Question

The fundamental period of the function $f(x)=4{\cos }^4\left(\frac{x-\pi }{4{\pi }^2}\right)-2\cos \left(\frac{x-\pi }{2{\pi }^2}\right)$ is equal to

• A. ${\pi }^3$
• B. $4{\pi }^2$
• C. $3{\pi }^2$
• D. $2{\pi }^3$

Answer 1

Answer: (D)

We have, $f(x)=4{\cos }^4\left(\frac{x-\pi }{4{\pi }^2}\right)-2\cos \left(\frac{x-\pi }{2{\pi }^2}\right)$
[13th, 31-07-2011, P-1]
$={\left(2{\cos }^2\left(\frac{x-\pi }{4{\pi }^2}\right)\right)}^2-2\cos \left(\frac{x-\pi }{2{\pi }^2}\right)={\left(1+\cos \left(\frac{x-\pi }{2{\pi }^2}\right)\right)}^2-2\cos \left(\frac{x-\pi }{2{\pi }^2}\right)=1+{\cos }^2\left(\frac{x-\pi }{2{\pi }^2}\right)$
Clearly, period of $f(x)=\frac{\pi }{\frac{1}{2{\pi }^2}}=2{\pi }^3$.