Question

f(x) = (x^2 + (m+1)x + m + 2) / (x^2 + x + m)

f(x) <= 2, x€R

m=?

Answer 1

$f(x) = \dfrac{x^2+(m+1)x+m+2}{x^2+x+m}\\ \\ f(x)\leq 2 \Rightarrow \dfrac{x^2+(m+1)x+m+2}{x^2+x+m} \leq 2\\ \\ \Rightarrow \dfrac{x^2+(m+1)x+m+2}{x^2+x+m} - 2 \leq 0 \\ \\ \Rightarrow \dfrac{x^2+(m+1)x+m+2 - 2x^2-2x-2m}{x^2+x+m}\leq 0\\ \\ \Rightarrow \dfrac{-x^2+(m-1)x-m+2}{x^2+x+m}\leq 0\\ \\ \Rightarrow \dfrac{x^2+(1-m)x+m-2}{x^2+x+m}\geq 0\\ \\ \\\boxed{1}\quad \begin{cases}x^2+(1-m)x+m-2 \geq 0 \\ x^2+x+m > 0 \end{cases} \Rightarrow$

$\Rightarrow \begin{cases}(1-m)^2 - 4(m-2)\leq 0\,\,\,\{\Delta \leq 0\} \\ 1 - 4m < 0\,\,\,\{\Delta < 0\}\end{cases} \Rightarrow \begin{cases}m^2-6m+9 \leq 0\\ m >\dfrac{1}{4}\end{cases} \Rightarrow \\ \\\Rightarrow (m-3)^2\leq 0 \Rightarrow m = 3>\dfrac{1}{4} \quad (A)\\ \\\boxed{2}\quad\begin{cases}x^2+(1-m)x+m-2 \leq 0 \\x^2+x+m < 0\end{cases}\quad (Fals,\quad \forall m\in \mathbb{R}) \\\\\\\\\text{Din }\boxed{1}\,\cup\,\boxed{2} \Rightarrow \boxed{m=3}$