新教材適用高中數(shù)學(xué)綜合測(cè)評(píng)新人教A版選擇性必修第二冊(cè)

綜合測(cè)評(píng)(時(shí)間:120分鐘滿分:150分)一、選擇題(本大題共8小題,每小題5分,共40分.在每小題給出的四個(gè)選項(xiàng)中,只有一項(xiàng)是符合題目要求的)1.對(duì)于數(shù)列1,37,314,321,…,398是這個(gè)數(shù)列的()A.不在此數(shù)列中 B.第13項(xiàng)C.第14項(xiàng) D.第15項(xiàng)答案:D2.已知等差數(shù)列{an},且a1+a2+a3+a4=10,a13+a14+a15+a16=70,則前16項(xiàng)的和等于()A.140 B.160 C.180 D.200解析:∵a1+a2+a3+a4+a13+a14+a15+a16=4(a1+a16)=80,∴a1+a16=20.∴所求和為16(a1答案:B3.若函數(shù)f(x)=13x3-f'(1)·x2-x,則f'(3)的值為(A.0 B.-1 C.8 D.-8解析:f'(x)=x2-2f'(1)·x-1,則f'(1)=12-2f'(1)·1-1,得f'(1)=0.故f'(x)=x2-1,從而f'(3)=8.答案:C4.設(shè)等比數(shù)列{an}的前6項(xiàng)和S6=6,且1-a22為a1,a3的等差中項(xiàng),則a7+a8+a9=(A.-2 B.8 C.10 D.14解析:由題意得2-a2=a1+a3,∴a1+a2+a3=2,又S6=6,∴a4+a5+a6=4.又{an}為等比數(shù)列,∴S3,S6-S3,S9-S6為等比數(shù)列,∴42=2(S9-S6),∴S9-S6=8,即a7+a8+a9=8.答案:B5.兩個(gè)等差數(shù)列{an}和{bn}的前n項(xiàng)和分別為Sn,Tn,且SnTn=7nA.94 B.37C.7914 D.解析:a2答案:D6.若函數(shù)f(x)=13x3-ax2+ax在區(qū)間(0,1)上有極大值,在區(qū)間(1,2)上有微小值,則實(shí)數(shù)a的取值范圍是(A.1,43C.(-∞,0)∪(1,+∞) D.0解析:f'(x)=x2-2ax+a,由題意知,f'(x)=0在區(qū)間(0,1),(1,2)上都有根,則f'(0)>0,f'(1)<0,f'(2)>0,即a>0,1-a答案:A7.設(shè)函數(shù)f'(x)是奇函數(shù)f(x)(x∈R)的導(dǎo)函數(shù),f(-1)=0,當(dāng)x>0時(shí),xf'(x)-f(x)<0,則使得f(x)>0成立的x的取值范圍是()A.(-∞,-1)∪(0,1) B.(-1,0)∪(1,+∞)C.(-∞,-1)∪(-1,0) D.(0,1)∪(1,+∞)解析:設(shè)F(x)=f(x)則F'(x)=xf'(x)-f(x)x2<0,∴F(∵f(x)為奇函數(shù),且f(-1)=0,∴f(1)=0,于是F(1)=0.∴在區(qū)間(0,1)上,F(x)>0;在區(qū)間(1,+∞)上,F(x)<0,即當(dāng)0<x<1時(shí),f(x)>0;當(dāng)x>1時(shí),f(x)<0.又f(x)為奇函數(shù),∴當(dāng)x∈(-∞,-1)時(shí),f(x)>0;當(dāng)x∈(-1,0)時(shí),f(x)<0.綜上可知,f(x)>0的解集為(-∞,-1)∪(0,1).故選A.答案:A8.已知函數(shù)f(x)=ax-1+lnx,若存在x0>0,使得f(x0)≤0有解,則實(shí)數(shù)a的取值范圍是(A.(2,+∞) B.(-∞,3) C.(-∞,1] D.[3,+∞)解析:函數(shù)f(x)的定義域是(0,+∞),不等式ax-1+lnx≤0有解,即a≤x-xlnx在區(qū)間(0,+∞)上有解.設(shè)h(x)=x-xlnx,則h'(x)=1-(lnx+1)=-lnx.令h'(x)=0,可得x=1.由h(x)的單調(diào)性可得,當(dāng)x=1時(shí),函數(shù)h(x)=x-xlnx取得最大值1.要使不等式a≤x-xlnx在(0,+∞)上有解,只要a小于等于h(x)的最大值,即a≤1.所以選C答案:C二、選擇題(本大題共4小題,每小題5分,共20分.在每小題給出的四個(gè)選項(xiàng)中,有多項(xiàng)符合題目要求.全部選對(duì)的得5分,部分選對(duì)的得2分,有選錯(cuò)的得0分)9.已知等差數(shù)列{an}的前n項(xiàng)和為Sn,公差d≠0,且a1d≤1.