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CA Foundation · Quantitative Aptitude · Differential and Integral Calculus

The function f(x) = 2x³ − 9x² + 12x + 1 has a local maximum at x equal to:

The local maximum occurs at x = 1. Setting f'(x) = 6(x − 1)(x − 2) to zero gives x = 1 and 2; the second derivative 12x − 18 is negative at x = 1, giving a maximum, and positive at x = 2, giving a minimum.

  1. Ax = 2
  2. Bx = 1Correct
  3. Cx = 3
  4. Dx = 0

Explanation

f'(x) = 6x² − 18x + 12 = 6(x − 1)(x − 2) = 0, so x = 1 or 2. f''(x) = 12x − 18; at x = 1 it is −6 < 0, so a local maximum at x = 1. At x = 2, f'' = 6 > 0, a minimum, so x = 2 is wrong.

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