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JEE MAIN 2025
23-01-2025 SHIFT-2
Question
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 :
Select the correct option:
A
9
B
11
C
8
D
10
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$f(x)=6+16\left(\frac{1}{4} \cos 3 x\right) \sin 3 x \cdot \cos 6 x$
$ =6+4 \cos 3 x \sin 3 x \cos 6 x $
$ =6+\sin 12 x $
Range of $\mathrm{f}(\mathrm{x})$ is [5,7]
$(\alpha, \beta) \equiv(5,7)$
distance $=\left|\frac{15+28+12}{5}\right|=11$
Question Tags
JEE Main
Mathematics
Easy
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