JEE Advance - Mathematics (2017 - Paper 2 Offline - No. 14)
If $$f(x) = \left| {\matrix{
{\cos 2x} & {\cos 2x} & {\sin 2x} \cr
{ - \cos x} & {\cos x} & { - \sin x} \cr
{\sin x} & {\sin x} & {\cos x} \cr
} } \right|$$,
then
then
f(x) attains its minimum at x = 0
f(x) attains its maximum at x = 0
f'(x) = 0 at more than three points in ($$-$$$$\pi $$, $$\pi $$)
f'(x) = 0 at exactly three points in ($$-$$$$\pi $$, $$\pi $$)
Explanation
$$f(x) = \left| {\matrix{
{\cos 2x} & {\cos 2x} & {\sin 2x} \cr
{ - \cos x} & {\cos x} & { - \sin x} \cr
{\sin x} & {\sin x} & {\cos x} \cr
} } \right|$$
$$\cos 2x({\cos ^2}x + {\sin ^2}x) - \cos 2x( - {\cos ^2}x + {\sin ^2}x) + \sin 2x( - \sin 2x)$$
$$ = \cos 2x + \cos 4x$$
$$f'(x) = - 2\sin 2x - 4\sin 4x$$
$$ = - 2\sin 2x(1 + 4\cos 2x)$$
At x = 0
f'(x) = 0 and f(x) = 2
Also, f'(x) = 0 sin2x = 0 or $$\cos 2x = {{ - 1} \over 4}$$
$$ \Rightarrow x = {{n\pi } \over 2}$$ or $$\cos 2x = - {1 \over 4}$$
$$\cos 2x({\cos ^2}x + {\sin ^2}x) - \cos 2x( - {\cos ^2}x + {\sin ^2}x) + \sin 2x( - \sin 2x)$$
$$ = \cos 2x + \cos 4x$$
$$f'(x) = - 2\sin 2x - 4\sin 4x$$
$$ = - 2\sin 2x(1 + 4\cos 2x)$$
At x = 0
f'(x) = 0 and f(x) = 2
Also, f'(x) = 0 sin2x = 0 or $$\cos 2x = {{ - 1} \over 4}$$
$$ \Rightarrow x = {{n\pi } \over 2}$$ or $$\cos 2x = - {1 \over 4}$$
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