2019高考数学二轮复习专题--数列课件及练习（与）2019高考数学二轮复习专题-函数与导数提分训练 本文简介：
2019高考数学二轮复习专题--数列课件及练习（与）2019高考数学二轮复习专题-函数与导数提分训练2019高考数学二轮复习专题--数列课件及练习 等差数列、等比数列的基本问题1.等差数列{an}前9项的和等于前4项的和,若a1=1,ak+a4=0,则k=.?2.已知在等比数列{an}中,a3=2,
2019高考数学二轮复习专题--数列课件及练习（与）2019高考数学二轮复习专题-函数与导数提分训练 本文内容：
2019高考数学二轮复习专题--数列课件及练习（与）2019高考数学二轮复习专题-函数与导数提分训练2019高考数学二轮复习专题--数列课件及练习 等差数列、等比数列的基本问题
1.等差数列{an}前9项的和等于前4项的和,若a1=1,ak+a4=0,则k=
.?
2.已知在等比数列{an}中,a3=2,a4a6=16,则(a_7
"-"
a_9)/(a_3
"-"
a_5
)=
.?
3.(2018江苏南通中学高三考前冲刺练习)已知等差数列{an}的公差d=3,Sn是其前n项和,若a1,a2,a9成等比数列,则S5的值为
.?
4.(2018南通高三第二次调研)设等比数列{an}的前n项和为Sn.若S3,S9,S6成等差数列,且a8=3,则a5=
.?
5.设数列{an}的首项a1=1,且满足a2n+1=2a2n-1与a2n=a2n-1+1,则数列{an}的前20项和为
.?
6.(2018江苏锡常镇四市高三教学情况调研(二))已知公差为d的等差数列{an}的前n项和为Sn,若S_10/S_5
=4,则(4a_1)/d=
.?
7.已知Sn为数列{an}的前n项和,若a1=2,且〖〖S_n〗_+〗_1=2Sn,设bn=log2an,则1/(b_1
b_2
)+1/(b_2
b_3
)+…+1/(b_10
b_11
)的值是
.?
8.(2018扬州高三第三次调研)已知实数a,b,c成等比数列,a+6,b+2,c+1成等差数列,则b的最大值为
.?
9.(2018扬州高三第三次调研)已知数列{an}满足an+1+(-1)nan=(n+5)/2(n∈N*),数列{an}的前n项和为Sn.
(1)求a1+a3的值;(2)若a1+a5=2a3.
①求证:数列{a2n}为等差数列;
②求满足S2p=4S2m(p,m∈N*)的所有数对(p,m).10.(2018苏锡常镇四市高三教学情况调研(二))已知等差数列{an}的首项为1,公差为d,数列{bn}的前n项和为Sn,若对任意的n∈N*,6Sn=9bn-an-2恒成立.
(1)如果数列{Sn}是等差数列,证明数列{bn}也是等差数列;
(2)如果数列{b_n+1/2}为等比数列,求d的值;
(3)如果d=3,数列{cn}的首项为1,cn=bn-bn-1(n≥2),证明数列{an}中存在无穷多项可表示为数列{cn}中的两项之和.
?
答案精解精析
1.答案 10
解析 S9=S4,则9a1+36d=4a1+6d,a1+6d=a7=0,则a4+a10=2a7=0,则k=10.
2.答案 4
解析 等比数列中奇数项符号相同,a3>0,则a5>0,又a4a6=〖a_5〗^2=16,则a5=4,从而a7=8,a9=16,则(a_7
"-"
a_9)/(a_3
"-"
a_5
)=("-"
8)/("-"
2)=4.
3.答案 65/2
解析 由题意可得a1a9=〖a_2〗^2,则由a1(a1+24)=(a1+3)2,解得a1=1/2,则S5=5×1/2+(5×4)/2×3=65/2.
4.答案 -6
解析 由S3,S9,S6成等差数列可得S3+S6=2S9,当等比数列{an}的公比q=1时不成立,则q≠1,(a_1
"("
1"-"
q^3
")"
)/(1"-"
q)+(a_1
"("
1"-"
q^6
")"
)/(1"-"
q)=2(a_1
"("
1"-"
q^9
")"
)/(1"-"
q),化简得2q6-q3-1=0,q3=-1/2(舍去1),则a5=a_8/q^3
=-6.
