Question

Given $f\left(x\right)={\cos }^{-1}\left(sgn\left(\frac{2\left|x\right|}{3x-\left|x\right|}\right)\right)$ where $sgn(.)$ denotes the signum function $[\cdot ]$ denotes the greatest integer function. Discuss the continuity and differentiablity at $x=\pm 1$

• A. $f$ is continuous and derivable at $x=-1$ but $f$ is continuous but not derivable at $x=1$
• B. $f$ is continuous but not derivable at $x=-1$ but $f$ is neither continuous nor derivable at $x=1$
• C. $f$ is continuous and derivable at $x=-1,1$
• D. $f$ is continuous and derivable at $x=-1$ but $f$ is neither continuous nor derivable at $x=1$

Answer 1

Answer: (D)

$f$ is continuous and derivable at $x=-1$ but $f$ is neither continuous nor derivable at $x=1$

$f\left(1^{-}\right)=\underset{h\to 0}{lim}{\cos }^{-1}\left(sgn\left(\frac{2[1-h]}{3(1-h)-[1-h]}\right)\right)=\frac{\pi }{2}$

$f\left(1^{+}\right)=\underset{h\to 0}{lim}{\cos }^{-1}\left(sgn\left(\frac{2[1+h]}{3(1+h)-[1+h]}\right)\right)=\underset{h\to 0}{lim}{\cos }^{-1}\left(sgn\left(\frac{2}{2}\right)\right)=0$

Hence $f\left(x\right)$ isnot continuous and not dirivable at $x=1$
Now at $x=-1$

$f\left({-1}^{-}\right)=\underset{h\to 0}{lim}{\cos }^{-1}\left(sgn\left(\frac{2[-1-h]}{3(-1-h)-[-1-h]}\right)\right)=\underset{h\to 0}{lim}{\cos }^{-1}\left(sgn\left(\frac{-4}{-3+2}\right)\right)={\cos }^{-1}1=0$

Also
$f\left({-1}^{+}\right)=\underset{h\to 0}{lim}{\cos }^{-1}\left(sgn\left(\frac{2[-1+h]}{3(-1+h)-[-1+h]}\right)\right)=\underset{h\to 0}{lim}{\cos }^{-1}\left(sgn\left(\frac{-2}{-3+1}\right)\right)={\cos }^{-1}1=0$

Hence $f\left(x\right)$ is continuous and differentiable at $x=-1$