Question

The maximum value of $f\left(x\right)=1+2sinx+2{\cos }^{2}x$ in $[0,\frac{\pi }{2}]$ is

• A. $\frac{7}{2}$
• B. $\frac{5}{2}$
• C. $\frac{3}{2}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $\frac{7}{2}$

$f\left(x\right)=3+2sinx-2si{n}^{2}x$

$=-(2si{n}^{2}x-2sinx-3)$

$=-2(si{n}^{2}x-sinx-\frac{3}{2}+\frac{1}{4}-\frac{1}{4})$

$=-2\left(\right(sinx-\frac{1}{2}{)}^2-\frac{7}{4})$

$=-2(sinx-\frac{1}{2}{)}^2+\frac{7}{2}$

$f\left(x\right)$ is max when $sinx=\frac{1}{2}$