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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) = , x≠5
(c) f(x) = ,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.

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  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.

Solution:
 at x≠2

Ex 5.1 Class 12 Maths Question 7.

Solution:

Ex 5.1 Class 12 Maths Question 8.
Test the continuity of the function f (x) at x = 0

Solution:
We have;
(LHL at x=0)

Ex 5.1 Class 12 Maths Question 9.

Solution:

Ex 5.1 Class 12 Maths Question 10.

Solution:

Ex 5.1 Class 12 Maths Question 11.

Solution:

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

Ex 5.1 Class 12 Maths Question 12.

Solution:

Ex 5.1 Class 12 Maths Question 13.
Is the function defined by  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.

Solution:

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.

Solution:

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.

Solution:

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

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 =  or a = b + , 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

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,

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

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

Ex 5.1 Class 12 Maths Question 23.
Find all points of discontinuity of f, where

Solution:
At x = 0

Ex 5.1 Class 12 Maths Question 24.
Determine if f defined by  is a continuous function?
Solution:
At x = 0

Ex 5.1 Class 12 Maths Question 25.
Examine the continuity of f, where f is defined by 
Solution:

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.

Solution:
At x = 
L.H.L = 

Ex 5.1 Class 12 Maths Question 27.

Solution:

Ex 5.1 Class 12 Maths Question 28.

Solution:

Ex 5.1 Class 12 Maths Question 29.

Solution:

Ex 5.1 Class 12 Maths Question 30.
Find the values of a and b such that the function defined by

to is a continuous function.
Solution:

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 32.
Show that the function defined by f (x) = |cos x| is a continuous function.
Solution:
Let g(x) =|x|and h (x) = cos x, f(x) = goh(x) = g (h (x)) = g (cosx) = |cos x |
Now g (x) = |x| and h (x) = cos x both are continuous for all values of x ∈ R.
∴ (goh) (x) is also continuous.
Hence, f (x) = goh (x) = |cos x| is continuous for all values of 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

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


= 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
∴

Putting the value of t, 

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

Ex 5.2 Class 12 Maths Question 4.
sec(tan(√x))
Solution:
let y = sec(tan(√x))
by chain rule

Ex 5.2 Class 12 Maths Question 5.

Solution:
y =  = 
u = sin(ax+b)

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

Ex 5.2 Class 12 Maths Question 7.

Solution:

Ex 5.2 Class 12 Maths Question 8.
cos(√x) = y(say)
Solution:
cos(√x) = y(say)

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

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

Find  in the following

Ex 5.3 Class 12 Maths Question 1.
2x + 3y = sinx
Solution:
2x + 3y = sinx
Differentiating w.r.t x,

=>

Ex 5.3 Class 12 Maths Question 2.
2x + 3y = siny
Solution:
2x + 3y = siny
Differentiating w.r.t x,

=>

Ex 5.3 Class 12 Maths Question 3.
ax + by² = cosy
Solution:
ax + by² = cosy
Differentiate w.r.t. x,

=>
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,

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

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

Ex 5.3 Class 12 Maths Question 8.
sin²x + cos²y = 1
Solution:
Given that
sin²x + cos²y = 1
Differentiating both sides, we get

Ex 5.3 Class 12 Maths Question 9.

Solution:

put x = tanθ

Ex 5.3 Class 12 Maths Question 10.

Solution:

put x = tanθ

Ex 5.3 Class 12 Maths Question 11.

Solution:

put x = tanθ

Ex 5.3 Class 12 Maths Question 12.

Solution:

put x = tanθ
we get

Ex 5.3 Class 12 Maths Question 13.

Solution:

put x = tanθ
we get

Ex 5.3 Class 12 Maths Question 14.

Solution:

put x = tanθ
we get

Ex 5.3 Class 12 Maths Question 15.

Solution:

put x = tanθ
we get

Differentiate the following w.r.t.x:

Ex 5.4 Class 12 Maths Question 1.

