Question

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

Answer 1

`x^2+2xsin(xy) +1 = 0`
`=>(x+sin(xy))^2 - sin^2(xy)+1 = 0`
`=>(x+sin(xy))^2 + cos^2(xy) = 0`
`=>x+sin(xy) = 0 and cos(xy) = 0`
When, `cos(xy) = 0, sin(xy) = sqrt(1-cos^2(xy)) = +-1`
When `sin(xy) = 1`
`x+sin(xy) = 0=>x+1 = 0 => x = -1`
`=>sin(-y) = 1 =>siny = -1 => y = (3pi)/2`
When `sin(xy) = -1`
`x+sin(xy) = 0=>x-1 = 0 => x = 1`
`=>sin(y) = -1 => y = (3pi)/2`
So, there are two ordered pairs, `(1,(3pi)/2)` and `(-1,(3pi)/2)` that satisfy the given equation.