“元三维大联考”2023级高三第一次诊断考试数学试题（含答案）

高中2023级第一次诊断性考试

数学参考答案及评分标准

一、选择题：本题共8小题，每小题5分，共40分．

1．C2．D3．B4．D5．A6．B7．A8．C

二、选择题：本大题共3小题，每小题6分，共18分．全部选对的得6分，选对但不

全的得部分分，有选错的得0分．

9．BCD10．ACD11．AC

三、填空题：本题共3个小题，每小题5分，共15分．

12．1；13．；14．(一1，0)U(0，1)

四、解答题：本题共5小题，第15题13分，第16、17小题15分，第18、19小题17

分，共77分．解答应写出文字说明、证明过程或演算步骤．

15．解：（1）∵f(x)=sin(①x+φ)的最小正周期为兀，

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

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

又兀)，则，······························································5分

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

（2）由题知：g=fsinsin···········8分

由x，则··················································10分

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

故g(x)的值域为．·····························································13分

数学参考答案与评分标准第1页，共6页

16．解：（1）当m=2时，f=x|x·········3分

故f(x)的单调递增区间为：(—∞,—2]，[—1，+∞)，单调递减区间为：[−2，−1]；

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

（2）由题知x∈[1，2]，f(x)≤2，

即对任意x∈[1，2]，xx+m≤2恒成立，············································7分

x+m，即—≤x+m≤,则——x≤m≤—x，

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

xx

令g=x易知g(x)在x∈[1，2]单调递增，故g(x)max=g(2)=1，

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

令h=x易知g(x)在x∈[1，s2]单调递减，在x∈[、i2，2]单调递增，

故分

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

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

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

17．解：（1）因为：当n≥2时，an=2an—1—n+2，

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

因为：a1—1=2，············································································4分

所以数列{an—n}是以2为首项，2为公比的等比数列，···························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

设23n①

Tn=2+2.2+3.2+…+n.2

则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)定义域为R，且图象关于点对称，

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

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

①b≥7时，f/(x)=b—(3x2—4x)≥0，此时f(x)在[−1，0]单调递增，

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

②b≤0时，f/(x)=b—(3x2—4x)≤0，此时f(x)在[−1，0]单调递减，

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

，2

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

且当x∈[—1，x0)时，f/(x)<0；当x∈(x0，0]时，f/(x)>0，

即f(x)在区间[—1，x0)单调递减，在(x0，0]单调递增，

〔4—b0<b≤3

,，

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

{l1，3<b<7，

综上：h(b···································································8分

当b∈(—∞,3]时，h(b)单调递减，此时h(b)的最小值为h(3)=1；

当b∈(3，+∞)时，h(b)=1，································································9分

综上所述：h(b)的最小值为1；··························································10分

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

2

a，x1，x2是函数g(x)=f(x)—ax—m的三个互异零点，即g(a)=g(x1)=g(x2)，

也即g(x)=g(a)的三个根是a，x1，x2，·················································11分

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

数学参考答案与评分标准第3页，共6页

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必然为方程x+(2a—2)x+2a—2a—b=0的两个相异实根，·······13分

则Δ=(2a—2)2—4(2a2—2a—b)=4(b+1—a2)>0，

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

又方程x2+(2a—2)x+2a2—2a—b=0两个根都不同于a，则b≠5a2—4a，

2

∴对于b>—1，a∈(—·、，·、)，且b≠5a—4a，使得a，x1，x2是函数

g(x)=f(x)—ax2—m的三个互异零点，····················································16分

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

19．解：（1）f(x)不为偶函数，

理由如下：若f(x)为偶函数，则只需要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，

而该等式显然对任意实数不恒成立，

故f(x)不为偶函数；········································································3分

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

又x>0，故f’’’(x)>0，∴f’’(x)在(0，+∞)上单调递增，······················4分

①当k时，f’’(x)>f’’(0)=1—2k≥0，对x∈(0，+∞)恒成立，

∴f’(x)在(0，+∞)上单调递增，

∴f’(x)>f’(0)=0，

数学参考答案与评分标准第4页，共6页

∴f(x)在(0，+∞)上单调递增，f(x)>0，

∴f(x)在(0，+∞)上无极值点，也没有零点，不满足题意；·····················5分

②当k时，f’’(0)=1—2k<0，又f’’(x)在(0，+∞)上单调递增，且当x→+∞,

f’’(x)→+∞,因此3x0>0，使f’’(x0)=0，

∴当，时，，单调递减，当时，，

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

单调递增，····························································································6分

∴f’(x0)<f’(0)=0又x→+∞时，f’(x)→+∞,

∴由零点存在性定理知：，使，

3x1∈(x0,+∞)f’(x1)=0

∴当，时，，单调递减，当，时，，

x∈(0x1)f’(x)<0f(x)x∈(x1+∞)f’(x)>0f(x)

单调递增，

∴在，有唯一的极值点，分

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

又f(x1)<f(0)=0且，当x→+∞时，f(x)→+∞,

由零点存在性定理知：，，使，

3x2∈(x1+∞)f(x2)=0

∴在，有唯一的零点，分

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

综上所述：k满足题意；···························································9分

3

（ii）要证：f<0，由f(x2)=0，

即证：f—f3<0，

即f—f3<0，······················10分

令t=x2—x1，由（i）知t>0，

即证当t>0时，f—ft3<0恒成立，······················11分

数学参考答案与评分标准第5页，共6页

令h=f—ft3

即证：h(t)<0在t∈(0，+∞)恒成立，注意到h(0)=0，