Solution:

Ex 5.4 Class 12 Maths Question 2.

Solution:


x=sint

Ex 5.4 Class 12 Maths Question 3.

Solution:

Ex 5.4 Class 12 Maths Question 4.

Solution:

Ex 5.4 Class 12 Maths Question 5.

Solution:

Ex 5.4 Class 12 Maths Question 6.
…
Solution:

…

Ex 5.4 Class 12 Maths Question 7.

Solution:
y = 

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

Ex 5.4 Class 12 Maths Question 9.

Solution:
let 

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

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

Ex 5.5 Class 12 Maths Question 2.

Solution:

taking log on both sides
log y = log 

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,
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

Ex 5.5 Class 12 Maths Question 6.

Solution:
let 
let 

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)

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

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

Ex 5.5 Class 12 Maths Question 10.

Solution:

y = u + v

Ex 5.5 Class 12 Maths Question 11.

Solution:

Let u = (x cosx)^x
logu = x log(x cosx)

Find  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,

Now u = x

Ex 5.5 Class 12 Maths Question 13.
y^x = x^y
Solution:
y = x
x logy = y logx

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)

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

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)

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

Ex 5.5 Class 12 Maths Question 18.
If u, v and w are functions of w then show that

in two ways-first by repeated application of product rule, second by logarithmic differentiation.
Solution:
Let y = u.v.w
=> y = u. (vw)

If x and y are connected parametrically by the equations given in Questions 1 to 10, without eliminating the parameter. Find .

Ex 5.6 Class 12 Maths Question 1.
x = 2at², y = at⁴
Solution:

Ex 5.6 Class 12 Maths Question 2.
x = a cosθ,y = b cosθ
Solution:

Ex 5.6 Class 12 Maths Question 3.
x = sin t, y = cos 2t
Solution:

Ex 5.6 Class 12 Maths Question 4.

Solution:

Ex 5.6 Class 12 Maths Question 5.
x = cos θ – cos 2θ, y = sin θ – sin 2θ
Solution:

Ex 5.6 Class 12 Maths Question 6.
x = a(θ – sinθ), y = a(1 + cosθ)
Solution:

Ex 5.6 Class 12 Maths Question 7.

Solution:

Ex 5.6 Class 12 Maths Question 8.

Solution:

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

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)

Ex 5.6 Class 12 Maths Question 11.
If  show that 
Solution:
Given that

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:

Ex 5.7 Class 12 Maths Question 2.
x²⁰ = y(say)
Solution:

Ex 5.7 Class 12 Maths Question 3.
x.cos x = y(say)
Solution:

Ex 5.7 Class 12 Maths Question 4.
log x = y (say)
Solution:

Ex 5.7 Class 12 Maths Question 5.
x³ log x = y (say)
Solution:
x³ log x = y

Ex 5.7 Class 12 Maths Question 6.
e^x sin5x = y
Solution:
e^x sin5x = y

Ex 5.7 Class 12 Maths Question 7.
e^6x cos3x = y
Solution:
e^6x cos3x = y

Ex 5.7 Class 12 Maths Question 8.
tan^-1 x = y
Solution:

Ex 5.7 Class 12 Maths Question 9.
log(logx) = y
Solution:
log(logx) = y

Ex 5.7 Class 12 Maths Question 10.
sin(log x) = y
Solution:
sin(log x) = y

Ex 5.7 Class 12 Maths Question 11.
If y = 5 cosx – 3 sin x, prove that 
Solution:



Hence proved

Ex 5.7 Class 12 Maths Question 12.
If y = cos^-1 x, Find  in terms of y alone.
Solution:

Ex 5.7 Class 12 Maths Question 13.
If y = 3 cos (log x) + 4 sin (log x), show that

Solution:
Given that
y = 3 cos (log x) + 4 sin (log x)

Ex 5.7 Class 12 Maths Question 14.

