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 :

Question

Aicharya · Accepted Answer

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

Let the range of the function $f(x)=6+16 \cos x \cdot \cos \left(\frac{\pi}{3}-x\right) \cdot \cos…

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