MCQ
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 $y =f( x )$ ?
- A$(2,4)$
- B$(1,2)$
- C$(4,17)$
- ✓$(6,8)$
$f(x)=A x-B$ $\dots(i)$
$A=1+\int\limits_{0}^{1} f(t) d t=1+\int\limits_{0}^{1}(A t-B) d t$
$\Rightarrow A=2(1-B)$ $\dots(ii)$
Also $B=\int\limits_{0}^{1} t f(t) d t=\int\limits_{0}^{1}\left(A t^{2}-B t\right) d t$
$A=\frac{9}{2} B$ $\dots(iii)$
From $(2),(3)$
$A=\frac{18}{13}, B=\frac{4}{13}$
so $f(6) =8$
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where $[x]$ is the greatest integer less than or equal to $x$, then the value of $\alpha$ is :