Question

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

• A. 0
• B. 1
• C. 2
• D. infinity

Answer 1

Answer: (B)

We have,
$\underset{0}{\overset{1}{\int }}xf\left(x\right)dx=\frac{1}{3}+\frac{1}{4}\underset{0}{\overset{1}{\int }}{\left(f\left(x\right)\right)}^{2}dx$
$\Rightarrow -\frac{1}{3}=\frac{1}{4}\left[\underset{0}{\overset{1}{\int }}{\left[(f\left(x\right)\right)}^2-4xf\left(x\right)\right]dx$
$\Rightarrow -\frac{1}{3}=\frac{1}{4}\left[\underset{0}{\overset{1}{\int }}{\left[f\left(x\right)-2x\right]}^{2}dx-\underset{0}{\overset{1}{\int }}4x^{2}dx\right]$
$\Rightarrow -\frac{4}{3}=\underset{0}{\overset{1}{\int }}{\left[f\left(x\right)-2x\right]}^{2}dx-{\left[\frac{4x^3}{3}\right]}_0^1$
$\Rightarrow \frac{4}{3}=\underset{0}{\overset{1}{\int }}{\left[\left(x\right)-2x\right]}^{2}dx-\frac{4}{3}$
$\underset{0}{\overset{1}{\int }}[f(x)-2x{]}^{2}dx=0$
$\therefore$ Only one continuous function