NTA Abhyas JEE Main2020MathematicsApplication of DerivativesPractice

A differentiable function f x satisfies f 0 = 0 and f 1 = sin ⁡ 1 , then (where f ' represents derivative of f )

Options

Af ' c = cos c ,   ∀ c ∈ 0 , 1
Bf ' c = cos c for some c ∈ 0 , 1
Cf ' c = − cos c ,   ∀ c ∈ 0 , 1
Df ' c = 2 cos c ,   ∀ c ∈ 0 , 1

Correct answer

B. f ' c = cos c for some c ∈ 0 , 1

Step-by-step solution

Let, g x = f x - sin ⁡ x Now, g 0 = f 0 - 0 = 0 & g 1 = f 1 - sin ⁡ 1 = 0 As g x is differentiable, so by Rolle’s theorem, g ′ c = 0 for some c ∈ 0,1 i.e. f ′ c − cos c = 0 ⇒ f ′ c = cos c   for   some   c ∈ 0 , 1

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