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Q.
Let $f$ be a real valued continuous function on $[0,1]$ and $f(x)=x+\int\limits_{0}^{1}(x-t) f(t) d t$. Then which of the following points $( x , y )$ lies on the curve $y =f( x )$ ?
$f(x)=\left(1+\int\limits_{0}^{1} f(t) d t\right) x-\int\limits_{0}^{1} tf ( t ) dt$
$f(x)=A x-B .....$ (i)
$A=1+\int\limits_{0}^{1} f(t) d t=1+\int\limits_{0}^{1}( At - B ) dt$
$\Rightarrow A =2(1- B ) .....$(ii)
Also $B =\int\limits_{0}^{1} tf ( t ) dt =\int\limits_{0}^{1}\left( At { }^{2}- Bt \right) dt$
$A =\frac{9}{2} B.....$(iii)
From $(2),(3)$
$A =\frac{18}{13}, B =\frac{4}{13}$
so $f(6) =8$