COMEDK2025MathematicsContinuity and DifferentiabilityActual
Let f(x) = x 4ax - x^2 , a > 0 then f(x) at x = 2a is
Options
- ADecreasing
- BIncreasing
- CDoes not exist
- 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 .