Question

If $f\left(x\right)={\cos }^{-1}\left(\cos x\right)$ then $f\left(x\right)$ is

• A. continuous at $x=\pi$
• B. discontinuous at $x=-\pi$
• C. differentiable at $x=0$
• D. non differentiable at $x=\pi$

Answer 1

Answer: (A), (D)

continuous at $x=\pi$
non differentiable at $x=\pi$

We know that

$f\left(x\right)={\cos }^{-1}\cos x=\{\begin{array}{c}2\pi +x;x\in [2\pi ,-\pi ]\\ -x;x\in [-\pi ,0]\\ x;x\in [0,\pi ]\\ 2\pi -x;x\in [\pi ,2\pi ]\\ -2\pi +x;x\in [2\pi ,3\pi ]\end{array}$

at $x=\pi$

${lim}_{x\to {\pi }^{-}}f\left(x\right)={lim}_{x\to {\pi }^{-}}x=\pi$

${lim}_{x\to {\pi }^{+}}f\left(x\right)={lim}_{x\to {\pi }^{+}}(2\pi -x)=2\pi -\pi =\pi$

LHL=RHL $\Rightarrow$continuous

$f^{\prime }\left(x\right)=\{\begin{array}{c}1;x\in [0,\pi ]\\ 0-1;x\in [\pi ,2\pi ]\end{array}$

at $x=\pi$

${lim}_{x\to {\pi }^{-}}{f}^{\prime }\left(x\right)=1$

${lim}_{x\to {\pi }^{+}}{f}^{\prime }\left(x\right)=-1$

LHL$\ne$RHL $\Rightarrow$Not differentiable