Consider a function `f: [0, 1] -> R` satisfying `int_0 ^1 (f(x)(x-f(x))dx = 1/12`, then, `f'(1)` is `k` where `[1/k]` is

Answer

2

Explanation

`int_(0)^(1)(f(x).x(f(x))^(2))dx=int_(0)^(1) (x^(2))/4 dx`
`int_(0)^(1)[(f(x))^(2)-x.f(x)+(x^(2))/4]dx=0`
`int_(0)^(1)(f(x)-x/2)^(2) dx=0`
`impliesf(x)=x/2`
`:.f^(')(x)=1/2`