | Edugrown

It is important that you have a strong foundation in Maths Solutions. This can help you immensely in exams. Because there is no paper pattern set for the board exams. Every year there are different twists and turns in questions asked in board exams. While preparing for maths it becomes to go through every bit of detail. Thus, NCERT Solutions Class 12 Maths can help you lay a strong foundation. There solutions of each and every question given in the NCERT 12 Textbook in this article. There are crucial chapters like integration, algebra, and calculus which you can solve easily using these Class 12 Maths solutions. Using these NCERT Maths Class 12 Solutions can give you an edge over the other students.

NCERT Solutions for Class 12 Maths Chapter : 7 Integrals

Class 12 Maths Chapter 7 Integrals Ex 7.1

Find an antiderivative (or integral) of the following by the method of inspection:

Ex 7.1 Class 12 Maths Question 1.
sin 2x
Solution:
$\int { sin2x\quad dx=-\frac { cos2x }{ 2 } +C }$

Ex 7.1 Class 12 Maths Question 2.
cos 3x
Solution:
$\int { cos3x\quad dx=\frac { sin3x }{ 3 } +C }$

Ex 7.1 Class 12 Maths Question 3.
${ e }^{ 2x }$
Solution:
$\int { { e }^{ 2x }dx=\frac { { e }^{ 2x } }{ 2 } +C }$

Ex 7.1 Class 12 Maths Question 4.
(ax + c)²
Solution:
$\int { { (ax+b) }^{ 2 }dx=\frac { { (ax+b) }^{ 3 } }{ 3a } } +C$

Ex 7.1 Class 12 Maths Question 5.
${ sin\quad 2x-4e }^{ 3x }$
Solution:
$\int { \left( { sin2x-4e }^{ 3x } \right) dx=-\frac { cos2x }{ 2 } -\frac { { 4e }^{ 3x } }{ 3 } +C }$

Find the following integrals in Exercises 6 to 20 :

Ex 7.1 Class 12 Maths Question 6.
$\int { \left( { 4e }^{ 3x }+1 \right) dx }$
Solution:
$=\int { { 4e }^{ 3x }dx+\int { dx=\frac { 4 }{ 3 } { e }^{ 3x }+x+c } }$

Ex 7.1 Class 12 Maths Question 7.
$\int { { x }^{ 2 }\left( 1-\frac { 1 }{ { x }^{ 2 } } \right) dx }$
Solution:
$=\int { { x }^{ 2 }\left( 1-\frac { 1 }{ { x }^{ 2 } } \right) dx } =\frac { { x }^{ 3 } }{ 3 } -x+C$

Ex 7.1 Class 12 Maths Question 8.
$\int { { (ax }^{ 2 }+bx+c)dx }$
Solution:
$=\frac { { ax }^{ 3 } }{ 3 } +\frac { { bx }^{ 2 } }{ 2 } +cx+d$

Ex 7.1 Class 12 Maths Question 9.
$\int { \left( { 2x }^{ 2 }+{ e }^{ x } \right) dx }$
Solution:
$=\frac { { 2x }^{ 3 } }{ 3 } +{ e }^{ x }+c$

Ex 7.1 Class 12 Maths Question 10.
$\int { { \left[ \sqrt { x } -\frac { 1 }{ \sqrt { x } } \right] }^{ 2 }dx }$
Solution:
$=\frac { { x }^{ 2 } }{ 2 } +logx-2x+C$

Ex 7.1 Class 12 Maths Question 11.
$\int { \frac { { x }^{ 3 }+{ 5x }^{ 2 }-4 }{ { x }^{ 2 } } dx }$
Solution:
$\int { \left( \frac { { x }^{ 3 } }{ { x }^{ 2 } } +\frac { { 5x }^{ 2 } }{ { x }^{ 2 } } -\frac { 4 }{ { x }^{ 2 } } \right) }$
$=\int { xdx+5\int { 1dx-4 } \int { { x }^{ 2 }dx } }$
$=\frac { { x }^{ 2 } }{ 2 } +5x+\frac { 4 }{ x } +c$

Ex 7.1 Class 12 Maths Question 12.
$\int { \frac { { x }^{ 3 }+3x+4 }{ \sqrt { x } } dx }$
Solution:
$=\int { \left( { x }^{ \frac { 5 }{ 2 } }+{ 3x }^{ \frac { 1 }{ 2 } }+4{ x }^{ -\frac { 1 }{ 2 } } \right) } dx$
$=\frac { 2 }{ 7 } { x }^{ \frac { 7 }{ 2 } }+{ 2x }^{ \frac { 3 }{ 2 } }+8\sqrt { x } +c$

Ex 7.1 Class 12 Maths Question 13.
$\int { \frac { { x }^{ 3 }-{ x }^{ 2 }+x-1 }{ x-1 } dx }$
Solution:
$=\int { \frac { { x }^{ 2 }(x-1)+(x-1) }{ x-1 } dx }$
$=\int { \left( { x }^{ 2 }+1 \right) dx } =\frac { { x }^{ 3 } }{ 3 } +x+c$

Ex 7.1 Class 12 Maths Question 14.
$\int { \left( 1-x \right) \sqrt { x } dx }$
Solution:
$=\int { { x }^{ \frac { 1 }{ 2 } }-{ x }^{ \frac { 3 }{ 2 } }dx\quad =\quad \frac { 2 }{ 3 } { x }^{ \frac { 3 }{ 2 } }-\frac { 2 }{ 5 } { x }^{ \frac { 5 }{ 2 } } }$

Ex 7.1 Class 12 Maths Question 15.
$\int { \sqrt { x } \left( { 3x }^{ 2 }+2x+3 \right) dx }$
Solution:
$=\int { \left( { 3x }^{ \frac { 5 }{ 2 } }+{ 2 }^{ \frac { 3 }{ 2 } }+{ 3x }^{ \frac { 1 }{ 2 } } \right) dx }$
$=\frac { 6 }{ 7 } { x }^{ \frac { 7 }{ 2 } }+\frac { 4 }{ 5 } { x }^{ \frac { 5 }{ 2 } }+\frac { 6 }{ 3 } { x }^{ \frac { 3 }{ 2 } }+c$

Ex 7.1 Class 12 Maths Question 16.
$\int { (2x-3cosx+{ e }^{ x })dx }$
Solution:
$=\frac { { 2x }^{ 2 } }{ 2 } -3sinx+{ e }^{ x }+c$
$={ x }^{ 2 }-3sinx+{ e }^{ x }+c$

Ex 7.1 Class 12 Maths Question 17.
$\int { \left( { 2x }^{ 2 }-3sinx+5\sqrt { x } \right) dx }$
Solution:
$=\frac { { 2x }^{ 3 } }{ 3 } +3cosx+5\frac { { x }^{ \frac { 3 }{ 2 } } }{ \frac { 3 }{ 2 } } +c$
$=\frac { 2 }{ 3 } { x }^{ 3 }+3cosx+\frac { 10 }{ 3 } { x }^{ \frac { 3 }{ 2 } }+c$

Ex 7.1 Class 12 Maths Question 18.
$\int { secx(secx+tanx)dx }$
Solution:
$=\int { { (sec }^{ 2 }x+secxtanx)dx }$
= tanx + secx + c

Ex 7.1 Class 12 Maths Question 19.
$\int { \frac { { sec }^{ 2 }x }{ { cosec }^{ 2 }x } dx }$
Solution:
$=\int { \frac { 1 }{ { cos }^{ 2 }x } } { sin }^{ 2 }xdx$
$=\int { tan } ^{ 2 }xdx\quad =\int { { (sec }^{ 2 }x-1)dx\quad =tanx-x+c }$

Ex 7.1 Class 12 Maths Question 20.
$\int { \frac { 2-3sinx }{ { cos }^{ 2 }x } dx }$
Solution:
$=\int { \left( \frac { 2 }{ { cos }^{ 2 }x } -3\frac { sinx }{ { cos }^{ 2 }x } \right) dx }$
$=\int { ({ 2sec }^{ 2 }x-3secxtanx)dx }$
= 2tanx – 3secx + c

Choose the correct answer in Exercises 21 and 22.

Ex 7.1 Class 12 Maths Question 21.
The antiderivative $\left( \sqrt { x } +\frac { 1 }{ \sqrt { x } } \right)$ equals
(a) $\frac { 1 }{ 3 } { x }^{ \frac { 1 }{ 3 } }+{ 2x }^{ \frac { 1 }{ 2 } }+c$
(b) $\frac { 2 }{ 3 } { x }^{ \frac { 2 }{ 3 } }+{ \frac { 1 }{ 2 } x }^{ 2 }+c$
(c) $\frac { 2 }{ 3 } { x }^{ \frac { 3 }{ 2 } }+{ 2x }^{ \frac { 1 }{ 2 } }+c$
(d) $\frac { 3 }{ 2 } { x }^{ \frac { 3 }{ 2 } }+\frac { 1 }{ 2 } { x }^{ \frac { 1 }{ 2 } }+c$
Solution:
(c) $\int { \left( \sqrt { x } +\frac { 1 }{ \sqrt { x } } \right) dx }$
$=\int { \left( { x }^{ \frac { 1 }{ 2 } }+{ x }^{ \frac { 1 }{ 2 } } \right) dx }$
$=\frac { 2 }{ 3 } { x }^{ \frac { 3 }{ 2 } }+{ 2x }^{ \frac { 1 }{ 2 } }+c$

Ex 7.1 Class 12 Maths Question 22.
If $\frac { d }{ dx } f(x)={ 4x }^{ 3 }-\frac { 3 }{ { x }^{ 4 } }$ such that f(2)=0 then f(x) is
(a) ${ x }^{ 4 }+\frac { 1 }{ { x }^{ 3 } } -\frac { 129 }{ 8 }$
(b) ${ x }^{ 3 }+\frac { 1 }{ { x }^{ 4 } } +\frac { 129 }{ 8 }$
(c) ${ x }^{ 4 }+\frac { 1 }{ { x }^{ 3 } } +\frac { 129 }{ 8 }$
(d) ${ x }^{ 3 }+\frac { 1 }{ { x }^{ 4 } } -\frac { 129 }{ 8 }$
Solution:
(a) $f(x)=\int { \left( { 4x }^{ 3 }-\frac { 3 }{ { x }^{ 4 } } \right) dx }$
$={ x }^{ 4 }+\frac { 1 }{ { x }^{ 3 } } +c$
$\therefore f(2)={ (2) }^{ 4 }+\frac { 1 }{ { (2) }^{ 3 } } +c=0=-\frac { 129 }{ 8 }$

Class 12 Maths Chapter 7 Integrals Ex 7.2

Integrate the functions in Exercises 1 to 37:

Ex 7.1 Class 12 Maths Question 1.
$\frac { 2x }{ 1+{ x }^{ 2 } }$
Solution:
Let 1+x² = t
⇒ 2xdx = dt
$\therefore \int { \frac { 2x }{ 1+{ x }^{ 2 } } dx\quad = } \int { \frac { dt }{ t } \quad =logt+C\quad =log(1+{ x }^{ 2 })+C }$

Ex 7.2 Class 12 Maths Question 2.
$\frac { { \left( logx \right) }^{ 2 } }{ x }$
Solution:
Let logx = t
⇒ $\frac { 1 }{ x }dx=dt$
$\therefore \int { \frac { { (logx) }^{ 2 } }{ x } dx } \quad =\int { { t }^{ 2 }dt } \quad =\frac { { t }^{ 3 } }{ 3 } +c\quad =\frac { 1 }{ 3 } { (logx) }^{ 3 }+c$

Ex 7.2 Class 12 Maths Question 3.
$\frac { 1 }{ x+xlogx }$
Solution:
Put 1+logx = t
∴ $\frac { 1 }{ x }dx=dt$
$\int { \frac { 1 }{ x(1+logx) } dx } =\int { \frac { 1 }{ t } dt } =log|t|+c$
= log|1+logx|+c

Ex 7.2 Class 12 Maths Question 4.
sinx sin(cosx)
Solution:
Put cosx = t, -sinx dx = dt
$\int { sinx\quad sin(cosx)dx } =-\int { sin(cosx) } (-sinx)dx$
$=-\int { sint\quad dt } \quad =cost+c\quad =cos(cosx)+c$

Ex 7.2 Class 12 Maths Question 5.
sin(ax+b) cos(ax+b)
Solution:
let sin(ax+b) = t
⇒ cos(ax+b)dx = dt
$\therefore \int { sin(ax+b)cos(ax+b)dx } =\frac { 1 }{ a } \int { t\quad dt }$
$=\frac { 1 }{ a } .\frac { { t }^{ 2 } }{ 2 } +c\quad =\frac { 1 }{ 2a } { sin }^{ 2 }(ax+b)+C$

Ex 7.2 Class 12 Maths Question 6.
$\sqrt { ax+b }$
Solution:
$\int { \sqrt { ax+b } dx } \quad =\frac { 2 }{ 3a } { (ax+b) }^{ \frac { 3 }{ 2 } }+C$

Ex 7.2 Class 12 Maths Question 7.
$x\sqrt { x+2 }$
Solution:
Let x+2 = t²
⇒ dx = 2t dt
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.2 Q7.1

Ex 7.2 Class 12 Maths Question 8.
$x\sqrt { 1+{ 2x }^{ 2 } }$
Solution:
let 1+2x² = t²
⇒ 4x dx = 2t dt
$x\quad dx=\frac { t }{ 2 } dt\int { x\sqrt { 1+{ 2x }^{ 2 } } dx }$
$=\frac { 1 }{ 2 } \int { { t }^{ 2 }dt } =\frac { { t }^{ 3 } }{ 6 } +c=\frac { 1 }{ 6 } { ({ 1+2x }^{ 2 }) }^{ \frac { 3 }{ 2 } }+c$

Ex 7.2 Class 12 Maths Question 9.
$(4x+2)\sqrt { { x }^{ 2 }+x+1 }$
Solution:
let x²+x+1 = t
⇒(2x+1)dx = dt
$\therefore \int { (4x+1)\sqrt { { x }^{ 2 }+x+1 } dx } =2\int { \sqrt { t } dt }$
$=\frac { { 2t }^{ \frac { 3 }{ 2 } } }{ ^{ \frac { 3 }{ 2 } } } +c\quad =\frac { 4 }{ 3 } { t }^{ \frac { 3 }{ 2 } }+c\quad =\frac { 4 }{ 3 } { ({ x }^{ 2 }+x+1) }^{ \frac { 3 }{ 2 } }+c$

Ex 7.2 Class 12 Maths Question 10.
$\frac { 1 }{ x-\sqrt { x } }$
Solution:
$\int { \frac { 1 }{ x-\sqrt { x } } dx } =\int { \frac { 1 }{ \sqrt { x } (\sqrt { x-1 } ) } dx } =I$
Let √x-1 = t
$\frac { 1 }{ 2 } { x }^{ -\frac { 1 }{ 2 } }dx=dt$
$I=2\int { \frac { dt }{ t } }$
= 2logt + c
= 2log(√x-1)+c

Ex 7.2 Class 12 Maths Question 11.
$\frac { x }{ \sqrt { x+4 } } ,x>0$
Solution:
let x+4 = t
⇒ dx = dt, x = t-4
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.2 Q11.1

Ex 7.2 Class 12 Maths Question 12.
${ { (x }^{ 3 }-1) }^{ \frac { 1 }{ 3 } }.{ x }^{ 5 }$
Solution:
$\int { { { (x }^{ 3 }-1) }^{ \frac { 1 }{ 3 } }.{ x }^{ 5 }.dx } \quad =\frac { 1 }{ 7 } { { (x }^{ 3 }-1) }^{ \frac { 7 }{ 3 } }+\frac { 1 }{ 4 } { { (x }^{ 3 }-1) }^{ \frac { 4 }{ 3 } }+c$

Ex 7.2 Class 12 Maths Question 13.
$\frac { { x }^{ 2 } }{ { { (2+3x }^{ 3 }) }^{ 3 } }$
Solution:
Let 2+3x³ = t
⇒ 9x² dx = dt
$\therefore \int { \frac { { x }^{ 2 } }{ { { (2+3x }^{ 3 }) }^{ 3 } } dx } =\frac { 1 }{ 9 } \int { \frac { dt }{ { t }^{ 3 } } =\frac { 1 }{ 9 } \int { { t }^{ -3 }dt } }$
$=-\frac { 1 }{ { 18t }^{ 2 } } +c\quad =-\frac { 1 }{ 18(2+{ 3x }^{ 3 })^{ 2 } } +c$

Ex 7.2 Class 12 Maths Question 14.
$\frac { 1 }{ x(logx)^{ m } } ,x>0$
Solution:
Put log x = t, so that $\frac { 1 }{ x }dx=dt$
$\therefore \int { \frac { 1 }{ { x(logx) }^{ m } } dx } =\int { \frac { dt }{ { t }^{ m } } =\frac { { t }^{ -m+1 } }{ -m+1 } +c }$
$=\frac { { (logx) }^{ 1-m } }{ 1-m } +c$

Ex 7.2 Class 12 Maths Question 15.
$\frac { x }{ 9-4{ x }^{ 2 } }$
Solution:
put 9-4x² = t, so that -8x dx = dt
$\therefore \int { \frac { x }{ 9-{ 4x }^{ 2 } } dx } =-\frac { 1 }{ 8 } \int { \frac { dt }{ t } } =-\frac { 1 }{ 8 } log|t|+c$
$=\frac { 1 }{ 8 } log\frac { 1 }{ |9-{ 4x }^{ 2 }| } +c$

Ex 7.2 Class 12 Maths Question 16.
${ e }^{ 2x+3 }$
Solution:
put 2x+3 = t
so that 2dx = dt
$\int { { e }^{ 2x+3 } } dx\quad =\frac { 1 }{ 2 } \int { { e }^{ t }dt } \quad =\frac { 1 }{ 2 } { e }^{ t }+c\quad =\frac { 1 }{ 2 } { e }^{ 2x+3 }+c$

Ex 7.2 Class 12 Maths Question 17.
$\frac { x }{ { e }^{ { x }^{ 2 } } }$
Solution:
Let x² = t
⇒ 2xdx = dt ⇒ $xdx=\frac { dt }{ 2 }$
$\therefore \int { \frac { x }{ { e }^{ { x }^{ 2 } } } dx } \quad =\frac { 1 }{ 2 } \int { \frac { dt }{ { e }^{ t } } \quad =\frac { 1 }{ 2 } \int { { e }^{ -t } } dt }$
$=-\frac { 1 }{ 2 } { e }^{ { -x }^{ 2 } }+c$

Ex 7.2 Class 12 Maths Question 18.
$\frac { { e }^{ { tan }^{ -1 }x } }{ 1+{ x }^{ 2 } }$
Solution:
$let\quad { tan }^{ -1 }x=t\Rightarrow \frac { 1 }{ 1+{ x }^{ 2 } } dx=dt$
$\therefore \int { \frac { { e }^{ { tan }^{ -1 }x } }{ 1+{ x }^{ 2 } } dx } \quad =\int { { e }^{ t }dt\quad ={ e }^{ { tan }^{ -1 }x }+c }$

Ex 7.2 Class 12 Maths Question 19.
$\frac { { e }^{ 2x }-1 }{ { e }^{ 2x }+1 }$
Solution:
$\int { \frac { { e }^{ 2x }-1 }{ { e }^{ 2x }+1 } dx\quad =\int { \frac { { e }^{ x }({ e }^{ x }-{ e }^{ -x }) }{ { e }^{ x }({ e }^{ x }+{ e }^{ -x }) } dx=I } }$
put e^x+e^-x = t
so that (e^x-e^-x)dx = dt
$\therefore I=\int { \frac { dt }{ t } =log|t|+c } =log|{ e }^{ x }+{ e }^{ -x }|+c$

Ex 7.2 Class 12 Maths Question 20.
$\frac { { e }^{ 2x }-{ e }^{ 2x } }{ { e }^{ 2x }+{ e }^{ -2x } }$
Solution:
put e^2x-e^-2x = t
so that (2e^2x-2e^-2x)dx = dt
$\therefore \int { \frac { { e }^{ 2x }-{ e }^{ 2x } }{ { e }^{ 2x }+{ e }^{ -2x } } } dx=\frac { 1 }{ 2 } \int { \frac { 1 }{ t } dt } =\frac { 1 }{ 2 } log|t|+c$
$=\frac { 1 }{ 2 } log+|{ e }^{ 2x }+{ e }^{ -2x }|+c$

Ex 7.2 Class 12 Maths Question 21.
tan²(2x-3)
Solution:
∫tan²(2x-3)dx = ∫[sec²(2x-3)-1]dx = I
put 2x-3 = t
so that 2dx = dt
I = $\frac { 1 }{ 2 }$ ∫sec²t dt-x+c
= $\frac { 1 }{ 2 }t-x+c$
= $\frac { 1 }{ 2 }tan(2x-3)-x+c$

Ex 7.2 Class 12 Maths Question 22.
sec²(7-4x)
Solution:
∫sec²(7-4x)dx
= $\frac { tan(7-4x) }{ -4 }+c$

Ex 7.2 Class 12 Maths Question 23.
$\frac { { sin }^{ -1 }x }{ \sqrt { 1-{ x }^{ 2 } } }$
Solution:
$let\quad { sin }^{ -1 }x=t\quad \Rightarrow \frac { 1dx }{ \sqrt { 1-{ x }^{ 2 } } } =dt$
$\int { \frac { { sin }^{ -1 }x }{ \sqrt { 1-{ x }^{ 2 } } } dx } =\int { t\quad dt } =\frac { 1 }{ 2 } { t }^{ 2 }+c=\frac { 1 }{ 2 } { { (sin }^{ -1 }x) }^{ 2 }+c$

Ex 7.2 Class 12 Maths Question 24.
$\frac { 2cosx-3sinx }{ 6cosx+4sinx }$
Solution:
put 2sinx+4cosx = t
⇒ (2cosx-3sinx)dx = dt
$\frac { 1 }{ 2 } \int { \frac { 2cosx-3sinx }{ 2sinx+3cosx } dx } =\frac { 1 }{ 2 } \int { \frac { dt }{ t } } =\frac { 1 }{ 2 } log|t|+c$
$\frac { 1 }{ 2 } log|2sinx+3cosx|+c$

Ex 7.2 Class 12 Maths Question 25.
$\frac { 1 }{ { cos }^{ 2 }x{ (1-tanx) }^{ 2 } }$
Solution:
put 1-tanx = t
so that -sec²x dx = dt
$\therefore \int { \frac { 1 }{ { cos }^{ 2 }x{ (1-tanx) }^{ 2 } } dx } =\int { \frac { { sec }^{ 2 }x }{ { (1-tanx) }^{ 2 } } dx }$
$=-\int { \frac { dt }{ { t }^{ 2 } } } =\frac { 1 }{ t } +c=\frac { 1 }{ (1-tanx) } +c$

Ex 7.2 Class 12 Maths Question 26.
$\frac { cos\sqrt { x } }{ \sqrt { x } }$
Solution:
$put\quad \sqrt { x } =t,so\quad that\frac { 1 }{ 2\sqrt { x } } dx=dt$
$\therefore \int { \frac { cos\sqrt { x } }{ \sqrt { x } } } dx\quad =\quad 2\quad =\int { cost\quad dt\quad = } 2sint+c$
= 2sin√x+c

Ex 7.2 Class 12 Maths Question 27.
$\sqrt { sin2x } cos2x$
Solution:
put sin2x = t²
⇒ cos2x dx = t dt
$\therefore \int { \sqrt { sin2x } .cos2x\quad dx } \quad =\int { t.tdt=\frac { { t }^{ 3 } }{ 3 } +c }$
$=\frac { { (sin2x) }^{ \frac { 3 }{ 2 } } }{ 3 } +c$

Ex 7.2 Class 12 Maths Question 28.
$\frac { cosx }{ \sqrt { 1+sinx } }$
Solution:
put 1+sinx = t²
⇒cosx dx = 2t dt
$\therefore \int { \frac { cosx }{ \sqrt { 1+sinx } } dx } =2\int { dt } =2t+c$
$=2\sqrt { 1+sinx } +c$

Ex 7.2 Class 12 Maths Question 29.
cotx log sinx
Solution:
put log sinx = t,
⇒ cot x dx = dt
$\therefore \int { cot\quad logsinx\quad dx } =\int { t } dt\quad =\frac { { t }^{ 2 } }{ 2 } +c$
$=\frac { 1 }{ 2 } { (log\quad sinx) }^{ 2 }+c$

Ex 7.2 Class 12 Maths Question 30.
$\frac { sinx }{ 1+cosx }$
Solution:
put 1+cosx = t
⇒ -sinx dx = dt
$\therefore \int { \frac { sinx }{ 1+cosx } dx } =\int { -\frac { dt }{ t } } =-logt+c$
=-log(1+cosx)+c

Ex 7.2 Class 12 Maths Question 31.
$\frac { sinx }{ { (1+cosx) }^{ 2 } }$
Solution:
put 1+cosx = t
so that -sinx dx = dt
$\therefore \int { \frac { sinx }{ { (1+cosx) }^{ 2 } } dx } =-\int { \frac { dt }{ { t }^{ 2 } } }$
$=\frac { 1 }{ t } +c=\frac { 1 }{ 1+cosx } +c$

Ex 7.2 Class 12 Maths Question 32.
$\frac { 1 }{ 1+cotx }$
Solution:
$\int { \frac { 1 }{ 1+\frac { cosx }{ sinx } } } dx=\frac { 1 }{ 2 } \int { \frac { 2sinx\quad dx }{ sinx+cosx } }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.2 Q32.1

Ex 7.2 Class 12 Maths Question 33.
$\frac { 1 }{ 1-tanx }$
Solution:
$\int { \frac { 1 }{ 1-tanx } } dx=\frac { 1 }{ 2 } \int { \frac { 2cosx\quad dx }{ cosx-sinx } }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.2 Q33.1

Ex 7.2 Class 12 Maths Question 34.
$\frac { \sqrt { tanx } }{ sinxcosx }$
Solution:
$\int { \frac { \sqrt { tanx } }{ sinxcosx } dx } =\int { \frac { \sqrt { tanx } }{ tanx } } .{ sec }^{ 2 }xdx$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.2 Q34.1

Ex 7.2 Class 12 Maths Question 35.
$\frac { { (1+logx) }^{ 2 } }{ x }$
Solution:
let 1+logx = t
⇒ $\frac { 1 }{ x }dx=dt$
$\int { \frac { { (1+logx) }^{ 2 } }{ x } dx } =\int { { t }^{ 2 }dt } =\frac { { t }^{ 3 } }{ 3 } +c$
$=\frac { 1 }{ 3 } { (1+logx) }^{ 3 }+c$

Ex 7.2 Class 12 Maths Question 36.
$\frac { (x+1){ (x+logx) }^{ 2 } }{ x }$
Solution:
put x+logx = t
$\left( \frac { x+1 }{ x } \right) dx=dt$
$\therefore \int { \frac { { (x+1)(x+logx) }^{ 2 } }{ x } } dx=\int { { t }^{ 2 }dt }$
$=\frac { { (x+logx) }^{ 3 } }{ 3 } +c$

Ex 7.2 Class 12 Maths Question 37.
$\frac { { x }^{ 3 }sin({ tan }^{ -1 }{ x }^{ 4 }) }{ 1+{ x }^{ 8 } } dx$
Solution:
$put\quad { tan }^{ -1 }{ x }^{ 4 }=t\quad so\quad that\frac { 1 }{ 1+{ x }^{ 8 } } .{ 4x }^{ 3 }dx=dt$
$\therefore \int { \frac { { x }^{ 3 }sin({ tan }^{ -1 }{ x }^{ 4 }) }{ 1+{ x }^{ 8 } } } dx=\frac { 1 }{ 4 } \int { sint\quad dt }$
$=\frac { 1 }{ 4 } (-cost)+c=-\frac { 1 }{ 4 } cos({ tan }^{ -1 }{ x }^{ 4 })+c$

Choose the correct answer in exercises 38 and 39

Ex 7.2 Class 12 Maths Question 38.
$\int { \frac { { 10x }^{ 9 }+{ 10 }^{ x }log{ e }^{ 10 } }{ { x }^{ 10 }+{ 10 }^{ x } } dx }$
(a) 10^x – x¹⁰ + C
(b) 10^x + x¹⁰ + C
(c) (10^x – x¹⁰) + C
(d) log (10^x + x¹⁰) + C
Solution:
(d) $\int { \frac { { 10x }^{ 9 }+{ 10 }^{ x }log{ e }^{ 10 } }{ { x }^{ 10 }+{ 10 }^{ x } } dx }$
= log (10^x + x¹⁰) + C

Ex 7.2 Class 12 Maths Question 39.
$\int { \frac { dx }{ { sin }^{ 2 }x{ \quad cos }^{ 2 }x } = }$
(a) tanx + cotx + c
(b) tanx – cotx + c
(c) tanx cotx + c
(d) tanx – cot2x + c
Solution:
(c) $\int { \frac { dx }{ { sin }^{ 2 }x{ \quad cos }^{ 2 }x } = }$
$=\int { \left( { sec }^{ 2 }x+{ cosec }^{ 2 }x \right) dx }$
= tanx – cotx + c

Class 12 Maths Chapter 7 Integrals Ex 7.3

Find the integrals of the functions in Exercises 1 to 22.

Ex 7.3 Class 12 Maths Question 1.
sin²(2x+5)
Solution:
∫sin²(2x+5)dx
= $\frac { 1 }{ 2 }$ ∫[1-cos2(2x+5)]dx
= $\frac { 1 }{ 2 }$ ∫[1-cos(4x+10)]dx
= $\frac { 1 }{ 2 } \left[ x-\frac { sin(4x+10) }{ 4 } \right] +c$

Ex 7.3 Class 12 Maths Question 2.
sin3x cos4x
Solution:
∫sin3x cos4x
= $\frac { 1 }{ 2 }$ ∫[sin(3x+4x)+cos(3x-4x)]dx
= $\frac { 1 }{ 2 }$ ∫[sin7x+sin(-x)]dx
= $-\frac { 1 }{ 14 } cos7x+\frac { 1 }{ 2 } cosx+c$

Ex 7.3 Class 12 Maths Question 3.
∫cos2x cos4x cos6x dx
Solution:
$\frac { 1 }{ 2 }$ ∫cos2x cos4x cos6x dx
= $\frac { 1 }{ 2 }$ ∫(cos6x+cos2x) cos6x dx
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.3 Q3.1

Ex 7.3 Class 12 Maths Question 4.
∫sin³(2x+1)dx
Solution:
= $\frac { 1 }{ 4 }$ ∫[3sin(2x+1)-sin3(2x+1)]dx
= $-\frac { 3 }{ 8 } cos(2x+1)+\frac { 1 }{ 24 } [4{ cos }^{ 3 }(2x+1)-3cos(2x+1)]+c$
= $-\frac { 1 }{ 2 } cos(2x+1)+\frac { 1 }{ 6 } { cos }^{ 3 }(2x+1)+c$

Ex 7.3 Class 12 Maths Question 5.
sin³x cos³x
Solution:
put sin x = t
⇒ cos x dx = dt
$\therefore \int { { sin }^{ 3 }x{ cos }^{ 3 }xdx } =\int { { t }^{ 3 }(1-{ t }^{ 2 })dt }$
$\frac { { t }^{ 4 } }{ 4 } -\frac { { t }^{ 6 } }{ 6 } +c=\frac { { (sinx) }^{ 4 } }{ 4 } -\frac { { (sinx) }^{ 6 } }{ 6 } +c$

Ex 7.3 Class 12 Maths Question 6.
sinx sin2x sin3x
Solution:
∫sinx sin2x sin3x dx
= $\frac { 1 }{ 2 }$ ∫ 2sin x sin 2x sin 3x dx
= $\frac { 1 }{ 2 }$ ∫ (cosx – cos3x)sin 3x dx
= $\frac { 1 }{ 2 }$ ∫ (sin 4x + sin 2x – sin 6x)dx
= $\frac { 1 }{ 4 } \left\{ \frac { -cos4x }{ 4 } -\frac { cos2x }{ 2 } +\frac { cos6x }{ 6 } \right\} +c$

Ex 7.3 Class 12 Maths Question 7.
sin 4x sin 8x
Solution:
$\frac { 1 }{ 2 }$ ∫sin 4x sin 8xdx
= $\frac { 1 }{ 2 }$ ∫(cos 4x – cos 12x)dx
= $\frac { 1 }{ 2 } \left[ \frac { sin4x }{ 4 } -\frac { sin12x }{ 12 } \right] +c$

Ex 7.3 Class 12 Maths Question 8.
$\frac { 1-cosx }{ 1+cosx }$
Solution:
$\int { \frac { 1-cosx }{ 1+cosx } dx }$
$\int { \frac { { 2sin }^{ 2 }\frac { x }{ 2 } }{ { 2cos }^{ 2 }\frac { x }{ 2 } } dx } =\int { { tan }^{ 2 }\frac { x }{ 2 } dx }$
$=\int { \left[ { sec }^{ 2 }\frac { x }{ 2 } -1 \right] } dx\quad =2tan\frac { x }{ 2 } -x+c$

Ex 7.3 Class 12 Maths Question 9.
$\frac { cosx }{ 1+cosx }$
Solution:
$\int { \frac { cosx }{ 1+cosx } dx }$
$=\int { 1 } dx-\int { \frac { 1 }{ 1+cosx } dx }$
$=x-\frac { 1 }{ 2 } \int { { sec }^{ 2 }\frac { x }{ 2 } dx+c\quad =x-tan\frac { x }{ 2 } +c }$

Ex 7.3 Class 12 Maths Question 10.
∫sinx⁴ dx
Solution:
$\int { { (\frac { 1-cos2x }{ 2 } ) }^{ 2 }dx } \quad =\frac { 1 }{ 4 } \int { \left( { 1+cos }^{ 2 }2x-2cos2x \right) dx }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.3 Q10.1

Ex 7.3 Class 12 Maths Question 11.
cos⁴ 2x
Solution:
∫ cos⁴ 2x dx
$\int { { \left( \frac { 1+cos4x }{ 2 } \right) }^{ 2 } } dx$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.3 Q11.1

Ex 7.3 Class 12 Maths Question 12.
$\frac { { sin }^{ 2 }x }{ 1+cosx }$
Solution:
$\int { \frac { { sin }^{ 2 }x }{ 1+cosx } } dx\quad =\int { \frac { 1-{ cos }^{ 2 }x }{ 1+cosx } } dx$
$\int { (1-cosx) } dx\quad =x-sinx+c$

Ex 7.3 Class 12 Maths Question 13.
$\frac { cos2x-cos2\alpha }{ cosx-cos\alpha }$
Solution:
let I = $\int { \frac { \left( { 2cos }^{ 2 }x-1 \right) -\left( { 2cos }^{ 2 }\alpha -1 \right) }{ cosx-cos\alpha } } dx$
$\int { \frac { 2\left( { cos }x-cos\alpha \right) -\left( { cos }x+cos\alpha \right) }{ cosx-cos\alpha } } dx$
= 2∫cos x dx + 2cos α∫dx
= 2(sinx+xcosα)+c

Ex 7.3 Class 12 Maths Question 14.
$\frac { cosx-sinx }{ 1+sin2x }$
Solution:
let I = $\int { \frac { cosx-sinx }{ 1+sin2x } } dx=\int { \frac { cosx-sinx }{ { (cosx+sinx) }^{ 2 } } dx }$
put cosx+sinx = t
⇒ (-sinx+cosx)dx = dt
$I=\int { \frac { dt }{ { t }^{ 2 } } } =-\frac { 1 }{ t } +c\quad =\frac { -1 }{ cosx+sinx } +c$

Ex 7.3 Class 12 Maths Question 15.
$\int { { tan }^{ 3 }2x\quad sec2x\quad dx=I }$
Solution:
I = ∫(sec²2x-1)sec2x tan 2xdx
put sec2x=t,2 sec2x tan2x dx=dt
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.3 Q15.1

Ex 7.3 Class 12 Maths Question 16.
tan⁴x
Solution:
let I = ∫tan⁴ dx
= ∫(sec²x-1)²dx
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.3 Q16.1
Ex 7.3 Class 12 Maths Question 17.
$\frac { { sin }^{ 3 }x+{ cos }^{ 3 }x }{ { sin }^{ 2 }x{ cos }^{ 2 }x }$
Solution:
$\int { \left( \frac { { sin }^{ 3 }x }{ { sin }^{ 2 }x{ cos }^{ 2 }x } +\frac { { cos }^{ 2 }x }{ sinx{ cos }^{ 2 }x } \right) dx }$
= secx-cosecx+c

Ex 7.3 Class 12 Maths Question 18.
$\frac { cos2x+{ 2sin }^{ 2 }x }{ { cos }^{ 2 }x }$
Solution:
$I=\int { \frac { \left( { cos }^{ 2 }x-{ sin }^{ 2 }x \right) +2{ sin }^{ 2 }x }{ { cos }^{ 2 }x } } dx$
$=\int { \frac { \left( { cos }^{ 2 }x-{ sin }^{ 2 }x \right) }{ { cos }^{ 2 }x } } dx\quad =\int { { sec }^{ 2 }xdx\quad =tanx+c }$

Ex 7.3 Class 12 Maths Question 19.
$\frac { 1 }{ sinx{ cos }^{ 3 }x }$
Solution:
$I=\int { \left( tanx+\frac { 1 }{ tanx } \right) } { sec }^{ 2 }xdx$
put tanx = t
so that sec²x dx = dt
$I=\int { \left( t+\frac { 1 }{ t } \right) } dt\quad =\frac { { t }^{ 2 } }{ 2 } +log|t|+c$
$=log|tanx|+\frac { 1 }{ 2 } { tan }^{ 2 }x+c$

Ex 7.3 Class 12 Maths Question 20.
$\frac { cos2x }{ { (cosx+sinx) }^{ 2 } }$
Solution:
$I=\int { \frac { { cos }^{ 2 }x-{ sin }^{ 2 }x }{ (cosx+sinx)^{ 2 } } dx } =\int { \frac { cosx-sinx }{ cosx+sinx } dx }$
put cosx+sinx=t
⇒(-sinx+cox)dx = dt
$I=\int { \frac { dt }{ t } } =log|t|+c\quad =log|cosx+sinx|+c$

Ex 7.3 Class 12 Maths Question 21.
sin^-1 (cos x)
Solution:
$\int { { sin }^{ -1 }(cosx)dx } \quad ={ sin }^{ -1 }\left[ sin\left( \frac { \pi }{ 2 } -x \right) \right] dx$
$\int { \left( \frac { \pi }{ 2 } -x \right) dx } \quad =\frac { \pi x }{ 2 } -\frac { { x }^{ 2 } }{ 2 } +c$

Ex 7.3 Class 12 Maths Question 22.
$\int { \frac { 1 }{ cos(x-a)cos(x-b) } dx }$
Solution:
$\frac { 1 }{ sin(a-b) } \int { \frac { sin[(x-b)-(x-a)] }{ cos(x-a)cos(x-b) } dx }$
$=\frac { 1 }{ sin(a-b) } \left[ \int { tan(x-b)dx-\int { tan(x-a)dx } } \right]$
$=\frac { 1 }{ sin(a-b) } log\left| \frac { cos(x-a) }{ cos(x-b) } \right| +c$

Ex 7.3 Class 12 Maths Question 23.
$\int { \frac { { sin }^{ 2 }x-{ cos }^{ 2 }x }{ { sin }^{ 2 }x{ cos }^{ 2 }x } } dx\quad is\quad equal\quad to$
(a) tanx+cotx+c
(b) tanx+cosecx+c
(c) -tanx+cotx+c
(d) tanx+secx+c
Solution:
(a) $\int { \frac { { sin }^{ 2 }x-{ cos }^{ 2 }x }{ { sin }^{ 2 }x{ cos }^{ 2 }x } } dx$
= ∫(sec²x-cosec²x)dx
= tanx+cotx+c

Ex 7.3 Class 12 Maths Question 24.
$\int { \frac { e^{ x }(1+x) }{ cos^{ 2 }({ e }^{ x }.{ x }) } } dx\quad is\quad equal\quad to$
(a) -cot(e.x^x)+c
(b) tan(xe^x)+c
(c) tan(e^x)+c
(d) cot e^x+c
Solution:
(b) $\int { \frac { e^{ x }(1+x) }{ cos^{ 2 }({ e }^{ x }.{ x }) } } dx$
= ∫sec²t dt
= tan t+c = tan(xe^x)+c

Class 12 Maths Chapter 7 Integrals Ex 7.4

Integrate the functions in exercises 1 to 23

Ex 7.4 Class 12 Maths Question 1.
$\frac { { 3x }^{ 2 } }{ { x }^{ 6 }+1 }$
Solution:
Let x³ = t ⇒ 3x²dx = dt
$\int { \frac { { 3x }^{ 2 } }{ { x }^{ 6 }+1 } dx } =\int { \frac { dt }{ { t }^{ 2 }+1 } } ={ tan }^{ -1 }t+c$
= tan^-1 (x³)+c

Ex 7.4 Class 12 Maths Question 2.
$\frac { 1 }{ \sqrt { 1+{ 4x }^{ 2 } } }$
Solution:
$\frac { 1 }{ 2 } \int { \frac { dx }{ \sqrt { \frac { 1 }{ 4 } +{ x }^{ 2 } } } } =\frac { 1 }{ 2 } \int { \frac { dx }{ \sqrt { { \left( \frac { 1 }{ 2 } \right) }^{ 2 }+{ x }^{ 2 } } } }$
$=\frac { 1 }{ 2 } log\left| 2x+\sqrt { 1+{ 4x }^{ 2 } } \right| +c$

Ex 7.4 Class 12 Maths Question 3.
$\frac { 1 }{ \sqrt { { (2-x) }^{ 2 }+1 } }$
Solution:
put (2-x)=t
so that -dx=dt
⇒ dx=-dt
$\int { \frac { dx }{ \sqrt { { (2-x) }^{ 2 }+1 } } } =-\int { \frac { dt }{ \sqrt { { t }^{ 2 }+1 } } } =-log|t+\sqrt { { t }^{ 2 }+1 } |+c$
$=log\left| \frac { 1 }{ (2-x)+\sqrt { { x }^{ 2 }-4x+5 } } \right| +c$

Ex 7.4 Class 12 Maths Question 4.
$\frac { 1 }{ \sqrt { 9-{ 25x }^{ 2 } } }$
Solution:
$\int { \frac { dx }{ \sqrt { 9-{ 25x }^{ 2 } } } } =\frac { 1 }{ 5 } \int { \frac { dx }{ \sqrt { { \left( \frac { 3 }{ 5 } \right) }^{ 2 }-{ x }^{ 2 } } } }$
$=\frac { 1 }{ 5 } { sin }^{ -1 }\left( \frac { x }{ \frac { 3 }{ 5 } } \right) +c\quad =\frac { 1 }{ 5 } { sin }^{ -1 }\left( \frac { 5x }{ 3 } \right) +c$

Ex 7.4 Class 12 Maths Question 5.
$\frac { 3x }{ 1+{ 2x }^{ 4 } }$
Solution:
Put x²=t,so that 2x dx=dt
⇒x dx = $\frac { dt }{ 2 }$
$\therefore \int { \frac { 3x }{ 1+{ 2x }^{ 4 } } dx } =\frac { 1 }{ 2 } \int { \frac { dt }{ 1+{ 2t }^{ 2 } } } =\frac { 3 }{ 4 } \int { \frac { dt }{ { \left( \frac { 1 }{ \sqrt { 2 } } \right) }^{ 2 }+{ t }^{ 2 } } }$
$=\frac { 3 }{ 2\sqrt { 2 } } { tan }^{ -1 }(\sqrt { 2t } )+c\quad =\frac { 3 }{ 2\sqrt { 2 } } { tan }^{ -1 }(\sqrt { { 2x }^{ 2 } } )+c$

Ex 7.4 Class 12 Maths Question 6.
$\frac { { x }^{ 2 } }{ 1-{ x }^{ 6 } }$
Solution:
put x³ = t,so that 3x²dx = dt
$\int { \frac { { x }^{ 2 } }{ 1-{ x }^{ 6 } } dx } \quad =\frac { 1 }{ 3 } \int { \frac { dt }{ 1-{ t }^{ 2 } } \quad =\frac { 1 }{ 6 } log } \left| \frac { 1+t }{ 1-t } \right| +c$
$=\frac { 1 }{ 6 } log\left| \frac { 1+{ x }^{ 3 } }{ 1-{ x }^{ 3 } } \right| +c$

Ex 7.4 Class 12 Maths Question 7.
$\frac { x-1 }{ \sqrt { { x }^{ 2 }-1 } }$
Solution:
$I=\int { \frac { x-1 }{ \sqrt { { x }^{ 2 }-1 } } dx } -\int { \frac { 1 }{ \sqrt { { x }^{ 2 }-1 } } dx } ,I={ I }_{ 1 }-{ I }_{ 2 }$
put x²-1 = t,so that 2x dx = dt
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 Q7.1

Ex 7.4 Class 12 Maths Question 8.
$\frac { { x }^{ 2 } }{ \sqrt { { x }^{ 6 }+{ a }^{ 6 } } }$
Solution:
put x³ = t
so that 3x²dx = dt
$I=\frac { 1 }{ 3 } \int { \frac { dt }{ { t }^{ 2 }+{ { (a }^{ 3 }) }^{ 2 } } =\frac { 1 }{ 3 } log\left| t+\sqrt { { t }^{ 2 }+{ a }^{ 6 } } \right| +c }$
$=\frac { 1 }{ 3 } log|{ x }^{ 3 }+\sqrt { { a }^{ 6 }+{ x }^{ 6 } } |+c$

Ex 7.4 Class 12 Maths Question 9.
$\frac { { sec }^{ 2 }x }{ \sqrt { { tan }^{ 2 }x+4 } }$
Solution:
let tanx = t
sec x²dx = dt
$I=\int { \frac { dt }{ \sqrt { { t }^{ 2 }+{ (2) }^{ 2 } } } } =log|t+\sqrt { { t }^{ 2 }+4 } |+c$
$=log|tanx+\sqrt { { tan }^{ 2 }x+4 } |+c$

Ex 7.4 Class 12 Maths Question 10.
$\frac { 1 }{ \sqrt { { x }^{ 2 }+2x+2 } }$
Solution:
$\int { \frac { 1 }{ \sqrt { { x }^{ 2 }+2x+2 } } dx } =\int { \frac { dx }{ \sqrt { { (x+1) }^{ 2 }+1 } } }$
$=log|(x+1)+\sqrt { { x }^{ 2 }+2x+2 } |+c$

Ex 7.4 Class 12 Maths Question 11.
$\frac { 1 }{ { 9x }^{ 2 }+6x+5 }$
Solution:
$\int { \frac { 1 }{ { 9x }^{ 2 }+6x+5 } } =\frac { 1 }{ 9 } \int { \frac { dx }{ { \left( x+\frac { 1 }{ 3 } \right) }^{ 2 }{ +\left( \frac { 2 }{ 3 } \right) }^{ 2 } } }$
$=\frac { 1 }{ 6 } { tan }^{ -1 }\left( \frac { 3x+1 }{ 2 } \right) +c$

Ex 7.4 Class 12 Maths Question 12.
$\frac { 1 }{ \sqrt { 7-6x-{ x }^{ 2 } } }$
Solution:
$I=\int { \frac { dx }{ \sqrt { { 4 }^{ 2 }-{ (x+3) }^{ 2 } } } } \quad ={ sin }^{ -1 }\left( \frac { x+3 }{ 4 } \right) +c$

Ex 7.4 Class 12 Maths Question 13.
$\frac { 1 }{ \sqrt { (x-1)(x-2) } }$
Solution:
$\int { \frac { 1 }{ \sqrt { (x-1)(x-2) } } dx } =\int { \frac { dx }{ \sqrt { { \left( x-\frac { 3 }{ 2 } \right) }^{ 2 }-{ \left( \frac { 1 }{ 2 } \right) }^{ 2 } } } }$
$=log\left| x-\frac { 3 }{ 2 } +\sqrt { { x }^{ 2 }-3x+2 } \right| +c$

