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 $f^{'} (\sqrt[5]{\frac{π}{2}})$ is equal to

5 {(\frac{π}{2})}^{\frac{4}{5}}

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- 5 {(\frac{π}{2})}^{\frac{4}{5}}

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Solution

The correct option is B $- 5 {(\frac{π}{2})}^{\frac{4}{5}}$
$f (x) = sin (\frac{π}{2} [x] - x^{5})$
because $\Rightarrow [x] = 1$

$\Rightarrow f (x) = sin (\frac{π}{2} - x^{5})$
$f^{'} (x) = cos (\frac{π}{2} - x^{5}) \times (- 5) x^{4}$
$f^{'} ({(\frac{π}{2})}^{1 / 5}) = 1 \times (- 5) \times {(\frac{π}{2})}^{4 / 5}$
$f^{'} ({(\frac{π}{2})}^{1 / 5}) = - 5 \times {(\frac{π}{2})}^{4 / 5}$

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