JEE Mathematics 2026 Question with Solution

Given: Find the least value of $f(\theta) = \cos^2 \theta - 6\sin \theta \cos \theta + 3\sin^2 \theta + 2$.

Find: The minimum value of $f(\theta)$.

Convert the squared terms to double-angle form:

$$\cos^2\theta = \frac{1+\cos2\theta}{2}, \qquad \sin^2\theta = \frac{1-\cos2\theta}{2}$$

Also,

$$2\sin\theta\cos\theta = \sin2\theta$$

So,

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

Expanding,

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

Combine the constants and coefficients:

$$\frac{1}{2} + \frac{3}{2} + 2 = 4, \qquad \frac{1}{2} - \frac{3}{2} = -1$$

Hence,

$$f(\theta) = 4 - \cos2\theta - 3\sin2\theta$$

Now use the result that the range of $a\cos x + b\sin x$ is from $-\sqrt{a^2+b^2}$ to $\sqrt{a^2+b^2}$. Therefore,

$$-\cos2\theta - 3\sin2\theta \in \left[-\sqrt{(-1)^2+(-3)^2}, \sqrt{(-1)^2+(-3)^2}\right] = [-\sqrt{10}, \sqrt{10}]$$

So the minimum value of $f(\theta)$ is

$$4 - \sqrt{10}$$

Therefore, the correct option is C.

Given: $\cos^2 \theta - 6\sin \theta \cos \theta + 3\sin^2 \theta + 2$

Find: Its least value.

After converting to double angles, the expression becomes

$$4 - \cos2\theta - 3\sin2\theta$$

For any expression of the form $a\cos x + b\sin x + c$, the minimum value is

$$c - \sqrt{a^2+b^2}$$

Here, $a = -1$, $b = -3$, and $c = 4$. Hence,

$$\min f(\theta) = 4 - \sqrt{(-1)^2+(-3)^2} = 4 - \sqrt{10}$$

Therefore, the correct option is C.