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JEE Main20207 Jan 2020Evening ShiftMathematicsApplication of DerivativesActual

The value of c , in the Lagrange’s mean value theorem for the function f x = x 3 - 4 x 2 + 8 x + 11 , when x ∈ 0,1 , is

Options

  1. A4 - 5 3
  2. B4 - 7 3
  3. C2 3
  4. D7 - 2 3

Correct answer

B. 4 - 7 3

Step-by-step solution

As f x , is polynomial function, so it is continuous and differentiable in 0,1 . Here f 0 = 11 , f 1 = 1 - 4 + 8 + 11 = 16 f ' x = 3 x 2 - 8 x + 8 ∴ f . c = f 1 - f 0 1 - 0 = 16 - 11 1 = 3 c 2 - 8 c + 8 3 c 2 - 8 c + 3 = 0 c = 8 ± 2 7 6 = 4 ± 7 3 ∴ c = 4 - 7 3 ∈ 0,1 .

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