Let the range of the function f(x)=6+16 cos x × cos (π/3-x ) × cos (π/3+x ) sin 3 x × cos 6 x, x in R be [α, β] . Then the distance of the point (α, β) from the line 3 x+4 y+12=0 is :

Mathematics · Functions

JEE Main 2025 — 23 January, Evening Shift — Question 12

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 :

$f(x)=6+16\left(\frac{1}{4} \cos 3 x\right) \sin 3 x \cdot \cos 6 x$

\begin{aligned} & =6+4 \cos 3 x \sin 3 x \cos 6 x \\& =6+\sin 12 x \end{aligned}

Range of $f(x)$ is $[5,7]$

$(\alpha, \beta) \equiv(5,7)$

distance $=\left|\frac{15+28+12}{5}\right|=11$

Answer key and solution verified before publishing.

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