InterviewSolution
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InterviewSolution
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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 |
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Answer» `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. |
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