# Doubt from area under the curve

Denominator should always be positive in such situation

Option a?

No B

Oh damn yes I missed one condition.. Basically a and b need to be positive for the first two functions to be bounded. For the third function since the numerator is always positive, the denominator should not become 0 at any point to prevent the function from becoming unbounded. So put discriminant of denominator<0 and you'll get the condition on a and b which is basically the locus of an ellipse whose total area is 2π but we'll only take the area in one quadrant since a and b need to be positive which is only possible in the first quadrant so you get the answer 2π/4 = π/2

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