5.答案 2056
解析 由题意可得奇数项构成等比数列,则a1+a3+…+a19=(1"-"
2^10)/(1"-"
2)=1023,偶数项a2+a4+…+a20=(a1+1)+(a3+1)+…+(a19+1)=1033,故数列{an}的前20项和为2056.
6.答案 2
解析 由〖S_1〗_0/S_5
=4得〖S_1〗_0=4S5,即10a1+45d=4(5a1+10d),则(4a_1)/d=2.
7.答案 19/10
解析 由〖〖S_n〗_+〗_1=2Sn,且S1=a1=2,得数列{Sn}是首项、公比都为2的等比数列,则Sn=2n.当n≥2时,an=Sn-Sn-1=2n-2n-1=2n-1,a1=2不适合,则an={■(2","
n=1","
@2^(n"-"
1)
","
n≥2","
)┤故bn={■(1","
n=1","
@n"-"
1","
n≥2","
)┤所以1/(b_1
b_2
)+1/(b_2
b_3
)+…+1/(b_10
b_11
)=1+1/(1×2)+1/(2×3)+…+1/(9×10)=1+(1"-"?
1/2)+(1/2
"-"?
1/3)+…+(1/9
"-"?
1/10)=2-1/10=19/10.
8.答案 3/4
解析 设等比数列a,b,c的公比为q(q≠0),则a=b/q,c=bq,又a+6=b/q+6,b+2,c+1=bq+1成等差数列,则(b/q+6)+(bq+1)=2(b+2),化简得b=3/2"-"
(q+1/q)
,当b最大时q<0,此时q+1/q≤-2,b=3/2"-"
(q+1/q)
≤3/4,当且仅当q=-1时取等号,故b的最大值为3/4.
9.解析 (1)由条件,得{■(a_2
"-"
a_1=3",
①"
@a_3+a_2=7/2
",②"
)┤
②-①得a1+a3=1/2.
(2)①证明:因为an+1+(-1)nan=(n+5)/2,
所以{■(a_2n
"-"
a_(2n"-"
1)=(2n+4)/2
",③"
@a_(2n+1)+a_2n=(2n+5)/2
",④"
)┤
④-③得a2n-1+a2n+1=1/2.
于是1=1/2+1/2=(a1+a3)+(a3+a5)=4a3,
所以a3=1/4,又由(1)知a1+a3=1/2,则a1=1/4.
所以a2n-1-1/4=-(a_(2n"-"
3)
"-"?
1/4)=…
=(-1)n-1(a_1
"-"?
1/4)=0,
所以a2n-1=1/4,将其代入③式,
得a2n=n+9/4.
所以a2(n+1)-a2n=1(常数),所以数列{a2n}为等差数列.
②易知a1=a2n+1,所以S2n=a1+a2+…+a2n
=(a2+a3)+(a4+a5)+…+(a2n+a2n+1)
=n^2/2+3n.
由S2p=4S2m知p^2/2+3p=4(m^2/2+3m).
所以(2m+6)2=(p+3)2+27,
即(2m+p+9)(2m-p+3)=27,又p,m
∈N*,
所以2m+p+9≥12且2m+p+9,2m-p+3均为正整数,所以{■(2m+p+9=27","
@2m"-"
p+3=1","
)┤解得p=10,m=4,
所以所求数对为(10,4).
10.解析 (1)证明:设数列{Sn}的公差为d",
∵6Sn=9bn-an-2,①
6Sn-1=9bn-1-an-1-2(n≥2),②
①-②得6(Sn-Sn-1)=9(bn-bn-1)-(an-an-1),③
即6d"=9(bn-bn-1)-d,所以bn-bn-1=(6d"""
+d)/9(n≥2)为常数,
所以{bn}为等差数列.
(2)由③得6bn=9bn-9bn-1-d,即3bn=9bn-1+d,
因为{b_n+1/2}为等比数列,所以(b_n+1/2)/(b_(n"-"
1)+1/2)=(3b_(n"-"
1)+d/3+1/2)/(b_(n"-"
1)+1/2)=(3(b_(n"-"
1)+1/2)+d/3
"-"
1)/(b_(n"-"
1)+1/2)=3+(d/3
"-"
1)/(b_(n"-"
1)+1/2)(n≥2)是与n无关的常数,
所以d/3-1=0或bn-1+1/2为常数.
