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设函数f1(x)=x2,f2(x)=3x+1,f3(x)=sinπx,xi=i9(i=0,1,2,…,9),记Ik=9i=1|fk(xi)-fk(xi-1)|,则()A.I1<I2<I3B.I2<I1<I3C.I3<I2<I1D.I1<I3<I2

题目详情

设函数f1(x)=x2,f2(x)=

3
x+1
,f3(x)=sinπx,xi=
i
9
(i=0,1,2,…,9),记Ik=
9
i=1
|fk(xi)-fk(xi-1)|,则(  )

A. I1<I2<I3

B. I2<I1<I3

C. I3<I2<I1

D. I1<I3<I2

▼优质解答
答案和解析
f1(x)=x2,在(0,1)是单调增函数,|f1(xi)-f1(xi-1)|=f1(xi)-f1(xi-1),
I1=|f1(x1)-f1(x0)|+|f1(x2)-f1(x1)丨+…+|f1(x9)-fk(x8)|,
=f1(x1)-f1(x0)+f1(x2)-f1(x1)+…+f1(x9)-f1(x8),
=f1(x9)-f1(x0),
=f1(1)-f1(0),
=1;
f2(x)=
3
x+1
,在(0,1)是单调减函数,|f1(xi)-f1(xi-1)|=f1(xi-1)-f1(xi),
I2=|f2(x1)-f2(x0)|+|f2(x2)-f2(x1)丨+…+|f2(x9)-f2(x8)|,
=f1(x0)-f1(x9),
=
3
2

f3(x)=sinπx,在(0,
1
2
)单调递增,在(
1
2
,1)单调递增,且图象关于x=
1
2
对称,
I3=|f3(x1)-f3(x0)|+|f3(x2)-f3(x1)丨+…+|f3(x9)-f3(x8)|,
=f3(x1)-f3(x0)+f3(x2)-f3(x1)+…+f3(x5)-f3(x4)+f3(x5)-f3(x6)+…+f3(x7)-f3(x9)+f3(x8)-f3(x9),
=f3(x5)-f3(x0)+f3(x5)-f3(x9),
=2.
故答案为:A.