Question

Hàm số \(f\left(x\right)=\left\{{}\begin{matrix}\frac{\sqrt{x+3}-2}{x-1}\left(x>1\right)\\m^2+m+\frac{1}{4}\left(x\le1\right)\end{matrix}\right.\) . Tất cả giá trị m để f(x) liên tục tại x = 1 là :

A. m = 0

B. \(m\in\left\{0;-1\right\}\)

C. m = 1

D. \(m\in\left\{0;1\right\}\)

Answer 1

\(\lim\limits_{x\rightarrow1^+}\frac{\sqrt{x+3}-2}{x-1}=\lim\limits_{x\rightarrow1^+}\frac{\left(\sqrt{x+3}-2\right)\left(\sqrt{x+3}+2\right)}{\left(x-1\right)\left(\sqrt{x+3}+2\right)}=\lim\limits_{x\rightarrow1^+}\frac{x-1}{\left(x-1\right)\left(\sqrt{x+3}+2\right)}\)

\(=\lim\limits_{x\rightarrow1^+}\frac{1}{\sqrt{x+3}+2}=\frac{1}{4}\)

Để hàm số liên tục tại \(x=1\)

\(\Leftrightarrow\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^-}f\left(x\right)=f\left(1\right)\)

\(\Leftrightarrow m^2+m+\frac{1}{4}=\frac{1}{4}\)

\(\Leftrightarrow m^2+m=0\Rightarrow\left[{}\begin{matrix}m=0\\m=-1\end{matrix}\right.\)

Đáp án B