Trigonometry – Max and Min Values – JEE Main 23 Jan 2026 Shift 1

Solution:

Simplify the trigonometric terms:
1) $\sin\left(\frac{7\pi}{2}-\theta\right) = \sin\left(3\pi + \frac{\pi}{2} – \theta\right) = -\cos\theta \Rightarrow \sin^4(…) = \cos^4\theta$.
2) $\sin(11\pi+\theta) = \sin(\pi+\theta) = -\sin\theta \Rightarrow \sin^4(…) = \sin^4\theta$.
3) $\sin\left(\frac{3\pi}{2}-\theta\right) = -\cos\theta \Rightarrow \sin^6(…) = \cos^6\theta$.
4) $\sin(9\pi-\theta) = \sin(\pi-\theta) = \sin\theta \Rightarrow \sin^6(…) = \sin^6\theta$.

Substitute these into $f(\theta)$:
$$ f(\theta) = 4(\cos^4\theta + \sin^4\theta) – 2(\cos^6\theta + \sin^6\theta) $$

Use the standard identities:
$a^4+b^4 = (a^2+b^2)^2 – 2a^2b^2 = 1 – 2\sin^2\theta\cos^2\theta$
$a^6+b^6 = (a^2+b^2)^3 – 3a^2b^2(a^2+b^2) = 1 – 3\sin^2\theta\cos^2\theta$

$$ f(\theta) = 4(1 – 2\sin^2\theta\cos^2\theta) – 2(1 – 3\sin^2\theta\cos^2\theta) $$

$$ f(\theta) = 4 – 8\sin^2\theta\cos^2\theta – 2 + 6\sin^2\theta\cos^2\theta $$

$$ f(\theta) = 2 – 2\sin^2\theta\cos^2\theta $$

Simplify using double angle $\sin 2\theta = 2\sin\theta\cos\theta$:
$$ f(\theta) = 2 – \frac{1}{2}(4\sin^2\theta\cos^2\theta) = 2 – \frac{1}{2}\sin^2(2\theta) $$

Find Max ($\alpha$) and Min ($\beta$):
Range of $\sin^2(2\theta)$ is $[0, 1]$.
Max value ($\alpha$): occurs when $\sin^2(2\theta) = 0$.
$$ \alpha = 2 – 0 = 2 $$

Min value ($\beta$): occurs when $\sin^2(2\theta) = 1$.
$$ \beta = 2 – \frac{1}{2} = \frac{3}{2} $$

Calculate $\alpha + 2\beta$:
$$ \alpha + 2\beta = 2 + 2\left(\frac{3}{2}\right) = 2 + 3 = 5 $$

Ans. (2)