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COMEDK2025MathematicsContinuity and DifferentiabilityActual

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

Options

  1. ADecreasing
  2. BIncreasing
  3. CDoes not exist
  4. DZero

Correct answer

B. Increasing

Step-by-step solution

f'(x) = d dx (x) 4ax-x^2 + x 4a-2x 2 4ax-x^2 = 4ax-x^2 + x(2a-x) 4ax-x^2 = (4ax-x^2) + x(2a-x) 4ax-x^2 = 6ax-2x^2 4ax-x^2 At x = 2a : f'(2a) = 6a(2a) - 2(2a)^2 4a(2a)-(2a)^2 = 12a^2 - 8a^2 8a^2-4a^2 = 4a^2 4a^2 = 4a^2 2a = 2a Since a > 0 , f'(2a) = 2a > 0 , so f(x) is increasing at x = 2a .

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