Question

12．若函數(shù)變?yōu)閒（x）=$\left\{\begin{array}{l}{xlnx-a，x＞0}\\{-{x}^{2}-2x-a，x≤0}\end{array}\right.$，若函數(shù)y=f（x）有三個(gè)零點(diǎn)，則實(shí)數(shù)a的取值范圍是（-$\frac{1}{e}$，1）．

Answer 1

分析 轉(zhuǎn)化為函數(shù)y=$\left\{\begin{array}{l}{xlnx，x＞0}\\{-{x}^{2}-2x，x≤0}\end{array}\right.$與y=a的圖象有3個(gè)交點(diǎn)；求導(dǎo)分析函數(shù)y=$\left\{\begin{array}{l}{xlnx，x＞0}\\{-{x}^{2}-2x，x≤0}\end{array}\right.$，從而作出函數(shù)的大致圖象，從而解得．

解答 解：函數(shù)f（x）=$\left\{\begin{array}{l}{xlnx-a，x＞0}\\{-{x}^{2}-2x-a，x≤0}\end{array}\right.$有三個(gè)零點(diǎn)，
即函數(shù)y=$\left\{\begin{array}{l}{xlnx，x＞0}\\{-{x}^{2}-2x，x≤0}\end{array}\right.$與y=a的圖象有3個(gè)交點(diǎn)；
當(dāng)x＞0時(shí)，y=xlnx，y′=lnx+1，
故當(dāng)lnx+1=0，即x=$\frac{1}{e}$時(shí)，y=xlnx有極小值-$\frac{1}{e}$；
當(dāng)x≤0時(shí)，y=-x²-2x在x=-1時(shí)有極大值1；
作函數(shù)y=$\left\{\begin{array}{l}{xlnx，x＞0}\\{-{x}^{2}-2x，x≤0}\end{array}\right.$的圖象如右圖，
由圖象可知，
當(dāng)-$\frac{1}{e}$＜a＜1時(shí)，函數(shù)y=$\left\{\begin{array}{l}{xlnx，x＞0}\\{-{x}^{2}-2x，x≤0}\end{array}\right.$與y=a的圖象有3個(gè)交點(diǎn)；
故答案為：（-$\frac{1}{e}$，1）．

點(diǎn)評 本題考查了導(dǎo)數(shù)的綜合應(yīng)用及數(shù)形結(jié)合的思想應(yīng)用，屬于中檔題．