高中2023級(jí)第一次診斷性考試

數(shù)學(xué)參考答案及評(píng)分標(biāo)準(zhǔn)

一、選擇題：本題共8小題，每小題5分，共40分．

1．C2．D3．B4．D5．A6．B7．A8．C

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

全的得部分分，有選錯(cuò)的得0分．

9．BCD10．ACD11．AC

三、填空題：本題共3個(gè)小題，每小題5分，共15分．

12．1；13．；14．(一1，0)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(yīng)寫(xiě)出文字說(shuō)明、證明過(guò)程或演算步驟．

15．解：（1）∵f(x)=sin(①x+φ)的最小正周期為兀，

∴①=2，·······················································································2分

又f，即sin=一，則sin，···························4分

又兀)，則，······························································5分

∴f=sin；····································································6分

（2）由題知：g=fsinsin···········8分

由x，則··················································10分

∴一sin································································12分

故g(x)的值域?yàn)椋ぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁ?3分

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16．解：（1）當(dāng)m=2時(shí)，f=x|x·········3分

故f(x)的單調(diào)遞增區(qū)間為：(—∞,—2]，[—1，+∞)，單調(diào)遞減區(qū)間為：[?2，?1]；

·····································································································6分

（2）由題知x∈[1，2]，f(x)≤2，

即對(duì)任意x∈[1，2]，xx+m≤2恒成立，············································7分

x+m，即—≤x+m≤,則——x≤m≤—x，

∴xm≤x·····································································9分

xx

令g=x易知g(x)在x∈[1，2]單調(diào)遞增，故g(x)max=g(2)=1，

∴—m≥1，則m≤—1，····································································11分

令h=x易知g(x)在x∈[1，s2]單調(diào)遞減，在x∈[、i2，2]單調(diào)遞增，

故分

h(x)min=h()=2·,·······························································13

∴—m≤2、，即m≥—2、，···························································14分

綜上：—2·≤m≤—1．··································································15分

17．解：（1）因?yàn)椋寒?dāng)n≥2時(shí)，an=2an—1—n+2，

所以：an—n=2[an—1—(n—1)](n≥2)，·················································3分

因?yàn)椋篴1—1=2，············································································4分

所以數(shù)列{an—n}是以2為首項(xiàng)，2為公比的等比數(shù)列，···························5分

n

所以：k=—1，b=0，an=2+n；·················································7分

2nn

（2）由（1）可知：bn=an—nan=4+n.2，·····································9分

設(shè)23n①

Tn=2+2.2+3.2+…+n.2

則234n+1②分

2Tn=2+2.2+3.2+…+n.2·············································10

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由②①整理得：23nn+1n+1，分

?Tn=—(2+2+2+…+2)+n.2=(n—1)2+2····13

∴23nn分

Sn=4+4+4+…+4+Tn2··························15

18．解：（1）f(x)定義域?yàn)镽，且圖象關(guān)于點(diǎn)對(duì)稱，

解得：c=1；····················································································3分

（2）f/(x)=—3x2+4x+b=b—(3x2—4x)，3x2—4x∈[0，7]，······················4分

①b≥7時(shí)，f/(x)=b—(3x2—4x)≥0，此時(shí)f(x)在[?1，0]單調(diào)遞增，

∴f(x)max=f(0)=1；········································································5分

②b≤0時(shí)，f/(x)=b—(3x2—4x)≤0，此時(shí)f(x)在[?1，0]單調(diào)遞減，

∴f(x)max=f(—1)=4—b；··································································6分

，2

③0<b<7時(shí)，存在x0∈(—10)，使得f/(x0)=b—(3x0—4x0)=0，

且當(dāng)x∈[—1，x0)時(shí)，f/(x)<0；當(dāng)x∈(x0，0]時(shí)，f/(x)>0，

即f(x)在區(qū)間[—1，x0)單調(diào)遞減，在(x0，0]單調(diào)遞增，

〔4—b0<b≤3

,，

此時(shí)f(x)max=max{f(—1)，f(0)}=max{1，4—b}=··················7分

{l1，3<b<7，

綜上：h(b···································································8分

當(dāng)b∈(—∞,3]時(shí)，h(b)單調(diào)遞減，此時(shí)h(b)的最小值為h(3)=1；

當(dāng)b∈(3，+∞)時(shí)，h(b)=1，································································9分

