Parents can use ISC Mathematics Class 12 Solutions Chapter 5 Continuity and Differentiability Ex 5.1 to provide additional support to their children.

## ML Aggarwal Class 12 Maths Solutions Section A Chapter 5 Continuity and Differentiability Ex 5.1

Question 1.
Examine the following functions for continuity at the inclined points:
(i) f(x) = $$\left\{\begin{array}{ccc} x^3+3 & , & x \neq 0 \\ 1 & , & x=0 \end{array}\right.$$ at x = 0 (NCERT)
(ii) f(x) = x3 + 2x2 – 1 at x = 1. (NCERT Exampler)
Solution:
(i) $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ (x3 + 3)
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ [h3 + 3] = 3
$$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ (x3 + 3)
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ [(- h)3 + 3] = 3
∴ $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ f(x) ≠ f(0)
[∵ f(0) = 1]
Thus f is continuous at x = 0.

(ii) Given f(x) = x3 + 2x2 – 1
∴ f(1) = 13 + 2 – 12 – 1
= 3 – 1 = 2
$$\underset{x \rightarrow 1}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 1}{\mathrm{Lt}}$$ (x3 + 2x2 – 1)
= 1 + 2 – 1 = 2
∴ $$\underset{x \rightarrow 1}{\mathrm{Lt}}$$ f(x) = f(1).
Thus f(x) is continuous at x = 1.

Question 2.
If f(x) = $$\frac{x^2-1}{x-1}$$ for x ≠ 1 annd f(x) = 2 when x = 1, show that the function is continuous at x = 1.
Solution:
(i) L.H.L. = $$\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}$$ f(x)
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ f (1 – h)
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ $$\frac{(1-h)^2-1}{1-h-1}$$
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ $$\frac{h^2-2 h}{-h}$$
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ $$\frac{h(h-2)}{-h}$$
= 2

R.H.L. = $$\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}$$ f(x)
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ f (1 + h)
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ $$\frac{(1+h)^2-1}{1+h-1}$$
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ $$\frac{h^2+2 h}{h}$$
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ $$\frac{h(h+2)}{h}$$
= 2
also, f(1) = 2
∴ $$\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}$$ f(x) = f(1) = 2.

Question 2(old).
Prove that the function f(x) = 5x – 3 is continuous at x = 0, at x = – 3 and at x = 5. (NCERT)
Solution:
Given, f(x) = 5x – 3

at x = 0 :
$$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ 5x – 3
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ 5 (0 – h) – 3
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ – 5h – 3 = – 3
$$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ 5x – 3
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ 5 (0 + h) – 3 = – 3
Also f(0) = 5 . 0 – 3
= – 3
Thus $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ f(x) = f(0)
∴ f is continuous at x = 0

at x = – 3
$$\underset{x \rightarrow 3^{-}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 3^{-}}{\mathrm{Lt}}$$ 5x – 3
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ 5 (- 3 – h) – 3
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ – 5h – 18
= – 18
$$\underset{x \rightarrow 3^{+}}{\mathrm{Lt}}$$ f(x) = $$\ {Lt}_{x \rightarrow-3^{+}}$$ 5x – 3
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ 5 (- 3 + h) – 3
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ (5h – 18) = 18
Also f(- 3) = 5 (- 3) – 3 = – 18
∴ $$\underset{x \rightarrow 3^{-}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 3^{+}}{\mathrm{Lt}}$$ f(x) = f(3)
Hence f(x) is continuous at x = – 3.

at x = 5
$$\underset{x \rightarrow 5^{-}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 5^{-}}{\mathrm{Lt}}$$ (5x – 3)
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ (5 (5 – h) – 3)
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ (22 – 5h)
= 22
$$\underset{x \rightarrow 5^{+}}{\mathrm{Lt}}$$] f(x) = $$\underset{x \rightarrow 5^{+}}{\mathrm{Lt}}$$ (5x – 3)
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ 5 (5 + h) – 3
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ (22 + 5h)
= 22
Also, f(5) = 5 × 5 – 3
= 22
∴ $$\underset{x \rightarrow 5^{-}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 5^{+}}{\mathrm{Lt}}$$ f(x) = f(5)
Hence f(x) is continuous at x = 5.

Question 3.
A function f is defined as f(x) = $$\left\{\begin{array}{cc} \frac{x^2-x-6}{x-3} & , x \neq 3 \\ 5 & , x=3 \end{array}\right.$$. show that f is continuous at x = 3.
Solution:
L.H.L. = $$\underset{x \rightarrow 3^{-}}{\mathrm{Lt}}$$ f(x)
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ f (3 – h)
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ $$\frac{(3-h)^2-(3-h)-6}{3-4-3}$$
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ $$\frac{h^2-5 h}{-h}$$
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ $$\frac{h(h-5)}{-h}$$
= 5
R.H.L. = $$\underset{x \rightarrow 3^{+}}{\mathrm{Lt}}$$ f(x)
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ f (3 + h)
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ $$\frac{(3+h)^2-(3+h)-6}{3+h-3}$$
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ $$\frac{h^2+5 h}{h}$$
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ $$\frac{h(h+5)}{h}$$
= + 5
∴ L.H.L. = R.H.L. and f(3) = 5.
Thus $$\underset{x \rightarrow 3^{-}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 3^{+}}{\mathrm{Lt}}$$ f(x) = 5 = f(3)
∴ f(x) is continuous at x = 3.

