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If $f (x) = \sin (\frac{π}{2} [x] - x^{5}), 1 < x < 2 and [x]$ denotes the greatest integer less than or equal to $' x'$ , then $f' (\sqrt[5]{\frac{π}{2}}) =$

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$- 5 {(\frac{π}{2})}^{\frac{4}{5}}$

$5 {(\frac{π}{2})}^{\frac{4}{5}}$

answer is B.

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Detailed Solution

Given $1 < x < 2$

$∴ [x] = 1$

$∴ f (x) = \sin (\frac{π}{2} - x^{5}) = \cos x^{5}$

$\Rightarrow f' (x) = - 5 x^{4} \sin x^{5}$

$\Rightarrow f' (\sqrt[5]{\frac{π}{2}}) = - 5 {(\frac{π}{2})}^{\frac{4}{5}} \cdot \sin \frac{π}{2}$

$= - 5 {(\frac{π}{2})}^{\frac{4}{5}}$ .

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