If $f(x)=\cos \left\{\frac{\pi}{2}[x]-x^3\right\}$, 1 < x < 2 (where [.] denotes the greatest integer function) then find $f'(x)$ at $x^3=\frac{\pi}{2}$

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$x=\left(\frac{\pi}{2}\right)^{1 / 3} \Rightarrow[x]=1$

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

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

$f'\left(\left(\frac{\pi}{2}\right)^{1 / 3}\right)=3\left(\frac{\pi}{2}\right)^{2 / 3} \cos \frac{\pi}{2}=0$

Hence (1) is the correct answer.