Question

If $f(x)=\frac{\sin ^2 x+4 \sin x+5}{2 \sin ^2 x+8 \sin x+8}$, then range of $f(x)$ is

Answer 1

SOLUTION : $y=\frac{\sin ^2 x+4 \sin x+4}{2 \sin ^2 x+8 \sin x+8}+\frac{1}{2 \sin ^2 x+8 \sin x+8}=\frac{1}{2}+\frac{1}{2(\sin x+2)^2}$

$y_{\max }=\frac{1}{2}+\frac{1}{2(-1+2)^2}=1 \quad ; \quad y_{\min }=\frac{1}{2}+\frac{1}{2(1+2)^2}=\frac{5}{9}$

$\therefore \quad$ range $=\left[\frac{5}{9}, 1\right]$