Ex 7.4 Class 12 Maths Question 14.
$\frac { 1 }{ \sqrt { 8+3x-{ x }^{ 2 } } }$
Solution:
$\int { \frac { dx }{ \sqrt { 8+3x-{ x }^{ 2 } } } } =\int { \frac { dx }{ \sqrt { 8-\left( { x }^{ 2 }-3x \right) } } }$
$=\int { \frac { dx }{ \sqrt { { \left( \frac { \sqrt { 41 } }{ 2 } \right) }^{ 2 }-{ \left( x-\frac { 3 }{ 2 } \right) }^{ 2 } } } } \quad ={ sin }^{ -1 }\left( \frac { 2x-3 }{ \sqrt { 41 } } \right) +c$

Ex 7.4 Class 12 Maths Question 15.
$\frac { 1 }{ \sqrt { (x-a)(x-b) } }$
Solution:
$\int { \frac { dx }{ \sqrt { (x-a)(x-b) } } } =\int { \frac { dx }{ { \left( x-\frac { a+b }{ 2 } \right) }^{ 2 }-{ \left( \frac { a-b }{ 2 } \right) }^{ 2 } } }$
$=log\left| \left( x-\frac { a+b }{ 2 } \right) +\sqrt { (x-a)(x-b) } \right| +c$

Ex 7.4 Class 12 Maths Question 16.
$\frac { 4x+1 }{ \sqrt { { 2x }^{ 2 }+x-3 } }$
Solution:
$let\quad I=\int { \frac { 4x+1 }{ \sqrt { { 2x }^{ 2 }+x-3 } } } dx$
put 2x²+x-3=t
so that (4x+1)dx=dt
$let\quad I=\int { \frac { 4x+1 }{ \sqrt { { 2x }^{ 2 }+x-3 } } } dx$
$\therefore I=\int { \frac { dt }{ \sqrt { t } } } ={ 2t }^{ \frac { 1 }{ 2 } }+c\quad =2\sqrt { { 2x }^{ 2 }+x-3 } +c$

Ex 7.4 Class 12 Maths Question 17.
$\frac { x+2 }{ \sqrt { { x }^{ 2 }-1 } }$
Solution:
$\int { \frac { x+2 }{ \sqrt { { x }^{ 2 }-1 } } dx } \quad =\int { \frac { x }{ \sqrt { { x }^{ 2 }-1 } } dx } +\int { \frac { 2 }{ \sqrt { { x }^{ 2 }-1 } } dx }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 Q17.1

Ex 7.4 Class 12 Maths Question 18.
$\frac { 5x-2 }{ 1+2x+{ 3x }^{ 2 } }$
Solution:
put 5x-2=A $\frac { d }{ dx }$ (1+2x+3x²)+B
⇒ 6A=5, A= $\frac { 5 }{ 6 }-2=2A+B$ , B= $-\frac { 11 }{ 3 }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 Q18.1

Ex 7.4 Class 12 Maths Question 19.
$\frac { 6x+7 }{ \sqrt { (x-5)(x-4) } }$
Solution:
$\int { \frac { 6x+7 }{ \sqrt { (x-5)(x-4) } } dx } =\int { \frac { (6x+7)dx }{ \sqrt { { x }^{ 2 }-9x+20 } } }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 Q19.1

Ex 7.4 Class 12 Maths Question 20.
$\frac { x+2 }{ \sqrt { 4x-{ x }^{ 2 } } }$
Solution:
$I=\int { \frac { x-2 }{ \sqrt { 4-{ (x-2) }^{ 2 } } } dx+4\int { \frac { dx }{ \sqrt { 4-{ (x-2) }^{ 2 } } } } }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 Q20.1

Ex 7.4 Class 12 Maths Question 21.
$\frac { x+2 }{ \sqrt { { x }^{ 2 }+2x+3 } }$
Solution:
$I=\frac { 1 }{ 2 } \int { \frac { 2x+2 }{ \sqrt { { x }^{ 2 }+2x+3 } } dx }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 Q21.1

Ex 7.4 Class 12 Maths Question 22.
$\frac { x+3 }{ { x }^{ 2 }-2x-5 }$
Solution:
$I=\frac { 1 }{ 2 } \int { \frac { 2x-2 }{ { x }^{ 2 }-2x-5 } dx } +\int { \frac { dx }{ { x }^{ 2 }-2x-5 } }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 Q22.1

Ex 7.4 Class 12 Maths Question 23.
$\frac { 5x+3 }{ \sqrt { { x }^{ 2 }+4x+10 } }$
Solution:
$I=\int { \frac { \frac { 5 }{ 2 } (2x+4)+(3-10) }{ \sqrt { { x }^{ 2 }+4x+10 } } dx }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.4 Q23.1

Ex 7.4 Class 12 Maths Question 24.
$\int { \frac { dx }{ { x }^{ 2 }+2x+2 } equals }$
(a) xtan^-1(x+1)+c
(b) (x+1)tan^-1x+c
(c) tan^-1(x+1)+c
(d) tan^-1x+c
Solution:
(b) $let\quad I=\int { \frac { dx }{ { x }^{ 2 }+2x+2 } } =\int { \frac { dx }{ (x+1)^{ 2 }+1 } }$
= (x+1)tan^-1x+c

Ex 7.4 Class 12 Maths Question 25.
$\int { \frac { dx }{ \sqrt { 9x-{ 4x }^{ 2 } } } equals }$
(a) $\frac { 1 }{ 9 } { sin }^{ -1 }\left( \frac { 9x-8 }{ 8 } \right) +c$
(b) $\frac { 1 }{ 2 } { sin }^{ -1 }\left( \frac { 8x-9 }{ 9 } \right) +c$
(c) $\frac { 1 }{ 3 } { sin }^{ -1 }\left( \frac { 9x-8 }{ 8 } \right) +c$
(d) ${ sin }^{ -1 }\left( \frac { 9x-8 }{ 9 } \right) +c$
Solution:
(b) $\int { \frac { dx }{ \sqrt { 9x-{ 4x }^{ 2 } } } } =\frac { 1 }{ 2 } \left[ \frac { dx }{ \sqrt { \left( \frac { 9 }{ 8 } \right) ^{ 2 }-\left[ { x }^{ 2 }-{ \frac { 9 }{ 4 } }x+\left( \frac { 9 }{ 8 } \right) ^{ 2 } \right] } } \right]$
$\frac { 1 }{ 2 } { sin }^{ -1 }\left( \frac { 8x-9 }{ 9 } \right) +c$

Class 12 Maths Chapter 7 Integrals Ex 7.5

Integrate the rational function in exercises 1 to 21

Ex 7.5 Class 12 Maths Question 1.
$\frac { x }{ (x+1)(x+2) }$
Solution:
let $\frac { x }{ (x+1)(x+2) }$ ≡ $\frac { A }{ x+1 } +\frac { B }{ x+2 }$
⇒ x ≡ A(x+2)+B(x+1)….(i)
putting x = -1 & x = -2 in (i)
we get A = 1,B = 2
$\therefore \int { \frac { 1 }{ (x+1)(x+2) } dx } =\int { \frac { -1 }{ x+1 } dx } +\int { \frac { 2 }{ x+2 } dx }$
=-log|x+1| + 2log|x+2|+c

Ex 7.5 Class 12 Maths Question 2.
$\frac { 1 }{ { x }^{ 2 }-9 }$
Solution:
let $\frac { 1 }{ { x }^{ 2 }-9 } =\frac { 1 }{ (x-3)(x+3) } \equiv \frac { A }{ x-3 } +\frac { B }{ x+3 }$
⇒ x ≡ A(x+3)+B(x-3)…(i)
put x = 3, -3 in (i)
we get $A=\frac { 1 }{ 6 }$ & $B=-\frac { 1 }{ 6 }$
$\therefore \int { \frac { 1 }{ { x }^{ 2 }-9 } dx } =\frac { 1 }{ 6 } \int { \left[ \frac { 1 }{ x-3 } -\frac { 1 }{ x+3 } \right] dx }$
$=\frac { 1 }{ 6 } log\left| \frac { x-3 }{ x+3 } \right| +c$

Ex 7.5 Class 12 Maths Question 3.
$\frac { 3x-1 }{ (x-1)(x-2)(x-3) }$
Solution:
Let $\frac { 3x-1 }{ (x-1)(x-2)(x-3) } =\frac { A }{ x-1 } +\frac { B }{ x-2 } +\frac { C }{ x-3 }$
⇒ 3x-1 = A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(-2)…..(i)
put x = 1,2,3 in (i)
we get A = 1,B = -5 & C = 4
$\therefore I=\int { \frac { 1 }{ x-1 } dx } -5\int { \frac { 1 }{ x-2 } dx } +4\int { \frac { 1 }{ x-3 } dx }$
=log|x-1| – 5log|x-2| + 4log|x+3| + C

Ex 7.5 Class 12 Maths Question 4.
$\frac { x }{ (x-1)(x-2)(x-3) }$
Solution:
let $\frac { x }{ (x-1)(x-2)(x-3) } =\frac { A }{ x-1 } +\frac { B }{ x-2 } +\frac { C }{ x-3 }$
⇒ x ≡ A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)…(i)
put x = 1,2,3 in (i)
$A=\frac { 1 }{ 2 } ,B=-2,C=\frac { 3 }{ 2 }$
$\therefore I=\frac { 1 }{ 2 } \int { \frac { dx }{ x-1 } } -2\int { \frac { dx }{ x-2 } } +\frac { 3 }{ 2 } \int { \frac { dx }{ x-3 } }$
$=\frac { 1 }{ 2 } log|x-1|-2log|x-2|+\frac { 3 }{ 2 } log|x-3|+c$

Ex 7.5 Class 12 Maths Question 5.
$\frac { 2x }{ { x }^{ 2 }+3x+2 }$
Solution:
let $\frac { 2x }{ { x }^{ 2 }+3x+2 } =\frac { 2x }{ (x+1)(x+2) } =\frac { A }{ x+1 } +\frac { B }{ x+2 }$
⇒ 2x = A(x+2)+B(x+1)…(i)
put x = -1, -2 in (i)
we get A = -2, B = 4
$\therefore \int { \frac { 2x }{ { x }^{ 2 }+3x+2 } dx } =-2\int { \frac { dx }{ x+1 } } +4\int { \frac { dx }{ x+2 } }$
=-2log|x+1|+4log|x+2|+c

Ex 7.5 Class 12 Maths Question 6.
$\frac { 1-{ x }^{ 2 } }{ x(1-2x) }$
Solution:
$\frac { 1-{ x }^{ 2 } }{ (x-2{ x }^{ 2 }) }$ is an improper fraction therefore we
convert it into a proper fraction. Divide 1 – x² by x – 2x² by long division.
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5 Q6.1

Ex 7.5 Class 12 Maths Question 7.
$\frac { x }{ \left( { x }^{ 2 }+1 \right) \left( x-1 \right) }$
Solution:
let $\frac { x }{ \left( { x }^{ 2 }+1 \right) \left( x-1 \right) } =\frac { A }{ x-1 } +\frac { Bx+C }{ { x }^{ 2 }+1 }$
⇒ x = A(x²+1)+(Bx+C)(x-1)
Put x = 1,0
⇒ $A=\frac { 1 }{ 2 } C=\frac { 1 }{ 2 } \Rightarrow B=-\frac { 1 }{ 2 }$
$\therefore I=\frac { 1 }{ 2 } \int { \frac { dx }{ x-1 } } -\frac { 1 }{ 2 } \int { \frac { x }{ { x }^{ 2 }+1 } dx } +\frac { 1 }{ 2 } \int { \frac { dx }{ { x }^{ 2 }+1 } }$
$=\frac { 1 }{ 2 } log(x-1)-\frac { 1 }{ 4 } log({ x }^{ 2 }+1)+\frac { 1 }{ 2 } { tan }^{ -1 }x+c$

Ex 7.5 Class 12 Maths Question 8.
$\frac { x }{ { \left( x-1 \right) }^{ 2 }\left( x+2 \right) }$
Solution:
$\frac { x }{ { \left( x-1 \right) }^{ 2 }\left( x+2 \right) } =\frac { A }{ x-1 } +\frac { B }{ { \left( x-1 \right) }^{ 2 } } +\frac { C }{ x+2 }$
⇒ x ≡ A(x-1)(x+2)+B(x+2)+C(x-1)² …(i)
put x = 1, -2
we get $B=\frac { 1 }{ 3 } ,C=\frac { -2 }{ 9 }$
$\therefore I=\frac { 2 }{ 9 } \int { \frac { 1 }{ x-1 } dx } +\frac { 1 }{ 3 } \int { \frac { 1 }{ { (x-1) }^{ 2 } } dx } -\frac { 2 }{ 9 } \int { \frac { 1 }{ x+2 } dx }$
$=\frac { 2 }{ 9 } log\left| \frac { x-1 }{ x+2 } \right| -\frac { 1 }{ 3\left( x-1 \right) } +c$

Ex 7.5 Class 12 Maths Question 9.
$\frac { 3x+5 }{ { x }^{ 3 }-{ x }^{ 2 }-x+1 }$
Solution:
let $\frac { 3x+5 }{ { x }^{ 2 }(x-1)-1(x-1) }$
$\frac { 3x+5 }{ (x-1)^{ 2 }(x+1) } =\frac { A }{ x-1 } +\frac { B }{ { (x-1) }^{ 2 } } +\frac { C }{ x+1 }$
⇒ 3x+5 = A(x-1)(x+1)+B(x+1)+C(x-1)
put x = 1,-1,0
we get $B=4,C=\frac { 1 }{ 2 } ,A=-\frac { 1 }{ 2 }$
$\therefore I=-\frac { 1 }{ 2 } \int { \frac { dx }{ (x-1) } } +4\frac { dx }{ { (x-1) }^{ 2 } } +\frac { 1 }{ 2 } \int { \frac { dx }{ x+1 } }$
$=\frac { 1 }{ 2 } log\left| \frac { x+1 }{ x-1 } \right| -\frac { 4 }{ x-1 } +c$

Ex 7.5 Class 12 Maths Question 10.
$\frac { 2x-3 }{ ({ x }^{ 2 }-1)(2x+3) }$
Solution:
$\frac { 2x-3 }{ ({ x }^{ 2 }-1)(2x+3) } =\frac { 2x-3 }{ (x-1)(x+1)(2x+3) }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5 Q10.1

Ex 7.5 Class 12 Maths Question 11.
$\frac { 5x }{ (x-1)({ x }^{ 2 }-4) }$
Solution:
let $\frac { 5x }{ (x-1)({ x }^{ 2 }-4) } =\frac { 5x }{ (x+1)(x+2)(x-2) }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5 Q11.1

Ex 7.5 Class 12 Maths Question 12.
$\frac { { x }^{ 3 }+x+1 }{ { x }^{ 2 }-1 }$
Solution:
$\frac { { x }^{ 3 }+x+1 }{ { x }^{ 2 }-1 } =x+\frac { 2x+1 }{ (x+1)(x-1) }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5 Q12.1

Ex 7.5 Class 12 Maths Question 13.
$\frac { 2 }{ (1-x)(1+{ x }^{ 2 }) }$
Solution:
$\frac { 2 }{ (1-x)(1+{ x }^{ 2 }) } =\frac { A }{ 1-x } +\frac { Bx+C }{ 1+{ x }^{ 2 } }$
⇒ 2 = A(1+x²) + (Bx+C)(1 -x) …(i)
Putting x = 1 in (i), we get; A = 1
Also 0 = A – B and 2 = A + C ⇒B = A = 1 & C = 1
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5 Q13.1

Ex 7.5 Class 12 Maths Question 14.
$\frac { 3x-1 }{ { (x+2) }^{ 2 } }$
Solution:
$\frac { 3x-1 }{ { (x+2) }^{ 2 } } \equiv \frac { A }{ x+1 } +\frac { B }{ { (x+2) }^{ 2 } }$
=>3x – 1 = A(x + 2) + B …(i)
Comparing coefficients A = -1 and B = -7
$\therefore \int { \frac { 3x-1 }{ { (x+2) }^{ 2 } } dx } =3\int { \frac { dx }{ x+2 } } -7\int { \frac { dx }{ { (x+2) }^{ 2 } } }$
$=3log|x+2|+\frac { 7 }{ x+2 } +c$

Ex 7.5 Class 12 Maths Question 15.
$\frac { 1 }{ { x }^{ 4 }-1 }$
Solution:
$\frac { 1 }{ { x }^{ 4 }-1 } =\frac { A }{ x+1 } +\frac { B }{ x-1 } +\frac { Cx+D }{ { x }^{ 2 }+1 }$
⇒ 1 ≡ A(x-1)(x²+1) + B(x+1)(x²+1) + (Cx+D)(x+1)(x-1) ….(i)
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5 Q15.1

Ex 7.5 Class 12 Maths Question 16.
$\frac { 1 }{ x({ x }^{ n }+1) }$
[Hint : multiply numerator and denominator by x^n-1 and put xⁿ = t ]
Solution:
$\frac { { x }^{ n-1 } }{ x.{ x }^{ n-1 }({ x }^{ n }+1) } =\frac { { x }^{ n-1 } }{ { x }^{ n }({ x }^{ n }+1) }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5 Q16.1

Ex 7.5 Class 12 Maths Question 17.
$\frac { cosx }{ (1-sinx)(2-sinx) }$
Solution:
put sinx = t
so that cosx dx = dt
$\therefore I=\int { \frac { 1 }{ (1-t)(2-t) } dt }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5 Q17.1

Ex 7.5 Class 12 Maths Question 18.
$\frac { \left( { x }^{ 2 }+1 \right) \left( { x }^{ 2 }+2 \right) }{ \left( { x }^{ 2 }+3 \right) \left( { x }^{ 2 }+4 \right) }$
Solution:
put x²=y
$I=1-\frac { 2(2y+5) }{ (y+3)(y+4) }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5 Q18.1

Ex 7.5 Class 12 Maths Question 19.
$\frac { 2x }{ ({ x }^{ 2 }+1)({ x }^{ 2 }+3) }$
Solution:
put x²=y
so that 2xdx = dy
$\therefore \int { \frac { 2x }{ ({ x }^{ 2 }+1)({ x }^{ 2 }+3) } dx } =\int { \frac { dy }{ (y+1)(y+3) } }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5 Q19.1

Ex 7.5 Class 12 Maths Question 20.
$\frac { 1 }{ x({ x }^{ 4 }-1) }$
Solution:
put x⁴ = t
so that 4x³ dx = dt
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5 Q20.1

Ex 7.5 Class 12 Maths Question 21.
$\frac { 1 }{ { e }^{ x }-1 }$
Solution:
Let e^x = t ⇒ e^x dx = dt
⇒ $dx=\frac { dt }{ t }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5 Q21.1

Ex 7.5 Class 12 Maths Question 22.
choose the correct answer in each of the following :
$\int { \frac { xdx }{ (x-1)(x-2) } equals }$
(a) $log\left| \frac { { (x-1) }^{ 2 } }{ x-2 } \right| +c$
(b) $log\left| \frac { { (x-2) }^{ 2 } }{ x-1 } \right| +c$
(c) $log\left| \left( \frac { x-{ 1 }^{ 2 } }{ x-2 } \right) \right| +c$
(d) log|(x-1)(x-2)|+c
Solution:
(b) $\int { \frac { x }{ (x-1)(x-2) } dx } =\int { \left[ \frac { -1 }{ x-1 } +\frac { 2 }{ x-2 } \right] dx }$
$log\left| \frac { { (x-2) }^{ 2 } }{ x-1 } \right| +c$

Ex 7.5 Class 12 Maths Question 23.
$\int { \frac { dx }{ x({ x }^{ 2 }+1) } equals }$
(a) $log|x|-\frac { 1 }{ 2 } log({ x }^{ 2 }+1)+c$
(b) $log|x|+\frac { 1 }{ 2 } log({ x }^{ 2 }+1)+c$
(c) $-log|x|+\frac { 1 }{ 2 } log({ x }^{ 2 }+1)+c$
(d) $\frac { 1 }{ 2 } log|x|+log({ x }^{ 2 }+1)+c$
Solution:
(a) let $\frac { 1 }{ x\left( { x }^{ 2 }+1 \right) } =\frac { A }{ x } +\frac { Bx+C }{ { x }^{ 2 }+1 }$
⇒ 1 = A(x²+1)+(Bx+C)(x)
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.5 Q23.1

Class 12 Maths Chapter 7 Integrals Ex 7.6

Integrate the functions in Exercises 1 to 22.

Ex 7.6 Class 12 Maths Question 1.
x sinx
Solution:
By part integration
∫x sinx dx = x(-cosx) – ∫1(-cosx)dx
=-x cosx + ∫cosxdx
=-x cosx + sinx + c

Ex 7.6 Class 12 Maths Question 2.
x sin3x
Solution:
∫x sin3x dx = $x\left( -\frac { cos3x }{ 3 } \right) -\int { 1 } .\left( \frac { -cos3x }{ 3 } \right) dx$
$=-\frac { 1 }{ 3 } x\quad cos3x+\frac { 1 }{ 9 } sin3x+c$

Ex 7.6 Class 12 Maths Question 3.
${ x }^{ 2 }{ e }^{ x }$
Solution:
$\int { { x }^{ 2 }{ e }^{ x } } dx={ x }^{ 2 }{ e }^{ x }-2{ x }{ e }^{ x }+2{ e }^{ x }+c$
$={ e }^{ x }\left( { x }^{ 2 }-2x+2 \right) +c$

Ex 7.6 Class 12 Maths Question 4.
x logx
Solution:
$\int { xlogx\quad dx } =logx\int { xdx } -\int { \left[ \frac { d }{ dx } (logx)\int { xdx } \right] dx }$
$=\frac { { x }^{ 2 } }{ 2 } logx-\frac { 1 }{ 2 } \int { x\quad dx } =\frac { { x }^{ 2 } }{ 2 } logx-\frac { 1 }{ 4 } { x }^{ 2 }+c$

Ex 7.6 Class 12 Maths Question 5.
x log2x
Solution:
$\int { x\quad log2xdx } =(log2x)\frac { { x }^{ 2 } }{ 2 } -\int { \frac { 1 }{ 2x } } .2\left( \frac { { x }^{ 2 } }{ 2 } \right) dx$
$=\frac { { x }^{ 2 } }{ 2 } log|2x|-\frac { 1 }{ 2 } \int { xdx } =\frac { { x }^{ 2 } }{ 2 } log|2x|-\frac { { x }^{ 2 } }{ 4 } +c$

Ex 7.6 Class 12 Maths Question 6.
${ x }^{ 2 }logx$
Solution:
$\int { { x }^{ 2 }logxdx } =log|x|\left( \frac { { x }^{ 3 } }{ 3 } \right) -\int { \frac { 1 }{ x } } \left( \frac { { x }^{ 3 } }{ 3 } \right) dx$
$=\frac { { x }^{ 3 } }{ 3 } log|x|-\frac { 1 }{ 3 } \int { { x }^{ 2 }dx } =\frac { { x }^{ 3 } }{ 3 } log|x|-\frac { { x }^{ 3 } }{ 9 } +c$

Ex 7.6 Class 12 Maths Question 7.
$x\quad { sin }^{ -1 }x$
Solution:
$I=x\quad { sin }^{ -1 }x.\left( \frac { { x }^{ 2 } }{ 2 } \right) -\int { \frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } } .\frac { { x }^{ 2 } }{ 2 } dx$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.6 Q7.1

Ex 7.6 Class 12 Maths Question 8.
$x\quad { tan }^{ -1 }x$
Solution:
$I=x\quad { tan}^{ -1 }x.\left( \frac { { x }^{ 2 } }{ 2 } \right) -\int { \frac { 1 }{ \sqrt { 1+{ x }^{ 2 } } } } .\frac { { x }^{ 2 } }{ 2 } dx$
$=\frac { { x }^{ 2 } }{ 2 } { tan }^{ -1 }x-\frac { 1 }{ 2 } \int { \left( 1-\frac { 1 }{ 1+{ x }^{ 2 } } \right) dx }$
$=\frac { { x }^{ 2 } }{ 2 } { tan }^{ -1 }x-\frac { 1 }{ 2 } x+\frac { 1 }{ 2 } { tan }^{ -1 }x+c$

Ex 7.6 Class 12 Maths Question 9.
$x\quad { cos }^{ -1 }x$
Solution:
let I = $\int { x } { cos }^{ -1 }xdx=\int { { cos }^{ -1 }x } .xdx$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.6 Q9.1

Ex 7.6 Class 12 Maths Question 10.
${ (sin }^{ -1 }{ x })^{ 2 }$
Solution:
$put\quad { sin }^{ -1 }x=\theta \Rightarrow x=sin\theta \Rightarrow dx=cos\theta d\theta$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.6 Q10.1

Ex 7.6 Class 12 Maths Question 11.
$\frac { x\quad { cos }^{ -1 }x }{ \sqrt { 1-{ x }^{ 2 } } }$
Solution:
$put\quad { cos }^{ -1 }x=t\quad so\quad that\frac { x\quad { cos }^{ -1 }x }{ \sqrt { 1-{ x }^{ 2 } } } dx=dt$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.6 Q11.1

Ex 7.6 Class 12 Maths Question 12.
x sec²x
Solution:
∫x sec²x dx =x(tanx)-∫1.tanx dx
= x tanx+log cosx+c

Ex 7.6 Class 12 Maths Question 13.
${ ta }n^{ -1 }x$
Solution:
$\int { { tan }^{ -1 }xdx } =x{ tan }^{ -1 }x-\frac { 1 }{ 2 } \int { \frac { 2x }{ 1+{ x }^{ 2 } } dx }$
$=x{ tan }^{ -1 }x-\frac { 1 }{ 2 } log|1+{ x }^{ 2 }|+c$

Ex 7.6 Class 12 Maths Question 14.
x(logx)²
Solution:
∫x(logx)² dx
$=\frac { { x }^{ 2 } }{ 2 } { (logx) }^{ 2 }-\left[ (logx)\frac { { x }^{ 2 } }{ 2 } -\int { \frac { 1 }{ x } \frac { { x }^{ 2 } }{ 2 } dx } \right]$
$=\frac { { x }^{ 2 } }{ 2 } { (logx) }^{ 2 }-\frac { { x }^{ 2 } }{ 2 } logx+\frac { 1 }{ 4 } { x }^{ 2 }+c$

Ex 7.6 Class 12 Maths Question 15.
(x²+1)logx
Solution:
∫(x²+1)logx dx
$=logx\left( \frac { { x }^{ 3 } }{ 3 } +x \right) -\int { \frac { 1 }{ x } \left( \frac { { x }^{ 3 } }{ 3 } +x \right) dx }$
$=\left( \frac { { x }^{ 3 } }{ 3 } +x \right) logx-\frac { { x }^{ 3 } }{ 9 } -x+c$

Ex 7.6 Class 12 Maths Question 16.
${ e }^{ x }(sinx+cosx)$
Solution:
$put\quad { e }^{ x }sinx=t\Rightarrow { e }^{ x }(sinx+cosx)dx=dt$
$\therefore \int { { e }^{ x }(sinx+cosx)dx } =\int { dt } =t+c$
$={ e }^{ x }sinx+c$

Ex 7.6 Class 12 Maths Question 17.
$\frac { { xe }^{ x } }{ { (1+x) }^{ 2 } }$
Solution:
$\int { \frac { { xe }^{ x } }{ { (1+x) }^{ 2 } } }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.6 Q17.1

Ex 7.6 Class 12 Maths Question 18.
$\frac { { e }^{ x }(1+sinx) }{ 1+cosx }$
Solution:
$I=\int { { e }^{ x } } \left[ \frac { 1+2sin\frac { x }{ 2 } cos\frac { x }{ 2 } }{ 2{ cos }^{ 2 }\frac { x }{ 2 } } \right] dx$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.6 Q18.1

Ex 7.6 Class 12 Maths Question 19.
${ e }^{ x }\left( \frac { 1 }{ x } -\frac { 1 }{ { x }^{ 2 } } \right)$
Solution:
put $\frac { { e }^{ x } }{ x } =t\Rightarrow { e }^{ x }\left( \frac { 1 }{ x } -\frac { 1 }{ { x }^{ 2 } } \right) dx=dt$
$\therefore I=\int { dt } =t+c=\frac { { e }^{ x } }{ x } +c$

Ex 7.6 Class 12 Maths Question 20.
$\frac { { (x-2)e }^{ x } }{ { (x-1) }^{ 3 } }$
Solution:
$I=\int { { e }^{ x }\left[ \frac { 1 }{ { (x-1) }^{ 2 } } -\frac { 2 }{ { (x-1) }^{ 3 } } \right] dx }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.6 Q20.1

Ex 7.6 Class 12 Maths Question 21.
${ e }^{ 2x }sinx$
Solution:
let $I=\int { { e }^{ 2x }sinx }$
$={ e }^{ 2x }(-cosx)-\int { 2{ e }^{ 2x }(-cosx)dx }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.6 Q21.1

Ex 7.6 Class 12 Maths Question 22.
${ sin }^{ -1 }\left( \frac { 2x }{ 1+{ x }^{ 2 } } \right)$
Solution:
Put x = tan t
so that dx = sec² t dt
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.6 Q22.1

Choose the correct answer in exercise 23 and 24

Ex 7.6 Class 12 Maths Question 23.
$\int { { x }^{ 2 }{ e }^{ { x }^{ 3 } } } dx\quad equals$
(a) $\frac { 1 }{ 3 } { e }^{ { x }^{ 3 } }+c$
(b) $\frac { 1 }{ 3 } +{ e }^{ { x }^{ 2 } }+c$
(c) $\frac { 1 }{ 2 } { e }^{ { x }^{ 3 } }+c$
(d) $\frac { 1 }{ 2 } { e }^{ { x }^{ 2 } }+c$
Solution:
(a) let x³ = t
⇒3x² dx = dt
$\therefore \int { { x }^{ 2 }{ e }^{ { x }^{ 3 } }dx } =\frac { 1 }{ 3 } \int { { e }^{ t }dt } =\frac { 1 }{ 3 } { e }^{ t }+c=\frac { 1 }{ 3 } { e }^{ { x }^{ 3 } }+c$

Ex 7.6 Class 12 Maths Question 24.
$\int { { e }^{ x }secx(1+tanx) } dx\quad equals$
(a) ${ e }^{ x }cosx+c$
(b) ${ e }^{ x }secx+c$
(c) ${ e }^{ x }sinx+c$
(d) ${ e }^{ x }tanx+c$
Solution:
(b) $\int { { e }^{ x }(secx+secx\quad tanx)dx } ={ e }^{ x }secx+c$

Class 12 Maths Chapter 7 Integrals Ex 7.7

Integral the function in exercises 1 to 9

Ex 7.7 Class 12 Maths Question 1.
$\sqrt { 4-{ x }^{ 2 } }$
Solution:
$let\quad I=\int { \sqrt { 4-{ x }^{ 2 } } } dx=\int { \sqrt { { (2) }^{ 2 }-{ x }^{ 2 } } dx }$
$=\frac { x\sqrt { 4-{ x }^{ 2 } } }{ 2 } +2{ sin }^{ -1 }\left( \frac { x }{ 2 } \right) +c$

Ex 7.7 Class 12 Maths Question 2.
$\sqrt { 1-{ 4x }^{ 2 } }$
Solution:
$\int { \sqrt { 1-{ 4x }^{ 2 } } } dx=2\int { \sqrt { { \left( \frac { 1 }{ 2 } \right) }^{ 2 }-{ x }^{ 2 } } } dx$
$=\frac { x\sqrt { 1-{ 4x }^{ 2 } } }{ 2 } +\frac { 1 }{ 4 } { sin }^{ -1 }(2x)+c$

Ex 7.7 Class 12 Maths Question 3.
$\sqrt { { x }^{ 2 }+4x+6 }$
Solution:
$\int { \sqrt { { x }^{ 2 }+4x+6 } } dx=\int { \sqrt { { (x+2) }^{ 2 }+{ (\sqrt { 2 } ) }^{ 2 } } } dx$
$=\frac { x+2 }{ 2 } \sqrt { { x }^{ 2 }+4x+6 } +log\left| (x+2)+\sqrt { { x }^{ 2 }+4x+6 } \right| +c$

Ex 7.7 Class 12 Maths Question 4.
$\sqrt { { x }^{ 2 }+4x+1 }$
Solution:
$\int { \sqrt { { x }^{ 2 }+4x+1 } } dx=\int { \sqrt { { (x+2) }^{ 2 }-{ (\sqrt { 3 } ) }^{ 2 } } } dx$
$=\frac { x+2 }{ 2 } \sqrt { { x }^{ 2 }+4x+1 } -\frac { 3 }{ 2 } log\left| (x+2)+\sqrt { { x }^{ 2 }+4x+1 } \right| +c$

Ex 7.7 Class 12 Maths Question 5.
$\sqrt { 1-4x-{ x }^{ 2 } }$
Solution:
$\int { \sqrt { 1-4x-{ x }^{ 2 } } } dx=\int { \sqrt { { (5) }^{ 2 }-{ (x+2) }^{ 2 } } dx }$
$=\frac { x+2 }{ 2 } \sqrt { 5-{ (x+2) }^{ 2 } } dx$

Ex 7.7 Class 12 Maths Question 6.
$\sqrt { { x }^{ 2 }+4x-5 }$
Solution:
$\int { \sqrt { { x }^{ 2 }+4x-5 } } dx=\int { \sqrt { { (x+2) }^{ 2 }-{ (3) }^{ 2 } } } dx$
$=\frac { x+2 }{ 2 } \sqrt { { x }^{ 2 }+4x-5 } -\frac { 9 }{ 2 } log|x+2+\sqrt { { x }^{ 2 }+4x-5 } |+c$

Ex 7.7 Class 12 Maths Question 7.
$\sqrt { 1+3x-{ x }^{ 2 } }$
Solution:
$\int { \sqrt { 1-\left( { x }^{ 2 }-3x \right) } } dx$
$=\int { \sqrt { { \left( \frac { \sqrt { 13 } }{ 2 } \right) }^{ 2 }-{ \left( x-\frac { 3 }{ 2 } \right) }^{ 2 } } } dx$
$=\frac { 2x-3 }{ 4 } \sqrt { 1+3x-{ x }^{ 2 } } +\frac { 13 }{ 8 } { sin }^{ -1 }\left[ \frac { 2x-3 }{ \sqrt { 3 } } \right] +c$

Ex 7.7 Class 12 Maths Question 8.
$\sqrt { { x }^{ 2 }+3x }$
Solution:
$\int { \sqrt { { x }^{ 2 }+3x } } dx=\int { \sqrt { { \left( x+\frac { 3 }{ 2 } \right) }^{ 2 }-{ \left( \frac { 3 }{ 2 } \right) }^{ 2 } } } dx$
$=\frac { 2x+3 }{ 4 } \sqrt { { x }^{ 2 }+3x } -\frac { 9 }{ 8 } log\left| x+\frac { 3 }{ 2 } +\sqrt { { x }^{ 2 }+3x } \right| +c$

Ex 7.7 Class 12 Maths Question 9.
$\sqrt { 1+\frac { { x }^{ 2 } }{ 9 } }$
Solution:
$\int { \sqrt { 1+\frac { { x }^{ 2 } }{ 9 } } } dx=\frac { 1 }{ 3 } \int { \sqrt { { x }^{ 2 }+{ 3 }^{ 2 } } }$
$=\frac { 1 }{ 6 } \left[ x\sqrt { { x }^{ 2 }+9 } +9log|x+\sqrt { { x }^{ 2 }+9 } | \right] +c$

Choose the correct answer in the Exercises 10 to 11:

Ex 7.7 Class 12 Maths Question 10.
$\int { \sqrt { 1+{ x }^{ 2 } } } dx\quad is\quad equal\quad to$
(a) $\frac { x }{ 2 } \sqrt { 1+{ x }^{ 2 } } +\frac { 1 }{ 2 } log|x+\sqrt { 1+{ x }^{ 2 } } |+c$
(b) $\frac { 2 }{ 3 } { \left( 1+{ x }^{ 2 } \right) }^{ \frac { 3 }{ 2 } }+c$
(c) $\frac { 2 }{ 3 } x{ \left( 1+{ x }^{ 2 } \right) }^{ \frac { 3 }{ 2 } }+c$
(d) $\frac { { x }^{ 2 } }{ 2 } \sqrt { 1+{ x }^{ 2 } } +\frac { 1 }{ 2 } { x }^{ 2 }log\left| x+\sqrt { 1+{ x }^{ 2 } } \right| +c$
Solution:
(a) $\int { \sqrt { 1+{ x }^{ 2 } } } dx$
$=\frac { x }{ 2 } \sqrt { 1+{ x }^{ 2 } } +\frac { 1 }{ 2 } log(x+\sqrt { 1+{ x }^{ 2 } } )+c$

Ex 7.7 Class 12 Maths Question 11.
$\int { \sqrt { { x }^{ 2 }-8x+7 } } dx\quad is\quad equal\quad to$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.7 Q11.1
Solution:
(d) $\int { \sqrt { { x }^{ 2 }-8x+7 } } dx=\int { \sqrt { { (x-4) }^{ 2 }-{ (3) }^{ 2 } } } dx$
$=\frac { x-4 }{ 2 } \sqrt { { x }^{ 2 }-8x+7 } -\frac { 9 }{ 2 } log\left| (x-4)+\sqrt { { x }^{ 2 }+8x+7 } \right| +c$

Class 12 Maths Chapter 7 Integrals Ex 7.8

Evaluate the following definite integral as limit of sums.

Ex 7.8 Class 12 Maths Question 1.
$\int _{ a }^{ b }{ x\quad dx }$
Solution:
on comparing
$\int _{ a }^{ b }{ x\quad dx } \quad with\quad \int _{ a }^{ b }{ f(x)dx }$
we have
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.8 Q1.1

Ex 7.8 Class 12 Maths Question 2.
$\int _{ 0 }^{ 5 }{ (x+1)dx }$
Solution:
on comparing
$\int _{ 0 }^{ 5 }{ (x+1)dx } \quad with\quad \int _{ 0 }^{ 5 }{ f(x)dx }$
we have f(x) = x+1, a = 0, b = 5
and nh = b-a = 5-0 = 5
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.8 Q2.1

Ex 7.8 Class 12 Maths Question 3.
$\int _{ 2 }^{ 3 }{ { x }^{ 2 } } dx$
Solution:
compare
$\int _{ 2 }^{ 3 }{ { x }^{ 2 } } dx\quad with\quad \int _{ a }^{ b }{ f({ x }) } dx$
we have
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.8 Q3.1

Ex 7.8 Class 12 Maths Question 4.
$\int _{ 1 }^{ 4 }{ ({ x }^{ 2 }-x) } dx$
Solution:
compare
$\int _{ 1 }^{ 4 }{ ({ x }^{ 2 }-x) } dx\quad with\quad \int _{ a }^{ b }{ f({ x }) } dx$
we have f(x) = x²-x and a = 1, b = 4
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.8 Q4.1

Ex 7.8 Class 12 Maths Question 5.
$\int _{ -1 }^{ 1 }{ { e }^{ x } } dx\quad$
Solution:
compare
$\int _{ -1 }^{ 1 }{ { e }^{ x } } dx\quad with\quad \int _{ a }^{ b }{ f({ x }) } dx$
we have
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.8 Q5.1

Ex 7.8 Class 12 Maths Question 6.
$\int _{ 0 }^{ 4 }{ { (x+e }^{ 2x }) } dx\quad$
Solution:
let f(x) = x + e^2x,
a = 0, b = 4
and nh = b – a = 4 – 0 = 4

Class 12 Maths Chapter 7 Integrals Ex 7.9

Evaluate the definite integrals in Exercise 1 to 20.

Ex 7.9 Class 12 Maths Question 1.
$\int _{ -1 }^{ 1 }{ { (x+1 }) } dx\quad$
Solution:
${ =\left[ \frac { { x }^{ 2 } }{ 2 } +x \right] }_{ -1 }^{ 1 }=\frac { 1 }{ 2 } (1-1)+(1+1)\quad =2$

Ex 7.9 Class 12 Maths Question 2.
$\int _{ 2 }^{ 3 }{ \frac { 1 }{ x } dx }$
Solution:
$={ \left[ log\quad x \right] }_{ 2 }^{ 3 }\quad =log3-log2\quad =log\frac { 3 }{ 2 }$

Ex 7.9 Class 12 Maths Question 3.
$\int _{ 1 }^{ 2 }{ \left( { 4x }^{ 3 }-{ 5x }^{ 2 }+6x+9 \right) dx }$
Solution:
$={ \left[ \frac { { 4x }^{ 4 } }{ 4 } -\frac { { 5x }^{ 3 } }{ 3 } +\frac { { 6x }^{ 2 } }{ 2 } +9x \right] }_{ 1 }^{ 2 }$
$={ \left[ { x }^{ 4 }-\frac { 5 }{ 3 } { x }^{ 3 }+{ 3x }^{ 2 }+9x \right] }_{ 1 }^{ 2 }\quad =\frac { 64 }{ 3 }$

Ex 7.9 Class 12 Maths Question 4.
$\int _{ 0 }^{ \frac { \pi }{ 4 } }{ sin2x\quad dx }$
Solution:
$={ \left[ -\frac { 1 }{ 2 } cos2x \right] }_{ 0 }^{ \frac { \pi }{ 4 } }\quad =\frac { 1 }{ 2 }$

Ex 7.9 Class 12 Maths Question 5.
$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ cos2x\quad dx }$
Solution:
$={ \left[ \frac { 1 }{ 2 } sin2x \right] }_{ 0 }^{ \frac { \pi }{ 2 } }\quad =0$

Ex 7.9 Class 12 Maths Question 6.
$\int _{ 4 }^{ 5 }{ { e }^{ x }dx }$
Solution:
$={ \left[ { e }^{ x } \right] }_{ 4 }^{ 5 }\quad ={ e }^{ 5 }-{ e }^{ 4 }$

Ex 7.9 Class 12 Maths Question 7.
$\int _{ 0 }^{ \frac { \pi }{ 4 } }{ tanx\quad dx }$
Solution:
$={ \left[ log\quad secx \right] }_{ 0 }^{ \frac { \pi }{ 4 } }\quad =\frac { 1 }{ 2 } log2$

Ex 7.9 Class 12 Maths Question 8.
$\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 4 } }{ cosec\quad xdx }$
Solution:
$=log{ \left( cosecx-cotx \right) }_{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 4 } }$
$=log(\sqrt { 2 } -1)-log(2-\sqrt { 3 } )\quad =log\left( \frac { \sqrt { 2 } -1 }{ 2-\sqrt { 3 } } \right)$

Ex 7.9 Class 12 Maths Question 9.
$\int _{ 0 }^{ 1 }{ \frac { dx }{ \sqrt { 1-{ x }^{ 2 } } } }$
Solution:
$={ sin }^{ -1 }(1)-{ sin }^{ -1 }(0)\quad =\frac { \pi }{ 2 }$

Ex 7.9 Class 12 Maths Question 10.
$\int _{ 0 }^{ 1 }{ \frac { dx }{ 1+{ x }^{ 2 } } }$
Solution:
$={ \left[ { tan }^{ -1 }x \right] }_{ 0 }^{ 1 }\quad ={ tan }^{ -1 }(1)-{ ta }n^{ -1 }(0)\quad =\frac { \pi }{ 4 }$

Ex 7.9 Class 12 Maths Question 11.
$\int _{ 2 }^{ 3 }{ \frac { dx }{ { x }^{ 2 }-1 } }$
Solution:
$={ \left[ \frac { 1 }{ 2 } log\left( \frac { x-1 }{ x+1 } \right) \right] }_{ 2 }^{ 3 }\quad =\frac { 1 }{ 2 } log\frac { 3 }{ 2 }$

Ex 7.9 Class 12 Maths Question 12.
$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { cos }^{ 2 } } xdx$
Solution:
$=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { \frac { 1+cos2x }{ 2 } } } dx=\frac { 1 }{ 2 } { \left[ x+\frac { sin2x }{ 2 } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }=\frac { \pi }{ 4 }$

Ex 7.9 Class 12 Maths Question 13.
$\int _{ 2 }^{ 3 }{ \frac { x }{ { x }^{ 2 }+1 } } dx$
Solution:
$=\frac { 1 }{ 2 } \int _{ 2 }^{ 3 }{ \frac { 2x }{ { x }^{ 2 }+1 } } dx\quad =\frac { 1 }{ 2 } { \left[ log\left( { x }^{ 2 }+1 \right) \right] }_{ 2 }^{ 3 }\quad =\frac { 1 }{ 2 } log2$

Ex 7.9 Class 12 Maths Question 14.
$\int _{ 0 }^{ 1 }{ \frac { 2x+3 }{ { 5x }^{ 2 }+1 } dx }$
Solution:
$=\frac { 1 }{ 5 } \int _{ 0 }^{ 1 }{ \frac { 10x }{ { 5x }^{ 2 }+1 } dx } +\frac { 3 }{ 5 } \int _{ 0 }^{ 1 }{ \frac { dx }{ { { x }^{ 2 }+\left[ \frac { 1 }{ \sqrt { 5 } } \right] }^{ 2 } } }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.9 Q14.1

Ex 7.9 Class 12 Maths Question 15.
$\int _{ 0 }^{ 1 }{ { xe }^{ { x }^{ 2 } }dx }$
Solution:
let x² = t ⇒ 2xdx = dt
when x = 0, t = 0 & when x = 1,t = 1
$\therefore I=\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ { e }^{ t }dt } \quad =\frac { 1 }{ 2 } { \left( { e }^{ t } \right) }_{ 0 }^{ 1 }\quad =\frac { 1 }{ 2 } [e-1]$

Ex 7.9 Class 12 Maths Question 16.
$\int _{ 1 }^{ 2 }{ \frac { { 5x }^{ 2 } }{ { x }^{ 2 }+4x+3 } dx }$
Solution:
$\int _{ 1 }^{ 2 }{ \left( 5-\frac { 20x+15 }{ { x }^{ 2 }+4x+3 } \right) dx }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.9 Q16.1

Ex 7.9 Class 12 Maths Question 17.
$\int _{ 0 }^{ \frac { \pi }{ 4 } }{ \left( { 2sec }^{ 2 }x+{ x }^{ 3 }+2 \right) dx }$
Solution:
$={ \left[ 2tanx+\frac { { x }^{ 4 } }{ 4 } +2x \right] }_{ 0 }^{ \frac { \pi }{ 4 } }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.9 Q17.1

Ex 7.9 Class 12 Maths Question 18.
$\int _{ 0 }^{ \pi }{ \left( { sin }^{ 2 }\frac { x }{ 2 } -{ cos }^{ 2 }\frac { x }{ 2 } \right) } dx$
Solution:
$=-\int _{ 0 }^{ \pi }{ cosx } dx\quad =-{ \left[ sinx \right] }_{ 0 }^{ \pi }-(0-0)\quad =0$

Ex 7.9 Class 12 Maths Question 19.
$\int _{ 0 }^{ 2 }{ \frac { 6x+3 }{ { x }^{ 2 }+4 } } dx$
Solution:
$=\int _{ 0 }^{ 2 }{ \frac { 6x }{ { x }^{ 2 }+4 } } dx+\int _{ 0 }^{ 2 }{ \frac { 3 }{ { x }^{ 2 }+4 } dx }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.9 Q19.1

Ex 7.9 Class 12 Maths Question 20.
$\int _{ 0 }^{ 1 }{ \left( { xe }^{ x }+sin\frac { \pi x }{ 4 } \right) dx }$
Solution:
$=\int _{ 0 }^{ 1 }{ { xe }^{ x }dx } +\int _{ 0 }^{ 1 }{ sin\frac { \pi x }{ 4 } } dx$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.9 Q20.1

Ex 7.9 Class 12 Maths Question 21.
$\int _{ 1 }^{ \sqrt { 3 } }{ \frac { dx }{ { 1+x }^{ 2 } } \quad equals }$
(a) $\frac { \pi }{ 3 }$
(b) $\frac { 2\pi }{ 3 }$
(c) $\frac { \pi }{ 6 }$
(d) $\frac { \pi }{ 12 }$
Solution:
(d) $\int _{ 1 }^{ \sqrt { 3 } }{ \frac { dx }{ { 1+x }^{ 2 } } } \quad ={ \left[ { tan }^{ -1 }x \right] }_{ 1 }^{ \sqrt { 3 } }\quad =\frac { \pi }{ 3 } -\frac { \pi }{ 4 } \quad =\frac { \pi }{ 12 }$

Ex 7.9 Class 12 Maths Question 22.
$\int _{ 0 }^{ \frac { 2 }{ 3 } }{ \frac { dx }{ 4+{ 9x }^{ 2 } } equals }$
(a) $\frac { \pi }{ 6 }$
(b) $\frac { \pi }{ 12 }$
(c) $\frac { \pi }{ 24 }$
(d) $\frac { \pi }{ 4 }$
Solution:
(c) $\int _{ 0 }^{ \frac { 2 }{ 3 } }{ \frac { dx }{ 4+{ 9x }^{ 2 } } } \quad =\frac { 1 }{ 9 } \int _{ 0 }^{ \frac { 2 }{ 3 } }{ \frac { dx }{ { \left( \frac { 2 }{ 3 } \right) }^{ 2 }+{ x }^{ 2 } } }$
$=\frac { 1 }{ 6 } { \left[ { tan }^{ -1 }\left( \frac { 3x }{ 2 } \right) \right] }_{ 0 }^{ \frac { 2 }{ 3 } }\quad =\frac { 1 }{ 6 } \times \frac { \pi }{ 4 } \quad =\frac { \pi }{ 24 }$

Class 12 Maths Chapter 7 Integrals Ex 7.10

Evaluate the integrals in Exercises 1 to 8 using substitution.

