Question

The maximum and minimum value of $6\sin x\cos x+4\cos 2x$ are respectively

• A. $5,5$
• B. $-5,5$
• C. $5,-5$
• D. None of these

Answer 1

Answer: (C)

$5,-5$

$6\sin x\cos x+4\cos 2x=3\sin 2x+4\cos 2x$ ($\because \sin 2x=2\cos x\sin x$)

we know that

$-\sqrt{a^2+b^2}\le a\sin x+b\cos x\le \sqrt{a^2+b^2}$

$\therefore -\sqrt{3^2+4^2}\le 3\sin 2x+4\cos 2x\le \sqrt{3^2+4^2}$

so, the minimum value of the given equation is $-5$ and the maximum value is $5$.