Question

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

• A. 0
• B. 1
• C. 2
• D. infinity

Answer 1

Answer: (B)

We have,
$\int\limits_{0}^{1} x f \left(x\right)dx =\frac{1}{3}+\frac{1}{4} \int\limits_{0}^{1}\left(f \left(x\right)\right)^{2}dx$
$\Rightarrow -\frac{1}{3}=\frac{1}{4}\left[\int\limits_{0}^{1}\left[(f\left(x\right)\right)^{2}-4xf\left(x\right)\right]dx$
$\Rightarrow -\frac{1}{3}=\frac{1}{4}\left[\int\limits_{0}^{1}\left[f\left(x\right)-2x\right]^{2}dx-\int\limits_{0}^{1}4x^{2}dx\right]$
$\Rightarrow -\frac{4}{3}=\int\limits_{0}^{1}\left[f\left(x\right)-2x\right]^{2}dx-\left[\frac{4x^{3}}{3}\right]^{1}_{0}$
$\Rightarrow \frac{4}{3}=\int\limits^{1}_0\left[\left(x\right)-2x\right]^2dx-\frac{4}{3}$
$\int\limits_0^1[f(x)-2x]^2dx=0$
$\therefore $ Only one continuous function