Question

If $x\in [0,3\pi ]$ and $r\in R$, then the number of pairs of $(r,x)$ satisfying $2\sin x=r^4-2r^2+3$ is

• A. $2$
• B. $3$
• C. $4$
• D. $6$

Answer 1

The correct option is C $4$

$2\sin x=r^4-2r^2+3\Rightarrow 2\sin x=(r^2-1{)}^2+2\Rightarrow 2\sin x=(r^2-1{)}^2+2$
We know that,
$2\sin x\le 2\text{and}(r^2-1{)}^2+2\ge 2$

So, both are equal only when
$2\sin x=2\Rightarrow \sin x=1\Rightarrow x=\frac{\pi }{2},\frac{5\pi }{2}$
And $(r^2-1{)}^2+2=2$
$\Rightarrow (r^2-1{)}^2=0\Rightarrow r=\pm 1$

$\therefore$ The required ordered pairs are
$(1,\frac{\pi }{2}),(-1,\frac{\pi }{2}),(1,\frac{5\pi }{2}),(-1,\frac{5\pi }{2})$

Hence, there are $4$ ordered pairs.