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AP EAMCET201825 Apr 2018Morning ShiftMathematicsApplication of DerivativesActual

If f ( x ) = ( x - 1 ) ( x - 2 ) ( x - 3 ) for x ∈ [ 0 , 4 ] , then the value of c ∈ ( 0 , 4 ) satisfying Lagrange's mean value theorem is

Options

  1. A3 ± 2 3
  2. B2 ± 2 3 3
  3. C2 ± 3 2
  4. D3 ± 3 3

Correct answer

B. 2 ± 2 3 3

Step-by-step solution

Given: f ( x ) = ( x - 1 ) ( x - 2 ) ( x - 3 ) ⇒ f x = x 3 - 6 x 2 + 11 x - 6 f ' x = 3 x 2 - 12 x + 11 Now, f x is continuous in R as it is a odd degree polynomial. From Lagrange's mean value theorem, if f x is continuous in a , b and differentiable in a , b then there exist c ∈ a , b such that f ' c = f b - f a b - a ⇒ 3 c 2 - 12 c + 11 = f 4 - f 0 4 - 0 ⇒ 3 c 2 - 12 c + 11 = 6 - - 6 4 - 0 ⇒ 3 c 2 - 12 c + 11 = 3 ⇒ 3 c 2 - 12 c + 8 = 0 ⇒ c = 12 ± 144 - 12

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