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JEE Main202427 Jan 2024Evening ShiftMathematicsContinuity and DifferentiabilityActual

Consider the function f : ( 0 , 2 ) → R defined by f ( x ) = x 2 + 2 x and the function g ( x ) defined by g x = min f ( t ) , 0 < t ≤ x and 0 < x ≤ 1 3 2 + x , 1 < x < 2 . Then

Options

  1. Ag is continuous but not differentiable at x = 1
  2. Bg is not continuous for all x ∈ ( 0 , 2 )
  3. Cg is neither continuous nor differentiable at x = 1
  4. Dg is continuous and differentiable for all x ∈ ( 0 , 2 )

Correct answer

A. g is continuous but not differentiable at x = 1

Step-by-step solution

Given, f : ( 0 , 2 ) → R ; f ( x ) = x 2 + 2 x ⇒ f ' x = 1 2 - 2 x 2 ⇒ f ' x = x 2 - 4 2 x 2 ∴ f x is decreasing in domain 0 , 2 . ⇒ g x = x 2 + 2 x 0 < x ≤ 1 3 2 + x 1 < x < 2 ⇒ g 1 = 1 2 + 2 1 = 5 2 ⇒ g 1 + = 3 2 + 1 = 5 2 ⇒ g x is continuous in 0 , 2 . ⇒ g ' x = f ' x , 0 < x ≤ 1 1 , 1 < x < 2 ⇒ g ' 1 = f ' 1 = - 3 2 ⇒ g ' 1 ≠ g ' 1 + So, g x is not differentiable at x = 1 .

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