Question

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

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

Answer 1

Answer: (A)

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

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

we know that $\pi =3.14\Rightarrow {\pi }^2=9.0595$

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

and $[-\pi {]}^2=-10$

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

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

$=\cos (2\pi +\frac{\pi }{4})+\cos (2\pi +\frac{\pi }{2})$

$=\cos \frac{\pi }{4}+\cos \frac{\pi }{2}=\frac{1}{\sqrt{2}}$

$f(-\pi )=\cos (-9\pi )+\cos (-10\pi )$

$=\cos 9\pi +\cos 10\pi$

$=-1+1=0$