Question

Find the range of $f\left(x\right)=\frac{2{\sin }^{2}x+2\sin x+3}{{\sin }^{2}x+\sin x+1}$

Answer 1

$f\left(x\right)=\frac{2{\sin }^{2}x+2\sin x+3}{{\sin }^{2}x+\sin x+1}$

$=\frac{2{\sin }^{2}x+2\sin x+2+1}{{\sin }^{2}x+\sin x+1}$

$=\frac{2({\sin }^{2}x+\sin x+1)+1}{{\sin }^{2}x+\sin x+1}$

$=2+\frac{1}{{\sin }^{2}x+\sin x+1}$

$=2+\frac{1}{{\sin }^{2}x+\sin x+\frac{1}{4}+\frac{3}{4}}$

$=2+\frac{1}{{(\sin x+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$

$f\left(x\right)$ is minimum when $\frac{1}{{(\sin x+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$ is maximum

$\Rightarrow \sin x$ is maximum at $x=\frac{\pi }{2}$

$f\left(x_{min}\right)=2+\frac{1}{{\left(\frac{3}{2}\right)}^2+\frac{3}{4}}$

$f\left(x_{min}\right)=2+\frac{1}{\frac{9}{4}+\frac{3}{4}}$

$f\left(x_{min}\right)=2+\frac{1}{\frac{12}{4}}$

$f\left(x_{min}\right)=2+\frac{1}{3}$

$f\left(x_{min}\right)=\frac{7}{3}$

$f\left(x\right)$ is maximum at $\frac{1}{{(\sin x+\frac{1}{2})}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}$ is minimum

Minimum value of $\sin x$ is when $\sin x+\frac{1}{2}=0$

$\therefore f_{max}=2+\frac{1}{\frac{3}{4}}=2+\frac{4}{3}=\frac{6+4}{3}=\frac{10}{3}$

Hence range of $f\left(x\right)\in [\frac{7}{3},\frac{10}{3}]$