Question

If $f\left(x\right)=\cos \left[{\pi }^2\right]x+\cos \left([-{\pi }^2]x\right)$, where $[.]$ denotes the step function, then

• A. $f\left(0\right)=1$
• B. $f\left(\frac{\pi }{4}\right)=2$
• C. $f\left(\frac{\pi }{2}\right)=-1$
• D. $f\left(\pi \right)=1$

Answer 1

The correct option is C $f\left(\frac{\pi }{2}\right)=-1$

${\pi }^2=9.87$

$\Rightarrow$$9<{\pi }^2<10$ and $-10<-{\pi }^2<-9$

$\therefore \left[{\pi }^2\right]=9$ and $[-{\pi }^2]=-10$

$\therefore f\left(x\right)=\cos 9x+\cos 10x$

$\therefore f\left(0\right)=1+1=2$

$\Rightarrow$$f\left(\frac{\pi }{4}\right)=\cos 9\times \frac{\pi }{4}+\cos 10\times \frac{\pi }{4}$

$\Rightarrow$$f\left(\frac{\pi }{4}\right)$$=-\frac{1}{\sqrt{2}}$

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

$\Rightarrow$$f\left(\frac{\pi }{2}\right)=$$-1$

$\Rightarrow$$f\left(\pi \right)=0$

Hence, option 'C' is correct.