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1南京市2025届高三年级学情调研数学参考答案2024.09一、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．12345678DDABACCB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．91011ABBCDACD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．12．24013．3π14．33四、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）假设H0：8点前到单位与方案选择无关，则χ2＝

100×(28×30－12×30)240×60×42×58······································································2分＝800203≈3.94＞3.841，··············

······························································4分所以有95%的把握认为8点前到单位与路线选择有关．············

··························6分（2）选择A方案上班，8点前到单位的概率为0.7，选择B方案上班，8点前到单位的概率为0.5．·······································

·········8分当X＝3时，则分两种情况：①若周一8点前到单位，则P1＝0.7×C24(1-0.5)2×0.52＝2180．·····························································10分

②若周一8点前没有到单位，则P2＝(1-0.7)×C34(1-0.5)×0.53＝680．··························································12分{#{QQABZYAEogCgAIBAARgCEwXKCkOQkACCAagG

BEAEoAABgQNABAA=}#}2DABCEFMNOxyz综上，P(X＝3)＝P1＋P2＝2780．···············································

······················13分16．(本小题满分15分)解：（1）因为E，F分别为线段AB，BC中点，所以EF∥AC．······································

···················································2分因为AM→＝2MD→，CN→＝2ND→，即DMDA＝DNDC＝13，所以MN∥AC，所以EF∥MN．·························

···········································4分又MN平面MNB，EF平面MNB，所以EF∥平面MNB．·····································································

············6分（2）取AC中点O，连接DO，OE．因为△ACD为正三角形，所以DO⊥AC．因为平面ACD⊥平面ABC，平面ACD∩平面ABC＝AC，DO平面ACD，所以DO⊥平面ABC．····

·············································································8分因为O，E分别为AC，AB中点，则OE∥BC．又因为AC⊥BC，所以OE⊥AC．以O为坐标原点，OE，OC，OD所在直线分

别为x，y，z轴建立空间直角坐标系，·······································································································10分则D(0，0，332)，B(3，32，

0)，M(0，－12，3)，N(0，12，3)，故→BM＝(－3，－2，3)，→MN＝(0，1，0)，→BD＝(－3，－32，332)．设平面MNB的法向量为n＝(x，y，z)，直线BD与平面MNB所成角为θ，则n·→BM＝0，n·→MN＝0，即

－3x－2y＋3z＝0，y＝0．取n＝(3，0，3)．···················································································12分则s

inθ＝|cos<→BD，n>|＝|→BD·n||→BD||n|＝|－33＋0＋932|9＋94＋274×3＋9＝33232×23＝28，所以BD与平面MNB所成角的正弦值为28．················································

··15分{#{QQABZYAEogCgAIBAARgCEwXKCkOQkACCAagGBEAEoAABgQNABAA=}#}317．(本小题满分15分)解：(1)因为an＝(－1)n＋2n，则a1＝1，a2＝5，a3＝7，a4＝17．又bn＝an＋1－λan，则b1＝a2－λa1＝5－λ

，b2＝a3－λa2＝7－5λ，b3＝a4－λa3＝17－7λ．·····················2分因为{bn}为等比数列，则b22＝b1·b3，所以(7－5λ)2＝(5－λ)(17－7λ)，…………………

4分整理得λ2－λ－2＝0，解得λ＝－1或2．因为λ＞0，故λ＝2．当λ＝2时，bn＝an＋1－2an＝(－1)n＋1＋2n＋1－2[(－1)n＋2n]＝(－1)×(－1)n＋2n＋1－2×(－1)n－2n＋1＝－3×(－

1)n．·····································6分则bn＋1bn＝－3×(－1)n＋1－3×(－1)n＝－1，故{bn}为等比数列，所以λ＝2符合题意．·····················································

························7分(2)bn·n2＝－3×(－1)n·n2，当n为偶数时，Tn＝－3×[－12＋22－32＋42－52＋62－…－(n－1)2＋n2]＝－3×(1＋2＋…＋n)＝－32n(n＋1)．··················

····································10分当n为奇数时，Tn＝Tn＋1－bn＋1(n＋1)2＝－32(n＋1)(n＋2)＋3(n＋1)2＝32n(n＋1)．·························

······························································12分综上，Tn＝32n(n＋1)，n为奇数，－32n(n＋1)，n为偶数．因为Ti·Ti＋2＞0，又Ti·Ti＋2＝15Ti＋1

，故Ti＋1＞0，所以i为偶数．···································································13分所以[－32i(i＋1)]·[－32(i＋2

)(i＋3)]＝15×32(i＋1)(i＋2)，整理得i2＋3i－10＝0，解得i＝2或i＝－5(舍)，所以i＝2．························································································15

分18．(本小题满分17分)解：（1）由题意可知c＝6，点T在C上，根据双曲线的定义可知|TF1|－|TF2|＝2a，即2a＝(36)2＋(10)2－(6)2＋(10)2＝4，所以a＝2，·········

··················2分{#{QQABZYAEogCgAIBAARgCEwXKCkOQkACCAagGBEAEoAABgQNABAA=}#}4则b2＝c2－a2＝2，所以C的方程为x24－y22＝1．······

·······························································3分（2）①设B(x0，y0)，DB→＝(x0－1，y0)．因为DA→＝3DB→，所以DA→＝(

3x0－3，3y0)，所以A点坐标为(3x0－2，3y0)，···································································5分因为A，B在双曲

线C上，所以x204－y202＝1，(3x0－2)24－(3y0)22＝1，解得x0＝3，y0＝±102，·································································

