Question

The range of $f\left(x\right)={\sin }^{2}x-2\sin x$ is

• A. $(0,3)$
• B. $[-1,3)$
• C. $[0,3]$
• D. $[-1,3]$

Answer 1

The correct option is D $[-1,3]$

We have,
$f\left(x\right)={\sin }^{2}x-2\sin x\Rightarrow f\left(x\right)={\sin }^{2}x-2\sin x+1-1\Rightarrow f\left(x\right)=(\sin x-1{)}^2-1$

We know that
$-1\le \sin x\le 1\Rightarrow -2\le \sin x-1\le 0\Rightarrow 0\le (\sin x-1{)}^2\le 4\Rightarrow -1\le (\sin x-1{)}^2-1\le 3$
Thus $f\left(x\right)\in [-1,3]$