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南京市 2026 届高三年级学情调研
数 学 答 案
2025.09
一、选择题：本大题共8小题，每小题5分，共40分．在每小题给出的四个选项中，只有
一项是符合题目要求的，请把答案填涂在答题卡相应位置上．
1．B 2．D 3．A 4．C 5．A 6．A 7．D 8．D
二、选择题：本大题共3小题，每小题6分，共18分．在每小题给出的四个选项中，有多
项符合题目要求，请把答案填涂在答题卡相应位置上．全部选对得6分，部分选对得
部分分，不选或有错选的得0分．
9．BD 10．ACD 11．ACD
三、填空题：本大题共3小题，每小题5分，共15分．请把答案填写在答题卡相应位置上．
12．10 13． 14．
四、解答题：本大题共5小题，共77分．请在答题卡指定区域内作答，解答时应写出必要
的文字说明，证明过程或演算步骤．
15．（本小题满分13分）
解：（1）法一：P(X≥1)＝1－P(X＝0)＝1－()2···················································3分
＝；·························································5分
法二：P(X≥1)＝P(X＝1)＋P(X＝2)＝××2＋×··················································3分
＝．····························································5分
（2）Y的可能取值为0，1，2，
法一：P(Y＝0)＝＝，···················································································7分
P(Y＝1)＝＝，····························································································9分
P(Y＝2)＝＝，···························································································11分
法二：P(Y＝0)＝＝，···················································································7分
P(Y＝1)＝＝，····························································································9分
P(Y＝2)＝＝，···························································································11分
则分布列为：
Y 0 1 2
P
所以E(Y)＝0×＋1×＋2×＝．······································································13分
16．（本小题满分15分）解：（1）由△a＝(n＋1)2＋(n＋1)＋1－(n2＋n＋1)··············································2分
n
＝2n＋2，·············································································4分
△a －△a＝[2(n＋1)＋2]－(2n＋2)＝2（定值），
n＋1 n
所以{△a}是等差数列．·······································································6分
n
（2）由△a＝＝2＋，············································································8分
n
得an＋1 －an ＝2＋＝2＋－，··································································10分
方法一：当n≥2时，
an ＝(a2 －a1)＋(a3 －a2)＋…＋(an －an－1)＋a1···········································12分
＝(2＋1－)＋(2＋－)＋…＋(2＋－)＋1
＝2(n－1)＋1－＋1＝2n－，····································································14分
又a1 ＝1符合上式，
所以a＝2n－．···················································································15分
n
方法二：an＋1 ＋＝an ＋＋2
令bn ＝an ＋，则数列{b
n
}是以2为公差，首项为2的等差数列························12分
所以bn ＝2＋2(n－1)＝2n，····································································14分
所以a＝2n－．···················································································15分
n
17．（本小题满分15分）
解：（1）法一：取AC 中点P，连接PN，AP，
1 1
因为M，N分别为AB和BC 的中点，
1 1
所以NP∥AB，且NP＝AB...........................................................................................2分
