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)$$\sin 3 x \cdot \cos 6 x, x \in R$ be $[\alpha, \beta]$. Then the distance of the point $(\alpha, \beta)$ from the line $3 x+4 y+12=0$ is : [JEE Main 2025, 23 Jan (Shift 2)]

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YESPYQ · Accepted Answer

Correct answer: a) 11. $\text{Since,}\cos x.\cos \left(\frac{\pi }{3}-x\right).\cos \left(\frac{\pi }{3}+x\right)\ =\frac{1}{4}\cos 3x\ f\left(x\right)=6+16\left(\frac{1}{4}\cos 3x\right)\sin 3x\cdot \cos 6x\ =6+4\cos 3x\sin 3x\cos 6x\ =6+\sin 12x\ \text{Since, }\sin \text{12x ∈ }\left[-1,1\right]\ f\left(x\right)\text{∈ }\left[5,7\right]\ \left(\alpha ,\beta \right)\equiv \left(5,7\right)\ \text{ distance }=\left|\frac{15+28+12}{5}\right|=11$

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}+…

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