Question

The function f (x) = sin⁻¹ (cos x) is

विकल्प

• discontinuous at x = 0

• continuous at x = 0

• differentiable at x = 0

• none of these

Answer 1

(b) continuous at x = 0 

Given:  

\[f(x) = \sin^{- 1} \left( \cos x \right) .\]

Continuity at x = 0: 

We have,
(LHL at x = 0) 

\[\lim_{x \to 0^-} f(x) \]
\[ = \lim_{h \to 0} \sin^{- 1} \left\{ \cos\left( 0 - h \right) \right\}\]
\[ = \lim_{h \to 0} \sin^{- 1} \left( \cos h \right)\]
\[ = \sin^{- 1} \left( 1 \right)\]
\[ = \frac{\pi}{2}\]

(RHL at x = 0)

\[\lim_{x \to 0^+} f\left( x \right)\]
\[ = \lim_{h \to 0} \sin^{- 1} \cos\left( 0 + h \right)\]
\[ = \lim_{h \to 0} \sin^{- 1} \left( \cos h \right)\]
\[ = \sin^{- 1} \left( 1 \right) \]
\[ = \frac{\pi}{2}\]

\[f(0) = \sin^{- 1} \left( \cos 0 \right) \]
\[ = \sin^{- 1} \left( 1 \right)\]
\[ = \frac{\pi}{2}\]

\[\lim_{x \to 0^-} \frac{f\left( x \right) - f\left( 0 \right)}{x - 0} \]
\[ = \lim_{h \to 0} \frac{\sin^{- 1} \cos\left( 0 - h \right) - \frac{\pi}{2}}{- h} \]
\[ = \lim_{h \to 0} \frac{\sin^{- 1} \cos\left( - h \right) - \frac{\pi}{2}}{- h}\]
\[ = \lim_{h \to 0} \frac{\sin^{- 1} \cos\left( h \right) - \frac{\pi}{2}}{- h}\]
\[ = \lim_{h \to 0} \frac{\sin^{- 1} \left\{ \sin \left( \frac{\pi}{2} - h \right) \right\} - \frac{\pi}{2}}{- h}\]
\[ = \lim_{h \to 0} \frac{- h}{- h}\]
\[ = 1\]

RHD at x = 0

\[\lim_{x \to 0^+} \frac{f\left( x \right) - f\left( 0 \right)}{x - 0} \]
\[ = \lim_{h \to 0} \frac{\sin^{- 1} \cos\left( 0 + h \right) - \frac{\pi}{2}}{h} \]
\[ = \lim_{h \to 0} \frac{\sin^{- 1} \cos\left( h \right) - \frac{\pi}{2}}{h}\]
\[ = \lim_{h \to 0} \frac{\sin^{- 1} \left\{ \sin \left( \frac{\pi}{2} - h \right) \right\} - \frac{\pi}{2}}{- h}\]
\[ = \lim_{h \to 0} \frac{- h}{h}\]
\[ = - 1\]

\[\therefore LHD \neq RHD\]

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