Question

If $f(x)=|\cos x-\sin x|$, then $f^{\prime }\left(\frac{\pi }{4}\right)$ is equal to

• A. $\sqrt{2}$
• B. $-\sqrt{2}$
• C. 0
• D. None of these

Answer 1

Answer: (D)

We have, $f(x)=|\cos x-\sin x|$
$\Rightarrow f(x)=\{\begin{array}{l}\cos x-\sin x,for0<x\le \frac{\pi }{4}\\ \sin x-\cos x,for\frac{\pi }{4}<x<\frac{\pi }{2}\end{array}$
Clearly, $L{f}^{\prime }\left(\frac{\pi }{4}\right)={\left\{\frac{d}{dx}(\cos x-\sin x)\right\}}_{atx=\frac{\pi }{4}}$ $=(-\sin x-\cos x{)}_{x-\frac{\pi }{4}}=\sqrt{2}$
and $R{f}^{\prime }\left(\frac{\pi }{4}\right)={\left\{\frac{d}{dx}(\sin x-\cos x)\right\}}_{\text{at }x=\frac{\pi }{4}}$ $=(\cos x+\sin x{)}_{x=\frac{\pi }{4}}=\sqrt{2}$
$\because L{f}^{\prime }\left(\frac{\pi }{4}\right)\ne R{f}^{\prime }\left(\frac{\pi }{4}\right)\therefore f^{\prime }\left(\frac{\pi }{4}\right)$ doesn't exist.