南京市2026屆高三年級(jí)學(xué)情調(diào)研一、選擇題：本大題共8小題，每小題5分，共40分．在每小題給出的四個(gè)選項(xiàng)中，只有一項(xiàng)是符合題目要求的，請(qǐng)把答案填涂在答題卡相應(yīng)位置上．二、選擇題：本大題共3小題，每小題6分，共18分．在每小題給出的四個(gè)選項(xiàng)中，有多項(xiàng)符合題目要求，請(qǐng)把答案填涂在答題卡相應(yīng)位置上．全部選對(duì)得6分，部分選對(duì)得部分分，不選或有錯(cuò)選的得0分．三、填空題：本大題共3小題，每小題5分，共15分．請(qǐng)把答案填寫(xiě)在答題卡相應(yīng)位置上．四、解答題：本大題共5小題，共77分．請(qǐng)?jiān)诖痤}卡指定區(qū)域內(nèi)作答，解答時(shí)應(yīng)寫(xiě)出必要的文字說(shuō)明，證明過(guò)程或演算步驟．15本小題滿(mǎn)分13分）（2）Y的可能取值為0，1，2，P················································································9分P·····················································································9分則分布列為：Y012P 28 2816本小題滿(mǎn)分15分）解1）由△an＝(n＋1)2＋(n＋1)＋1－(n2＋n＋1)················································2分=2n＋2，··············································································4分△an＋1-△an＝[2(n＋1)＋2]－(2n＋2)＝2（定值所以{△an}是等差數(shù)列．········································································6分由△an＝2n·····················································8分得an＋1－an＝2+····················································10分方法一：當(dāng)n≥2時(shí)，又a＝1符合上式，1方法二：an＋1＋an所以b＝2＋2(n－1)＝2n，······································································14分n17本小題滿(mǎn)分15分）解1）法一：取A1C1中點(diǎn)P，連接PN，AP，因?yàn)镸，N分別為AB和B1C1的中點(diǎn)，所以NP∥A1B1，且NPA1B1........................................................................................2分又因?yàn)槔庵鵄BC－A1B1C1，所以AB∥A1B1，且AB＝A1B1，則AM∥A1B1，且AMA1B1，所以AM∥NP，且AM＝NP，所以四邊形AMNP為平行四邊形，所以MN∥AP，······················································································4分因?yàn)锳Pc平面ACC1A1，MN丈平面ACC1A1，所以MN∥平面ACC1A1．·········································································6分法二：取BC中點(diǎn)Q，連結(jié)MQ，NQ，因?yàn)镸為AB中點(diǎn)，所以MQ∥AC，1分因?yàn)锳Cc平面ACC1A1，MQ丈平面ACC1A1，所以MQ∥平面ACC1A1．·········································································2分因?yàn)槔庵鵄BC－A1B1C1，N為B1C1中點(diǎn)，所以CQ∥C1N且CQ＝C1N，所以四邊形CQNC1為平行四邊形，所以NQ∥CC1，因?yàn)镃C1c平面ACC1A1，NQ丈平面ACC1A1，所以NQ∥平面ACC1A1．·········································································4分因?yàn)镹Q，MQc平面MNQ，NQ∩MQ＝Q，所以平面MNQ∥平面ACC1A1，又MNc平面MNQ，所以MN∥平面ACC1A1．·········································································6分法三：因?yàn)椤ぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁぁ?分→→→，CA共面．