Question

Sum of the maximum and minimum values of $12{\cos }^{2}x-6\sin x\cos x+2{\sin }^{2}x$ is

• A. $0$
• B. $7$
• C. $14$
• D. $15$

Answer 1

The correct option is C $14$

$S=12{\cos }^{2}x-6\sin x\cos x+2{\sin }^{2}x$

$=2({\cos }^{2}x+{\sin }^{2}x)+10{\cos }^{2}x-6\sin x\cos x$

$=2+10{\cos }^{2}x-3\left(2\sin xcosx\right)$

$=2+5(2{\cos }^{2}x-1)+5-3\left(2\sin x\cos x\right)$

$=2+5\cos 2x+5-3\left(\sin 2x\right)$

$=7+5\cos 2x-3sin2x$

$=7+(\frac{5\cos 2x}{\sqrt{5^3+3^2}}-\frac{3\sin 2x}{\sqrt{5^2+3^2}})\sqrt{5^2+3^2}$

$S=7+\sqrt{34}cos\alpha$

$S_{min}=7-\sqrt{34}$

$S_{max}=7+\sqrt{34}$

$S_{min}+S_{max}=14$