If ∫0x f(t)dt=x+∫x1 tf(t)dt then the value of f(1), is 

If 0xf(t)dt=x+x1tf(t)dt then the value of f(1), is 

  1. A

    1/2

  2. B

    0

  3. C

    1

  4. D

    -1/2

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    Solution:

    We have , 

    0xf(t)dt=x+x1tf(t)dt

    ddx0xf(t)dt=ddxx+x1tf(t)dt

     f(x)=1+0xf(x)     [Using Leibnitz's rule] 

     f(x)=1-x f(x)

    f(x)=1x+1f(1)=12

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