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JEE Main 2024
Trigonometry
Trigonometric Ratio and Identites
Medium

Question

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 :

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Solution

f(x)=6+16(14cos⁡3x)sin⁡3x⋅cos⁡6x=6+4cos⁡3xsin⁡3xcos⁡6x=6+sin⁡12x Range of f(x) is [5, 7] (α,β)≡(5,7) distance =∣15+28+125∣=11\begin{aligned} &\begin{aligned} 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 \end{aligned}\\ &\text { Range of } f(x) \text { is [5, 7] }\\ &\begin{aligned} & (\alpha, \beta) \equiv(5,7) \\ & \text { distance }=\left|\frac{15+28+12}{5}\right|=11 \end{aligned} \end{aligned}

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