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JEE Advanced2023MathematicsApplication of DerivativesActual

Let S be the set of all twice differentiable functions f from ℝ to ℝ such that d 2 f d x 2 x > 0 for all x ∈ - 1 , 1 . For f ∈ S , let X f be the number of points x ∈ - 1 , 1 for which f x = x . Then which of the following statements is(are) true?

Options

  1. AThere exists a function f ∈ S such that X f = 0
  2. BFor every function f ∈ S , we have X f ≤ 2
  3. CThere exists a function f ∈ S , such that X f = 2.
  4. DThere does NOT exist any function f in S such that X f = 1

Correct answer

A. There exists a function f ∈ S such that X f = 0

Step-by-step solution

Given, f " ( x ) > 0     &     f ( x ) - x = 0 Now let, g x = f x − x Now differentiating above equation we get, ⇒ g ' x = f ' x − 1 Again differentiating we get, g " ( x ) = f " ( x ) > 0 ⇒  concave up  So, possible graphs will be, Here in above graph X f = 0 as there is no point for which f x = x Now here X f ≤ 2 as there will be minimum two points when y = x will cut the given graph, Now in above graph X f = 1 so the

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