The extreme values of $4 \cos \left(x^2\right) \cos \left(\frac{\pi}{3}+x^2\right) \cos…
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
- −1, 1
- − 2 2,
- − 3, 3
- − 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
$\begin{aligned}
f(0) & =\cos (0)=1 \\
f(\pi) & =\cos (\pi)=-1
\end{aligned}$
Asked in: AP EAMCET 2005
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