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

• Correct answer
  A

  −1, 1
• B

  − 2 2,
• C

  − 3, 3
• D

  − 4, 4

Solution

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

Hence, the correct answer is option: 'A'.