Question

The maximum value of 7 ${\sin }^{2}2x+6{\cos }^{2}2x+\sin x+\cos x$ is

• A. $13$
• B. $12$
• C. $7+\sqrt{2}$
• D. $6+\sqrt{2}$

Answer 1

Answer: (C)

$7+\sqrt{2}$

$E=7{\sin }^{2}2x+6{\cos }^{2}2x+\sin x+\cos x$
$E=6+{\sin }^{2}2x+\sin x+\cos x$ \sin ce $6{\sin }^{2}2x+6{\cos }^{2}2x=6$
$\therefore E=6+{\{{(\sin x+\cos x)}^2-1\}}^2+\sin x+\cos x$ \sin ce we write ${\sin }^{2}2x$ as $4{\sin }^{2}x\times 4{\cos }^{2}x$
If $\sin x+\cos x$ is maximum, then $E$ is maximum.
We know that the maximum value of $\sin x+\cos x$ is $\sqrt{2}$
$\therefore$ maximum of $E=6+\left(1\right)+\sqrt{2}$
$=7+\sqrt{2}$
Hence, option 'C' is correct.