Let the range of the function $f(x)=6+16\cos x\cdot\cos\left(\frac\pi3-x\right)\cdot\cos\left(\frac\pi3+x\right)\cdot\sin3x\cdot\cos6x$, $x\in\mathbf R$ be $[\alpha,\beta]$. Then the distance of the point $(\alpha,\beta)$ from the line $3x+4y+12=0$ is :

Question

Dr. Abubacker · Accepted Answer

[Solution by Dr. Abubacker] $\cos x\cos\left(\frac\pi3-x\right)\cos\left(\frac\pi3+x\right)=\frac{\cos3x}4$, so $f=6+4\cos3x\sin3x\cos6x=6+2\sin6x\cos6x=6+\sin12x$. Range $[5,7]$, so $(\alpha,\beta)=(5,7)$. Distance $=\frac{|15+28+12|}5=11$. Answer (1).

Q. Let the range of the function $f(x)=6+16\cos x\cdot\cos\left(\frac\pi3-x\right)\cdot\cos\left(\frac\pi3+x\right)\cdot\sin3x\cdot\cos6x$, $x\in\mathbf R$ be $[\alpha,\beta]$. Then the distance of the point $(\alpha,\beta)$ from the line $3x+4y+12=0$ is :

Explanation

Reviewed by Dr. Abubacker

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