Question

The number of ordered pairs $(x,y)$ satisfying the equation $x^2+2x\sin \left(xy\right)+1=0$ is
$(\text{where}y\in [0,2\pi \left]\right)$

• A. $1$
• B. $2$
• C. $3$
• D. $0$

Answer 1

The correct option is B $2$

Given: $x^2+2x\sin \left(xy\right)+1=0$
$\Rightarrow x^2+1=-2x\sin \left(xy\right)\Rightarrow x+\frac{1}{x}=-2\sin \left(xy\right)$
We know that,
$x+\frac{1}{x}\in (-\infty ,-2]\cup [2,\infty )-2\sin \left(xy\right)\in [-2,2]$
So, solution is possible when
$x+\frac{1}{x}=-2=-2\sin \left(xy\right)\Rightarrow x=-1\Rightarrow \sin (-y)=1\Rightarrow \sin y=-1\Rightarrow y=\frac{3\pi }{2}$

$x+\frac{1}{x}=2=-2\sin \left(xy\right)\Rightarrow x=1\Rightarrow \sin y=-1\Rightarrow y=\frac{3\pi }{2}$

Hence, there are $2$ ordered pairs $(-1,\frac{3\pi }{2})$ and $(1,\frac{3\pi }{2})$.