Question

The number of ordered pairs which satisfy the equation $x^2+2x\sin \left(xy\right)+1=0$ are $\left(yϵ[0,2\pi ]\right)$

• A. $1$
• B. $2$
• C. $3$
• D. $0$

Answer 1

Answer: (B)

$2$

$x^2-2x\sin \left(xy\right)+1=0$ ...(1)
Since the roots of the equation are real
$\therefore 4{\sin }^2\left(xy\right)-4\ge 0\Rightarrow {\sin }^2\left(xy\right)\ge 1$
Since ${\sin }^2\left(xy\right)\le 1$
$\Rightarrow {\sin }^2\left(xy\right)=1\therefore \sin \left(xy\right)=\pm 1$
Substitute $\sin \left(xy\right)=\pm 1$ in equation (1), we get
$x^2\pm 2x+1=0\Rightarrow x=\pm 1$
For $x=-1$
$\sin \left(xy\right)=1\Rightarrow \sin y=-1\Rightarrow y=\frac{3\pi }{2}$
and $x=1$
$\sin \left(xy\right)=-1\Rightarrow \sin y=-1\Rightarrow y=\frac{3\pi }{2}$
Therefore number of ordered pairs of x & y is 2
Ans: B