Q36 / 75 · JEE Main 2025 (23 Jan Shift 2) · Mathematics

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

  • Correct answer
    A
    11
  • B
    8
  • C
    10
  • D
    9

Solution

$\begin{aligned} & f(x)=6+16 \cos x \cdot \cos \left(\frac{\pi}{3}-x\right) \\ & \qquad \cos \left(\frac{\pi}{3}+x\right) \cdot \sin 3 x \cdot \cos 6 x \\ & f(x)=6+4 \cos 3 x \cdot \sin 3 x \cdot \cos 6 x \\ & \therefore \quad f(x)=6+\sin 12 x \\ & \therefore \quad \text { Range of } f(x)=[5,7] \\ & \therefore \quad[\alpha, \beta]=[5,7] \end{aligned}$ $\therefore$ Distance of point from $3 x+4 y+12=0$ $\begin{aligned} & =\left|\frac{3.5+4.7+12}{\sqrt{3^2+4^2}}\right| \\ & =11 \text { units } \end{aligned}$

Hence, the correct answer is option: 'A'.

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