·········································································4分又因?yàn)镸N丈平面ACC1A1；所以MN∥平面ACC1A1．·········································································6分因?yàn)橹比庵鵄BC－A1B1C1，所以C1C⊥平面ABC，以C為坐標(biāo)原點(diǎn)，CA，CB，CC1分別為x，y，z軸建立如圖空間直角坐標(biāo)系，設(shè)CC1=t(t＞0)，→→→所以CA1＝(2V2，0，t)，MN＝(－V2，0，t)，CM＝(V2，1，0→→→→→→則MN＝(0，2)，設(shè)平面CMN的一個(gè)法向量為m＝(x，y，z)，→＝(2，0，2)，設(shè)A1C與平面CMN所成角為θ,則sinθ=|cos18本小題滿(mǎn)分17分）解1）由題意，C方程為a_xF_2－y則c2＝a2＋a2＝2a2，即c＝a．所以雙曲線(xiàn)C的方程為x·····························································2分所以ΔABF1的周長(zhǎng)為|＋|BF1|＋|AB|＝(|AF2|＋2V2)＋(|BF2|＋2V2)＋|AB|＝4V2＋2|AB|．已知ΔABF1的周長(zhǎng)為16V2，則4V2＋2|AB|＝16V2，解得|AB|＝6V2．·············4分法一：設(shè)直線(xiàn)l的方程為x＝my＋2，A(x1，y1)，B(x2，y2)．聯(lián)立消去x得(my＋2)2－y2＝2，即(m2－1)y2＋4my＋2＝0． 當(dāng)m2＝1時(shí)，方程化為一次方程，不符合有兩個(gè)交點(diǎn)的條件，所以m2≠1．此時(shí)△=16m2－8(m2－1)＝8(m2＋1)＞0，y1＋y2y1y2=································································6分由于A，B均在C的右支上，所以y1y2＜0，即m2－1＜0，所以m2＜1．－4y1y2所以2，解得m2即m所以直線(xiàn)l的方程為v2x－y－2v2＝0或\x＋y－2v2＝0．···························10分（若設(shè)直線(xiàn)l的方程為y＝k(x－2)，x1＋x2＝x1x法二：設(shè)直線(xiàn)l的方程為y＝k(x－2)，A(x1，y1)，B(x2，y2)．聯(lián)立消去y得(1－k2)x2＋4k2x－4k2－2＝0則1－k2≠0，△=16k4＋4(1－k2)(4k2＋2)＝8(1＋k2)＞0，x1＋x2=.······················································································6分由統(tǒng)一定義得，準(zhǔn)線(xiàn)方程為x＝1，AF2＝v2|x1－1|＝v2(x1－1)，BF2＝v2|x2－1|＝v2(x2－1)，解得k=±v2，所以直線(xiàn)l的方程為v2x－y－2v2＝0或\x＋y－2v2＝0．···························10分（3）設(shè)M(t，0)，法一i）當(dāng)直線(xiàn)l與x軸不重合時(shí)，設(shè)直線(xiàn)l的方程為x＝my＋2，由可得y1＋y2y1y2=x1＝my1＋2，x2＝my2＋2．→→則MA＝(x1－t，y1)，MB＝(x2－t，y2)．→·→MAMB＝(x1－t)(x2－t)＋y1y2＝(my1＋2－t)(my2＋2－t)＋→·→=(m2＋1)y1y2＋m(2－t)(y1＋y2)＋(2－t)2··········································13分設(shè)·=λ,整理得(t2－2-λ)m2＋2+λ-(t－2)2＝0．解得t＝1，λ=-1．（ii）當(dāng)直線(xiàn)l與x軸重合時(shí)，A，B的坐標(biāo)分別為(v2，0)，(－v2，0)．→·→若t＝1，則MAMB＝(V2－1)(－\－1)→·→→·→綜上，存在x軸上的定點(diǎn)M(1，0)，使MAMB為定值－1．→·→法二i）當(dāng)直線(xiàn)l的斜率存在時(shí)，設(shè)直線(xiàn)l的方程為y＝k(x－2)，A(x1，y1)，B(x2，y2)．聯(lián)立消去y得(1－k2)x2＋4k2x－4k2－2＝0．則1－k2≠0，△=16k4＋4(1－k2)(4k2＋2)＝8(1＋k2)＞0，y1＝k(x1－2)，y2＝k(x2－2)．→→則MA＝(x1－t，y1)，MB＝(x2－t，y2)．1－t)(x2－t)＋y1y2＝(x1－t)(x2－t)＋k2(x1－2)(x2－2)=(1＋k2)x1x2－(t＋2k2)(x1＋x2)＋t2＋4k2設(shè)·=λ,整理得[λ+2－(t－2)2]k2＋t2－2-λ=0．解得t＝1，λ=-1．（ii）當(dāng)直線(xiàn)l⊥x軸時(shí)，A，B的坐標(biāo)分別為(2，v2)，(2，－\)．