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JEE MATHEMATICS MCQ: RATE MEASURE

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Application +4 | -1
If y=x+c1+x2y = \frac{{x + c}}{{1 + {x^2}}} where cc is a constant, then when yy is stationary, xyxy is equal to
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Correct option: A
12\frac{1}{2}
The correct answer is 12\frac{1}{2} because it is derived by finding the stationary points of the function and then calculating the value of xyxy at those points.

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