記b1=S2,bn+1=S2n+2-S2n,n∈N*,則下列等式肯定成立的是(A.2a4=a2+a6 B.2b4=b2+b6 C.a42=a2a8 D.b42解析:A.由等差數(shù)列的性質(zhì)可知2a4=a2+a6,故A肯定成立;B.b4=S8-S6=a7+a8,b2=S4-S2=a3+a4,b6=S12-S10=a11+a12,又由題意可得2(a7+a8)=a3+a4+a11+a12,所以2b4=b2+b6,故B肯定成立;C.a42=a2a8?(a1+3d)2=(a1+d)(a1+7d),整理可得a1D.b8=S16-S14=a15+a16,當(dāng)b42=b2b8時(shí),(a7+a8)2=(a3+a4)(a15+a16),即(2a1+13d)2=(2a1+5d)·(2a1+29d),得2a1=3d,這與已知a答案:AB10.下列函數(shù)中,存在極值點(diǎn)的是()A.y=2|x| B.y=-2x3-xC.y=xlnx D.y=xsinx解析:對(duì)于A,x=0是函數(shù)的微小值點(diǎn);對(duì)于B,y'=-6x2-1<0恒成立,函數(shù)在R上單調(diào)遞減,所以函數(shù)無(wú)極值點(diǎn);對(duì)于C,y'=1+lnx.令y'=0,解得x=1e.當(dāng)x∈0,1e時(shí),y'<0,函數(shù)單調(diào)遞減;當(dāng)x∈1e,+∞時(shí),y'>0,函數(shù)單調(diào)遞增.所以x=1e是函數(shù)的微小值點(diǎn);對(duì)于D,y'=sinx+xcosx.當(dāng)x∈-π2,0時(shí),y'<0,函數(shù)單調(diào)遞減;當(dāng)x∈0,答案:ACD11.已知函數(shù)y=mex的圖象與直線y=x+2m有兩個(gè)交點(diǎn),則實(shí)數(shù)m的取值可以是()A.-1 B.1 C.2 D.3解析:設(shè)f(x)=mex-x-2m,則f'(x)=mex-1.要使函數(shù)y=mex的圖象與直線y=x+2m有兩個(gè)交點(diǎn),需f(x)有兩個(gè)零點(diǎn).當(dāng)m≤0時(shí),f'(x)=mex-1<0,函數(shù)f(x)在R上單調(diào)遞減,不行能有兩個(gè)零點(diǎn),不符合題意,舍去.當(dāng)m>0時(shí),由f'(x)=0得x=ln1m當(dāng)x∈ln1m,+∞時(shí),f'(x)>0,f(x)單調(diào)遞增;當(dāng)x∈-∞,ln1m時(shí),f'(x)<0,f(x)單調(diào)遞減,且f(0)=-m<0又當(dāng)x?-∞,f(x)?+∞,x?+∞時(shí),f(x)?+∞,所以當(dāng)m>0時(shí),函數(shù)f(x)有兩個(gè)零點(diǎn),即函數(shù)y=mex的圖象與直線y=x+2m有兩個(gè)交點(diǎn),視察各選項(xiàng),知m的取值可以是1,2,3.故選BCD.答案:BCD12.已知函數(shù)f(x)=exlnxA.f(x)的定義域是(0,+∞)B.當(dāng)x∈(0,1)時(shí),f(x)的圖象位于x軸下方C.f(x)存在單調(diào)遞增區(qū)間D.f(x)有且僅有兩個(gè)極值點(diǎn)解析:∵f(x)=exlnx,∴l(xiāng)nx≠0,∴x>0,且x≠1,即f(x)的定義域?yàn)?0,1)∪(1,+∞).故A錯(cuò)誤;當(dāng)x∈(0,1)時(shí),lnx<0,∴f(x)<0.故B正確;由f(x)=exlnx,得f'(x)=ex(xlnx-1)x(lnx)2.當(dāng)0<x<1時(shí),f'(x)<0,∴f(x)在區(qū)間(0,1)上單調(diào)遞減.設(shè)g(x)=xlnx-1,則g'(x)=lnx+1.當(dāng)x>1時(shí),g'(x)>0,則g(x)在區(qū)間(1,+∞)上單調(diào)遞增.又g(1)=-1<0,g(2)=2ln2-1>0,∴存在x0∈(1,2)使g(x0)=0,即f'(x0)=0.∴當(dāng)1<x<x0時(shí),g(x)<0,即f'(x)<0,當(dāng)x>x0時(shí),g(x)>0,即f'(x)>0,∴f(x故選BC.答案:BC三、填空題(本大題共4小題,每小題5分,共20分.把答案寫在題中的橫線上)13.若等差數(shù)列{an}滿意a7+a8+a9>0,a7+a10<0,則當(dāng)n=時(shí),{an}的前n項(xiàng)和最大.

