Trigonometry – Range – JEE Main 2025 Shift 2

Solution:

We use the trigonometric identity:
$$4 \cos x \cos(60^\circ – x) \cos(60^\circ + x) = \cos 3x$$

The given function is:
$$f(x) = 6 + 16 \cos x \cos\left(\frac{\pi}{3}-x\right) \cos\left(\frac{\pi}{3}+x\right) \sin 3x \cos 6x$$
$$f(x) = 6 + 4 \left[ 4 \cos x \cos\left(\frac{\pi}{3}-x\right) \cos\left(\frac{\pi}{3}+x\right) \right] \sin 3x \cos 6x$$

Substitute the identity:
$$f(x) = 6 + 4 (\cos 3x) \sin 3x \cos 6x$$
$$f(x) = 6 + 2 (2 \sin 3x \cos 3x) \cos 6x$$

Using $\sin 2A = 2 \sin A \cos A$:
$$f(x) = 6 + 2 (\sin 6x) \cos 6x$$
$$f(x) = 6 + \sin 12x$$

The range of $\sin \theta$ is $[-1, 1]$.
Therefore, the range of $f(x)$ is $[6-1, 6+1] = [5, 7]$.
So, $[\alpha, \beta] = [5, 7] \Rightarrow \alpha = 5, \beta = 7$.

We need to find the perpendicular distance of point $(5, 7)$ from the line $3x + 4y + 12 = 0$.
$$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$$
$$d = \frac{|3(5) + 4(7) + 12|}{\sqrt{3^2 + 4^2}}$$
$$d = \frac{|15 + 28 + 12|}{\sqrt{9 + 16}}$$
$$d = \frac{|55|}{\sqrt{25}} = \frac{55}{5} = 11$$

Ans. (1)