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JEE Main 202523 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
23 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Functions

JEE Main 2025 · 23 January · Evening Shift · Mathematics · Functions

Let the range of the function f(x)=6+16 x · ((π)/(3)-x) ·...

Let the range of the function f(x)=6+16cos⁡x⋅cos⁡(π3−x)⋅cos⁡(π3+x)⋅sin⁡3x⋅cos⁡6x,x∈Rf(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 [α,β][\alpha, \beta]. Then the distance of the point (α,β)(\alpha, \beta) from the line 3x+4y+12=03 x+4 y+12=0 is :

Options

  1. A11
  2. B8
  3. C10
  4. D9

Correct answer

A. 11

Step-by-step solution

f(x)=6+16cos⁡x⋅cos⁡(π3−x)cos⁡(π3+x)⋅sin⁡3x⋅cos⁡6xf(x)=6+4cos⁡3x⋅sin⁡3x⋅cos⁡6x∴f(x)=6+sin⁡12x∴ Range of  f(x)=[5,7]∴[α,β]=[5,7]\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\text{ of } } f(x)=[5,7] \\ & \therefore \quad[\alpha, \beta]=[5,7] \end{aligned}\thereforeDistanceofpointfromDistance of point from3 x+4 y+12=0 =∣3.5+4.7+1232+42∣=11 units \begin{aligned} & =\left|\frac{3.5+4.7+12}{\sqrt{3^2+4^2}}\right| \\ & =11 \text{ units } \end{aligned}$
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