Question 4.
Is the function f defined by f(x) = $$\left\{\begin{array}{lll} x, & \text { if } & x \leq 1 \\ 5, & \text { if } & x>1 \end{array}\right.$$ continuous at
(i) x = 0
(ii) x = 1
(iii) x = 2 ? (NCERT)
Solution:
Given, f(x) = $$\left\{\begin{array}{lll} x, & \text { if } & x \leq 1 \\ 5, & \text { if } & x>1 \end{array}\right.$$
at x = 0;
$$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ x = 0
and $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ x = 0
∴ $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ f(x) = f(0)
∴ f is continuous at x = 0

at x = 1 ;
$$\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}$$ 5 = 5
and $$\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}$$ x = 1
∴ $$\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}$$ f(x) ≠ $$\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}$$ f(x)
∴ f is not continuous at x = 1.

at x = 2 ;
$$\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}$$ 5 = 5
and $$\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}$$ 5 = 5
∴ $$\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}$$ f(x) = 5 = f(2)
∴ f is continuous at x = 2.

Question 5.
Is the function f defined by f(x) = $$\left\{\begin{array}{ccc} 3 x+5 & , \text { if } & x \geq 2 \\ x^2 & , \text { if } & x<2 \end{array}\right.$$ continuous at x = 2 ? (NCERT Exampler)
Solution:
Here f(2) = 3 × 2 + 5
= 6 + 5
= 11
$$\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}$$ 3x + 5
= 6 + 5 = 11
$$\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}$$ x2
= 22 = 4
∴ $$\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}$$ f(x) ≠ $$\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}$$ f(x)
Thus f is not continuous at x = 2.

Question 6.
If the function f defined by f(x) = $$\left\{\begin{array}{ccc} \frac{2 x^2-3 x-2}{x-5} & , \text { if } & x \neq 2 \\ 5 & , \text { if } & x=2 \end{array}\right.$$ continuous at x = 2? (NCERT Exampler)
Solution:
Here, $$\underset{x \rightarrow 2}{\mathrm{Lt}}$$ f(x)
= $$\underset{x \rightarrow 2}{\mathrm{Lt}}$$ $$\frac{2 x^2-3 x-2}{x-2}$$
= $$\underset{x \rightarrow 2}{\mathrm{Lt}}$$ $$\frac{(x-2)(2 x+1)}{x-2}$$
= $$\underset{x \rightarrow 2}{\mathrm{Lt}}$$ (2x + 1)
= 4 + 1 = 5
Here f(2) = 5
[since f(x) = 5 at x = 2]
∴ $$\underset{x \rightarrow 2}{\mathrm{Lt}}$$ f(x) = f(2)
Thus, f is continuous at x = 2.

Question 7.
If f(x) = $$\left\{\begin{array}{cc} k x^2+5, & x \leq 1 \\ 2, & x>1 \end{array}\right.$$, find k so that f may be continuous at x = 1.
Solution:
$$\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}$$ 2 = 2
$$\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}$$ (kx2 + 5)
= k × 12 + 5
= k + 5
Now f is given to be continuous at x = 1.
∴ $$\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}$$ f(x) = f(1)
k + 5 = 2
k = – 3.

Question 7 (old).
Is the function f defined by f(x) = $$\left\{\begin{array}{ccc} \frac{x}{\sin 2 x} & \text {, when } & x \neq 0 \\ 2 & \text {, when } & x=0 \end{array}\right.$$ continuous at x = 0?
Solution:
$$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ 
= $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ $$\frac{x}{\sin 2 x}$$
= $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ $$\frac{1}{2}\left(\frac{2 x}{\sin 2 x}\right)$$
= $$\frac{1}{2}$$ × 1
= $$\frac{1}{2}$$
[∵ $$\ {Lt}_{\theta \rightarrow 0} \frac{\theta}{\sin \theta}$$]
also, f(0) = 2
∴ $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ f(x) ≠ f(0)
Thus, f(x) is discontinuous at x = 0.