Ex 7.10 Class 12 Maths Question 1.
$\int _{ 0 }^{ 1 }{ \frac { x }{ { x }^{ 2 }+1 } } dx=I$
Solution:
Let x² + 1 = t
⇒2xdx = dt
when x = 0, t = 1 and when x = 1, t = 2
$\therefore I=\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ \frac { dt }{ t } } ={ \left[ \frac { 1 }{ 2logt } \right] }_{ 1 }^{ 2 }\quad =\frac { 1 }{ 2 } log2$

Ex 7.10 Class 12 Maths Question 2.
$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sqrt { sin\phi } { cos }^{ 5 }\phi d\phi =I }$
Solution:
$I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \sqrt { sin\phi } { (1-{ sin }^{ 2 }) }^{ 2 }cos\phi d\phi }$
put sinφ = t,so that cosφdφ = dt
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.10 Q2.1

Ex 7.10 Class 12 Maths Question 3.
$\int _{ 0 }^{ 1 }{ { sin }^{ -1 } } \left( \frac { 2x }{ 1+{ x }^{ 2 } } \right) dx=I$
Solution:
let x = tanθ =>dx = sec²θ dθ
when x = 0 => θ = 0
and when x = 1 => $\theta \frac { \pi }{ 4 }$
$\frac { 1 }{ 2 }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.10 Q3.1

Ex 7.10 Class 12 Maths Question 4.
$\int _{ 0 }^{ 2 }{ x\sqrt { x+2 } } dx=I(say)(put\quad x+2={ t }^{ 2 })$
Solution:
let x+2 = t =>dx = dt
when x = 0,t = 2 and when x = 2, t = 4
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.10 Q4.1

Ex 7.10 Class 12 Maths Question 5.
$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { sinx\quad dx }{ 1+{ cos }^{ 2 }x } =I }$
Solution:
put cosx = t
so that -sinx dx = dt
when x = 0, t = 1; when $x=\frac { \pi }{ 2 }$ , t = 0
$\therefore I=\int _{ 1 }^{ 0 }{ \frac { -dt }{ 1+{ t }^{ 2 } } =-{ \left[ { tan }^{ -1 }t \right] }_{ 1 }^{ 0 } } =\frac { \pi }{ 4 }$

Ex 7.10 Class 12 Maths Question 6.
$\int _{ 0 }^{ 2 }{ \frac { dx }{ x+4-{ x }^{ 2 } } =I }$
Solution:
$\int _{ 0 }^{ 2 }{ \frac { dx }{ x+4-{ x }^{ 2 } } =I }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.10 Q6.1

Ex 7.10 Class 12 Maths Question 7.
$\int _{ -1 }^{ 1 }{ \frac { dx }{ { x }^{ 2 }+2x+5 } =I }$
Solution:
$I=\int _{ -1 }^{ 1 }{ \frac { dx }{ { (x+1) }^{ 2 }+{ 2 }^{ 2 } } } =\frac { 1 }{ 2 } { \left[ { tan }^{ -1 }\frac { x+1 }{ 2 } \right] }_{ -1 }^{ 1 }\quad =\frac { \pi }{ 8 }$

Ex 7.10 Class 12 Maths Question 8.
$\int _{ 1 }^{ 2 }{ \left[ \frac { 1 }{ x } -\frac { 1 }{ { 2x }^{ 2 } } \right] { e }^{ 2x }dx } =I$
Solution:
let 2x = t ⇒ 2dx = dt
when x = 1, t = 2 and when x = 2, t = 4
$I=\int _{ 2 }^{ 4 }{ e } ^{ t }\left( \frac { 1 }{ t } -\frac { 1 }{ { t }^{ 2 } } \right) dt\quad ={ e }^{ t }{ \left[ \frac { 1 }{ t } \right] }_{ 2 }^{ 4 }\quad =\frac { e^{ 2 } }{ 2 } \left[ \frac { { e }^{ 2 } }{ 2 } -1 \right]$

Choose the correct answer in Exercises 9 and 10

Ex 7.10 Class 12 Maths Question 9.
The value of integral $\int _{ \frac { 1 }{ 3 } }^{ 1 }{ \frac { { { (x-x }^{ 3 }) }^{ \frac { 1 }{ 3 } } }{ { x }^{ 4 } } dx }$ is
(a) 6
(b) 0
(c) 3
(d) 4
Solution:
(a) let I = $\int _{ \frac { 1 }{ 3 } }^{ 1 }{ \frac { { { (x-x }^{ 3 }) }^{ \frac { 1 }{ 3 } } }{ { x }^{ 4 } } dx } \quad =\int _{ \frac { 1 }{ 3 } }^{ 1 }{ \frac { { x }^{ \frac { 1 }{ 3 } }(1-{ x }^{ 2 })^{ \frac { 1 }{ 3 } } }{ { x }^{ 4 } } dx }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.10 Q9.1

Ex 7.10 Class 12 Maths Question 10.
$If\quad f(x)=\int _{ 0 }^{ x }{ tsint,\quad then\quad { f }^{ \prime }(x)\quad is }$
(a) cosx+xsinx
(b) xsinx
(c) xcosx
(d) sinx+xcosx
Solution:
(b) $f(x)=\int _{ 0 }^{ x }{ tsint\quad dt }$
$=t(-cost)-\int { 1{ \left[ (-cost)dt \right] }_{ 0 }^{ x } }$
=-x cox+sinx

Class 12 Maths Chapter 7 Integrals Ex 7.11

By using the properties of definite integrals, evaluate the integrals in Exercises 1 to 19.

Ex 7.11 Class 12 Maths Question 1.
$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ { cos }^{ 2 }x\quad dx } =I$
Solution:
$I=\frac { 1 }{ 2 } \int _{ 0 }^{ \frac { \pi }{ 2 } }{ (1+cos2x)dx } =\frac { 1 }{ 2 } { \left[ x+\frac { sin2x }{ 2 } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }\quad =\frac { \pi }{ 4 }$

Ex 7.11 Class 12 Maths Question 2.
$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sqrt { sinx } }{ \sqrt { sinx } +\sqrt { cosx } } dx }$
Solution:
let I = $\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { \sqrt { sinx } }{ \sqrt { sinx } +\sqrt { cosx } } dx }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11 Q2.1

Ex 7.11 Class 12 Maths Question 3.
$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { { sin }^{ \frac { 3 }{ 2 } }xdx }{ { sin }^{ \frac { 3 }{ 2 } }x+{ cos }^{ \frac { 3 }{ 2 } }dx } dx }$
Solution:
let I = $\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { { sin }^{ \frac { 3 }{ 2 } }xdx }{ { sin }^{ \frac { 3 }{ 2 } }x+{ cos }^{ \frac { 3 }{ 2 } }dx } dx }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11 Q3.1

Ex 7.11 Class 12 Maths Question 4.
$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { { cos }^{ 5 }xdx }{ { sin }^{ 5 }x+{ cos }^{ 5 }x } }$
Solution:
let I = $\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { { cos }^{ 5 }xdx }{ { sin }^{ 5 }x+{ cos }^{ 5 }x } }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11 Q4.1

Ex 7.11 Class 12 Maths Question 5.
$\int _{ -5 }^{ 5 }{ \left| x+2 \right| dx=I }$
Solution:
$I=\int _{ -5 }^{ 5 }{ \left| x+2 \right| dx+\int _{ -2 }^{ 5 }{ \left| x+2 \right| dx } }$
at x = – 5, x + 2 < 0; at x = – 2, x + 2 = 0; at x = 5, x + 2>0;x + 2<0, x + 2 = 0, x + 2>0
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11 Q5.1

Ex 7.11 Class 12 Maths Question 6.
$\int _{ 2 }^{ 8 }{ |x-5|dx } =I$
Solution:
$\int _{ 2 }^{ 8 }{ |x-5|dx } =I$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11 Q6.1

Ex 7.11 Class 12 Maths Question 7.
$\int _{ 0 }^{ 1 }{ x(1-x)^{ n }dx } =I$
Solution:
$\int _{ 0 }^{ 1 }{ x(1-x)^{ n }dx } =I$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11 Q7.1

Ex 7.11 Class 12 Maths Question 8.
$\int _{ 0 }^{ \frac { \pi }{ 4 } }{ log(1+tanx)dx }$
Solution:
let I = $\int _{ 0 }^{ \frac { \pi }{ 4 } }{ log(1+tanx)dx }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11 Q8.1

Ex 7.11 Class 12 Maths Question 9.
$\int _{ 0 }^{ 2 }{ x\sqrt { 2-x } dx=I }$
Solution:
let 2-x = t
⇒ – dx = dt
when x = 0, t = 2 and when x = 2,t = 0
$\frac { 1 }{ 2 }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11 Q9.1

Ex 7.11 Class 12 Maths Question 10.
$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( 2logsinx-logsin2x \right) dx=I }$
Solution:
$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left( 2logsinx-logsin2x \right) dx=I }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11 Q10.1

Ex 7.11 Class 12 Maths Question 11.
$\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ { sin }^{ 2 } } xdx$
Solution:
Let f(x) = sin² x
f(-x) = sin² x = f(x)
∴ f(x) is an even function
$\therefore \int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ { sin }^{ 2 } } xdx\quad =2\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \left[ \frac { 1-cos2x }{ 2 } \right] dx }$
$={ \left[ x-\frac { sin2x }{ x } \right] }_{ 0 }^{ \frac { \pi }{ 2 } }\therefore I=\frac { \pi }{ 2 }$

Ex 7.11 Class 12 Maths Question 12.
$\int _{ 0 }^{ \pi }{ \frac { xdx }{ 1+sinx } }$
Solution:
let I = $\int _{ 0 }^{ \pi }{ \frac { xdx }{ 1+sinx } }$ …(i)
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11 Q12.1

Ex 7.11 Class 12 Maths Question 13.
$\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ { sin }^{ 7 } } xdx$
Solution:
Let f(x) = sin⁷ xdx
⇒ f(-x) = -sin⁷ x = -f(x)
⇒ f(x) is an odd function of x
⇒ $\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ { sin }^{ 7 } } xdx=0$

Ex 7.11 Class 12 Maths Question 14.
$\int _{ 0 }^{ 2\pi }{ { cos }^{ 5 } } xdx$
Solution:
let f(x) = cos⁵ x
⇒ f(2π – x) = cos⁵ x
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11 Q14.1

Ex 7.11 Class 12 Maths Question 15.
$\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { sinx-cosx }{ 1+sinx\quad cosx } dx }$
Solution:
let I = $\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \frac { sinx-cosx }{ 1+sinx\quad cosx } dx }$ …(i)
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11 Q15.1

Ex 7.11 Class 12 Maths Question 16.
$\int _{ 0 }^{ \pi }{ log(1+cosx)dx }$
Solution:
let I = $\int _{ 0 }^{ \pi }{ log(1+cosx)dx }$
then I = $\int _{ 0 }^{ \pi }{ log[1+cos(\pi -x)]dx }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11 Q16.1

Ex 7.11 Class 12 Maths Question 17.
$\int _{ 0 }^{ a }{ \frac { \sqrt { x } }{ \sqrt { x } +\sqrt { a-x } } dx }$
Solution:
let I = $\int _{ 0 }^{ a }{ \frac { \sqrt { x } }{ \sqrt { x } +\sqrt { a-x } } dx }$ …(i)
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11 Q17.1

Ex 7.11 Class 12 Maths Question 18.
$\int _{ 0 }^{ 4 }{ \left| x-1 \right| dx=I }$
Solution:
$I=-\int _{ 0 }^{ 1 }{ (x-1)dx } +\int _{ 1 }^{ 4 }{ (x-1)dx }$
$=-{ \left[ \frac { { x }^{ 2 } }{ 2 } -x \right] }_{ 0 }^{ 1 }+{ \left[ \frac { { x }^{ 2 } }{ 2 } -x \right] }_{ 1 }^{ 4 }=5$

Ex 7.11 Class 12 Maths Question 19.
show that $4\int _{ 0 }^{ a }{ f(x)g(x)dx } =2\int _{ 0 }^{ a }{ f(x)dx }$ if f and g are defined as f(x)=f(a-x) and g(x)+g(a-x)=4
Solution:
let I = $\int _{ 0 }^{ a }{ f(x)g(x)dx }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11 Q19.1

Ex 7.11 Class 12 Maths Question 20.
The value of $\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \left( { x }^{ 3 }+xcosx+{ tan }^{ 5 }x+1 \right) dx }$ is
(a) 0
(b) 2
(c) π
(d) 1
Solution:
(c) let I = $\int _{ \frac { -\pi }{ 2 } }^{ \frac { \pi }{ 2 } }{ \left( { x }^{ 3 }+xcosx+{ tan }^{ 5 }x+1 \right) dx }$ is
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11 Q20.1

Ex 7.11 Class 12 Maths Question 21.
The value of $\int _{ 0 }^{ \frac { \pi }{ 2 } }{ log\left[ \frac { 4+3sinx }{ 4+3sinx } \right] dx }$ is
(a) 2
(b) $\frac { 3 }{ 4 }$
(c) 0
(d) -2
Solution:
let I = $\int _{ 0 }^{ \frac { \pi }{ 2 } }{ log\left[ \frac { 4+3sinx }{ 4+3sinx } \right] dx }$
NCERT Solutions for Class 12 Maths Chapter 7 Integrals Ex 7.11 Q21.1

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Class 12th Chapter -6 Application of Derivatives | NCERT Maths Solution | NCERT Solution | Edugrown

NCERT Solutions for Class 12 Maths Chapter : 6 Application of Derivatives

Class 12 Maths Chapter 6 Application of Derivatives Ex 6.1

Ex 6.1 Class 12 Maths Question 1.
Find the rate of change of the area of a circle with respect to its radius r when (a) r = 3 cm (b) r = 4 cm
Solution:
Let A be the area of the circle
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.1 Q1.1

Ex 6.1 Class 12 Maths Question 2.
The volume of a cube is increasing at the rate of 8 cm³/s. How fast is the surface area increasing when the length of an edge is 12 cm?
Solution:
Let x be the length of the cube volume V = x³,
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.1 Q2.1

Ex 6.1 Class 12 Maths Question 3.
The radius of a circle is increasing uniformly at the rate of 3 cm/s. Find the rate at which the area of the circle is increasing when the radius is 10 cm.
Solution:
Let r be the radius of the circle.
Area of circle = πr² = A also $\frac { dr }{ dt }$ = 3 cm/ sec.
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.1 Q3.1

Ex 6.1 Class 12 Maths Question 4.
An edge of a variable cube is increasing at the rate of 3 cm/s. How fast is the volume of the cube increasing when the edge is 10 cm long?
Solution:
Let the edge of the cube = x cm
∴ $\frac { dx }{ dt }$ = 3
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.1 Q4.1

Ex 6.1 Class 12 Maths Question 5.
A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?
Solution:
Let r be the radius of a wave circle: $\frac { dx }{ dt }$ = 5cm/sec.
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.1 Q5.1

Ex 6.1 Class 12 Maths Question 6.
The radius of a circle is increasing at the rate of 0.7 cm/s. What is the rate of increase of its circumference?
Solution:
The rate of change of circle w.r.t time t is given
to be 0.7 cm/sec. i.e. $\frac { dr }{ dt }$ = 0.7 cm/sec.
Now, circumference of the circle is c = 2πr
$\therefore \frac { dc }{ dt } =\left[ \frac { d }{ dr } \left( 2\pi r \right) .\frac { dr }{ dt } \right] =2\pi \frac { dr }{ dt } =1.4\pi cm/sec$

Ex 6.1 Class 12 Maths Question 7.
The length x of a rectangle is decreasing at the rate of 5 cm/minute and the width y is increasing at the rate of 4 cm/minute. When x = 8 cm and y = 6 cm, find the rates of change of
(a) the perimeter, and
(b) the area of the rectangle.
Solution:
(a) The length x of a rectangle is decreasing at dx the rate of 5cm/min. => $\frac { dx }{ dt }$ = – 5cm min …(i)
The width y is increasing at the rate of 4cm/min.
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.1 Q7.1

Ex 6.1 Class 12 Maths Question 8.
A balloon, which always remains spherical on inflation, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is 15 cm.
Solution:
Volume of the spherical balloon = V = $\frac { 4 }{ 3 } \pi { r }^{ 3 }$
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.1 Q8.1

Ex 6.1 Class 12 Maths Question 9.
A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is 10 cm.
Solution:
Let r be the variable radius of the balloon which is in the form of sphere Vol. of the sphere
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.1 Q9.1

Ex 6.1 Class 12 Maths Question 10.
A ladder 5 m long is leaning against a wall. The bottom of the ladder is pulled along the ground, away from the wall, at the rate of 2cm/s. How fast is its height on the wall decreasing when the foot of the ladder is 4m away from die wall ?
Solution:
Let AB be the ladder and OB be the wall. At an instant,
let OA = x, OB = y,
x² + y² = 25 …(i)
On differentiating,
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.1 Q10.1

Ex 6.1 Class 12 Maths Question 11.
A particle moves along the curve 6y = x³ + 2. Find the points on the curve at which the y-coordinate is changing 8 times as fast as the x-coordinate.
Solution:
We have
6y = x³ + 2…(i)
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.1 Q11.1

Ex 6.1 Class 12 Maths Question 12.
The radius of an air bubble is increasing at the rate of $\frac { 1 }{ 2 }$ cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?
Solution:
Let r be the radius then V = $\frac { 4 }{ 3 } \pi { r }^{ 3 }$
$\frac { dr }{ dt } =\frac { 1 }{ 2 } cm/sec$
$\frac { dv }{ dt } =\frac { d }{ dt } \left( \frac { 4 }{ 3 } \pi { r }^{ 3 } \right) =\frac { 4 }{ 3 } { \pi .3r }^{ 2 }.\frac { dr }{ dt } ={ 2\pi r }^{ 2 }$
Hence, the rate of increase of volume when radius is 1 cm = 2π x 1² = 2π cm³/sec.

Ex 6.1 Class 12 Maths Question 13.
A balloon, which always remains spherical, has a variable diameter $\frac { 3 }{ 2 }(2x+1)$ . Find the rate of change of its volume with respect to x.
Solution:
Dia of sphere = $\frac { 3 }{ 2 }(2x+1)$
∴ Radius = $\frac { 3 }{ 4 }(2x+1)$
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.1 Q13.1

Ex 6.1 Class 12 Maths Question 14.
Sand is pouring from a pipe at the rate of 12 cm³/s. The falling sand forms a cone on the ground in such a way that the height of the cone is always one-sixth of the radius of the base. How fast is the height of the sand cone increasing when the height is 4 cm?
Solution:
Let r and h be the radius and height of the sand
– cone at time t respectively, h = $\frac { r }{ 6 }$ …(i)
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.1 Q14.1

Ex 6.1 Class 12 Maths Question 15.
The total cost C (x) in Rupees associated with the production of x units of an item is given by C (x) = 0.007 x³ – 0.003 x² + 15x + 4000. Find the marginal cost when 17 units are produced.
Solution:
Marginal cost MC = Instantaneous rate of change
of total cost at any level of out put = $\frac { dC }{ dx }$
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.1 Q15.1

Ex 6.1 Class 12 Maths Question 16.
The total revenue in Rupees received from the sale of x units of a product is given by
R (x) = 13x² + 26x +15. Find the marginal revenue when x = 7.
Solution:
Marginal Revenue (MR)
= Rate of change of total revenue w.r.t. the
number of items sold at an instant = $\frac { dR }{ dx }$
We know R(x) = 13x² + 26x + 15,
MR = $\frac { dR }{ dx }$ = 26x + 26 = 26(x+1)
Now x = 7, MR = 26 (x + 1) = 26 (7 + 1) = 208
Hence, Marginal Revenue = Rs 208.

Choose the correct answer in the Exercises 17 and 18.

Ex 6.1 Class 12 Maths Question 17.
The rate of change of the area of a circle with respect to its radius r at r = 6 cm is
(a) 10π
(b) 12π
(c) 8π
(d) 11π
Solution:
(b) ∵ A = πr² => $\frac { dA }{ dr }$ = 2π x 6 = 12π cm²/radius

Ex 6.1 Class 12 Maths Question 18.
The total revenue in Rupees received from the sale of x units of a product is given by R (x) = 3x² + 36x + 5. The marginal revenue, when x = 15 is
(a) 116
(b) 96
(c) 90
(d) 126
Solution:
(d) R(x) = 3x² + 36x + 5 ,
MR = $\frac { dR }{ dx }$ = 6x + 36 ,
At x = 15; $\frac { dR }{ dx }$ = 6 x 15 + 36 = 90 + 36 = 126

Class 12 Maths Chapter 6 Application of Derivatives Ex 6.2

Ex 6.2 Class 12 Maths Question 1.
Show that the function given by f (x) = 3x+17 is strictly increasing on R.
Solution:
f(x) = 3x + 17
∴ f’ (x) = 3>0 ∀ x∈R
⇒ f is strictly increasing on R.

Ex 6.2 Class 12 Maths Question 2.
Show that the function given by f (x) = e^2x is strictly increasing on R.
Solution:
We have f (x) = e^2x
⇒ f’ (x) = 2e^2x
Case I When x > 0, then f’ (x) = 2e^2x
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.2 Q2.1

Ex 6.2 Class 12 Maths Question 3.
Show that the function given by f (x) = sin x is
(a) strictly increasing in $\left( 0,\frac { \pi }{ 2 } \right)$
(b) strictly decreasing in $\left( \frac { \pi }{ 2 } ,\pi \right)$
(c) neither increasing nor decreasing in (0, π)
Solution:
We have f(x) = sinx
∴ f’ (x) = cosx
(a) f’ (x) = cos x is + ve in the interval $\left( 0,\frac { \pi }{ 2 } \right)$
⇒ f(x) is strictly increasing on $\left( 0,\frac { \pi }{ 2 } \right)$
(b) f’ (x) = cos x is a -ve in the interval $\left( \frac { \pi }{ 2 } ,\pi \right)$
⇒ f (x) is strictly decreasing in $\left( \frac { \pi }{ 2 } ,\pi \right)$
(c) f’ (x) = cos x is +ve in the interval $\left( 0,\frac { \pi }{ 2 } \right)$
while f’ (x) is -ve in the interval $\left( \frac { \pi }{ 2 } ,\pi \right)$
∴ f(x) is neither increasing nor decreasing in (0,π)

Ex 6.2 Class 12 Maths Question 4.
Find the intervals in which the function f given by f(x) = 2x² – 3x is
(a) strictly increasing
(b) strictly decreasing
Solution:
f(x) = 2x² – 3x
⇒ f’ (x) = 4x – 3
⇒ f’ (x) = 0 at x = $\frac { 3 }{ 4 }$
The point $x=\frac { 3 }{ 4 }$ divides the real
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.2 Q4.1

Ex 6.2 Class 12 Maths Question 5.
Find the intervals in which the function f given by f (x) = 2x³ – 3x² – 36x + 7 is
(a) strictly increasing
(b) strictly decreasing
Solution:
f(x) = 2x³ – 3x² – 36x + 7;
f (x) = 6 (x – 3) (x + 2)
⇒ f’ (x) = 0 at x = 3 and x = – 2
The points x = 3, x = – 2, divide the real line into three disjoint intervals viz. (-∞,-2), (-2,3), (3,∞)
Now f’ (x) is +ve in the intervals (-∞, -2) and (3,∞). Since in the interval (-∞, -2) each factor x – 3, x + 2 is -ve.
⇒ f’ (x) = + ve.
(a) f is strictly increasing in (-∞, -2)∪(3,∞)
(b) In the interval (-2,3), x+2 is +ve and x-3 is -ve.
f (x) = 6(x – 3)(x + 2) = + x – = -ve
∴ f is strictly decreasing in the interval (-2,3).

Ex 6.2 Class 12 Maths Question 6.
Find the intervals in which the following functions are strictly increasing or decreasing:
(a) x² + 2x – 5
(b) 10 – 6x – 2x²
(c) – 2x³ – 9x² – 12x + 1
(d) 6 – 9x – x²
(e) (x + 1)³(x – 3)³
Solution:
(c) Let f(x) = – 2x³ – 9x² – 12x + 1
∴ f’ (x) = – 6x² – 18x – 12
= – 6(x² + 3x + 2)
f'(x) = – 6(x + 1)(x + 2), f’ (x) = 0 gives x = -1 or x = -2
The points x = – 2 and x = – 1 divide the real line into three disjoint intervals namely ( – ∞, – 2) ( – 2, – 1) and( – 1 ∞).
In the interval (-∞,-2) i.e.,-∞<x<-2 (x+ 1) (x+2) are -ve.
∴f’ (x) = (-) (-) (- ) = – ve.
⇒ f (x) is decreasing in (-∞,-2)
In the interval (-2, -1) i.e., – 2 < x < -1,
(x + 1) is -ve and (x + 2) is + ve.
∴ f'(x) = (-)(-) (+) = + ve.
⇒ f (x) is increasing in (-2, -1)
In the interval (-1,∞) i.e.,-1 <x<∞,(x + 1) and (x + 2) are both positive. f’ (x) = (-) (+) (+) = -ve.
⇒ f (x) is decreasing in (-1, ∞)
Hence, f (x) is increasing for – 2 < x < – 1 and decreasing for x<-2 and x>-1.

Ex 6.2 Class 12 Maths Question 7.
Show that $y=log(1+x)-\frac { 2x }{ 2+x } x>-1$ , is an increasing function of x throughout its domain.
Solution:
let $f(x)=log(1+x)-\frac { 2x }{ 2+x } x>-1$
f’ (x) = $\frac { { x }^{ 2 } }{ { (x+1)(x+2) }^{ 2 } }$
For f (x) to be increasing f’ (x) > 0
$\Rightarrow \frac { 1 }{ x+1 } >0\Rightarrow x>-1$
Hence, $y=log(1+x)-\frac { 2x }{ 2+x }$ is an increasing function of x for all values of x > – 1.

Ex 6.2 Class 12 Maths Question 8.
Find the values of x for which y = [x (x – 2)]² is an increasing function.
Solution:
y = x⁴ – 4x³ + 4x²
∴ $\frac { dy }{ dx }$ = 4x³ – 12x² + 8x
For the function to be increasing $\frac { dy }{ dx }$ >0
4x³ – 12x² + 8x>0
⇒ 4x(x – 1)(x – 2)>0
For 0 < x < 1, $\frac { dy }{ dx }$ = (+)(-)(-) = +ve and for x > 2, $\frac { dy }{ dx }$ = (+) (+) (+) = +ve
Thus, the function is increasing for 0 < x < 1 and x > 2.

Ex 6.2 Class 12 Maths Question 9.
Prove that $y=\frac { 4sin\theta }{ (2+cos\theta ) } -\theta$ is an increasing function of θ in $\left[ 0,\frac { \pi }{ 2 } \right]$
Solution:
$\frac { dy }{ dx } =\frac { 8cos\theta +4 }{ { (2+cos\theta ) }^{ 2 } } -1=\frac { cos\theta (4-cos\theta ) }{ { (2+cos\theta ) }^{ 2 } }$
For the function to be increasing $\frac { dy }{ dx }$ > 0
⇒ cosθ(4-cos²θ)>0
⇒ cosθ>0
⇒ θ∈ $\left[ 0,\frac { \pi }{ 2 } \right]1$

Ex 6.2 Class 12 Maths Question 10.
Prove that the logarithmic function is strictly increasing on (0, ∞).
Solution:
Let f (x) = log x
Now, f’ (x) = $\frac { 1 }{ x }$ ; When takes the
values x > 0, $\frac { 1 }{ x }$ > 0, when x > 0,
∵ f’ (x) > 0
Hence, f (x) is an increasing function for x > 0 i.e

Ex 6.2 Class 12 Maths Question 11.
Prove that the function f given by f (x) = x² – x + 1 is neither strictly increasing nor strictly decreasing on (-1,1).
Solution:
Given
f (x) = x² – x + 1
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.2 Q11.1
∴ f (x) is neither increasing nor decreasing on (-1,1).

Ex 6.2 Class 12 Maths Question 12.
Which of the following functions are strictly decreasing on $\left[ 0,\frac { \pi }{ 2 } \right]$
(a) cos x
(b) cos 2x
(c) cos 3x
(d) tan x
Solution:
(a) We have f (x) = cos x
∴ f’ (x) = – sin x < 0 in $\left[ 0,\frac { \pi }{ 2 } \right]$
∴ f’ (x) is a decreasing function.
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.2 Q12.1

Ex 6.2 Class 12 Maths Question 13.
On which of the following intervals is the function f given by f (x )= x¹⁰⁰ + sin x – 1 strictly decreasing ?
(a) (0,1)
(b) $\left[ \frac { \pi }{ 2 } ,\pi \right]$
(c) $\left[ 0,\frac { \pi }{ 2 } \right]$
(d) none of these
Solution:
(d) f(x) = x¹⁰⁰ + sin x – 1
∴ f’ (x)= 100x⁹⁹+ cos x
(a) for(-1, 1)i.e.,- 1 <x< 1,-1 <x⁹⁹< 1
⇒ -100<100x⁹⁹<100;
Also 0 ⇒ f’ (x) can either be +ve or -ve on(-1, 1)
∴ f (x) is neither increasing nor decreasing on (-1,1).
(b) for (0,1) i.e. 0<x< 1 x⁹⁹ and cos x are both +ve ∴ f’ (x) > 0
⇒ f (x) is increasing on(0,1)
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.2 Q13.1

Ex 6.2 Class 12 Maths Question 14.
Find the least value of a such that the function f given by f (x) = x² + ax + 1 is strictly increasing on (1,2).
Solution:
We have f (x) = x² + ax + 1
∴ f’ (x) = 2x + a.
Since f (x) is an increasing function on (1,2)
f’ (x) > 0 for all 1 < x < 2 Now, f” (x) = 2 for all x ∈ (1,2) ⇒ f” (x) > 0 for all x ∈ (1,2)
⇒ f’ (x) is an increasing function on (1,2)
⇒ f’ (x) is the least value of f’ (x) on (1,2)
But f’ (x)>0 ∀ x∈ (1,2)
∴ f’ (1)>0 =>2 + a>0
⇒ a > – 2 : Thus, the least value of a is – 2.

Ex 6.2 Class 12 Maths Question 15.
Let I be any interval disjoint from (-1,1). Prove that the function f given by $f(x)=x+\frac { 1 }{ x }$ is strictly increasing on I.
Solution:
Given
$f(x)=x+\frac { 1 }{ x }$
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.2 Q15.1
Hence, f’ (x) is strictly increasing on I.

Ex 6.2 Class 12 Maths Question 16.
Prove that the function f given by f (x) = log sin x is strictly increasing on $\left( 0,\frac { \pi }{ 2 } \right)$ and strictly decreasing on
$\left( \frac { \pi }{ 2 } ,\pi \right)$
Solution:
f’ (x) = $\frac { 1 }{ sin\quad x } .cos\quad x\quad cot\quad x\quad$
when 0 < x < $\frac { \pi }{ 2 }$ , f’ (x) is +ve; i.e., increasing
When $\frac { \pi }{ 2 }$ < x < π, f’ (x) is – ve; i.e., decreasing,
∴ f (x) is decreasing. Hence, f is increasing on (0, π/2) and strictly decreasing on (π/2, π).

Ex 6.2 Class 12 Maths Question 17.
Prove that the function f given by f(x) = log cos x is strictly decreasing on $\left( 0,\frac { \pi }{ 2 } \right)$ and strictly increasing on $\left( \frac { \pi }{ 2 } ,\pi \right)$
Solution:
$f(x)=log\quad cosx$
f’ (x) = $\frac { 1 }{ cosx } (-sinx)=-tanx$
In the interval $\left( 0,\frac { \pi }{ 2 } \right)$ ,f’ (x) = -ve
∴ f is strictly decreasing.
In the interval $\left( \frac { \pi }{ 2 } ,\pi \right)$ , f’ (x) is + ve.
∴ f is strictly increasing in the interval.

Ex 6.2 Class 12 Maths Question 18.
Prove that the function given by
f (x) = x³ – 3x² + 3x -100 is increasing in R.
Solution:
f’ (x) = 3x² – 6x + 3
= 3 (x² – 2x + 1)
= 3 (x -1 )²
Now x ∈ R, f'(x) = (x – 1)²≥0
i.e. f'(x)≥0 ∀ x∈R; hence, f(x) is increasing on R.

Ex 6.2 Class 12 Maths Question 19.
The interval in which y = x² e^-x is increasing is
(a) (-∞,∞)
(b) (-2,0)
(c) (2,∞)
(d) (0,2)
Solution:
(d) f’ (x) = 2xe^-x+ x²( – e^-x) = xe^-x(2-x) = e^-xx(2-x)
Now e^-x is positive for all x ∈ R f’ (x) = 0 at x = 0,2
x = 0, x = 2 divide the number line into three disjoint intervals, viz. (-∞, 0), (0,2), (2, ∞)
(a) Interval (-∞,0) x is +ve and (2-x) is +ve
∴ f’ (x) = e^-xx (2- x)=(+)(-) (+) = -ve
⇒ f is decreasing in (-∞,0)
(b) Interval (0,2) f’ (x) = e^-x x (2 – x)
= (+)(+)(+) = +ve
⇒ f is increasing in (0,2)
(c) Interval (2, ∞) f’ (x) = e^-x x (2 – x) = (+) (+) (-)
= – ve
⇒ f is decreasing in the interval (2, ∞)

Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3

Ex 6.3 Class 12 Maths Question 1.
Find the slope of the tangent to the curve y = 3x⁴ – 4x at x = 4.
Solution:
The curve is y = 3x⁴ – 4x
∴ $\frac { dy }{ dx }$ = 12x³ – 4
∴Req. slope = ${ \left( \frac { dy }{ dx } \right) }_{ x=4 }$
= 12 x 4³ – 4 = 764.

Ex 6.3 Class 12 Maths Question 2.
Find the slope of the tangent to the curve $y=\frac { x-1 }{ x-2 } ,x\neq 2$ at x = 10.
Solution:
The curve is $y=\frac { x-1 }{ x-2 }$
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3 Q2.1

Ex 6.3 Class 12 Maths Question 3.
Find the slope of the tangent to curve y = x³ – x + 1 at the point whose x-coordinate is 2.
Solution:
The curve is y = x³ – x + 1
$\frac { dy }{ dx }$ = 3x² – 1
∴slope of tangent = ${ \left( \frac { dy }{ dx } \right) }_{ x=2 }$
= 3 x 2² – 1
= 11

Ex 6.3 Class 12 Maths Question 4.
Find the slope of the tangent to the curve y = x³ – 3x + 2 at the point whose x-coordinate is 3.
Solution:
The curve is y = x³ – 3x + 2
$\frac { dy }{ dx }$ = 3x² – 3
∴slope of tangent = ${ \left( \frac { dy }{ dx } \right) }_{ x=3 }$
= 3 x 3² – 3
= 24

Ex 6.3 Class 12 Maths Question 5.
Find the slope of the normal to the curve x = a cos³ θ, y = a sin³ θ at θ = $\frac { \pi }{ 4 }$ .
Solution:
$\frac { dx }{ d\theta } =-3a\quad { cos }^{ 2 }\theta sin\theta ,\frac { dy }{ d\theta } =3a\quad { sin }^{ 2 }\theta cos\theta$
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3 Q5.1

Ex 6.3 Class 12 Maths Question 6.
Find the slope of the normal to the curve x = 1 – a sin θ, y = b cos² θ at θ = $\frac { \pi }{ 2 }$
Solution:
$\frac { dx }{ d\theta } =-a\quad cos\theta \quad \& \quad \frac { dy }{ d\theta } =2b\quad cos\theta (-sin\theta )$
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3 Q6.1

Ex 6.3 Class 12 Maths Question 7.
Find points at which the tangent to the curve y = x³ – 3x² – 9x + 7 is parallel to the x-axis.
Solution:
Differentiating w.r.t. x; $\frac { dy }{ dx }$ = 3 (x – 3) (x + 1)
Tangent is parallel to x-axis if the slope of tangent = 0
or $\frac { dy }{ dx }=0$
⇒3(x + 3)(x + 1) = 0
⇒x = -1, 3
when x = -1, y = 12 & When x = 3, y = – 20
Hence the tangent to the given curve are parallel to x-axis at the points (-1, -12), (3, -20)

Ex 6.3 Class 12 Maths Question 8.
Find a point on the curve y = (x – 2)² at which the tangent is parallel to the chord joining the points (2,0) and (4,4).
Solution:
The equation of the curve is y = (x – 2)²
Differentiating w.r.t x
$\frac { dy }{ dx }=2(x-2)$
The point A and B are (2,0) and (4,4) respectively.
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3 Q8.1
Slope of AB = $\frac { { y }_{ 2 }-{ y }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { 4-0 }{ 4-2 } =\frac { 4 }{ 2 }$ = 2 …(i)
Slope of the tangent = 2 (x – 2) ….(ii)
from (i) & (ii) 2 (x – 2)=2
∴ x – 2 = 1 or x = 3
when x = 3,y = (3 – 2)² = 1
∴ The tangent is parallel to the chord AB at (3,1)

Ex 6.3 Class 12 Maths Question 9.
Find the point on the curve y = x³ – 11x + 5 at which the tangent is y = x – 11.
Solution:
Here, y = x³ – 11x + 5
⇒ $\frac { dy }{ dx }$ = 3x² – 11
The slope of tangent line y = x – 11 is 1
∴ 3x² – 11 = 1
⇒ 3x² = 12
⇒ x² = 4, x = ±2
When x = 2, y = – 9 & when x = -2,y = -13
But (-2, -13) does not lie on the curve
∴ y = x – 11 is the tangent at (2, -9)

Ex 6.3 Class 12 Maths Question 10.
Find the equation of all lines having slope -1 that are tangents to the curve $y=\frac { 1 }{ x-1 }$ , x≠1
Solution:
Here
$y=\frac { 1 }{ x-1 }$
⇒ $\frac { dy }{ dx } =\frac { -1 }{ { (x-1) }^{ 2 } }$
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3 Q10.1

Ex 6.3 Class 12 Maths Question 11.
Find the equation of ail lines having slope 2 which are tangents to the curve $y=\frac { 1 }{ x-3 }$ , x≠3.
Solution:
Here
$y=\frac { 1 }{ x-3 }$
$\frac { dy }{ dx } ={ (-1)(x-3) }^{ -2 }=\frac { -1 }{ { (x-3) }^{ 2 } }$
∵ slope of tangent = 2
$\frac { -1 }{ { (x-3) }^{ 2 } } =2\Rightarrow { (x-3) }^{ 2 }=-\frac { 1 }{ 2 }$
Which is not possible as (x – 3)² > 0
Thus, no tangent to $y=\frac { 1 }{ x-3 }$ has slope 2.

Ex 6.3 Class 12 Maths Question 12.
Find the equations of all lines having slope 0 which are tangent to the curve $y=\frac { 1 }{ { x }^{ 2 }-2x+3 }$
Solution:
Let the tangent at the point (x₁, y₁) to the curve
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3 Q12.1

Ex 6.3 Class 12 Maths Question 13.
Find points on the curve $\frac { { x }^{ 2 } }{ 9 } +\frac { { y }^{ 2 } }{ 16 } =1$ at which the tangents are
(a) parallel to x-axis
(b) parallel to y-axis
Solution:
The equation of the curve is $\frac { { x }^{ 2 } }{ 9 } +\frac { { y }^{ 2 } }{ 16 } =1$ …(i)
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3 Q13.1

Ex 6.3 Class 12 Maths Question 14.
Find the equations of the tangent and normal to the given curves at the indicated points:
(i) y = x⁴ – 6x³ + 13x² – 10x + 5 at (0,5)
(ii) y = x⁴ – 6x³ + 13x² – 10x + 5 at (1,3)
(iii) y = x³ at (1, 1)
(iv) y = x² at (0,0)
(v) x = cos t, y = sin t at t = $\frac { \pi }{ 4 }$
Solution:
$\frac { dy }{ dx } ={ 4x }^{ 3 }-18{ x }^{ 2 }+26x-10$
Putting x = 0, $\frac { dy }{ dx }$ at (0,5) = – 10
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3 Q14.1

Ex 6.3 Class 12 Maths Question 15.
Find the equation of the tangent line to the curve y = x² – 2x + 7 which is
(a) parallel to the line 2x – y + 9 = 0
(b) perpendicular to the line 5y – 15x = 13.
Solution:
Equation of the curve is y = x² – 2x + 7 …(i)
$\frac { dy }{ dx }$ = 2x – 2 = 2(x – 1)
(a) Slope of the line 2x – y + 9 = 0 is 2
⇒ Slope of tangent = $\frac { dy }{ dx }$ = 2(x – 1) = 2
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3 Q15.1

Ex 6.3 Class 12 Maths Question 16.
Show that the tangents to the curve y = 7x³ + 11 at the points where x = 2 and x = – 2 are parallel.
Solution:
Here, y = 7x³ + 11
=> x $\frac { dy }{ dx }$ = 21 x²
Now m₁ = slope at x = 2 is ${ \left( \frac { dy }{ dx } \right) }_{ x=2 }$ = 21 x 2² = 84
and m₂ = slope at x = -2 is ${ \left( \frac { dy }{ dx } \right) }_{ x=-2 }$ = 21 x (-2)² = 84
Hence, m₁ = m₂ Thus, the tangents to the given curve at the points where x = 2 and x = – 2 are parallel

Ex 6.3 Class 12 Maths Question 17.
Find the points on the curve y = x³ at which the slope of the tangent is equal to the y-coordinate of the point
Solution:
Let P (x₁, y₁) be the required point.
The given curve is: y = x³
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3 Q17.1

Ex 6.3 Class 12 Maths Question 18.
For the curve y = 4x³ – 2x⁵, find all the points at which the tangent passes through the origin.
Solution:
Let (x₁, y₁) be the required point on the given curve y = 4x³ – 2x⁵, then y₁ = 4x₁³ – 2x₁⁵ …(i)
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3 Q18.1

Ex 6.3 Class 12 Maths Question 19.
Find the points on the curve x² + y² – 2x – 3 = 0 at which the tangents are parallel to the x-axis.
Solution:
Here, x² + y² – 2x – 3 = 0
=> $\frac { dy }{ dx } =\frac { 1-x }{ y }$
Tangent is parallel to x-axis, if $\frac { dy }{ dx }=0$ i.e.
if 1 – x = 0
⇒ x = 1
Putting x = 1 in (i)
⇒ y = ±2
Hence, the required points are (1,2), (1, -2) i.e. (1, ±2).

Ex 6.3 Class 12 Maths Question 20.
Find the equation of the normal at the point (am², am³) for the curve ay² = x³.
Solution:
Here, ay² = x³
$2ay\frac { dy }{ dx } ={ 3x }^{ 2 }\Rightarrow \frac { dy }{ dx } =\frac { { 3x }^{ 2 } }{ 2ay }$
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3 Q20.1

Ex 6.3 Class 12 Maths Question 21.
Find the equation of the normal’s to the curve y = x³ + 2x + 6 which are parallel to the line x + 14y + 4 = 0.
Solution:
Let the required normal be drawn at the point (x₁, y₁)
The equation of the given curve is y = x³ + 2x + 6 …(i)
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3 Q21.1

Ex 6.3 Class 12 Maths Question 22.
Find the equations of the tangent and normal to the parabola y² = 4ax at the point (at²,2at).
Solution:
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3 Q22.1

Ex 6.3 Class 12 Maths Question 23.
Prove that the curves x = y² and xy = k cut at right angles if 8k² = 1.
Solution:
The given curves are x = y² …(i)
and xy = k …(ii)
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3 Q23.1

Ex 6.3 Class 12 Maths Question 24.
Find the equations of the tangent and normal to the hyperbola $\frac { { x }^{ 2 } }{ { a }^{ 2 } } -\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1$ at the point (x₀ ,y₀).
Solution:
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3 Q24.1

Ex 6.3 Class 12 Maths Question 25.
Find the equation of the tangent to the curve $y=\sqrt { 3x-2 }$ which is parallel to the line 4x – 2y + 5 = 0.
Solution:
Let the point of contact of the tangent line parallel to the given line be P (x₁, y₁) The equation of the curve is $y=\sqrt { 3x-2 }$
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.3 Q25.1

Choose the correct answer in Exercises 26 and 27.

Ex 6.3 Class 12 Maths Question 26.
The slope of the normal to the curve y = 2x² + 3 sin x at x = 0 is
(a) 3
(b) $\frac { 1 }{ 3 }$
(c) -3
(d) $-\frac { 1 }{ 3 }$
Solution:
(d) ∵ y = 2x² + 3sinx
∴ $\frac { dy }{ dx }=4x+3cosx$ at
x = 0, $\frac { dy }{ dx }=3$
∴ slope = 3
⇒ slope of normal is = $\frac { 1 }{ 3 }$

Ex 6.3 Class 12 Maths Question 27.
The line y = x + 1 is a tangent to the curve y² = 4x at the point
(a) (1,2)
(b) (2,1)
(c) (1,-2)
(d) (-1,2)
Solution:
(a) The curve is y² = 4x,
∴ $\frac { dy }{ dx } =\frac { 4 }{ 2y } =\frac { 2 }{ y }$
Slope of the given line y = x + 1 is 1 ∴ $\frac { 2 }{ y }=1$
y = 2 Putting y= 2 in y² = 4x 2² = 4x
⇒ x = 1
∴ Point of contact is (1,2)

Class 12 Maths Chapter 6 Application of Derivatives Ex 6.

