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NTA Abhyas JEE Main2020MathematicsApplication of DerivativesPractice

Let f b be the minimum value of the expression y = x 2 - 2 x + b 3 - 3 b 2 + 4 ∀ x ∈ R . Then, the maximum value of f b as b varies from 0 to 4 is

Options

  1. A20
  2. B19
  3. C63
  4. D64

Correct answer

B. 19

Step-by-step solution

y = x - 1 2 + b 3 - 3 b 2 + 3 ∴ y m i n = f b = b 3 - 3 b 2 + 3 Now, f ′ b = 3 b 2 − 6 b = 3 b b - 2 Also, f 0 = 3 f 2 = - 1 f 4 = 19 ∴ max ⁡ f b = 19

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