Biết rằng đồ thị hàm số bậc 4: y = f(x) được cho như hình vẽ sau: Tìm số

* Hướng dẫn giải

Đáp án A

Đồ thị hàm số y = f(x) cắt trục hoành tại 4 điểm phân biệt nên

$$\begin{gathered} f\left(x\right)=a\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)\left(x-x_4\right) \\ f\text{'}\left(x\right)=a\left(x-x_2\right)\left(x-x_3\right)\left(x-x_4\right)+a\left(x-x_1\right)\left(x-x_3\right)\left(x-x_4\right) \\ +a\left(x-x_1\right)\left(x-x_2\right)\left(x-x_4\right)+a\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right) \end{gathered}$$

Rõ ràng

$$\begin{gathered} f\left(x\right)=0\Rightarrow \left[\begin{array}{c}x=x_1\\ x=x_2\\ x=x_3\\ x=x_4\end{array}\right.\Rightarrow f\text{'}\left(x\right)\ne 0 \\ \Rightarrow g\left(x\right)={\left[f\text{'}\left(x\right)\right]}^2-f\text{'}\text{'}\left(x\right).f\left(x\right)\ne 0 \end{gathered}$$

nên ta chỉ xét $f\left(x\right)\ne 0$

$\Rightarrow \frac{f\text{'}\left(x\right)}{f\left(x\right)}=\frac{1}{x-x_1}+\frac{1}{x-x_2}+\frac{1}{x-x_3}+\frac{1}{x-x_4},\forall x\in \left\{x_1;x_2;x_3;x_4\right\}$

Đặt $h\left(x\right)=\frac{f\text{'}\left(x\right)}{f\left(x\right)}=\frac{1}{x-x_1}+\frac{1}{x-x_2}+\frac{1}{x-x_3}+\frac{1}{x-x_4},\forall x\in \left\{x_1;x_2;x_3;x_4\right\}$

Ta có: $h\text{'}\left(x\right)=\frac{f\text{'}\text{'}\left(x\right).f\left(x\right)-{\left[f\text{'}\left(x\right)\right]}^2}{f^2\left(x\right)}$

$$\begin{gathered} =\frac{-1}{{\left(x-x_1\right)}^2}+\frac{-1}{{\left(x-x_2\right)}^2}+\frac{-1}{{\left(x-x_3\right)}^2}+\frac{-1}{{\left(x-x_4\right)}^2}<0,\forall x\in \left\{x_1;x_2;x_3;x_4\right\} \\ \Rightarrow f\text{'}\text{'}\left(x\right).f\left(x\right)-{\left[f\text{'}\left(x\right)\right]}^2<0,\forall x\in \left\{x_1;x_2;x_3;x_4\right\} \\ \Rightarrow g\left(x\right)=-f\text{'}\text{'}\left(x\right).f\left(x\right)+{\left[f\text{'}\left(x\right)\right]}^2<0,\forall x\in \left\{x_1;x_2;x_3;x_4\right\} \end{gathered}$$

Vậy đồ thị hàm số $y=g\left(x\right)={\left[f\text{'}\left(x\right)\right]}^2-f\text{'}\text{'}\left(x\right).f\left(x\right)$ không cắt trục Ox