Trigonometric Functions – Maximum and Minimum Values – JEE Main 23 Jan 2026 Shift 2

Solution:

Let $f(\theta) = \cos^2\theta – 6\sin\theta\cos\theta + 3\sin^2\theta + 2$.

Using double angle formulas $\cos^2\theta = \frac{1+\cos2\theta}{2}$ and $\sin^2\theta = \frac{1-\cos2\theta}{2}$:
$$ f(\theta) = \frac{1+\cos2\theta}{2} – 3\sin2\theta + 3\left(\frac{1-\cos2\theta}{2}\right) + 2 $$

$$ f(\theta) = \frac{1}{2} + \frac{1}{2}\cos2\theta – 3\sin2\theta + \frac{3}{2} – \frac{3}{2}\cos2\theta + 2 $$

$$ f(\theta) = ( \frac{1}{2} + \frac{3}{2} + 2 ) – 3\sin2\theta – \cos2\theta $$

$$ f(\theta) = 4 – (3\sin2\theta + \cos2\theta) $$

The range of $a\sin x + b\cos x$ is $[-\sqrt{a^2+b^2}, \sqrt{a^2+b^2}]$.
Here $a=3, b=1$.
$$ \sqrt{3^2 + 1^2} = \sqrt{10} $$

So, $-\sqrt{10} \le 3\sin2\theta + \cos2\theta \le \sqrt{10}$.

Minimum value of $f(\theta)$ occurs when $(3\sin2\theta + \cos2\theta)$ is maximum:
$$ f(\theta)_{min} = 4 – \sqrt{10} $$

Ans. (3)