The least value of (² θ - 6 \;θ θ + 3 ² θ + 2) is [23 jan 2026 Shift 2]

Correct Answer is : 4−104 - {10}4−10

Correct Answer is : 4−104 - {10}4−10

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Application of Derivatives Part 3

Section: Mathematics

Question : 7 of 34

Marks: +1, -0

The least value of

(\cos^2 \theta - 6 \sin\;\theta \cos \theta + 3 \sin^2 \theta + 2)

Solution:

Concept:

Express the given trigonometric expression in the form

A - (a \cos 2\theta + b \sin 2\theta)

and use the known range of

a \cos x + b \sin x

Explanation:

We start with

f(\theta) = \cos^2 \theta - 6 \sin\theta \cos\theta + 3 \sin^2 \theta + 2

Use identities:

\cos^2\theta = \frac{1+\cos2\theta}{2}

\sin^2\theta = \frac{1-\cos2\theta}{2}

, and

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

Thus

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

Simplify:

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

Combine constants:

\frac{1}{2}+\frac{3}{2}+2 = 4

Combine cosine terms:

\frac{\cos2\theta}{2} - \frac{3\cos2\theta}{2} = -\cos2\theta

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

Let

g(\theta) = \cos2\theta + 3\sin2\theta

. Its range is

\left[-\sqrt{1^2+3^2}, \sqrt{1^2+3^2}\right] = [-\sqrt{10}, \sqrt{10}]

For the least value of

f(\theta) = 4 - g(\theta)

, we need the maximum of

g(\theta)

, which is

\sqrt{10}

Hence least value is

4 - \sqrt{10}

Answer:

4 - \sqrt{10}

(Option A).

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