当d/3-1=0时,d=3,符合题意;
当bn-1+1/2为常数时,
在6Sn=9bn-an-2中,令n=1,则6b1=9b1-a1-2,又a1=1,解得b1=1,
所以bn-1+1/2=b1+1/2=3/2(n≥2),
此时3+(d/3
"-"
1)/(b_(n"-"
1)+1/2)=3+(d/3
"-"
1)/(3/2)=1,
解得d=-6.
综上,d=3或d=-6.
(3)证明:当d=3时,an=3n-2.
由(2)得数列{b_n+1/2}是以3/2为首项,3为公比的等比数列,所以bn+1/2=3/2×3n-1=1/2?3n.即bn=1/2(3n-1).
当n≥2时,cn=bn-bn-1=1/2(3n-1)-1/2(3n-1-1)=3n-1;
当n=1时,也满足上式,
所以cn=3n-1(n≥1).
设an=ci+cj(1≤i
∈N*),则3n-2=3i-1+3j-1,即3n-(3i-1+3j-1)=2,
如果i≥2,因为3n为3的倍数,3i-1+3j-1为3的倍数,所以3n-(3i-1+3j-1)为3的倍数,
则2也为3的倍数,矛盾.
所以i=1,则3n=3+3j-1,即n=1+3j-2(j=2,3,4,…).
所以数列{an}中存在无穷多项可表示为数列{cn}中的两项之和.
2019高考数学二轮复习专题-函数与导数提分训练函数的图象与性质
1.设集合A=[-1,0],B={y|y=(1/2)^(x^2
"-"
1)
","
x"∈"
R},则A∪B=
.
2.函数f(x)=ln(1-√(3"-"
x))的定义域为
.
3.已知函数f(x)是定义在R上的周期为2的奇函数,当019/3)的值为
.
4.(2018南京师大附中高三年级模拟)“a=1”是“函数f(x)=(x+1)/x+sinx-a2为奇函数”的
条件.(填“充分不必要”“必要不充分”“充要”或“既不充分也不必要”)
5.(2018江苏扬州中学高三年级第四次模拟)若函数f(x)=xln(x+√(a+x^2
))为偶函数,则a=
.
6.(2018江苏扬州高三第一次模拟)已知函数f(x)={■(log_(1/2)
"(-"
x+1")-"
1","
x"∈[-"
1","
k"],"
@"-"
2"|"
x"-"
1"|,"
x"∈("
k","
a"],"
)┤若存在实数k使得该函数的值域为[-2,0],则实数a的取值范围是
.
7.(2018江苏苏中地区四校高三联考)已知函数f(x)=x2-2|x|+4的定义域为[a,b],其中a.
8.(2017无锡高三调研)已知函数f(x)={■((x^2+2x"-"
1)/x^2
","
x≤"-"
1/2
","
@log_(1/2)
(1+x)/2
","
x>"-"
1/2
","
)┤g(x)=-x2-2x-2.若存在a∈R,使得f(a)+g(b)=0,则实数b的取值范围是
.
9.(2018江苏苏州中学高三检测)已知函数f(x)=a+1/(4^x+1)的图象过点(1",-"
3/10).
(1)判断函数f(x)的奇偶性,并说明理由;
(2)若-1/6≤f(x)≤0,求实数x的取值范围.10.(2018江苏如东高级中学高三上学期期中)已知函数f(x)=ax2-2ax+2+b(a≠0)在区间[2,3]上有最大值5,最小值2.
(1)求a,b的值;
(2)若b<1,g(x)=f(x)-2mx在[2,4]上是单调函数,求实数m的取值范围.
答案精解精析
1.答案 [-1,2]
解析 因为x2-1≥-1,所以0<(1/2)^(x^2
"-"
1)≤2,则B=(0,2],所以A∪B=[-1,2].
2.答案 (2,3]
解析 要使函数f(x)=ln(1-√(3"-"
x))有意义,则{■(1"-"
√(3"-"
x)>0","
@3"-"
x≥0","
)┤解得23.答案 -2
解析 f("-"
19/3)=f("-"
19/3+6)=f("-"
1/3)=-f(1/3)=-2.
