Question

If $f\left(x\right)=\cos \left({\pi }^2\right]x+\cos \left(-{\pi }^2\right]x,$ where [x] stands for the greatest integer function, then

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

Answer 1

Answer: (A), (C)

The correct options are
A $f\left(\frac{\pi }{2}\right)=-1$
C $f(-\pi )=0$

$f\left(x\right)=\cos \left({\pi }^2\right]x+\cos \left(-{\pi }^2\right]x\mathrm{We}\mathrm{know},9<{\pi }^2<10\mathrm{and}-10<-{\pi }^2<-9\left[{\pi }^2\right]=9,[-{\pi }^2]=-10\therefore f\left(x\right)=\cos 9x+\cos (-10x)\Rightarrow f\left(x\right)=\cos \left(9x\right)+\cos \left(10x\right)(\because \cos (-x)=\cos (x\left)\right)\mathrm{If}x=\frac{\pi }{2},f\left(\frac{\pi }{2}\right)=\cos \left(\frac{9\pi }{2}\right)+\cos \left(5\pi \right)=-1\left(\mathrm{True}\right)\mathrm{If}x=\pi ,f\left(\pi \right)=\cos \left(9\pi \right)+\cos \left(10\pi \right)=0\left(\mathrm{False}\right)\mathrm{If}x=-\pi ,f(-\pi )=\cos \left(-9\pi \right)+\cos \left(-10\pi \right)=0\left(\mathrm{True}\right)\mathrm{If}x=\frac{\pi }{4},f\left(\frac{\pi }{4}\right)=\cos \left(\frac{9\pi }{4}\right)+\cos \left(\frac{10\pi }{4}\right)=\frac{1}{\sqrt{2}}\left(\mathrm{False}\right)$