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  4. If f(x) = sin((π/2)[x]-x5),1 < x <2 where [x] denotes greater integer less than or equal to x, then f'(√[5](π/2)) =

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Q. If $f\left(x\right) = \sin\left(\frac{\pi}{2}\left[x\right]-x^{5}\right),1 < x <2$ where $[x]$ denotes greater integer less than or equal to $x,$ then $ f'\left(\sqrt[5]{\frac{\pi}{2}}\right) = $

COMEDKCOMEDK 2010Continuity and Differentiability

Solution:

$f\left(x\right) = \sin\left(\frac{\pi}{2} \left[x\right] -x^{5}\right)$
$ = \sin \left(\frac{\pi}{2}-x^{5}\right)$        $ \left[ \because 1 < x < 2 \Rightarrow \left[x\right]=1\right]$
$\therefore \:\:\: f'\left(x\right) = \cos\left(\frac{\pi}{2}-x^{5}\right)\left(-5x^{4}\right) $
$f'\left(\sqrt[5]{\frac{\pi}{2}}\right) =\cos \left\{\frac{\pi}{2} -\left(\sqrt[5]{\frac{\pi}{2}}\right)^{5}\right\}\left\{-5\left(\sqrt[5]{\frac{\pi}{2}}\right)^{4}\right\}$
$ = \cos \left(\frac{\pi}{2} -\frac{\pi}{2}\right)\left(-5\left(\frac{\pi}{2}\right)^{\frac{4}{5}}\right) $
$= -5\left(\frac{\pi}{2}\right)^{\frac{4}{5}} \left(\cos 0\right) = -5\left(\frac{\pi}{2}\right)^{\frac{4}{5}}$