Bài tập 9 trang 91 SBT Toán 11 Tập 1 Chân trời sáng tạo - CTST

Ta có: $y=f\left(x\right)=\left\{\begin{array}{l}{x}^2+ax+b\text{ khi }\left|x\right|<2\\ x\left(2-x\right)\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}khi }\left|x\right|\ge 2\end{array}\right.$

Suy ra: $y=f\left(x\right)=\left\{\begin{array}{l}{x}^2+ax+b\text{ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}khi \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}-2<x<2\\ x\left(2-x\right)\text{\hspace{0.17em} \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}khi \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}x}\le -2;x\ge 2\end{array}\right..$

⦁ $\underset{x\to -2^{-}}{\text{lim}}f\left(x\right)=\underset{x\to -2^{-}}{\text{lim}}\left[x\left(2-x\right)\right]=-2\cdot \left(2+2\right)=-8=f\left(-2\right);$

⦁ $\underset{x\to -2^{+}}{\text{lim}}f\left(x\right)=\underset{x\to -2^{+}}{\text{lim}}\left(x^2+ax+b\right)=4-2a+b;$

⦁ $\underset{x\to 2^{-}}{\text{lim}}f\left(x\right)=\underset{x\to 2^{-}}{\text{lim}}\left(x^2+ax+b\right)=4+2a+b;$

⦁ $\underset{x\to 2^{+}}{\text{lim}}f\left(x\right)=\underset{x\to 2^{+}}{\text{lim}}\left[x\left(2-x\right)\right]=2\cdot \left(2-2\right)=0=f\left(2\right)$.

Hàm số liên tục tại x = ‒ 2 và x = 2 khi và chỉ khi:

$\left\{\begin{array}{l}\underset{x\to -2^{-}}{\text{lim}}f\left(x\right)=\underset{x\to -2^{+}}{\text{lim}}f\left(x\right)=f\left(-2\right)\\ \underset{x\to 2^{-}}{\text{lim}}f\left(x\right)=\underset{x\to 2^{+}}{\text{lim}}f\left(x\right)=f\left(2\right)\end{array}\right.$

$\iff \left\{\begin{array}{l}4-2a+b=-8\\ 4+2a+b=0\end{array}\iff \left\{\begin{array}{l}-2a+b=-12\\ 2a+b=-4\end{array}\iff \left\{\begin{array}{l}a=2\\ b=-8.\end{array}\right.\right.\right.$

Vậy a = 2, b = ‒ 8 là các giá trị cần tìm.

-- Mod Toán 11 HỌC247