The number of ordered pairs which satisfy the equation `x^2+2xsin(x y)+1=0` are (where `y in [0,2pi]` ) (a) 1 (b) 2 (c) 3 (d) 0

Options

A. 1
B. 2
C. 3
D. 0

Answer

B

Explanation

Given `x^(2) +2x sin (xy) +1=0`
or `[x+ sin (xy) ]^(2)+[1-sin^(2) (xy)]=0`
or `x+sin (xy)=0 and sin^(2) (xy)=1`
`sin^(2) (xy)=1` gives `sin (xy)=1 or -1`
if `sin (xy)=1" "rArr x=-1`
`rArr sin (-y)=1" "rArr sin y =-1`
Then the ordered pair is `(1, 3pi//2)`.
if `sin (xy)=-1" "rArr x=1`
`rArr sin y=-1" "rArr (-1, 3pi//2)`
Thus, there are two ordered pairs.