Let f be a real valued continuous function on [0,1] and f(x)=x+∫ _0^1(x-t)f(t)dt. Then which of the following points (x,y) lies on the curve y=f(x)?

Given,

$$f(x)=x(1+{\int }_{0}^{1}f(t)dt)-{\int }_{0}^{1}tf(t)dt$$f(x)=x(1+∫01​f(t)dt)−∫01​tf(t)dt

$$f(x)=Ax-B...(i)$$f(x)=Ax−B...(i)

$$A=1+{\int }_{0}^{1}f(t)dt=1+{\int }_{0}^{1}(At-B)dt$$A=1+∫01​f(t)dt=1+∫01​(At−B)dt

$$\Rightarrow A=2(1-B)...(ii)$$⇒A=2(1−B)...(ii)

Also $\displaystyle B={\int }_{0}^{1}tf(t)dt={\int }_{0}^{1}(A{t}^{2}-Bt)dt$B=∫01​tf(t)dt=∫01​(At2−Bt)dt

$$A=\frac{9}{2}B...(iii)$$A=29​B...(iii)

From $(ii),(iii)$(ii),(iii)

$$A=\frac{18}{13},B=\frac{4}{13}$$A=1318​,B=134​

So, $f(x)=\frac{18}{13}x-\frac{4}{13}$f(x)=1318​x−134​

$$\Rightarrow f(6)=\frac{18\times 6-4}{13}=8$$⇒f(6)=1318×6−4​=8