·······7分所以A点坐标为(7，±3102)，所以SΔF1F2A＝12|yA|×|F1F2|＝12×3102×26＝315．···································8分

②当直线l与y轴垂直时，此时PQ＝4不满足条件．设直线l的方程为x＝ty＋1，A(x1，y1)，B(x2，y2)，P(xP，0)，Q(xQ，0)．直线l与C联立x24－y22＝1，x＝ty＋1，消去x，得(t2－2)y2＋2ty－3＝0，所以y

1＋y2＝－2tt2－2，y1y2＝－3t2－2．·····················································10分由Δ＝4t2＋12(t2－2)＞0，t2－2≠0

．，得t2＞32且t2≠2．以AB为直径的圆方程为(x－x1)(x－x2)＋(y－y1)(y－y2)＝0，令y＝0，可得x2－(x1＋x2)x＋x1x2＋y1y2＝0，则xP，xQ为方程的两个根，所以xP＋xQ＝x1＋x2，xPxQ＝

x1x2＋y1y2，···················································13分所以PQ＝|xP－xQ|＝(xP＋xQ)2－4xPxQ＝(x1＋x2)2－4(x1x2＋y1y2)＝(x1－x2)2－4

y1y2＝t2(y1－y2)2－4y1y2＝t2(y1＋y2)2－4(t2＋1)y1y2＝4t4(t2－2)2＋12(t2＋1)t2－2＝16t4－12t2－24(t2－2)2＝2．···············································

············15分解得t2＝－2(舍)或t2＝53，即t＝±153，所以直线l的方程为：3x±15y－3＝0．··················································17分{#{QQABZY

AEogCgAIBAARgCEwXKCkOQkACCAagGBEAEoAABgQNABAA=}#}519．(本小题满分17分)解：（1）当a＝1时，f(x)＝ex－1＋x2－3x＋1，则f'(x)＝ex－1＋2x－3，所以曲线y＝f(x)在x＝1处

切线的斜率k＝f'(1)＝0．又因为f(1)＝0，所以曲线y＝f(x)在x＝1处切线的方程为y＝0．···················································3分

（2）f(1)＝e1－a－2a＋1，f'(x)＝ex－a＋2ax－3a，则f'(1)＝e1－a－a，当a＞1时，f''(x)＝ex－a＋2a＞0，则f'(x)在(1，＋∞)上单调递增．因为f'(1)＝e1－a－a＜e1－1

－1＝0，f'(a)＝1＋2a2－3a＝(2a－1)(a－1)＞0，所以存在唯一的x0∈(1，a)，使得f'(x0)＝0．·······················································5分当x∈(1，x0)时，f'(x)＜0，所以f(x)在[1，x

0)上单调递减；当x∈(x0，＋∞)时，f'(x)＞0，所以f(x)在(x0，＋∞)上单调递增．又因为f(1)＝e1－a－2a＋1＜e0－2＋1＝0，所以f(x0)＜f(1)＜0．又因为f(3)＝e3－a＋1＞0，所以当a＞1时，f(x)在[

1，＋∞)上有且只有一个零点．···································8分（3）①当a＞1时，f(1)＝e1－a－2a＋1＜e0－2＋1＝0，与当x≥0时，f(x)≥0矛盾，所

以a＞1不满足题意．···········································································9分②当a≤1时，f(0)＝e－a＋1＞0，f'(x)＝ex－a＋2ax－3a，f''(x)＝ex－a＋2a，f''(

0)＝e－a＋2a．记函数q(x)＝e－x＋2x，x≤1，则q'(x)＝－e－x＋2，当x∈(－ln2，1)时，q'(x)＞0，所以q(x)在(－ln2，1)单调递增；当x∈(－∞，－ln2)时，q'(x)＜0，所以q(x)在(－∞，－ln2)单调递减，所以q(x)≥q

(－ln2)＝2－2ln2＞0，所以f''(0)＞0．又因为f''(x)在[0，＋∞)上单调递增，所以f''(x)≥f''(0)＞0，所以f'(x)在[0，＋∞)上单调递增．························

········11分(i)若f'(0)＝e－a－3a≥0，则f'(x)≥f'(0)≥0，所以f(x)在[0，＋∞)上单调递增，则f(x)≥f(0)＞0，符合题意；·························································

··········13分(ii)若f'(0)＝e－a－3a＜0，可得a＞0，则0＜a≤1．因为f'(1)＝e1－a－a≥0，且f'(x)在[0，＋∞)上单调递增，{#{QQABZYAEogCgAIBAARgCEwXKCkOQkACCAagGBEAEoAABgQNABAA=}#}6所以存

在唯一的x1∈(0，1]，使得f'(x1)＝0．当x∈(0，x1)时，f'(x)＜0，所以f(x)在(0，x1)上单调递减，当x∈(x1，＋∞)时，f'(x)＞0，所以f(x)在(x1，＋∞)上单调递增，其中x1∈(0，1]，且ex1－a＋2ax1－3

a＝0．························································15分所以f(x)≥f(x1)＝ex1－a＋ax12－3ax1＋1＝3a－2ax1＋ax12－3ax1＋1＝ax12－5ax1＋3a＋1＝a(x12－5x1＋3)＋

1，因为x1∈(0，1]，所以x12－5x1＋3∈[－1，3)．又因为a∈(0，1]，所以a(x12－5x1＋3)≥－1，所以f(x)≥0，满足题意．结合①②可知，当a≤1时，满足题意．综上，a的取值范围为(－∞，1]．···············

··············································17分{#{QQABZYAEogCgAIBAARgCEwXKCkOQkACCAagGBEAEoAABgQNABAA=}#}