Question

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 $f^{\prime }\left(\sqrt[5]{\frac{\pi }{2}}\right)=$

• A. $5{\left(\frac{\pi }{2}\right)}^{\frac{4}{5}}$
• B. $-5{\left(\frac{\pi }{2}\right)}^{\frac{4}{5}}$
• C. 0
• D. none of these

Answer 1

Answer: (B)

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