2025届南京零模数学试题+答案

1南京市2025届高三年级学情调研数学参考答案2024.0912345678DDABACCB9ABACD 解1）假设H0：8点前到单位与方案选择无关，则······································································2分≈3.94＞3.841，·····································································（2）选择A方案上班，8点前到单位的概率为0.7，当X＝3时，则分两种情况：则P2＝(1-0.7)×C(1-0.5)×0.53=.··························································12分2综上，P(X＝3)＝P1＋P2=.············································解1）因为E，F分别为线段AB，BC中点， →→→→DMDN1因为AM＝2MD，CN＝2ND，即DA＝DC＝3，所以MN∥AC，所以EF∥MN.·····························································又MN平面MNB，EF丈平面MNB，所以EF∥平面MNB.··················zDMNzDMN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=AO································AC，DO平面ACD，AO································EFExBxB·······································································································10分 2所以BD与平面MNB所成角的正弦值为··················································15分2所以BD与平面MNB所成角的正弦值为··················································15分3－λan，2＝(－1)×(－1)n＋2n＋1－2×(－1)n－2n＋1＝－3×(－1)n．············所以λ＝2符合题意．···························································＝－当n为偶数时，Tn＝－3×[－12＋22－32＋42－52＋62－…－(n－1)2＋n2]3×(1＋2＋…＋n)n(n＋1)．········＝n(n＋1)．···································································综上，＞0，所以i为偶数．···································································13分＝－所以i＝2．···························· 4→→→→解得x0＝3，y0=±2， A.···································8分②当直线l与y轴垂直时，此时PQ＝4不满足条件.所以y1＋y2y1y2=-.·····················································10分所以xP＋xQ＝x1＋x2，xPxQ＝x1x2＋y1y2，································P=t2(y1＋y2)2－4(t2＋1)y1y2=+＝－＝－5当a＞1时，f''(x)＝ex－a＋2a＞0，则f'(x)在(1，+∞)上单调递增.当x∈(x0，+∞)时，f'(x)＞0，所以f(x)在(x0，+∞)上单调递增.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分所以a＞1不满足题意.·········································································＝－当x∈(-∞,-ln2)时，q'(x)＜0，所以q(x)在(-∞,-ln2)单调递减，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.································1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；··············································因为f'(1)＝e1－a－a≥0，且f'(x)在[0，+∞)上单调递增，6当x∈(x1，+∞)时，f'