The function f (x) = sin−1 (cos x) is

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(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) = $\pi$/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) = $\pi$/2 f(0) = \sin^{- 1} \left( \cos 0 \right) = \sin^{- 1} \left( 1 \right) = $\pi$/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) - $\pi$/2}{- h} = \lim_{h \to 0} \frac{\sin^{- 1} \cos\left( - h \right) - $\pi$/2}{- h} = \lim_{h \to 0} \frac{\sin^{- 1} \cos\left( h \right) - $\pi$/2}{- h} = \lim_{h \to 0} \frac{\sin^{- 1} \left\{ \sin \left( $\pi$/2 - h \right) \right\} - $\pi$/2}{- h} = \lim_{h \to 0} - 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) - $\pi$/2}{h} = \lim_{h \to 0} \frac{\sin^{- 1} \cos\left( h \right) - $\pi$/2}{h} = \lim_{h \to 0} \frac{\sin^{- 1} \left\{ \sin \left( $\pi$/2 - h \right) \right\} - $\pi$/2}{- h} = \lim_{h \to 0} - h/h = - 1 \therefore LHD \neq RHD Hence, the function is not differentiable at x = 0 but is continuous at x = 0.

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