JEE MATHEMATICS MCQ: MAXIMA & MINIMA

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Application +4 | -1

a, b>0

, then the minimum value

y=\frac{b^2}{a-x}+\frac{a^2}{x}, 0 < x < a

View answer and explanation

Correct option: D

\frac{(a+b)^2}{a}

Option D (

\frac{(a+b)^2}{a}

) is correct because by finding the critical points of the function and checking the second derivative, we can determine the minimum value.

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