Let f be a real valued continuous function on [0,1] and $\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}+{\int }_0^1 (\mathrm{x}-\mathrm{t})\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}$Then which of the following points (x,y) lies on the curve y = f (x)?

$\begin{array}{l}\mathrm{f}\left(\mathrm{x}\right)=\left(1+{\int }_0^1 \mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}\right)\mathrm{x}-{\int }_0^1 \mathrm{tf}\left(\mathrm{t}\right)\mathrm{dt}\\ \mathrm{f}\left(\mathrm{x}\right)=\mathrm{Ax}-\mathrm{B}...\left(\mathrm{i}\right)\\ \mathrm{A}=1+{\int }_0^1 \mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}=1+{\int }_0^1 (\mathrm{At}-\mathrm{B})\mathrm{dt}\\ \Rightarrow \mathrm{A}=2(1-\mathrm{B})...\left(\mathrm{ii}\right)\end{array}$
Also$\mathrm{B}={\int }_0^1 \mathrm{tf}\left(\mathrm{t}\right)\mathrm{dt}={\int }_0^1 \left({\mathrm{At}}^2-\mathrm{Bt}\right)\mathrm{dt}$
$\mathrm{A}=\frac{9}{2}\mathrm{B}$ ...(iii)
From (2), (3)
$\mathrm{A}=\frac{18}{13},\mathrm{B}=\frac{4}{13}$
so f(6) = 8