The value of c in Roll’s theorem for the function f(x)=x(x+1)ex defined on [−1,0], is

The value of c in Roll's theorem for the function f(x)=x(x+1)ex defined on [1,0], is

  1. A

    0.5

  2. B

    1+52

  3. C

    152

  4. D

    -0.5

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    Solution:

    We have, f(x)=x(x+1)ex,x[1,0]

     f(x)=x2x1ex,f(1)=0,f(0)=0

    Hence, f(x)=f(0)f(1)0(1)

     x2x1ex=0x2x1=0x=1±52

    Hence, c=152

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