Question

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

• A. (2, 4)
• B. (1, 2)
• C. (4, 17)
• D. (6, 8)

Answer 1

Answer: (D)

$f(x)=\left(1+\underset{0}{\overset{1}{\int }}f(t)dt\right)x-\underset{0}{\overset{1}{\int }}tf(t)dt$
$f(x)=Ax-B.....$ (i)
$A=1+\underset{0}{\overset{1}{\int }}f(t)dt=1+\underset{0}{\overset{1}{\int }}(At-B)dt$
$\Rightarrow A=2(1-B).....$(ii)
Also $B=\underset{0}{\overset{1}{\int }}tf(t)dt=\underset{0}{\overset{1}{\int }}\left(At{}^2-Bt\right)dt$
$A=\frac{9}{2}B.....$(iii)
From $(2),(3)$
$A=\frac{18}{13},B=\frac{4}{13}$
so $f(6)=8$