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

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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) = \(\underset{x \rightarrow 0^{-}}{\mathrm{Lt}}\) e^1/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) = \(\underset{x \rightarrow 0^{+}}{\mathrm{Lt}}\) e^1/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.