文档内容
无锡市第六高级中学 2025年10月高三教学质量调研数学试卷
答案和解析
一、单选题
1. 2. 3. 4. D 5. 6. 7. 8. A
B D C
二、多选题
9. 10. 11.
ACD AC
三、填空题
e2
12． ❑√2 13． ❑√3 14．[ −1,+∞)
4
四、解答题
15.【答案】解：
(1)
由题意，设⃗ ，
b=(x,y)
因为|⃗b|=2，所以√ x2+y2=2，即x2+y2=4，①......................................2分
又因为向量a⃗,⃗b的夹角为60°，
⃗ ⃗
⟨⃗ ⃗⟩ a·b y 1
所以cos a,b = = = ，解得y=1， ......................................4分
|⃗||⃗| 2 2
a b
将y=1代入①，解得x=±√ 3，
所以⃗ 或⃗ ． ......................................6分
b=(√ 3,1) b=(−√ 3,1)
因为 → → → → ，
(2)
(a+b)⊥(a−b),λ∈R
所以 ( ⃗ a+ ⃗ b)·( ⃗ a− ⃗ b)=0 ，即 a⃗ 2=⃗b2 ，
所以 ， ......................................8分
|a⃗|=|⃗b|=1
所以 |a⃗+λ⃗b|=√ (a⃗+λ⃗b) 2 =√ a⃗ 2+2λa⃗·⃗b+λ2⃗b2=√ λ2+λ+1=
√
(λ+
1
)
2
+
3， ......................................10分
2 4
所以当λ=− 1 时， ⃗ ⃗ 有最小值 √ 3 ． ......................................13分
2 |a+λb| 2
A A
16.【答案】（1）在△ABC中，由asinB=❑√3bsin 及正弦定理，得sinAsinB=❑√3sinBsin ，···········2
2 2
分
A A A A
则2sin cos sinB=❑√3sinBsin ，而A,B∈(0,π)，sinBsin >0，
2 2 2 2
A ❑√3 A π π
因此cos = ，解得 = ，所以A= . ·········································································5分
2 2 2 6 3
π 2❑√7 ❑√21
（2）由（1）知A= ，由cosB= ，得sinB=❑√1−cos2B= ，·································7分
3 7 7
❑√3 2❑√7 1 ❑√21 3❑√21
sinC=sin(A+B)=sinAcosB+cosAsinB= × + × = ，····························11分
2 7 2 7 14
3❑√21
4×
bsinC 14 1
由正弦定理得c= = =6，而⃗AD= (⃗AB+⃗AC)，
sinB ❑√21 2
7
1 1 √ 1
所以|⃗AD|= ❑√⃗AB2+⃗AC2+2⃗AB⋅⃗AC= ❑62+42+2×6×4× =❑√19. ····························15分
2 2 2
17.【答案】解：(1)由于g(x)为奇函数，且定义域为R，
40−n
∴g(0)=0，即 =0，n=1． ······························2分
20
4x−1
当n=1时，g(x)= =2x−2−x,
2x
g(−x)=2−x−2x=−g(x)，
∴n=1时g(x)为奇函数． ......................................4分
∵f(x)=log (4x+1)+mx，
4∴f(−x)=log (4−x+1)−mx=log (4x+1)−(m+1)x，
4 4
∵f(x)是偶函数，∴f(−x)=f(x)，
m=−m−1，
1
得到m=− ， ......................................6分
2
1
由此可得：m+n的值为 ． ·································7分
2
1
(2)∵ℎ(x)=f(x)+ x=log (4x+1)，∴ℎ[log (2a+1)]=log (2a+2)， ·································9分
2 4 4 4
又∵g(x)=2x−2−x在区间[1,+∞)上是增函数，
3
∴当x≥1时，g(x) =g(1)= ， ·································11分
min 2
3
{
2a+2<4❑2
1
由题意得 2a+1>0 ,∴− 0
1
综上，a的取值范围{a|− 0，即a<3，
1 2
所以 …………………………(8分)
因为f(x)＋f(x)＝(x＋3x＋ax＋1)＋(x＋3x＋ax＋1)
1 2 1 2
＝(x＋x)[(x＋x)2－3xx]＋3[(x＋x)2－2xx]＋a(x＋x)＋2
1 2 1 2 1 2 1 2 1 2 1 2
＝(－2)[(－2)2－3×]＋3[(－2)2－2×]＋a×(－2)＋2＝6－2a，
又f(x)＋f(x)≤5，所以6－2a≤5，解得a≥.
1 2
综上，≤a<3. …………………………(12分)
19.【答案】解
1 2 (x−2)(x−1)
（1）因为a=1，所以f(x)=2lnx+ x2−3x，f '(x)= +x−3= (x>0)，··············2分
2 x x
令f '(x)>0，解得02，令f '(x)<0，解得10
∴ x +x =2a>0 ，解得：a>1，···············································································7分
1 2
x x =a>0
1 2
( a a )
∴g(x )+g(x )−4a= −x +2alnx + −x +2alnx −4a
1 2 x 1 1 x 2 2
1 2
a(x +x )
= 1 2 −(x +x )+2aln(x x )−4a=2alna−4a；····················································9分
x x 1 2 1 2
1 2
令ℎ(a)=2alna−4a(a>1)，则ℎ'(a)=2lna−2，
∴当a∈(1,e)时，ℎ'(a)<0；当a∈(e,+∞)时，ℎ'(a)>0；
∴ℎ(a)在(1,e)上单调递减，在(e,+∞)上单调递增；·························································13分
∴ℎ(a) = ℎ(e)=2e−4e=−2e，
min
即g(x )+g(x )−4a的最小值为−2e. ··········································································15分
1 2
