Question

The extreme values of $4 \cos \left(x^{2}\right) \cos \left(\frac{\pi}{3}+x^{2}\right) \cos \left(\frac{\pi}{3}-x^{2}\right)$ over $R$, are

• A. -1.1
• B. -22
• C. -3,3
• D. -4.4

Answer 1

Answer: (A)

Let $f(x)=4 \cos \left(x^{2}\right) \cos \left(\frac{\pi}{3}+x^{2}\right) \cos \left(\frac{\pi}{3}-x^{2}\right)$
$= 2 \cos \left(x^{2}\right)\left[\cos \left(\frac{2 \pi}{3}\right)+\cos \left(2 x^{2}\right)\right] $
$[\because 2 \cos A \cos B=\cos (A+B)+\cos (A-B)] $
$= 2 \cos \left(x^{2}\right)\left[-\frac{1}{2}+\cos \left(2 x^{2}\right)\right] $
$=-\cos \left(x^{2}\right)+2 \cos \left(x^{2}\right) \cos \left(2 x^{2}\right) $
$=-\cos \left(x^{2}\right)+\cos \left(3 x^{2}\right)+\cos \left(x^{2}\right)$
$\Rightarrow f(x)=\cos \left(3 x^{2}\right)\,\,\,...(i)$
On differentiating w.r.t. $x$, we get
$f^{'}(x)=-\sin \left(3 x^{2}\right)(6 x)$
For extremum, put $f^{'}(x)=0$
$\Rightarrow -\sin \left(3 x^{2}\right)(6 x)=0$
$\Rightarrow x=0, \pi$
Put $x=0, \pi$, in equation (i), we get
$f(0)=\cos (0)=1 $
$f(\pi)=\cos (\pi)=-1$