4.答案 充分不必要
解析 若f(x)为奇函数,则f(-1)=-f(1),即sin(-1)-a2=-2-sin1+a2,a2=1,a=±1,故“a=1”是“函数f(x)=(x+1)/x+sinx-a2为奇函数”的充分不必要条件.
5.答案 1
解析 由题意知f(x)=f(-x),
即xln(x+√(a+x^2
))=-xln[(-x)+√(a+"(-"
x")"
^2
)],
所以xln(x+√(a+x^2
))+xln(-x+√(a+x^2
))=0,
所以xln(x2+a-x2)=0,所以xlna=0,则a=1.
6.答案 (1/2
","
2]
解析 作出函数f(x)的图象(图略),由图可得实数a的取值范围是(1/2
","
2].
7.答案 (1,4)
解析 因为f(x)=(|x|-1)2+3≥3,所以3a≥3,a≥1,则函数f(x)=(x-1)2+3,x∈[a,b]单调递增,所以f(a)=3a,f(b)=3b,则a,b是方程f(x)=x2-2x+4=x的两根,且a8.答案 (-2,0)
解析 当x≤-1/2时,令1/x=t,t∈[-2,0),则y=-t2+2t+1∈[-7,1),当x>-1/2时,(1+x)/2>1/4,
f(x)=log_(1/2)
(1+x)/2<2,所以f(x)的值域是(-∞,2).存在a∈R,使得f(a)+g(b)=0,则-g(b)=f(a)<2,即b2+2b+2<2,
解得-29.解析 (1)f(x)是奇函数.理由:因为f(x)的图象过点(1",-"
3/10),所以a+1/5=-3/10,解得a=-1/2,所以f(x)=1/(4^x+1)-1/2=(1"-"
4^x)/(2"("
4^x+1")"
),f(x)的定义域为R.因为f(-x)=1/(4^("-"
x)+1)-1/2=4^x/(4^x+1)-1/2=(4^x
"-"
1)/(2"("
4^x+1")"
)=-f(x),所以f(x)是奇函数.
(2)因为-1/6≤f(x)≤0,所以-1/6≤1/(4^x+1)-1/2≤0,
所以1/3≤1/(4^x+1)≤1/2,
所以2≤4x+1≤3,所以1≤4x≤2,解得0≤x≤1/2.
10.解析 (1)f(x)=a(x-1)2+2+b-a.
①当a>0时,f(x)在[2,3]上为增函数,
故{■(f"("
3")"
=5","
@f"("
2")"
=2","
)┤所以{■(9a"-"
6a+2+b=5","
@4a"-"
4a+2+b=2","
)┤解得{■(a=1","
@b=0"."
)┤
②当a<0时,f(x)在[2,3]上为减函数,
故{■(f"("
3")"
=2","
@f"("
2")"
=5","
)┤所以{■(9a"-"
6a+2+b=2","
@4a"-"
4a+2+b=5","
)┤解得{■(a="-"
1","
@b=3"."
)┤
故{■(a=1","
@b=0)┤或{■(a="-"
1","
@b=3"."
)┤
(2)因为b<1,所以a=1,b=0,即f(x)=x2-2x+2,
g(x)=x2-2x+2-2mx=x2-(2+2m)x+2.
若g(x)在[2,4]上单调,则(2+2^m)/2≤2或(2+2^m)/2≥4,
所以2m≤2或2m≥6,即m≤1或m≥log26.
故实数m的取值范围是(-∞,1]∪[log26,+∞).2019高考数学二轮复习专题--数列课件及练习（2）
等差数列、等比数列的基本问题
1.等差数列{an}前9项的和等于前4项的和,若a1=1,ak+a4=0,则k=
.?
2.已知在等比数列{an}中,a3=2,a4a6=16,则(a_7
"-"
a_9)/(a_3
"-"
a_5
)=
.?
3.(2018江苏南通中学高三考前冲刺练习)已知等差数列{an}的公差d=3,Sn是其前n项和,若a1,a2,a9成等比数列,则S5的值为
.?
4.(2018南通高三第二次调研)设等比数列{an}的前n项和为Sn.若S3,S9,S6成等差数列,且a8=3,则a5=
.?
5.设数列{an}的首项a1=1,且满足a2n+1=2a2n-1与a2n=a2n-1+1,则数列{an}的前20项和为
.?