Ex 6.5 Class 12 Maths Question 1.
Find the maximum and minimum values, if any, of the following functions given by
(i) f(x) = (2x – 1)² + 3
(ii) f(x) = 9x² + 12x + 2
(iii) f(x) = – (x – 1)² + 10
(iv) g(x) = x³ + 1
Solution:
(i) Minimum value of (2x – 1)² is zero.
Minimum value of (2x – 1)² + 3 is 3
Clearly it does not have maximum value,
(ii) f(x) = 9x² + 12x + 2
⇒ f(x) = (3x + 2)² – 2
Minimum value of (3 + 2)² is zero.
∴ Min.value of (3x + 2)² – 2 = 9x² + 12x + 2 is – 2
f (x) does not have finite maximum value
(iii) f(x) = – (x – 1)² + 10
Maximum value of – (x – 1)² is zero
Maximum valuer f f(x) = – (x – 1)² + 10 is 10
f (x) does not have finite minimum value.
(iv) As x—»∞,g(x)—»∞;Also x—»-∞,g(x)—»-∞
Thus there is no maximum or minimum value of f(x)

Ex 6.5 Class 12 Maths Question 2.
Find the maximum and minimum values, if any, of the following functions given by
(i) f(x) = |x + 2| – 1
(ii) g(x) = -|x + 1| + 3
(iii) h (x) = sin 2x + 5
(iv) f(x) = |sin(4x + 3)|
(v) h(x) = x + 1,x∈(-1,1)
Solution:
(i) We have :f(x) = |x + 2 |-1 ∀x∈R
Now |x + 2|≥0∀x∈R
|x + 2| – 1 ≥ – 1 ∀x∈R ,
So -1 is the min. value of f(x)
now f(x) = -1
⇒ |x + 2|-1
⇒ |x + 2| = 0
⇒ x = – 2
(ii) We have g(x) = -|x + 1| + 3 ∀x∈R
Now | x + 1| ≥ 0 ∀x∈R
-|x+ 1| + 3 ≤3 ∀x∈R
So 3 is the minimum value of f(x).
Now f(x) = 3
⇒ -|x+1| + 3
⇒ |x+1| = 0
⇒ x = – 1.
(iii) Thus maximum value of f(x) is 6 and minimum value is 4.
(iv) Let f(x) = |sin4x + 3|
Maximum value of sin 4x is 1
∴ Maximum value of |sin(4x+3)| is |1+3| = 4
Minimum value of sin 4x is -1
∴ Minimum value of f(x) is |-1+3| = |2|= 2
(v) Greatest value of f (x) is 2 and least value is 0.

Ex 6.5 Class 12 Maths Question 3.
Find the local maxima and local minima, if any, of the following functions. Find also the local maximum and the local minimum values, as the case may be:
(i) f(x) = x²
(ii) g(x) = x³ – 3x
(iii) h(x) = sinx+cosx,0<x< $\frac { \pi }{ 4 }$
(iv) f(x) = sin⁴x + cos⁴x,0<x< $\frac { \pi }{ 2 }$
(v) f(x) = x³– 6x² + 9x:+15
(vi) g(x) = $\frac { x }{ 2 } +\frac { 2 }{ x }$ , x>0
(vii) g(x) = $\frac { 1 }{ { x }^{ 2 }+2 }$ , x>0
(viii) f(x) = $x\sqrt { 1-x }$ , x>0
Solution:
(i) Let f(x) = x² ⇒ f’(x) = 2x
Now f'(x) = 0 ⇒ 2x = 0 i.e., x = 0
At x = 0; When x is slightly < 0, f’ (x) is -ve When x is slightly > 0, f(x) is +ve
∴ f(x) changes sign from -ve to +ve as x increases through 0.
⇒ f’ (x) has a local minimum at x = 0 local minimum value f(0) = 0.
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.5 Q3.1

Ex 6.5 Class 12 Maths Question 4.
Prove that the following functions do not have maxima or minima:
(i) f(x) = e^x
(ii) f(x) = log x
(iii) h(x) = x³ + x² + x + 1
Solution:
(i) f'(x) = e^x;
Since f’ (x) ≠ 0 for any value of x.
So f(x) = e^x does not have a max. or min.
(ii) f’ (x) = $\frac { 1 }{ x }$ ; Clearly f’ (x) ≠ 0 for any value of x.
So,f’ (x) = log x does not have a maximum or a minimum.
(iii) We have f(x) = x³ + x² + x + 1
⇒f’ (x) = 3x² + 2x + 1
Now, f’ (x) = 0 => 3x² + 2x + 1 = 0
$x=\frac { -2\pm \sqrt { 4-12 } }{ 6 } =\frac { -1+\sqrt { -2 } }{ 3 }$
i.e. f'(x) = 0 at imaginary points
i.e. f'(x) ≠ 0 for any real value of x
Hence, there is neither max. nor min.

Ex 6.5 Class 12 Maths Question 5.
Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals:
(i) f(x) = x³, x∈ [-2,2]
(ii) f(x) = sin x + cos x, x ∈ [0, π]
(iii) f(x) = $4x-\frac { 1 }{ 2 } { x }^{ 2 },x\in \left[ -2,\frac { 9 }{ 2 } \right]$
(iv) f(x) = ${ (x-1) }^{ 2 }+3,x\in \left[ -3,1 \right]$
Solution:
(i) We have f’ (x) = x³ in [ -2,2]
∴ f'(x) = 3x²; Now, f’ (x) = 0 at x = 0, f(0) = 0
Now, f(-2) = (-2)³ = – 8; f(0) = (0)² = 0 and f(0) = (2) = 8
Hence, the absolute maximum value of f (x) is 8 which it attained at x = 2 and absolute minimum value of f(x) = – 8 which is attained at x = -2.
(ii) We have f (x) = sin x + cos x in [0, π]
f’ (x) = cos x – sin x for extreme values f’ (x) = 0
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.5 Q5.1

Ex 6.5 Class 12 Maths Question 6.
Find the maximum profit that a company can make, if the profit function is given by p(x) = 41 – 24x – 18x²
Solution:
Profit function in p(x) = 41 – 24x – 18x²
∴ p'(x) = – 24 – 36x = – 12(2 + 3x)
for maxima and minima, p'(x) = 0
Now, p'(x) = 0
⇒ – 12(2 + 3x) = 0
⇒ x = $-\frac { 2 }{ 3 }$ ,
p'(x) changes sign from +ve to -ve.
⇒ p (x) has maximum value at x = $-\frac { 2 }{ 3 }$
Maximum Profit = 41 + 16 – 8 = 49.

Ex 6.5 Class 12 Maths Question 7.
Find both the maximum value and the minimum value of 3x⁴ – 8x³ + 12x² – 48x + 25 on the interval [0,3].
Solution:
Let f(x) = 3x⁴ – 8x³ + 12x² – 48x + 25
∴f'(x) = 12x³ – 24x² + 24x – 48
= 12(x² + 2)(x – 2)
For maxima and minima, f'(x) = 0
⇒ 12(x² + 2)(x – 2) = 0
⇒ x = 2
Now, we find f (x) at x = 0,2 and 3, f (0) = 25,
f (2) = 3 (2⁴) – 8 (2³) + 12 (2²) – 48 (2) + 25 = – 39
and f (3) = (3⁴) – 8 (3³) + 12 (3²) – 48 (3) + 25
= 243 – 216 + 108 – 144 + 25 = 16
Hence at x = 0, Maximum value = 25
at x = 2, Minimum value = – 39.

Ex 6.5 Class 12 Maths Question 8.
At what points in the interval [0,2π], does the function sin 2x attain its maximum value?
Solution:
We have f (x) = sin 2x in [0,2π], f’ (x) = 2 cos 2 x
For maxima and minima f’ (x) = 0 => cos 2 x = 0
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.5 Q8.1

Ex 6.5 Class 12 Maths Question 9.
What is the maximum value of the function sin x + cos x?
Solution:
Consider the interval [0, 2π],
Let f(x) = sinx + cosx,
f’ (x) = cosx – sinx
For maxima and minima, f’ (x) = 0
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.5 Q9.1

Ex 6.5 Class 12 Maths Question 10.
Find the maximum value of 2x³ – 24x + 107 in the interval [1,3]. Find the maximum value of the same function in [-3, -1].
Solution:
∵ f(x) = 2x³ – 24x + 107
∴f(x) = 6x² – 24 ,
For maxima and minima f'(x) = 0;⇒ x = ±2
For the interval [ 1,3], we find the values of f (x)
at x = 1,2,3; f(1) = 85, f(2) = 75, f(3) = 89
Hence, maximum f (x) = 89 at x = 3
For the interval [-3, -1], we find the values of f(x) at x = – 3, – 2, – 1;
f(-3) = 125;
f(-2) = 139
f(-1) = 129
∴ max.f(x) = 139 at x = – 2.

Ex 6.5 Class 12 Maths Question 11.
It is given that at x = 1, the function x⁴ – 62x² + ax + 9 attains its maximum value, on the interval [0,2]. Find the value of a.
Solution:
∵ f(x) = x⁴ – 62x² + ax + 9
∴ f’ (x) = 4x³ – 124x + a
Now f’ (x) = 0 at x = 1
⇒ 4 – 124 + a = 0
⇒ a = 120
Now f” (x) = 12x² – 124:
At x = 1 f” (1) = 12 – 124 = – 112 < 0
⇒ f(x) has a maximum at x = 1 when a = 120.

Ex 6.5 Class 12 Maths Question 12.
Find the maximum and minimum values of x + sin 2x on [0,2π]
Solution:
∴f(x) = x + sin2x on[0,2π]
∴f’ (x) = 1+2 cos2x
For maxima and minima f’ (x) = 0
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.5 Q12.1

Ex 6.5 Class 12 Maths Question 13.
Find two numbers whose sum is 24 and whose product is as large as possible.
Solution:
Let the required numbers hex and (24-x)
∴Their product,p = x(24 – x) = 24x – x²
Now $\frac { dp }{ dx }$ = 0 ⇒24 – 2x = 0 ⇒ x = 12
Also $\frac { { d }^{ 2 }p }{ { dx }^{ 2 } }$ = -2<0: ⇒ p is max at x = 12
Hence, the required numbers are 12 and (24-12)i.e. 12.

Ex 6.5 Class 12 Maths Question 14.
Find two positive numbers x and y such that x + y = 60 and xy³ is maximum.
Solution:
We have x + y = 60
⇒ y = 60 – x …(i)
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.5 Q14.1
Hence, the req. numbers are 15 and (60 -15) i.e. 15 and 45.

Ex 6.5 Class 12 Maths Question 15.
Find two positive numbers x and y such that their sum is 35 and the product x² y⁵ is a maximum.
Solution:
We have x + y = 35 ⇒ y = 35 – x
Product p = x² y⁵
= x² (35 – x)⁵
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.5 Q15.1

Ex 6.5 Class 12 Maths Question 16.
Find two positive numbers whose sum is 16 and the sum of whose cubes is minimum.
Solution:
Let two numbers be x and 16 – x
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.5 Q16.1
Hence, the required numbers are 8 and (16-8) i.e. 8 and 8.

Ex 6.5 Class 12 Maths Question 17.
A square piece of tin of side 18 cm is to be made into a box without top, by cutting a square from each corner and folding up the flaps to form the box. What should be the side of the square to be cut off so that the volume of the box is the maximum possible.
Solution:
Let each side of the square to be cut off be x cm.
∴ for the box length = 18 – 2x: breadth = 18 – 2x and height = x
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.5 Q17.1

Ex 6.5 Class 12 Maths Question 18.
A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top, by cutting off square from each comer and folding up the flaps. What should be the side of the square to be cut off so that the volume of the box is maximum?
Solution:
Let each side of the square cut off from each comer be x cm.
∴ Sides of the rectangular box are (45 – 2x), (24 – 2x) and x cm.
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.5 Q18.1

Ex 6.5 Class 12 Maths Question 19.
Show that of all the rectangles inscribed in a given fixed circle, the square has the maximum area.
Solution:
Let the length and breadth of the rectangle inscribed in a circle of radius a be x and y respectively.
∴ x² + y² = (2a)² => x² + y² = 4a² …(i)
∴ Perimeter = 2 (x + y)
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.5 Q19.1

Ex 6.5 Class 12 Maths Question 20.
Show that the right circular cylinder of given surface and maximum volume is such that its height is equal to the diameter of the base.
Solution:
Let S be the given surface area of the closed cylinder whose radius is r and height h let v be the its Volume. Then
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.5 Q20.1

Ex 6.5 Class 12 Maths Question 21.
Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic centimeters, find the dimensions of the can which has the minimum surface area ?
Solution:
Let r be the radius and h be the height of cylindrical can.
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.5 Q21.1

Ex 6.5 Class 12 Maths Question 22.
A wire of length 28 m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a circle. What should be the length of the two pieces so that the combined area of the square and the circle is minimum ?
Solution:
Let one part be of length x, then the other part = 28 – x
Let the part of the length x be converted into a circle of radius r.
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.5 Q22.1

Ex 6.5 Class 12 Maths Question 23.
Prove that the volume of the largest cone that can be inscribed in a sphere of radius R is $\frac { 8 }{ 27 }$ of the volume of the sphere.
Solution:
Let a cone. VAB of greatest volume be inscribed in the sphere let AOC = θ
∴ AC, radius of the base of the cone = R sin θ
and VC = VO + OC = R(1 +cosθ)
= R + Rcosθ
= height of the cone.,
V, the volume of the cone.
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.5 Q23.1

Ex 6.5 Class 12 Maths Question 24.
Show that die right circular cone of least curved surface and given volume has an altitude equal to √2 time the radius of the base.
Solution:
Let r and h be the radius and height of the cone.
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.5 Q24.1

Ex 6.5 Class 12 Maths Question 25.
Show that the semi-vertical angle of the cone of the maximum volume and of given slant height is tan^-1 √2.
Solution:
Let v be the volume, l be the slant height and 0 be the semi vertical angle of a cone.
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.5 Q25.1

Ex 6.5 Class 12 Maths Question 26.
Show that semi-vertical angle of right circular cone of given surface area and maximum volume is ${ sin }^{ -1 }\left( \frac { 1 }{ 3 } \right)$
Solution:
Let r be radius, l be the slant height and h be the height of the cone of given surface area s.Then
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.5 Q26.1

Choose the correct answer in the Exercises 27 and 29.

Ex 6.5 Class 12 Maths Question 27.
The point on die curve x² = 2y which is nearest to the point (0,5) is
(a) (2 √2,4)
(b) (2 √2,0)
(c) (0,0)
(d) (2,2)
Solution:
(a) Let P (x, y) be a point on the curve The other point is A (0,5)
Z = PA² = x² + y² + 25 – 10y [∵ x² = 2y]
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.5 Q27.1

Ex 6.5 Class 12 Maths Question 28.
For all real values of x, the minimum value of $\frac { 1-x+{ x }^{ 2 } }{ 1+x+{ x }^{ 2 } }$
(a) 0
(b) 1
(c) 3
(d) $\frac { 1 }{ 3 }$
Solution:
(d) Let $y=\frac { 1-x+{ x }^{ 2 } }{ 1+x+{ x }^{ 2 } }$
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.5 Q28.1

Ex 6.5 Class 12 Maths Question 29.
The maximum value of ${ \left[ x\left( x-1 \right) +1 \right] }^{ \frac { 1 }{ 3 } },0\le x\le 1$ is
(a) ${ \left( \frac { 1 }{ 3 } \right) }^{ \frac { 1 }{ 3 } }$
(b) $\frac { 1 }{ 2 }$
(c) 1
(d) 0
Solution:
(c) Let y = ${ \left[ x\left( x-1 \right) +1 \right] }^{ \frac { 1 }{ 3 } },0\le x\le 1$
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.5 Q29.1

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Class 12th Chapter -5 Continuity and Differentiability | NCERT Maths Solution | NCERT Solution | Edugrown

NCERT Solutions for Class 12 Maths Chapter : 5 Continuity and Differentiability

Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1

Ex 5.1 Class 12 Maths Question 1.
Prove that the function f (x) = 5x – 3 is continuous at x = 0, at x = – 3 and at x = 5.
Solution:
(i) At x = 0. lim_x–>0 f (x) = lim_x–>0 (5x – 3) = – 3 and
f(0) = – 3
∴f is continuous at x = 0
(ii) At x = – 3, lim_x–>3 f(x)= lim_x–>-3 (5x – 3) = – 18
and f( – 3) = – 18
∴ f is continuous at x = – 3
(iii) At x = 5, lim_x–>5 f(x) = lim_x–>5 (5x – 3) = 22 and
f(5) = 22
∴ f is continuous at x = 5

Ex 5.1 Class 12 Maths Question 2.
Examine the continuity of the function f(x) = 2x² – 1 at x = 3.
Solution:
lim_x–>3 f(x) = lim_x–>3 (2x² – 1) = 17 and f(3)= 17
∴ f is continuous at x = 3

Ex 5.1 Class 12 Maths Question 3.
Examine the following functions for continuity.
(a) f(x) = x – 5
(b) f(x) = $\\ \frac { 1 }{ x-5 }$ , x≠5
(c) f(x) = $\frac { { x }^{ 2 }-25 }{ x+5 }$ ,x≠5
(d) f(x) = |x – 5|
Solution:
(a) f(x) = (x-5) => (x-5) is a polynomial
∴it is continuous at each x ∈ R.
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q3.1

Ex 5.1 Class 12 Maths Question 4.
Prove that the function f (x) = xⁿ is continuous at x = n, where n is a positive integer.
Solution:
f (x) = xⁿ is a polynomial which is continuous for all x ∈ R.
Hence f is continuous at x = n, n ∈ N.

Ex 5.1 Class 12 Maths Question 5.
Is the function f defined by $f(x)=\begin{cases} x,ifx\le 1 \\ 5,ifx>1 \end{cases}$ continuous at x = 0? At x = 1? At x = 2?
Solution:
(i) At x = 0
lim_x–>0- f(x) = lim_x–>0- x = 0 and
lim_x–>0+ f(x) = lim_x–>0+ x = 0 => f(0) = 0
∴ f is continuous at x = 0
(ii) At x = 1
lim_x–>1- f(x) = lim_x–>1- (x) = 1 and
lim_x–>1+ f(x) = lim_x–>1+(x) = 5
∴ lim_x–>1- f(x) ≠ lim_x–>1+ f(x)
∴ f is discontinuous at x = 1
(iii) At x = 2
lim_x–>2 f(x) = 5, f(2) = 5
∴ f is continuous at x = 2

Find all points of discontinuity off, where f is defined by

Ex 5.1 Class 12 Maths Question 6.
$f(x)=\begin{cases} 2x+3,if\quad x\le 2 \\ 2x-3,if\quad x>2 \end{cases}$
Solution:
$f(x)=\begin{cases} 2x+3,if\quad x\le 2 \\ 2x-3,if\quad x>2 \end{cases}$ at x≠2
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q6.1

Ex 5.1 Class 12 Maths Question 7.
$f(x)=\begin{cases} |x|+3,if\quad x\le -3 \\ -2x,if\quad -3<x<3 \\ 6x+2,if\quad x\ge 3 \end{cases}$
Solution:
$f(x)=\begin{cases} |x|+3,if\quad x\le -3 \\ -2x,if\quad -3<x<3 \\ 6x+2,if\quad x\ge 3 \end{cases}$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q7.1

Ex 5.1 Class 12 Maths Question 8.
Test the continuity of the function f (x) at x = 0
$f(x)=\begin{cases} \frac { |x| }{ x } ;x\neq 0 \\ 0;x=0 \end{cases}$
Solution:
We have;
(LHL at x=0)
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q8.1

Ex 5.1 Class 12 Maths Question 9.
$f(x)=\begin{cases} \frac { x }{ |x| } ;if\quad x<0 \\ -1,if\quad x\ge 0 \end{cases}$
Solution:
$f(x)=\begin{cases} \frac { x }{ |x| } ;if\quad x<0 \\ -1,if\quad x\ge 0 \end{cases}$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q9.1

Ex 5.1 Class 12 Maths Question 10.
$f(x)=\begin{cases} x+1,if\quad x\ge 1 \\ { x }^{ 2 }+1,if\quad x<1 \end{cases}$
Solution:
$f(x)=\begin{cases} x+1,if\quad x\ge 1 \\ { x }^{ 2 }+1,if\quad x<1 \end{cases}$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q10.1

Ex 5.1 Class 12 Maths Question 11.
$f(x)=\begin{cases} { x }^{ 3 }-3,if\quad x\le 2 \\ { x }^{ 2 }+1,if\quad x>2 \end{cases}$
Solution:
$f(x)=\begin{cases} { x }^{ 3 }-3,if\quad x\le 2 \\ { x }^{ 2 }+1,if\quad x>2 \end{cases}$
At x = 2, L.H.L. lim_x–>2- (x³ – 3) = 8 – 3 = 5
R.H.L. = lim_x–>2+ (x² + 1) = 4 + 1 = 5
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q11.1

Ex 5.1 Class 12 Maths Question 12.
$f(x)=\begin{cases} { x }^{ 10 }-1,if\quad x\le 1 \\ { x }^{ 2 },if\quad x>1 \end{cases}$
Solution:
$f(x)=\begin{cases} { x }^{ 10 }-1,if\quad x\le 1 \\ { x }^{ 2 },if\quad x>1 \end{cases}$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q12.1

Ex 5.1 Class 12 Maths Question 13.
Is the function defined by $f(x)=\begin{cases} x+5,if\quad x\le 1 \\ x-5,if\quad x>1 \end{cases}$ a continuous function?
Solution:
At x = 1,L.H.L.= lim_x–>1- f(x) = lim_x–>1- (x + 5) = 6,
R.HL. = lim_x–>1+ f(x) = lim_x–>1+ (x – 5) = – 4
f(1) = 1 + 5 = 6,
f(1) = L.H.L. ≠ R.H.L.
=> f is not continuous at x = 1
At x = c < 1, lim_x–>c (x + 5) = c + 5 = f(c)
At x = c > 1, lim_x–>c (x – 5) = c – 5 = f(c)
∴ f is continuous at all points x ∈ R except x = 1.

Discuss the continuity of the function f, where f is defined by

Ex 5.1 Class 12 Maths Question 14.
$f(x)=\begin{cases} 3,if\quad 0\le x\le 1 \\ 4,if\quad 1<x<3 \\ 5,if\quad 3\le x\le 10 \end{cases}$
Solution:
$f(x)=\begin{cases} 3,if\quad 0\le x\le 1 \\ 4,if\quad 1<x<3 \\ 5,if\quad 3\le x\le 10 \end{cases}$
In the interval 0 ≤ x ≤ 1,f(x) = 3; f is continuous in this interval.
At x = 1,L.H.L. = lim f(x) = 3,
R.H.L. = lim_x–>1+ f(x) = 4 => f is discontinuous at
x = 1
At x = 3, L.H.L. = lim_x–>3- f(x)=4,
R.H.L. = lim_x–>3+ f(x) = 5 => f is discontinuous at
x = 3
=> f is not continuous at x = 1 and x = 3.

Ex 5.1 Class 12 Maths Question 15.
$f(x)=\begin{cases} 2x,if\quad x<0 \\ 0,if\quad 0\le x\le 1 \\ 4x,if\quad x>1 \end{cases}$
Solution:
$f(x)=\begin{cases} 2x,if\quad x<0 \\ 0,if\quad 0\le x\le 1 \\ 4x,if\quad x>1 \end{cases}$
At x = 0, L.H.L. = lim_x–>0- 2x = 0 ,
R.H.L. = lim_x–>0+ (0)= 0 , f(0) = 0
=> f is continuous at x = 0
At x = 1, L.H.L. = lim_x–>1- (0) = 0,
R.H.L. = lim_x–>1+ 4x = 4
f(1) = 0, f(1) = L.H.L.≠R.H.L.
∴ f is not continuous at x = 1
when x < 0 f (x) = 2x, being a polynomial, it is
continuous at all points x < 0. when x > 1. f (x) = 4x being a polynomial, it is
continuous at all points x > 1.
when 0 ≤ x ≤ 1, f (x) = 0 is a continuous function
the point of discontinuity is x = 1.

Ex 5.1 Class 12 Maths Question 16.
$f(x)=\begin{cases} -2,if\quad x\le -1 \\ 2x,if\quad -1<x\le 1 \\ 2,if\quad x>1 \end{cases}$
Solution:
$f(x)=\begin{cases} -2,if\quad x\le -1 \\ 2x,if\quad -1<x\le 1 \\ 2,if\quad x>1 \end{cases}$
At x = – 1,L.H.L. = lim_x–>1- f(x) = – 2, f(-1) = – 2,
R.H.L. = lim_x–>1+ f(x) = – 2
=> f is continuous at x = – 1
At x= 1, L.H.L. = lim_x–>1- f(x) = 2,f(1) = 2
∴ f is continuous at x = 1,
R.H.L. = lim_x–>1+ f(x) = 2
Hence, f is continuous function.

Ex 5.1 Class 12 Maths Question 17.
Find the relationship between a and b so that the function f defined by
$f(x)=\begin{cases} ax+1,if\quad x\le 3 \\ bx+3,if\quad x>3 \end{cases}$
is continuous at x = 3
Solution:
At x = 3, L.H.L. = lim_x–>3- (ax+1) = 3a+1 ,
f(3) = 3a + 1, R.H.L. = lim_x–>3+ (bx+3) = 3b+3
f is continuous ifL.H.L. = R.H.L. = f(3)
3a + 1 = 3b + 3 or 3(a – b) = 2
a – b = $\\ \frac { 2 }{ 3 }$ or a = b + $\\ \frac { 2 }{ 3 }$ , for any arbitrary value of b.
Therefore the value of a corresponding to the value of b.

Ex 5.1 Class 12 Maths Question 18.
For what value of λ is the function defined by
$f(x)=\begin{cases} \lambda ({ x }^{ 2 }-2x),if\quad x\le 0 \\ 4x+1,if\quad x>0 \end{cases}$
continuous at x = 0? What about continuity at x = 1?
Solution:
At x = 0, L.H.L. = lim_x–>0- λ (x² – 2x) = 0 ,
R.H.L. = lim_x–>0+ (4x+ 1) = 1, f(0)=0
f (0) = L.H.L. ≠ R.H.L.
=> f is not continuous at x = 0,
whatever value of λ ∈ R may be
At x = 1, lim_x–>1 f(x) = lim_x–>1 (4x + l) = f(1)
=> f is not continuous at x = 0 for any value of λ but f is continuous at x = 1 for all values of λ.

Ex 5.1 Class 12 Maths Question 19.
Show that the function defined by g (x) = x – [x] is discontinuous at all integral points. Here [x] denotes the greatest integer less than or equal to x.
Solution:
Let c be an integer, [c – h] = c – 1, [c + h] = c, [c] = c, g(x) = x – [x].
At x = c, lim_x–>c- (x – [x]) = lim_h–>0 [(c – h) – (c – 1)]
= lim_h–>0 (c – h – (c – 1)) = 1[∵ [c – h] = c – 1]
R.H.L. = lim_x–>c+ (x – [x])= lim_h–>0 (c + h – [c + h])
= lim_h–>0 [c + h – c] = 0
f(c) = c – [c] = 0,
Thus L.H.L. ≠ R.H.L. = f (c) => f is not continuous at integral points.

Ex 5.1 Class 12 Maths Question 20.
Is the function defined by f (x) = x² – sin x + 5 continuous at x = π?
Solution:
Let f(x) = x² – sinx + 5,
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q20.1

Ex 5.1 Class 12 Maths Question 21.
Discuss the continuity of the following functions:
(a) f (x) = sin x + cos x
(b) f (x) = sin x – cos x
(c) f (x) = sin x · cos x
Solution:
(a) f(x) = sinx + cosx
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q21.1

Ex 5.1 Class 12 Maths Question 22.
Discuss the continuity of the cosine, cosecant, secant and cotangent functions.
Solution:
(a) Let f(x) = cosx
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q22.1

Ex 5.1 Class 12 Maths Question 23.
Find all points of discontinuity of f, where
$f(x)=\begin{cases} \frac { sinx }{ x } ,if\quad x<0 \\ x+1,if\quad x\ge 0 \end{cases}$
Solution:
At x = 0
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q23.1

Ex 5.1 Class 12 Maths Question 24.
Determine if f defined by $f(x)=\begin{cases} { x }^{ 2 }sin\frac { 1 }{ x } ,if\quad x\neq 0 \\ 0,if\quad x=0 \end{cases}$ is a continuous function?
Solution:
At x = 0
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q24.1

Ex 5.1 Class 12 Maths Question 25.
Examine the continuity of f, where f is defined by $f(x)=\begin{cases} sinx-cosx,if\quad x\neq 0 \\ -1,if\quad x=0 \end{cases}$
Solution:
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q25.1

Find the values of k so that the function is continuous at the indicated point in Questions 26 to 29.

Ex 5.1 Class 12 Maths Question 26.
$f(x)=\begin{cases} \frac { k\quad cosx }{ \pi -2x } ,\quad if\quad x\neq \frac { \pi }{ 2 } \quad at\quad x=\frac { \pi }{ 2 } \qquad \\ 3,if\quad x=\frac { \pi }{ 2 } \quad at\quad x=\frac { \pi }{ 2 } \end{cases}$
Solution:
At x = $\frac { \pi }{ 2 }$
L.H.L = $\underset { x\rightarrow { \left( \frac { \pi }{ 2 } \right) }^{ - } }{ lim } \frac { k\quad cosx }{ \pi -2x }$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q26.1

Ex 5.1 Class 12 Maths Question 27.
$f(x)=\begin{cases} { kx }^{ 2 },if\quad x\le 2\quad at\quad x=2 \\ 3,if\quad x>2\quad at\quad x=2 \end{cases}$
Solution:
$f(x)=\begin{cases} { kx }^{ 2 },if\quad x\le 2\quad at\quad x=2 \\ 3,if\quad x>2\quad at\quad x=2 \end{cases}$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q27.1

Ex 5.1 Class 12 Maths Question 28.
$f(x)=\begin{cases} kx+1,if\quad x\le \pi \quad at\quad x=\pi \\ cosx,if\quad x>\pi \quad at\quad x=\pi \end{cases}$
Solution:
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q28.1

Ex 5.1 Class 12 Maths Question 29.
$f(x)=\begin{cases} kx+1,if\quad x\le 5\quad at\quad x=5 \\ 3x-5,if\quad x>5\quad at\quad x=5 \end{cases}$
Solution:
$f(x)=\begin{cases} kx+1,if\quad x\le 5\quad at\quad x=5 \\ 3x-5,if\quad x>5\quad at\quad x=5 \end{cases}$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q29.1

Ex 5.1 Class 12 Maths Question 30.
Find the values of a and b such that the function defined by
$f(x)=\begin{cases} 5,if\quad x\le 2 \\ ax+b,if\quad 2<x<10 \\ 21,if\quad x\ge 10 \end{cases}$
to is a continuous function.
Solution:
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q30.1

Ex 5.1 Class 12 Maths Question 31.
Show that the function defined by f(x)=cos (x²) is a continuous function.
Solution:
Now, f (x) = cosx², let g (x)=cosx and h (x) x²
∴ goh(x) = g (h (x)) = cos x²
Now g and h both are continuous ∀ x ∈ R.
f (x) = goh (x) = cos x² is also continuous at all x ∈ R.

Ex 5.1 Class 12 Maths Question 33.
Examine that sin |x| is a continuous function.
Solution:
Let g (x) = sin x, h (x) = |x|, goh (x) = g (h(x))
= g(|x|) = sin|x| = f(x)
Now g (x) = sin x and h (x) = |x| both are continuous for all x ∈ R.
∴f (x) = goh (x) = sin |x| is continuous at all x ∈ R.

Ex 5.1 Class 12 Maths Question 34.
Find all the points of discontinuity of f defined by f(x) = |x|-|x+1|.
Solution:
f(x) = |x|-|x+1|, when x< – 1,
f(x) = -x-[-(x+1)] = – x + x + 1 = 1
when -1 ≤ x < 0, f(x) = – x – (x + 1) = – 2x – 1,
when x ≥ 0, f(x) = x – (x + 1) = – 1
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 Q34.1

Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.2

Differentiate the functions with respect to x in Questions 1 to 8.

Ex 5.2 Class 12 Maths Question 1.
sin(x² + 5)
Solution:
Let y = sin(x2 + 5),
put x² + 5 = t
y = sint
t = x²+5
$\frac { dy }{ dx } =\frac { dy }{ dt } .\frac { dt }{ dx }$
$\frac { dy }{ dx } =cost.\frac { dt }{ dx } =cos({ x }^{ 2 }+5)\frac { d }{ dx } ({ x }^{ 2 }+5)$
= cos (x² + 5) × 2x
= 2x cos (x² + 5)

Ex 5.2 Class 12 Maths Question 2.
cos (sin x)
Solution:
let y = cos (sin x)
put sinx = t
∴ y = cost,
t = sinx
∴ $\frac { dy }{ dx } =-sin\quad t,\frac { dt }{ dx } =cos\quad x$
$\frac { dy }{ dx } =\frac { dy }{ dt } .\frac { dt }{ dx } =(-sint)\times cosx$
Putting the value of t, $\frac { dy }{ dx } =-sin(sinx)\times cosx$
$\frac { dy }{ dx } =-[sin(sinx)]cosx$

Ex 5.2 Class 12 Maths Question 3.
sin(ax+b)
Solution:
let = sin(ax+b)
put ax+bx = t
∴ y = sint
t = ax+b
$\frac { dy }{ dt } =cost,\frac { dt }{ dx } =\frac { d }{ dx } (ax+b)=a$
$Now\frac { dy }{ dx } =\frac { dy }{ dt } .\frac { dt }{ dx } =cost\times a=acos\quad t$
$\frac { dy }{ dx } =acos(ax+b)$

Ex 5.2 Class 12 Maths Question 4.
sec(tan(√x))
Solution:
let y = sec(tan(√x))
by chain rule
$\frac { dy }{ dx } =sec(tan\sqrt { x } )tan(tan\sqrt { x } )\frac { d }{ dx } (tan\sqrt { x } )$
$\frac { dy }{ dx } =sec(tan\sqrt { x } ).tan(tan\sqrt { x } ){ sec }^{ 2 }\sqrt { x } .\frac { 1 }{ 2\sqrt { x } }$

Ex 5.2 Class 12 Maths Question 5.
$\\ \frac { sin(ax+b) }{ cos(cx+d) }$
Solution:
y = $\\ \frac { sin(ax+b) }{ cos(cx+d) }$ = $\\ \frac { v }{ u }$
u = sin(ax+b)
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.2 Q5.1

Ex 5.2 Class 12 Maths Question 6.
cos x³ . sin²(x⁵) = y(say)
Solution:
Let u = cos x³ and v = sin² x⁵,
put x³ = t
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.2 Q6.1

Ex 5.2 Class 12 Maths Question 7.
$2\sqrt { cot({ x }^{ 2 }) } =y(say)$
Solution:
$2\sqrt { cot({ x }^{ 2 }) } =y(say)$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.2 Q7.1

Ex 5.2 Class 12 Maths Question 8.
cos(√x) = y(say)
Solution:
cos(√x) = y(say)
$\frac { dy }{ dx } =\frac { d }{ dx } cos\left( \sqrt { x } \right) =-sin\sqrt { x } .\frac { d\sqrt { x } }{ dx }$
$=-sin\sqrt { x } .\frac { 1 }{ 2 } { (x) }^{ -\frac { 1 }{ 2 } }=\frac { -sin\sqrt { x } }{ 2\sqrt { x } }$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.2 Q8.1

Ex 5.2 Class 12 Maths Question 9.
Prove that the function f given by f (x) = |x – 1|,x ∈ R is not differential at x = 1.
Solution:
The given function may be written as
$f(x)=\begin{cases} x-1,\quad if\quad x\ge 1 \\ 1-x,\quad if\quad x<1 \end{cases}$
$R.H.D\quad at\quad x=1\quad =\underset { h\rightarrow 0 }{ lim } \frac { f(1+h)-f(1) }{ h }$

Ex 5.2 Class 12 Maths Question 10.
Prove that the greatest integer function defined by f (x)=[x], 0 < x < 3 is not differential at x = 1 and x = 2.
Solution:
(i) At x = 1
$R.H.D=\underset { h\rightarrow 0 }{ lim } \frac { f(1+h)-f(1) }{ h }$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.2 Q10.1

Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3

Find $\\ \frac { dy }{ dx }$ in the following

Ex 5.3 Class 12 Maths Question 1.
2x + 3y = sinx
Solution:
2x + 3y = sinx
Differentiating w.r.t x,
$2+3\frac { dy }{ dx } =cosx$
=> $\frac { dy }{ dx } =\frac { 1 }{ 3 } (cosx-2)$

Ex 5.3 Class 12 Maths Question 2.
2x + 3y = siny
Solution:
2x + 3y = siny
Differentiating w.r.t x,
$2+3.\frac { dy }{ dx } =cosy\frac { dy }{ dx }$
=> $\frac { dy }{ dx } =\frac { 2 }{ cosy-3 }$

Ex 5.3 Class 12 Maths Question 3.
ax + by² = cosy
Solution:
ax + by² = cosy
Differentiate w.r.t. x,
$a+2\quad by\quad \frac { dy }{ dx } =-siny\frac { dy }{ dx }$
=> $or\quad (2b+siny)\frac { dy }{ dx } =-a=>\frac { dy }{ dx } =-\frac { a }{ 2b+siny }$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 Q3.1 Ex 5.3 Class 12 Maths Question 4.
xy + y² = tan x + y
Solution:
xy + y² = tanx + y
Differentiating w.r.t. x,

Ex 5.3 Class 12 Maths Question 5.
x² + xy + y² = 100
Solution:
x² + xy + xy = 100
Differentiating w.r.t. x,
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 Q5.1

Ex 5.3 Class 12 Maths Question 6.
x³ + x²y + xy² + y³ = 81
Solution:
Given that
x³ + x²y + xy² + y³ = 81
Differentiating both sides we get
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 Q6.1

Ex 5.3 Class 12 Maths Question 7.
sin² y + cos xy = π
Solution:
Given that
sin² y + cos xy = π
Differentiating both sides we get
$2\quad sin\quad y\frac { d\quad siny }{ dx } +(-sinxy)\frac { d(xy) }{ dx } =0$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 Q7.1

Ex 5.3 Class 12 Maths Question 8.
sin²x + cos²y = 1
Solution:
Given that
sin²x + cos²y = 1
Differentiating both sides, we get
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 Q8.1

Ex 5.3 Class 12 Maths Question 9.
$y={ sin }^{ -1 }\left( \frac { 2x }{ { 1+x }^{ 2 } } \right)$
Solution:
$y={ sin }^{ -1 }\left( \frac { 2x }{ { 1+x }^{ 2 } } \right)$
put x = tanθ
$y={ sin }^{ -1 }\left( \frac { 2tan\theta }{ { 1+tan }^{ 2 }\theta } \right) ={ sin }^{ -1 }(sin2\theta )=2\theta$
$y={ 2sin }^{ -1 }x\quad \therefore \frac { dy }{ dx } =\frac { 2 }{ 1+{ x }^{ 2 } }$

Ex 5.3 Class 12 Maths Question 10.
$y={ tan }^{ -1 }\left( \frac { { 3x-x }^{ 3 } }{ { 1-3x }^{ 2 } } \right) ,-\frac { 1 }{ \sqrt { 3 } } <x<\frac { 1 }{ \sqrt { 3 } }$
Solution:
$y={ tan }^{ -1 }\left( \frac { { 3x-x }^{ 3 } }{ { 1-3x }^{ 2 } } \right)$
put x = tanθ
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 Q10.1

Ex 5.3 Class 12 Maths Question 11.
$y={ cos }^{ -1 }\left( \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \right) ,0<x<1$
Solution:
$y={ cos }^{ -1 }\left( \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \right) ,0<x<1$
put x = tanθ
$y={ cos }^{ -1 }\left( \frac { 1-tan^{ 2 }\quad \theta }{ 1+{ tan }^{ 2 }\quad \theta } \right) ={ cos }^{ -1 }(cos2\theta )=2\theta$
$y={ 2tan }^{ -1 }x\quad \therefore \frac { dy }{ dx } =\frac { 2 }{ 1+{ x }^{ 2 } }$

Ex 5.3 Class 12 Maths Question 12.
$y={ sin }^{ -1 }\left( \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \right) ,0<x<1$
Solution:
$y={ sin }^{ -1 }\left( \frac { 1-{ x }^{ 2 } }{ 1+{ x }^{ 2 } } \right) ,0<x<1$
put x = tanθ
we get
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 Q12.1

Ex 5.3 Class 12 Maths Question 13.
$y={ cos }^{ -1 }\left( \frac { 2x }{ 1+{ x }^{ 2 } } \right) ,-1<x<1$
Solution:
$y={ cos }^{ -1 }\left( \frac { 2x }{ 1+{ x }^{ 2 } } \right) ,-1<x<1$
put x = tanθ
we get
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.3 Q13.1

Ex 5.3 Class 12 Maths Question 14.
$y=sin^{ -1 }\left( 2x\sqrt { 1-{ x }^{ 2 } } \right) ,-\frac { 1 }{ \sqrt { 2 } } <x<\frac { 1 }{ \sqrt { 2 } }$
Solution:
$y=sin^{ -1 }\left( 2x\sqrt { 1-{ x }^{ 2 } } \right) ,-\frac { 1 }{ \sqrt { 2 } } <x<\frac { 1 }{ \sqrt { 2 } }$
put x = tanθ
we get
$y=sin^{ -1 }\left( 2sin\quad \theta \sqrt { 1-{ x }^{ 2 } } \right)$
$y=sin^{ -1 }\left( 2sin\theta \quad cos\theta \right) \quad ={ sin }^{ -1 }(sin2\theta )\quad =2\theta$
$y=2sin^{ -1 }x\quad \therefore \frac { dy }{ dx } =\frac { 2 }{ \sqrt { { 1-x }^{ 2 } } }$

Ex 5.3 Class 12 Maths Question 15.
$y=sin^{ -1 }\left( \frac { 1 }{ { 2x }^{ 2 }-1 } \right) ,0<x<\frac { 1 }{ \sqrt { 2 } }$
Solution:
$y=sin^{ -1 }\left( \frac { 1 }{ { 2x }^{ 2 }-1 } \right) ,0<x<\frac { 1 }{ \sqrt { 2 } }$
put x = tanθ
we get
$y=sec^{ -1 }\left( \frac { 1 }{ { 2cos }^{ 2 }\theta -1 } \right) ={ sec }^{ -1 }\left( \frac { 1 }{ cos2\theta } \right)$
$y=sec^{ -1 }(sec2\theta )=2\theta ,\quad y=2{ cos }^{ -1 }x$
$\therefore \frac { dy }{ dx } =\frac { -2 }{ \sqrt { { 1-x }^{ 2 } } }$

Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.4

Differentiate the following w.r.t.x:

Ex 5.4 Class 12 Maths Question 1.
$\frac { { e }^{ x } }{ sinx }$
Solution:
$y=\frac { { e }^{ x } }{ sinx }$
$for\quad y=\frac { u }{ v } ,$
$\frac { dy }{ dx } =\frac { { e }^{ x }{ sin }x-{ e }^{ x }cosx }{ { sin }^{ 2 }x }$
$or\frac { dy }{ dx } =\frac { { e }^{ x }{ sin }x-{ e }^{ x }cosx }{ { sin }^{ 2 }x } ,where\quad x\neq n\pi ,x\in z$

Ex 5.4 Class 12 Maths Question 2.
${ e }^{ { sin }^{ -1 }x }$
Solution:
${ e }^{ { sin }^{ -1 }x }$
$y={ e }^{ { sin }^{ -1 }x }$
x=sint
$\therefore y={ e }^{ t },\frac { dt }{ dx } =\frac { 1 }{ \sqrt { 1-{ x }^{ 2 } } } ,\frac { dy }{ dt } ={ e }^{ t }$
$\therefore \frac { dy }{ dx } =\frac { dy }{ dt } .\frac { dt }{ dx } ={ e }^{ t }.\frac { 1 }{ \sqrt { { 1- }x^{ 2 } } } =\frac { { e }^{ { sin }^{ -1 }x } }{ \sqrt { 1-{ x }^{ 2 } } }$

Ex 5.4 Class 12 Maths Question 3.
${ e }^{ { x }^{ 3 } }=y$
Solution:
${ e }^{ { x }^{ 3 } }=y$
$Put\quad { x }^{ 3 }=t\quad \therefore \quad y={ e }^{ t },\frac { dy }{ dt } ={ e }^{ t },\frac { dt }{ dx } ={ 3x }^{ 2 }$
$\therefore \frac { dy }{ dx } =\frac { dy }{ dt } \times \frac { dt }{ dx } ={ e }^{ t }\times { 3x }^{ 2 }={ 3x }^{ 2 }{ e }^{ { x }^{ 3 } }$

Ex 5.4 Class 12 Maths Question 4.
$sin\left( { tan }^{ -1 }{ e }^{ -x } \right) =y$
Solution:
$sin\left( { tan }^{ -1 }{ e }^{ -x } \right) =y$
$\frac { dy }{ dx } =cos\left( { tan }^{ -1 }{ e }^{ -x } \right) \frac { d }{ dx } \left( { tan }^{ -1 }{ e }^{ -x } \right)$
$=cos\left( { tan }^{ -1 }{ e }^{ -x } \right) \frac { 1 }{ 1+{ e }^{ -2x } } \frac { d }{ dx } \left( { e }^{ -x } \right)$
$=-cos\left( { tan }^{ -1 }{ e }^{ -x } \right) \frac { 1 }{ 1+{ e }^{ -2x } } .\left( { e }^{ -x } \right)$

Ex 5.4 Class 12 Maths Question 5.
$log(cos\quad { e }^{ x })=y$
Solution:
$\frac { dy }{ dx } =\frac { 1 }{ cos\quad { e }^{ x } } \left( -sin{ e }^{ x } \right) .{ e }^{ x }\quad =-tan\left( { e }^{ x } \right)$

Ex 5.4 Class 12 Maths Question 6.
${ e }^{ x }+{ e }^{ { x }^{ 2 } }+$ … $+{ e }^{ { x }^{ 5 } }=y(say)$
Solution:
$let\quad u={ e }^{ { x }^{ n } },put\quad { x }^{ n }=t,u={ e }^{ t },t={ x }^{ n }$
${ e }^{ x }+{ e }^{ { x }^{ 2 } }+$ … $+{ e }^{ { x }^{ 5 } }=y(say)$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.4 Q6.1

Ex 5.4 Class 12 Maths Question 7.
$\sqrt { { e }^{ \sqrt { x } } } ,x>0$
Solution:
y = $\sqrt { { e }^{ \sqrt { x } } } ,x>0$
$y=\sqrt { { e }^{ \sqrt { x } } } ,let\quad y=\sqrt { s } ,s={ e }^{ t },t=\sqrt { x }$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.4 Q7.1

Ex 5.4 Class 12 Maths Question 8.
log(log x),x>1
Solution:
y = log(log x),
put y = log t, t = log x,
differentiating
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.4 Q8.1

Ex 5.4 Class 12 Maths Question 9.
$\frac { cosx }{ logx } =y(say),x>0$
Solution:
let $y=\frac { cosx }{ logx }$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.4 Q9.1

Ex 5.4 Class 12 Maths Question 10.
cos(log x+e^x),x>0
Solution:
y = cos(log x+e^x),x>0
put y = cos t,t = log x+e^x
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.4 Q10.1

Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5

Differentiate the functions given in Questions 1 to 11 w.r.to x

Ex 5.5 Class 12 Maths Question 1.
cos x. cos 2x. cos 3x
Solution:
Let y = cos x. cos 2x . cos 3x,
Taking log on both sides,
log y = log (cos x. cos 2x. cos 3x)
log y = log cos x + log cos 2x + log cos 3x,
Differentiating w.r.t. x, we get
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5 Q1.1

Ex 5.5 Class 12 Maths Question 2.
$\sqrt { \frac { (x-1)(x-2) }{ (x-3)(x-4)(x-5) } }$
Solution:
$y=\sqrt { \frac { (x-1)(x-2) }{ (x-3)(x-4)(x-5) } }$
taking log on both sides
log y = log $\sqrt { \frac { (x-1)(x-2) }{ (x-3)(x-4)(x-5) } }$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5 Q2.1

Ex 5.5 Class 12 Maths Question 3.
(log x)^cosx
Solution:
let y = (log x)^cosx
Taking log on both sides,
log y = log (log x)^cosx
log y = cos x log (log x),
Differentiating w.r.t. x,
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5 Q3.1 Ex 5.5 Class 12 Maths Question 4.
x – 2^sinx
Solution:
let y = x – 2^sinx,
y = u – v

Ex 5.5 Class 12 Maths Question 5.
(x+3)².(x + 4)³.(x + 5)⁴
Solution:
let y = (x + 3)².(x + 4)³.(x + 5)⁴
Taking log on both side,
logy = log [(x + 3)² • (x + 4)³ • (x + 5)⁴]
= log (x + 3)² + log (x + 4)³ + log (x + 5)⁴
log y = 2 log (x + 3) + 3 log (x + 4) + 4 log (x + 5)
Differentiating w.r.t. x, we get
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5 Q5.1

Ex 5.5 Class 12 Maths Question 6.
${ \left( x+\frac { 1 }{ x } \right) }^{ x }+{ x }^{ \left( 1+\frac { 1 }{ x } \right) }$
Solution:
let $y={ \left( x+\frac { 1 }{ x } \right) }^{ x }+{ x }^{ \left( 1+\frac { 1 }{ x } \right) }$
let $u={ \left( x+\frac { 1 }{ x } \right) }^{ x }and\quad v={ x }^{ \left( 1+\frac { 1 }{ x } \right) }$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5 Q6.1

Ex 5.5 Class 12 Maths Question 7.
(log x)^x + x^logx
Solution:
let y = (log x)^x + x^logx = u+v
where u = (log x)^x
∴ log u = x log(log x)
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5 Q7.1

Ex 5.5 Class 12 Maths Question 8.
(sin x)^x+sin^-1 √x
Solution:
Let y = (sin x)^x+ sin^-1√x
let u = (sin x)x, v = sin^-1 √x
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5 Q8.1

Ex 5.5 Class 12 Maths Question 9.
x^sinx + (sin x)^cosx
Solution:
let y = x^sinx + (sin x)^cosx = u+v
where u = x^sinx
log u = sin x log x
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5 Q9.1

Ex 5.5 Class 12 Maths Question 10.
${ x }^{ x\quad cosx }+\frac { { x }^{ 2 }+1 }{ { x }^{ 2 }-1 }$
Solution:
$y={ x }^{ x\quad cosx }+\frac { { x }^{ 2 }+1 }{ { x }^{ 2 }-1 }$
y = u + v
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5 Q10.1

Ex 5.5 Class 12 Maths Question 11.
${ (x\quad cosx) }^{ x }+{ (x\quad sinx) }^{ \frac { 1 }{ x } }$
Solution:
$y={ (x\quad cosx) }^{ x }+{ (x\quad sinx) }^{ \frac { 1 }{ x } }$
Let u = (x cosx)^x
logu = x log(x cosx)
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5 Q11.1

Find $\\ \frac { dy }{ dx }$ of the functions given in Questions 12 to 15.

