# The function

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

Solution:

(b) continuous at x = 0

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

Continuity at x = 0:

We have,

$(\mathrm{LHL}$ at $x=0)$

$\lim _{x \rightarrow 0^{-}} f(x)$

$=\lim _{h \rightarrow 0} \sin ^{-1}\{\cos (0-h)\}$

$=\lim _{h \rightarrow 0} \sin ^{-1}(\cos h)$

$=\sin ^{-1}(1)$

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

(RHL at x = 0)

$\lim _{x \rightarrow 0^{+}} f(x)$

$=\lim _{h \rightarrow 0} \sin ^{-1} \cos (0+h)$

$=\lim _{h \rightarrow 0} \sin ^{-1}(\cos h)$

$=\sin ^{-1}(1)$

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

$f(0)=\sin ^{-1}(\cos 0)$

$=\sin ^{-1}(1)$

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

Differentiability at $x=0$ :

(LHD at $x=0$ )

$\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}$

$=\lim _{h \rightarrow 0} \frac{\sin ^{-1} \cos (0-h)-\frac{\pi}{2}}{-h}$

$=\lim _{h \rightarrow 0} \frac{\sin ^{-1} \cos (-h)-\frac{\pi}{2}}{-h}$

$=\lim _{h \rightarrow 0} \frac{\sin ^{-1} \cos (h)-\frac{\pi}{2}}{-h}$

$=\lim _{h \rightarrow 0} \frac{\sin ^{-1}\left\{\sin \left(\frac{\pi}{2}-h\right)\right\}-\frac{\pi}{2}}{-h}$

$=\lim _{h \rightarrow 0} \frac{-h}{-h}$

$=1$

RHD at x = 0

$\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}$

$=\lim _{h \rightarrow 0} \frac{\sin ^{-1} \cos (0+h)-\frac{\pi}{2}}{h}$

$=\lim _{h \rightarrow 0} \frac{\sin ^{-1} \cos (h)-\frac{\pi}{2}}{h}$

$=\lim _{h \rightarrow 0} \frac{\sin ^{-1}\left\{\sin \left(\frac{\pi}{2}-h\right)\right\}-\frac{\pi}{2}}{-h}$

$=\lim _{h \rightarrow 0} \frac{-h}{h}$

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

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

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