Question 8.
If f(x) = $$\left\{\begin{array}{ccc} 3 x-8 & , & x \leq 5 \\ 2 k & , & x>5 \end{array}\right.$$, find k so that may be continuous at x = 5.
Solution:
$$\underset{x \rightarrow 5^{+}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 5^{+}}{\mathrm{Lt}}$$ 2k = 2k
$$\underset{x \rightarrow 5^{-}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 5^{-}}{\mathrm{Lt}}$$ (3x – 8)
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ 3 (5 – h) – 8
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ (7 – 3h)
= 7
Since f(x) is continuous at x = 5
∴ $$\underset{x \rightarrow 5^{+}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 5^{-}}{\mathrm{Lt}}$$ f(x) = f(5)
⇒ 2k = 7
⇒ k = $$\frac{7}{2}$$

Question 8 (old).
If f(x) = $$\left\{\begin{array}{cc} k x^2 & , x \leq 2 \\ 3 & , x>2 \end{array}\right.$$, find k so that f may be continuous atx = 2. (NCERT)
Solution:
l.H.L. = $$\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}$$ f(x)
= $$\underset{x \rightarrow 2}{\mathrm{Lt}}$$ kx2 = 4k
and R.H.L. = $$\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}$$ f(x)
= $$\underset{x \rightarrow 2}{\mathrm{Lt}}$$ 3 = 3
and f(2) = 4k
Since f(x) is continuous at x = 2
∴ $$\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}$$ f(x) = f(2)
⇒ 4k = 3
⇒ k = $$\frac{3}{4}$$.

Question 9.
If f(x) = $$\begin{cases}3 x-4, & 0 \leq x \leq 2 \\ 2 x+\lambda, & 2<x \leq 5\end{cases}$$
∴ L.H.L. = $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ f(x)
= $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ – x = 0
R.H.L. = $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ f(x)
= $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ x = 0
f(0) = 0
∴ $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$
f(x) = f(0)
Thus, f is continuous at x = 0.

Question 10.
(i) f(x) = $$\frac{x^2-9}{x-3}$$ is not defined at x = 3. What value should be assigned to f(3) for continuity of f(x) at x = 3?
(ii) If f(x) = $$\left\{\begin{array}{cc} \frac{(x+3)^2-36}{x-3}, & x \neq 3 \\ k, & x=3 \end{array}\right.$$, find k so that the function f may be continuous at x = – 1 (ISC 2018)
(iii) Determine the value of ‘k’for which the following function is continuous at x = 3 ; f(x) = $$\left\{\begin{array}{cc} \frac{(x+3)^2-36}{x-3}, & x \neq 3 \\ k, & x=3 \end{array}\right.$$.
(iv) If f(x) = $$\left\{\begin{array}{ccc} \frac{x^2-x-6}{x^2-2 x-3} & , & x \neq 3 \\ k & , & x=3 \end{array}\right.$$, find k so that the function f may be continuous at x = 3.
Solution:
L.H.L. = $$\underset{x \rightarrow 3^{-}}{\mathrm{Lt}}$$ f(x)
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ f (3 – h)
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ $$\frac{(3-h)^2-9}{3-h-3}$$
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ $$\frac{h^2-6 h+9-9}{-h}$$
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ $$\frac{h(h-6)}{-h}$$
= 6
R.H.L. = $$\underset{x \rightarrow 3^{+}}{\mathrm{Lt}}$$ f(x)
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ f (3 + h)
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ $$\frac{(3+h)^2-9}{3+h-3}$$
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ $$\frac{9+h^2+6 h-9}{h}$$
= $$\underset{h \rightarrow 0}{\mathrm{Lt}}$$ $$\frac{h(h+6)}{h}$$
= 0 + 6 = 6
∴ L.H.L. = R.H.L.
Thus $$\underset{x \rightarrow 3}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 3}{\mathrm{Lt}}$$ $$\frac{x^2-9}{x-3}$$ = 6
Now f(x) is continuous at x = 3 if $$\underset{x \rightarrow 3}{\mathrm{Lt}}$$ f(x) = f(3) i.e., if f(3) = 6.

(ii) Given f(x) = $$\left\{\begin{array}{cc} \frac{(x+3)^2-36}{x-3}, & x \neq 3 \\ k, & x=3 \end{array}\right.$$
 f(x) = $$\underset{x \rightarrow-1}{\mathrm{Lt}$$ $$\frac{x^2-2 x-3}{x+1}$$
= $$\underset{x \rightarrow-1}{\mathrm{Lt}}$$ $$\frac{(x+1)(x-3)}{x+1}$$
= $$\underset{x \rightarrow-1}{\mathrm{Lt}}$$ x – 3
= – 1 – 3 = – 4
also f (- 1) = k
Since f may be continuous at x = – 1
∴ $$\underset{x \rightarrow-1}{\mathrm{Lt}}$$ f(x) = f(- 1)
⇒ – 4 = k

