Question

The range of $f\left(x\right)=\sqrt{{\sin }^{2}x-6\sin x+9}+3$ is $[a,b]$, then $b-a$ is equal to

Answer 1

$f\left(x\right)=\sqrt{{\sin }^{2}x-6\sin x+9}+3$
$=\sqrt{(\sin x-3{)}^2}+3$
$=|\sin x-3|+3$

Now, $-1\le \sin x\le 1$
$\Rightarrow -4\le \sin x-3\le -2$
$\Rightarrow 2\le |\sin x-3|\le 4$
$\Rightarrow 5\le |\sin x-3|+3\le 7$

Hence ,
$a=5,b=7\therefore b-a=2$