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JEE Main20198 Apr 2019Evening ShiftMathematicsLimitsActual

Let f : R → R be a differentiable function satisfying f ' 3 + f ' 2 = 0 . Then l i m x → 0 1 + f 3 + x - f 3 1 + f 2 - x - f 2 1 x is equal to

Options

  1. A1
  2. Be
  3. Ce 2
  4. De - 1

Correct answer

A. 1

Step-by-step solution

This limit is 1 ∞ form ∴ l i m x → 0 1 + f 3 + x - f 3 1 + f 2 - x - f 2 1 x = e l i m x → 0 1 x 1 + f 3 + x - f 3 1 + f 2 - x - f 2 - 1 = e l i m x → 0 1 x f 3 + x - f 3 - f 2 - x + f 2 1 + f 2 - x - f 2 0 0   f o r m = e l i m x → 0   f ' 3 + x + f ' 2 - x x - f ' 2 - x + 1 + f 2 - x - f 2 = e 0 = 1 .

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