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MHT CET202410 May 2024Evening ShiftMathematicsApplication of DerivativesActual

Let f (x)=(x-1)(x-2)(x-3), x [0,4] , Values of C will be _______ [if L.M.V.T. (Lagrange's Mean Value Theorem) can be applied].

Options

  1. A4-2 3 3 , 4+2 3 3
  2. B6-2 3 3 , 6+2 3 3
  3. C6- 3 3 , 6+ 3 3
  4. D2- 3 , 2+ 3

Correct answer

B. 6-2 3 3 , 6+2 3 3

Step-by-step solution

aligned & Let y=(x-1)(x-2)(x-3) & y= (x-1)+ (x-2)+ (x-3) aligned Differentiating w.r.t. x , we get 1 y ~d y ~d x = 1 x-1 + 1 x-2 + 1 x-3 aligned & f ^ (x)= d y ~d x =(x-2)(x-3)+(x-1)(x-3) &+(x-1)(x-2) aligned array ll & f ^ ( c )= f (4)- f (0) 4-0 = 6-(-6) 4 =3, for c [0,4] & ( c -2)( c -3)+( c -1)( c -3)+( c -1)( c -2)=3 & 3 c ^2-12 c +11=3 & 3 c ^2-12 c +8=0 & c = 12 144-96 6 = 6 2 3 3 array

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