Ex 5.5 Class 12 Maths Question 12.
x^y + y^x = 1
Solution:
x^y + y^x = 1
let u = x^y and v = y^x
∴ u + v = 1,
$\frac { du }{ dx } +\frac { dv }{ dx }=0$
Now u = x
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5 Q12.1

Ex 5.5 Class 12 Maths Question 13.
y^x= x^y
Solution:
y = x
x logy = y logx
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5 Q13.1

Ex 5.5 Class 12 Maths Question 14.
(cos x)^y = (cos y)^x
Solution:
We have
(cos x)^y = (cos y)^x
=> y log (cosx) = x log (cosy)
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5 Q14.1

Ex 5.5 Class 12 Maths Question 15.
xy = e^(x-y)
Solution:
log(xy) = log e^(x-y)
=> log(xy) = x – y
=> logx + logy = x – y
$=>\frac { 1 }{ x } +\frac { 1 }{ y } \frac { dy }{ dx } =1-\frac { dy }{ dx } =>\frac { dy }{ dx } =\frac { y(x-1) }{ x(y+1) }$

Ex 5.5 Class 12 Maths Question 16.
Find the derivative of the function given by f (x) = (1 + x) (1 + x²) (1 + x⁴) (1 + x⁸) and hence find f'(1).
Solution:
Let f(x) = y = (1 + x)(1 + x²)(1 + x⁴)(1 + x⁸)
Taking log both sides, we get
logy = log [(1 + x)(1 + x²)(1 + x⁴)(1 + x⁸)]
logy = log(1 + x) + log (1 + x²) + log(1 + x⁴) + log(1 + x⁸)
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5 Q16.1

Ex 5.5 Class 12 Maths Question 17.
Differentiate (x² – 5x + 8) (x³ + 7x + 9) in three ways mentioned below:
(i) by using product rule
(ii) by expanding the product to obtain a single polynomial.
(iii) by logarithmic differentiation.
Do they all give the same answer?
Solution:
(i) By using product rule
f’ = (x² – 5x + 8) (3x² + 7) + (x³ + 7x + 9) (2x – 5)
f = 5x⁴ – 20x³ + 45x² – 52x + 11.
(ii) By expanding the product to obtain a single polynomial, we get
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5 Q17.1

Ex 5.5 Class 12 Maths Question 18.
If u, v and w are functions of w then show that
$\frac { d }{ dx } (u.v.w)=\frac { du }{ dx } v.w+u.\frac { dv }{ dx } .w+u.v\frac { dw }{ dx }$
in two ways-first by repeated application of product rule, second by logarithmic differentiation.
Solution:
Let y = u.v.w
=> y = u. (vw)
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.5 Q18.1

Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.6

If x and y are connected parametrically by the equations given in Questions 1 to 10, without eliminating the parameter. Find $\\ \frac { dy }{ dx }$ .

Ex 5.6 Class 12 Maths Question 1.
x = 2at², y = at⁴
Solution:
$\frac { dx }{ dt } =4at,\frac { dy }{ dt } ={ 4at }^{ 3 }\quad \therefore \frac { dy }{ dx } =\frac { \frac { dy }{ dt } }{ \frac { dx }{ dt } } =\frac { { 4at }^{ 3 } }{ { 4at } } ={ t }^{ 2 }$

Ex 5.6 Class 12 Maths Question 2.
x = a cosθ,y = b cosθ
Solution:
$\frac { dx }{ d\theta } =-asin\theta ,\frac { dy }{ d\theta } =-sinb\quad sin\theta =>\frac { dy }{ dx } =\frac { b }{ a }$

Ex 5.6 Class 12 Maths Question 3.
x = sin t, y = cos 2t
Solution:
$\therefore \frac { dx }{ dt } =cos\quad t\quad and\frac { dy }{ dt } =-sin2t.2=-2sin2t$
$\frac { dy }{ dx } =\frac { -2sin2t }{ cost } =\frac { -2.2sintcost }{ cost } =-4sint$

Ex 5.6 Class 12 Maths Question 4.
$x=4t,y=\frac { 4 }{ t }$
Solution:
$\frac { dx }{ dt } =4;\frac { dy }{ dt } =\frac { -4 }{ { t }^{ 2 } } =>\frac { dy }{ dx } =\frac { -4 }{ { t }^{ 2 } } \times \frac { 1 }{ 4 } =\frac { -1 }{ { t }^{ 2 } }$

Ex 5.6 Class 12 Maths Question 5.
x = cos θ – cos 2θ, y = sin θ – sin 2θ
Solution:
$\frac { dx }{ d\theta } =-sin\theta -(-sin2\theta ).2=2sin2\theta -sin\theta$
$\frac { dy }{ d\theta } =cos\theta -2cos2\theta \quad \therefore \frac { dy }{ dx } =\frac { cos\theta -2cos2\theta }{ 2sin2\theta -sin\theta }$

Ex 5.6 Class 12 Maths Question 6.
x = a(θ – sinθ), y = a(1 + cosθ)
Solution:
$\frac { dx }{ d\theta } =a\left[ 1-cos\theta \right] \& \frac { dy }{ d\theta } =-asin\theta$
$\frac { dy }{ dx } =\frac { -asin\theta }{ a(1-cos\theta ) } =\frac { -2sin\frac { \theta }{ 2 } .cos\frac { \theta }{ 2 } }{ 2{ sin }^{ 2 }\frac { \theta }{ 2 } } =-cot\frac { \theta }{ 2 }$

Ex 5.6 Class 12 Maths Question 7.
$x=\frac { { sin }^{ 3 }t }{ \sqrt { cos2t } } \& y=\frac { { cos }^{ 3 }t }{ \sqrt { cos2t } }$
Solution:
$x=\frac { { sin }^{ 3 }t }{ \sqrt { cos2t } } \& y=\frac { { cos }^{ 3 }t }{ \sqrt { cos2t } }$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.6 Q7.1

Ex 5.6 Class 12 Maths Question 8.
$x=a\left( cost+log\quad tan\frac { t }{ 2 } \right) ,y=a\quad sint$
Solution:
$x=a\left( cost+log\quad tan\frac { t }{ 2 } \right) ,y=a\quad sint$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.6 Q8.1

Ex 5.6 Class 12 Maths Question 9.
x = a sec θ,y = b tan θ
Solution:
x = a sec θ,y = b tan θ
$\frac { dx }{ d\theta } =a\quad sec\theta \quad tan\theta \quad and\frac { dy }{ d\theta } =b{ sec }^{ 2 }\theta$
$\frac { dy }{ dx } =\frac { { bsec }^{ 2 }\theta }{ asec\theta tan\theta } \frac { b }{ a } cosec\theta$

Ex 5.6 Class 12 Maths Question 10.
x = a(cosθ+θsinθ), y = a(sinθ-θcosθ)
Solution:
x = a(cosθ+θsinθ), y = a(sinθ-θcosθ)
$\frac { dx }{ d\theta } =a\left[ -sin\theta +\theta .cos\theta +sin\theta \right] =a\theta cos\theta$
$\frac { dy }{ d\theta } =a\theta sin\theta =>\frac { dy }{ dx } =\frac { a\theta sin\theta }{ a\theta cos\theta } =tan\theta$

Ex 5.6 Class 12 Maths Question 11.
If $x=\sqrt { { a }^{ { sin }^{ -1 }t } } ,y=\sqrt { { a }^{ { cos }^{ -1 }t } }$ show that $\frac { dy }{ dx } =-\frac { y }{ x }$
Solution:
Given that
$x=\sqrt { { a }^{ { sin }^{ -1 }t } } ,y=\sqrt { { a }^{ { cos }^{ -1 }t } }$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.6 Q11.1

Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.7

Find the second order derivatives of the functions given in Questions 1 to 10.

Ex 5.7 Class 12 Maths Question 1.
x² + 3x + 2 = y(say)
Solution:
$\frac { dy }{ dx } =2x+3\quad and\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =2$

Ex 5.7 Class 12 Maths Question 2.
x²⁰ = y(say)
Solution:
$\frac { dy }{ dx } ={ 20 }x^{ 19 }\quad =>\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =20\times { 19x }^{ 18 }={ 380 }x^{ 18 }\qquad$

Ex 5.7 Class 12 Maths Question 3.
x.cos x = y(say)
Solution:
$\frac { dy }{ dx } =x(-sinx)+cosx.1,=-xsinx+cosx$
$\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =-xcosx-sinx-sinx=-xcosx-2sinx$

Ex 5.7 Class 12 Maths Question 4.
log x = y (say)
Solution:
$\frac { dy }{ dx } =\frac { 1 }{ x } =>\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =-\frac { 1 }{ { x }^{ 2 } }$

Ex 5.7 Class 12 Maths Question 5.
x³ log x = y (say)
Solution:
x³ log x = y
$=>\frac { dy }{ dx } ={ x }^{ 3 }.\frac { 1 }{ x } +logx\times { 3x }^{ 2 }={ x }^{ 2 }+{ 3x }^{ 2 }logx$
$\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =2x+{ 3x }^{ 2 }.\frac { 1 }{ x } +logx.6x=x(5+6logx)$

Ex 5.7 Class 12 Maths Question 6.
e^x sin5x = y
Solution:
e^x sin5x = y
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.7 Q6.1

Ex 5.7 Class 12 Maths Question 7.
e^6x cos3x = y
Solution:
e^6x cos3x = y
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.7 Q7.1

Ex 5.7 Class 12 Maths Question 8.
tan^-1 x = y
Solution:
$\frac { dy }{ dx } =\frac { 1 }{ 1+{ x }^{ 2 } } =>\frac { { d }^{ 2y } }{ { dx }^{ 2 } } =\frac { -2x }{ { ({ 1+x }^{ 2 }) }^{ 2 } }$

Ex 5.7 Class 12 Maths Question 9.
log(logx) = y
Solution:
log(logx) = y
$\frac { dy }{ dx } =\frac { 1 }{ logx } .\frac { 1 }{ x }$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.7 Q9.1

Ex 5.7 Class 12 Maths Question 10.
sin(log x) = y
Solution:
sin(log x) = y
$\frac { dy }{ dx } =\frac { cos(logx) }{ x }$
$and\quad \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =\frac { x.\left[ -sin(logx) \right] .\frac { 1 }{ x } -cos(logx).1 }{ { x }^{ 2 } }$
$=\frac { \left[ sin(logx)+cos(logx) \right] }{ { x }^{ 2 } }$

Ex 5.7 Class 12 Maths Question 11.
If y = 5 cosx – 3 sin x, prove that $\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +y=0$
Solution:
$\frac { dy }{ dx } =-5sinx-3cosx$
$\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =-5cosx+3sinx=-y$
$\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +y=0$
Hence proved

Ex 5.7 Class 12 Maths Question 12.
If y = cos^-1 x, Find $\frac { { d }^{ 2 }y }{ { dx }^{ 2 } }$ in terms of y alone.
Solution:
$\frac { dy }{ dx } =-{ \left( { 1-x }^{ 2 } \right) }^{ -\frac { 1 }{ 2 } }$
$\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =\frac { -cosy }{ { \left( { sin }^{ 2 }y \right) }^{ \frac { 3 }{ 2 } } } =-coty\quad { cosec }^{ 2 }y$

Ex 5.7 Class 12 Maths Question 13.
If y = 3 cos (log x) + 4 sin (log x), show that
${ x }^{ 2 }{ y }_{ 2 }+{ xy }_{ 1 }+y=0$
Solution:
Given that
y = 3 cos (log x) + 4 sin (log x)
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.7 Q13.1

Ex 5.7 Class 12 Maths Question 14.
$If\quad y=A{ e }^{ mx }+B{ e }^{ nx },\quad show\quad that\quad \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -(m+n)\frac { dy }{ dx } +mny=0$
Solution:
Given that
$\quad y=A{ e }^{ mx }+B{ e }^{ nx },\quad$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.7 Q14.1

Ex 5.7 Class 12 Maths Question 15.
If y = 500e^7x + 600e^-7x, show that $\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =49y$ .
Solution:
we have
y = 500e^7x + 600e^-7x
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.7 Q15.1

Ex 5.7 Class 12 Maths Question 16.
If e^y(x+1) = 1,show that $\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } ={ \left( \frac { dy }{ dx } \right) }^{ 2 }$
Solution:
${ e }^{ y }(x+1)=1=>{ e }^{ y }=\frac { 1 }{ x+1 }$
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.7 Q16.1

Ex 5.7 Class 12 Maths Question 17.
If y=(tan^-1 x)² show that (x²+1)²y₂+2x(x²+1)y₁=2
Solution:
we have
y=(tan^-1 x)²
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.7 Q17.1

Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.8

Ex 5.8 Class 12 Maths Question 1.
Verify Rolle’s theorem for the function
f(x) = x² + 2x – 8,x∈ [-4,2]
Solution:
Now f(x) = x² + 2x – 8 is a polynomial
∴ it is continuous and derivable in its domain x∈R.
Hence it is continuous in the interval [-4,2] and derivable in the interval (- 4,2)
f(-4) = (-4)² + 2(-4) – 8 = 16 – 8 – 8 = 0,
f(2) = 2² + 4 – 8 = 8 – 8 = 0
Conditions of Rolle’s theorem are satisfied.
f'(x) = 2x + 2
∴ f’ (c) = 2c + 2 = 0
or c = – 1, c = – 1 ∈ [-4,2]
Thus f’ (c) = 0 at c = – 1.

Ex 5.8 Class 12 Maths Question 2.
Examine if Rolle’s theorem is applicable to any of the following functions. Can you say some thing about the converse of Rolle’s theorem from these example?
(i) f(x) = [x] for x ∈ [5,9]
(ii) f (x) = [x] for x ∈ [-2,2]
(iii) f (x) = x² – 1 for x ∈ [1,2]
Solution:
(i) In the interval [5, 9], f (x) = [x] is neither continuous nor derivable at x = 6,7,8 Hence Rolle’s theorem is not applicable
(ii) f (x) = [x] is not continuous and derivable at -1, 0, 1. Hence Rolle’s theorem is not applicable.
(iii) f(x) = (x² – 1),f(1) = 1 – 1 = 0,
f(2) = 22 – 1 = 3
f(a)≠f(b)
Though it is continous and derivable in the interval [1,2].
Rolle’s theorem is not applicable.
In case of converse if f (c)=0, c ∈ [a, b] then conditions of rolle’s theorem are not true.
(i) f (x) = [x] is the greatest integer less than or equal to x.
∴f(x) = 0, But fis neither continuous nor differentiable in the interval [5,9].
(ii) Here also, theough f (x) = 0, but f is neither continuous nor differentiable in the interval [-2,2].
(iii) f (x)=x² – 1, f'(x)=2x. Here f'(x) is not zero in the [1,2], So f (2) ≠ f’ (2).

Ex 5.8 Class 12 Maths Question 3.
If f: [-5,5] –>R is a differentiable function and if f (x) does not vanish anywhere then prove that f (- 5) ≠ f (5).
Solution:
For Rolle’s theorem
If (i) f is continuous in [a, b]
(ii) f is derivable in [a, b]
(iii) f (a) = f (b)
then f’ (c)=0, c e (a, b)
∴ f is continuous and derivable
but f (c) ≠ 0 =>f(a) ≠ f(b) i.e., f(-5)≠f(5)

Ex 5.8 Class 12 Maths Question 4.
Verify Mean Value Theorem, if
f (x) = x² – 4x – 3 in the interval [a, b], where a = 1 and b = 4.
Solution:
f (x) = x² – 4x – 3. It being a polynomial it is continuous in the interval [1,4] and derivable in (1,4), So all the condition of mean value theorem hold.
then f’ (x) = 2x – 4,
f’ (c) = 2c – 4
f(4)= 16 – 16 – 3 = – 3,
f(1)= 1 – 4 – 3 = – 6
Then there exist a value c such that
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.8 Q4.1

Ex 5.8 Class 12 Maths Question 5.
Verify Mean Value Theorem, if f (x)=x³ – 5x² – 3x in the interval [a, b], where a = 1 and b = 3. Find all c ∈ (1,3) for which f’ (c) = 0.
Solution:
f (x)=x³ – 5x² – 3x,
It is a polynomial. Therefore it is continuous in the interval [1,3] and derivable in the interval (1,3)
Also, f'(x)=3x²-10x-3
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.8 Q5.1

Ex 5.8 Class 12 Maths Question 6.
Examine the applicability of Mean Value theroem for all three functions given in the above Question 2.
Solution:
(i) F (x)= [x] for x ∈ [5,9], f (x) = [x] in the interval [5, 9] is neither continuous, nor differentiable.
(ii) f (x) = [x], for x ∈ [-2,2],
Again f (x) = [x] in the interval [-2,2] is neither continous, nor differentiable.
(iii) f(x) = x²-1 for x ∈ [1,2], It is a polynomial. Therefore it is continuous in the interval [1,2] and differentiable in the interval (1,2)
f (x) = 2x, f(1) = 1 – 1 = 0 ,
f(2) = 4 – 1 = 3, f'(c) = 2c
NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.8 Q6.1

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Class 12th Chapter -4 Determinants | NCERT Maths Solution | NCERT Solution | Edugrown

NCERT Solutions for Class 12 Maths Chapter : 4 Determinants

Class 12 Maths Chapter 4 Determinants Ex 4.1

Ex 4.1 Class 12 Maths Question 1.
Evaluate the following determinant:
$\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix}$
Solution:
$\begin{vmatrix} 2 & 4 \\ -5 & -1 \end{vmatrix}$
= 2x(-1)-(-5)x(4)
=-2+20
=18

Ex 4.1 Class 12 Maths Question 2.
(i) $\begin{vmatrix} cos\theta & \quad -sin\theta \\ sin\theta & \quad cos\theta \end{vmatrix}$
(ii) $\begin{vmatrix} { x }^{ 2 }-x+1 & x-1 \\ x+1 & x+1 \end{vmatrix}$
Solution:
(i) $\begin{vmatrix} cos\theta & \quad -sin\theta \\ sin\theta & \quad cos\theta \end{vmatrix}$
= cosθ cosθ – (sinθ)(-sinθ)
= cos²θ + sin²θ
= 1
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.1 Q2.1

Ex 4.1 Class 12 Maths Question 3.
If $A=\begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}$ then show that |2A|=|4A|
Solution:
$A=\begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix}$
=> $2A=\begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix}$
L.H.S = |2A|
= $2A=\begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix}$
= – 24
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.1 Q3.1

Ex 4.1 Class 12 Maths Question 4.
$A=\left[ \begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{matrix} \right]$ , then show that |3A| = 27|A|
Solution:
3A = $3\left[ \begin{matrix} 1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{matrix} \right]$
= $3\left[ \begin{matrix} 3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12 \end{matrix} \right]$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.1 Q4.1

Ex 4.1 Class 12 Maths Question 5.
Evaluate the following determinant:
(i) $\left| \begin{matrix} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{matrix} \right|$
(ii) $\left| \begin{matrix} 3 & -4 & 5 \\ 1 & 1 & -2 \\ 2 & 3 & 1 \end{matrix} \right|$
(iii) $\left| \begin{matrix} 0 & 1 & 2 \\ -1 & 0 & -3 \\ -2 & 3 & 0 \end{matrix} \right|$
(iv) $\left| \begin{matrix} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{matrix} \right|$
Solution:
(i) $\left| \begin{matrix} 3 & -1 & -2 \\ 0 & 0 & -1 \\ 3 & -5 & 0 \end{matrix} \right|$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.1 Q5.1 >

Ex 4.1 Class 12 Maths Question 6.
If $\left[ \begin{matrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{matrix} \right]$ , find |A|
Solution:
|A| = $\left[ \begin{matrix} 1 & 1 & -2 \\ 2 & 1 & -3 \\ 5 & 4 & -9 \end{matrix} \right]$
= 1(-9+12)-1(-18+15)-2(8-5)
= 0

Ex 4.1 Class 12 Maths Question 7.
Find the values of x, if
(i) $\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix}=\begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}$
(ii) $\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix}=\begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}$
Solution:
(i) $\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix}=\begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}$
=> 2 – 20 = 2x² – 24
=> x² = 3
=> x = ±√3
(ii) $\begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix}=\begin{vmatrix} x & 3 \\ 2x & 5 \end{vmatrix}$
or
2 × 5 – 4 × 3 = 5 × x – 2x × 3
=>x = 2

Ex 4.1 Class 12 Maths Question 8.
If $\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix}=\begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix}$ , then x is equal to
(a) 6
(b) +6
(c) -6
(d) 0
Solution:
(b) $\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix}=\begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix}$
=> x² – 36 = 36 – 36
=> x² = 36
=> x = ± 6

Class 12 Maths Chapter 4 Determinants Ex 4.2

Ex 4.2 Class 12 Maths Question 1.
$\left| \begin{matrix} x & a & x+a \\ y & b & y+b \\ z & c & z+c \end{matrix} \right| =0$
Solution:
L.H.S = $\left| \begin{matrix} x & a & x \\ y & b & y \\ z & c & z \end{matrix} \right| +\left| \begin{matrix} x & a & a \\ y & b & b \\ z & c & c \end{matrix} \right|$
(C1 = C3 and C2 = C3)
= 0 + 0
= 0
= R.H.S

Ex 4.2 Class 12 Maths Question 2.
$\left| \begin{matrix} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{matrix} \right| =0$
Solution:
L.H.S = $\left| \begin{matrix} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{matrix} \right| =0$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.2 Q2.1

Ex 4.2 Class 12 Maths Question 3.
$\left| \begin{matrix} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{matrix} \right| =0$
Solution:
$\left| \begin{matrix} 2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86 \end{matrix} \right| =\left| \begin{matrix} 2 & 7 & 0 \\ 3 & 8 & 0 \\ 5 & 9 & 0 \end{matrix} \right|$
${ C }_{ 3 }\rightarrow { C }_{ 3 }-{ C }_{ 1 }-{ 9C }_{ 2 }=0$

Ex 4.2 Class 12 Maths Question 4.
$\left| \begin{matrix} 1 & bc & a(b+c) \\ 1 & ca & b(c+a) \\ 1 & ab & c(a+b) \end{matrix} \right| =0$
Solution:
L.H.S = $\left| \begin{matrix} 1 & bc & a(b+c) \\ 1 & ca & b(c+a) \\ 1 & ab & c(a+b) \end{matrix} \right|$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.2 Q4.1

Ex 4.2 Class 12 Maths Question 5.
$\left| \begin{matrix} b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{matrix} \right| =2\left| \begin{matrix} a & p & x \\ b & q & y \\ c & r & z \end{matrix} \right|$
Solution:
L.H.S = ∆ = $\left| \begin{matrix} b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{matrix} \right|$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.2 Q5.1

By using properties of determinants in Q 6 to 14, show that

Ex 4.2 Class 12 Maths Question 6.
$\left| \begin{matrix} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{matrix} \right| =0$
Solution:
L.H.S = ∆ = $\left| \begin{matrix} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{matrix} \right|$ …(i)
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.2 Q6.1

Ex 4.2 Class 12 Maths Question 7.
$\left| \begin{matrix} { -a }^{ 2 } & ab & ac \\ ba & { -b }^{ 2 } & bc \\ ac & cb & { -c }^{ 2 } \end{matrix} \right| ={ 4a }^{ 2 }{ b }^{ 2 }{ c }^{ 2 }$
Solution:
L.H.S = $\left| \begin{matrix} { -a }^{ 2 } & ab & ac \\ ba & { -b }^{ 2 } & bc \\ ac & cb & { -c }^{ 2 } \end{matrix} \right|$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.2 Q7.1

Ex 4.2 Class 12 Maths Question 8.
(a) $\left| \begin{matrix} 1 & a & { a }^{ 2 } \\ 1 & b & { b }^{ 2 } \\ 1 & c & { c }^{ 2 } \end{matrix} \right| =(a-b)(b-c)(c-a)$
(b) $\left| \begin{matrix} 1 & 1 & 1 \\ a & b & c \\ { a }^{ 3 } & { b }^{ 3 } & { c }^{ 3 } \end{matrix} \right| =(a-b)(b-c)(c-a)(a+b+c)$
Solution:
(a) L.H.S = $\left| \begin{matrix} 1 & a & { a }^{ 2 } \\ 1 & b & { b }^{ 2 } \\ 1 & c & { c }^{ 2 } \end{matrix} \right|$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.2 Q8.1

Ex 4.2 Class 12 Maths Question 9.
$\left| \begin{matrix} x & x^{ 2 } & yx \\ y & { y }^{ 2 } & zx \\ z & { z }^{ 2 } & xy \end{matrix} \right| =(x-y)(y-z)(z-x)(xy+yz+zx)$
Solution:
Let ∆ = $\left| \begin{matrix} x & x^{ 2 } & yx \\ y & { y }^{ 2 } & zx \\ z & { z }^{ 2 } & xy \end{matrix} \right|$
Applying R1–>R1 – R2, R2–>R2 – R3
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.2 Q9.1

Ex 4.2 Class 12 Maths Question 10.
(a) $\left| \begin{matrix} x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{matrix} \right| =(5x+4){ (4-x) }^{ 2 }$
(b) $\left| \begin{matrix} y+x & y & y \\ y & y+k & y \\ y & y & y+k \end{matrix} \right| ={ k }^{ 2 }(3y+k)$
Solution:
(a) L.H.S = $\left| \begin{matrix} x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{matrix} \right|<br />$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.2 Q10.1

Ex 4.2 Class 12 Maths Question 11.
(a) $\left| \begin{matrix} a-b-c & \quad 2a & \quad 2a \\ 2b & \quad b-c-a & \quad 2b \\ 2c & 2c & \quad c-a-b \end{matrix} \right| ={ (a+b+c) }^{ 3 }$
(b) $\left| \begin{matrix} x+y+2z & \quad z & \quad z \\ x & \quad y+z+2x & \quad x \\ y & y & \quad z+x+2y \end{matrix} \right| ={ 2(x+y+z) }^{ 3 }$
Solution:
(a) L.H.S = $\left| \begin{matrix} a-b-c & \quad 2a & \quad 2a \\ 2b & \quad b-c-a & \quad 2b \\ 2c & 2c & \quad c-a-b \end{matrix} \right|$
= $\left( a+b+c \right) \left| \begin{matrix} 1 & \quad 1 & \quad 1 \\ 2b & \quad b-c-a & \quad 2b \\ 2c & \quad 2c & \quad c-a-b \end{matrix} \right|$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.2 Q11.1

Ex 4.2 Class 12 Maths Question 12.
$\left| \begin{matrix} 1 & \quad x & { \quad x }^{ 2 } \\ { x }^{ 2 } & \quad 1 & x \\ x & { \quad x }^{ 2 } & 1 \end{matrix} \right| ={ { (1-x }^{ 3 }) }^{ 2 }$
Solution:
L.H.S = $\left| \begin{matrix} 1 & \quad x & { \quad x }^{ 2 } \\ { x }^{ 2 } & \quad 1 & x \\ x & { \quad x }^{ 2 } & 1 \end{matrix} \right|$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.2 Q12.1

Ex 4.2 Class 12 Maths Question 13.
$\left| \begin{matrix} 1+{ a }^{ 2 }-{ b }^{ 2 } & \quad 2ab & \quad -2b \\ 2ab & \quad 1-{ a }^{ 2 }+{ b }^{ 2 } & \quad 2a \\ 2b & \quad -2a & \quad 1-{ a }^{ 2 }+{ b }^{ 2 } \end{matrix} \right| ={ (1+{ a }^{ 2 }+{ b }^{ 2 }) }^{ 3 }$
Solution:
L.H.S = $\left| \begin{matrix} 1+{ a }^{ 2 }-{ b }^{ 2 } & \quad 2ab & \quad -2b \\ 2ab & \quad 1-{ a }^{ 2 }+{ b }^{ 2 } & \quad 2a \\ 2b & \quad -2a & \quad 1-{ a }^{ 2 }+{ b }^{ 2 } \end{matrix} \right|$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.2 Q13.1

Ex 4.2 Class 12 Maths Question 14.
$\left| \begin{matrix} { a }^{ 2 }+1 & \quad ab & \quad ac \\ ab\quad & \quad b^{ 2 }+1 & \quad bc \\ ca\quad & \quad cb & \quad { c }^{ 2 }+1 \end{matrix} \right| =1+{ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }$
Solution:
Let ∆ = $\left| \begin{matrix} { a }^{ 2 }+1 & \quad ab & \quad ac \\ ab\quad & \quad b^{ 2 }+1 & \quad bc \\ ca\quad & \quad cb & \quad { c }^{ 2 }+1 \end{matrix} \right|$
$\left| \begin{matrix} { a }^{ 2 }+1 & \quad ab+0 & \quad ac+0 \\ ab+0\quad & \quad b^{ 2 }+1 & \quad bc+0 \\ ca+0\quad & \quad cb+0 & \quad { c }^{ 2 }+1 \end{matrix} \right|$
This may be expressed as the sum of 8 determinants
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.2 Q14.1

Ex 4.2 Class 12 Maths Question 15.
If A be a square matrix of order 3×3, then | kA | is equal to
(a) k|A|
(b) k² |A|
(c) k³ |A|
(d) 3k|A|
Solution:
Option (c) is correct.

Ex 4.2 Class 12 Maths Question 16.
Which of the following is correct:
(a) Determinant is a square matrix
(b) Determinant is a number associated to a matrix
(c) Determinant is a number associated to a square matrix
(d) None of these
Solution:
Option (c) is correct

Class 12 Maths Chapter 4 Determinants Ex 4.3

Ex 4.3 Class 12 Maths Question 1.
Find the area of the triangle with vertices at the point given in each of the following:
(i) (1,0), (6,0) (4,3)
(ii) (2,7), (1,1), (10,8)
(iii) (-2,-3), (3,2), (-1,-8)
Solution:
(i) Area of triangle = $\frac { 1 }{ 2 } \left| \begin{matrix} 1\quad & 0 & \quad 1 \\ 6\quad & 0 & \quad 1 \\ 4\quad & 3 & \quad 1 \end{matrix} \right|$
= $\\ \frac { 1 }{ 2 }$ [1(0-3)+1(18-0)]
= 7.5 sq units
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.3 Q1.1

Ex 4.3 Class 12 Maths Question 2.
Show that the points A (a, b + c), B (b, c + a) C (c, a+b) are collinear.
Solution:
The vertices of ∆ABC are A (a, b + c), B (b, c + a) and C (c, a + b)
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.3 Q2.1

Ex 4.3 Class 12 Maths Question 3.
Find the value of k if area of triangle is 4 square units and vertices are
(i) (k, 0), (4,0), (0,2)
(ii) (-2,0), (0,4), (0, k).
Solution:
(i) Area of ∆ = 4 (Given)
$\frac { 1 }{ 2 } \left| \begin{matrix} k\quad & 0 & \quad 1 \\ 4\quad & 0 & \quad 1 \\ 0\quad & 2 & \quad 1 \end{matrix} \right|$
= $\\ \frac { 1 }{ 2 }$ [-2k+8]
= -k+4
Case (a): -k + 4 = 4 ==> k = 0
Case(b): -k + 4 = -4 ==> k = 8
Hence, k = 0,8
(ii) The area of the triangle whose vertices are (-2,0), (0,4), (0, k)
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.3 Q3.1

Ex 4.3 Class 12 Maths Question 4.
(i) Find the equation of line joining (1, 2) and (3,6) using determinants.
(ii) Find the equation of line joining (3,1), (9,3) using determinants.
Solution:
(i) Given: Points (1,2), (3,6)
Equation of the line is
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.3 Q4.1

Ex 4.3 Class 12 Maths Question 5.
If area of triangle is 35 sq. units with vertices (2, – 6), (5,4) and (k, 4). Then k is
(a) 12
(b) – 2
(c) -12,-2
(d) 12,-2
Solution:
(d) Area of ∆ = $\frac { 1 }{ 2 } \left| \begin{matrix} 2\quad & -6 & \quad 1 \\ 5\quad & 4 & \quad 1 \\ k\quad & 4 & \quad 1 \end{matrix} \right|$
= $\\ \frac { 1 }{ 2 }$ [50 – 10k] = 25 – 5k
∴ 25-5k = 35 or 25-5k = -35
-5k = 10 or 5k = 60
=> k = -2 or k = 12

Class 12 Maths Chapter4 Determinants Ex 4.4

Ex 4.4 Class 12 Maths Question 1.
Write the minors and cofactors of the elements of following determinants:
(i) $\begin{vmatrix} 2 & -4 \\ 0 & 3 \end{vmatrix}$
(ii) $\begin{vmatrix} a & c \\ b & d \end{vmatrix}$
Solution:
(i) Let A = $\begin{vmatrix} 2 & -4 \\ 0 & 3 \end{vmatrix}$
M₁₁ = 3, M₁₂ = 0, M₂₁ = – 4, M₂₂= 2
For cofactors

Ex 4.4 Class 12 Maths Question 2.
Write Minors and Cofactor of elements of following determinant
(i) $\left| \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right|$
(ii) $\left| \begin{matrix} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{matrix} \right|$
Solution:
(i) Minors M₁₁ = $\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}$ = 1
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.4 Q2.1

Ex 4.4 Class 12 Maths Question 3.
Using cofactors of elements of second row, evaluate
$\Delta =\left| \begin{matrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{matrix} \right|$
Solution:
Given
$\Delta =\left| \begin{matrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{matrix} \right|$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.4 Q3.1

Ex 4.4 Class 12 Maths Question 4.
Using Cofactors of elements of third column, evaluate
$\Delta =\left| \begin{matrix} 1 & x & yz \\ 1 & y & zx \\ 1 & z & xy \end{matrix} \right|$
Solution:
Elements of third column are yz, zx, xy
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.4 Q4.1

Ex 4.4 Class 12 Maths Question 5.
If $\Delta =\left| \begin{matrix} { a }_{ 11 } & { a }_{ 12 } & { a }_{ 13 } \\ { a }_{ 21 } & { a }_{ 22 } & { a }_{ 23 } \\ { a }_{ 31 } & { a }_{ 32 } & { a }_{ 33 } \end{matrix} \right|$ and Aij is the cofactors of aij? then value of ∆ is given by
(a) a₁₁A₃₁+a₁₂A₃₂+a₁₃A₃₃
(b) a₁₁A₁₁+a₁₂A₂₁+a₁₃A₃₁
(c) a₂₁A₁₁+a₂₂A₁₂+a₂₃A₁₃
(d) a₁₁A₁₁+a₂₁A₂₁+a₃₁A₃₁
Solution:
Option (d) is correct.

Class 12 Maths Chapter 4 Determinants Ex 4.5

Find the adjoint of each of the matrices in Questions 1 and 2.

Ex 4.5 Class 12 Maths Question 1.
$\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}=A(say)$
Solution:
Let C_ij be cofactor of a_ij in A. Then, the cofactors of elements of A are given by
C₁₁ = (-1)¹⁺¹ (4) = 4; C₁₂ = (-1)¹⁺²(3) = -3
C₂₁ = (-1)²⁺¹ (2)= – 2; C₂₂ = (-1)²⁺² (1) = -1
Adj A = $\begin{bmatrix} 4 & -3 \\ -2 & 1 \end{bmatrix}$
= $\begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix}$

Ex 4.5 Class 12 Maths Question 2.
$\left[ \begin{matrix} 1 & -1 & 2 \\ 2 & 3 & 5 \\ -2 & 0 & 1 \end{matrix} \right] =A(say)$
Solution:
${ A }_{ 11 }={ (-1) }^{ 1+1 }M_{ 11 }=\begin{vmatrix} 3 & 5 \\ 0 & 1 \end{vmatrix}=3$
Similarly,
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 Q2.1

Verify A (adjA) = (adjA) •A = |A| I in Qs. 3 and 4.
Ex 4.5 Class 12 Maths Question 3.
$\begin{bmatrix} 2 & 3 \\ -4 & 6 \end{bmatrix}=A(say)$
Solution:
|A| = 24
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 Q3.1

Ex 4.5 Class 12 Maths Question 4.
$\left[ \begin{matrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{matrix} \right] =A(say)$
Solution:
A₁₁ = 0, A₁₂= – 11, A₁₃= 0,
A₂₁= – 3, A₂₂= 1, A₂₃= 1, A₃₁ = – 2
A₃₂= 8, A₃₃= – 3
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 Q4.1

Find the inverse of each of the matrices (if it exists) given in Questions 5 to 11:

Ex 4.5 Class 12 Maths Question 5.
$\begin{bmatrix} 2 & -2 \\ 4 & 3 \end{bmatrix}=A(say)$
Solution:
$\left| A \right| =\begin{vmatrix} 2 & -2 \\ 4 & 3 \end{vmatrix}=6+8=14\neq 0$
So, A is a non-singular matrix and therefore it is invertible. Let c_ij be cofactor of a_ij in A. Then, the cofactors of elements of A are given by
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 Q5.1

Ex 4.5 Class 12 Maths Question 6.
$\begin{bmatrix} -1 & 5 \\ -3 & 2 \end{bmatrix}=A(say)$
Solution:
$\left| A \right| =\begin{vmatrix} -1 & 5 \\ -3 & 2 \end{vmatrix}=-2+15=13\neq 0$
So, A is a non-singular matrix and therefore it is invertible. Let c_ij be cofactor of a_ij in A. Then, the cofactors of elements of A are given by
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 Q6.1

Ex 4.5 Class 12 Maths Question 7.
$\left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{matrix} \right] =A$
Solution:
|A| = 10
$\left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 2 & 4 \\ 0 & 0 & 5 \end{matrix} \right] =A$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 Q7.1

Ex 4.5 Class 12 Maths Question 8.
$\left[ \begin{matrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{matrix} \right] =A$
Solution:
$\left| A \right| =\left| \begin{matrix} 1 & 0 & 0 \\ 3 & 3 & 0 \\ 5 & 2 & -1 \end{matrix} \right| =1\left| \begin{matrix} 3 & 0 \\ 2 & -1 \end{matrix} \right| =-3\neq 0$
So, A is a non-singular matrix and therefore it is invertible. Let c_ij be cofactor of a_ijin A. Then, the cofactors of elements of A are given by
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 Q8.1

Ex 4.5 Class 12 Maths Question 9.
$\left[ \begin{matrix} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{matrix} \right] =A$
Solution:
|A| = $\left[ \begin{matrix} 2 & 1 & 3 \\ 4 & -1 & 0 \\ -7 & 2 & 1 \end{matrix} \right] =A$
= 2(-1-0)-1(4-0)+3(8-3)
So, A is non-singular matrix and therefore, it is invertible.
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 Q9.1

Ex 4.5 Class 12 Maths Question 10.
$\left[ \begin{matrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{matrix} \right] =A$
Solution:
|A| = $\left[ \begin{matrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{matrix} \right] =A$
= 1(8-6)+1(0+9)+2(0-6)
= 2+9-12
= -1≠0
∴A is invertible and
${ A }^{ -1 }=\frac { Adj\quad A }{ |A| }$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 Q10.1

Ex 4.5 Class 12 Maths Question 11.
$\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & cos\alpha & sin\alpha \\ 0 & sin\alpha & -cos\alpha \end{matrix} \right]$
Solution:
A = $\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & cos\alpha & sin\alpha \\ 0 & sin\alpha & -cos\alpha \end{matrix} \right]$
adj A = $\left[ \begin{matrix} -1 & 0 & 0 \\ 0 & -cos\alpha & -sin\alpha \\ 0 & -sin\alpha & cos\alpha \end{matrix} \right]$
First find |A| = -cos²α-sin²α
=-1≠0
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 Q11.1

Ex 4.5 Class 12 Maths Question 12.
Let $A=\begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix},B=\begin{bmatrix} 6 & 8 \\ 7 & 9 \end{bmatrix}$ , verify that (AB)^-1 = B^-1A^-1
Solution:
Here |A| = $A=\begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}$
= 15-14
= 1≠0
$Adj A=\begin{bmatrix} 5 & -7 \\ -2 & 3 \end{bmatrix}$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 Q12.1

Ex 4.5 Class 12 Maths Question 13.
If $A=\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$ show that A² – 5A + 7I = 0,hence find A^-1
Solution:
A = $\begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix}$
A² = $\begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix}$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 Q13.1

Ex 4.5 Class 12 Maths Question 14.
For the matrix A = $\begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}$ find file numbers a and b such that A²+aA+bI²=0. Hence, find A^-1.
Solution:
A = $\begin{bmatrix} 3 & 2 \\ 1 & 1 \end{bmatrix}$
A²+aA+bI²=0
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 Q14.1

Ex 4.5 Class 12 Maths Question 15.
For the matrix $A=\left[ \begin{matrix} 1 & 1 & 1 \\ 1 & 2 & -3 \\ 2 & -1 & 3 \end{matrix} \right]$ Show that A³-6A²+5A+11I₃=0.Hence find A^-1
Solution:
A² = $\left[ \begin{matrix} 4 & 2 & 1 \\ -3 & 8 & -14 \\ 7 & -3 & 14 \end{matrix} \right]$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 Q15.1

Ex 4.5 Class 12 Maths Question 16.
If $A=\left[ \begin{matrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{matrix} \right]$ show that A³-6A²+9A-4I=0 and hence, find A^-1
Solution:
We have
$A=\left[ \begin{matrix} 2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2 \end{matrix} \right]$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 Q16.1

Ex 4.5 Class 12 Maths Question 17.
Let A be a non-singular square matrix of order 3×3. Then | Adj A | is equal to:
(a) | A |
(b) | A |²
(c) | A |³
(d) 3 | A |
Solution:
Let A = $\left[ \begin{matrix} { a }_{ 11 } & { a }_{ 12 } & { a }_{ 13 } \\ { a }_{ 21 } & { a }_{ 22 } & { a }_{ 23 } \\ { a }_{ 31 } & { a }_{ 32 } & { a }_{ 33 } \end{matrix} \right]$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.5 Q17.1
Dividing by | A |, |Adj. A| = | A |²
Hence, Part (b) is the correct answer.

Ex 4.5 Class 12 Maths Question 18.
If A is an invertible matrix of order 2, then det. (A^-1) is equal to:
(a) det. (A)
(b) $\\ \frac { 1 }{ det.(A) }$
(c) 1
(d) 0
Solution:
|A|≠0
=> A^-1 exists => AA^-1 = I
|AA^-1| = |I| = I
=> |A||A^-1| = I
$|{ A }^{ -1 }|=\frac { 1 }{ |A| }$
Hence option (b) is correct.

Class 12 Maths Chapter 4 Determinantsc Ex 4.6

Examine the consistency of the system of equations in Questions 1 to 6Ex 4.6 Class 12 Maths Question 1.
x + 2y = 2
2x + 3y = 3
Solution:
x + 2y = 2,
2x + 3y = 3
=> $\begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}\left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 2 \\ 3 \end{matrix} \right]$
=> AX = B
Now |A| = $\begin{vmatrix} 1 & 2 \\ 2 & 3 \end{vmatrix}$ = 3 – 4
= – 1 ≠ 0.
Hence, equations are consistent.

Ex 4.6 Class 12 Maths Question 2.
2x – y = 5
x + y = 4
Solution:
2x – y = 5,
x + y = 4
=> $\begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}\left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 5 \\ 4 \end{matrix} \right]$
=> AX = B
Now |A| = $\begin{vmatrix} 2 & -1 \\ 1 & 1 \end{vmatrix}$
= 2 + 1
= 3 ≠ 0.
Hence, equations are consistent.

Ex 4.6 Class 12 Maths Question 3.
x + 3y = 5,
2x + 6y = 8
Solution:
x + 3y = 5,
2x + 6y = 8
=> $\begin{bmatrix} 1 & 3 \\ 2 & 6 \end{bmatrix}\left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 5 \\ 8 \end{matrix} \right]$
=> AX = B
Now |A| = $\begin{vmatrix} 1 & 3 \\ 2 & 6 \end{vmatrix}$
= 6 – 6
= 0.
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 Q3.1
Hence, equations are consistent with no solution

Ex 4.6 Class 12 Maths Question 4.
x + y + z = 1
2x + 3y + 2z = 2
ax + ay + 2az = 4
Solution:
x + y + z = 1
2x + 3y + 2z = 2
x + y + z = $\\ \frac { 4 }{ a }$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 Q4.1 Ex 4.6 Class 12 Maths Question 5.
3x – y – 2z = 2
2y – z = – 1
3x – 5y = 3
Solution:
$\left[ \begin{matrix} 3 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 2 \\ -1 \\ 3 \end{matrix} \right]$
=> AX = B
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 Q5.1

Ex 4.6 Class 12 Maths Question 6.
5x – y + 4z = 5
2x + 3y + 5z = 2
5x – 2y + 6z = -1
Solution:
Given
5x – y + 4z = 5
2x + 3y + 5z = 2
5x – 2y + 6z = -1
$\left[ \begin{matrix} 5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 5 \\ 2 \\ -1 \end{matrix} \right]$
$AX=B|A|=\left[ \begin{matrix} 5 & -1 & 4 \\ 2 & 3 & 5 \\ 5 & -2 & 6 \end{matrix} \right]$
= 5(18 + 10)+1(12 – 25)+4(-4-15)
= 140-13-76
= 51 ≠ 0
Hence equations are consistent with a unique
solution.

