A problem related to $\int_{0}^{1} f(x)(x-f(x))dx=1/12$

Question

Consider a differentiable function satisfying $$\int_{0}^{1} f(x)(x-f(x))dx=1/12$$ Then find the nearest integer less than or equal to $\frac{1}{f'(1)}$.

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Let $$F(x)=\int_{0}^{x}f(x)(x-f(x))$$ we notice $F(1)=1/12$ , $F(0)=0$ and $$F'(1)=f(1)-{f(1)}^2$$ What do i do next? I have tried using LMVT and Rolle's but not getting anything nice.

Answer 1

We can manipulate $$\int_{0}^{1} (4xf(x)-4f^2(x)) \, dx=1/3$$

$$\int_{0}^{1} (4xf(x)-4f^2(x)-x^2) \, dx=1/3 - \int_0^1 x^2 dx$$

$$\int_{0}^{1} (2f(x)-x)^2 \, dx=0 $$

$$\Rightarrow f(x)=x/2 $$

Answer 2

Thanks to @cosmo5 for inspiration we have in general $$ab\le \frac{{(a+b)}^2}{4}$$ $$\int_{0}^{1}f(x)((x-f(x))\le \int_{0}^1 \frac{{(f(x)+x-f(x))}^2}{4}=\int_{0}^{1}\frac{x^2}{4}=1/12$$ For equality to hold $$f(x)=x-f(x)\Rightarrow f(x)=\frac{x}{2}$$