NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 5.1.

• Continuity and Differentiability Class 12 Ex 5.2
• Continuity and Differentiability Class 12 Ex 5.3
• Continuity and Differentiability Class 12 Ex 5.4
• Continuity and Differentiability Class 12 Ex 5.5
• Continuity and Differentiability Class 12 Ex 5.6
• Continuity and Differentiability Class 12 Ex 5.7
• Continuity and Differentiability Class 12 Ex 5.8

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Maths
Chapter	Chapter 5
Chapter Name	Continuity and Differentiability
Exercise	Ex 5.1
Number of Questions Solved	34
Category	NCERT Solutions

NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Exc  5.1

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

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

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.

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.

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

Question 6.

Solution:
at x≠2

Question 7.

Solution:

Question 8.
Test the continuity of the function f (x) at x = 0

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

Question 9.

Solution:

Question 10.

Solution:

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

Question 12.

Solution:

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

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.

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.

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.

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.

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.

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.

Question 20.
Is the function defined by f (x) = x² – sin x + 5 continuous at x = π?
Solution:
Let f(x) = x² – sinx + 5,

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

Question 22.
Discuss the continuity of the cosine, cosecant, secant and cotangent functions.
Solution:
(a) Let f(x) = cosx

Question 23.
Find all points of discontinuity of f, where

Solution:
At x = 0

Question 24.
Determine if f defined by is a continuous function?
Solution:
At x = 0

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.

Question 26.

Solution:
At x =
L.H.L =

Question 27.

Solution:

Question 28.

Solution:

Question 29.

Solution:

Question 30.
Find the values of a and b such that the function defined by

to is a continuous function.
Solution:

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.

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.

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.

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

 

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