Solve system of linear equations using matrix method in Questions 7 to 14:

Ex 4.6 Class 12 Maths Question 7.
5x + 2y = 4
7x + 3y = 5
Solution:
The given system of equations can be written as
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 Q7.1

Ex 4.6 Class 12 Maths Question 8.
2x – y = – 2
3x + 3y = 3
Solution:
The given system of equations can be written
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 Q8.1

Ex 4.6 Class 12 Maths Question 9.
4x – 3y = 3
3x – 5y = 7
Solution:
The given system of equations can be written as
$\begin{bmatrix} 4 & -3 \\ 3 & -5 \end{bmatrix}\left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 3 \\ 7 \end{matrix} \right] i.e,,AX=B$
where $A=\begin{bmatrix} 4 & -3 \\ 3 & -5 \end{bmatrix}$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 Q9.1

Ex 4.6 Class 12 Maths Question 10.
5x + 2y = 3
3x + 2y = 5
Solution:
The given system of equations can be written as
$\begin{bmatrix} 5 & 2 \\ 3 & 2 \end{bmatrix}\left[ \begin{matrix} x \\ y \end{matrix} \right] =\left[ \begin{matrix} 3 \\ 5 \end{matrix} \right] i.e,,AX=B$
where $A=\begin{bmatrix} 5 & 2 \\ 3 & 2 \end{bmatrix}$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 Q10.1

Ex 4.6 Class 12 Maths Question 11.
2x + y + z = 1,
x – 2y – z = 3/2
3y – 5z = 9
Solution:
The given system of equations are
2x + y + z = 1,
x – 2y – z = 3/2,
3y – 5z = 9
We know AX = B => X = A^-1B
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 Q11.1

Ex 4.6 Class 12 Maths Question 12.
x – y + z = 4
2x + y – 3z = 0
x + y + z = 2.
Solution:
The given system of equations can be written
$\left[ \begin{matrix} 1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 4 \\ 0 \\ 2 \end{matrix} \right] i.e,,AX=B$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 Q12.1

Ex 4.6 Class 12 Maths Question 13.
2x + 3y + 3z = 5
x – 2y + z = – 4
3x – y – 2z = 3
Solution:
The given system of equations can be written as:
$\left[ \begin{matrix} 2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 5 \\ -4 \\ 3 \end{matrix} \right] i.e,,AX=B$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 Q13.1

Ex 4.6 Class 12 Maths Question 14.
x – y + 2z = 7
3x + 4y – 5z = – 5
2x – y + 3z = 12.
Solution:
The given system of equations can be written
$\left[ \begin{matrix} 1 & -1 & 2 \\ 3 & 4 & -5 \\ 2 & -1 & 3 \end{matrix} \right] \left[ \begin{matrix} x \\ y \\ z \end{matrix} \right] =\left[ \begin{matrix} 7 \\ -5 \\ 12 \end{matrix} \right] i.e,,AX=B$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 Q14.1

Ex 4.6 Class 12 Maths Question 15.
If A = $\left[ \begin{matrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{matrix} \right]$ Find A^-1. Using A^-1. Solve the following system of linear equations 2x – 3y + 5z = 11,3x + 2y – 4z = – 5, x + y – 2z = – 3
Solution:
We have AX = B
where $A=\left[ \begin{matrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{matrix} \right] ,X=\left[ \begin{matrix} x \\ y \\ z \end{matrix} \right]$
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 Q15.1

Ex 4.6 Class 12 Maths Question 16.
The cost of 4 kg onion, 3 kg wheat and 2 kg rice is Rs. 69. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is Rs. 90. The cost of 6 kg onion, 2 kg wheat and 3 kg rice is Rs. 70. Find the cost of each item per kg by matrix method
Solution:
Let cost of 1 kg onion = Rs x
and cost of 1 kg wheat = Rs y
and cost of 1 kg rice = Rs z
4x+3y+2z=60
2x+4y+6z=90
6x+2y+3z=70
NCERT Solutions for Class 12 Maths Chapter 4 Determinants Ex 4.6 Q16.1

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Class 12th Chapter -3 Matrices | NCERT Maths Solution | NCERT Solution | Edugrown

NCERT Solutions for Class 12 Maths Chapter : 3 Matrices

Class 12 Maths Chapter 3 Matrices Ex 3.1

Ex 3.1 Class 12 Maths Question 1.
In the matrix $A=\left[ \begin{matrix} 2 \\ 35 \\ \sqrt { 3 } \end{matrix}\begin{matrix} 5 \\ -2 \\ 1 \end{matrix}\begin{matrix} 19 \\ 5/2 \\ -5 \end{matrix}\begin{matrix} -7 \\ 12 \\ 17 \end{matrix} \right]$
(i) The order of the matrix
(ii) The number of elements
(iii) Write the elements a₁₃, a₂₁, a₃₃, a₂₄, a₂₃
Solution:
(i) The matrix A has three rows and 4 columns.
The order of the matrix is 3 x 4.
(ii) There are 3 x 4 = 12 elements in the matrix A
(iii) a₁₃ = 19, a₂₁ = 35, a₃₃ = – 5, a₂₄ = 12, a₂₃ = $\\ \frac { 5 }{ 2 }$

Ex 3.1 Class 12 Maths Question 2.
If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?
Solution:
(i) 24 = 1 x 24 = 2 x 12 = 3 x 8 = 4 x 6
Thus there are 8 matrices having 24 elements their order are (1 x 24), (24 x 1), (2 x 12), (12 x 2),(3 x 8), (8 x 3), (4 x 6), (6 x 4).
(ii) 13 = 1 x 13,
There are 2 matrices of 13 elements of order (1 x 13) and (13 x 1).

Ex 3.1 Class 12 Maths Question 3.
If a matrix has 18 elements, what are the possible orders it can have ? What, if it has 5 elements.
Solution:
We know that if a matrix is of order m × n, it has mn elements.
=> 18 = 1 x 18 = 2 x 9 = 3 x 6
Thus, all possible ordered pairs of the matrix
having 18 elements are:
(1,18), (18,1), (2,9), (9,2), (3,6), (6,3)
If it has 5 elements, then possible order are: (1,5), (5,1)

Ex 3.1 Class 12 Maths Question 4.
Construct a 2 x 2 matrix, A= [a_ij] whose elements are given by:
$(i)\quad { a }_{ ij }=\frac { { (i+j) }^{ 2 } }{ 2 }$
$(ii)\quad { a }_{ ij }=\frac { i }{ j }$
$(iii)\quad { a }_{ ij }=\frac { { (i+2j) }^{ 2 } }{ 2 }$
Solution:
$A={ \left[ { a }_{ ij } \right] }_{ 2\times 2 }=\begin{bmatrix} { a }_{ 11 } & { a }_{ 12 } \\ { a }_{ 21 } & { a }_{ 22 } \end{bmatrix}$
$(i)\quad { a }_{ ij }=\frac { { (i+j) }^{ 2 } }{ 2 }$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.1 Q4.1

Ex 3.1 Class 12 Maths Question 5.
Construct a 3 x 4 matrix , whose elements are given by:
$(i){ a }_{ ij }=\frac { 1 }{ 2 } \left| -3i+j \right|$
$(ii){ a }_{ ij }=2i-j$
Solution:
$A={ \left[ { a }_{ ij } \right] }_{ 3\times 4 }=\left[ \begin{matrix} { a }_{ 11 } \\ { a }_{ 21 } \\ { a }_{ 31 } \end{matrix}\begin{matrix} { a }_{ 12 } \\ { a }_{ 22 } \\ { a }_{ 32 } \end{matrix}\begin{matrix} { a }_{ 13 } \\ { a }_{ 23 } \\ { a }_{ 33 } \end{matrix}\begin{matrix} { a }_{ 14 } \\ { a }_{ 24 } \\ { a }_{ 34 } \end{matrix} \right]$
$(i){ a }_{ ij }=\frac { 1 }{ 2 } \left| -3i+j \right|$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.1 Q5.1

Ex 3.1 Class 12 Maths Question 6.
Find the values of x, y, z from the following equations:
$(i)\begin{bmatrix} 4 & 3 \\ x & 5 \end{bmatrix}=\begin{bmatrix} y & z \\ 1 & 5 \end{bmatrix}$
$(ii)\begin{bmatrix} x+y & 2 \\ 5+z & xy \end{bmatrix}=\begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix}$
$(iii)\left[ \begin{matrix} \begin{matrix} x+ & y+ & z \end{matrix} \\ \begin{matrix} x & +y \end{matrix} \\ \begin{matrix} y & +z \end{matrix} \end{matrix} \right] =\left[ \begin{matrix} 9 \\ 5 \\ 7 \end{matrix} \right]$
Solution:
$(i)\begin{bmatrix} 4 & 3 \\ x & 5 \end{bmatrix}=\begin{bmatrix} y & z \\ 1 & 5 \end{bmatrix}$
Clearly x = 1,y = 4,z = 3
$(ii)\begin{bmatrix} x+y & 2 \\ 5+z & xy \end{bmatrix}=\begin{bmatrix} 6 & 2 \\ 5 & 8 \end{bmatrix}$
Now 5 + z = 5 => z = 0
Now x + y = 6 and xy = 8
∴ y = 6 – x and x(6 – x) = 8
6x – x² = 8
x² – 6x + 8 = 0
(x – 4)(x – 2) = 0
=>x = 2,4
When x = 2, y = 6 – 2 = 4
and when x = 4,y = 6 – 4 = 2
Hence x = 2,y = 4,z = 0 or x = 4,y = 2,z = 0.
(iii) Equating the corresponding elements.
=> x+y+z=9 …..(i)
x+z = 5 …(ii)
y+ z = 7 …(iii)
Adding eqs. (ii) & (iii)
x + y + 2z = 12
=> (x+y+z) + z = 12,
9+z = 12 (from equ (i))
z = 3
x + z = 5
=>x + 3 = 5 => x = 2
and y+z = 7
=>y+3 = 7
=> y = 4
=> x = 2, y = 4 and z = 3

Ex 3.1 Class 12 Maths Question 7.
Find the values of a,b,c and d from the equation:
$\begin{bmatrix} a-b & 2a+c \\ 2a-b & 3c+d \end{bmatrix}=\begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix}$
Solution:
$\begin{bmatrix} a-b & 2a+c \\ 2a-b & 3c+d \end{bmatrix}=\begin{bmatrix} -1 & 5 \\ 0 & 13 \end{bmatrix}$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.1 Q7.1

Ex 3.1 Class 12 Maths Question 8.
A = [a_ij]_m×n is a square matrix, if
(a) m < n (b) n > n
(c) m = n
(d) none of these
Solution:
For a square matrix m=n.
Thus option (c) m = n, is correct.

Ex 3.1 Class 12 Maths Question 9.
Which of the given values of x and y make the following pairs of matrices equal:
$\begin{bmatrix} 3x+7 & 5 \\ y+1 & 2-3x \end{bmatrix},\begin{bmatrix} 0 & y-2 \\ 8 & 4 \end{bmatrix}$
(a) $x=\frac { -1 }{ 3 } ,y=7$
(b) Not possible to find
(c) $y=7,x=\frac { -2 }{ 3 }$
(d) $x=\frac { -1 }{ 3 } ,y=\frac { -2 }{ 3 }$
Solution:
$\begin{bmatrix} 3x+7 & 5 \\ y+1 & 2-3x \end{bmatrix},\begin{bmatrix} 0 & y-2 \\ 8 & 4 \end{bmatrix}$
(a) $x=\frac { -1 }{ 3 } ,y=7$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.1 Q9.1

Ex 3.1 Class 12 Maths Question 10.
The number of all possible matrices of order 3×3 with each entry 0 or 1 is
(a) 27
(b) 18
(c) 81
(d) 512
Solution:
There are 3 x 3 matrix or 9 entries in matrix each place can be filled with 0 or 1
∴ 9 Places can be filled in 2⁹ = 512 ways
Number of such matrices = 512
Option (d) is correct.

Class 12 Maths Chapter 3 Matrices Ex 3.2

Ex 3.2 Class 12 Maths Question 1.
Let $A=\begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix},B=\begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix},C=\begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}\qquad$
Find each of the following:
(i) A + B
(ii) A – B
(iii) 3A – C
(iv) AB
(v) BA
Solution:
Let $A=\begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix},B=\begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix},C=\begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}\qquad$
(i) A + B
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 Q1.1

Ex 3.2 Class 12 Maths Question 2.
Compute the following:
$(i)\begin{bmatrix} a & \quad b \\ -b & \quad a \end{bmatrix}+\begin{bmatrix} a & \quad b \\ b & \quad a \end{bmatrix}$
$(ii)\begin{bmatrix} { a }^{ 2 }+{ b }^{ 2 } & \quad { b }^{ 2 }+{ c }^{ 2 } \\ { a }^{ 2 }+{ c }^{ 2 } & \quad { a }^{ 2 }+{ b }^{ 2 } \end{bmatrix}+\begin{bmatrix} 2ab & \quad 2bc \\ -2ac & \quad -2ab \end{bmatrix}$
$(iii)\left[ \begin{matrix} \begin{matrix} -1 \\ 8 \\ 2 \end{matrix} & \begin{matrix} 4 \\ 5 \\ 8 \end{matrix} & \begin{matrix} -6 \\ 16 \\ 5 \end{matrix} \end{matrix} \right] +\left[ \begin{matrix} \begin{matrix} 12 \\ 8 \\ 3 \end{matrix} & \begin{matrix} 7 \\ 0 \\ 2 \end{matrix} & \begin{matrix} 6 \\ 5 \\ 4 \end{matrix} \end{matrix} \right]$
$(iv)\begin{bmatrix} { cos }^{ 2 }x & \quad { sin }^{ 2 }x \\ { sin }^{ 2 }x & { \quad cos }^{ 2 }x \end{bmatrix}+\begin{bmatrix} { sin }^{ 2 }x & \quad { cos }^{ 2 }x \\ { cos }^{ 2 }x & { \quad sin }^{ 2 }x \end{bmatrix}$
Solution:
$(i)\begin{bmatrix} a & \quad b \\ -b & \quad a \end{bmatrix}+\begin{bmatrix} a & \quad b \\ b & \quad a \end{bmatrix}$
$=\begin{bmatrix} 2a & \quad 2b \\ 0 & \quad 2a \end{bmatrix}$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 Q2.1

Ex 3.2 Class 12 Maths Question 3.
Compute the indicated products.
(i) $\begin{bmatrix} a & \quad b \\ -b & \quad a \end{bmatrix}\begin{bmatrix} a & \quad -b \\ b & \quad \quad a \end{bmatrix}$
(ii) $\left[ \begin{matrix} 1 \\ 2 \\ 3 \end{matrix} \right] \left[ \begin{matrix} 2 & 3 & 4 \end{matrix} \right]$
(iii) $\begin{bmatrix} 1 & -2 \\ 2 & \quad 3 \end{bmatrix}\left[ \begin{matrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{matrix} \right]$
(iv) $\left[ \begin{matrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{matrix} \right] \left[ \begin{matrix} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{matrix} \right]$
(v) $\left[ \begin{matrix} 2 \\ 3 \\ -1 \end{matrix}\begin{matrix} 1 \\ 2 \\ 1 \end{matrix} \right] \left[ \begin{matrix} \begin{matrix} 1 & 0 & 1 \end{matrix} \\ \begin{matrix} -1 & 2 & 1 \end{matrix} \end{matrix} \right]$
(vi) $\left[ \begin{matrix} \begin{matrix} 3 & -1 & 3 \end{matrix} \\ \begin{matrix} -1 & 0 & 2 \end{matrix} \end{matrix} \right] \left[ \begin{matrix} \begin{matrix} 2 \\ 1 \\ 3 \end{matrix} & \begin{matrix} -3 \\ 0 \\ 1 \end{matrix} \end{matrix} \right]$
Solution:
(i) $\begin{bmatrix} a & \quad b \\ -b & \quad a \end{bmatrix}\begin{bmatrix} a & \quad -b \\ b & \quad \quad a \end{bmatrix}$
= $\begin{bmatrix} { a }^{ 2 }+{ b }^{ 2 } & 0 \\ 0 & { b }^{ 2 }+{ a }^{ 2 } \end{bmatrix}$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 Q3.1

Ex 3.2 Class 12 Maths Question 4.
If $A=\left[ \begin{matrix} 1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1 \end{matrix} \right] ,B=\left[ \begin{matrix} 3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3 \end{matrix} \right] ,C=\left[ \begin{matrix} 4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3 \end{matrix} \right]$
then compute (A + B) and (B – C). Also verify that A + (B – C) = (A + B) – C.
Solution:
Given
$A=\left[ \begin{matrix} 1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1 \end{matrix} \right] ,B=\left[ \begin{matrix} 3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3 \end{matrix} \right] ,C=\left[ \begin{matrix} 4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3 \end{matrix} \right]$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 Q4.1

Ex 3.2 Class 12 Maths Question 5.
If $A=\left[ \begin{matrix} \frac { 2 }{ 3 } & 1 & \frac { 5 }{ 3 } \\ \frac { 1 }{ 3 } & \frac { 2 }{ 3 } & \frac { 4 }{ 3 } \\ \frac { 7 }{ 3 } & 2 & \frac { 2 }{ 3 } \end{matrix} \right] and\quad B=\left[ \begin{matrix} \frac { 2 }{ 5 } & \frac { 3 }{ 5 } & 1 \\ \frac { 1 }{ 5 } & \frac { 2 }{ 5 } & \frac { 4 }{ 5 } \\ \frac { 7 }{ 5 } & \frac { 6 }{ 5 } & \frac { 2 }{ 5 } \end{matrix} \right] ,$
then compute 3A – 5B.
Solution:
$3A-5B=3\left[ \begin{matrix} \frac { 2 }{ 3 } & 1 & \frac { 5 }{ 3 } \\ \frac { 1 }{ 3 } & \frac { 2 }{ 3 } & \frac { 4 }{ 3 } \\ \frac { 7 }{ 3 } & 2 & \frac { 2 }{ 3 } \end{matrix} \right] -5\left[ \begin{matrix} \frac { 2 }{ 5 } & \frac { 3 }{ 5 } & 1 \\ \frac { 1 }{ 5 } & \frac { 2 }{ 5 } & \frac { 4 }{ 5 } \\ \frac { 7 }{ 5 } & \frac { 6 }{ 5 } & \frac { 2 }{ 5 } \end{matrix} \right] ,$
= $\left[ \begin{matrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{matrix} \right] -\left[ \begin{matrix} 2 & 3 & 5 \\ 1 & 2 & 4 \\ 7 & 6 & 2 \end{matrix} \right] =\left[ \begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} \right]$

Ex 3.2 Class 12 Maths Question 6.
Simplify:
$cos\theta \begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}+sin\theta \begin{bmatrix} sin\theta & -cos\theta \\ cos\theta & sin\theta \end{bmatrix}$
Solution:
$cos\theta \begin{bmatrix} cos\theta & sin\theta \\ -sin\theta & cos\theta \end{bmatrix}+sin\theta \begin{bmatrix} sin\theta & -cos\theta \\ cos\theta & sin\theta \end{bmatrix}$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 Q6.1

Ex 3.2 Class 12 Maths Question 7.
Find X and Y if
$(i)\quad X+Y=\begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix}and\quad X-Y=\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$
$(ii)\quad 2X+3Y=\begin{bmatrix} 2 & 0 \\ 4 & 0 \end{bmatrix}and\quad 3X+2Y=\begin{bmatrix} 2 & -2 \\ -1 & 5 \end{bmatrix}$
Solution:
$(i)\quad X+Y=\begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix}and\quad X-Y=\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 Q7.1

Ex 3.2 Class 12 Maths Question 8.
Find
$X\quad if\quad Y=\begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}and\quad 2X+Y=\begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix}$
Solution:
$Y=\begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}$
We are given that
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 Q8.1

Ex 3.2 Class 12 Maths Question 9.
Find x and y, if $2\begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix}+\begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix}=\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$
Solution:
$2\begin{bmatrix} 1 & 3 \\ 0 & x \end{bmatrix}+\begin{bmatrix} y & 0 \\ 1 & 2 \end{bmatrix}=\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$
=> $\begin{bmatrix} 2+y & \quad 6 \\ 1 & \quad 2x+2 \end{bmatrix}=\begin{bmatrix} 5 & 6 \\ 1 & 8 \end{bmatrix}$
=> 2+y = 5 and 2x+2 = 8
=> y=3 and x=3
Hence x=3 and y=3

Ex 3.2 Class 12 Maths Question 10.
Solve the equation for x,y,z and t, if
$2\begin{bmatrix} x & z \\ y & t \end{bmatrix}+3\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix}=3\begin{bmatrix} 3 & 5 \\ 4 & 6 \end{bmatrix}$
Solution:
$2\begin{bmatrix} x & z \\ y & t \end{bmatrix}+3\begin{bmatrix} 1 & -1 \\ 0 & 2 \end{bmatrix}=3\begin{bmatrix} 3 & 5 \\ 4 & 6 \end{bmatrix}$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 Q10.1

Ex 3.2 Class 12 Maths Question 11.
If $x\left[ \begin{matrix} 2 \\ 3 \end{matrix} \right] +y\left[ \begin{matrix} -1 \\ 1 \end{matrix} \right] =\left[ \begin{matrix} 10 \\ 5 \end{matrix} \right]$ then find the values of x and y
Solution:
$x\left[ \begin{matrix} 2 \\ 3 \end{matrix} \right] +y\left[ \begin{matrix} -1 \\ 1 \end{matrix} \right] =\left[ \begin{matrix} 10 \\ 5 \end{matrix} \right]$
=> $\left[ \begin{matrix} 2x-y \\ 3x+y \end{matrix} \right] =\left[ \begin{matrix} 10 \\ 5 \end{matrix} \right]$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 Q11.1

Ex 3.2 Class 12 Maths Question 12.
Given
$3\begin{bmatrix} x & \quad y \\ z & \quad w \end{bmatrix}=\begin{bmatrix} x & \quad 6 \\ -1 & \quad 2w \end{bmatrix}+\begin{bmatrix} 4 & \quad x+y \\ z+w & 3 \end{bmatrix}$
find the values of x,y,z and w.
Solution:
$3\begin{bmatrix} x & \quad y \\ z & \quad w \end{bmatrix}=\begin{bmatrix} x & \quad 6 \\ -1 & \quad 2w \end{bmatrix}+\begin{bmatrix} 4 & \quad x+y \\ z+w & 3 \end{bmatrix}$
=> $\begin{bmatrix} 3x & \quad 3y \\ 3z & \quad 3w \end{bmatrix}=\begin{bmatrix} x+4 & \quad 6+x+y \\ -1+z+w & \quad 2w+3 \end{bmatrix}$
=> 3x = x + 4 => x = 2
and 3y = 6 + x + y => y = 4
Also, 3w = 2w + 3 => w = 3
Again, 3z = – 1 + z + w
=> 2z = – 1 + 3
=> 2z = 2
=> z = 1
Hence x = 2 ,y = 4, z = 1, w = 3.

Ex 3.2 Class 12 Maths Question 13.
If F(x) = $\left[ \begin{matrix} cosx & -sinx & 0 \\ sinx & cosx & 0 \\ 0 & 0 & 1 \end{matrix} \right]$
then show that F(x).F(y) = F(x+y)
Solution:
F(x) = $\left[ \begin{matrix} cosx & -sinx & 0 \\ sinx & cosx & 0 \\ 0 & 0 & 1 \end{matrix} \right]$
∴ F(y) = $\left[ \begin{matrix} cosy & -siny & 0 \\ siny & cosy & 0 \\ 0 & 0 & 1 \end{matrix} \right]$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 Q13.1

Ex 3.2 Class 12 Maths Question 14.
Show that
$(i)\begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix}\begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}\neq \begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix}$
$(ii)\left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{matrix} \right] \left[ \begin{matrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{matrix} \right] \neq \left[ \begin{matrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4 \end{matrix} \right] \left[ \begin{matrix} 1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \end{matrix} \right]$
Solution:
$(i)L.H.S=\begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix}\begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}=\begin{bmatrix} 7 & 1 \\ 33 & 34 \end{bmatrix}$
$R.H.S=\begin{bmatrix} 2 & 1 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 5 & -1 \\ 6 & 7 \end{bmatrix}=\begin{bmatrix} 16 & 5 \\ 39 & 25 \end{bmatrix}$
L.H.S≠R.H.S
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 Q14.1

Ex 3.2 Class 12 Maths Question 15.
Find A² – 5A + 6I, if A = $\left[ \begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{matrix} \right]$
Solution:
A² – 5A + 6I = $\left[ \begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{matrix} \right] \left[ \begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{matrix} \right] -5\left[ \begin{matrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{matrix} \right] +6\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right]$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 Q15.1

Ex 3.2 Class 12 Maths Question 16.
If A = $\left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right]$ Prove that A³-6A²+7A+2I = 0
Solution:
We have
A² = A x A
= $\left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right] \times \left[ \begin{matrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{matrix} \right] =\left[ \begin{matrix} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{matrix} \right]$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 Q16.1

Ex 3.2 Class 12 Maths Question 17.
If $A=\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix},I=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ then find k so that A²=kA-2I
Solution:
Given
$A=\begin{bmatrix} 3 & -2 \\ 4 & -2 \end{bmatrix},I=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
Required: To find the value of k
Now A²=kA-2I
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 Q17.1

Ex 3.2 Class 12 Maths Question 18.
If $A=\begin{bmatrix} 0 & -tan\frac { \alpha }{ 2 } \\ tan\frac { \alpha }{ 2 } & 0 \end{bmatrix}$ and I is the identity matrix of order 2,show that
$I+A=I-A\begin{bmatrix} cos\alpha & \quad -sin\alpha \\ sin\alpha & \quad cos\alpha \end{bmatrix}$
Solution:
L.H.S= $I+A=\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}+\begin{bmatrix} 0 & -tan\frac { \alpha }{ 2 } \\ tan\frac { \alpha }{ 2 } & 0 \end{bmatrix}$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 Q18.1

Ex 3.2 Class 12 Maths Question 19.
A trust has Rs 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year and second bond pays 7% interest per year. Using matrix multiplication, determine how to divide Rs 30,000 among the two types of bond if the trust fund obtains an annual total interest of
(a) Rs 1800
(b) Rs 2000
Solution:
Let Rs 30,000 be divided into two parts and Rs x and Rs (30,000-x)
Let it be represented by 1 x 2 matrix [x (30,000-x)]
Rate of interest is 005 and 007 per rupee.
It is denoted by the matrix R of order 2 x 1.
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 Q19.1

Ex 3.2 Class 12 Maths Question 20.
The book-shop of a particular school has 10 dozen Chemistry books, 8 dozen Physics books, 10 dozen Economics books. Their selling price are Rs 80, Rs 60 and Rs 40 each respectively. Find die total amount the book-shop will receive from selling all the books using matrix algebra.
Solution:
Number of Chemistry books = 10 dozen books
= 120 books
Number of Physics books = 8 dozen books = 96 books
Number of Economics books = 10 dozen books
= 120 books
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.2 Q20.1

Assuming X, Y, Z, W and P are the matrices of order 2 x n, 3 x k, 2 x p, n x 3 and p x k respectively. Choose the correct answer in Question 21 and 22.

Ex 3.2 Class 12 Maths Question 21.
The restrictions on n, k and p so that PY + WY will be defined are
(a) k = 3 ,p = n
(b) k is arbitrary,p = 2
(c) pis arbitrary, k = 3
(d) k = 2,p = 3
Solution:
Given : x_2xn, y_3xn, z_2xp, w_nx3,P_pxk
Now py +wy = P_pxk x y_3+k x w_nx3 x y_3xk
Clearly, k = 3 and p = n
Hence, option (a) is correct p x 2.

Ex 3.2 Class 12 Maths Question 22.
If n = p, then the order of the matrix 7X – 5Z is:
(a) p x 2
(b) 2 x n
(c) n x 3
(d) p x n.
Solution:
7X – 5Z = 7X_2xn – 5X_2xp
∴ We can add two matrices if their order is same n = P
∴ Order of 7X – 5Z is 2 x n.
Hence, option (b) is correct 2 x n.

Class 12 Maths Chapter 3 Matrices Ex 3.3

Ex 3.3 Class 12 Maths Question 1.
Find the transpose of each of the following matrices:
(i) $\left[ \begin{matrix} 5 \\ \frac { 1 }{ 2 } \\ -1 \end{matrix} \right]$
(ii) $\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}$
(iii) $\left[ \begin{matrix} -1 & 5 & 6 \\ \sqrt { 3 } & 5 & 6 \\ 2 & 3 & -1 \end{matrix} \right]$
Solution:
(i) let A = $\left[ \begin{matrix} 5 \\ \frac { 1 }{ 2 } \\ -1 \end{matrix} \right]$
∴ transpose of A = A’ = $\left[ \begin{matrix} 5 & \frac { 1 }{ 2 } & -1 \end{matrix} \right]$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.3 Q1.1

Ex 3.3 Class 12 Maths Question 2.
If $A=\left[ \begin{matrix} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{matrix} \right] ,B=\left[ \begin{matrix} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{matrix} \right]$
then verify that:
(i) (A+B)’=A’+B’
(ii) (A-B)’=A’-B’
Solution:
$A=\left[ \begin{matrix} -1 & 2 & 3 \\ 5 & 7 & 9 \\ -2 & 1 & 1 \end{matrix} \right] ,B=\left[ \begin{matrix} -4 & 1 & -5 \\ 1 & 2 & 0 \\ 1 & 3 & 1 \end{matrix} \right]$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.3 Q2.1

Ex 3.3 Class 12 Maths Question 3.
If $A'=\left[ \begin{matrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{matrix} \right] ,B=\left[ \begin{matrix} -1 & 2 & 1 \\ 1 & 2 & 3 \end{matrix} \right]$
then verify that:
(i) (A+B)’ = A’+B’
(ii) (A-B)’ = A’-B’
Solution:
$A'=\left[ \begin{matrix} 3 & 4 \\ -1 & 2 \\ 0 & 1 \end{matrix} \right] ,B=\left[ \begin{matrix} -1 & 2 & 1 \\ 1 & 2 & 3 \end{matrix} \right]$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.3 Q3.1

Ex 3.3 Class 12 Maths Question 4.
If $A'=\begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix},B=\begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix}$
then find (A+2B)’
Solution:
$A'=\begin{bmatrix} -2 & 3 \\ 1 & 2 \end{bmatrix},B=\begin{bmatrix} -1 & 0 \\ 1 & 2 \end{bmatrix}$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.3 Q4.1

Ex 3.3 Class 12 Maths Question 5.
For the matrices A and B, verify that (AB)’ = B’A’, where
$(i)\quad A=\left[ \begin{matrix} 1 \\ -4 \\ 3 \end{matrix} \right] ,B=\left[ \begin{matrix} -1 & 2 & 1 \end{matrix} \right]$
$(ii)\quad A=\left[ \begin{matrix} 0 \\ 1 \\ 2 \end{matrix} \right] ,B=\left[ \begin{matrix} 1 & 5 & 7 \end{matrix} \right]$
Solution:
$(i)\quad A=\left[ \begin{matrix} 1 \\ -4 \\ 3 \end{matrix} \right]$
$A'=\left[ \begin{matrix} 1 & -4 & 3 \end{matrix} \right]$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.3 Q5.1

Ex 3.3 Class 12 Maths Question 6.
If (i) $A=\begin{bmatrix} cos\alpha & \quad sin\alpha \\ -sin\alpha & \quad cos\alpha \end{bmatrix}$ ,the verify that A’A=I
If (ii) $A=\begin{bmatrix} sin\alpha & \quad cos\alpha \\ -cos\alpha & \quad sin\alpha \end{bmatrix}$ ,the verify that A’A=I
Solution:
(i) $A=\begin{bmatrix} sin\alpha & \quad cos\alpha \\ -sin\alpha & \quad cos\alpha \end{bmatrix}$
$A'=\begin{bmatrix} cos\alpha & \quad -sin\alpha \\ sin\alpha & \quad cos\alpha \end{bmatrix}$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.3 Q6.1

Ex 3.3 Class 12 Maths Question 7.
(i) Show that the matrix $A=\left[ \begin{matrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{matrix} \right]$ is a symmetric matrix.
(ii) Show that the matrix $A=\left[ \begin{matrix} 0 & 1 & -1 \\ -1 & 0 & 1 \\ 1 & -1 & 0 \end{matrix} \right]$ is a skew-symmetric matrix.
Solution:
(i) For a symmetric matrix a_ij = a_ji
Now,
$A=\left[ \begin{matrix} 1 & -1 & 5 \\ -1 & 2 & 1 \\ 5 & 1 & 3 \end{matrix} \right]$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.3 Q7.1

Ex 3.3 Class 12 Maths Question 8.
For the matrix, $A=\begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix}$
(i) (A+A’) is a symmetric matrix.
(ii) (A-A’) is a skew-symmetric matrix.
Solution:
$A=\begin{bmatrix} 1 & 5 \\ 6 & 7 \end{bmatrix}$
=> $A'=\begin{bmatrix} 1 & 6 \\ 5 & 7 \end{bmatrix}$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.3 Q8.1

Ex 3.3 Class 12 Maths Question 9.
Find $\\ \frac { 1 }{ 2 } (A+A')$ and $\\ \frac { 1 }{ 2 } (A-A')$ ,when
$A=\left[ \begin{matrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{matrix} \right]$
Solution:
$A=\left[ \begin{matrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{matrix} \right]$
$A'=\left[ \begin{matrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{matrix} \right]$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.3 Q9.1

Ex 3.3 Class 12 Maths Question 10.
Express the following matrices as the sum of a symmetric and a skew-symmetric matrix.
(i) $\begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}$
(ii) $\left[ \begin{matrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{matrix} \right]$
(iii) $\left[ \begin{matrix} 3 & 3 & -1 \\ -2 & -2 & 1 \\ -4 & -5 & 2 \end{matrix} \right]$
(iv) $\begin{bmatrix} 1 & 5 \\ -1 & 2 \end{bmatrix}$
Solution:
(i) let $A=\begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}$
=> $A'=\begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix}$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.3 Q10.1

Ex 3.3 Class 12 Maths Question 11.
Choose the correct answer in the following questions:
If A, B are symmetric matrices of same order then AB-BA is a
(a) Skew – symmetric matrix
(b) Symmetric matrix
(c) Zero matrix
(d) Identity matrix
Solution:
Now A’ = B, B’ = B
(AB-BA)’ = (AB)’-(BA)’
= B’A’ – A’B’
= BA-AB
= – (AB – BA)
AB – BA is a skew-symmetric matrix Hence, option (a) is correct.

Ex 3.3 Class 12 Maths Question 12.
If $A=\begin{bmatrix} cos\alpha & \quad -sin\alpha \\ sin\alpha & \quad cos\alpha \end{bmatrix}$ then A+A’ = I, if the
value of α is
(a) $\frac { \pi }{ 6 }$
(b) $\frac { \pi }{ 3 }$
(c) π
(d) $\frac { 3\pi }{ 2 }$
Solution:
Now
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.3 Q12.1
Thus option (b) is correct.

Class 12 Maths Chapter 3 Matrices Ex 3.4

Ex 3.4 Class 12 Maths Question 1.
$\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}$
Solution:
Let $A=\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}$
We know that
A = IA
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Q1.1

Ex 3.4 Class 12 Maths Question 2.
$\begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}$
Solution:
Let $A=\begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}$
We know that
A = IA
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Q2.1

Ex 3.4 Class 12 Maths Question 3.
$\begin{bmatrix} 1 & 3 \\ 2 & 7 \end{bmatrix}$
Solution:
Let $A=\begin{bmatrix} 1 & 3 \\ 2 & 7 \end{bmatrix}$
We know that
A = IA
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Q3.1

Ex 3.4 Class 12 Maths Question 4.
$\begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix}$
Solution:
Let $A=\begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix}$
We know that
A = IA
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Q4.1

Ex 3.4 Class 12 Maths Question 5.
$\begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix}$
Solution:
Let $A=\begin{bmatrix} 2 & 1 \\ 7 & 4 \end{bmatrix}$
We know that
A = IA
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Q5.1

Ex 3.4 Class 12 Maths Question 6.
$\begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}$
Solution:
Let $A=\begin{bmatrix} 2 & 5 \\ 1 & 3 \end{bmatrix}$
We know that
A = IA
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Q6.1

Ex 3.4 Class 12 Maths Question 7.
$\begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix}$
Solution:
Let $A=\begin{bmatrix} 3 & 1 \\ 5 & 2 \end{bmatrix}$
We know that
A = IA
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Q7.1

Ex 3.4 Class 12 Maths Question 8.
$\begin{bmatrix} 4 & 5 \\ 3 & 4 \end{bmatrix}$
Solution:
Let $A=\begin{bmatrix} 4 & 5 \\ 3 & 4 \end{bmatrix}$
We know that
A = IA
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Q8.1

Ex 3.4 Class 12 Maths Question 9.
$\begin{bmatrix} 3 & 10 \\ 2 & 7 \end{bmatrix}$
Solution:
Let $A=\begin{bmatrix} 3 & 10 \\ 2 & 7 \end{bmatrix}$
We know that
A = IA
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Q9.1

Ex 3.4 Class 12 Maths Ex 3.4 Class 12 Maths Question 10.
$\begin{bmatrix} 3 & -1 \\ -4 & 2 \end{bmatrix}$
Solution:
Let $A=\begin{bmatrix} 3 & -1 \\ -4 & 2 \end{bmatrix}$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Q10.1

Ex 3.4 Class 12 Maths Question 11.
$\begin{bmatrix} 2 & -6 \\ 1 & -2 \end{bmatrix}$
Solution:
Let $A=\begin{bmatrix} 2 & -6 \\ 1 & -2 \end{bmatrix}$
We know that
A = IA
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Q11.1

Ex 3.4 Class 12 Maths Question 12.
$\begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}$
Solution:
Let $A=\begin{bmatrix} 6 & -3 \\ -2 & 1 \end{bmatrix}$
We know that
A = IA
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Q12.1

Ex 3.4 Class 12 Maths Question 13.
$\begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}$
Solution:
Let $A=\begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}$
We know that
A = IA
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Q13.1

Ex 3.4 Class 12 Maths Question 14.
$\begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix}$
Solution:
Let $A=\begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix}$
We know that
A = IA
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Q14.1

Ex 3.4 Class 12 Maths Question 15.
$\left[ \begin{matrix} 2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2 \end{matrix} \right]$
Solution:
Let $A=\left[ \begin{matrix} 2 & -3 & 3 \\ 2 & 2 & 3 \\ 3 & -2 & 2 \end{matrix} \right]$
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Q15.1

Ex 3.4 Class 12 Maths Question 16.
$\left[ \begin{matrix} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 2 \end{matrix} \right]$
Solution:
Let $A=\left[ \begin{matrix} 1 & 3 & -2 \\ -3 & 0 & -5 \\ 2 & 5 & 2 \end{matrix} \right]$
We know that
A = IA
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Q16.1

Ex 3.4 Class 12 Maths Question 17.
$\left[ \begin{matrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{matrix} \right]$
Solution:
Let $A=\left[ \begin{matrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{matrix} \right]$
We know that
A = IA
NCERT Solutions for Class 12 Maths Chapter 3 Matrices Ex 3.4 Q17.1

Ex 3.4 Class 12 Maths Question 18.
Choose the correct answer in the following question:
Matrices A and B will be inverse of each other only if
(a) AB = BA
(b) AB = BA = 0
(c) AB = 0,BA = 1
(d) AB = BA = I
Solution:
Choice (d) is correct
i.e., AB = BA = I

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Class 12th Chapter -2 Inverse Trigonometric Functions | NCERT Maths Solution | NCERT Solution | Edugrown

NCERT Solutions for Class 12 Maths Chapter : 2 Inverse Trigonometric

Class 12 Maths Chapter 2 Inverse Trigonometric Ex 2.1

Ex 2.1 Class 12 Maths Question 1-10.
Find the principal values of the following:
(1) $\sin ^{ -1 }{ \left( -\frac { 1 }{ 2 } \right) }$
(2) $\cos ^{ -1 }{ \left( \frac { \sqrt { 3 } }{ 2 } \right) }$
(3) $\csc ^{ -1 }{ (2) }$
(4) $\tan ^{ -1 }{ \left( -\sqrt { 3 } \right) }$
(5) $\cos ^{ -1 }{ \left( -\frac { 1 }{ 2 } \right) }$
(6) $\tan ^{ -1 }{ (-1) }$
(7) $\sec ^{ -1 }{ \left( \frac { 2 }{ \sqrt { 3 } } \right) }$
(8) $\cot ^{ -1 }{ \left( \sqrt { 3 } \right) }$
(9) $\cos ^{ -1 }{ \left( -\frac { 1 }{ \sqrt { 2 } } \right) }$
(10) $\csc ^{ -1 }{ \left( -\sqrt { 2 } \right) }$
Solution:
(1) Let $\sin ^{ -1 }{ \left( -\frac { 1 }{ 2 } \right) }$ = y
∴ $sin\quad y=-\frac { 1 }{ 2 } =-sin\frac { \pi }{ 6 } =sin\left( -\frac { \pi }{ 6 } \right)$
the range of principal value of sin^-1 is
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.1 Q1.1

Ex 2.1 Class 12 Maths Question 11-12.
Find the principal values of the following:
(11) $\tan ^{ -1 }{ (1) } +\cos ^{ -1 }{ \left( -\frac { 1 }{ 2 } \right) } +\sin ^{ -1 }{ \left( -\frac { 1 }{ 2 } \right) }$
(12) $\cos ^{ -1 }{ \left( \frac { 1 }{ 2 } \right) } +2\sin ^{ -1 }{ \left( \frac { 1 }{ 2 } \right) }$
Solution:
(11) $\tan ^{ -1 }{ (1) } +\cos ^{ -1 }{ \left( -\frac { 1 }{ 2 } \right) } +\sin ^{ -1 }{ \left( -\frac { 1 }{ 2 } \right) }$
Now tan^-1 (1) = $\frac { \pi }{ 4 }$
∴the range of principal value branch of
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.1 Q2.1

Ex 2.1 Class 12 Maths Question 13.
If sin^-1 x = y, then
(a) 0 ≤ y ≤ π
(b) $-\frac { \pi }{ 2 } \le y\le \frac { \pi }{ 2 }$
(c) 0 < y < π
(d) $-\frac { \pi }{ 2 } <y<\frac { \pi }{ 2 }$
Solution:
The range of principal value of sin is $\left[ -\frac { \pi }{ 2 } ,\frac { \pi }{ 2 } \right]$
∴ if sin^-1 x = y then
$-\frac { \pi }{ 2 } \le y\le \frac { \pi }{ 2 }$
Option (b) is correct

Ex 2.1 Class 12 Maths Question 14.
$\tan ^{ -1 }{ \sqrt { 3 } -\sec ^{ -1 }{ (-2) } }$ is equal to
(a) π
(b) $-\frac { \pi }{ 3 }$
(c) $\frac { \pi }{ 3 }$
(d) $\frac { 2\pi }{ 3 }$
Solution:
$\tan ^{ -1 }{ \sqrt { 3 } =\frac { \pi }{ 3 } ,\sec ^{ -1 }{ (-2) } } =\pi -\frac { \pi }{ 3 } =\frac { 2\pi }{ 3 }$
∴ Principal values of sec^-1 is [0,π] – $\left\{ \frac { \pi }{ 2 } \right\}$
$\tan ^{ -1 }{ \sqrt { 3 } - } \sec ^{ -1 }{ (-2) } =\frac { \pi }{ 3 } -\frac { 2\pi }{ 3 } =-\frac { \pi }{ 3 }$
Option (b) is correct

Class 12 Maths Chapter 2 Inverse Trigonometric Ex 2.2

Ex 2.2 Class 12 Maths Question 1.
$3\sin ^{ -1 }{ x=\sin ^{ -1 }{ (3x-4x^{ 3 });x\in \left[ -\frac { 1 }{ 2 } ,\frac { 1 }{ 2 } \right] } }$
Solution:
Let sin^-1 x = θ
sin θ = x sin 3θ = 3 sin θ – 4 sin³ θ
sin 3θ = 3x – 4x³
3θ = sin^-1(3x – 4x³)
or $3\sin ^{ -1 }{ x=\sin ^{ -1 }{ (3x-4x^{ 3 });x\in \left[ -\frac { 1 }{ 2 } ,\frac { 1 }{ 2 } \right] } }$

Ex 2.2 Class 12 Maths Question 2.
$3\cos ^{ -1 }{ x } =\cos ^{ -1 }{ \left( { 4x }^{ 3 }-3x \right) ,x\in \left[ \frac { 1 }{ 2 } ,1 \right] }$
Solution:
Let cos^-1 x = θ
x = cos θ
R.H.S= cos^-1 (4x³ – 3cosx)
= cos^-1 (4 cos³θ – 3 cosθ)
= cos^-1 (cos 3θ) [∴ cos 3θ = 4 cos³ θ – 3 cos θ]
= 3θ
= 3 cos^-1 x
= L.H.S.