6.(2018江苏锡常镇四市高三教学情况调研(二))已知公差为d的等差数列{an}的前n项和为Sn,若S_10/S_5
=4,则(4a_1)/d=
.?
7.已知Sn为数列{an}的前n项和,若a1=2,且〖〖S_n〗_+〗_1=2Sn,设bn=log2an,则1/(b_1
b_2
)+1/(b_2
b_3
)+…+1/(b_10
b_11
)的值是
.?
8.(2018扬州高三第三次调研)已知实数a,b,c成等比数列,a+6,b+2,c+1成等差数列,则b的最大值为
.?
9.(2018扬州高三第三次调研)已知数列{an}满足an+1+(-1)nan=(n+5)/2(n∈N*),数列{an}的前n项和为Sn.
(1)求a1+a3的值;(2)若a1+a5=2a3.
①求证:数列{a2n}为等差数列;
②求满足S2p=4S2m(p,m∈N*)的所有数对(p,m).10.(2018苏锡常镇四市高三教学情况调研(二))已知等差数列{an}的首项为1,公差为d,数列{bn}的前n项和为Sn,若对任意的n∈N*,6Sn=9bn-an-2恒成立.
(1)如果数列{Sn}是等差数列,证明数列{bn}也是等差数列;
(2)如果数列{b_n+1/2}为等比数列,求d的值;
(3)如果d=3,数列{cn}的首项为1,cn=bn-bn-1(n≥2),证明数列{an}中存在无穷多项可表示为数列{cn}中的两项之和.
?
答案精解精析
1.答案 10
解析 S9=S4,则9a1+36d=4a1+6d,a1+6d=a7=0,则a4+a10=2a7=0,则k=10.
2.答案 4
解析 等比数列中奇数项符号相同,a3>0,则a5>0,又a4a6=〖a_5〗^2=16,则a5=4,从而a7=8,a9=16,则(a_7
"-"
a_9)/(a_3
"-"
a_5
)=("-"
8)/("-"
2)=4.
3.答案 65/2
解析 由题意可得a1a9=〖a_2〗^2,则由a1(a1+24)=(a1+3)2,解得a1=1/2,则S5=5×1/2+(5×4)/2×3=65/2.
4.答案 -6
解析 由S3,S9,S6成等差数列可得S3+S6=2S9,当等比数列{an}的公比q=1时不成立,则q≠1,(a_1
"("
1"-"
q^3
")"
)/(1"-"
q)+(a_1
"("
1"-"
q^6
")"
)/(1"-"
q)=2(a_1
"("
1"-"
q^9
")"
)/(1"-"
q),化简得2q6-q3-1=0,q3=-1/2(舍去1),则a5=a_8/q^3
=-6.
5.答案 2056
解析 由题意可得奇数项构成等比数列,则a1+a3+…+a19=(1"-"
2^10)/(1"-"
2)=1023,偶数项a2+a4+…+a20=(a1+1)+(a3+1)+…+(a19+1)=1033,故数列{an}的前20项和为2056.
6.答案 2
解析 由〖S_1〗_0/S_5
=4得〖S_1〗_0=4S5,即10a1+45d=4(5a1+10d),则(4a_1)/d=2.
7.答案 19/10
解析 由〖〖S_n〗_+〗_1=2Sn,且S1=a1=2,得数列{Sn}是首项、公比都为2的等比数列,则Sn=2n.当n≥2时,an=Sn-Sn-1=2n-2n-1=2n-1,a1=2不适合,则an={■(2","
n=1","
@2^(n"-"
1)
","
n≥2","
)┤故bn={■(1","
n=1","
@n"-"
1","
n≥2","
)┤所以1/(b_1
b_2
)+1/(b_2
b_3
)+…+1/(b_10
b_11
)=1+1/(1×2)+1/(2×3)+…+1/(9×10)=1+(1"-"?
1/2)+(1/2
"-"?
1/3)+…+(1/9
"-"?
1/10)=2-1/10=19/10.
8.答案 3/4
解析 设等比数列a,b,c的公比为q(q≠0),则a=b/q,c=bq,又a+6=b/q+6,b+2,c+1=bq+1成等差数列,则(b/q+6)+(bq+1)=2(b+2),化简得b=3/2"-"
(q+1/q)
,当b最大时q<0,此时q+1/q≤-2,b=3/2"-"
(q+1/q)
≤3/4,当且仅当q=-1时取等号,故b的最大值为3/4.
