Question

The number of continuous functions $f:[0,1]\to R$ that satisfy ${\int }_0^{1}xf\left(x\right)dx=\frac{1}{3}+\frac{1}{4}{\int }_0^1\left(f\right(x){)}^{2}dx$ is

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

Answer 1

Answer: (B)

$1$

$0={\int }_0^1{x}^{2}dx+\frac{1}{4}{\int }_0^1\left(f\right(x){)}^{2}dx-{\int }_0^{1}xf(x\left)dx\right(\because {\int }_0^1{x}^{2}dx=\frac{1}{3})$

$⟹0={\int }_0^1\left[\right(\frac{f\left(x\right)}{2}{)}^2-2\ast \frac{f\left(x\right)}{2}\ast x+\left(x{)}^2\right]dx⟹{\int }_0^1(\frac{f\left(x\right)}{2}-x{)}^{2}dx=0⟹\frac{f\left(x\right)}{2}=x(\because (\frac{f\left(x\right)}{2}-x{)}^2>0)⟹f\left(x\right)=2x$

Hence there exists one such function.