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JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Trigonometric Ratios and Identities

Let

f (θ) = 3 (sin^{4} (\frac{3 π}{2} - θ) + sin^{4} (3 π + θ)) - 2 (1 - sin^{2} 2 θ)

and

S = {θ \in [0, π] : f^{'} (θ) = - \frac{3}{2}}

. If

4 β = \sum_{θ \in S} θ

, then

f (β)

is equal to

(A) $\frac{11}{8}$
(B) $\frac{5}{4}$ ✓ Correct
(C) $\frac{9}{8}$
(D) $\frac{3}{2}$

Answer: (B)

Solution

Using

sin (\frac{3 π}{2} - θ) = - cos θ

and

sin (3 π + θ) = - sin θ

, together with

1 - sin^{2} 2 θ = cos^{2} 2 θ

,

f (θ) = 3 (cos^{4} θ + sin^{4} θ) - 2 cos^{2} 2 θ = 3 (1 - \frac{1}{2} sin^{2} 2 θ) - 2 cos^{2} 2 θ = \frac{5}{4} - \frac{cos 4 θ}{4},

so

f^{'} (θ) = sin 4 θ

. On

[0, π]

the equation

sin 4 θ = - \frac{3}{2}

has roots

θ = \frac{π}{3}, \frac{5 π}{12}, \frac{5 π}{6}, \frac{11 π}{12}

, whose sum is

\frac{5 π}{2}

; with

4 β

equal to this sum,

β = \frac{5 π}{8}

. Since

cos (4 \cdot \frac{5 π}{8}) = cos \frac{5 π}{2} = 0

, the value is

f (β) = \frac{5}{4}

.

JEE Main · 2023 · 29 Jan 2023 · Shift 1 · Single correct option

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