解析:∵a7+a8+a9=3a8>0,∴a8>0.∵a7+a10=a8+a9<0,∴a9<-a8<0.∴數(shù)列{an}的前8項(xiàng)和最大.答案:814.已知周長(zhǎng)為20cm的矩形,繞一條邊旋轉(zhuǎn)成一個(gè)圓柱,則該圓柱體積的最大值為cm3.

解析:設(shè)矩形相鄰兩邊長(zhǎng)分別為x(0<x<10)cm,(10-x)cm,繞長(zhǎng)為(10-x)cm的一邊旋轉(zhuǎn)得到的圓柱的體積V(x)=πx2(10-x)=10πx2-πx3,則V'(x)=20πx-3πx2.令V'(x)=0,解得x=0(舍去)或x=203.當(dāng)x∈0,203時(shí),V'(x)>0,V(當(dāng)x∈203,10時(shí),V'(x)<0,V(x因此當(dāng)x=203時(shí),V(x)取得最大值為4000π27答案:400015.設(shè)直線y=m與曲線C:y=x(x-2)2的三個(gè)交點(diǎn)分別為A(a,m),B(b,m),C(c,m),其中a<b<c,則實(shí)數(shù)m的取值范圍是,a2+b2+c2的值為.

解析:設(shè)f(x)=x(x-2)2,則f'(x)=3x2-8x+4.令f'(x)=0,解得x=23或x=2由f(x)的單調(diào)性,得f(x)的極大值為f23=3227,微小值為f若直線y=m與曲線C:y=x(x-2)2有三個(gè)交點(diǎn),則0<m<3227,即m的取值范圍為0設(shè)g(x)=f(x)-m=x(x-2)2-m=x3-4x2+4x-m.若直線y=m與曲線C:y=x(x-2)2有三個(gè)交點(diǎn),且其坐標(biāo)分別為A(a,m),B(b,m),C(c,m),則方程x3-4x2+4x-m=0有三個(gè)根,分別為a,b,c,即x3-4x2+4x-m=(x-a)(x-b)(x-c)=x3-(a+b+c)x2+(ab+ac+bc)x-abc.故a+b+c=4,ab+bc+ac=4,于是a2+b2+c2=(a+b+c)2-2(ab+bc+ac)=8.答案:0,3216.數(shù)列{an}滿意:nan+2+(n+1)an=(2n+1)·an+1-1,a1=1,a2=6,令cn=ancosnπ2,數(shù)列{cn}的前n項(xiàng)和為Sn,則S4n=解析:∵nan+2+(n+1)an=(2n+1)an+1-1,∴nan+2-nan+1=(n+1)an+1-(n+1)an-1,∴an∴a2-a11a3a4……an-an上述n-1個(gè)式子相加得5-an+1-an即an+1-an=4n+1(n≥2).又當(dāng)n=1時(shí),a2-a1=4×1+1=5也成立,∴an+1-an=4n+1.∴a2-a1=4×1+1,a3-a2=4×2+1,a4-a3=4×3+1,……an-an-1=4(n-1)+1(n≥2),上述n-1個(gè)式子相加得an-1=(n-1)(2n+1),即an=n(2n-1)(n≥2).又當(dāng)n=1時(shí),a1=1×(2×1-1)=1也成立,∴an=n(2n-1).∵cn=ancosnπ∴c4k-3+c4k-2+c4k-1+c4k=0-(4k-2)(8k-5)+0+4k(8k-1)=32k-10(k∈N*).∴S4n=(c1+c2+c3+c4)+(c5+c6+c7+c8)+…+(c4n-3+c4n-2+c4n-1+c4n)=(32×1-10)+(32×2-10)+…+(32n-10)=16n2+6n.答案:16n2+6n四、解答題(本大題共6小題,共70分.解答應(yīng)寫出文字說(shuō)明、證明過(guò)程或演算步驟)17.(10分)已知{an}是等差數(shù)列,{bn}是等比數(shù)列,且b2=3,b3=9,a1=b1,a14=b4.(1)求{an}的通項(xiàng)公式;(2)記cn=an+bn,求數(shù)列{cn}的前n項(xiàng)和.解:(1)設(shè)等比數(shù)列{bn}的公比為q.