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 $f(x)=4\cos \left(x^2\right)\cos \left(\frac{\pi }{3}+x^2\right)\cos \left(\frac{\pi }{3}-x^2\right)$
$=2\cos \left(x^2\right)\left[\cos \left(\frac{2\pi }{3}\right)+\cos \left(2x^2\right)\right]$
$[\because 2\cos A\cos B=\cos (A+B)+\cos (A-B)]$
$=2\cos \left(x^2\right)\left[-\frac{1}{2}+\cos \left(2x^2\right)\right]$
$=-\cos \left(x^2\right)+2\cos \left(x^2\right)\cos \left(2x^2\right)$
$=-\cos \left(x^2\right)+\cos \left(3x^2\right)+\cos \left(x^2\right)$
$\Rightarrow f(x)=\cos \left(3x^2\right)...(i)$
On differentiating w.r.t. $x$, we get
$f^{{}^{\prime }}(x)=-\sin \left(3x^2\right)(6x)$
For extremum, put $f^{{}^{\prime }}(x)=0$
$\Rightarrow -\sin \left(3x^2\right)(6x)=0$
$\Rightarrow x=0,\pi$
Put $x=0,\pi$, in equation (i), we get
$f(0)=\cos (0)=1$
$f(\pi )=\cos (\pi )=-1$