Question

Cho lim \(\frac{x^2+ax+b}{x^2-1}=-\frac{1}{2}\) ( a , b \(\in\) R ) ( x\(\rightarrow\) 1) . Tổng \(S=a^2+b^2\) bằng

A. S = 13

B. S = 9

C. S = 4

D. S = 1

Answer 1

\(\lim\limits_{x\rightarrow1}\frac{x^2+ax+b}{\left(x-1\right)\left(x+1\right)}=-\frac{1}{2}\) hữu hạn

\(\Rightarrow\) phương trình \(x^2+ax+b=0\) có 1 nghiệm bằng 1

\(\Leftrightarrow1+a+b=0\Rightarrow b=-a-1\)

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

\(\Rightarrow\frac{a+2}{2}=-\frac{1}{2}\Rightarrow a=-3\Rightarrow b=2\)

\(\Rightarrow a^2+b^2=\left(-3\right)^2+2^2=13\)