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AP EAMCET201823 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

We have, aligned & f(x)=(x-1)(x-2)(x-3) & f(x)=x^3-6 x^2+11 x-6 aligned Given function is algebraic function so. it is continuous and difference in [0,4] Now, aligned f(4) & =(4)^3-6(4)^2+11(4)-6 & =64-96+44-6=6 f(0) & =-6 aligned Now, f^ (x)=3 x^2-12 x+11 by Lagrange's mean value theorem aligned f^ (c) & = f(4)-f(0) 4-0 3 c^2-12 c+11 & = 6-(-6) 4 3 c^2-12 c+11 & = 12 4 3 c^2-12 c+11-3 & =0 3 c^2-12 c+8 & =0 c & = 12 (-12)^2-4(3)(8) 2 3 & = 12 144-96 6 = 12 4 3 6 c & =2 2 3 3 [0,4) aligned

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