Question

Let $f\left(x\right)=1-|\cos x|\forall x\in R,$ then which among the following options is/are correct?

• A. $f^{\prime }\left(\frac{\pi }{2}\right)$ does not exist
• B. $f\left(x\right)$ is continuous everywhere
• C. $f\left(x\right)$ is not differentiable everywhere
• D. $\underset{x\to {\frac{\pi }{2}}^{+}}{lim}f\left(x\right)=1$

Answer 1

Answer: (A), (B), (C), (D)

$\underset{x\to {\frac{\pi }{2}}^{+}}{lim}f\left(x\right)=1$


$f\left(x\right)=1-|\cos x|$ is continuous as $\cos x$ is continuous for all $x$ which gives $|\cos x|$ is also continuous.
Hence, $1-|\cos x|$ is continuous.

Also, $\underset{x\to {\frac{\pi }{2}}^{+}}{lim}f\left(x\right)=\underset{x\to {\frac{\pi }{2}}^{+}}{lim}1-|\cos x|=1$

Also, since $f\left(x\right)$ has sharp corner at $x=\frac{\pi }{2}$. Hence, $f^{\prime }\left(\frac{\pi }{2}\right)$ does not exist.

Similar sharp corners occur at all the points where $\cos x=0$ i.e. at $x=(2n+1)\frac{\pi }{2},n\in Z.$
Hence, $f\left(x\right)$ is not differentiable everywhere.