Question

13．已知函數(shù)f（x）=$\left\{\begin{array}{l}{{e}^{x}，x≤0}\\{{x}^{2}-2x+1，x＞0}\end{array}\right.$，則方程f²（x）-3f（x）+2=0的根的個(gè)數(shù)是（　�。�

• A． 3
• B． 4
• C． 5
• D． 6

Answer 1

分析 求解方程f²（x）-3f（x）+2=0，得f（x）=1或f（x）=2，畫(huà)出函數(shù)f（x）=$\left\{\begin{array}{l}{{e}^{x}，x≤0}\\{{x}^{2}-2x+1，x＞0}\end{array}\right.$的圖象，數(shù)形結(jié)合得答案．

解答 解：由f²（x）-3f（x）+2=0，得f（x）=1或f（x）=2．
畫(huà)出函數(shù)f（x）=$\left\{\begin{array}{l}{{e}^{x}，x≤0}\\{{x}^{2}-2x+1，x＞0}\end{array}\right.$的圖象如圖：

由圖可知，方程f（x）=1有1根，方程f（x）=2有2根．
∴方程f²（x）-3f（x）+2=0的根的個(gè)數(shù)是3．
故選：A．

點(diǎn)評(píng) 本題考查根的存在性及根的個(gè)數(shù)判斷，考查數(shù)學(xué)轉(zhuǎn)化思想方法與數(shù)形結(jié)合的解題思想方法，是中檔題．