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प्रश्न
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
उत्तर
(b) continuous at x = 0
Given:
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
\[\lim_{x \to 0^+} \frac{f\left( x \right) - f\left( 0 \right)}{x - 0} \]
\[ = \lim_{h \to 0} \frac{\sin^{- 1} \cos\left( 0 + h \right) - \frac{\pi}{2}}{h} \]
\[ = \lim_{h \to 0} \frac{\sin^{- 1} \cos\left( h \right) - \frac{\pi}{2}}{h}\]
\[ = \lim_{h \to 0} \frac{\sin^{- 1} \left\{ \sin \left( \frac{\pi}{2} - h \right) \right\} - \frac{\pi}{2}}{- h}\]
\[ = \lim_{h \to 0} \frac{- h}{h}\]
\[ = - 1\]
Hence, the function is not differentiable at x = 0 but is continuous at x = 0.
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