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AP EAMCET202124 Aug 2021Evening ShiftMathematicsApplication of DerivativesActual

The value of c of the Lagrange's mean value theorem for f(x)= x^2-x , x [1,4] is

Options

  1. A4 3
  2. B3 2
  3. C5 4
  4. D3

Correct answer

B. 3 2

Step-by-step solution

We have, aligned & f(x)= x^2-x , x [1,4] f(1)= 1-1 =0 & f(4)= 16-4 = 12 =2 3 f^ (c)= 2 c-1 2 c^2-c aligned By Lagrange's mean value theorem f^ (c)= f(b)-f(a) b-a 2 c-1 2 c^2-c = f(4)-f(1) 4-1 = 2 3 -0 3 2 c-1 2 c^2-c = 2 3 3 (2 c-1)=4 c^2-c 3(2 c-1)^2=16 c^2-16 c aligned & 3 (4 c^2-4 c+1 )=16 c^2-16 c & 12 c^2-12 c+3=16 c^2-16 c 4 c^2-4 c-3=0 & (2 c-3)(2 c+1)=0 c=3 / 2 (1,4) aligned c=3 / 2

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