$\begin{aligned}
& f(x)=6+16 \cos x \cdot \cos \left(\frac{\pi}{3}-x\right) \
& \qquad \cos \left(\frac{\pi}{3}+x\right) \cdot \sin 3 x \cdot \cos 6 x \
& f(x)=6+4 \cos 3 x \cdot \sin 3 x \cdot \cos 6 x \
& \therefore \quad f(x)=6+\sin 12 x \
& \therefore \quad \text { Range of } f(x)=[5,7] \
& \therefore \quad[\alpha, \beta]=[5,7]
\end{aligned}\thereforeDistanceofpointfrom3 x+4 y+12=0\begin{aligned}
& =\left|\frac{3.5+4.7+12}{\sqrt{3^2+4^2}}\right| \
& =11 \text { units }
\end{aligned}$
Let the range of the function f(x)=6+16 cos x · cos ( π 3-x ) · cos ( π 3+x ) · sin 3 x · cos 6 x, x ∈ R be [α, β]. Then the distance of the point…
JEE Main 2025 — Mathematics Algebra
Let the range of the function f(x)=6+16cosx⋅cos(3π−x)⋅cos(3π+x)⋅sin3x⋅cos6x,x∈R be [α,β]. Then the distance of the point (α,β) from the line 3x+4y+12=0 is :
Held on 23 Jan 2025 · Verified 6 Jul 2026.
Options
- A
11
- B
8
- C
10
- D
9
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