Let $\alpha$ and $\beta$ respectively be the maximum and the minimum values of the function $$f(\theta)=4\left(\sin ^{4}\left(\frac{7 \pi}{2}-\theta\right)+\sin ^{4}(11 \pi+\theta)\right)-2\left(\sin ^{6}\left(\frac{3 \pi}{2}-\theta\right)+\sin ^{6}(9 \pi-\theta)\right), \theta \in \mathbf{R}$$ Then $\alpha+2 \beta$ is equal to:

JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Trigonometric Ratios and Identities

Let

α

and

β

respectively be the maximum and the minimum values of the function

f (θ) = 4 (sin^{4} (\frac{7 π}{2} - θ) + sin^{4} (11 π + θ)) - 2 (sin^{6} (\frac{3 π}{2} - θ) + sin^{6} (9 π - θ)), θ \in R

Then

α + 2 β

is equal to:

Answer: (B)

Simplifying using periodicity and co-function identities:

sin (\frac{7 π}{2} - θ) = cos θ

sin (11 π + θ) = - sin θ

sin (\frac{3 π}{2} - θ) = - cos θ

sin (9 π - θ) = sin θ

.

So

f (θ) = 4 (cos^{4} θ + sin^{4} θ) - 2 (cos^{6} θ + sin^{6} θ)

.

Using

cos^{4} θ + sin^{4} θ = 1 - 2 sin^{2} θ cos^{2} θ

and

cos^{6} θ + sin^{6} θ = 1 - 3 sin^{2} θ cos^{2} θ

f (θ) = 4 (1 - 2 sin^{2} θ cos^{2} θ) - 2 (1 - 3 sin^{2} θ cos^{2} θ) = 2 - 2 sin^{2} θ cos^{2} θ = 2 - \frac{s i n ^{2} ( 2 θ )}{2}

.

Maximum

α = 2

(when

sin^{2} (2 θ) = 0

), minimum

β = \frac{3}{2}

(when

sin^{2} (2 θ) = 1

).

Thus

α + 2 β = 2 + 3 = 5

JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Single correct option