(iii) $$\underset{x \rightarrow 3}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 3}{\mathrm{Lt}}$$ $$\frac{(x+3)^2-36}{x-3}$$
= $$\underset{x \rightarrow 3}{\mathrm{Lt}}$$ $$\frac{(x+3)^2-6^2}{x-3}$$
= $$\underset{x \rightarrow 3}{\mathrm{Lt}}$$ $$\frac{(x+3-6)(x+3+6)}{x-3}$$
= $$\underset{x \rightarrow 3}{\mathrm{Lt}}$$ $$\frac{(x-3)(x+9)}{x-3}$$
= $$\underset{x \rightarrow 3}{\mathrm{Lt}}$$ (x + 9) = 12
Also f(3) = k
Since f is continuous at x = 3
∴ $$\underset{x \rightarrow 3}{\mathrm{Lt}}$$ f(x) = f(3)
⇒ 12 = k

(iv) $$\underset{x \rightarrow 3}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 3}{\mathrm{Lt}}$$ $$\frac{x^2-x-6}{x^2-2 x-3}$$
= $$\underset{x \rightarrow 3}{\mathrm{Lt}}$$ $$\frac{(x-3)(x+2)}{(x-3)(x+1)}$$
= $$\underset{x \rightarrow 3}{\mathrm{Lt}}$$ $$\frac{x+2}{x+1}$$
= $$\frac{3+2}{3+1}$$
= $$\frac{5}{4}$$
Hence f(3) = k (given)
Since f is continuous at x = 3
∴ $$\underset{x \rightarrow 3}{\mathrm{Lt}}$$ f(x) = f(3)
⇒ $$\frac{5}{4}$$ = k
Hence the required value of k be $$\frac{5}{4}$$.

Question 11.
If f(x) = $$\left\{\begin{array}{cc} \frac{\tan 3 x}{k x}, & x \neq 0 \\ 1 & , x=0 \end{array}\right.$$, find k so that the function f may be continuous at x = 0.
Solution:
Here, $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ $$\frac{\tan 3 x}{k x}$$
= $$\frac{3}{k}$$ $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ $$\frac{\tan 3 x}{k x}$$
= $$\frac{3}{k} {Lt}_{x \rightarrow 0} \frac{\tan 3 x}{3 x}$$
= $$\frac{3}{k} \times 1$$
= $$\frac{3}{k}$$
[∵ $$\ {Lt}_{\theta \rightarrow 0} \frac{\tan \theta}{\theta}$$ = 1]
Also, given f(0) = 1
Since f is continuous at x = 0
∴ $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ f(x) = f(0)
⇒ $$\frac{3}{k}$$ = 1
⇒ k = 3.

Question 12.
Is the function f defined by f(x) = tan x continuous at x = $$\frac{\pi}{2}$$?
Solution:
Given f(x) = tan x
Since $$\underset{x \rightarrow \frac{\pi^{-}}{2}}{\ {Lt}}$$ f(x) = $$\underset{x \rightarrow \frac{\pi^{-}}{2}}{\ {Lt}}$$ tan x → ∞
and $$\underset{x \rightarrow \frac{\pi^{+}}{2}}{\ {Lt}}$$ f(x) = $$\underset{x \rightarrow \frac{\pi^{+}}{2}}{\ {Lt}}$$ + tan x → – ∞
∴ $$\mathrm{Lt}_{x \rightarrow \frac{\pi}{2}}$$ f(x) does not exists.
Hence f(x) is not continuous at x = $$\frac{\pi}{2}$$.

Question 13.
Is the function f defined by f(x) = |x| continuous at x = 0? (NCERT)
Solution:
f(x) = |x|
= \left\{\begin{aligned} x & ; \quad x>0 \\ 0 ; & x=0 \\ -x ; & x<0 \end{aligned}\right.
∴ L.H.L. = $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ f(x)
= $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ – x = 0
R.H.L. = $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ f(x)
= $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ x = 0
f(0) = 0
∴ $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ f(x) = f(0)
Thus, f is continuous at x = 0.

Question 14.
Is the function f defined by f(x) = |x – 1| continuous at x = 1?
Solution:
Given f(x) = |x – 1|
= $$\left\{\begin{array}{rll} x-1 & ; & x>1 \\ -(x-1) & ; & x<1 \\ 0 & ; & x=1 \end{array}\right.$$
∴ L.H.L. = $$\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}$$ f(x)
= $$\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}$$ – (x – 1)
= – (1 – 1) = 0
R.H.L. = $$\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}$$ f(x)
= $$\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}$$ x – 1
= 1 – 1 = 0
and f(1) = 1 – 1 = 0
∴ L.H.L = R.H.L.
= f(1) = 0
Thus f is continuous at x = 1.

