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If int0^xf(t) dt=x+intx^1 tf(t)dt, then ...

`If int_0^xf(t) dt=x+int_x^1 tf(t)dt,` then the value of `f(1)`

A

`1//2`

B

`0`

C

`1`

D

`-1//2`

Text Solution

Verified by Experts

The correct Answer is:
A
`int_(0)^(x)f(t)dt=x+int_(x)^(1)tf(t)dt`
or `d/(dx) (int_(0)^(x)f(t)dt)=d/(dx) (x+int_(x)^(1)tf(t)dt)` [Using Leibniz's rule]
or `f(x)=10-xf(x)`
`=1-xf(x)`
`=f(1)-1/(1+x)`
`:f(1)=1/1`
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