Question

設(shè)函數(shù)f(x)滿足x2f′(x)＋2xf(x)＝，f(2)＝，則x＞0時(shí)，f(x)(　　)．

A．有極大值，無(wú)極小值

B．有極小值，無(wú)極大值

C．既有極大值又有極小值

D．既無(wú)極大值也無(wú)極小值

 

Answer 1

D

【解析】由x2f′(x)＋2xf(x)＝，得f′(x)＝，令g(x)＝ex－2x2f(x)，x＞0，則g′(x)＝ex－2x2f′(x)－4xf(x)＝ex－2·＝.令g′(x)＝0，得x＝2.當(dāng)x＞2時(shí)，g′(x)＞0；0＜x＜2時(shí)，g′(x)＜0，∴g(x)在x＝2時(shí)有最小值g(2)＝e2－8f(2)＝0，從而當(dāng)x＞0時(shí)，f′(x)≥0，則f(x)在(0，＋∞)上是增函數(shù)，所以函數(shù)f(x)無(wú)極大值，也無(wú)極小值．