Question 15.
Is the function f defined by f(x) = x – |x| continuous at x = 0?
Solution:
Given f(x) = x – |x|
= $$\left\{\begin{array}{rc} x-x ; & x \geq 0 \\ x-(-x) & ; \quad x<0 \end{array}\right.$$
= \left\{\begin{aligned} 0 & ; \quad x \geq 0 \\ 2 x & ; \quad x<0 \end{aligned}\right.
∴ L.H.L. = $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ f(x)
= $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ 2x
= 2 × 0 = 0
R.H.L. = $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ f(x)
= $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ 0 = 0
Hence, f is continuous at x = 0.

Question 16.
Examine the following functions for continuity at the inclined points :
(i) f(x) = $$\begin{cases}x^2 & , \quad x \geq 0 \\ -x & , \quad x<0\end{cases}$$ at x = 0
(ii) f(x) = $$\left\{\begin{array}{cl} 5 x-4 & , \text { if } x<1 \\ 4 x^2-3 x & , \text { if } x \geq 1 \end{array}\right.$$ at x = 1
Solution:
(i) L.H.L. = $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ f(x)
= $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ – x = 0
R.H.L. = $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ f(x)
= $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ x2 = 0
and f(0) = 02 = 0
∴ L.H.L. = R.H.L. = f(0)
Thus, f is continuous at x = 0.

(ii) L.H.L. = $$\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}$$ f(x)
= $$\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}$$ 5x – 4
= 5 × 1 – 4 = 1
and R.H.L. = $$\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}$$ f(x)
= $$\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}$$ 4x2 – 3x
= 4 – 3 = 1
also f(1) = 4 × 12 – 3 × 1 = 1
∴ L.H.L. = R.H.L. = f(1)
Thus, f is continuous at x = 1.

Question 17.
Is the function f defined by f(x) = $$f(x)=\left\{\begin{array}{ccc} 2 x^2-3 & , \text { if } & x \leq 1 \\ 5 & , \text { if } & x>1 \end{array}\right.$$ continuous at x = 0? what about its continuity at x = 1 and x = 2?
Solution:
L.H.L. = $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ f(x)
= $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ (2x2 – 3)
= 0 – 3
[as x → 0 ⇒ x < 0]
= – 3
R.H.L. = $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ f(x)
= $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ (2x2 – 3)
= 2 × 02 – 3
= – 3
[as x → 0+
⇒ x > 0 but x < 1
∴ f(x) = 2x2 – 3]
∴ L.H.L. = R.H.L. = f(0)
Thus, f is continuous at x = 0.

at x = 1 :
L.H.L. = $$\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}$$ f(x)
= $$\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}$$ (2x2 – 3)
= 2 × 12 – 3
= 2 – 3 = – 1
R.H.L. = $$\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}$$ f(x)
= $$\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}$$ 5 = 5
∴ L.H.L. ≠ R.H.L.
Thus, f is not continuous at x = 1.

at x = 2 :
When x = 2 > 1
∴ f(x) = 5
i.e. when x → 2+ or x → 2
∴ f(x) = 5
∴ L.H.L. = R.H.L. = 5 = f(2)
Thus f is continuous at x = 2.

Question 18.
Examine the following functions for continuity :
(i) f(x) = $$\left\{\begin{array}{cc} \frac{x}{|x|}, & x \neq 0 \\ 0, & x=0 \end{array}\right.$$ at x = 0
(ii) f(x) = $$\left\{\begin{array}{cc} \frac{x}{|x|}, & x \neq 0 \\ 0, & x=0 \end{array}\right.$$ at x = 4
Solution:
(i) L.H.L. = $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ f(x)
= $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ $$\frac{x}{|x|}$$
= $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ $$\frac{x}{-x}$$
= – 1
[∵ x → 0
⇒ x < 0
⇒ |x| = – x]
R.H.L. = $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ f(x)
= $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ $$\frac{x}{|x|}$$
= $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ $$\frac{x}{x}$$
= 1
[∵ x → 0+
⇒ x > 0
⇒ |x| = x]
∴ L.H.L. ≠ R.H.L.
Thus, f is not continuous at x = 0.

(ii) L.H.L. = $$\underset{x \rightarrow 4^{-}}{\mathrm{Lt}}$$ f(x)
= $$\underset{x \rightarrow 4^{-}}{\mathrm{Lt}}$$ $$\frac{x-4}{2|x-4|}$$
= $$\underset{x \rightarrow 4^{-}}{\mathrm{Lt}}$$ $$\frac{x-4}{-2(x-4)}=-\frac{1}{2}$$
[∵ x → 4
⇒ x < 4
⇒ x – 4 < 0
∴ |x – 4| = – (x – 4)]
and R.H.L. = $$\underset{x \rightarrow 4^{+}}{\mathrm{Lt}}$$ f(x)
= $$\underset{x \rightarrow 4^{+}}{\mathrm{Lt}}$$ $$\frac{x-4}{2|x-4|}$$
= $$\underset{x \rightarrow 4^{+}}{\mathrm{Lt}}$$ $$\frac{x-4}{2(x-4)}=\frac{1}{2}$$
[∵ x → 4+
⇒ x > 4
⇒ x – 4 > 0
∴ |x – 4| = -x – 4]
∴ $$\underset{x \rightarrow 4^{-}}{\mathrm{Lt}}$$ f(x) ≠ $$\underset{x \rightarrow 4^{+}}{\mathrm{Lt}}$$ f(x)
Hence f is not continuous at x = 4.

