Question

Cho hàm số \(f\left(x\right)=\left\{{}\begin{matrix}\dfrac{x-\sqrt{x+2}}{x^2-4}\left(x>2\right)\\x^2+3b\left(x< 2\right)\\2a+b-6\left(x=2\right)\end{matrix}\right.\) liên tục tại x=2. Tính I=a+b

Answer 1

\(\lim\limits f\left(x\right)_{x\rightarrow2^+}=\lim\limits_{x\rightarrow2^+}\dfrac{x-\sqrt{x+2}}{x^2-4}=\lim\limits_{x\rightarrow2^+}\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)\left(x+\sqrt{x+2}\right)}=\dfrac{3}{16}\)

\(\lim\limits_{x\rightarrow2^-}f\left(x\right)=\lim\limits_{x\rightarrow2^-}\left(x^2+3b\right)=4+3b\)

\(f\left(2\right)=2a+b-6\)

Để hàm số liên tục tại \(x=2\Rightarrow\lim\limits_{x\rightarrow2^+}f\left(x\right)=\lim\limits_{x\rightarrow2^-}f\left(x\right)=f\left(2\right)\)

\(\Leftrightarrow4+3b=2a+b-6=\dfrac{3}{16}\)

\(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{179}{48}\\b=\dfrac{-61}{48}\end{matrix}\right.\) \(\Rightarrow I=\dfrac{59}{24}\)