JEE Main 2026 Jan 23 Shift 1, Mathematics Q18: Maxima & Minima

JEE Main2026Jan 23, Shift 1Mathematics

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 :

Solution

Reduce the trigonometric arguments:

$sin (\frac{7 π}{2} - θ) = - cos θ$ , $sin (11 π + θ) = - sin θ$ , $sin (\frac{3 π}{2} - θ) = - cos θ$ , $sin (9 π - θ) = sin θ$ .

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

Use $sin^{4} + cos^{4} = 1 - 2 sin^{2} θ cos^{2} θ$ and $sin^{6} + cos^{6} = 1 - 3 sin^{2} θ cos^{2} θ$ :

$f = 4 (1 - 2 s) - 2 (1 - 3 s) = 2 - 2 s$ where $s = sin^{2} θ cos^{2} θ = \frac{sin ^{2} 2 θ}{4}$ .

$f (θ) = 2 - \frac{sin ^{2} 2 θ}{2}$ . So $α = f_{m a x} = 2$ (at $sin 2 θ = 0$ ); $β = f_{m i n} = \frac{3}{2}$ (at $sin^{2} 2 θ = 1$ ).

$α + 2 β = 2 + 3 = 5$ .

Concept behind this question

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