Question 19.
Examine the following functions for continuity :
(i) f(x) = $$\left\{\begin{array}{cc} \frac{x}{\sin 3 x} & , \text { when } x \neq 0 \\ 3 & \text {, when } x=0 \end{array}\right.$$ at x = 0
(ii) f(x) = $$\left\{\begin{array}{c} \frac{\sin 2 x}{\sin 3 x}, \text { when } x \neq 0 \\ 2, \text { when } x=0 \end{array}\right.$$ at x = 1
Solution:
(i) $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ $$\frac{x}{\sin 3 x}$$
= $$\frac{1}{3}$$ $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ $$\frac{3 x}{\sin 3 x}$$
= $$\frac{1}{3}$$ × 1
= $$\frac{1}{3}$$
[∵ $$\underset{\theta \rightarrow 0}{\mathrm{Lt}}$$ $$\frac{\theta}{\sin \theta}$$ = 1]
also f(0) = 3
∴ $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ f(x) ≠ f(0)
Thus, f is discontinuous at x = 0

(ii) $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ $$\frac{\sin 2 x}{\sin 3 x}$$
= $$\frac{\operatorname{Lt}_{x \rightarrow 0} \frac{\sin 2 x}{2 x} \times 2 x}{\operatorname{Lt}_{x \rightarrow 0} \frac{\sin 3 x}{3 x} \times 3 x}$$
= $$\frac{2}{3} \times \frac{1}{1}=\frac{2}{3}$$
[∵ $$\underset{\theta \rightarrow 0}{\mathrm{Lt}}$$ $$\frac{\theta}{\sin \theta}$$ = 1 and
$$\operatorname{Lt}_{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{\operatorname{Lt}_{x \rightarrow a} f(x)}{\operatorname{Lt}_{x \rightarrow a} g(x)}$$]
also, f(0) = 2
∴ $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ f(x) ≠ f(0)
Thus, f is discontinuous at x = 0.

Question 20.
(i) If f(x) = $$\left\{\begin{array}{lll} k x+1 & \text {, if } & x \leq 5 \\ 3 x-5 & \text { if } & x>5 \end{array}\right.$$ is continuous at x = 5, find the value of k.
(ii) Find the value of k so that the function f(x) = $$\left\{\begin{array}{lll} k x+1 & \text {, if } & x \leq 5 \\ 3 x-5 & \text {, if } & x>5 \end{array}\right.$$ is continuous at x = 5. (NCERT)
(iii) For what value of k is the following function continuous at x = 2? f(x) = $$\left\{\begin{array}{cc} 2 x+1, & x<2 \\ k, & x=2 \\ 3 x-1, & x>2 \end{array}\right.$$
(iv) Find the value of k so that the function f defined by f(x) = $$\begin{cases}k x+1, & \text { if } x \leq \pi \\ \cos x & , \text { if } x>\pi\end{cases}$$ is continuous at x = π.
(v) For what value of k is the function f(x) = $$\left\{\begin{array}{cc} \frac{\tan 5 x}{\sin 2 x}, & x \neq 0 \\ k & , x=0 \end{array}\right.$$ continuous at x = 0?
(vi) Find the value of the constant k so that the function f(x) = $$\left\{\begin{array}{cc} \frac{1-\cos 4 x}{8 x^2} & , x \neq 0 \\ k & , x=0 \end{array}\right.$$ is continuous at x = 0. (NCERT Exampler)
(vii) Determine the value of constant ‘k’ so that the function f(x) = $$\left\{\begin{array}{cc} \frac{k x}{|x|}, & \text { if } x<0 \\ 3, & \text { if } x \geq 0 \end{array}\right.$$ is continuous at x = 0.
Solution:
(i) Since f(x) is continuous at x = 5.
∴ $$\underset{x \rightarrow 5}{\mathrm{Lt}}$$ f(x) = f(5)
⇒ $$\underset{x \rightarrow 5}{\mathrm{Lt}}$$ $$\frac{x^2-25}{x-5}$$ = k
⇒ $$\underset{x \rightarrow 5}{\mathrm{Lt}}$$ $$\frac{(x-5)(x+5)}{x-5}$$ = k
⇒ $$\underset{x \rightarrow 5}{\mathrm{Lt}}$$ (x + 5) = kx
⇒ 5 + 5 = k
⇒ k = 10
Thus f is continuous at x = 5 if k = 10.

