Question

The range of the function $f\left(x\right)=\mathrm{sgn}\left(\frac{{\sin }^{2}x+2\sin x+4}{{\sin }^{2}x+2\sin x+3}\right)$ is (where sgn(.) denotes signum function)

• A. $\{-1,0,1\}$
• B. $\{-1,0\}$
• C. $\left\{1\right\}$
• D. $\{0,1\}$

Answer 1

The correct option is C $\left\{1\right\}$

We know that $f\left(x\right)=Sgn\left(y\right)=\{\begin{array}{cc}1& f\left(x\right)>0\\ 0& f\left(x\right)=0\\ -1& f\left(x\right)<1\end{array}$

Given, $y=\left(\frac{{\sin }^{2}x+2\sin x+4}{{\sin }^{2}x+2\sin x+3}\right)$

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

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

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

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

$⟹0\le \sin x+1\le 2$

$⟹0\le (\sin x+1{)}^2\le 4$

$⟹2\le (\sin x+1{)}^2+2\le 6$

$⟹\frac{1}{2}\ge \frac{1}{(\sin x+1{)}^2+2}\ge \frac{1}{6}$

$⟹\frac{3}{2}\ge 1\frac{1}{(\sin x+1{)}^2+2}\ge \frac{7}{6}$

Thus, $y\in [\frac{7}{6},\frac{3}{2}]$

$f\left(x\right)=Sgn\left(y\right)=1$

$Range=\left\{1\right\}$