JEE Main 2026 — 23 January, Evening Shift — Question 13

Let $f(\theta) = \cos^2 \theta - 6 \sin \theta \cos \theta + 3 \sin^2 \theta + 2$. Use identities: $\cos^2 \theta = \frac{1+\cos 2\theta}{2}$, $\sin^2 \theta = \frac{1-\cos 2\theta}{2}$, $\sin \theta \cos \theta = \frac{\sin 2\theta}{2}$. Then $f(\theta) = \frac{1+\cos 2\theta}{2} - 6\cdot\frac{\sin 2\theta}{2} + 3\cdot\frac{1-\cos 2\theta}{2} + 2$. Simplify: $f(\theta) = \frac{1+\cos 2\theta - 6\sin 2\theta + 3 - 3\cos 2\theta}{2} + 2 = \frac{4 - 6\sin 2\theta - 2\cos 2\theta}{2} + 2$. Thus $f(\theta) = 2 - 3\sin 2\theta - \cos 2\theta + 2 = 4 - 3\sin 2\theta - \cos 2\theta$. Write as $f(\theta) = 4 - (3\sin 2\theta + \cos 2\theta)$.

The expression $3\sin 2\theta + \cos 2\theta$ has range $[-\sqrt{10}, \sqrt{10}]$. Hence $f(\theta) \in [4-\sqrt{10}, 4+\sqrt{10}]$.

The least value is $4-\sqrt{10}$.