→→綜上，存在x軸上的定點(diǎn)M(1，0)，使MA·MB為定值－1．··························17分19本小題滿(mǎn)分17分）解1）若a＝1，則f＝lnxx－1，f'＝x定義域?yàn)?0，+∞)．因?yàn)閥x＋b是曲線(xiàn)y＝f(x)的切線(xiàn)，設(shè)切點(diǎn)為(x0，f(x0))，所以f'(x0)1，即0－x整理得2x02＋x0－1＝0，則fln因?yàn)榍悬c(diǎn)在切線(xiàn)yx＋b上，所以b＝2－ln2．··························4分法一：①當(dāng)a＜0時(shí)，定義域?yàn)?-∞,0)，此時(shí)ax2＋x－1＜0，則f'(x)＜0，所以f(x)在(-∞,0)上單調(diào)遞減；································································6分-1－v1＋4a②當(dāng)a＞0時(shí)，定義域?yàn)?0，+∞)，此時(shí)方程ax2＋x－1＝0的兩根-1－v1＋4a2a1＜0＜x2.··································································7分所以當(dāng)0＜x＜x2，f'(x)＜0，f(x)在(0，)上單調(diào)遞減當(dāng)x＞x2，f'(x)＞0，f(x)在(，+∞)上單調(diào)遞增．···········································································································9分綜上：當(dāng)a＜0時(shí)，f(x)在(-∞,0)上單調(diào)遞減；當(dāng)a＞0時(shí)，f(x)在(0，)上單調(diào)遞減，在(，+∞)上單調(diào)遞增．法二：令g(x)＝ax2＋x－1，則△=1＋4a，對(duì)稱(chēng)軸為x①當(dāng)a＜0時(shí)，定義域?yàn)?-∞,0).當(dāng)a時(shí)，△≤0，g(x)≤0，則f'(x)≤0，所以f(x)在(-∞,0)上單調(diào)遞減；······5分4(ii)當(dāng)－1＜a＜0時(shí)，△>0，對(duì)稱(chēng)軸為x則g(x)在(-∞,0)單調(diào)遞增，4則g(x)＜g(0)1，所以g(x)＜0，則f'(x)＜0，所以f(x)在(-∞,0)上單調(diào)遞減；·············································································································6分②當(dāng)a＞0時(shí)，定義域?yàn)?0，+∞).△=1＋4a＞0，x則g(x)在(0，+∞)單調(diào)遞增，且g(0)1.令g(x)＝ax2＋x－1＝0，則x＝.····························7分當(dāng)0＜xf'(x)＜0，f(x)在(0，)上單調(diào)遞減當(dāng)xf'(x)＞0，f(x)在(，+∞)上單調(diào)遞增．···········································································································9分綜上：當(dāng)a＜0時(shí)，f(x)在(-∞,0)上單調(diào)遞減；當(dāng)a＞0時(shí)，f(x)在(0，a)上單調(diào)遞減，在上單調(diào)遞增．（3）①當(dāng)a＜0時(shí)，法一：當(dāng)a＜0時(shí)，由（2）知f(x)在(-∞,0)上單調(diào)遞減，注意到f若f(x)＜1有且僅有一個(gè)整數(shù)解，則該整數(shù)解必為－1．法二：當(dāng)a＜0時(shí)，由（2）知f(x)在(-∞,0)上單調(diào)遞減.若f(x)＜1有且僅有一個(gè)整數(shù)解，則該整數(shù)解必為－1．令－a＝t＞0，因?yàn)閔(t)在(0，+∞)單調(diào)遞增，且h則t≥,即a≤-.····························10分令φ(t)＝lnt＋2t－2＜0，因?yàn)棣?t)在(0，+∞)單調(diào)遞增，且h(1)＝0.則t＜1，即a＞－1.所以－1＜a≤-······················································································11分②當(dāng)a＞0時(shí)，由（2）知f(x)在(0，a)上單調(diào)遞減，在(，+∞)上單調(diào)遞增.注意到f（i）當(dāng)a，即a=2時(shí)，fmin＝f此時(shí)f(x)＜1無(wú)解，舍去.簡(jiǎn)得這與a＞2矛盾，舍去.·······················································14分若f(x)＜1