2003 Vietnamese Mathematical Olympiad

We have: $$f(\cot x)=\sin 2x+\cos 2x\forall x\in (0;\pi )$$ $$\iff f(\cot x)=\frac{2\cot x}{{\cot }^{2}x+1}+\frac{{\cot }^{2}x-1}{{\cot }^{2}x+1}=\frac{{\cot }^{2}x+2\cot x-1}{{\cot }^{2}x+1}\forall x\in (0;\pi ).$$ Therefore, remarking that for every $t\in \mathbb{R}$ there exists $x\in (0;\pi )$ such that $\cot x=t$, we get: $$f(t)=\frac{t^2+2t-1}{t^2+1}\forall t\in \mathbb{R}.$$ It implies $$g(x)=f(x)\cdot f(1-x)=\frac{x^2(1-x{)}^2+8x(1-x)-2}{x^2(1-x{)}^2-2x(1-x)+2}\forall x\in \mathbb{R}.(\ast )$$ Put $u=x(1-x)$. It is easy to see that when $x$ runs through $[-1;1]$, $u$ runs through $[-2;1/4]$. So, from (*) we have: $$\begin{array}{l}\min g(x)=\min h(u)\text{and}\max g(x)=\max h(u),\\ -1\le x\le 1-2\le u\le 1/4-1\le x\le 1-2\le u\le 1/4\end{array}$$ where $h(u)=\frac{u^2+8u-2}{u^2-2u+2}$. By studying the sign of $h^{\prime }(u)=\frac{2(-5u^2+4u+6)}{(u^2-2u+2{)}^2}$ on $[-2;1/4]$, we get: $$\underset{-2\le u\le 1/4}{\min }h(u)=h((2-\sqrt{34})/5)=4-\sqrt{34},\text{ and}$$ $$\underset{-2\le u\le 1/4}{\max }h(u)=\max \{h(-2),h(1/4)\}=\max \{-7/5,1/25\}=1/25.$$ So, on $[-1;1]$, $\min g(x)=4-\sqrt{34}$, $\max g(x)=1/25$.