Mathematics - Functions Question with Solution | TestHub

$9<{\pi }^2<10\Rightarrow \left[{\pi }^2\right]=9$

$$\begin{array}{rl}& -10<-{\pi }^2<-9\Rightarrow \left[-{\pi }^2\right]=-10\\ & f(x)=\cos 9x+\cos (-10)x=\cos 9x+\cos 10x\\ & f\left(\frac{\pi }{2}\right)=\cos 5\pi =-1f(\pi )=0f(-\pi )=0f\left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}\end{array}$$