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JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Determinants

The maximum value of

f (x) = sin^{2} x 1 + sin^{2} x sin^{2} x 1 + cos^{2} x cos^{2} x cos^{2} x cos 2 x cos 2 x sin 2 x, x \in R

is :

(A) $\frac{3}{4}$
(B)5
(C) $5$ ✓ Correct
(D) $7$

Answer: (C)

Solution

Applying

R_{1} \to R_{1} - R_{3}

and

R_{2} \to R_{2} - R_{3}

and writing

k = cos 2 x - sin 2 x

,

f (x) = 01 sin^{2} x 10 cos^{2} x k k sin 2 x

Expanding along the first row,

f (x) = - (sin 2 x - k sin^{2} x) + k cos^{2} x = - sin 2 x + k (sin^{2} x + cos^{2} x) = cos 2 x - 2 sin 2 x

An expression

a cos θ + b sin θ

has greatest value

a^{2} + b^{2}

, so the maximum value of

f

is

1 + 4 = 5

.

JEE Main · 2021 · 16 Mar 2021 · Shift 2 · Single correct option

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