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JEE Advanced2017MathematicsApplication of DerivativesActual

If f : R → R is a twice differentiable function such that f ′ ′ x > 0 for all x ∈ R and f 1 2 = 1 2 , f 1 = 1 then

Options

  1. A0 < f &prime; 1 ≤ 1 2
  2. Bf &prime; 1 ≤ 0
  3. Cf ′ 1 > 1
  4. D1 2 < f &prime; 1 ≤ 1

Correct answer

C. f ′ 1 > 1

Step-by-step solution

Using LMVT on f x for x ∈ 1 2 , 1 f 1 - f 1 2 1 - 1 2 = f ' c , where c ∈ 1 2 , 1 1 - 1 2 1 2 = f ' c ⇒ f ' c = 1 , where c ∈ 1 2 , 1 ∵ f ″ x > 0 ∀ x ∈ R ∴ f ' x is an increasing function ∀ x ∈ R ∴ f ′ 1 > 1

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