JEE Main 2025 Jan 23 Shift 2, Mathematics Q12: Trigonometric Functions

JEE Main2025Jan 23, Shift 2Mathematics

Let the range of the function $f (x) = 6 + 16 cos x \cdot cos (\frac{π}{3} - x) \cdot cos (\frac{π}{3} + x) \cdot sin 3 x \cdot cos 6 x$ , $x \in R$ be $[α, β]$ . Then the distance of the point $(α, β)$ from the line $3 x + 4 y + 12 = 0$ is :

A
11
B
8
C
10
D
9

Solution

Use the identity $cos x \cdot cos (\frac{π}{3} - x) \cdot cos (\frac{π}{3} + x) = \frac{1}{4} cos 3 x$ .

So $f (x) = 6 + 16 \cdot \frac{1}{4} cos 3 x \cdot sin 3 x \cdot cos 6 x = 6 + 4 sin 3 x cos 3 x \cdot cos 6 x$ .

Using $2 sin 3 x cos 3 x = sin 6 x$ : $f (x) = 6 + 2 sin 6 x cos 6 x = 6 + sin 12 x$ .

$sin 12 x \in [- 1, 1]$ , so the range is $[α, β] = [5, 7]$ , i.e. $(α, β) = (5, 7)$ .

Distance from $(5, 7)$ to $3 x + 4 y + 12 = 0$ :

$d = \frac{∣3 ( 5 ) + 4 ( 7 ) + 12∣}{9 + 16} = \frac{∣15 + 28 + 12∣}{5} = \frac{55}{5} = 11.$

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JEE Main 2025 Jan 23 Shift 2, Mathematics Q12: Trigonometric Functions

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