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JEE Main20231 Feb 2023Morning ShiftMathematicsApplication of DerivativesActual

Let f x = 2 x + tan - 1 x and g x = log e 1 + x 2 + x , x ∈ 0 , 3 . Then

Options

  1. AThere exists x ∈ 0 , 3 such that f ' x < g ' x
  2. Bmax   f x > max   g x
  3. CThere exist 0 < x 1 < x 2 < 3 such that f x < g x , ∀ x ∈ x 1 , x 2
  4. Dmin   f ' x = 1 + max   g ' x

Correct answer

B. max   f x > max   g x

Step-by-step solution

Given: f x = 2 x + tan - 1 x ⇒ f ' x = 2 + 1 1 + x 2 And, g x = ln 1 + x 2 + x ⇒ g ' x = 1 1 + x 2 Now, 0 ≤ x ≤ 3 ⇒ 0 ≤ x 2 ≤ 9 ⇒ 1 ≤ 1 + x 2 ≤ 10 So, 1 10 ≤ 1 1 + x 2 ≤ 1 ⇒ 2 + 1 10 ≤ 2 + 1 1 + x 2 ≤ 3 ⇒ 2 + 1 10 ≤ f ' x ≤ 3 ⇒ 21 10 ≤ f ' x ≤ 3 And, 1 10 ≤ g ' x ≤ 1 So, min   f ' x = 21 10 ≠ 1 + max   g ' x Option 4 is incorrect From a

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