綿陽市高中2023級第一次診斷性考試

數學參考答案及評分標準

一、選擇題：本題共8小題，每小題5分，共40分．

1．B2．D3．C4．A5．B6．A7．D8．C

二、選擇題：本大題共3小題，每小題6分，共18分．全部選對的得6分，選對但不

全的得部分分，有選錯的得0分．

9．ACD10．AC11．ABD

三、填空題：本題共3個小題，每小題5分，共15分．

12．；13．2；14．(—1，0)U(0，1)

四、解答題：本題共5小題，第15題13分，第16、17小題15分，第18、19小題17

分，共77分．解答應寫出文字說明、證明過程或演算步驟．

15．解：（1）∵f=cos則cos······························2分

兀

又φ∈(0，兀)，∴,·································································4分

∴f=cos·······································································6分

（2）由題知：g=fcossinx，

∴h.g=2sinxcos············································7分

∴hsin2xcos2xsin···························9分

∵y=sinx的單調遞增區(qū)間為，k∈Z，···················11分

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k兀≤2x2k兀，k∈Z·············································10分

k?！躼k兀，k∈Z······················································12分

∴函數h(x)的單調遞增區(qū)間為k兀k兀]，k∈Z．···················13分

16．解：（1）∵f(x)為R上的奇函數，

∴必有f(0)=—b=0，則b=0，·························································2分

∴f(x)=x|x+a|，同理由f(1)=—f(—1)，∴|1+a|=|—1+a|，故a=0，

此時，f(x)=x|x|，對任意實數x，都滿足f(x)=—f(—x)，···················4分

∴f(x)為R上的奇函數，

∴f(x)=xx；················································································6分

（2）思路一：∵當x≥1時，f(x)>0，

∴f(1)=|a+1|—(a2+1)>0，解得：0<a<1，·······································9分

∴f(x)=x(x+a)—(a2+1)=x2+ax—a2—1，

易知f(x)在[1，+∞)上單調遞增，······················································12分

∴當0<a<1時，f(x)≥f(1)=a(1—a)>0，········································14分

∴實數a的取值范圍(0，1)．·····························································15分

思路二：∵當x≥1時，f(x)>0，且b=a2+1，即對任意x≥1，都有x|x+a|>a2+1

恒成立，即|x+a·····································································7分

∴x+a或x+a對任意的x≥1恒成立，·······················8分

∴x或x+a對任意的x≥1恒成立，·················10分

令g=xa，易知g(x)在[1，+∞)單調遞增，························11分

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2

故g(x)min=g(1)=a—a>0，···························································12分

∴0<a<1，··················································································13分

令h=xa，易知h(x)在[1，+∞)無最大值，故不滿足h(x)<0恒成立，

···································································································14分

綜上：0<a<1．···········································································15分

17．解：（1）設公差為d，則由題意可得：···········3分

解得：=1d=所以：································6分

a1，，an

（2）不能構成等比數列，··································································7分

其理由如下：在數列{an}中任取三項分別為：=m+1—

am·、·、，

=n+1—=·t+1—·················································9分

an·、·、，at·、，

2

若am，an，at成等比數列，則an=am.at，

即：(·、n+1—·、)2=(·、m+1—·、).(·、+1—·、)，····························10分

整理得：3n2—3mt—6n+3(m+t)=·(m+t—2n)，······························11分

因為m，n，t為正整數，所以：·················12分

化簡整理得：(m—t)2=0，所以m=n=t與題意矛盾，·························14分

所以，在數列{an}中取三個不同的項，均不能構成等比數列．··················15分

18．解：（1）f(x)=—x3+ax2+x=—x(x2—ax—1)，

可知有一個零點一定是0，且對于方程：x2—ax—1=0，Δ=a2+4>0，且0一定不

是方程x2—ax—1=0的根，

∴f(x)有3個相異零點；··································································3分

（2）f,(x)=—3x2+2ax+1，其中Δ=4a2+12>0，

2

故x1，x2是方程—3x+2ax+1=0的兩根，················································4分

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222

由韋達定理可得：，故x1+x2=—2x1x······5分

3232

f(x1)+f(x2)=—x1+ax1+x1—x2+ax2+x2

2222

—(x1+x2)(x1—x1x2+x2)+a(x1+x2)+(x1+x2)，··············7分

帶入得解得：a=1；······9分

另解：由f,(x)=—3x2+2ax+1，f,,(x)=—6x+2a，

令f,,(x)=—6x+2a=0，解得：x

a

又f故函數f(x)圖像關于點(，)成中心對稱，又三次函數

3

的極值點關于對稱中心對稱，故f+f解得：a=1；

（3）由三次函數圖象可知，f(x)=f(x2)有且僅有兩根為x2，m，

則m<x1<0<x2，

3232

即—x+ax+x=—x2+ax2+x2，有且僅有兩根為x2，m，

22

整理得：(x—x2)[x+x2x+x2—a(x+x2)—1]=0，

22

所以x3是方程x+x2x+x2—a(x+x2)—1=0的根，······························12分

2

又x2是方程—3x+2ax+1=0的根，故a，·································13分

222

代入上式整理得到：2x2x—(x2—1)x—x2(x2+1)=0，

2

即(x—x2)(2x2x+x2+1)=0，·······························································15分

故m······················································16分

故m的最大值為?1．·······································································17分

19．解：（1）證明：f,(x)=—a(ex—1)+(1—ax)ex—1，注意到f,(0)=0，

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f(x)=(—ax+1—2a)ex，x≥0．························································2分

因為a，則f(x)≤0，

因此f(x)在[0，+∞)單調遞減，故f(x)≤f(0)=0，·····························3分

故f(x)在[0，+∞)單調遞減，因此f(x)≤f(0)=0；·······························4分

（2）（i）證明：g故g(x)在點An處的切線方程為y+lnean，

·······························································································5分

與ex—1聯(lián)立，可得ex，

h(x)=lnan=0

令ex，則ex

u=lnanu=

故u(x)在(0，1—lnan)單調遞減，在(1—lnan，+∞)單調遞增，··················6分

因為an<1，則1—lnan>1，且ulnanln

an—1

而u(an)=e—1—lnan>an—1—lnan>0，

故u(x)在(an，1)上存在唯一零點，即為bn，故an<bn<1，······················7分

b—1b—1

同理，在點處的切線方程為enen，分

h(x)Bny=(x—bn)+·····················8

b—1

與聯(lián)立，有en，

g(x)(x+1—bn)—lnx—1=0

令ebn—1，ebn

v(x)=(x+1—bn)—lnx—1v=

1—b1—bn

則v(x)在(0，en)單調遞減，在(e，+∞)單調遞增，

1—b

因為，故en，分

0<bn<1>1······························································9

b—1x—1x—1

en．考慮e，e，

v(1)=(2—bn)—1w(x)=(2—x)—1w(x)=(1—x)

則w(x)在(0，1)單調遞增，故w(x)<w(1)=0，故v(1)<0，

且ebn—1，

v(bn)=—lnbn—1>bn—lnbn—1>0

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故v(x)在(bn，1)存在唯一零點，即an+1，故bn<an+1<1，

因此an<bn<an+1<1，故an<an+1；··················································