If $f(x)=\cos \left\{\frac{\pi}{2}[x]-x^3\right\}, 1<x<2$, and [x] denotes the greatest integer less than or equal to x, then the value of $f^{\prime}\left(\sqrt[3]{\frac{\pi}{2}}\right)$, is

0

In the neighbourhood of $x=\sqrt[3]{\frac{\pi}{2}}$, we have [x] = 1

∴   $f(x)=\cos \left(\frac{\pi}{2}-x^3\right)$

$\Rightarrow f(x)=\sin x^3$ for all x in the neighbourhood of $\sqrt[3]{\frac{\pi}{2}}$

$\Rightarrow f'(x)=3 x^2 \cos x^3$

$\Rightarrow f'\left(\sqrt[3]{\frac{\pi}{2}}\right)=3\left(\frac{\pi}{2}\right)^{2 / 3} \cos \frac{\pi}{2}=0$