China Mathematical Competition

$f^{\prime }(x)=3a{x}^2+2bx+c$. We have $$\{\begin{array}{l}{f}^{\prime }(0)=c,\\ f^{\prime }\left(\frac{1}{2}\right)=\frac{3}{4}a+b+c,\\ f^{\prime }(1)=3a+2b+c.\end{array}$$ Then $$3a=2f^{\prime }(0)+2f^{\prime }(1)-4f^{\prime }\left(\frac{1}{2}\right).$$ We get $$\begin{array}{rl}3|a|& =\left|2f^{\prime }(0)+2f^{\prime }(1)-4f^{\prime }\left(\frac{1}{2}\right)\right|\\ & \le 2|f^{\prime }(0)|+2|f^{\prime }(1)|+4\left|f^{\prime }\left(\frac{1}{2}\right)\right|\le 8.\end{array}$$ Therefore, $a\le \frac{8}{3}$. Furthermore, it is easy to find that $f(x)=\frac{8}{3}{x}^3-4x^2+x+m$ (where $m$ is any constant) satisfies the given condition. Therefore, the maximum value of $a$ is $\frac{8}{3}$.

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Alternative solution.

Let $g(x)=f^{\prime }(x)+1$. Then $0\le g(x)\le 2$ for $0\le x\le 1$. Let $z=2x-1$. Then $x=\frac{z+1}{2}$ and $-1\le z\le 1$. Let $$h(z)=g\left(\frac{z+1}{2}\right)=\frac{3a}{4}{z}^2+\frac{3a+2b}{2}z+\frac{3a}{4}+b+c+1.$$ It is easy to check that $0\le h(z)\le 2$ and $0\le h(-z)\le 2$ for $-1\le z\le 1$. Therefore, $0\le \frac{h(z)+h(-z)}{2}\le 2$ for $-1\le z\le 1$. And that is $$0\le \frac{3a}{4}{z}^2+\frac{3a}{4}+b+c+1\le 2.$$ Then we have $\frac{3a}{4}+b+c+1\ge 0$ and $\frac{3a}{4}{z}^2\le 2$. From $0\le z^2\le 1$ we get $a\le \frac{8}{3}$. As $f(x)=\frac{8}{3}{x}^3-4x^2+x+m$ (where $m$ is any constant) satisfies the given condition. We obtain that the maximum value of $a$ is $\frac{8}{3}$. $\square$