綜上所述：h(b)的最小值為1；··························································10分

（3）g(x)=f(x)—ax2—m=—x3—(a—2)x2+bx+1—m，

2

a，x1，x2是函數(shù)g(x)=f(x)—ax—m的三個(gè)互異零點(diǎn)，即g(a)=g(x1)=g(x2)，

也即g(x)=g(a)的三個(gè)根是a，x1，x2，·················································11分

代入得：—x3—(a—2)x2+bx+1—m=—a3—(a—2)a2+ba+1—m，

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整理得：x3—a3+(a—2)(x2—a2)—b(x—a)=0，

∴(x—a)(x2+ax+a2)+(a—2)(x—a)(x+a)—b(x—a)=0，

22

即：(x—a)[x+(2a—2)x+2a—2a—b]=0的三根是a，x1，x2，

22

所以x1，x2必然為方程x+(2a—2)x+2a—2a—b=0的兩個(gè)相異實(shí)根，·······13分

則Δ=(2a—2)2—4(2a2—2a—b)=4(b+1—a2)>0，

所以b>a2—1，則b>—1，································································15分

又方程x2+(2a—2)x+2a2—2a—b=0兩個(gè)根都不同于a，則b≠5a2—4a，

2

∴對(duì)于b>—1，a∈(—·、，·、)，且b≠5a—4a，使得a，x1，x2是函數(shù)

g(x)=f(x)—ax2—m的三個(gè)互異零點(diǎn)，····················································16分

∴b的取值范圍為(—1，+∞)．····························································17分

19．解：（1）f(x)不為偶函數(shù)，

理由如下：若f(x)為偶函數(shù)，則只需要f(x)=f(—x)，

即exx3—kx2—x—1=e—xx3—kx2+x—1恒成立，···························1分

即ex—e—xx3—2x=0恒成立，

而該等式顯然對(duì)任意實(shí)數(shù)不恒成立，

故f(x)不為偶函數(shù)；········································································3分

∴f’(0)=0且f’’(x)=ex—2k—x，則f’’’(x)=ex—1，

又x>0，故f’’’(x)>0，∴f’’(x)在(0，+∞)上單調(diào)遞增，······················4分

①當(dāng)k時(shí)，f’’(x)>f’’(0)=1—2k≥0，對(duì)x∈(0，+∞)恒成立，

∴f’(x)在(0，+∞)上單調(diào)遞增，

∴f’(x)>f’(0)=0，

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∴f(x)在(0，+∞)上單調(diào)遞增，f(x)>0，

∴f(x)在(0，+∞)上無(wú)極值點(diǎn)，也沒(méi)有零點(diǎn)，不滿足題意；·····················5分

②當(dāng)k時(shí)，f’’(0)=1—2k<0，又f’’(x)在(0，+∞)上單調(diào)遞增，且當(dāng)x→+∞,

f’’(x)→+∞,因此3x0>0，使f’’(x0)=0，

∴當(dāng)，時(shí)，，單調(diào)遞減，當(dāng)時(shí)，，

x∈(0x0)f’’(x)<0f’(x)x∈(x0,+∞)f’’(x)>0f’(x)

單調(diào)遞增，····························································································6分

∴f’(x0)<f’(0)=0又x→+∞時(shí)，f’(x)→+∞,

∴由零點(diǎn)存在性定理知：，使，

3x1∈(x0,+∞)f’(x1)=0

∴當(dāng)，時(shí)，，單調(diào)遞減，當(dāng)，時(shí)，，

x∈(0x1)f’(x)<0f(x)x∈(x1+∞)f’(x)>0f(x)

單調(diào)遞增，

∴在，有唯一的極值點(diǎn)，分

f(x)(0+∞)x1················································7

又f(x1)<f(0)=0且，當(dāng)x→+∞時(shí)，f(x)→+∞,

由零點(diǎn)存在性定理知：，，使，

3x2∈(x1+∞)f(x2)=0

∴在，有唯一的零點(diǎn)，分

f(x)(0+∞)x2··················································8

綜上所述：k滿足題意；···························································9分

3

（ii）要證：f<0，由f(x2)=0，

即證：f—f3<0，

即f—f3<0，······················10分

令t=x2—x1，由（i）知t>0，

即證當(dāng)t>0時(shí)，f—ft3<0恒成立，······················11分

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令h=f—ft3

即證：h(t)<0在t∈(0，+∞)恒成立，注意到h(0)=0，