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ExamsJEE AdvancedMaths

Let f(x) be a differentiable function on [2, 5] such that f(2) = 1/5 and f(5) = 1/2. Then there exists x0 with 2 < x0 < 5 such that f'(x0) equals:

  1. 1/15
  2. 1/10
  3. 1/5
  4. 1/2

Correct answer: 1/10

Solution

By the Mean Value Theorem: if f is continuous on [2,5] and differentiable on (2,5), there exists x0 in (2,5) such that f'(x0) = (f(5)-f(2))/(5-2) = (1/2 - 1/5)/3 = (5/10-2/10)/3 = (3/10)/3 = 1/10.

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