Question

The number of continuous functions $f:[0,1]\to R$ that satisfy
$\underset{0}{\overset{1}{\int }}xf\left(x\right)dx=\frac{1}{3}+\frac{1}{4}\underset{0}{\overset{1}{\int }}\left(f\right(x){)}^{2}dx$

• A. $0$
• B. $1$
• C. $2$
• D. Infinity

Answer 1

The correct option is B $1$

$\underset{0}{\overset{1}{\int }}xf\left(x\right)dx=\frac{1}{3}+\frac{1}{4}\underset{0}{\overset{1}{\int }}\left(f\right(x){)}^{2}dx\Rightarrow -\frac{1}{3}=\underset{0}{\overset{1}{\int }}({\left(\frac{f\left(x\right)}{2}\right)}^2-xf(x\left)\right)dx\Rightarrow \underset{0}{\overset{1}{\int }}{(\frac{f\left(x\right)}{2}-x)}^{2}dx-\underset{0}{\overset{1}{\int }}{x}^{2}dx=-\frac{1}{3}\Rightarrow \underset{0}{\overset{1}{\int }}{(\frac{f\left(x\right)}{2}-x)}^{2}dx=0$
$\Rightarrow {(\frac{f\left(x\right)}{2}-x)}^2=0$
$\therefore f\left(x\right)=2x$
Only one function exists.