(ii) L.H.L. = $$\underset{x \rightarrow 5^{-}}{\mathrm{Lt}}$$ f(x)
= $$\underset{x \rightarrow 5}{\mathrm{Lt}}$$ kx + 1
= 5k + 1
R.H.L. = $$\underset{x \rightarrow 5^{+}}{\mathrm{Lt}}$$ f(x)
= $$\underset{x \rightarrow 5}{\mathrm{Lt}}$$ 3x – 5
= 15 – 5
= 10
[as x → 5
⇒ x > 5
∴ f(x) = 3x – 5]
also f(5) = 5k + 1
Since f(x) is continuous at x = 5.
∴ L.H.L. = R.H.L. = f(5)
⇒ 5k + 1 = 10
⇒ 5k = 9
⇒ k = $$\frac{9}{5}$$.

(iii) $$\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}$$ 2x + 1
= 2 × 2 + 1 = 5
$$\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}$$ (3x – 1)
= 6 – 1 = 5
also f(2) = 5
Since f is continuous at x = 2.
∴ $$\underset{x \rightarrow 2^{-}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 2^{+}}{\mathrm{Lt}}$$ f(x) = f(2)
Thus, 5 = k.

(iv) L.H.L = $$\underset{x \rightarrow \pi^{-}}{\mathrm{Lt}}$$ f(x)
= $$\underset{x \rightarrow \pi}{\mathrm{Lt}}$$ kx + 1
= kπ + 1
R.H.L. = $$\underset{x \rightarrow \pi^{+}}{\mathrm{Lt}}$$ f(x)
= $$\underset{x \rightarrow \pi}{\mathrm{Lt}}$$ cos x
= cos π
= – 1
and f(π) = kπ + 1
Since f(x) is continuous at x = π
∴ L.H.L. = R.H.L. = f(π)
⇒ kπ + 1 = – 1
⇒ k = – $$\frac{2}{\pi}$$.
Thus f(x) is continuous at x = π if k = – $$\frac{2}{\pi}$$

(v)

Also, f(0) = k
Since f is continuous at x = 0.
∴ $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ f(x) = f(0)
⇒ $$\frac{5}{2}$$ = k

(vi) Since f(x) is continuous at x = 0.
∴ $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ f(x) = f(0)
⇒ $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ $$\frac{1-\cos 4 x}{8 x^2}$$ = k
⇒ $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ $$\frac{2 \sin ^2 2 x}{8 x^2}$$ = k
⇒ $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ $$\frac{\sin ^2 2 x}{(2 x)^2}$$ = k
⇒ $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ $$\left(\frac{\sin 2 x}{2 x}\right)^2$$ = k
[∵ $$\underset{\theta \rightarrow 0}{\mathrm{Lt}}$$ = $$\frac{\sin \theta}{\theta}$$ = 1]
Thus f(x) is continuous at x = 0 if k = 1.

(vii) $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ $$\frac{k x}{|x|}$$
= $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ $$\frac{k x}{-x}$$ = – k
[as x → 0
⇒ x < 0 ⇒ |x| = – x] $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ 3 = 3
and f(0) = 3
Since f is continuous at x = 0
∴ $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$
f(x) = f(0) – k = 3
⇒ k = – 3.

Question 21.
Find the relation between a and b so that the function f defined by f(x) = $$\left\{\begin{array}{l} a x+1, \text { if } x \leq 3 \\ b x+3, \text { if } x>3 \end{array}\right.$$ is continuous at x = 3.
Solution:
L.H.L. = $$\underset{x \rightarrow 3^{-}}{\mathrm{Lt}}$$ f(x)
= $$\underset{x \rightarrow 3}{\mathrm{Lt}}$$ ax + 1
= 3a + 1
and R.H.L. = $$\underset{x \rightarrow 3^{+}}{\mathrm{Lt}}$$ f(x)
= $$\underset{x \rightarrow 3}{\mathrm{Lt}}$$ bx + 3
= 3b + 3
also f(3) = 3a + 1
Since f(x) is given to be continuous at x = 3.
∴ L.H.L. = R.H.L. = f(3)
⇒ 3a + 1 = 3b + 3
⇒ 3a – 3b = 2