Ex 2.2 Class 12 Maths Question 3.
$\tan ^{ -1 }{ \frac { 2 }{ 11 } } +\tan ^{ -1 }{ \frac { 7 }{ 24 } } =\tan ^{ -1 }{ \frac { 1 }{ 2 } }$
Solution:
L.H.S = $\tan ^{ -1 }{ \frac { 2 }{ 11 } } +\tan ^{ -1 }{ \frac { 7 }{ 24 } }$
= $\tan ^{ -1 }{ \left[ \frac { \frac { 2 }{ 11 } +\frac { 7 }{ 24 } }{ 1-\frac { 2 }{ 11 } \times \frac { 7 }{ 24 } } \right] }$
= $\tan ^{ -1 }{ \left[ \frac { 1 }{ 2 } \right] }$
= R.H.S

Ex 2.2 Class 12 Maths Question 4.
$2\tan ^{ -1 }{ \frac { 1 }{ 2 } } +\tan ^{ -1 }{ \frac { 1 }{ 7 } } =\tan ^{ -1 }{ \frac { 31 }{ 17 } }$
Solution:
L.H.S =
$2\tan ^{ -1 }{ \frac { 1 }{ 2 } } +\tan ^{ -1 }{ \frac { 1 }{ 7 } }$
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2 Q4.1

Ex 2.2 Class 12 Maths Question 5.
Write the function in the simplest form
$\tan ^{ -1 }{ \left( \frac { \sqrt { 1+{ x }^{ 2 }-1 } }{ x } \right) } ,x\neq 0$
Solution:
Putting x = θ
∴ θ = tan^-1 x
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2 Q5.1

Ex 2.2 Class 12 Maths Question 6.
$\tan ^{ -1 }{ \left( \frac { 1 }{ \sqrt { { x }^{ 2 }-1 } } \right) ,\left| x \right| } >1$
Solution:
Given expression
$\tan ^{ -1 }{ \left( \frac { 1 }{ \sqrt { { x }^{ 2 }-1 } } \right) ,\left| x \right| } >1$
Let x = secθ
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2 Q6.1

Ex 2.2 Class 12 Maths Question 7.
$\tan ^{ -1 }{ \left( \sqrt { \frac { 1-cosx }{ 1+cosx } } \right) } ,0<x<\pi$
Solution:
$\tan ^{ -1 }{ \left( \sqrt { \frac { 1-cosx }{ 1+cosx } } \right) } ,0<x<\pi$
= $\tan ^{ -1 }{ \left[ \sqrt { \frac { { 2sin }^{ 2 }\frac { x }{ 2 } }{ { 2cos }^{ 2 }\frac { x }{ 2 } } } \right] }$
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2 Q7.1

Ex 2.2 Class 12 Maths Question 8.
$\tan ^{ -1 }{ \left( \frac { cosx-sinx }{ cosx+sinx } \right) ,0<x<\pi }$
Solution:
$\tan ^{ -1 }{ \left( \frac { cosx-sinx }{ cosx+sinx } \right) ,0<x<\pi }$
Dividing numerator and denominator by cos x
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2 Q8.1

Ex 2.2 Class 12 Maths Question 9.
$\tan ^{ -1 }{ \left( \frac { x }{ \sqrt { { a }^{ 2 }-{ x }^{ 2 } } } \right) ,\left| x \right| } <a$
Solution:
Let x = a sinθ
=> $\\ \frac { x }{ a }$ = sinθ
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2 Q9.1

Ex 2.2 Class 12 Maths Question 10.
$\tan ^{ -1 }{ \left[ \frac { { 3a }^{ 2 }-{ x }^{ 3 } }{ { a }^{ 3 }-{ 3ax }^{ 2 } } \right] ,a>0;\frac { -a }{ \sqrt { 3 } } <x,<\frac { a }{ \sqrt { 3 } } }$
Solution:
Put x = a tanθ,
we get
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2 Q10.1

Ex 2.2 Class 12 Maths Question 11.
Find the value of the following
$\tan ^{ -1 }{ \left[ 2cos\left( 2\sin ^{ -1 }{ \frac { 1 }{ 2 } } \right) \right] }$
Solution:
$\tan ^{ -1 }{ \left[ 2cos\left( 2\sin ^{ -1 }{ \frac { 1 }{ 2 } } \right) \right] }$
= $\tan ^{ -1 }{ \left[ 2cos2.\frac { \pi }{ 6 } \right] }$
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2 Q11.1

Ex 2.2 Class 12 Maths Question 12.
cot[tan^-1 a + cot^-1 a]
Solution:
Given
cot[tan^-1 a + cot^-1 a]
= $cot\left( \tan ^{ -1 }{ a } +\tan ^{ -1 }{ \frac { 1 }{ a } } \right)$
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2 Q12.1

Ex 2.2 Class 12 Maths Question 13.
$tan\frac { 1 }{ 2 } \left[ \sin ^{ -1 }{ \frac { 2x }{ 1+{ x }^{ 2 } } +\cos ^{ -1 }{ \frac { 1-{ y }^{ 2 } }{ 1+{ y }^{ 2 } } } } \right] \left| x \right| <1,y>0\quad and\quad xy<1$
Solution:
Putting x = tanθ
=> tan^-1 x = θ
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2 Q13.1

Ex 2.2 Class 12 Maths Question 14.
If $sin\left( \sin ^{ -1 }{ \frac { 1 }{ 5 } } +\cos ^{ -1 }{ x } \right) =1$ then find the value of x
Solution:
$sin\left( \sin ^{ -1 }{ \frac { 1 }{ 5 } } +\cos ^{ -1 }{ x } \right) =sin\frac { \pi }{ 2 }$
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2 Q14.1

Ex 2.2 Class 12 Maths Question 15.
If $\tan ^{ -1 }{ \frac { x-1 }{ x-2 } } +\tan ^{ -1 }{ \frac { x+1 }{ x+2 } } =\frac { \pi }{ 4 }$ then find the value of x
Solution:
L.H.S
$\tan ^{ -1 }{ \frac { x-1 }{ x-2 } } +\tan ^{ -1 }{ \frac { x+1 }{ x+2 } } =\frac { \pi }{ 4 }$
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2 Q15.1

Ex 2.2 Class 12 Maths Question 16.
$\sin ^{ -1 }{ \left( sin\frac { 2\pi }{ 3 } \right) }$
Solution:
$\sin ^{ -1 }{ \left( sin\frac { 2\pi }{ 3 } \right) }$
= $\sin ^{ -1 }{ \left( sin\left( \pi -\frac { \pi }{ 3 } \right) \right) }$
= $\sin ^{ -1 }{ \left( sin\left( \frac { \pi }{ 3 } \right) \right) } =\frac { \pi }{ 3 }$

Ex 2.2 Class 12 Maths Question 17.
$\tan ^{ -1 }{ \left( tan\frac { 3\pi }{ 4 } \right) }$
Solution:
$\tan ^{ -1 }{ \left( tan\frac { 3\pi }{ 4 } \right) }$
= $\tan ^{ -1 }{ \left( sin\frac { 3\pi }{ 4 } \right) }$
= $\tan ^{ -1 }{ tan\left( \pi -\frac { \pi }{ 4 } \right) }$
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2 Q17.1

Ex 2.2 Class 12 Maths Question 18.
$tan\left( \sin ^{ -1 }{ \frac { 3 }{ 5 } +\cot ^{ -1 }{ \frac { 3 }{ 2 } } } \right)$
Solution:
$tan\left( \sin ^{ -1 }{ \frac { 3 }{ 5 } +\cot ^{ -1 }{ \frac { 3 }{ 2 } } } \right)$
Let $\sin ^{ -1 }{ \frac { 3 }{ 5 } = } \theta$
sinθ = $\\ \frac { 3 }{ 5 }$
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2 Q18.1

Ex 2.2 Class 12 Maths Question 19.
$\cos ^{ -1 }{ \left( cos\frac { 7\pi }{ 6 } \right) }$ is equal to
(a) $\frac { 7\pi }{ 6 }$
(b) $\frac { 5\pi }{ 6 }$
(c) $\frac { \pi }{ 5 }$
(d) $\frac { \pi }{ 6 }$
Solution:
$\cos ^{ -1 }{ \left( cos\frac { 7\pi }{ 6 } \right) }$
= $\cos ^{ -1 }{ cos\left( \pi +\frac { \pi }{ 6 } \right) }$
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2 Q19.1

Ex 2.2 Class 12 Maths Question 20.
$sin\left[ \frac { \pi }{ 3 } -\sin ^{ -1 }{ \left( -\frac { 1 }{ 2 } \right) } \right]$ is equal to
(a) $\\ \frac { 1 }{ 2 }$
(b) $\\ \frac { 1 }{ 3 }$
(c) $\\ \frac { 1 }{ 4 }$
(d) 1
Solution:
$sin\left[ \frac { \pi }{ 3 } -\sin ^{ -1 }{ \left( -\frac { 1 }{ 2 } \right) } \right]$
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2 Q20.1

Ex 2.2 Class 12 Maths Question 21.
$\tan ^{ -1 }{ \sqrt { 3 } -\cot ^{ -1 }{ \left( -\sqrt { 3 } \right) } }$ is equal to
(a) π
(b) $-\frac { \pi }{ 2 }$
(c) 0
(d) 2√3
Solution:
$\tan ^{ -1 }{ \sqrt { 3 } -\cot ^{ -1 }{ \left( -\sqrt { 3 } \right) } }$
NCERT Solutions for Class 12 Maths Chapter 2 Inverse Trigonometric Functions Ex 2.2 Q21.1

Class 12 Maths Chapter 2 Inverse Trigonometric Miscellaneous Exercise

Page No: 51

Miscellaneous Exercise on Chapter 2

Find the value of the following:

Question: 1

Answer

Question: 2

Answer

Prove that

Question: 3

Answer

Question: 4

Answer

Question: 5

Answer

Question: 6

Answer

Question: 7

Answer

Question: 8

Answer

Page No. 52

Prove that

Question: 9

Answer

Question: 10

Answer

Question: 11

Answer

Question: 12

Answer

Solve the following equations:

Question: 13

Answer

Question: 14

Answer

Question: 15

Answer

The correct option is D.

Question: 16

sin^–1(1 – x) – 2 sin^–1x = π/2, then x is equal to

(A) 0, 1/2

(B) 1, 1/2

(C) 0

(D) 1/2

Answer

Given that sin^–1(1 − x) − 2sin^–1x = π/2

Let x = sin y

∴ sin^–1(1 − sin y) − 2y = π/2

⇒ sin^–1(1 − sin y) = π/2 + 2y

⇒ 1 − sin y = sin (π/2 + 2y)

⇒ 1 − sin y = cos 2y

⇒ 1 − sin y = 1 − 2sin²y [as cos²y = 1−2sin²y]

⇒ 2sin²y − sin y = 0

⇒ 2x² − x = 0 [as x = sin y]

⇒ x(2x − 1) = 0

⇒ x = 0 or, x = 1/2

But x = 1/2 does not satisfy the given equation.

∴ x = 0 is the solution of the given equation.

The correct option is C.

Question: 17

tan^–1(x/y) − tan^–1(x-y/x+y) is equal to

(A) π/2

(B) π/3

(C) π/4

(D) -3π/4

Answer

The correct option is C.

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Class 12th Chapter -1 Relations and Functions | NCERT Maths Solution | NCERT Solution | Edugrown

NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions

Class 12 Maths Chapter 1 Relations and Functions Ex 1.1

Ex 1.1 Class 12 Maths Question 1.
Determine whether each of the following relations are reflexive, symmetric and transitive:
(i) Relation R in the set A= {1,2,3,….13,14} defined as R={(x,y):3x – y = 0}
(ii) Relation R in the set N of natural numbers defined as R= {(x, y): y = x + 5 and x < 4}
(iii) Relation R in the set A= {1,2,3,4,5,6} as R= {(x, y): y is divisible by x}
(iv) Relation R in the set Z of all integers defined as R = {(x, y): x – y is an integer}
(v) Relation R in the set A of human beings in a town at a particular time given by
(a) R = {(x, y): x and y work at the same place}
(b) R = {(x, y) : x and y live in the same locality}
(c) R = {(x, y) : x is exactly 7 cm taller than y}
(d) R={(x,y) : x is wife of y}
(e) R= {(x, y): x is father of y}
Solution:
(i) Relation R in the set A = {1, 2,….,14} defined as R= {(x,y): 3x -y = 0}
(a) Put y = x, 3x – x ≠ 0 => R is not reflexive.
(b) If 3x – y = 0, then 3y – x ≠ 0, R is not symmetric
(c) If 3x – y = 0, 3y – z = 0,then 3x – z ≠ 0,R is not transitive.

(ii) Relations in the set N of natural numbers in defined by R = {(x, y): y = x + 5 and x < 4}
(a) Putting y = x, x ≠ x + 5, R is not reflexive
(b) Putting y = x + 5, then x ≠ y + 5,R is not symmetric.
(c) If y = x + 5, z = y + 5, then z ≠ x + 5 =>R is not transitive.

(iii) Relation R in the set A = {1,2,3,4,5,6} asR = {(x, y): y is divisible by x}
(a) Putting y = x, x is divisible by x => R is reflexive.
(b) If y is divisible by x, then x is not divisible by y when x ≠ y => R is not symmetric.
(c) If y is divisible by x and z is divisible by y then z is divisible by x e.g., 2 is divisible by 1,4 is divisible by 2.
=> 4 is divisible by 1 => R is transitive.

(iv) Relation R in Z of all integers defined as R = {(x, y): x – y is an integer}
(a) x – x=0 is an integer => R is reflexive
(b) x – y is an integer so is y – x => R is transitive.
(c) x – y is an integer, y- z is an integer and x – z is also an integer => R is transitive.

(v) R is a set of human beings in a town at a particular time.
(a) R = {(x, y)} : x and y work at the same place. It is reflexive as x works at the same place. It is symmetric since x and y or y and x work at same place.
It is transitive since X, y work at the same place and if y, z work at the same place,u then x and z also work at the same place.
(b) R : {(x, y) : x and y line in the same locality}
With similar reasoning as in part (a), R is reflexive, symmetrical and transitive.
(c) R: {(x, y)}: x is exactly 7 cm taller than y it is not reflexive: x cannot 7 cm taller than x. It is not symmetric: x is exactly 7 cm taller than y, y cannot be exactly 7 cm taller than x. It is not transitive: If x is exactly 7 cm taller than y and if y is exactly taller than z, then x is not exactly 7 cm taller than z.
(d) R = {(x, y): x is wife of y}
R is not reflexive: x cannot be wife of x. R is not symmetric: x is wife of y but y is not wife of x.
R is not transitive: if x is a wife of y then y cannot be the wife of anybody else.
(e) R= {(x, y): x is a father of y}
It is not reflexive: x cannot be father of himself. It is not symmetric: x is a father of y but y cannot be the father of x.
It is not transitive: x is a father of y and y is a father of z then x cannot be the father of z.

Ex 1.1 Class 12 Maths Question 2.
Show that the relation R in the set R of real numbers, defined as
R = {(a, b) : a ≤, b²} is neither reflexive nor symmetric nor transitive.
Solution:
(i) R is not reflexive, a is not less than or equal to a² for all a ∈ R, e.g., $\\ \frac { 1 }{ 2 }$ is not less than $\\ \frac { 1 }{ 4 }$ .
(ii) R is not symmetric since if a ≤ b² then b is not less than or equal to a² e.g. 2 < 5² but 5 is not less than 2².
(iii) R is not transitive: If a ≤ b², b ≤ c², then a is not less than c², e.g. 2 < (-2)², -2 < (-1)², but 2 is not less than (-1)².

Ex 1.1 Class 12 Maths Question 3.
Check whether the relation R defined in the set {1,2,3,4,5,6} as
R={(a, b): b = a+1} is reflexive, symmetric or transitive.
Solution:
(i) R is not reflexive a ≠ a + 1.
(ii) R is not symmetric if b = a + 1, then a ≠ b+ 1
(iii) R is not transitive if b = a + 1, c = b + 1 then c ≠ a + 1.

Ex 1.1 Class 12 Maths Question 4.
Show that the relation R in R defined as R={(a, b): a ≤ b}, is reflexive and transitive but not symmetric.
R = {(a,b):a≤b}
Solution:
(i) R is reflexive, replacing b by a, a ≤ a =>a = a is true.
(ii) R is not symmetric, a ≤ b, and b ≤ a which is not true 2 < 3, but 3 is not less than 2.
(iii) R is transitive, if a ≤ b and b ≤ c, then a ≤ c, e.g. 2 < 3, 3 < 4 => 2 < 4.

Ex 1.1 Class 12 Maths Question 5.
Check whether the relation R in R defined by R = {(a, b): a ≤ b³} is reflexive, symmetric or transitive.
Solution:
(i) R is not reflexive.
(ii) R is not symmetric.
(iii) R is not transitive.

Ex 1.1 Class 12 Maths Question 6.
Show that the relation R in the set {1,2,3} given by R = {(1, 2), (2,1)} is symmetric but neither reflexive nor transitive.
Solution:
(i) (1, 1),(2, 2),(3, 3) do not belong to relation R
∴ R is not reflexive.
(ii) It is symmetric (1,2) and (2,1) belong to R.
(iii) there are only two element 1 and 2 in this relation and there is no third element c in it =>R is not transitive

Ex 1.1 Class 12 Maths Question 7.
Show that the relation R in the set A of all the books in a library of a college, given by R= {(x, y): x and y have same number of pages} is an equivalence relation.
Solution:
(i) The number of pages in a book remain the same
=> Relation R is reflexive.
(ii) The book x has the same number of pages as the book y.
=> Book y has the same number of pages as the book x.
=> The relation R is symmetric.
(iii) Book x and y have the same number of pages. Also book y and z have the same number of pages.
=> Books x and z also have the same number of pages.
R is transitive also Thus, R is an equivalence relation.

Ex 1.1 Class 12 Maths Question 8.
Show that the relation R in the set A= {1,2,3,4,5} given by R = {(a, b) : |a – b| is even}, is an equivalence relation. Show that all the elements of {1,3,5} are related to each other and all the elements of {2,4} are related to each other. But no element of {1,3,5} is related to any element of {2,4}.
Solution:
A= {1,2,3,4,5} and R= {(a,b): |a – b| is even}
R= {(1,3), (1,5), (3,5), (2,4)}
(a) (i) Let us take any element of a set A. then |a – a| = 0 which is even.
=> R is reflexive.
(ii) If |a – b| is even, then |b – a| is also even, where,
R = {(a, b) : |a – b| is even} => R is symmetric.
(iii) Further a – c = a – b + b – c
If |a – b| and |b – c| are even, then their sum |a – b + b – c| is also even.
=> |a – c| is even,
∴ R is transitive. Hence R is an equivalence relation.

(b) Elements of {1,3,5} are related to each other.
Since|1 – 3| = 2, |3 – 5| = 2, |1 – 5| = 4. All are even numbers.
= Elements of {1, 3, 5} are related to each other. Similarly elements of {2,4} are related to each other. Since |2 – 4| = 2 an even number.
No element of set {1, 3, 5} is related to any element of {2,4}.

Ex 1.1 Class 12 Maths Question 9.
Show that each of the relation R in the set A = {x ∈ Z: 0 ≤ x ≤ 12}, given by
(i) R={(a,b) : |a – b|is a multiple of 4}
(ii) R={(a, b) : a = b}
is an equivalence relation. Find the set of all elements related to 1 in each case.
Solution:
The set A ∴ {x ∈ Z : 0 ≤ x ≤ 12} = {0, 1, 2,…..12}
(i) R= {(a, b): |a – b| is a multiple of 4}
|a – b| = 4k on b = a + 4k.
∴R ={(1,5),(1,9),(2, 6), (2, 10), (3, 7), (3, 11) , (4,8), (4,12), (5,9), (6,10), (7,11), (8.12) ,(0,0),(1,1), (2,2),…..(12,12)
(a) (a – a) = 0 = 4k,where k = 0=>(a,a)∈R
∴ R is reflexive.
(b) If |a – b| = 4k,then|b – a| = 4k i.e. (a,b) and (b, a) both belong to R. R is symmetric.
(c) a – c = a – b + b – c
when a – b and b – c are both multiples of 4 then a – c is also a multiple of 4. This shows if (a,b)(b,c) ∈ R then a – c also ∈ R
∴ R is an equivalence relation. The sets related 1 are {(1,5), (1,9)}.

(ii) R= {(a, b):a = b}
{(0,0), (1,1), (2,2)…..(12,12)}
(a) a = a => (a, a) ∈R
R is reflexive.
(b) Again if (a, b) ∈R, then (b, a) also ∈R
Since a = b and (a, b) ∈ R => R is symmetric.
(c) If(a, b) ∈R,then (b, c) ∈R =>a = b = c
∴ a = c => (a, c) ∈R, Hence, R is transitive set related is {1}.

Ex 1.1 Class 12 Maths Question 10.
Give an example of a relation which is
(i) Symmetric but neither reflexive nor transitive.
(ii) Transitive but neither reflexive nor symmetric.
(iii) Reflexive and symmetric but not transitive.
(iv) Reflexive and transitive but not symmetric.
(v) Symmetric and transitive but not reflexive.
Solution:
Let A = set of straight lines in a plane
(i) R: {(a, b): a is perpendicular to b} let a, b be two perpendicular lines.

(ii) Let A = set of real numbers R= {(a,b):a>b}
(a) An element is not greater than itself
∴R is not reflexive.
(b) If a > b than b is not greater than a => R is not symmetric
(c) If a > b also b > c, then a > c thus R is transitive
Hence, R is transitive but neither reflexive nor symmetric.

(iii) The relation R in the set {1,2,3} is given by R= {(a, b) :a + b≤4}
R= {(1,1), (1,2), (2,1), (1,3), (3,1), (2,2)} (1,1), (2,2) ∈ R => R is reflexive
(1,2), (2,1), (1,3), (3,1) => R is symmetric
But it is not transitive, since (2,1) ∈R, (1,3) ∈R but (2,3)∉R.

(iv) The relation R in the set {1,2,3} given by R = {(a, b): a≤b} = {(1,2), (2,2), (3,3), (2,3), 0,3)}
(a) (1,1), (2,2), (3,3) ≤R => R is reflexive
(b) (1, 2) ∈R, but (2, 1) ∉R => R is not symmetric
(c) (1,2) ∈R, (2,3) ∈R,Also (1,3) ∈R=>R is transitive.

(v) The relation R in the set {1,2,3} given by R= {(a, b): 0 < |a – b| ≤ 2} = {(1,2), (2,1), (1, 3), (3,1), (2,3), (3,2)}
(a) R is not reflexive
∵ (1,1),(2,2), (3,3) do not belong to R
(b) R is symmetric
∵ (1,2), (2,1), (1,3), (3,1), (2,3), (3,2) ∈R
(c) R is transitive (1,2) ∈R, (2,3) ∈R, Also (1,3) ∈R

Ex 1.1 Class 12 Maths Question 11.
Show that the relation R in the set A of punts in a plane given by R = {(P, Q) : distance of the point P from the origin is same as the distance of the punt Q from the origin}, is an equivalence relation. Further, show that the set of all punts related to a point P ≠ (0,0) is the circle passing through P with origin as centre.
Solution:
Let O be the origin then the relation
R={(P,Q):OP=OQ}
(i) R is reflexive. Take any distance OP,
OP = OP => R is reflexive.
(ii) R is symmetric, if OP = OQ then OQ = OP
(iii) R is transitive, let OP = OQ and OQ = OR =>OP=OR
Hence, R is an equivalence relation.
Since OP = K (constant) => P lies on a circle with centre at the origin.

Ex 1.1 Class 12 Maths Question 12.
Show that the relation R defined in the set A of all triangles as R= {(T1, T2): T1 is similar to T2}, is equivalence relation. Consider three right angle triangles T1 with sides 3,4,5, T2 with sides 5,12,13 and T3 with sides 6,8,10. Which triangles among T1, T2 and T3 are related?
Solution:
(i) In a set of triangles R = {(T1, T2) : T1 is similar T2}
(a) Since A triangle T is similar to itself. Therefore (T, T) ∈ R for all T ∈ A.
∴ Since R is reflexive
(b) If triangle T1 is similar to triangle T2 then T2 is similar triangle T1
∴ R is symmetric.
(c) Let T1 is similar to triangle T2 and T2 to T3 then triangle T1 is similar to triangle T3,
∴ R is transitive.
Hence, R is an equivalence relation.
(ii) Two triangles are similar if their sides are proportional now sides 3,4,5 of triangle T1 are proportional to the sides 6, 8, 10 of triangle T3.
∴ T1 is related to T3.

Ex 1.1 Class 12 Maths Question 13.
Show that the relation R defined in the set A of all polygons as R = {(P1, P2) : P1 and P2 have same number of sides}, is an equivalence relation. What is the set of all elements in A related to the right angle triangle T with sides 3,4 and 5?
Solution:
Let n be the number of sides of polygon P1.
R= {(P1, P2): P1 and P2 are n sides polygons}
(i) (a) Any polygon P1 has n sides => R is reflexive
(b) If P1 has n sides, P2 also has n sides then if P2 has n sides P1 also has n sides.
=> R is symmetric.
(c) Let P1, P2; P2, P3 are n sided polygons. P1 and P3 are also n sided polygons.
=> R is transitive. Hence R is an equivalence relation.
(ii) The set A = set of all the triangles in a plane.

Ex 1.1 Class 12 Maths Question 14.
Let L be the set of all lines in XY plane and R be the relation in L defined as R={(L1, L2): L1 is parallel to L2}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x+4.
Solution:
L = set of all the lines in XY plane, R= {(L1,L2) : L1 is parallel to L2}
(i) (a) L1 is parallel to itself => R is reflexive.
(b) L1 is parallel to L2 => L2 is parallel to L1 R is symmetric.
(c) Let L1 is parallel to L2 and L2 is parallel to L3 and L1 is parallel to L3 => R is transitive.
Hence, R is an equivalence relation.
(ii) Set of parallel lines related to y = 2x + 4 is y = 2x + c, where c is an arbitrary constant.

Ex 1.1 Class 12 Maths Question 15.
Let R be the relation in the set {1,2,3,4} given by R={(1,2), (2,2), (1,1), (4,4), (1,3), (3,3), (3,2) }. Choose the correct answer.
(a) R is reflexive and symmetric but not transitive.
(b) R is reflexive and transitive but not symmetric.
(c) R is symmetric and transitive but not reflexive.
(d) R is an equivalence relation.
Solution:
(b)

Ex 1.1 Class 12 Maths Question 16.
Let R be the relation in the set N given by R = {(a, b): a=b – 2, b > 6}. Choose the correct answer.
(a)(2,4)∈R
(b)(3,8)∈R
(c)(6,8)∈R
(d)(8,7)∈R
Solution:
Option (c) satisfies the condition that a = b – 2
i. e. 6 = 8 – 2 and b > 6, i.e. b = 8
=> option (c) is correct.

Class 12 Maths Chapter 1 Relations and Functions Ex 1.2

Ex 1.2 Class 12 Maths Question 1.
Show that the function f: R —> R defined by f (x) = $\\ \frac { 1 }{ x }$ is one-one onto, where R is the set of all non-zero real numbers. Is the result true, if the domain R is replaced by N with co-domain being same as R ?
Solution:
(a) We observe the following properties of f:
(i) f(x) = $\\ \frac { 1 }{ x }$ if f(x₁) = f(x₂)
$\frac { 1 }{ { x }_{ 1 } } =\frac { 1 }{ { x }_{ 2 } }$
=> x₁ = x₂
Each x ∈ R has a unique image in codomain
=> f is one-one.
(ii) For each y belonging codomain then
$y= \frac { 1 }{ x }$ or $x= \frac { 1 }{ y }$ there is a unique pre image of y.
=> f is onto.

(b) When domain R is replaced by N. codomain remaining the same, then f: N—> R If f(x₁) = f(x₂)
=> $\frac { 1 }{ { n }_{ 1 } } =\frac { 1 }{ { n }_{ 2 } }$ => n1 = n2 where n1; n2 ∈ N
=> f is one-one.
But for every real number belonging to codomain may not have a pre-image in N.
eg: $\frac { 1 }{ 2 } ,\frac { 3 }{ 2 } ,N$
∴ f is not onto.

Ex 1.2 Class 12 Maths Question 2.
Check the injectivity and surjectivity of the following functions:
(i) f: N -> N given by f (x) = x²
(ii) f: Z -> Z given by f (x) = x²
(iii) f: R -> R given by f (x) = x²
(iv) f: N -> N given by f (x) = x³
(v) f: Z ->Z given by f (x) = x³
Solution:
(i) f: N —> N given by f (x) = x²
(a) f(x₁) =>f(x₂)
=>x₁² = x₂² =>x₁ = x₂
∴ f is one-one i.e. it is injective.
(b) There are such member of codomain which have no image in domain N.
e.g. 3 ∈ codomain N. But there is no pre-image in domain of f.
=> f is not onto i.e. not surjective.
(ii) f: z —> z given by f(x) = x²
(a) f (-1) = f (1) = 1 => -1 and 1 have the same image.
∴ f is not one-one i.e. not injective.
(b) There are many such elements belonging to codomain have no pre-image in its codomain z.
e.g. 3 ∈ codomain z but √3 ∉ domain z of f,
∴ f is not onto i.e. not surjective
(iii) f: R->R, given by f(x) = x²
(a) f is not one-one since f(-1) = f(1) = 1
– 1 and 1 have the same image i.e., f is not injective
(b) – 2∈ codomain R off but √-2 does not belong to domain R of f.
=> f is not onto i.e. f is not surjective.
(iv) Injective but not surjective.
(v) Injective but not surjective.

Ex 1.2 Class 12 Maths Question 3.
Prove that the Greatest Integer Function f: R->R given by f (x)=[x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.
Solution:
f: R—> R given by f (x) = [x]
(a) f(1. 2) = 1, f(1. 5) = 1 => f is not one-one
(b) All the images of x e R belonging to its domain have integers as the images in codomain. But no fraction proper or improper belonging to codomain of f has any pre-image in its domain.
=> f is not onto.

Ex 1.2 Class 12 Maths Question 4.
Show that the Modulus Function f: R -> R given by f (x) = |x|, is neither one-one nor onto, where |x| is x, if x is positive or 0 and |x| is- x, if xis negative.
Solution:
f: R->R given by f(x) = |x|
(a) f(-1) = |-1| = 1,f(1) = |1| = 1
=> -1 and 1 have the same image
∴ f is not one-one
(b) No negative value belonging to codomain of f has any pre-image in its domain
∴ f is not onto. Hence, f is neither one-one nor onto.

Ex 1.2 Class 12 Maths Question 5.
Show that the Signum Function f: R–>R given by
f(x) = 1, if x > 0
f(x) = 0, if x = 0
f(x) = – 1, if x < 0
is neither one-one nor onto.
Solution:
f: R–>R given by
f(x) = 1, if x > 0
f(x) = 0, if x = 0
f(x) = -1, if x < 0
(a) f(x₁) = f(x₂) = 1
∴ 1 and 2 have the same image i.e.
f(x₁) = f(x₂) = 1 for x>0
=> x₁≠x₂
Similarly f(x₁) = f(x₂) = – 1, for x<0 where x₁ ≠ x₂ => f is not one-one.
(b) Except – 1,0,1 no other member of codomain of f has any pre-image in its domain.
∴ f is not onto.
=> f is neither into nor onto.

Ex 1.2 Class 12 Maths Question 6.
Let A= {1,2,3}, B = {4,5,6,7} and let f = {(1,4), (2,5), (3,6)} be a function from A to B. Show that f is one-one.
Solution:
A= {1,2,3},B= {4,5,6,7} f= {(1,4), (2,5), (3,6) }.
Every member of A has a unique image in B
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.2 Q6.1
∴ f is one – one.

Ex 1.2 Class 12 Maths Question 7.
In each of die following cases, state whether the function is one-one, onto or bijective. Justify your answer.
(i) f: R–>R defined by f(x) = 3 – 4x
(ii) f: R–>R defined by f(x) = 1 + x²
Solution:
(i) f: R —> R defined by 3 – 4x,
f (x₁) = 3 – 4x₁, f(x₂) = 3 – 4x₂
(a) f(x₁) = f(x₂) =>3 – 4x₁ = 3 – 4x₂
=> x₁ = x₂. This shows that f is one-one
(b) f(x) = y = 3 – 4x
$x= \frac { 3-y }{ 4 }$
For every value of y belonging to its codomain. There is a pre-image in its domain => f is onto.
Hence, f is one-one onto

(ii) f: R—>R given by f(x)= 1 + x²
(a) f(1) = 1 + 1 = 2,f(-1) = 1 +1 = 2
∴ f (-1) = f (1) = 2 i.e.-1 and 1 have the same image 2.
=> f is not one-one.
(b) No negative number belonging to its codomain has its pre-image in its domain
=> f is not onto. Thus f is neither one- one nor onto.

Ex 1.2 Class 12 Maths Question 8.
Let A and B be sets. Show that f:A x B –>B x A such that f (a, b) = (b, a) is bijective function.
Solution:
We have f: (A x B) —> B x A such that f (a, b) = b, a
(a) ∴ f(a1, b1)= (b1, a1) f(a2, b2) = (b2, a2) f(a1, b1) = f(a2, b2)
=>(b1, a1)
= (b2, a2)
=> b1 = b2 and a1 = a2 f is one-one
(b) Every member (p, q) belonging to its codomain has its pre-image in its domain as (q, p) f is onto. Thus, f is one-one and onto i.e. it is bijective.

Ex 1.2 Class 12 Maths Question 9.
Let f: N —> N be defined by
f (n) = $\frac { n+1 }{ 2 }$ ,if n is odd
f (n) = $\frac { n }{ 2 }$ ,if n is even
for all n∈N
State whether the function f is bijective. Justify your answer.
Solution:
f: N —> N, defined by
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.2 Q9.1
The elements 1, 2 belonging to domain of f have the same image 1 in its codomain
=> f is not one-one.
∴ it is not injective,
(b) Every member of codomain has pre-image in its domain e.g. 1 has two pre-images 1 and 2
=> f is onto. Thus f is not one-one but it is onto
=> f is not bijective.

Ex 1.2 Class 12 Maths Question 10.
Let A = R-{3} and B = R-{1}. consider the function f: A -> B defined by f (x) = $\left( \frac { x-2 }{ x-3 } \right)$
Solution:
Is f one-one and onto? Justify your answer.
f: A –> B where A = R – {3}, B = R – {1} f is defined by
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.2 Q10.1

Ex 1.2 Class 12 Maths Question 11.
Let f: R -> R be defined as f (x)=x⁴. Choose the correct answer.
(a) f is one-one onto
(b) f is many-one onto
(c) f is one-one but not onto
(d) f is neither one-one nor onto
Solution:
f(-1) = (-1)⁴ = 1,f(1) = 1⁴ = 1
∴ – 1, 1 have the same image 1 => f is not one- one
Further – 2 in the codomain of f has no pre-image in its domain.
∴ f is not onto i.e. f is neither one-one nor onto Option (d) is correct.

Question 12.
Let f: R –> R be defined as f (x)=3x. Choose the correct answer.
(a) f is one-one onto
(b) f is many-one onto
(c) f is one-one but not onto
(d) f is neither one-one nor onto
Solution:
f: R –> R is defined by f (x) = 3x
(a) f(x₁) = 3x₁, f(x₂) = 3x₂
=> f(x₁) = f(x₂)
=> 3x₁ = 3x₂
=> x₁ = x₂
=> f is one-one
(b) for every member y belonging to co-domain has pre-image x in domain of f.
∵ y = 3x
=> $x= \frac { y }{ 3 }$
f is onto
f is one-one and onto. Option (a) is correct.

Class 12 Maths Chapter 1 Relations and Functions Ex 1.3

Ex 1.3 Class 12 Maths Question 1.
Let f: {1,3,4} –> {1,2, 5} and g : {1, 2,5} –> {1,3} be given by f = {(1, 2), (3,5), (4,1) and g = {(1,3), (2,3), (5,1)}. Write down g of.
Solution:
f= {(1,2),(3,5),(4,1)}
g= {(1,3),(2,3),(5,1)}
f(1) = 2, g(2) = 3 => gof(1) = 3
f(3) = 5, g(5)= 1 =>gof(3) = 1
f(4) = 1, g(1) = 3 => gof(4) = 3
=> gof= {(1,3), (3,1), (4,3)}

Ex 1.3 Class 12 Maths Question 2.
Let f, g and h be functions from R to R. Show that (f + g) oh = foh + goh, (f • g) oh = (foh) • (goh)
Solution:
f + R –> R, g: R –> R, h: R –> R
(i) (f+g)oh(x)=(f+g)[h(x)]
= f[h(x)]+g[h(x)]
={foh} (x)+ {goh} (x)
=>(f + g) oh = foh + goh
(ii) (f • g) oh (x) = (f • g) [h (x)]
= f[h (x)] • g [h (x)]
= {foh} (x) • {goh} (x)
=> (f • g) oh = (foh) • (goh)

Ex 1.3 Class 12 Maths Question 3.
Find gof and fog, if
(i) f (x) = |x| and g (x) = |5x – 2|
(ii) f (x) = 8x³ and g (x) = ${ x }^{ 1/3 }$ .
Solution:
(i) f(x) = |x|, g(x) = |5x – 2|
(a) gof(x) = g(f(x)) = g|x|= |5| x | – 2|
(b) fog(x) = f(g (x)) = f(|5x – 2|) = ||5 x – 2|| = |5x-2|
(ii) f(x) = 8x³ and g(x) = ${ x }^{ 1/3 }$
(a) gof(x) = g(f(x)) = g(8x³) = ${ { (8x }^{ 3 }) }^{ 1/3 }$ = 2x
(b) fog (x) = f(g (x))=f( ${ x }^{ 1/3 }$ ) = 8.( ${ x }^{ 1/3 }$ )³ = 8x

Ex 1.3 Class 12 Maths Question 4.
If $f(x)=\frac { 4x+3 }{ 6x-4 } x\neq \frac { 2 }{ 3 }$ , show that fof (x) = x, for all $x\neq \frac { 2 }{ 3 }$ . What is the inverse of f?
Solution:
$f(x)=\frac { 4x+3 }{ 6x-4 } x\neq \frac { 2 }{ 3 }$
(a) fof (x) = f(f(x)) = $f\frac { 4x+3 }{ 6x-4 }$
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.3 Q4.1

Ex 1.3 Class 12 Maths Question 5.
State with reason whether following functions have inverse
(i) f: {1,2,3,4}–>{10} with f = {(1,10), (2,10), (3,10), (4,10)}
(ii) g: {5,6,7,8}–>{1,2,3,4} with g = {(5,4), (6,3), (7,4), (8,2)}
(iii) h: {1,2,3,4,5}–>{7,9,11,13} with h = {(2,7), (3,9), (4,11), (5,13)}
Solution:
f: {1,2,3,4} –> {10} with f = {(1,10), (2,10), (3,10), (4,10)}
(i) f is not one-one since 1,2,3,4 have the same image 4.
=> f has no inverse.
(ii) g: {5,6,7,8} –> {1,2,3,4} with g = {(5,4), (6,3) , (7,4), (8,2)}
Here also 5 and 7 have the same image
∴ g is not one-one. Therefore g is not invertible.
(iii) f has an inverse

Ex 1.3 Class 12 Maths Question 6.
Show that f: [-1,1] –> R, given by f(x) = $\frac { x }{ (x+2) }$ is one-one. Find the inverse of the function f: [-1,1] –> Range f.
Hint – For y ∈ Range f, y = f (x) = $\frac { x }{ (x+2) }$ for some x in [- 1,1], i.e., x = $\frac { 2y }{ (1-y) }$
Solution:
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.3 Q6.1 Ex 1.3 Class 12 Maths Question 7.
Consider f: R –> R given by f (x) = 4x + 3. Show that f is invertible. Find the inverse of f.
Solution:
f: R—>R given by f(x) = 4x + 3
f (x₁) = 4x₁ + 3, f (x₂) = 4x₂ + 3
If f(x₁) = f(x₂), then 4x₁ + 3 = 4x₂ + 3
or 4x₁ = 4x₂ or x₁ = x₂
f is one-one
Also let y = 4x + 3, or 4x = y – 3
∴ $x=\frac { y-3 }{ 4 }$
For each value of y ∈ R and belonging to co-domain of y has a pre-image in its domain.
∴ f is onto i.e. f is one-one and onto
f is invertible and f^-1 (y) = g (y) = $\frac { y-3 }{ 4 }$

Ex 1.3 Class 12 Maths Question 8.
Consider f: R₊ –> [4, ∞] given by f (x) = x² + 4. Show that f is invertible with the inverse f^-1 of f given by f^-1 (y) = √y-4 , where R₊ is the set of all non-negative real numbers.
Solution:
f(x₁) = x₁² + 4 and f(x₂) = x₂² + 4
f(x₁) = f(x₂) => x₁² + 4 = x₂² + 4
or x₁² = x₂² => x₁ = x₂ As x ∈ R
∴ x>0, x₁² = x₂² => x₁ = x₂ =>f is one-one
Let y = x² + 4 or x² = y – 4 or x = ±√y-4
x being > 0, -ve sign not to be taken
x = √y – 4
∴ f^-1 (y) = g(y) = √y-4 ,y ≥ 4
For every y ≥ 4, g (y) has real positive value.
∴ The inverse of f is f^-1 (y) = √y-4

Ex 1.3 Class 12 Maths Question 9.
Consider f: R₊ –> [- 5, ∞) given by f (x) = 9x² + 6x – 5. Show that f is invertible with
${ f }^{ -1 }(y)=\left( \frac { \left( \sqrt { y+6 } \right) -1 }{ 3 } \right)$
Solution:
Let y be an arbitrary element in range of f.
Let y = 9x² + 6x – 5 = 9x² + 6x + 1 – 6
=> y = (3x + 1)² – 6
=> y + 6 = (3x + 1)²
=> 3x + 1 = √y + 6
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.3 Q9.1

Ex 1.3 Class 12 Maths Question 10.
Let f: X –> Y be an invertible function. Show that f has unique inverse.
Hint – suppose g₁ and g₂ are two inverses of f. Then for all y∈Y, fog₁(y)=I_y(y)=fog₂(y).Use one-one ness of f.
Solution:
If f is invertible gof (x) = I_x and fog (y) = I_y
∴ f is one-one and onto.
Let there be two inverse g₁ and g₂
fog₁ (y) = I_y, fog₂ (y) = I_y
I_y being unique for a given function f
=> g₁ (y) = g₂ (y)
f is one-one and onto
f has a unique inverse.

Ex 1.3 Class 12 Maths Question 11.
Consider f: {1,2,3} –> {a, b, c} given by f (1) = a, f (2)=b and f (3)=c. Find f^-1 and show that (f^-1)f^-1=f.
Solution:
f: {1,2, 3,} –> {a,b,c} so that f(1) = a, f(2) = b, f(3) = c
Now let X = {1,2,3}, Y = {a,b,c}
∴ f: X –> Y
∴ f^-1: Y –> X such that f^-1 (a)= 1, f^-1(b) = 2; f^-1(c) = 3
Inverse of this function may be written as
(f^-1)^-1 : X –> Y such that
(f^-1)^-1 (1) = a, (f^-1)^-1 (2) = b, (f^-1)^-1 (3) = c
We also have f: X –> Y such that
f(1) = a,f(2) = b,f(3) = c => (f^-1)^-1= f

Ex 1.3 Class 12 Maths Question 12.
Let f: X –> Y be an invertible function. Show that the inverse of f^-1 is f, i.e., (f^-1)^-1 = f.
Solution:
f: X —> Y is an invertible function
f is one-one and onto
=> g : Y –> X, where g is also one-one and onto such that
gof (x) = I_x and fog (y) = I_y => g = f^-1
Now f^-1 o (f^-1)^-1 = I
and fo[f^-1o (f^-1)^-1] =fol
or (fof^-1)^-1 o (f^-1)^-1 = f
=> Io (f^-1)^-1 = f
=> (f^-1)^-1 = f

Ex 1.3 Class 12 Maths Question 13.
If f: R –> R be given by f(x) = ${ \left( 3-{ x }^{ 3 } \right) }^{ \frac { 1 }{ 3 } }$ , then fof (x) is
(a) ${ x }^{ \frac { 1 }{ 3 } }$
(b) x³
(c) x
(d) (3 – x³)
Solution:
f: R-> R defined by f(x) = ${ \left( 3-{ x }^{ 3 } \right) }^{ \frac { 1 }{ 3 } }$
fof (x) = f[f(x)] = ${f{ \left( 3-{ x }^{ 3 } \right) }^{ \frac { 1 }{ 3 } }}$
= ${ \left[ 3-{ \left\{ { \left( 3-{ x }^{ 3 } \right) }^{ \frac { 1 }{ 3 } } \right\} }^{ 3 } \right] }^{ \frac { 1 }{ 3 } }$
= ${ \left[ 3-{ \left\{ { \left( 3-{ x }^{ 3 } \right) }^{ \frac { 1 }{ 3 } } \right\} } \right] }$
= ${ \left( { x }^{ 3 } \right) }^{ \frac { 1 }{ 3 } }$
= x

Ex 1.3 Class 12 Maths Question 14.
Let f: $R-\left\{ -\frac { 4 }{ 3 } \right\} \rightarrow R$ be a function defined as f (x) = $\frac { 4x }{ 3x+4 }$ . The inverse of f is the map g: Range f–> $R-\left\{ -\frac { 4 }{ 3 } \right\} \rightarrow R$ given by
(a) $g(y)=\frac { 3y }{ 3-4y }$
(b) $g(y)=\frac { 4y }{ 4-3y }$
(c) $g(y)=\frac { 4y }{ 3-4y }$
(d) $g(y)=\frac { 3y }{ 4-3y }$
Solution:
(b)

Class 12 Maths Chapter 1 Relations and Functions Ex 1.4

Ex 1.4 Class 12 Maths Question 1.
Determine whether or not each of the definition of * given below gives a binary operation. In the event that * is not a binary operation, give justification for this.
(i) On Z⁺,define * by a * b = a – b
(ii) On Z⁺, define * by a * b = ab
(iii) On R, define * by a * b = ab²
(iv) On Z⁺,define * by a * b = |a – b|
(v) On Z⁺, define * by a * b = a
Solution:
(i) If a > b, a * b = a – b > 0, which belongs to Z⁺.
But if a < b, a * b = a – b < 0, which does not belong to Z⁺
=> * given operation is not a binary operation.
(ii) For all a and b belonging to Z^-1, ab also belongs to Z⁺.
∴ The operation *, defined by a * b = ab is a binary operation.
(iii) For all a and b belonging to R, ab² also belongs to R.
∴ The operation * defined by a * b = ab² is binary operation.
(iv) For all a and b belonging to Z⁺, |a – b| also belongs to Z⁺¹
∴ The operation a * b = |a – b| is a binary operation.
(v) On Z⁺ defined by a * b = a
a, b ∈ Z⁺ = a ∈ Z⁺
∴ The operation * is a binary operation.

Ex 1.4 Class 12 Maths Question 2.
For each binary operation * defined below, determine whether * is commutative or associative.
(i) OnZ, define a * b = a – b
(ii) OnQ, define a * b = ab + 1
(iii) On Q, define a * b = $\\ \frac { ab }{ 2 }$
(iv) On Z⁺, define a * b = 2ab
(v) On Z⁺, define a * b = ab
(vi) On R- {-1}, define a * b = $\\ \frac { a }{ b+1 }$
Solution:
(i) On Z, operation * is defined as
(a) a * b = a – b => b * a = b – a
But a – b ≠ b – a ==> a * b ≠ b * a
Defined operation is not commutative
(b) a – (b – c) ≠ (a – b) – c
Binary operation * as defined is not associative.

(ii) On Q, Operation * is defined as a * b = ab + 1
(a) ab + 1 = ba + 1, a * b = b * a
Defined binary operation is commutative.
(b) (a*b)*c = (ab + 1)*c = (ab + 1)c + 1 = abc + c + 1
a * (b * c) = a * (bc + 1) = a(bc + 1)+ 1 = abc + a+ 1
=> a * (b * c) ≠ (a * b) * c
∴ Binary operation defined is not associative.

(iii) (a) On Q, operation * is defined as a * b = $\\ \frac { ab }{ 2 }$
∴ a * b = b * a
∴ Operation binary defined is commutative.
be abc
(b) a * (b * c) = a * $\\ \frac { bc }{ 2 }$ = $\\ \frac { abc }{ 4 }$ and
(a * b) * c = $\\ \frac { bc }{ 2 }$ * c $\\ \frac { abc }{ 4 }$
=> (a * b) * c = $\\ \frac { bc }{ 2 }$ * c $\\ \frac { abc }{ 4 }$
Defined binary operation is associative.

(iv) On Z⁺ operation * is defined as a * b = 2^ab
(a) a * b = 2^ab, b * a = 2^ba = 2^ab
=> a * b = b * a .
∴ Binary operation defined as commutative.
(b) a * (b * c) = a * 2^ba = 2^a.2bc
(a * b) * c = 2^ab * c = 2^2ab
Thus (a * b) * c ≠ a * (b * c)
∴ Binary operation * as defined as is not associative.

(v) On Z⁺, a * b = ab
(a) b * a = b^a .
∴ a^b ≠ b^a = a * b ≠ b * a.
* is not commutative.
(b) (a*b)*c = a^b*c = (a^b)^c = a^bc a*(b* c)
= a*b^c = a^bc
Thus (a * b) * c ≠ (a * b * c)
∴ Operation * is not associative.

(vi) Neither commutative nor associative.

Ex 1.4 Class 12 Maths Question 3.
Consider the binary operation ^ on the set {1, 2, 3, 4, 5} defined by a ^ b=min {a, b}. Write the operation table of the operation ^.
Solution:
Operation ^ table on the set {1, 2, 3, 4, 5} is as follows.
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.4 Q3.1

Ex 1.4 Class 12 Maths Question 4.
Consider a binary operation * on the set {1, 2, 3, 4, 5} given by the following multiplication table.
(i) Compute (2 * 3) * 4 and 2 * (3 * 4)
(ii) Is * commutative?
(iii) Compute (2 * 3) * (4 * 5).
Hint – use the following table)
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.4 Q4.1
Solution:
(i) From the given table, we find
2*3 = 1, 1*4 = 1
(a) (2*3)*4 = 1 * 4 = 1
(b) 2*(3*4) = 2 * 1 = 1
(ii) Let a, b ∈ {1,2,3,4,5} From the given table, we find
a*a = a
a*b = b*a = 1 when a or b or are odd and a b.
2 * 4 = 4 * 2 = 2, when a and b are even and a ≠ b
Thus a * b = b * c
∴Binary operation * given is commutative.
(iii) Binary operation * given is commutative (2 * 3) * (4 * 5) = 1 * 1 = 1.

Ex 1.4 Class 12 Maths Question 5.
Let *’ be the binary operation on the set {1,2,3,4,5} defined by a *’ b=H.C.F. of a and b. Is the operation *’ same as the operation * defined in the exp no. 4 above? Justify your answer.
Solution:
The set is {1,2,3,4, 5} and a * b = HCF of a and b.
Let us prepare the table of operation *.
NCERT Solutions for Class 12 Maths Chapter 1 Relations and Functions Ex 1.4 Q5.1

Ex 1.4 Class 12 Maths Question 6.
Let * be the binary operation on N given by a * b = L.C.M. of a and b.Find
(i) 5 * 7, 20 * 16
(ii) Is * commutative?
(iii) Is * associative?
(iv) Find the identity of * in N
Solution:
Binary operation * defined as a * b = 1 cm. of a and b.
(i) 5 * 7 = 1 cm of 5 and 7 = 35
20 * 16= 1 cm of 20 and 16 = 80
(ii) a * b= 1 cm of a and b b * a = 1 cm of b and a
=> a * b = b * a, 1 cm of a, b and b, a are equal
∴ Binary operation * is commutative.
(iii) a * (b * c) = 1 cm of a, b, c and (a * b) * c = 1 cm of a, b, c
=> a * (b * c) = (a * b) * c
=> Binary operation * given is associative.
(iv) Identity of * in N is 1
1 * a = a * 1 = a = 1 cm of 1 and a.
(v) Let * : N x N—> N defined as a * b = 1.com of (a, b)
For a = 1, b = 1, a * b = l b * a
Otherwise a * b ≠ 1
∴ Binary operation * is not invertible
=> 1 is invertible for operation *

Ex 1.4 Class 12 Maths Question 7.
Is * defined on the set {1, 2, 3, 4, 5} by a * b = L.C.M. of a and b a binary operation? Justify your answer.
Solution:
The given set = {1,2,3,4,5} Binary operation is defined as a * b = 1 cm of a and b. 4 * 5 = 20 which does not belong to the given set {1, 2, 3, 4, 5}.
It is not a binary operation.

