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COMEDK · Maths · 25. Continuity and Differentiability

Let \(f(x) = x\sqrt{4ax - x^2}, a > 0\) then \(f(x)\) at \(x = 2a\) is

  1. A Decreasing
  2. B Increasing
  3. C Does not exist
  4. D Zero
Verified Solution

Answer & Solution

Correct Answer

(B) Increasing

Step-by-step Solution

Detailed explanation

\(f'(x) = \dfrac{d}{dx}(x) \cdot \sqrt{4ax-x^2} + x \cdot \dfrac{4a-2x}{2\sqrt{4ax-x^2}}\) \(= \sqrt{4ax-x^2} + \dfrac{x(2a-x)}{\sqrt{4ax-x^2}} = \dfrac{(4ax-x^2) + x(2a-x)}{\sqrt{4ax-x^2}} = \dfrac{6ax-2x^2}{\sqrt{4ax-x^2}}\) At \(x = 2a\):…