Solution:
Given that

Ex 5.7 Class 12 Maths Question 15.
If y = 500e^7x + 600e^-7x, show that .
Solution:
we have
y = 500e^7x + 600e^-7x

Ex 5.7 Class 12 Maths Question 16.
If e^y(x+1) = 1,show that 
Solution:

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

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

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

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

Ex 4.1 Class 12 Maths Question 1.
Evaluate the following determinant:

Solution:

= 2x(-1)-(-5)x(4)
=-2+20
=18

Ex 4.1 Class 12 Maths Question 2.
(i) 
(ii) 
Solution:
(i) 
= cosθ cosθ – (sinθ)(-sinθ)
= cos²θ + sin²θ
= 1

Ex 4.1 Class 12 Maths Question 3.
If  then show that |2A|=|4A|
Solution:

=> 
L.H.S = |2A|
= 
= – 24

Ex 4.1 Class 12 Maths Question 4.
 , then show that |3A| = 27|A|
Solution:
3A = 
= 

Ex 4.1 Class 12 Maths Question 5.
Evaluate the following determinant:
(i) 
(ii) 
(iii) 
(iv) 
Solution:
(i) 
>

Ex 4.1 Class 12 Maths Question 6.
If , find |A|
Solution:
|A| = 
= 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) 
(ii)
Solution:
(i) 
=> 2 – 20 = 2x² – 24
=> x² = 3
=> x = ±√3
(ii)
or
2 × 5 – 4 × 3 = 5 × x – 2x × 3
=>x = 2

Ex 4.1 Class 12 Maths Question 8.
If , then x is equal to
(a) 6
(b) +6
(c) -6
(d) 0
Solution:
(b) 
=> x² – 36 = 36 – 36
=> x² = 36
=> x = ± 6

Ex 4.2 Class 12 Maths Question 1.

Solution:
L.H.S = 
(C1 = C3 and C2 = C3)
= 0 + 0
= 0
= R.H.S

Ex 4.2 Class 12 Maths Question 2.

Solution:
L.H.S = 

Ex 4.2 Class 12 Maths Question 3.

Solution:

Ex 4.2 Class 12 Maths Question 4.

Solution:
L.H.S = 

Ex 4.2 Class 12 Maths Question 5.

Solution:
L.H.S = ∆ = 

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

Ex 4.2 Class 12 Maths Question 6.

Solution:
L.H.S = ∆ =  …(i)

Ex 4.2 Class 12 Maths Question 7.

Solution:
L.H.S = 

Ex 4.2 Class 12 Maths Question 8.
(a) 
(b) 
Solution:
(a) L.H.S = 

Ex 4.2 Class 12 Maths Question 9.

Solution:
Let ∆ = 
Applying R1–>R1 – R2, R2–>R2 – R3

Ex 4.2 Class 12 Maths Question 10.
(a) 
(b) 
Solution:
(a) L.H.S = 

Ex 4.2 Class 12 Maths Question 11.
(a) 
(b) 
Solution:
(a) L.H.S = 
= 

Ex 4.2 Class 12 Maths Question 12.

Solution:
L.H.S = 

Ex 4.2 Class 12 Maths Question 13.

Solution:
L.H.S = 

Ex 4.2 Class 12 Maths Question 14.

Solution:
Let ∆ = 

This may be expressed as the sum of 8 determinants

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

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 = 
=  [1(0-3)+1(18-0)]
= 7.5 sq units

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)

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)

=  [-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)

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

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 ∆ = 
=  [50 – 10k] = 25 – 5k
∴ 25-5k = 35 or 25-5k = -35
-5k = 10 or 5k = 60
=> k = -2 or k = 12

Ex 4.4 Class 12 Maths Question 1.
Write the minors and cofactors of the elements of following determinants:
(i) 
(ii) 
Solution:
(i) Let A = 
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) 
(ii) 
Solution:
(i) Minors M₁₁ =  = 1

Ex 4.4 Class 12 Maths Question 3.
Using cofactors of elements of second row, evaluate

Solution:
Given

Ex 4.4 Class 12 Maths Question 4.
Using Cofactors of elements of third column, evaluate

Solution:
Elements of third column are yz, zx, xy

Ex 4.4 Class 12 Maths Question 5.
If  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.

