設(shè)f(x)=xlogax+(1-x)loga(1-x)(a>1)
(1)判斷f(x)的單調(diào)性;
(2)已知m+n=4,且m>0,n>0,求mlog4m+nlog4n的最小值.
解:(1)由f′(x)=log
ax+
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-log
a(1-x)-
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=log
ax-log
a(1-x)=log
a
令f′(x)=0得x=
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∵a>1,∴當(dāng)0<x<
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時,
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,f′(x)<0;同理,當(dāng)
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時,f′(x)>0
∴f(x)在(0,
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)上遞減,在(
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,1)上遞增…(6分)
(2)由(1)知,f(x)在x=
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處取最小值,f(x)
min=f(
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)=log
a
令m=4m
1,n=4n
1,則m
1+n
1=1,所以mlog
4m+nlog
4n=4[1+m
1log
4m
1+(1-m
1)log
4(1-m
1)]≥4(1+log
4
)=2
∴mlog
4m+nlog
4n的最小值為2.…(12分)
分析:(1)求導(dǎo)函數(shù),令f′(x)=log
a
=0得x=
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,利用導(dǎo)數(shù)的正負(fù),即可確定函數(shù)的單調(diào)性;
(2)由(1)知,f(x)在x=
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處取最小值,f(x)
min=f(
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)=log
a
,令m=4m
1,n=4n
1,則m
1+n
1=1,所以mlog
4m+nlog
4n=4[1+m
1log
4m
1+(1-m
1)log
4(1-m
1)],由此可得結(jié)論.
點(diǎn)評:本題考查導(dǎo)數(shù)知識的運(yùn)用,考查函數(shù)的單調(diào)性,考查最值的求解,解題的關(guān)鍵是求導(dǎo)確定函數(shù)的最值.