If f(x)

Given: $f(x)=\cos \left[\pi^{2}\right] x+\cos \left[-\pi^{2}\right] x$

$f(x)=\cos \left[\pi^{2}\right] x+\cos \left[-\pi^{2}\right] x$

$=\cos 9 x+\cos (-10 x) \quad\left(\because \pi^{2}=9.85\right.$ approx $)$

$=\cos 9 x+\cos 10 x$

$\Rightarrow f\left(\frac{\pi}{2}\right)=\cos 9\left(\frac{\pi}{2}\right)+\cos 10\left(\frac{\pi}{2}\right)$

$=\cos \frac{9 \pi}{2}+\cos 5 \pi$

$=0-1$

$=-1$

Hence, $f\left(\frac{\pi}{2}\right)=\underline{-1}$