9.解析 (1)由条件,得{■(a_2
"-"
a_1=3",
①"
@a_3+a_2=7/2
",②"
)┤
②-①得a1+a3=1/2.
(2)①证明:因为an+1+(-1)nan=(n+5)/2,
所以{■(a_2n
"-"
a_(2n"-"
1)=(2n+4)/2
",③"
@a_(2n+1)+a_2n=(2n+5)/2
",④"
)┤
④-③得a2n-1+a2n+1=1/2.
于是1=1/2+1/2=(a1+a3)+(a3+a5)=4a3,
所以a3=1/4,又由(1)知a1+a3=1/2,则a1=1/4.
所以a2n-1-1/4=-(a_(2n"-"
3)
"-"?
1/4)=…
=(-1)n-1(a_1
"-"?
1/4)=0,
所以a2n-1=1/4,将其代入③式,
得a2n=n+9/4.
所以a2(n+1)-a2n=1(常数),所以数列{a2n}为等差数列.
②易知a1=a2n+1,所以S2n=a1+a2+…+a2n
=(a2+a3)+(a4+a5)+…+(a2n+a2n+1)
=n^2/2+3n.
由S2p=4S2m知p^2/2+3p=4(m^2/2+3m).
所以(2m+6)2=(p+3)2+27,
即(2m+p+9)(2m-p+3)=27,又p,m∈N*,
所以2m+p+9≥12且2m+p+9,2m-p+3均为正整数,所以{■(2m+p+9=27","
@2m"-"
p+3=1","
)┤解得p=10,m=4,
所以所求数对为(10,4).
10.解析 (1)证明:设数列{Sn}的公差为d",∵6Sn=9bn-an-2,①
6Sn-1=9bn-1-an-1-2(n≥2),②
①-②得6(Sn-Sn-1)=9(bn-bn-1)-(an-an-1),③
即6d"=9(bn-bn-1)-d,所以bn-bn-1=(6d"""
+d)/9(n≥2)为常数,
所以{bn}为等差数列.
(2)由③得6bn=9bn-9bn-1-d,即3bn=9bn-1+d,
因为{b_n+1/2}为等比数列,所以(b_n+1/2)/(b_(n"-"
1)+1/2)=(3b_(n"-"
1)+d/3+1/2)/(b_(n"-"
1)+1/2)=(3(b_(n"-"
1)+1/2)+d/3
"-"
1)/(b_(n"-"
1)+1/2)=3+(d/3
"-"
1)/(b_(n"-"
1)+1/2)(n≥2)是与n无关的常数,
所以d/3-1=0或bn-1+1/2为常数.
当d/3-1=0时,d=3,符合题意;
当bn-1+1/2为常数时,
在6Sn=9bn-an-2中,令n=1,则6b1=9b1-a1-2,又a1=1,解得b1=1,
所以bn-1+1/2=b1+1/2=3/2(n≥2),
此时3+(d/3
"-"
1)/(b_(n"-"
1)+1/2)=3+(d/3
"-"
1)/(3/2)=1,
解得d=-6.
综上,d=3或d=-6.
(3)证明:当d=3时,an=3n-2.
由(2)得数列{b_n+1/2}是以3/2为首项,3为公比的等比数列,所以bn+1/2=3/2×3n-1=1/2?3n.即bn=1/2(3n-1).
当n≥2时,cn=bn-bn-1=1/2(3n-1)-1/2(3n-1-1)=3n-1;
当n=1时,也满足上式,
所以cn=3n-1(n≥1).
设an=ci+cj(1≤i∈N*),则3n-2=3i-1+3j-1,即3n-(3i-1+3j-1)=2,
如果i≥2,因为3n为3的倍数,3i-1+3j-1为3的倍数,所以3n-(3i-1+3j-1)为3的倍数,
则2也为3的倍数,矛盾.
所以i=1,则3n=3+3j-1,即n=1+3j-2(j=2,3,4,…).
所以数列{an}中存在无穷多项可表示为数列{cn}中的两项之和.
2019高考数学二轮复习专题--数列课件及练习（与）2019高考数学二轮复习专题-函数与导数提分训练 本文关键词：复习，高考数学，导数，专题，数列
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