1 1 1 1
又因为棱柱ABC－ABC ，
1 1 1
所以AB∥AB，且AB＝AB，
1 1 1 1
则AM∥AB，且AM＝AB，
1 1 1 1
所以AM∥NP，且AM＝NP，
所以四边形AMNP为平行四边形，
所以MN∥AP，·····················································································4分
因为AP平面ACC A，MN平面ACC A，
1 1 1 1
所以MN∥平面ACC A．········································································6分
1 1
法二：取BC中点Q，连结MQ，NQ，因为M为AB中点，所以MQ∥AC， 1分
因为AC平面ACC A，MQ平面ACC A，
1 1 1 1
所以MQ∥平面ACC A．········································································2分
1 1
因为棱柱ABC－ABC ，N为BC 中点，
1 1 1 1 1
所以CQ∥C N且CQ＝C N，
1 1
所以四边形CQNC 为平行四边形，所以NQ∥CC ，
1 1
因为CC 平面ACC A，NQ平面ACC A，
1 1 1 1 1
所以NQ∥平面ACC A．·········································································4分
1 1
因为NQ，MQ平面MNQ，NQ∩MQ＝Q，
所以平面MNQ∥平面ACC A，
1 1
又MN平面MNQ，
所以MN∥平面ACC A．········································································6分
1 1
法三：
因为MN＝CC1－CA，············································································2分
所以MN，CC1，CA共面．······································································4分
又因为MN平面ACC A；
1 1
所以MN∥平面ACC A．········································································6分
1 1
（2）因为AC＝2，AB＝2，∠ACB＝90°，
所以BC＝＝＝2．
因为直三棱柱ABC－ABC ，所以C C⊥平面ABC，
1 1 1 1
以C为坐标原点，CA，CB，CC 分别为x，y，z轴建立如图空间直角坐标系，设CC
1 1
＝t(t＞0)，
则A(2，0，0)，B(0，2，0)，C(0，0，0)，M(，1，0)，N(0，1，t)，A(2，0，t)，8分
1
所以\s\up6(→)＝(2，0，t)，\s\up6(→)＝(－，0，t)，\s\up6(→)＝(，1，0)，
因为MN⊥CA ，所以\s\up6(→)·\s\up6(→)＝0，即－4＋t2＝0，所以t＝2．········10分
1
则\s\up6(→)＝(－，0，2)，设平面CMN的一个法向量为m＝(x，y，z)，
则\s\up6(→)\s\up6(→)即取x＝，则y＝－2，z＝1，
则m＝(，－2，1)．··············································································12分
又\s\up6(→)＝(2，0，2)，设AC与平面CMN所成角为θ，
1
则sinθ＝|cos＜\s\up6(→)，m＞|＝\s\up6(→)\s\up6(→)＝＝，
所以AC与平面CMN所成角的正弦值为．················································15分
118．（本小题满分17分）
解：（1）由题意，C方程为－＝1，则c2＝a2＋a2＝2a2，即c＝a．
因为|FF|＝4，所以2c＝4，则c＝2，所以a＝，
1 2
所以双曲线C的方程为－＝1．·································································2分
（2）由双曲线定义知|AF|－|AF|＝2，|BF|－|BF|＝2．
1 2 1 2
所以ΔABF 的周长为
1
|AF|＋|BF|＋|AB|＝(|AF|＋2)＋(|BF|＋2)＋|AB|＝4＋2|AB|．
1 1 2 2
已知ΔABF 的周长为16，则4＋2|AB|＝16，解得|AB|＝6．··························4分
1
法一：设直线l的方程为x＝my＋2，A(x，y)，B(x，y)．
1 1 2 2
联立，消去x得(my＋2)2－y2＝2，即(m2－1)y2＋4my＋2＝0．
当m2＝1时，方程化为一次方程，不符合有两个交点的条件，所以m2≠1．
此时△＝16m2－8(m2－1)＝8(m2＋1)＞0，
y＋y＝－，yy＝．···············································································6分
1 2 1 2
由于A，B均在C的右支上，所以yy＜0，即m2－1＜0，所以m2＜1．
1 2
因为|AB|＝·
＝·＝2·································································································8分
所以2·＝6，解得m2＝，即m＝±，
所以直线l的方程为x－y－2＝0或x＋y－2＝0．·········································10分
（若设直线l的方程为y＝k(x－2)，x＋x＝，xx＝．
1 2 1 2
|AB|＝2·，参照给分）
法二：设直线l的方程为y＝k(x－2)，A(x，y)，B(x，y)．
1 1 2 2