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

Ex 4.5 Class 12 Maths Question 1.

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 = 
= 

Ex 4.5 Class 12 Maths Question 2.

Solution:

Similarly,

Verify A (adjA) = (adjA) •A = |A| I in Qs. 3 and 4.
Ex 4.5 Class 12 Maths Question 3.

Solution:
|A| = 24

Ex 4.5 Class 12 Maths Question 4.

Solution:
A₁₁ = 0, A₁₂ = – 11, A₁₃ = 0,
A₂₁ = – 3, A₂₂ = 1, A₂₃ = 1, A₃₁ = – 2
A₃₂ = 8, A₃₃ = – 3

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.

Solution:

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

Ex 4.5 Class 12 Maths Question 6.

Solution:

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

Ex 4.5 Class 12 Maths Question 7.

Solution:
|A| = 10

Ex 4.5 Class 12 Maths Question 8.

Solution:

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

Ex 4.5 Class 12 Maths Question 9.

Solution:
|A| = 
= 2(-1-0)-1(4-0)+3(8-3)
So, A is non-singular matrix and therefore, it is invertible.

Ex 4.5 Class 12 Maths Question 10.

Solution:
|A| = 
= 1(8-6)+1(0+9)+2(0-6)
= 2+9-12
= -1≠0
∴A is invertible and

Ex 4.5 Class 12 Maths Question 11.

Solution:
A = 
adj A = 
First find |A| = -cos²α-sin²α
=-1≠0

Ex 4.5 Class 12 Maths Question 12.
Let , verify that (AB)^-1 = B^-1A^-1
Solution:
Here |A| = 
= 15-14
= 1≠0

Ex 4.5 Class 12 Maths Question 13.
If  show that A² – 5A + 7I = 0,hence find A^-1
Solution:
A = 
A² = 

Ex 4.5 Class 12 Maths Question 14.
For the matrix A =  find file numbers a and b such that A²+aA+bI²=0. Hence, find A^-1.
Solution:
A = 
A²+aA+bI²=0

Ex 4.5 Class 12 Maths Question 15.
For the matrix  Show that A³-6A²+5A+11I₃=0.Hence find A^-1
Solution:
A² = 

Ex 4.5 Class 12 Maths Question 16.
If  show that A³-6A²+9A-4I=0 and hence, find A^-1
Solution:
We have

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 = 

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) 
(c) 1
(d) 0
Solution:
|A|≠0
=> A^-1 exists => AA^-1 = I
|AA^-1| = |I| = I
=> |A||A^-1| = I

Hence option (b) is correct.

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
=> 
=> AX = B
Now |A| = = 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
=> 
=> AX = B
Now |A| = 
= 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
=> 
=> AX = B
Now |A| = 
= 6 – 6
= 0.

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 = 
Ex 4.6 Class 12 Maths Question 5.
3x – y – 2z = 2
2y – z = – 1
3x – 5y = 3
Solution:

=> AX = B

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


= 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

Ex 4.6 Class 12 Maths Question 8.
2x – y = – 2
3x + 3y = 3
Solution:
The given system of equations can be written

Ex 4.6 Class 12 Maths Question 9.
4x – 3y = 3
3x – 5y = 7
Solution:
The given system of equations can be written as

where 

Ex 4.6 Class 12 Maths Question 10.
5x + 2y = 3
3x + 2y = 5
Solution:
The given system of equations can be written as

where 

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

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

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:

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

Ex 4.6 Class 12 Maths Question 15.
If A =  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 

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

Ex 3.1 Class 12 Maths Question 1.
In the matrix 
(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 = 

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:



Solution:

Ex 3.1 Class 12 Maths Question 5.
Construct a 3 x 4 matrix , whose elements are given by:


Solution:

Ex 3.1 Class 12 Maths Question 6.
Find the values of x, y, z from the following equations:



Solution:

Clearly x = 1,y = 4,z = 3

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:

Solution:

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:

(a) 
(b) Not possible to find
(c) 
(d) 
Solution:

(a) 

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.

