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# 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

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Solution

## (b) continuous at x = 0 Given: $f\left(x\right)={\mathrm{sin}}^{-1}\left(\mathrm{cos}x\right).$ Continuity at x = 0: We have, (LHL at x = 0) $\underset{x\to {0}^{-}}{\mathrm{lim}}f\left(x\right)\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}{\mathrm{sin}}^{-1}\left\{\mathrm{cos}\left(0-h\right)\right\}\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}{\mathrm{sin}}^{-1}\left(\mathrm{cos}h\right)\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{-1}\left(1\right)\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{2}$ (RHL at x = 0) $\underset{x\to {0}^{+}}{\mathrm{lim}}f\left(x\right)\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}{\mathrm{sin}}^{-1}\mathrm{cos}\left(0+h\right)\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}{\mathrm{sin}}^{-1}\left(\mathrm{cos}h\right)\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{-1}\left(1\right)\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{2}\phantom{\rule{0ex}{0ex}}$ $f\left(0\right)={\mathrm{sin}}^{-1}\left(\mathrm{cos}0\right)\phantom{\rule{0ex}{0ex}}={\mathrm{sin}}^{-1}\left(1\right)\phantom{\rule{0ex}{0ex}}=\frac{\mathrm{\pi }}{2}$ Differentiability at x = 0: (LHD at x = 0) $\underset{x\to {0}^{-}}{\mathrm{lim}}\frac{f\left(x\right)-f\left(0\right)}{x-0}\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\frac{{\mathrm{sin}}^{-1}\mathrm{cos}\left(0-h\right)-\frac{\mathrm{\pi }}{2}}{-h}\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\frac{{\mathrm{sin}}^{-1}\mathrm{cos}\left(-h\right)-\frac{\mathrm{\pi }}{2}}{-h}\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\frac{{\mathrm{sin}}^{-1}\mathrm{cos}\left(h\right)-\frac{\mathrm{\pi }}{2}}{-h}\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\frac{{\mathrm{sin}}^{-1}\left\{\mathrm{sin}\left(\frac{\mathrm{\pi }}{2}-h\right)\right\}-\frac{\mathrm{\pi }}{2}}{-h}\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\frac{-h}{-h}\phantom{\rule{0ex}{0ex}}=1\phantom{\rule{0ex}{0ex}}$ RHD at x = 0 $\underset{x\to {0}^{+}}{\mathrm{lim}}\frac{f\left(x\right)-f\left(0\right)}{x-0}\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\frac{{\mathrm{sin}}^{-1}\mathrm{cos}\left(0+h\right)-\frac{\mathrm{\pi }}{2}}{h}\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\frac{{\mathrm{sin}}^{-1}\mathrm{cos}\left(h\right)-\frac{\mathrm{\pi }}{2}}{h}\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\frac{{\mathrm{sin}}^{-1}\left\{\mathrm{sin}\left(\frac{\mathrm{\pi }}{2}-h\right)\right\}-\frac{\mathrm{\pi }}{2}}{-h}\phantom{\rule{0ex}{0ex}}=\underset{h\to 0}{\mathrm{lim}}\frac{-h}{h}\phantom{\rule{0ex}{0ex}}=-1\phantom{\rule{0ex}{0ex}}$ $\therefore \mathrm{LHD}\ne \mathrm{RHD}$ Hence, the function is not differentiable at x = 0 but is continuous at x = 0.

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