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AP EAMCET · Maths · Application of Derivatives

If \(a, b>0\), then minimum value of \(y=\frac{b^2}{a-x}+\frac{a^2}{x}, 0 < x < a\) is

  1. A \(\frac{(a+b)^2}{a}\)
  2. B \(\frac{(a+b)^2}{b}\)
  3. C \(\frac{(a-b)^2}{a}\)
  4. D \(\frac{(a-b)^2}{b}\)
Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

Given, \(y=\frac{b^2}{a-x}+\frac{a^2}{x}\) \(\therefore \quad \frac{d y}{d x}=\frac{-b^2(-1)}{(a-x)^2}+\left(\frac{-a^2}{x^2}\right) \Rightarrow \frac{d y}{d x}=\frac{b^2}{(a-x)^2}-\frac{a^2}{x^2}\) \(\because \quad \frac{d y}{d x}=0, x=\frac{a^2}{a \pm b}\)…