Question 22.
If the function f is defined by f(x) = $$\left\{\begin{array}{ccc} 3 a x+b & \text {, if } & x>1 \\ 11 & \text {, if } & x=1 \\ 5 a x-2 b & \text {, if } & x<1 \end{array}\right.$$ is continuous at x = 1, find the values of a and b. Solution: Given f(x) = $$\left\{\begin{array}{ccc} 3 a x+b & \text {, if } & x>1 \\ 11 & \text {, if } & x=1 \\ 5 a x-2 b & \text {, if } & x<1 \end{array}\right.$$
∴ $$\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}$$ (3ax + b)
Put x = 1 + h
as x → 1+
⇒ h → 0+
= $$\underset{h \rightarrow 0^{+}}{\mathrm{Lt}}$$ 3a (1 + h) + b
= 3a + b
$$\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}$$ (5ax – 2b)
Put x = 1 – h
as x → 1
⇒ h → 0+
= $$\underset{h \rightarrow 0^{+}}{\mathrm{Lt}}$$ 5a (1 – h) – 2b
= 5a – 2b
also f(1) = 11
Since f(x) is continuous at x = 1
∴ $$\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}$$ f(x) = f(1)
∴ 3a + b = 5a – 2b = 11
i.e. 3a+ b = 11 ………..(1)
5a – 2b = 11 …………..(2)
On solving (1) and (2) ;
11a = 33
⇒ a = 3 ; b = 2.

Question 23.
Consider the function defined as follows, for x ≠ 0. In each case, what choice (if any) of f(0) will make the function continuous at x = 0?
(i) f(x) = $$\frac{\sqrt[3]{1+x}-1}{x}$$
(ii) f(x) = $$\frac{5|x|-3 \tan x+\sin 2 x}{x}$$.
Solution:
$$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ $$\frac{\sqrt[3]{1+x}-1}{x}$$

For f(x) is to be continuous at x = 0
we have $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ f(x) = f(0)
i.e. f(0) = $$\frac{1}{3}$$

(ii) Now, $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ f(x)

Thus, f is discontinuous at x = 0, irrespective of the choice of f(0).
Hence, there is no choice of f(0) that makes f(x) is continuous at x = 0.

Question 24.
Examine the function f(x) = $$\left\{\begin{array}{cc} x \sin \frac{1}{x}, & x \neq 0 \\ 0 & , x=0 \end{array}\right.$$, for continuity at x = 0. (NCERT Exampler)
Solution:
Let g(x) = x
and h(x) = sin $$\frac{1}{x}$$
∴ $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ g(x) = $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ x = 0
and |h(x)| = |sin $$\frac{1}{x}$$| ≤ 1 ∀ x ∈ R, x ≠ 0
[∵ |sin t| ≤ 1 ∀ x ∈ R]
i.e. h (x) is bounded in the deleted neighbourhood of 0.
∴ $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ g(x) h(x) = 0
⇒ $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ x sin $$\frac{1}{x}$$ = 0
⇒ $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ f(x) = 0, also f(0) = 0
[∵ f(x) = 0 at x = 0
∴ f(0) = 0]
∴ $$\underset{x \rightarrow 0}{\mathrm{Lt}}$$ f(x) = f(0)
Thus, f is continuous at x = 0.

Question 25.
Show that the functionf(x) = |x| + |x – 1|, x ∈ R, is continuous both at x = 0 and x = 1. (NCERT Exampler)
Solution:
Given f(x) = |x| + |x – 1|
∴ f(x) = $$\left\{\begin{array}{ccc} -x-(x-1) & \text { if } & x<0 \\ x-(x-1) & \text { if } & 0 \leq x<1 \\ x+x-1 & \text { if } & x \geq 1 \end{array}\right.$$
i.e. f(x) = $$\left\{\begin{array}{ccc} -2 x+1 & ; & x<0 \\ 1 & ; & 0 \leq x<1 \\ 2 x-1 & ; & x \geq 1 \end{array}\right.$$

Continuity at x = 0:
L.H.L. = $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ f(x)
= $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ (- 2x + 1)
= 0 + 1 = 1
and R.H.L.
= $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ f(x)
= $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ 1 = 1
also f(0) = 1
∴ $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ f(x)
= f(0) = 1
Thus, f(x) is continuous at x = 0.

Continuity at x = 1 :
$$\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}$$ 1 = 1
$$\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}$$ 2x – 1
= 2 × 1 – 1 = 1
also f(1) = 2 × 1 – 1 = 1
∴ $$\underset{x \rightarrow 1^{-}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 1^{+}}{\mathrm{Lt}}$$ f(x)
= f(1) = 1
Thus, f(x) is continuous at x = 1.

Question 26.
Examine the function f(x) = $$\left\{\begin{array}{cc} e^{1 / x}, & x \neq 0 \\ 1, & x=0 \end{array}\right.$$ for continuity at x = 0.
Solution:
$$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ e1/x
Put x = 0 – h
as x → 0
⇒ h → 0+
= $$\underset{h \rightarrow 0^{+}}{\mathrm{Lt}}$$ e– 1/h
= e-∞
= $$\frac{1}{\infty}$$ = 0
$$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ f(x) = $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ e1/𝜏
= e+ ∞ → + ∞
Thus $$\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}$$ f(x) ≠ $$\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}$$ f(x)
Hence function f(x) is discontinuous at x = 0.