Question

3．設函數f（x）=$\left\{\begin{array}{l}{{x}^{2}+bx+c，x≤0}\\{2，x＞0}\end{array}\right.$，若f（-4）=2，f（-2）=-2，則關于x的方程f（x）=x的解的個數為（　�。�

• A． 1
• B． 2
• C． 3
• D． 4

Answer 1

分析 求出f（x）的解析式，解方程f（x）=x，根據解得個數得出結論．

解答 解：∵f（-4）=2，f（-2）=-2，
∴$\left\{\begin{array}{l}{16-4b+c=2}\\{4-2b+c=-2}\end{array}\right.$，解得：$\left\{\begin{array}{l}{b=4}\\{c=2}\end{array}\right.$，
∴f（x）=$\left\{\begin{array}{l}{{x}^{2}+4x+2，x≤0}\\{2，x＞0}\end{array}\right.$，
令f（x）=x得$\left\{\begin{array}{l}{{x}^{2}+4x+2=x}\\{x≤0}\end{array}\right.$或$\left\{\begin{array}{l}{x=2}\\{x＞0}\end{array}\right.$，
解得x=-1或x=-2或x=2．
∴f（x）=x有3解，
故選C．

點評 本題考查了方程的解得個數判斷，屬于基礎題．