Question

$f(x)=\sin |x|$ . Then $f(x)$ is not differentiable at

• A. $x=0$ only
• B. all $x$
• C. multiples of $\pi$
• D. multiples of $\frac{\pi }{2}$

Answer 1

Answer: (A)

Any function $f(x)$ is derivable at $x=a$, if ${\lim }_{h\to 0^{+}}\frac{f(a+h)-f(a)}{h}={\lim }_{h\to 0^{-}}\frac{f(a-h)-f(a)}{-h}$ Given,
$<br/>\begin{array}{l}<br/>f(x)=\sin |x|\Rightarrow f(x)=\{\begin{array}{cl}<br/>\sin x& x>0\\ <br/>0& x=0RHD={\lim }_{h\to 0}\frac{\sin |(0+h)|-\sin (0)}{-}\\ <br/>-\sin x& x<0<br/>\end{array}\\ <br/>RHD={\lim }_{h\to 0}\frac{\sin ((0+h)\mid -\sin (0)}{h}\\ <br/>={\lim }_{h\to 0}\frac{\sinh }{h}=1\\ <br/>LHD={\lim }_{h\to 0}\frac{\sin |(0-h)|-\sin (0)}{-h}\\ <br/>=\frac{-\sin h}{h}=-1\\ <br/>\therefore LHD\ne RHD\text{ at }x=0\\ <br/>\therefore f(x)\text{ is not derivable at }x=0.<br/>\end{array}<br/>$