Ex 3.2 Class 12 Maths Question 1.
Let 
Find each of the following:
(i) A + B
(ii) A – B
(iii) 3A – C
(iv) AB
(v) BA
Solution:
Let 
(i) A + B

Ex 3.2 Class 12 Maths Question 2.
Compute the following:




Solution:

Ex 3.2 Class 12 Maths Question 3.
Compute the indicated products.
(i) 
(ii) 
(iii) 
(iv) 
(v) 
(vi) 
Solution:
(i) 
= 

Ex 3.2 Class 12 Maths Question 4.
If 
then compute (A + B) and (B – C). Also verify that A + (B – C) = (A + B) – C.
Solution:
Given

Ex 3.2 Class 12 Maths Question 5.
If 
then compute 3A – 5B.
Solution:

= 

Ex 3.2 Class 12 Maths Question 6.
Simplify:

Solution:

Ex 3.2 Class 12 Maths Question 7.
Find X and Y if


Solution:

Ex 3.2 Class 12 Maths Question 8.
Find

Solution:

We are given that

Ex 3.2 Class 12 Maths Question 9.
Find x and y, if 
Solution:

=> 
=> 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

Solution:

Ex 3.2 Class 12 Maths Question 11.
If  then find the values of x and y
Solution:

=> 

Ex 3.2 Class 12 Maths Question 12.
Given

find the values of x,y,z and w.
Solution:

=> 
=> 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) = 
then show that F(x).F(y) = F(x+y)
Solution:
F(x) = 
∴ F(y) = 

Ex 3.2 Class 12 Maths Question 14.
Show that


Solution:


L.H.S≠R.H.S

Ex 3.2 Class 12 Maths Question 15.
Find A² – 5A + 6I, if A = 
Solution:
A² – 5A + 6I = 

Ex 3.2 Class 12 Maths Question 16.
If A =  Prove that A³-6A²+7A+2I = 0
Solution:
We have
A² = A x A
= 

Ex 3.2 Class 12 Maths Question 17.
If  then find k so that A²=kA-2I
Solution:
Given

Required: To find the value of k
Now A²=kA-2I

Ex 3.2 Class 12 Maths Question 18.
If  and I is the identity matrix of order 2,show that

Solution:
L.H.S=

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.

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

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.

Ex 3.3 Class 12 Maths Question 1.
Find the transpose of each of the following matrices:
(i) 
(ii) 
(iii) 
Solution:
(i) let A = 
∴ transpose of A = A’ = 

Ex 3.3 Class 12 Maths Question 2.
If 
then verify that:
(i) (A+B)’=A’+B’
(ii) (A-B)’=A’-B’
Solution:

Ex 3.3 Class 12 Maths Question 3.
If 
then verify that:
(i) (A+B)’ = A’+B’
(ii) (A-B)’ = A’-B’
Solution:

Ex 3.3 Class 12 Maths Question 4.
If 
then find (A+2B)’
Solution:

Ex 3.3 Class 12 Maths Question 5.
For the matrices A and B, verify that (AB)’ = B’A’, where


Solution:

Ex 3.3 Class 12 Maths Question 6.
If (i)  ,the verify that A’A=I
If (ii) ,the verify that A’A=I
Solution:
(i) 

Ex 3.3 Class 12 Maths Question 7.
(i) Show that the matrix  is a symmetric matrix.
(ii) Show that the matrix  is a skew-symmetric matrix.
Solution:
(i) For a symmetric matrix a_ij = a_ji
Now,

