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JEE Main202430 Jan 2024Morning ShiftMathematicsApplication of DerivativesActual

Let g : R → R be a non constant twice differentiable such that g ' 1 2 = g ' 3 2 . If a real valued function f is defined as f ( x ) = 1 2 [ g ( x ) + g ( 2 - x ) ] , then

Options

  1. Af " ( x ) = 0 for atleast two x in ( 0 , 2 )
  2. Bf " ( x ) = 0 for exactly one x in ( 0 , 1 )
  3. Cf " ( x ) = 0 for no x in ( 0 , 1 )
  4. Df ' 3 2 + f ' 1 2 = 1

Correct answer

A. f " ( x ) = 0 for atleast two x in ( 0 , 2 )

Step-by-step solution

Given: f x = 1 2 g x + g 2 - x ⇒ f ' x = 1 2 g ' x - g ' 2 - x ⇒ f ' 1 2 = 1 2 g ' 1 2 - g ' 3 2 ⇒ f ' 1 2 = 0 Also, f ' 3 2 = 1 2 g ' 3 2 - g ' 1 2 ⇒ f ' 3 2 = 0 Now, f ' 1 = 1 2 g ' 1 - g ' 1 ⇒ f ' 1 = 0 So, f ' x has three roots and thus f " x will have atleast two roots in 0 , 2

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