Ex 1.4 Class 12 Maths Question 8.
Let * be the binary operation on N defined by a * b = H.C.F. of a and b. Is * commutative? Is * associative? Does there exist identity for this binary operation on N?
Solution:
Binary operation on set N is defined as a * b = HCF of a and b
(a) We know HCF of a, b = HCF of b, a
∴ a * b = b * a
∴ Binary operation * is commutative.
(b) a*(b*c) = a* (HCF of b, c) = HCF of a and (HCF of b, c) = HCF of a, b and c
Similarly (a * b) * c = HCF of a, b, and c
=> (a * b) * c = a * (b * c)
Binary operation * as defined above is associative.
(c) 1 * a = a * 1 = 1 ≠ a
∴ There does not exists any identity element.

Ex 1.4 Class 12 Maths Question 9.
Let * be a binary operation on the set Q of rational numbers as follows:
(i) a * b = a – b
(ii) a * b = a² + b²
(iii) a * b = a + ab
(iv) a * b = (a – b)²
(v) a * b = $\\ \frac { ab }{ 2 }$
(vi) a * b = ab²
Find which of the binary operations are commutative and which are associative.
Solution:
Operation is on the set Q.,,
(i) defined as a * b = a – b
(a) Now b * a = b – a
But a – b ≠ b – a
∴ a * b ≠ b * a
∴ Operation * is not commutative.
(b) a * (b * c) = a * (b – c) = a – (b – c) = a – b + c (a * b) * c = (a – b) * c = a – b – c
Thus a * (b * c)¹ (a * b) * c = (a² + b²)² + c²
=> a * (b * c) ≠ (a * b) * c
∴ The operation * as defined is not associative.

(ii) (a) a * b = a² + b * a = b² + a² = a² + b².
a * b = b * a
This binary operation is commutative,
(b) a*(b*c) = a*(b² + c²) = a² + (b²)² + c²)²
=> (a*b)*c = (a² + b²)*c = (a² + b²) + c²
Thus a * (b*c) (a*b) * c
The operation * given is not associative.

(iii) Operation * is defined as a * b = a + ab
(a) b * a = b + ba
a * b ≠ b * a
The operation is not commutative.
(b) a*(b*c) = a*(b + bc)
= a + a(b + bc)
= a + ab + abc (a * b) * c
= (a + ab) * c
= (a + ab) + (a + ab) • c
= a + ab + ac + abc
=> a * (b * c) ≠ (a * b) * c
=> The binary operation is not associative.

(iv) The binary operation is defined as a * b = (a – b)²
(a) b*a = (b – a)² = (a – b)² => a*b = b*a
.’. This binary operation * is commutative.
(b) a*(b*c) = a*(b – c)² = [a – (b – c)²]² (a * b) * c
= (a – b)² * c
= [(a – b)² – c]²
=> a * (b * c) ≠ (a * b) * c
the operation * is not associative.

(v) Commutative and associative.

(vi) Neither commutative nor associative.

Ex 1.4 Class 12 Maths Question 10.
Show that none of the operations given above has identity.
Solution:
The binary operation * on set Q is
(i) defined as a*b = a – b
For identity element e, a*e = e*a = a
But a*e = a – e≠a and e*a = e – a≠a
There is no identity element for this operation
(ii) Binary operation * is defined as a * b = a² + b² ≠ a
This operation * has no identity.
(iii) The binary operation is defined as a*b = a+ab
Putting b = e, a + e = a + eb ≠ a
There is no identity element.
(iv) The binary operation is defined as a * b = (a – b)²
Put b = e, a * e = (a – e)² ≠ a for any value of
e∈Q
=> there is no Identity Element.
(v) The operation is a * b = $\\ \frac { ab }{ 4 }$
∴ a * e = $\\ \frac { ae }{ 4 }$ ≠ a for any value of e ∈ Q
∴ Operation * has no identity
(vi) The operation * is a * b = ab² Put b = e, a
*e = ae² and e * a = ea² ≠ a for any value of e∈Q
=> There is no Identity Element. Thus, these operations have no Identity.

Ex 1.4 Class 12 Maths Question 11.
Let A=N x N and * be the binary operation on A defined by (a,b)*(c,d)=(a+c,b+d)
Show that * is commutative and associative. Find the identity element for * on A, if any.
Solution:
A = N x N Binary operation * is defined as (a, b) * (c, d) = (a + c, b + d)
(a) Now (c, d) * (a,b) = (c+a, d+b) = (a+c,b+d)
=> (a, b) * (c, d) = (c, d) * (a, b)
∴ This operation * is commutative
(b) Next(a,b)* [(c,d)*(e,f)]=(a,b)*(c+e,d+f) = ((a + c + e), (b + d + f))
and [(a, b) * (c, d)] * (e, f)=(a+c, b+d) * (e, f) = ((a + c + e, b+d + f))
=> (a, b) * [(c, d) * (e, f)] = [(a, b) * (c, d)] * (e,f)
∴ The binary operation given is associative
(c) Identity element does not exists.

Ex 1.4 Class 12 Maths Question 12.
State whether the following statements are true or false. Justify.
(i) For an arbitrary binary operation * on a set N,
a*a=a∀a∈N.
(ii) If * is a commutative binary operation on N, then
a * (b * c) = (c * b) * a
Solution:
(i) A binary operation on N is defined as
a*a=a∀a∈N.
Here operation * is not defined.
∴ Given statement is false.
(ii) * is a binary commutative operation on N. c
* b = b * c
∵ * is commutative
∵ (c * b) * a = (b * c) * a = a * (b * c)
∴ Thus a * (b * c) = (c * b) * a
This statement is true.

Ex 1.4 Class 12 Maths Question 13.
Consider a binary operation * on N defined as a * b = a³ + b³. Choose the correct answer.
(a) Is * both associative and commutative?
(b) Is * commutative but not associative?
(c) Is * associative but not commutative?
(d) Is * neither commutative nor associative?
Solution:
(b)

Class 12 Maths Chapter 1 Relations and Functions Miscellaneous Exercise

Exercise 1.3

1. Let f : R → R be defined as f(x) = 10x + 7. Find the function g : R → R such that gof = fog = I_R .

Answer

It is given that f: R → R is defined as f(x) = 10x + 7.

For one – one

Let f(x) = f(y), where x, y ∈ R.

⇒ 10x + 7 = 10y + 7

⇒ x = y

∴ f is a one – one function.

For onto

For y ∈ R, let y = 10x + 7.

∴ f is onto.

Therefore, f is one–one and onto.

Thus, f is an invertible function.

Let us define g: R → R as g(y) = y−7/10

Now, we have

2. Let f : W → W be defined as f(n) = n–1, if n is odd and f(n) = n+1, if n is even. Show that f is invertible. Find the inverse of f. Here, W is the set of all whole numbers.

Answer

It is given that:

For one–one

Let f(n) = f(m).

It can be observed that if n is odd and m is even, then we will have n−1 = m + 1.

⇒ n − m = 2

However, this is impossible.

Similarly, the possibility of n being even and m being odd can also be ignored

under a similar argument.

∴ Both n and m must be either odd or even. Now, if both n and m are odd,

Then, we have

f(n) = f(m)

⇒ n − 1 = m – 1

⇒ n = m

Again, if both n and m are even,

Then, we have

f(n) = f(m)

⇒ n + 1 = m + 1

⇒ n = m

∴ f is one–one.

For onto:

It is clear that any odd number 2r + 1 in co-domain N is the image of 2r in domain

N and any even number 2r in co-domain N is the image of 2r + 1 in domain N.

∴ f is onto.

Hence, f is an invertible function.

Now, when n is odd

gof(n) = g(f(n)) = g(n − 1) = n − 1 + 1 = n and

When n is even

gof(n) = g(f(n)) = g(n + 1) = n + 1 − 1 = n

Similarly,

When m is odd

fog(m) = f(g(m)) = f(m − 1) = m − 1 + 1 = m and

When m is even

fog(m) = f(g(m)) = f(m + 1) = m + 1 − 1 = m

∴ gof = I_W and fog = I_W

Thus, f is invertible and the inverse of f is given by f^-1 = g, which is the same as f. Hence, the inverse of f is f itself.

3. If f : R → R is defined by f(x) = x² – 3x + 2, find f (f(x)).

Answer

It is given that f: R → R is defined as f(x) = x² − 3x + 2.

f(f(x)) = f(x² − 3x + 2)

= (x² − 3x + 2)² − 3(x² − 3x + 2) + 2

= (x⁴ + 9x² + 4 − 6x³ − 12x + 4x²) + (−3x² + 9x − 6) + 2

= x⁴ − 6x³ + 10x² − 3x

4. Show that the function f : R → {x ∈ R : –1 < x < 1} defined by

, x ∈ R is one one and onto function.

Answer

For one – one

Suppose f(x) = f(y), where x, y ∈ R.

Since, x is positive and y is negative

x > y ⇒ x − y > 0

But, 2xy is negative.

Then 2xy ≠ x − y

Thus, the case of x being positive and y being negative can be ruled out.

Under a similar argument, x being negative and y being positive can also be ruled out.

∴ x and y have to be either positive or negative.

When x and y are both positive, we have

∴ f is one – one.

For onto

Now, let y ∈ R such that −1 < y < 1.

If y is negative, then, there exists x =y/1+y ∈ R such that

∴ f is onto.

Hence, f is one–one and onto.

5. Show that the function f : R → R given by f(x) = x³ is injective.

Answer

f: R → R is given as f(x) = x3.

For one – one

Suppose f(x) = f(y), where x, y ∈ R.

⇒ x³ = y³ ….(1)

Now, we need to show that x = y.

Suppose x ≠ y, their cubes will also not be equal.

⇒ x³ ≠ y³

However, this will be a contradiction to (1).

∴ x = y

Hence, f is injective.

6. Give examples of two functions f : N → Z and g : Z → Z such that g o f is injective but g is not injective.

(Hint : Consider f(x) = x and g (x) = | x |).

Answer

Define f: N → Z as f(x) = x and g: Z → Z as g (x) = |x|.

We first show that g is not injective.

It can be observed that

g (−1) =|−1| = 1

g (1) = |1| = 1

∴ g (−1) = g (1), but −1 ≠ 1.

∴ g is not injective.

Now, g of: N → Z is defined as gof(x) = g(f(x)) = g(x) = |x|.

Let x, y ∈ N such that g of(x) = g of(y).

⇒ |x| = |y|

Since x and y ∈ N, both are positive.

∴ |x| = |y| ⇒ x = y

Hence, gof is injective.

7. Give examples of two functions f : N → N and g : N → N such that g o f is onto but f is not onto.

Answer

Define f: N → N by f(x) = x + 1 and,

We first show that g is not onto.

For this, consider element 1 in co-domain N. It is clear that this element is not an image of any of the elements in domain N.

∴ f is not onto.

Now, gof: N → N is defined by

gof(x) = g(f(x)) = g(x + 1) = x + 1 − 1 = x [x ∈ N ⇒ x + 1 > 1]

Then, it is clear that for y ∈ N, there exists x = y ∈ N such that gof(x) = y.

Hence, gof is onto.

8. Given a non empty set X, consider P(X) which is the set of all subsets of X.

Define the relation R in P(X) as follows:

For subsets A, B in P(X), ARB if and only if A ⊂ B. Is R an equivalence relation on P(X)? Justify your answer.

Answer

Since every set is a subset of itself, ARA for all A ∈ P(X).

∴ R is reflexive.

Let ARB ⇒ A ⊂ B.

This cannot be implied to B ⊂ A.

For instance, if A = {1, 2} and B = {1, 2, 3}, then it cannot be implied that B is

related to A.

∴ R is not symmetric.

Further, if ARB and BRC, then A ⊂ B and B ⊂ C.

⇒ A ⊂ C

⇒ ARC

∴ R is transitive.

Hence, R is not an equivalence relation as it is not symmetric.

Page No. 30

9. Given a non-empty set X, consider the binary operation ∗ : P(X) × P(X) → P(X) given by A ∗ B = A ∩ B ∀ A, B in P(X), where P(X) is the power set of X. Show that X is the identity element for this operation and X is the only invertible element in P(X) with respect to the operation ∗.

Answer

It is given the binary operation *:

P(X) × P(X) → P(X) given by A * B = A ∩ B ∀ A, B in P(X)

We know that A ∩ X = A = X ∩ A for all A ∈ P(X)

⇒ A * X = A = X * A for all A ∈ P(X)

Thus, X is the identity element for the given binary operation *.

Now, an element A ∈ P(X) is invertible if there exists B ∈ P(X) such that

A * B = X = B * A [As X is the identity element]

A ∩ B = X = B ∩ A

This case is possible only when A = X = B.

Thus, X is the only invertible element in P(X) with respect to the given operation*.

Hence, the given result is proved.

10. Find the number of all onto functions from the set {1, 2, 3, … , n} to itself.

Answer

Onto functions from the set {1, 2, 3, … , n} to itself is simply a permutation on n symbols 1, 2, …, n.

Thus, the total number of onto maps from {1, 2, … , n} to itself is the same as the total number of permutations on n symbols 1, 2, …, n, which is n.

11. Let S = {a, b, c} and T = {1, 2, 3}. Find F^-1 of the following functions F from S to T, if it exists.

(i) F = {(a, 3), (b, 2), (c, 1)}

(ii) F = {(a, 2), (b, 1), (c, 1)}

Answer

S = {a, b, c}, T = {1, 2, 3}

(i) F: S → T is defined as F = {(a, 3), (b, 2), (c, 1)}

⇒ F (a) = 3, F (b) = 2, F(c) = 1

Therefore, F^-1 : T → S is given by F^-1 = {(3, a), (2, b), (1, c)}.

(ii) F: S → T is defined as F = {(a, 2), (b, 1), (c, 1)}

Since F (b) = F (c) = 1, F is not one – one.

Hence, F is not invertible i.e., F^-1 does not exist.

12. Consider the binary operations ∗ : R × R → R and o : R × R → R defined as a ∗b = |a – b| and a o b = a, ∀ a, b ∈ R. Show that ∗ is commutative but not associative, o is associative but not commutative. Further, show that ∀ a, b, c ∈ R, a ∗ (b o c) = (a ∗ b) o (a ∗ c). [If it is so, we say that the operation ∗ distributes over the operation o]. Does o distribute over ∗? Justify your answer.

Answer

It is given that *: R × R → and o: R × R → R is defined as a ∗ b = |a − b| and

aob = a, ∀ a, b ∈ R

For a, b ∈ R, we have

a ∗ b = |a − b| and b ∗ a = |b − a| = |−(a − b)| = |a − b|

∴ a * b = b * a

Hence, the operation * is commutative.

It can be observed that

(1 ∗ 2) ∗ 3 = (|1 − 2|) ∗ 3 = 1 ∗ 3 = |1 − 3| = 2

and

1 ∗ (2 ∗ 3) = 1 ∗ (|2 − 3|) = 1 ∗ 1 = |1 − 1| = 0

∴ (1 ∗ 2) ∗ 3 ≠ 1 ∗ (2 ∗ 3) where 1, 2, 3 ∈ R.

Hence, the operation * is not associative.

Now, consider the operation o

It can be observed that 1 o 2 = 1 and 2 o 1 = 2.

∴ 1 o 2 ≠ 2 o 1 where 1, 2 ∈ R.

Hence, the operation o is not commutative.

Let a, b, c ∈ R. Then, we have

(a o b) o c = a o c = a

and

a o (b o c) = a o b = a

∴ (a o b) o c = a o (b o c), where a, b, c ∈ R

Hence, the operation o is associative.

Now, let a, b, c ∈ R, then we have

a * (b o c) = a * b =|a − b|

(a * b) o (a * c) = (|a − b|)o(|a − c|) = |a − b|

Hence, a * (b o c) = (a * b) o (a * c).

Now,

1 o (2 ∗ 3) = 1o(|2 − 3|) = 1o1 = 1

(1 o 2) * (1 o 3) = 1 * 1 = |1 − 1| = 0

∴ 1 o (2 * 3) ≠ (1 o 2) * (1 o 3) where 1, 2, 3 ∈ R

Hence, the operation o does not distribute over *.

13. Given a non-empty set X, let ∗ : P(X) × P(X) → P(X) be defined as A * B = (A–B) ∪ (B–A), ∀ A, B ∈ P(X). Show that the empty set φ is the identity for the operation ∗ and all the elements A of P(X) are invertible with A^-1 = A. (Hint : (A–φ) ∪ (φ–A) = A and (A–A) ∪ (A–A) = A ∗ A = φ).

Answer

It is given that *: P(X) × P(X) → P(X) is defined

as A * B = (A − B) ∪ (B − A) ∀ A, B ∈ P(X).

Let A ∈ P(X). Then, we have

A * Φ = (A − Φ) ∪ (Φ − A) = A ∪ Φ = A

Φ * A = (Φ − A) ∪ (A − Φ) = Φ ∪ A = A

∴ A * Φ = A = Φ * A for all A ∈ P(X)

Thus, Φ is the identity element for the given operation*.

Now, an element A ∈ P(X) will be invertible if there exists B ∈ P(X) such that

A * B = Φ = B * A. [As Φ is the identity element]

Now, we observed that,

A * A = ( A – A) ∪ (A – A) = Φ ∪ Φ = Φ for all A ∈ P(X).

Hence, all the elements A of P(X) are invertible with A^-1 = A.

14. Define a binary operation ∗ on the set {0, 1, 2, 3, 4, 5} as

Show that zero is the identity for this operation and each element a ≠ 0 of the set is invertible with 6–a being the inverse of a

Answer

Let X = {0, 1, 2, 3, 4, 5}.

An element e ∈ X is the identity element for the operation *, if

a ∗ e = a = e ∗ a for all a ∈ X

For a ∈ X, we have

a ∗ 0 = a + 0 = a [a ∈ X ⇒ a + 0 < 6]

0 ∗ a = 0 + a = a [a ∈ X ⇒ 0 + a < 6]

∴ a ∗ 0 = a = 0 ∗ a for all a ∈ X

Thus, 0 is the identity element for the given operation *.

An element a ∈ X is invertible if there exists b ∈ X such that a * b = 0 = b * a.

⇒ a = −b or b = 6 − a

But, X = {0, 1, 2, 3, 4, 5} and a, b ∈ X. Then, a ≠ −b.

∴ b = 6 − a is the inverse of a for all a ∈ X.

Hence, the inverse of an element a ∈X, a ≠ 0 is 6 − a i.e., a^-1 = 6 − a.

15. Let A = {– 1, 0, 1, 2}, B = {– 4, – 2, 0, 2} and f, g : A → B be functions defined by

Justify your answer. (Hint: One may note that two functions f : A → B and g : A → B such that f(a) = g (a) ∀ a ∈ A, are called equal functions).

Answer

It is given that A = {−1, 0, 1, 2}, B = {−4, −2, 0, 2}.

Also, it is given that f, g: A → B are defined by

∴ f(a) = g(a) for all a ∈ A

Hence, the functions f and g are equal.

16. Let A = {1, 2, 3}. Then number of relations containing (1, 2) and (1, 3) which are

reflexive and symmetric but not transitive is

(A) 1

(B) 2

(C) 3

(D) 4

Answer

The given set is A = {1, 2, 3}.

The smallest relation containing (1, 2) and (1, 3) which is reflexive and symmetric, but not transitive is given by:

R = {(1, 1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 1), (3, 1)}

This is because relation R is reflexive as (1, 1), (2, 2), (3, 3) ∈ R.

Relation R is symmetric since (1, 2), (2, 1) ∈ R and (1, 3), (3, 1) ∈ R.

But relation R is not transitive as (3, 1), (1, 2) ∈ R, but (3, 2) ∉ R.

Now, if we add any two pairs (3, 2) and (2, 3) (or both) to relation R, then relation R will become transitive.

Hence, the total number of desired relations is one.

The correct answer is A.

17. Let A = {1, 2, 3}. Then number of equivalence relations containing (1, 2) is

(A) 1

(B) 2

(C) 3

(D) 4

Answer

It is given that A = {1, 2, 3}.

The smallest equivalence relation containing (1, 2) is given by,

R1 = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 1)}

Now, we are left with only four pairs i.e., (2, 3), (3, 2), (1, 3), and (3, 1).

If we odd any one pair [say (2, 3)] to R1, then for symmetry we must add (3, 2).

Also, for transitivity we are required to add (1, 3) and (3, 1).

Hence, the only equivalence relation (bigger than R1) is the universal relation.

This shows that the total number of equivalence relations containing (1, 2) is two.

The correct answer is B.

Page No. 31

18. Let f : R → R be the Signum Function defined as

and g : R → R be the Greatest Integer Function given by g (x) = [x], where [x] is greatest integer less than or equal to x. Then, does fog and gof coincide in (0, 1]?

Answer

It is given that,

Also, g: R → R is defined as g(x) = [x], where [x] is the greatest integer less than or equal to x.

Now, let x (0, 1].

Then, we have

[x] = 1 if x = 1 and [x] = 0 if 0 < x < 1.

19. Number of binary operations on the set {a, b} are

(A) 10

(B) 16

(C) 20

(D ) 8

Answer

A binary operation * on {a, b} is a function from {a, b} × {a, b} → {a, b}

i.e., * is a function from {(a, a), (a, b), (b, a), (b, b)} → {a, b}.

Hence, the total number of binary operations on the set {a, b} is 2⁴ i.e., 16.

The correct answer is B.

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CLASS 11TH -पाठ 12 – देव |अंतरा भाग-2 हिंदी |NCERT SOLUTION| EDUGROWN

Class 11 NCERT Solutions for Hindi Antra provides you an idea of the language and helps you understand the subject better. We have explained NCERT Solutions for Class 11th Hindi Antra.

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NCERT Solutions for Class 11th -पाठ 12 - देव -अंतरा भाग-2 हिंदी

प्रश्न-अभ्यास

1. ‘हँसी की चोट’ सवैये में कवि ने किन पंच तत्त्वों का वर्णन किया है तथा वियोग में वे किस प्रकार विदा होते हैं?

उत्तर

‘हँसी की चोट’ सवैये में कवि ने पाँच तत्वों आकाश, अग्नि, वायु, भूमि तथा जल का वर्णन किया गया है। गोपी द्वारा तेज़-तेज़ साँस लेने-छोड़ने से वायु तत्व चला गया है। अत्यधिक रोने से जल तत्व आँसुओं के रूप में विदा हो गया है। तन में व्याप्त गर्मी के जाने से अग्नि तत्व समाप्त हो गया है। वियोग में कमज़ोर होने के कारण भूमि तत्व चला गया है।

2. नायिका सपने में क्यों प्रसन्न थी और वह सपना कैसे टूट गया?

उत्तर

नायिका ने सपने में देखा कि कृष्ण उसके पास आते हैं और उसे झूला-झूलने का निमंत्रण देते हैं। यह उसके लिए बहुत प्रसन्नता की बात थी। उसे सपने में ही सही कृष्ण का साथ मिला था। वह जैसे ही प्रसन्नतापूर्वक कृष्ण के साथ चलने के लिए उठती है, इस बीच उसकी नींद उचट जाती है। नींद उचटने से उसका सपना टूट जाता है और कृष्ण का साथ भी छूट जाता है।

3. ‘सपना’ कवित्त का भाव-सौंदर्य लिखिए।

उत्तर

‘सपना’ कवित्त में गोपी का श्रीकृष्ण के प्रति अगाध प्रेम और मिलन की इच्छा का भाव व्यक्त हुआ है। सपने में नायिका कृष्ण का साथ पाती है। वह जैसे ही इस साथ को और आगे तक ले जाना चाहती है नींद खुलने के कारण छूट जाता है। सपना टूटने से कृष्ण का साथ छूट जाता है और वह दुखी हो जाती है।अनुप्रास तथा पुनरुक्ति प्रकाश अलंकार के प्रयोग को देखकर ‘सपना’ कवित्त में कवि के शिल्प सौंदर्य की अद्भुत क्षमता का पता चलता है। इसने कवित्त के भाव सौंदर्य को निखारने में सोने पर सुहागा जैसा काम किया है।

4. ‘दरबार’ सवैये में किस प्रकार के वातावरण का वर्णन किया गया है?

उत्तर

‘दरबार’ सवैये को पढ़कर ही पता चलता है कि इसमें दरबार के विषय में कहा गया है। उस समय दरबार में कला की कमी थी। भोग तथा विलास दरबार की पहचान बनती जा रही थी। कर्म का अभाव दरबारियों में था।

5. दरबार में गुणग्राहकता और कला की परख को किस प्रकार अनदेखा किया जाता है?

उत्तर

दरबार में गुणग्राहकता और कला की परख को चाटुकारों की बातें सुनकर अनदेखा किया गया है। यही कारण है कि वहाँ पर कला को अनदेखा किया जाता है। कला की परख करना, तो उन्हें आता ही नहीं है। चाटुकारों द्वारा की गई चापलूसी से भरी कविताओं को मान मिलता है। राजा तथा दरबारी भोग-विलास के कारण अंधे बन गए हैं। ऐसे वातावरण में कला का कोई महत्व नहीं होता है।

6. भाव स्पष्ट कीजिए-

(क) हेरि हियो जु लियो हरि जू हरि।

उत्तर

नायक ने जब से नायिका को हँसकर देखा है तब से नायिका को ऐसा लगता है जैसे उस नायक ने हँसकर देखने मात्र से ही उस का हृदय चुरा लिया है। वह नायक से मिलने के लिए व्याकुल रहने लगती है और निरंतर उससे नहीं मिल पाने की वियोगाग्नि में जलती रहती है।

(ख) सोए गए भाग मेरे जानि वा जगन में।

उत्तर

गोपी कृष्ण से मिलन का सपना देख रही थी। कृष्ण ने उसे अपने साथ झूला झूलने का निमंत्रण दिया था, वह इससे प्रसन्न थी। कृष्ण के साथ जाने के लिए वह उठने ही वाली थी कि उसकी नींद टूट गई। इसलिए वह कहती है कि उसका जागना उसके भाग्य को सुला गया यानी उसके नींद से जागने के कारण कृष्ण का साथ छूट गया। यह जागना उसके लिए दुर्भाग्य के समान है।

(ग) वेई छाई बूंदें मेरे आँसु है दुगन में।।

उत्तर

श्रीकृष्ण ने गोपी को उसके सपने में जब झूले पर झूलने का आग्रह किया था तब बाहर रिमझिम बारिश की झड़ी लगी हुई थी। गोपी की नींद खुलते ही उसे वास्तविकता का पता चला कि वह तो सपना देख रही थी। न तो बाहर वर्षा हो रही थी और न ही श्रीकृष्ण वहाँ थे। इस कारण उसकी आँखों से आँसू बह निकले। गोपी को लगा कि वही वर्षा की बूंदें उसकी आँखों में आँसू की बूंदों के रूप में दिखाई देने लगी हैं।

(घ) साहिब अंध, मुसाहिब मूक, सभा बहिरी।

उत्तर

देव दरबारी वातावरण का वर्णन कर रहे हैं। वह कहते हैं कि दरबार का राजा अँधा हो गया है। दरबारी गूँगे तथा बहरे हो गए हैं। वे भोग-विलास में इतना लिप्त हैं कि उन्हें कुछ भी सुनाई दिखाई नहीं देता है। इसलिए वे बोलने में भी असमर्थ हैं।

7. देव ने दरबारी चाटुकारिता और दंभपूर्ण वातावरण पर किस प्रकार व्यंग्य किया है?

उत्तर

देव दरबार के दंभपूर्ण वातावरण का वर्णन करते हुए बताते हैं कि दरबार में राजा तथा लोग भोग विलास में लिप्त रहते हैं। दरबारियों के साथ-साथ राजा भी अंधा है, जो कुछ देख नहीं पा रहा है। यही कारण है कि कला तथा सौंदर्य का उन्हें ज्ञान नहीं रह गया है। अहंकार उन पर इतना हावी है कि कोई किसी की बात सुनने या मानने को राज़ी नहीं है।

8. निम्नलिखित पद्यांशों की सप्रसंग व्याख्या करिए-

(क) साँसनि ही ………तनुता करि।

उत्तर

प्रसंग- प्रस्तुत पंक्ति देव द्वारा रचित रचना ‘हँसी की चोट’ से ली गई है। इसमें एक गोपी के विरह का वर्णन है। कृष्ण की उपेक्षा पूर्ण व्यवहार उसे दुखी कर गया है।

व्याख्या- गोपी कहती है कि कृष्ण की उपेक्षित द़ृष्टि के कारण उसकी दशा बहुत खराब है। वह विरह की अग्नि में जल रही है। विरह में तेज़-तेज़ साँसें छोड़ने से वायु तत्व चला गया है। अत्यधिक रोने से जल तत्व आँसुओं के रूप में विदा हो गया है। तन में व्याप्त गर्मी के जाने से अग्नि तत्व समाप्त हो गया है और वियोग में कमज़ोर होने के कारण भूमि तत्व भी चला गया है।

(ख) झहरि ……… गगन में।

उत्तर

प्रसंग- प्रस्तुत पंक्ति देव द्वारा रचित रचना ‘सपना’ से ली गई है। इसमें वर्षा ऋतु का वर्णन है। आकाश में बादल छाए हैं और बूँदे बरस रही हैं।

व्याख्या- कवि कहता है कि वर्षा ऋतु के समय बारिश की बूँदे झर रही हैं। आकाश में काली घटाएँ छा गई हैं।

(ग) साहिब अंधा ……… बाच्यो।

उत्तर

प्रसंग- प्रस्तुत पंक्ति देव द्वारा रचित रचना ‘दरबार’ से ली गई है। इसमें कवि राज दरबार में स्थित राजा और सभासदों के व्यवहार का वर्णन करता है।

व्याख्या- देव दरबार के दंभपूर्ण वातावरण का वर्णन करते हुए बताते हैं कि दरबार में राजा तथा लोग भोग-विलास में लिप्त रहते हैं। दरबारियों के साथ-साथ राजा भी अंधा है, जो कुछ देख नहीं पा रहा है। यही कारण है कि कला तथा सौंदर्य का उन्हें ज्ञान नहीं रह गया है। दरबारियों पर अहंकार इतना हावी है कि कोई किसी की बात सुनने या मानने को राज़ी नहीं है। भोग-विलास के कारण वे काम नहीं रह गए हैं|

9. देव के अलंकार प्रयोग और भाषा प्रयोग के कुछ उदाहरण पठित पदों से लिखिए।

उत्तर

‘हेरि हियो जु लियो हरि जू हरि’ में अनुप्रास और यमक अलंकार है।
‘झहरि-झहरि’, ‘घहरि-घहरि’ आदि में पुनरुक्ति प्रकाश अलंकार है।
घहरि-घहरि घटा घेरी में अनुप्रास अलंकार का प्रयोग है।
‘सोए गए भाग मेरे जानि व जगन में’ विरोधाभास अलंकार का सुंदर उदाहरण है।
‘रंग रीझ को माच्यो’ में अनुप्रास अलंकार है।
‘फूली न समानी’ ‘सोए गए भाग’ और मुहावरों का सटीक प्रयोग किया गया है।

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CLASS 11TH – पाठ 11 – सूरदास |अंतरा भाग-2 हिंदी |NCERT SOLUTION| EDUGROWN

Class 11 NCERT Solutions for Hindi Antra provides you an idea of the language and helps you understand the subject better. We have explained NCERT Solutions for Class 11th Hindi Antra.

NCERT Solutions for Class 11th - पाठ 11 - सूरदास -अंतरा भाग-2 हिंदी

प्रश्न-अभ्यास

1. ‘खेलन में को काको गुसैयाँ’ पद में कृष्ण और सुदामा के बीच किस बात पर तकरार हुई?

उत्तर

कृष्ण और सुदामा के खेल-खेल में रूठने और फिर खुद मान जाने के स्वाभाविक प्रसंग का वर्णन किया गया है। श्रीकृष्ण खेल में हार गए थे और श्रीदामा जीत गए थे, पर श्रीकृष्ण अपनी हार मानने को तैयार नहीं थे। खेल रुक गया। श्रीकृष्ण अभी और खेलना चाहते थे, इसलिए उन्होंने नंद बाबा की दुहाई देते हुए अपनी हार मान ली।

2. खेल में रूठनेवाले साथी के साथ सब क्यों नहीं खेलना चाहते?

उत्तर

खेल में रूठनेवाले साथी से सभी परेशान हो जाते हैं। खेल में सभी बराबर होते हैं। अतः जो हारता है, उसे दूसरों को बारी देनी होती है। जो अपनी बारी नहीं देता है और रूठा रहता है, उसे कोई पसंद नहीं करता है। सभी खेलना चाहते हैं। अतः ऐसे साथी से सभी दूर रहते हैं।

3. खेल में कृष्ण के रूठने पर उनके साथियों ने उन्हें डाँटते हुए क्या-क्या तर्क दिए?

उत्तर

खेल में कृष्ण के रूठने पर उनके साथियों ने डाँटते हुए ये तर्क दिए-
• तुम्हारी हार हुई है और तुम नाराज़ हो रहे हो। यह गलत है।
• तुम्हारी और हमारी जाति सबकी समान है। खेल में सभी समान होते हैं।
• तुम हमारे पालक नहीं हो। इसलिए तुम्हें हमें यह अकड़ नहीं दिखानी चाहिए।
• तुम यदि खेलते समय बेईमानी करोगे, तो कोई तुम्हारे साथ नहीं खेलेगा।

4. कृष्ण ने नंद बाबा की दुहाई देकर दाँव क्यों दिया?

उत्तर

कृष्ण ने नंद बाबी की दुहाई देकर यह निश्चित किया कि वह अपनी बारी देंगे और सबको हारकर ही रहेंगे। नंद उनके पिता है। इसलिए पिता का नाम लेकर वह झूठ नहीं बोलेंगे और सब उनकी बात मान जाएँगे। इसलिए उन्होंने नंद बाबा की दुहाई दी।

5. इस पद से बाल-मनोविज्ञान पर क्या प्रकाश पड़ता है?

उत्तर

इस पद से बाल-मनोविज्ञान पर प्रकाश पड़ता है कि बच्चे हमेशा जीतना चाहते हैं| उनके अनुसार हमेशा जीत जरूरी होती है| वे हर बात का सूक्ष्म अध्ययन करते हैं। वह ऊँच-नीच, बड़ा-छोटा, अच्छा-बुरा सब समझते है। हालांकि उनके बीच के मनमुटाव क्षणिक होते हैं| थोड़ी देर में वह फिर एक हो जाते हैं।

6. ‘गिरिधर नार नवावति’ से सखी का क्या आशय है?

उत्तर

ऐसा कहकर गोपियाँ कृष्ण पर व्यंग्य कसती हैं। वे कहती हैं कि कृष्ण प्रेम के वशीभूत होकर एक साधारण बाँसुरी को बजाते समय अपनी गर्दन झुका देते हैं। चूँकि गोपियाँ चूंकि बाँसुरी से सौत के समान ईर्ष्या रखती हैं। इसलिए वे बाँसुरी को औरत के रूप में देखते हुए उन पर व्यंग्य कसती हैं। वे नहीं चाहती कि कृष्ण बाँसुरी को इस प्रकार अपने होटों से लगाए।

7. कृष्ण के अधरों की तुलना सेज से क्यों की गई है?

उत्तर

कृष्ण के अधरों की तुलना निम्नलिखित कारणों से की गई हैं।-
• कृष्ण के अधर सेज के समान कोमल हैं।
• जिस प्रकार सेज सोने के काम आती है, वैसे ही कृष्ण बाँसुरी को बजाने के लिए अपने अधर रूपी सेज में रखते हैं। ऐसा लगता है मानो बाँसुरी सो रही है।

8. पठित पदों के आधार पर सूरदास के काव्य की विशेषताएँ बताइए।

उत्तर

सूरदास श्रीकृष्ण भक्त हैं जिन्होनें अपनी पदों में श्रीकृष्ण के प्रति अपनी अगाध भक्तिभावना को प्रकट किया है। उन्होंने पहले पद में बाल-लीलाओं का सुंदर चित्रण किया है। बालकों के बीच अक्सर होते मनमुटाव और फिर कुछ देर में सुलह का बड़ा ही मनोहारी चित्रण किया है| इससे पता लगता है की सूरदास बाल मनोविज्ञान को अच्छी तरह से समझते हैं| दूसरे पद में उन्होंने स्त्रियों की मनोदशा को बहुत अच्छी तरह से दिखाया है| किस तरह उनका कोमल हृदय अपने प्रिय से मिलने को तरसता है इसलिए वे बाँसुरी को भला-बुरा कहती हैं क्योंकि वह श्रीकृष्ण और उनके बीच की एक बाधा बन रही थी| वात्सल्य और श्रृंगार रसों का पूर्ण रूप से प्रयोग किया है| पदों में उत्प्रेक्षा, उपमा तथा अनुप्रास अलंकार का सुंदर चित्रण है। ब्रजभाषा का प्रयोग हुआ है। पदों में गेयता का गुण विद्यमान है।

9. निम्नलिखित पद्यांशों की संदर्भ सहित व्याख्या कीजिए-

(क) जाति-पाँति…”तुम्हारै गैयाँ।

उत्तर

प्रसंग- प्रस्तुत पंक्ति सूरदास द्वारा लिखित ग्रंथ सूरसागर से ली गई हैं। इस पंक्ति में कृष्ण द्वारा बारी न दिए जाने पर ग्वाले कृष्ण को नाना प्रकार से समझाते हुए अपनी बारी देने के लिए विवश करते हैं।

व्याख्या- ‘कृष्ण’ गोपियों से हारने पर नाराज़ होकर बैठ जाते हैं। उनके मित्र उन्हें उदाहरण देकर समझाते हैं। वे कहते हैं कि तुम जाति-पाति में हमसे बड़े नहीं हो, तुम हमारा पालन-पोषण भी नहीं करते हो। अर्थात तुम हमारे समान ही हो। इसके अतिरिक्त यदि तुम्हारे पास हमसे अधिक गाएँ हैं और तुम इस अधिकार से हम पर अपनी चला रहे हो, तो यह उचित नहीं कहा जाएगा। अर्थात खेल में सभी समान होते हैं। जाति, धन आदि के कारण किसी को खेल में विशेष अधिकार नहीं मिलता है। खेलभावना को इन सब बातों से अलग रखकर खेलना चाहिए।

(ख) सुनि री”.”नवावति।

उत्तर

प्रसंग- प्रस्तुत पंक्ति सूरदास द्वारा लिखित ग्रंथ सूरसागर से ली गई हैं। इस पंक्ति में गोपियों की जलन का पता चलता है। वह कृष्ण द्वारा बजाई जाने वाली बाँसुरी से सौत की सी ईर्ष्या रखती हैं।

व्याख्या- एक गोपी अन्य गोपी से कहती है कि हे सखी! सुन यह बाँसुरी तो श्रीकृष्ण से अत्यंत अपमानजनक व्यवहार करती है, फिर भी वह उन्हें अच्छी लगती है। यह नंदलाल को अनेक भाँति से नचाती है। उन्हें एक ही पाँव पर खड़ा करके रखती है और अपना बहुत अधिक अधिकार जताती है। कृष्ण का शरीर कोमल है ही, वह उनसे अपनी आज्ञा का पालन करवाती है और इसी कारण से उनकी कमर टेढ़ी हो जाती है| यह बाँसुरी ऐसे कृष्ण को अपना कृतज्ञ बना देती है, जो स्वयं चतुर हैं। इसने गोर्वधन पर्वत उठाने वाले कृष्ण तक को अपने सम्मुख झुक जाने पर विवश कर दिया है। असल में बाँसुरी बजाते समय के साड़ी मुद्राओं को देखकर गोपियों को लगता है कि कृष्ण हमारी कुछ नहीं सुनते हैं। जब बाँसुरी बजाने की बारी आती है, तो कृष्ण इसके कारण हमें भूल जाते हैं।

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CLASS 11TH – पाठ 10 – कबीर |अंतरा भाग-2 हिंदी |NCERT SOLUTION| EDUGROWN

Class 11 NCERT Solutions for Hindi Antra provides you an idea of the language and helps you understand the subject better. We have explained NCERT Solutions for Class 11th Hindi Antra.

NCERT Solutions for Class 11th - पाठ 10 - कबीर -अंतरा भाग-2 हिंदी

प्रश्न-अभ्यास

1. ‘अरे इन दोहुन राह न पाई’ से कबीर का क्या आशय है और वे किस राह की बात कर रहे हैं?

उत्तर

कबीर ने इस पंक्ति में कहा है कि हिन्दू और मुसलमान धार्मिक आडंबरों में उलझे हुए हैं| इन्हें सच्ची भक्ति का अर्थ नहीं मालूम है। धार्मिक आंडबरों को धर्म मानकर चलते हैं। कबीर के अनुसार ये दोनों भटके हुए हैं।

2. इस देश में अनेक धर्म, जाति, मजहब और संप्रदाय के लोग रहते थे किंतु कबीर हिंदू और मुसलमान की ही बात क्यों करते हैं?

उत्तर

कबीर ने हिंदू और मुसलमान की बात इसलिए की है क्योंकि उस समय भारत में हिंदू और मुस्लिम दो धर्म सबसे ज्यादा प्रचलित थे। जैन, बौद्ध आदि धर्म हिन्दू धर्म की ही शाखाएँ हैं। इसलिए उन्होंने उस समय कबीर ने अलग-अलग करके नहीं देखा था। इन दो धर्मों के बीच ही लड़ाई होती रहती थी| उन्होंने दोनों की भक्ति विधि का खंडन करते हुए उन्हें संमार्ग पर चलने के लिए प्रेरित किया है।

3. ‘हिंदुन की हिंदुवाई देखी तुरकन की तुरकाई’ के माध्यम से कबीर क्या कहना चाहते हैं? वे उनकी किन विशेषताओं की बात करते हैं?

उत्तर

कबीर कहते हैं कि दोनों ही धर्मों में अनेक प्रकार के आडंबर प्रचलित है। दोनों स्वयं को श्रेष्ठ बताकर आपस में लड़ते हैं। हिन्दू छुआछूत में भरोसा रखते हैं और दूसरी ओर वेश्यावृत्ति में लिप्त हैं परन्तु अपवित्र नहीं होते हैं। इसलिए इनकी शुद्धता और श्रेष्ठा बेकार है। वे मुसलमानों के बारे में कहते हैं कि वे जीव हत्या करते हैं और उसे मिल-जुलकर खाते हैं और सगे-संबंधियों से विवाह करते हैं। इसलिए हिंदू मुसलमान दोनों ही एक जैसे हैं।

4. ‘कौन राह है जाई’ का प्रश्न कबीर के सामने भी था। क्या इस तरह का प्रश्न आज समाज में मौजूद है? उदाहरण सहित स्पष्ट कीजिए।

उत्तर

प्राचीनकाल से लेकर अभी तक मनुष्य इसी दुविधा में फँसा हुआ है कि वह किस राह को चुने। आज के समाज में भी यह प्रश्न सभी के सामने है। भारत जैसे देश में तो हिन्दू, मुस्लिम, सिख, ईसाई, बौद्ध, जैन इत्यादि धर्म प्रचलित हैं। सब स्वयं को अच्छा और श्रेष्ठ बताते हैं। सबकी अपनी मान्यताएँ हैं। मनुष्य इनके मध्य उलझकर रह गया है। उसे समझ ही नहीं आता है कि वह किसे अपनाए, जिससे उसे जीवन की सही राह मिले।

5. ‘बालम आवो हमारे गेह रे’ में कवि किसका आह्वान कर रहा है और क्यों?

उत्तर

प्रस्तुत पंक्ति में कबीर भगवान का आह्वान कर रहे हैं। वे अपने भगवान के दर्शन के प्यासे हैं। अपने भगवान के दर्शन पाने के लिए उन्हें अपने पास बुला रहे हैं।

6. ‘अन्न न भावै नींद न आवै’ का क्या कारण है? ऐसी स्थिति क्यों हो गई है?

उत्तर

अपने नायक के वियोग में जिस तरह नायिका को कुछ भी अच्छा नहीं लगता। वह खाना-पीना छोड़ देती है और उसे नींद भी नहीं आती। उसी तरह से कबीर की जीवात्मा को भी परमात्मा रूपी प्रियतम के वियोग में खाना-पीना अच्छा नहीं लगता। वह निरंतर उसी के चिंतन में डूबे रहते हैं, इसलिए उसे नोंद भी नहीं आती है। उसकी यह स्थिति परमात्मा रूपी प्रियतम से नहीं मिलने के कारण हो गई है।

7. ‘कामिन को है बालम प्यारा, ज्यों प्यासे को नौर रे’ से कवि का क्या आशय है? स्पष्ट कीजिए।

उत्तर

कबीर कहते हैं कि कामिनी औरत को प्रियतम (बालम) बहुत प्रिय होता है। प्यास से व्याकुल व्यक्ति को पानी बहुत प्रिय होता है। ऐसे ही भक्त को अपने भगवान प्रिय होते हैं। कबीर को भी अपने भगवान प्रिय हैं और वे उनके लिए व्याकुल हो रहे हैं।

8. कबीर निर्गुण संत परंपरा के कवि हैं और यह पद (बालम आवो हमारे गेह रे) साकार प्रेम की ओर संकेत करता है। इस संबंध में आप अपने विचार लिखिए।

उत्तर

कबीर निर्गुण संत परंपरा के कवि हैं। वे ईश्वर के मूर्ति रूप को नहीं मानते हैं परन्तु सांसारिक संबंधों को अवश्य मानते हैं। उनका प्रेम में अटूट विश्वास है। प्रेम कभी साकार या निराकार नहीं होता। बल्कि यह एक भावना है| संतों ने परमात्मा को पति और जीवात्मा को पत्नी के प्रतीक के रूप में दर्शाया है। परमात्मा रूपी पति को न मिलने से पत्नी रूपी जीवात्मा की प्रेम-भावना तड़प उठती है। इसलिए यह पद प्रतीत तो साकार प्रेम की तरह हो रहा है लेकिन सत्य यह है कि वह निर्गुण रूप ही है।

9. उदाहरण देते हुए दोनों पदों का भाव-सौंदर्य और शिल्प-सौंदर्य लिखिए।

उत्तर

प्रथम पद में कबीर ने व्यंग्य शैली को अपनाया है। विभिन्न उदाहरणों द्वारा उन्होंने हिन्दुओं तथा मुस्लमानों के धार्मिक आंडबरों पर करारा व्यंग्य किया है। दोनों के बीच की लड़ाई को भी दर्शाया है| भाषा बहुत ही सरल तथा सुबोध है। अनुप्रास अलंकार का प्रयोग है तद्भव शब्दावली का प्रयोग किया गया है और प्रतीकात्मकता विद्यमान है|

दूसरे पद में कबीर ने परमात्मा के प्रति अपने प्रेम को दर्शाया है| उन्होंने जीवात्मा को पत्नी और परमात्मा को पति के प्रतीक के रूप में बताकर उनसे मिलने की तड़प को दिखाया है| यहाँ पर प्रियतम और प्रिया के साकार प्रेम को माध्यम बनाया गया है। विरह उसकी साधना में बाधक के स्थान पर मार्ग बनाने का कार्य करती है। इस पद की भाषा भी सरल और सधुक्कड़ी है। परमात्मा को प्रियतम और स्वयं को प्रिया दिखाने के कारण प्रतीकात्मकता का सुंदर प्रयोग हुआ है। भक्ति रस की प्रधानता है|

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