已知函數(shù)f(x)=x2-2(-1)klnx(k∈N*)存在極值,則k的取值集合是( )
A.{2,4,6,8,…}
B.{0,2,4,6,8,…}
C.{l,3,5,7,…}
D.N*
【答案】
分析:對(duì)k分奇偶討論,對(duì)原函數(shù)求導(dǎo),進(jìn)而探求在導(dǎo)數(shù)為0的左右附近,導(dǎo)數(shù)符號(hào)的改變,從而確定是否存在極值點(diǎn).
解答:解:∵k∈N
*,
①當(dāng)k的取值集合是{2,4,6,8,…}時(shí),函數(shù)f(x)=x
2-2lnx,
∴f'(x)=2x-
=
,由f'(x)=0得x=-1,或x=1.
當(dāng)x∈(-∞,-1)或x∈(1,+∞)時(shí),y′>0;
當(dāng)x∈(-1,1)時(shí),y′<0
∴當(dāng)x=-1和x=1是函數(shù)的極值點(diǎn).
②當(dāng)k的取值集合是{l,3,5,7,…}時(shí),函數(shù)f(x)=x
2+2lnx,
∴f'(x)=2x+
=
,由f'(x)=0得x∈∅.故此時(shí)原函數(shù)不存在極值點(diǎn).
故選A.
點(diǎn)評(píng):本題以函數(shù)為載體,考查導(dǎo)數(shù)的運(yùn)用,考查函數(shù)的極值,關(guān)鍵是求導(dǎo)函數(shù),并注意在導(dǎo)數(shù)為0的左右附近,導(dǎo)數(shù)符號(hào)的改變.