联立，消去y 得(1－k2)x2＋4k2x－4k2－2＝0
则1－k2≠0，△＝16k4＋4(1－k2)(4k2＋2)＝8(1＋k2)＞0，
x＋x＝．····························································································6分
1 2
由统一定义得，准线方程为x＝1，
AF＝|x－1|＝(x－1)，BF＝|x－1|＝(x－1)，
2 1 1 2 2 2
AB＝AF＋BF＝(x＋x)－2＝－2＝6，·······················································8分
2 2 1 2
解得k＝±，
所以直线l的方程为x－y－2＝0或x＋y－2＝0．·········································10分
（3）设M(t，0)，
法一： （i）当直线l与x轴不重合时，设直线l的方程为x＝my＋2，
由（2）可得y＋y＝－，yy＝．
1 2 1 2x＝my＋2，x＝my＋2．
1 1 2 2
则\s\up6(→)＝(x－t，y)，\s\up6(→)＝(x－t，y)．
1 1 2 2
\s\up6(→)·\s\up6(→)＝(x－t)(x－t)＋yy＝(my＋2－t)(my＋2－t)＋yy
1 2 1 2 1 2 1 2
＝(m2＋1)yy＋m(2－t)(y＋y)＋(2－t)2
1 2 1 2
＝(m2＋1)·＋m(2－t)·(－)＋(2－t)2．··························································13分
设\s\up6(→)·\s\up6(→)＝λ，整理得(t2－2－λ)m2＋2＋λ－(t－2)2＝0．
因为上式对任意m≠±1，m∈R都成立，所以．·········································15分
解得t＝1，λ＝－1．
（ii）当直线l与x轴重合时，A，B的坐标分别为(，0)，(－，0)．
若t＝1，则\s\up6(→)·\s\up6(→)＝(－1)(－－1)＝－1．
综上，存在x轴上的定点M(1，0)，使\s\up6(→)·\s\up6(→)为定值－1．··········17分
法二：（i）当直线l的斜率存在时，
设直线l的方程为y＝k(x－2)，A(x，y)，B(x，y)．
1 1 2 2
联立，消去y 得(1－k2)x2＋4k2x－4k2－2＝0．
则1－k2≠0，△＝16k4＋4(1－k2)(4k2＋2)＝8(1＋k2)＞0，
x＋x＝，xx＝．
1 2 1 2
y＝k(x－2)，y＝k(x－2)．
1 1 2 2
则\s\up6(→)＝(x－t，y)，\s\up6(→)＝(x－t，y)．
1 1 2 2
\s\up6(→)·\s\up6(→)＝(x－t)(x－t)＋yy＝(x－t)(x－t)＋k2(x－2)(x－2)
1 2 1 2 1 2 1 2
＝(1＋k2)xx－(t＋2k2)(x＋x)＋t2＋4k2
1 2 1 2
＝(1＋k2)·－(t＋2k2)·＋t2＋4k2．·······························································13分
设\s\up6(→)·\s\up6(→)＝λ，整理得[λ＋2－(t－2)2]k2＋t2－2－λ＝0．
因为上式对任意k≠±1，k∈R都成立，所以．···········································15分
解得t＝1，λ＝－1．
（ii）当直线l⊥x轴时，A，B的坐标分别为(2，)，(2，－)．
若t＝1，则\s\up6(→)·\s\up6(→)＝(2－1)×(2－1)＋×(－)＝－1．
综上，存在x轴上的定点M(1，0)，使\s\up6(→)·\s\up6(→)为定值－1．··········17分
19．（本小题满分17分）
解：（1）若a＝1，则f(x)＝lnx＋＋x－1，f '(x)＝－＋1，定义域为(0，＋∞)．
·················································································································1分
因为y＝－x＋b是曲线y＝f(x)的切线，设切点为(x，f(x))，
0 0
所以f '(x)＝－1，即－＋1＝－1，整理得2x2＋x－1＝0，
0 0 0解得x＝－1(舍)，或x＝．·········································································3分
0 0
则f()＝－ln2＋．
因为切点(，－ln2＋)在切线y＝－x＋b上，所以b＝2－ln2．·····························4分
（2）f '(x)＝－＋a＝，
法一：①当a＜0时，定义域为(－∞，0)，此时ax2＋x－1＜0，则f '(x)＜0，
所以f(x)在(－∞，0)上单调递减；································································6分
②当a＞0时，定义域为(0，＋∞)，此时方程ax2＋x－1＝0的两根x＝，
1
x＝，且x＜0＜x .···················································································7分
2 1 2
所以当0＜x＜x，f '(x)＜0，f(x)在(0，)上单调递减
2
当x＞x，f '(x)＞0，f(x)在(，＋∞)上单调递增．
2
···········································································································9分
综上：当a＜0时，f(x)在(－∞，0)上单调递减；
当a＞0时，f(x)在(0，)上单调递减，在(，＋∞)上单调递增．
法二：令g(x)＝ax2＋x－1，则△＝1＋4a， 对称轴为x＝－．
①当a＜0时，定义域为(－∞，0).
