设函数f(x)定义在R上,f(x+1)=f(1-x),且满足x≥1,f(x)=lnx,则()A.f(13)<f(2)<f(12)B.f(12)<f(2)<f(13)C.f(12)<f(13)<f(2)

题目简介

设函数f(x)定义在R上,f(x+1)=f(1-x),且满足x≥1,f(x)=lnx,则()A.f(13)<f(2)<f(12)B.f(12)<f(2)<f(13)C.f(12)<f(13)<f(2)

题目详情

设函数f(x)定义在R上,f(x+1)=f(1-x),且满足x≥1,f(x)=lnx,则(  )
A.f(
1
3
)<f(2)<f(
1
2
B.f(
1
2
)<f(2)<f(
1
3
C.f(
1
2
)<f(
1
3
)<f(2)
D.f(2)<f(
1
2
)<f(
1
3
题型:单选题难度:中档来源:不详

答案

由f(x+1)=f(1-x),得f(class="stub"1
2
)=f(1-class="stub"1
2
)=f(1+class="stub"1
2
)=f(class="stub"3
2
),f(class="stub"1
3
)=f(1-class="stub"2
3
)=f(1+class="stub"2
3
)=f(class="stub"5
3
),
因为x≥1时,f(x)=lnx,且1<class="stub"3
2
<class="stub"5
3
<2
,所以f(class="stub"3
2
)=lnclass="stub"3
2
,f(class="stub"5
3
)=lnclass="stub"5
3
,f(2)=ln2,
又f(x)=lnx在定义域内递增,1<class="stub"3
2
<class="stub"5
3
<2

所以f(class="stub"3
2
)<f(class="stub"5
3
)<f(2),即f(class="stub"1
2
)<f(class="stub"1
3
)<f(2),
故选C.

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