Ex 3.3 Class 12 Maths Question 8.
For the matrix, 
(i) (A+A’) is a symmetric matrix.
(ii) (A-A’) is a skew-symmetric matrix.
Solution:

=> 

Ex 3.3 Class 12 Maths Question 9.
Find  and ,when

Solution:

Ex 3.3 Class 12 Maths Question 10.
Express the following matrices as the sum of a symmetric and a skew-symmetric matrix.
(i)
(ii)
(iii)
(iv)
Solution:
(i) let 
=> 

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  then A+A’ = I, if the
value of α is
(a) 
(b) 
(c) π
(d) 
Solution:
Now

Thus option (b) is correct.

Ex 3.4 Class 12 Maths Question 1.

Solution:
Let 
We know that
A = IA

Ex 3.4 Class 12 Maths Question 2.

Solution:
Let 
We know that
A = IA

Ex 3.4 Class 12 Maths Question 3.

Solution:
Let 
We know that
A = IA

Ex 3.4 Class 12 Maths Question 4.

Solution:
Let 
We know that
A = IA

Ex 3.4 Class 12 Maths Question 5.

Solution:
Let 
We know that
A = IA

Ex 3.4 Class 12 Maths Question 6.

Solution:
Let 
We know that
A = IA

Ex 3.4 Class 12 Maths Question 7.

Solution:
Let 
We know that
A = IA

Ex 3.4 Class 12 Maths Question 8.

Solution:
Let 
We know that
A = IA

Ex 3.4 Class 12 Maths Question 9.

Solution:
Let 
We know that
A = IA

Ex 3.4 Class 12 Maths Ex 3.4 Class 12 Maths Question 10.

Solution:
Let 

Ex 3.4 Class 12 Maths Question 11.

Solution:
Let 
We know that
A = IA

Ex 3.4 Class 12 Maths Question 12.

Solution:
Let 
We know that
A = IA

Ex 3.4 Class 12 Maths Question 13.

Solution:
Let 
We know that
A = IA

Ex 3.4 Class 12 Maths Question 14.

Solution:
Let 
We know that
A = IA

Ex 3.4 Class 12 Maths Question 15.

Solution:
Let 

Ex 3.4 Class 12 Maths Question 16.

Solution:
Let 
We know that
A = IA

Ex 3.4 Class 12 Maths Question 17.

Solution:
Let 
We know that
A = IA

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

Ex 2.1 Class 12 Maths Question 1-10.
Find the principal values of the following:
(1) 
(2) 
(3) 
(4) 
(5) 
(6) 
(7) 
(8)
(9) 
(10) 
Solution:
(1) Let  = y
∴ 
the range of principal value of sin^-1 is

Ex 2.1 Class 12 Maths Question 11-12.
Find the principal values of the following:
(11) 
(12) 
Solution:
(11) 
Now tan^-1 (1) = 
∴the range of principal value branch of

Ex 2.1 Class 12 Maths Question 13.
If sin^-1 x = y, then
(a) 0 ≤ y ≤ π
(b) 
(c) 0 < y < π
(d) 
Solution:
The range of principal value of sin is 
∴ if sin^-1 x = y then

Option (b) is correct

Ex 2.1 Class 12 Maths Question 14.
 is equal to
(a) π
(b) 
(c) 
(d) 
Solution:

∴ Principal values of sec^-1 is [0,π] – 

Option (b) is correct

Ex 2.2 Class 12 Maths Question 1.

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 

Ex 2.2 Class 12 Maths Question 2.

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.

Solution:
L.H.S = 
= 
= 
= R.H.S

Ex 2.2 Class 12 Maths Question 4.

Solution:
L.H.S =

Ex 2.2 Class 12 Maths Question 5.
Write the function in the simplest form

Solution:
Putting x = θ
∴ θ = tan^-1 x

Ex 2.2 Class 12 Maths Question 6.

Solution:
Given expression

Let x = secθ

Ex 2.2 Class 12 Maths Question 7.

