Question

The function $f\left(x\right)={sin}^{-1}\left(cosx\right)$ is :

• A. Discontinuous at x = 0
• B. Continuous at x = 0
• C. Differentiable at x = 0
• D. None of these

Answer 1

The correct option is A Discontinuous at x = 0

$f\left(x\right)={\sin }^{-1}\left(\cos x\right)$

LHL :

$li{m}_{x\to 0^{-}}{\sin }^{-1}\left(\cos x\right)=li{m}_{h\to 0}{\sin }^{-1}\left(\cos \right(0-h\left)\right)$

$li{m}_{h\to 0}{\sin }^{-1}\left(\cos \right(-h\left)\right)=li{m}_{h\to 0}{\sin }^{-1}\left(\cos 0\right)=\frac{\pi }{2}$

RHL:

$li{m}_{x\to 0^{+}}{\sin }^{-1}\left(\cos x\right)$

$=li{m}_{h\to 0}{\sin }^{-1}\cos (0+h)$

$={\sin }^{-1}\cos 0$

$=\frac{\pi }{2}$

Thus, $LHL=RHL=f\left(0\right)=\frac{\pi }{2}$

RHD :

$li{m}_{h\to 0}\frac{f(x+h)-f\left(x\right)}{h}=\frac{si{n}^{-1}\left(\cos h\right)-1}{h}$

$li{m}_{h\to 0}\frac{{\sin }^{-1}\left(\cos h\right)-1}{h}=\frac{1-\sin h}{\sqrt{1-{\cos }^{2}h}}=\frac{-\sin h}{\sin h}=-1$

LHD :

$li{m}_{h\to 0}\frac{f(x-h)-f\left(x\right)}{-h}=li{m}_{h\to 0}\frac{{\sin }^{-1}(\cos -h)-1}{-h}$

$li{m}_{h\to 0}\frac{{\sin }^{-1}\left(\cos h\right)-1}{-h}=\frac{-\sin h}{-\sin h}=1$

$LHD\ne RHD$

Thus, function is not differentiable at $x=0$.