Question

The maximum value of $1+2\sin x+3{\cos }^{2}x$ is

• A. $\frac{3}{13}$
• B. $\frac{13}{3}$
• C. $\frac{2}{13}$
• D. $\frac{13}{2}$

Answer 1

The correct option is B

$\frac{13}{3}$

We will first make the expression in terms of $\sin x$ only. Then we will try to convert it into perfect square. Once we have it in the form $k\pm a(\sin x\pm b{)}^2$, we can easily find its range.

$1+2\sin x+3{\cos }^{2}x$
$=1+2\sin x+3(1-{\sin }^{2}x)$
$=1+2\sin x+3-3{\sin }^{2}x$
$=4+2\sin x-3{\sin }^{2}x$
$=4-3({\sin }^{2}x-\frac{2}{3}\sin x+\frac{1}{9}-\frac{1}{9})$
$=4-3(\sin x-\frac{1}{3}{)}^2+\frac{1}{3}$
$=\frac{13}{3}-3{(\sin x-\frac{1}{3})}^2$
The expression attains maximum value if ${(\sin x-\frac{1}{3})}^2=0$, which is possible if $\sin x=\frac{1}{3}$.
$\Rightarrow$ Maximum value = $\frac{13}{3}$.