Question

If $f\left(x\right)=\frac{{\sin }^{2}x+4\sin x+5}{2{\sin }^{2}x+8\sin x+8}$, then :

• A. minimum value of $f\left(x\right)$ is $\frac{5}{9}$
• B. minimum value of $f\left(x\right)$ is $\frac{3}{4}$
• C. minimum value of $f\left(x\right)$ is $\frac{4}{9}$
• D. maximum value of $f\left(x\right)$ is $1$

Answer 1

Answer: (A), (D)

minimum value of $f\left(x\right)$ is $\frac{5}{9}$
maximum value of $f\left(x\right)$ is $1$

$f\left(x\right)=\frac{{\sin }^{2}x+4\sin x+5}{2{\sin }^{2}x+8\sin x+8},$

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

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

Since, $-1\le \sin x\le 1$

$\Rightarrow 1\le (\sin x+2{)}^2\le 9$

$\Rightarrow \frac{1}{18}\le \frac{1}{2(\sin x+2{)}^2}\le \frac{1}{2}$

$\Rightarrow \frac{1}{2}+\frac{1}{18}\le \frac{1}{2}+\frac{1}{2(\sin x+2{)}^2}\le \frac{1}{2}+\frac{1}{2}$

$\Rightarrow \frac{5}{9}\le f\left(x\right)\le 1$