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EAMCET Engineering 2005 Solved Paper

Section: Mathematics

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Question : 104 of 160

Marks: +1, -0

The extreme values of

4 \cos (x^2) \cos \left( \; \frac{\pi}{3}+x^2 \right) \cos \left( \; \frac{\pi}{3}-x^2 \right)

R

$-1,1$
$-2,2$
$-3,3$
$-4,4$

Solution:

Let

f(x)=4 \cos (x^2) \cos \left( \; \frac{\pi}{3}+x^2 \right) \cos \left( \; \frac{\pi}{3}-x^2 \right)

= \; 2 \cos (x^2) \left[ \cos \left( \; \frac{2\pi}{3} \right) + \cos (2x^2) \right]

\; \left[ \because 2 \cos A \cos B = \cos (A+B) + \cos (A-B) \right]

= \; 2 \cos (x^2) \left[ -\; \frac{1}{2} + \cos (2x^2) \right]

= \; -\cos (x^2) + 2 \cos (x^2) \cos (2x^2)

= \; -\cos (x^2) + \cos (3x^2) + \cos (x^2)

\Rightarrow f(x)= \; \cos (3x^2)

. . . (i)

On differentiating w.r.t.

x

, we get

f'(x) = -\sin \left( \; 3x^2 \right) (6x)

For extremum, put

f'(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

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