JEE MAIN - Mathematics (2025 - 23rd January Evening Shift - No. 9)
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 :
11
10
8
9
Explanation
$$\begin{aligned}
&\begin{aligned}
f(x) & =6+16\left(\frac{1}{4} \cos 3 x\right) \sin 3 x \cdot \cos 6 x \\
& =6+4 \cos 3 x \sin 3 x \cos 6 x \\
& =6+\sin 12 x
\end{aligned}\\
&\text { Range of } f(x) \text { is [5, 7] }\\
&\begin{aligned}
& (\alpha, \beta) \equiv(5,7) \\
& \text { distance }=\left|\frac{15+28+12}{5}\right|=11
\end{aligned}
\end{aligned}$$
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