則q=b3b2=3,于是b1=b2q=1,b4=b3q=9設(shè)等差數(shù)列{an}的公差為d.已知a1=b1=1,a14=b4=27,由a14=a1+13d,得d=27-113=2.∴an=a1+(n-1)d=2(2)由(1)知an=2n-1,bn=3n-1,∴cn=an+bn=(2n-1)+3n-1.∴{cn}的前n項(xiàng)和Sn=1+3+…+(2n-1)+1+3+…+3n-1=n(1+2n-118.(12分)已知在等差數(shù)列{an}中,a1=-60,a17=-12,求數(shù)列{|an|}的前n項(xiàng)和.解:由a1=-60,a17=-12知,等差數(shù)列{an}的公差d=a17-a所以an=a1+(n-1)d=-60+(n-1)×3=3n-63.由an≤0,即3n-63≤0,得n≤21,即{an}中前20項(xiàng)是負(fù)數(shù),從第21項(xiàng)起為非負(fù)數(shù).設(shè)Sn和Sn'分別表示{an}和{|an|}的前n項(xiàng)和.當(dāng)n≤20時(shí),Sn'=-Sn=--60n+3n(n-1)2當(dāng)n>20時(shí),Sn'=-S20+(Sn-S20)=Sn-2S20=-60n+3n(n-1)2-2-綜上,Sn'=-19.(12分)已知函數(shù)f(x)=x3-2ax2+bx+c,(1)當(dāng)c=0時(shí),f(x)在點(diǎn)P(1,3)處的切線平行于直線y=x+2,求a,b的值;(2)若f(x)在點(diǎn)A(-1,8),B(3,-24)處有極值,求f(x)的解析式.解:(1)當(dāng)c=0時(shí),f(x)=x3-2ax2+bx,則f'(x)=3x2-4ax+b.由題意得f(1)=3,f'(1)=1,即1解得a(2)因?yàn)閒(x)=x3-2ax2+bx+c,所以f'(x)=3x2-4ax+b.由題意知-1,3是方程3x2-4ax+b=0的兩根,所以-解得a=32,b=-9由f(-1)=-1-2a-b+c=8,a=32,b=-可得c=3,所以f(x)=x3-3x2-9x+3.檢驗(yàn)知,符合題意.20.(12分)已知成等差數(shù)列的三個(gè)正數(shù)的和等于15,并且這三個(gè)數(shù)分別加上2,5,13后成為等比數(shù)列{bn}中的b3,b4,b5.(1)求數(shù)列{bn}的通項(xiàng)公式;(2)數(shù)列{bn}的前n項(xiàng)和為Sn,求證:數(shù)列Sn+(1)解:設(shè)成等差數(shù)列的三個(gè)正數(shù)分別為a-d,a,a+d.依題意得a-d+a+a+d=15,解得a=5.所以數(shù)列{bn}中的b3,b4,b5依次為7-d,10,18+d.依題意得(7-d)(18+d)=100,解得d=2或d=-13(舍去).故數(shù)列{bn}是第3項(xiàng)為5,公比為2的等比數(shù)列.所以其通項(xiàng)公式為bn=b3·qn-3=5·2n-3.(2)證明:數(shù)列{bn}的前n項(xiàng)和Sn=54(1-2n)1-即Sn+54=5·2n-2所以S1+54=5因此Sn+5421.(12分)設(shè)函數(shù)f(x)=a2lnx-x2+ax(a>0).(1)求f(x)的單調(diào)區(qū)間;(2)求全部使e-1≤f(x)≤e2對(duì)x∈[1,e]恒成立的a的值.(注:e為自然對(duì)數(shù)的底數(shù))解:(1)函數(shù)f(x)=a2lnx-x2+ax(a>0)的定義域?yàn)?0,+∞),導(dǎo)數(shù)f'(x)=a2x-2x+a=-由于a>0,故當(dāng)x∈(0,a)時(shí),f'(x)>0,于是f(x)的單調(diào)遞增區(qū)間為(0,a);當(dāng)x∈(a,+∞)時(shí),f'(x)<0,于是f(x)的單調(diào)遞減區(qū)間為(a,+∞).(2)由題意得f(1)=a-1≥e-1,則a≥e.由(1)