(i)当a≤－\f(1,4时，△≤0，g(x)≤0，则f '(x)≤0，所以f(x)在(－∞，0)上单调递减；·5分
(ii)当－\f(1,4＜a＜0时，△＞0， 对称轴为x＝－＞0，则g(x)在(－∞，0)单调递增，
则g(x)＜g(0)＝－1，所以g(x)＜0，则f '(x)＜0，所以f(x)在(－∞，0)上单调递减；
·············································································································6分
②当a＞0时，定义域为(0，＋∞). △＝1＋4a＞0，x＝－＜0，则g(x)在(0，＋∞)单调递增，
且g(0)＝－1. 令g(x)＝ax2＋x－1＝0，则x＝. ····················································7分
当0＜x＜，f '(x)＜0，f(x)在(0，)上单调递减
当x＞，f '(x)＞0，f(x)在(，＋∞)上单调递增．
···········································································································9分
综上：当a＜0时，f(x)在(－∞，0)上单调递减；
当a＞0时，f(x)在(0，)上单调递减，在(，＋∞)上单调递增．
（3）①当a＜0时，
法一：当a＜0时，由（2）知f(x)在(－∞，0)上单调递减，注意到f()＝1，
若f(x)＜1有且仅有一个整数解，则该整数解必为－1．
则－2≤＜－1，所以－1＜a≤－；··································································11分
法二：当a＜0时，由（2）知f(x)在(－∞，0)上单调递减.若f(x)＜1有且仅有一个整数解，则该整数解必为－1．
则即，即 .
令－a＝t＞0，
令h(t)＝ln(2t)＋3t－\f(3,2≥0，
因为h(t)在(0，＋∞)单调递增，且h(\f(1,2)＝0. 则t≥\f(1,2，即a≤－\f(1,2.·············10分
令φ(t)＝lnt＋2t－2＜0，因为φ(t)在(0，＋∞)单调递增，且h(1)＝0. 则t＜1，即a＞－1.
所以－1＜a≤－·······················································································11分
②当a＞0时，
法一：当a＞0时，【讨论极值点与显零点的大小关系】
由（2）知f(x)在(0，)上单调递减，在(，＋∞)上单调递增.
注意到f()＝1.
（i）当＝，即a=2时，f(x) ＝f(\f(1,2)＝1，此时f(x)＜1无解，舍去.
min
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（ii）当＜，即a＞2时，此时＜\f(1,2.
若f(x)＜1有且仅有一个整数解，则该整数解只能为1．从而即化简得这与a＞2矛盾，舍
去.············································································································14分
（iii）当＞，即0＜a＜2时，此时＞\f(1,2.
若f(x)＜1有且仅有一个整数解，则该整数解必为1．从而1＜≤2，所以≤a＜1.
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综上，－1＜a≤－或≤a＜1．·····································································17分
法二：当a＞0时，【关注极值点与1的大小关系】
由（2）知f(x)在(0，)上单调递减，在(，＋∞)上单调递增.
f '(x)＝－＋a＝，x＞0.
令g(x)＝ax2＋x－1，g(0)＝－1＜0，g(1)＝a＞0, 则0＜＜1.·······························12分
【另解：因为＜＝＝1】
所以f(x)在[1，＋∞)上单调递增．
若f(x)＜1有且仅有一个整数解，则该整数解只能为1．
从而即化简得··························································································14分
设g(a)＝ln(2a)＋a，因为g'(a)＝＋1＞0，所以g(a)在(0，＋∞)上单调递增．
又因为g()＝，所以g(a)≥的解集为a≥．
故a的取值范围是[，1)．···········································································16分综上，－1＜a≤－或≤a＜1．·····································································17分
法三：当a＞0时，【关注极值点与和1的大小关系】
由（2）知f(x)在(0，)上单调递减，在(，＋∞)上单调递增，
f(1)＝lna＋1·····························································································12分
（i）当0＜a＜1时，f(1)＝lna＋1＜1
则1为f(x)＜1的整数解. 注意到f()＝1，＜，
若f(x)＜1有且仅有一个整数解，则1＜≤2，所以≤a＜1；·······························14分
(ii)当a≥1时，＜1，
由（2）知定义域为(0，＋∞)且f(x)在(1，＋∞)上单调递增，
又因为f(1)＝lna＋1≥1，所以f(x)＜1无整数解．············································16分
综上，－1＜a≤－或≤a＜1．·····································································17分
