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Let $\mathrm{f}$ be a real valued continuous function on $\mathrm{\left [ 0,1 \right ]}$ and $\mathrm{f(x)=x+\int_{0}^{1}(x-t) f(t) d t}$ . Then which of the following points $\mathrm{(x,y)}$ lies on the curve $\mathrm{y=f(x)}$ ?
Option: 1
$\mathrm{(2,4)}$

Option: 2
$\mathrm{(1,2)}$

Option: 3
$\mathrm{(4,17)}$

Option: 4
$\mathrm{(6,8)}$

Answers (1)

best_answer

$\mathrm{f(x) =x+x \int_{0}^{1} f(t) d t-\int_{0}^{1} t f(t) d t} \\$

$\mathrm{=x+A x-B=(A+1) x-B }\\$

$\mathrm{\text { where } A =\int_{0}^{1} f(t) d t, B=\int_{0}^{1} t f(t) d t}$

$\mathrm{A =\int_{0}^{1} f(t) d t} \\$

$\mathrm{\Rightarrow A =\int_{0}^{1}((A+1) t-B) d t} \\$

$\mathrm{\Rightarrow A =(A+1) \frac{t^{2}}{2}-\left.B t\right|_{0} ^{1} }$

$\mathrm{\Rightarrow A=\frac{A+1}{2}-B} \\$

$\mathrm{\Rightarrow \frac{A}{2}=\frac{1}{2}-B} \\$

$\mathrm{\Rightarrow A=1-2 B} \\$ .....(i)

$\mathrm{\text { Also } B=\int_{0}^{1} t f(t) d t}$

$\mathrm{B=\int_{0}^{1} t((A+1) t-B) d t} \\$

$\mathrm{B=\int_{0}^{1}\left((A+1) t^{2}-B t\right) d t }\\$

$\mathrm{B=(A+1) \frac{t^{3}}{3}-\left.\frac{B t^{2}}{2}\right|_{0} ^{1}}$

$\mathrm{B=\frac{A+1}{3}-\frac{B}{2}} \\$

$\mathrm{6 B=2 A+2-3 B} \\$

$\mathrm{9 B=2 A+2}$ .............(ii)

From (i) and (ii)

$\mathrm{B =\frac{4}{13}, A=\frac{5}{13}} \\$

$\mathrm{\therefore \quad f(x) =(A+1) x-B }\\$

$\mathrm{=\frac{18}{13} x-\frac{4}{13}}$

Checking option 4 lies on it

Hence answer is option 4

Posted by

shivangi.bhatnagar

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