The extreme values of 4 cos left(x^2right) co | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $f(x) = 4 \cos \left(x^2\right) \cos \left(\frac{\pi}{3}+x^2\right) \cos \left(\frac{\pi}{3}-x^2\right)$.Using the identity $2 \cos A \cos B =

Vedclass · Accepted Answer

$-1, 1$

Vedclass · Answer

(A) Let $f(x) = 4 \cos \left(x^2ight) \cos \left(\frac{\pi}{3}+x^2ight) \cos \left(\frac{\pi}{3}-x^2ight)$.Using the identity $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$,we have:$f(x) = 2 \cos \left(x^2ight) \left[ 2 \cos \left(\frac{\pi}{3}+x^2ight) \cos \left(\frac{\pi}{3}-x^2ight) ight]$$f(x) = 2 \cos \left(x^2ight) \left[ \cos \left(\frac{2\pi}{3}ight) + \cos \left(2x^2ight) ight]$Since $\cos \left(\frac{2\pi}{3}ight) = -\frac{1}{2}$,we get:$f(x) = 2 \cos \left(x^2ight) \left[ -\frac{1}{2} + \cos \left(2x^2ight) ight]$$f(x) = -\cos \left(x^2ight) + 2 \cos \left(x^2ight) \cos \left(2x^2ight)$Using $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$ again:$f(x) = -\cos \left(x^2ight) + \cos \left(3x^2ight) + \cos \left(x^2ight)$$f(x) = \cos \left(3x^2ight) \quad \dots (i)$Since the range of $\cos(	heta)$ is $[-1, 1]$,the extreme values of $f(x) = \cos \left(3x^2ight)$ are $-1$ and $1$.

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 $\mathbb{R}$ are

Similar Questions

$\operatorname{coth}^2 x - \tanh^2 x =$

The value of $\frac{\sin^2 \theta}{\sin \theta - \cos \theta} - \frac{\sin \theta + \cos \theta}{\tan^2 \theta - 1}$ for all permissible values of $\theta$ is:

Let the range of the function $f(x) = 6 + 16 \cos x \cdot \cos \left(\frac{\pi}{3} - x\right) \cdot \cos \left(\frac{\pi}{3} + x\right) \sin 3x \cdot \cos 6x$,where $x \in R$,be $[\alpha, \beta]$. Then the distance of the point $(\alpha, \beta)$ from the line $3x + 4y + 12 = 0$ is:

If $\sum_{r=1}^{50} \tan ^{-1} \frac{1}{2 r^2}=p$,then $\tan p$ is

$\tan 9^{\circ}-\tan 27^{\circ}-\tan 63^{\circ}+\tan 81^{\circ}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module