Solution:

= 

Ex 2.2 Class 12 Maths Question 8.

Solution:

Dividing numerator and denominator by cos x

Ex 2.2 Class 12 Maths Question 9.

Solution:
Let x = a sinθ
=>  = sinθ

Ex 2.2 Class 12 Maths Question 10.

Solution:
Put x = a tanθ,
we get

Ex 2.2 Class 12 Maths Question 11.
Find the value of the following

Solution:

= 

Ex 2.2 Class 12 Maths Question 12.
cot[tan^-1 a + cot^-1 a]
Solution:
Given
cot[tan^-1 a + cot^-1 a]
= 

Ex 2.2 Class 12 Maths Question 13.

Solution:
Putting x = tanθ
=> tan^-1 x = θ

Ex 2.2 Class 12 Maths Question 14.
If  then find the value of x
Solution:

Ex 2.2 Class 12 Maths Question 15.
If  then find the value of x
Solution:
L.H.S

Ex 2.2 Class 12 Maths Question 16.

Solution:

= 
= 

Ex 2.2 Class 12 Maths Question 17.

Solution:

= 
= 

Ex 2.2 Class 12 Maths Question 18.

Solution:

Let 
sinθ = 

Ex 2.2 Class 12 Maths Question 19.
 is equal to
(a) 
(b) 
(c) 
(d) 
Solution:

= 

Ex 2.2 Class 12 Maths Question 20.
 is equal to
(a) 
(b) 
(c) 
(d) 1
Solution:

Ex 2.2 Class 12 Maths Question 21.
 is equal to
(a) π
(b) 
(c) 0
(d) 2√3
Solution:

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.,  is not less than .
(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.

Ex 1.2 Class 12 Maths Question 1.
Show that the function f: R —> R defined by f (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) =  if f(x₁) = f(x₂)

=> x₁ = x₂
Each x ∈ R has a unique image in codomain
=> f is one-one.
(ii) For each y belonging codomain then
 or  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)
=>  => 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:
∴ 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

∴ 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

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) =  ,if n is odd
f (n) =  ,if n is even
for all n∈N
State whether the function f is bijective. Justify your answer.
Solution:
f: N —> N, defined by

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) = 
Solution:
Is f one-one and onto? Justify your answer.
f: A –> B where A = R – {3}, B = R – {1} f is defined by

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
=>
f is onto
f is one-one and onto. Option (a) is correct.

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) = .
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) = 
(a) gof(x) = g(f(x)) = g(8x³) =  = 2x
(b) fog (x) = f(g (x))=f() = 8.()³ = 8x

Ex 1.3 Class 12 Maths Question 4.
If , show that fof (x) = x, for all . What is the inverse of f?
Solution:

(a) fof (x) = f(f(x)) = 

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) =  is one-one. Find the inverse of the function f: [-1,1] –> Range f.
Hint – For y ∈ Range f, y = f (x) =  for some x in [- 1,1], i.e., x = 
Solution:
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
∴ 
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) = 

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

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

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) = , then fof (x) is
(a) 
(b) x³
(c) x
(d) (3 – x³)
Solution:
f: R-> R defined by f(x) = 
fof (x) = f[f(x)] = 
= 
= 
= 
= x

Ex 1.3 Class 12 Maths Question 14.
Let f:  be a function defined as f (x) =  . The inverse of f is the map g: Range f–>  given by
(a) 
(b) 
(c) 
(d) 
Solution:
(b)

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 = 
(iv) On Z⁺, define a * b = 2ab
(v) On Z⁺, define a * b = ab
(vi) On R- {-1}, define a * b = 
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 = 
∴ a * b = b * a
∴ Operation binary defined is commutative.
be abc
(b) a * (b * c) = a *  =  and
(a * b) * c =  * c 
=> (a * b) * c =  * c 
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.

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)

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 *.

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 = 
(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 = 
∴ a * e =  ≠ 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)