Question

Let $f\left(x\right)=cox\left[{\pi }^2\right]x+\cos \left[-{\pi }^2\right]x,$ where $\left[x\right]$ is the greatest integer function, then find$f\left(\frac{\pi }{2}\right)$ and $f\left(\pi \right)$

Answer 1

We know that $\pi =3.14$

Then, $\left(\pi \right)\left(\pi \right)=9$ (something), i.e., $9<{\pi }^2<10$

Now. the greatest integer of ${\pi }^2$ = greatest integer of $9=9$

Greatest integer of -${\pi }^2$=greatest integer of -9=10

Substituting these values in $f\left(x\right)$, we get

$f\left(x\right)=\cos \left(9x\right)+\cos (-10x)f\left(x\right)=\cos \left(9x\right)+\cos \left(10x\right)[As\cos (-x)=\cos \left(x\right)]f\left(\frac{\pi }{2}\right)=\cos \left(\frac{9\pi }{2}\right)+\cos \left(\frac{10\pi }{2}\right)=\cos \left(\frac{9\pi }{2}\right)+\cos \left(5\pi \right)=\cos (4\pi +\frac{\pi }{2})+\cos (4\pi +\pi )=\cos \left(\frac{\pi }{2}\right)+\cos \left(\pi \right)=0-1=-1Similarly,f\left(\pi \right)=\cos (-9\pi )+\cos (-10\pi )=\cos \left(9\pi \right)+\cos \left(10\pi \right)=\cos (8\pi +\pi )+\cos \left(10\pi \right)=\cos \left(\pi \right)+\cos \left(0\right)=-1+1=0$