The number of continuous functions $f:[0,1]\to\mathbb R$ that satisfy $\int_0^1xf(x)\,dx=\frac13+\frac14\int_0^1(f(x))^2\,dx$

Question

90) The number of continuous functions $f:[0,1]\to\mathbb R$ that satisfy $$\int_0^1xf(x)\,dx=\frac13+\frac14\int_0^1(f(x))^2\,dx$$ is
A) 0
B) 1
C) 2
D) $\infty$

How to approach this sum? I thought of using Newton-Leibniz but the limits are constants, so that approach failed.

Answer 1

Another way:

$$\int^{1}_{0}4xf(x)dx-\int^{1}_{0}(f(x))^2dx=\frac{4}{3}$$

$$\int^{1}_{0}f(x)\bigg(4x-f(x)\bigg)dx\leq \frac{1}{4}\int^{1}_{0}\bigg[f(x)+4x-f(x)\bigg]^2dx=\frac{4}{3}.$$

Equality hold when $f(x)=2x$

In $2$ line earlilier i have used the inequality $$ab\leq \frac{(a+b)^2}{4}$$ equality hold when $a=b.$

Answer 2

$$\int^{1}_{0}\bigg(\frac{f(x)}{2}\bigg)^2dx-\int^{1}_{0} xf(x)dx+\int^{1}_{0}x^2dx=0$$

$$\int^{1}_{0}\bigg(\frac{f(x)}{2}-x\bigg)^2dx=0$$

$$\frac{f(x)}{2}-x=0$$ so $f(x)=2x$

Answer 3

Using $$\displaystyle \int^1_0 f(x)\bigg(4x-f(x)\bigg)dx=\frac{4}{3}$$

Let $\displaystyle g(x)=f(x)\bigg(4x-f(x)\bigg)$

$\displaystyle \bigg(f(x)\bigg)^2-4xf(x)+g(x)=0$

For real roots, We have $\displaystyle D\geq 0 $

So we have

$\displaystyle 16x^2-4g(x)\geq 0\Longrightarrow g(x)\leq 4x^2$

So $$f(x)\bigg(4x-f(x)\bigg)\leq (2x)^2$$

$$\displaystyle \bigg(f(x)-2x\bigg)^2\leq 0\Longrightarrow \bigg(f(x)-2x\bigg)^2= 0$$

$$\Longrightarrow f(x)=2x$$