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)

$6 \sin x \cos x+4 \cos 2 x=3 \sin 2 x+4 \cos 2 x$
$\therefore -\sqrt{3^{2}+4^{2}} \leq 3 \sin 2 x+4 \cos 2 x \leq \sqrt{3^{2}+4^{2}}$
$\Rightarrow -5 \leq 3 \sin 2 x+4 \cos 2 x \leq 5$
Thus, maximum value is 5 and minimum value is $-5 .$