Question

The function f (x) = sin⁻¹ (cos x) is
(a) discontinuous at x = 0
(b) continuous at x = 0
(c) differentiable at x = 0
(d) none of these

Answer 1

(b) continuous at x = 0

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

Continuity at x = 0:

We have,
(LHL at x = 0)

$$\begin{gathered} \underset{x\to 0^{-}}{\lim }f\left(x\right) \\ =\underset{h\to 0}{\lim }{\sin }^{-1}\left\{\cos \left(0-h\right)\right\} \\ =\underset{h\to 0}{\lim }{\sin }^{-1}\left(\cos h\right) \\ ={\sin }^{-1}\left(1\right) \\ =\frac{\mathrm{\pi }}{2} \end{gathered}$$

(RHL at x = 0)

$$\begin{gathered} \underset{x\to 0^{+}}{\lim }f\left(x\right) \\ =\underset{h\to 0}{\lim }{\sin }^{-1}\cos \left(0+h\right) \\ =\underset{h\to 0}{\lim }{\sin }^{-1}\left(\cos h\right) \\ ={\sin }^{-1}\left(1\right) \\ =\frac{\mathrm{\pi }}{2} \end{gathered}$$

$$\begin{gathered} f\left(0\right)={\sin }^{-1}\left(\cos 0\right) \\ ={\sin }^{-1}\left(1\right) \\ =\frac{\mathrm{\pi }}{2} \end{gathered}$$

Differentiability at x = 0:
(LHD at x = 0)

$$\begin{gathered} \underset{x\to 0^{-}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0} \\ =\underset{h\to 0}{\lim }\frac{{\sin }^{-1}\cos \left(0-h\right)-{\displaystyle \frac{\mathrm{\pi }}{2}}}{-h} \\ =\underset{h\to 0}{\lim }\frac{{\sin }^{-1}\cos \left(-h\right)-{\displaystyle \frac{\mathrm{\pi }}{2}}}{-h} \\ =\underset{h\to 0}{\lim }\frac{{\sin }^{-1}\cos \left(h\right)-{\displaystyle \frac{\mathrm{\pi }}{2}}}{-h} \\ =\underset{h\to 0}{\lim }\frac{{\sin }^{-1}\left\{\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}-h}\right)\right\}-{\displaystyle \frac{\mathrm{\pi }}{2}}}{-h} \\ =\underset{h\to 0}{\lim }\frac{-h}{-h} \\ =1 \end{gathered}$$

RHD at x = 0

$$\begin{gathered} \underset{x\to 0^{+}}{\lim }\frac{f\left(x\right)-f\left(0\right)}{x-0} \\ =\underset{h\to 0}{\lim }\frac{{\sin }^{-1}\cos \left(0+h\right)-{\displaystyle \frac{\mathrm{\pi }}{2}}}{h} \\ =\underset{h\to 0}{\lim }\frac{{\sin }^{-1}\cos \left(h\right)-{\displaystyle \frac{\mathrm{\pi }}{2}}}{h} \\ =\underset{h\to 0}{\lim }\frac{{\sin }^{-1}\left\{\sin \left({\displaystyle \frac{\mathrm{\pi }}{2}-h}\right)\right\}-{\displaystyle \frac{\mathrm{\pi }}{2}}}{-h} \\ =\underset{h\to 0}{\lim }\frac{-h}{h} \\ =-1 \end{gathered}$$

$\therefore \mathrm{LHD}\ne \mathrm{RHD}$

Hence, the function is not differentiable at x = 0 but is continuous at x = 0.