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.
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. limx–>0 f (x) = limx–>0 (5x – 3) = – 3 and
f(0) = – 3
∴f is continuous at x = 0
(ii) At x = – 3, limx–>3 f(x)= limx–>-3 (5x – 3) = – 18
and f( – 3) = – 18
∴ f is continuous at x = – 3
(iii) At x = 5, limx–>5 f(x) = limx–>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:
limx–>3 f(x) = limx–>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) =
(c) f(x) = 
(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 
Solution:
(i) At x = 0
limx–>0- f(x) = limx–>0- x = 0 and
limx–>0+ f(x) = limx–>0+ x = 0 => f(0) = 0
∴ f is continuous at x = 0
(ii) At x = 1
limx–>1- f(x) = limx–>1- (x) = 1 and
limx–>1+ f(x) = limx–>1+(x) = 5
∴ limx–>1- f(x) ≠ limx–>1+ f(x)
∴ f is discontinuous at x = 1
(iii) At x = 2
limx–>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:
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. limx–>2- (x³ – 3) = 8 – 3 = 5
R.H.L. = limx–>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 
Solution:
At x = 1,L.H.L.= limx–>1- f(x) = limx–>1- (x + 5) = 6,
R.HL. = limx–>1+ f(x) = limx–>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, limx–>c (x + 5) = c + 5 = f(c)
At x = c > 1, limx–>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. = limx–>1+ f(x) = 4 => f is discontinuous at
x = 1
At x = 3, L.H.L. = limx–>3- f(x)=4,
R.H.L. = limx–>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. = limx–>0- 2x = 0 ,
R.H.L. = limx–>0+ (0)= 0 , f(0) = 0
=> f is continuous at x = 0
At x = 1, L.H.L. = limx–>1- (0) = 0,
R.H.L. = limx–>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. = limx–>1- f(x) = – 2, f(-1) = – 2,
R.H.L. = limx–>1+ f(x) = – 2
=> f is continuous at x = – 1
At x= 1, L.H.L. = limx–>1- f(x) = 2,f(1) = 2
∴ f is continuous at x = 1,
R.H.L. = limx–>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. = limx–>3- (ax+1) = 3a+1 ,
f(3) = 3a + 1, R.H.L. = limx–>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 = 
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. = limx–>0- λ (x² – 2x) = 0 ,
R.H.L. = limx–>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, limx–>1 f(x) = limx–>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, limx–>c- (x – [x]) = limh–>0 [(c – h) – (c – 1)]
= limh–>0 (c – h – (c – 1)) = 1[∵ [c – h] = c – 1]
R.H.L. = limx–>c+ (x – [x])= limh–>0 (c + h – [c + h])
= limh–>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 
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,
![\frac { dy }{ dx } =-[sin(sinx)]cosx](https://s0.wp.com/latex.php?latex=%5Cfrac+%7B+dy+%7D%7B+dx+%7D+%3D-%5Bsin%28sinx%29%5Dcosx&bg=ffffff&fg=000&s=2 "\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
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 
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+ex),x>0
Solution:
y = cos(log x+ex),x>0
put y = cos t,t = log x+ex
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 – 2sinx
Solution:
let y = x – 2sinx,
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 + xlogx
Solution:
let y = (log x)x + xlogx = 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.
xsinx + (sin x)cosx
Solution:
let y = xsinx + (sin x)cosx = u+v
where u = xsinx
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
Ex 5.5 Class 12 Maths Question 12.
xy + yx = 1
Solution:
xy + yx = 1
let u = xy and v = yx
∴ u + v = 1,
Now u = x
Ex 5.5 Class 12 Maths Question 13.
yx = xy
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:
![\frac { dx }{ d\theta } =a\left[ 1-cos\theta \right] \& \frac { dy }{ d\theta } =-asin\theta](https://s0.wp.com/latex.php?latex=%5Cfrac+%7B+dx+%7D%7B+d%5Ctheta+%7D+%3Da%5Cleft%5B+1-cos%5Ctheta+%5Cright%5D+%5C%26+%5Cfrac+%7B+dy+%7D%7B+d%5Ctheta+%7D+%3D-asin%5Ctheta+&bg=ffffff&fg=000&s=2 "\frac { dx }{ d\theta } =a\left[ 1-cos\theta \right] \& \frac { dy }{ d\theta } =-asin\theta")
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)
![\frac { dx }{ d\theta } =a\left[ -sin\theta +\theta .cos\theta +sin\theta \right] =a\theta cos\theta](https://s0.wp.com/latex.php?latex=%5Cfrac+%7B+dx+%7D%7B+d%5Ctheta+%7D+%3Da%5Cleft%5B+-sin%5Ctheta+%2B%5Ctheta+.cos%5Ctheta+%2Bsin%5Ctheta+%5Cright%5D+%3Da%5Ctheta+cos%5Ctheta+&bg=ffffff&fg=000&s=2 "\frac { dx }{ d\theta } =a\left[ -sin\theta +\theta .cos\theta +sin\theta \right] =a\theta cos\theta")
Ex 5.6 Class 12 Maths Question 11.
If 
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.
ex sin5x = y
Solution:
ex sin5x = y
Ex 5.7 Class 12 Maths Question 7.
e6x cos3x = y
Solution:
e6x 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
![and\quad \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =\frac { x.\left[ -sin(logx) \right] .\frac { 1 }{ x } -cos(logx).1 }{ { x }^{ 2 } }](https://s0.wp.com/latex.php?latex=and%5Cquad+%5Cfrac+%7B+%7B+d+%7D%5E%7B+2+%7Dy+%7D%7B+%7B+dx+%7D%5E%7B+2+%7D+%7D+%3D%5Cfrac+%7B+x.%5Cleft%5B+-sin%28logx%29+%5Cright%5D+.%5Cfrac+%7B+1+%7D%7B+x+%7D+-cos%28logx%29.1+%7D%7B+%7B+x+%7D%5E%7B+2+%7D+%7D+&bg=ffffff&fg=000&s=2 "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 } }](https://s0.wp.com/latex.php?latex=%3D%5Cfrac+%7B+%5Cleft%5B+sin%28logx%29%2Bcos%28logx%29+%5Cright%5D+%7D%7B+%7B+x+%7D%5E%7B+2+%7D+%7D+&bg=ffffff&fg=000&s=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
Solution:
Hence proved
Ex 5.7 Class 12 Maths Question 12.
If y = cos-1 x, Find 
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 = 500e7x + 600e-7x, show that 
Solution:
we have
y = 500e7x + 600e-7x
Ex 5.7 Class 12 Maths Question 16.
If ey(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