JEE Main 2026 Jan 23 Shift 2, Mathematics Q13: Trigonometric Functions

JEE Main2026Jan 23, Shift 2Mathematics

The least value of $(cos^{2} θ - 6 sin θ cos θ + 3 sin^{2} θ + 2)$ is

Solution

Use $cos^{2} θ = \frac{1 + c o s 2 θ}{2}$ , $sin^{2} θ = \frac{1 - c o s 2 θ}{2}$ , $2 sin θ cos θ = sin 2 θ$ .

$f (θ) = \frac{1 + cos 2 θ}{2} - 3 sin 2 θ + 3 \cdot \frac{1 - cos 2 θ}{2} + 2 = 4 - 3 sin 2 θ - cos 2 θ$ .

$3 sin 2 θ + cos 2 θ$ has amplitude $9 + 1 = 10$ , so it lies in $[- 10, 10]$ .

$∴ f (θ) \in [4 - 10, 4 + 10]$ . Least value $= 4 - 10$ .

Concept behind this question

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