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

Class 12: Maths Chapter 5 solutions. Complete Class 12 Maths Chapter 5 Notes.

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

Class 12: Maths Chapter 5 solutions. Complete Class 12 Maths Chapter 5 Notes.

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 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

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

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

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

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 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 continuous, nor differentiable.
(iii) f(x) = x²-1 for x ∈ [1,2], It is a polynomial. Therefore it is continuous in the interval [1,2] and differentiable in the interval (1,2)
f (x) = 2x, f(1) = 1 – 1 = 0 ,
f(2) = 4 – 1 = 3, f'(c) = 2c

NCERT Solutions for 12th Class Maths: Chapter 5: Download PDF

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

Download PDF: NCERT Solutions for 12th Class Maths: Chapter 5-Continuity and Differentiability PDF

Chapterwise NCERT Solutions for Class 12 Maths :

• Chapter 1 – Relations and Functions
• Chapter 2 – Inverse Trigonometric Functions.
• Chapter 3 – Matrices
• Chapter 4 – Determinants.
• Chapter 5 – Continuity and Differentiability.0.0
• Chapter 6 – Application of Derivatives.
• Chapter 7 – Integrals.
• Chapter 8 – Application of Integrals.
• Chapter 9: Differential Equations
• Chapter 10: Vector Algebra
• Chapter 11: Three Dimensional Geometry
• Chapter 12: Linear Programming
• Chapter 13: Probability

About NCERT

The National Council of Educational Research and Training is an autonomous organization of the Government of India which was established in 1961 as a literary, scientific, and charitable Society under the Societies Registration Act. Its headquarters are located at Sri Aurbindo Marg in New Delhi. Visit the Official NCERT website to learn more.