Exam

JEE Main

Subject

Mathematics

Chapter

Trigonometric Equations

JEE Main · Mathematics · Trigonometric Equations

The extreme values of 4 (x^2) ((π)/(3)+x^2) ((π)/(3)- x^2) over R,...

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

Step-by-step 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$$\begin{aligned}f(0) & =\cos (0)=1 \\ f(\pi)& =\cos (\pi)=-1 \end{aligned}$$