Question

Which of the following function is not differentiable at $x=0?$
(i) $f(x)=\left(x^2-1\right)|(x-1)(x-2)|$
(ii) $f(x)=\sin (|x-1|)-|x-1|$
(iii) $f(x)=\tan (|x-1|)+|x-1|$

• A. only (i)
• B. only (iii)
• C. Both (ii) and (i)
• D. Both (ii) and (iii)

Answer 1

Answer: (C)

(i) $f(x)=\left(x^2-1\right)|(x-1)(x-2)|$
$f^{\prime }\left(1^{+}\right)=\underset{h\to 0}{\lim }\frac{\left((1+h{)}^2-1\right)|h(1+h-2)|-0}{h}=0$ $f^{\prime }\left(1^{-}\right)=\underset{h\to 0}{\lim }\frac{\left((1-h{)}^2-1\right)|-h(1-h-2)|-0}{-h}=0$
Hence, it is differentiable at $x=0$.
(ii) $f(x)=\sin (|x-1|)-|x-1|$
$f^{\prime }\left(0^{+}\right)=\underset{h\to 0}{\lim }\frac{\sin h-h-0}{h}=0$
$f^{\prime }\left(0^{-}\right)=\underset{h\to 0}{\lim }\frac{\sin |-h|-|-h|}{-h}=\underset{h\to 0}{\lim }\frac{\sin h-h}{-h}=0$
Hence, $f(x)$ is differentiable at $x=0$
(iii) $f(x)=\tan (|x-1|)+|x-1|$
$f^{\prime }\left(0^{+}\right)=\underset{h\to 0}{\lim }\frac{\tan h+h-0}{h}=2$
$f^{\prime }\left(0^{-}\right)=\underset{h\to 0}{\lim }\frac{\tan |-h|+\mid -h}{-h}=\underset{h\to 0}{\lim }\frac{\tan h+h}{-h}=-2$
Hence, $f(x)$ is non-differentiable at $x=0$