廈門一中2023-2024學年度高2025屆高二下學期強化限時訓練03 數(shù)學答案

1項是符合題目要求的.2.【答案】C【詳解】依題可知，f′(x)=aex 1?xax 1a,即a≥1e3．【答案】B【詳解】若存款利率為x，則存款量是kx2，銀行支付的利息是kx3，獲得的貸款利息∴當x=0.0324時，y取得最大值，即當存款利率為0.0324時，銀行獲得最大收益.t上，可得tttf(t)遞減，所以a，直線y=b與曲線y=f(t)的圖象有兩個交點，則maxa的圖象如下：當0<b<ea時，直線y=b與y=f(t)圖象有兩個交點.法二：畫出函數(shù)曲線y=ex的圖象如圖所示，根據直觀即可判定點(a,b)在曲線下方和x軸上方時才可以作出兩條切線.由此可知0<b<ea.5．【答案】D【詳解】連接BF,AF，延長交拋物線于C,D兩點，不妨假設A在B的下方，根據對稱性可知，F(xiàn)B,FA與y軸正方向所夾的角為 p, p42理，得AB==.36.【答案】A【詳解】由雙曲線方程知，右焦點F(2,0)，M(4,3)在雙曲線上，因為直線MF的4+02又=2,=0，故MN的中點為F，所以以MN4+02又3?02=7，連接NQ,NP,PF，因為Q在以MN為直徑的圓上，所以MQ⊥NQ，PQ=PNcos(π?∠MPN)，=?2?所以PM?PQ=?2??+?+2=2?2=49?2，---min因為P為雙曲線上點，所以PF=1，且此時P在圓內，所以PM?PQ≤49?1=48.---min7.【答案】D【詳解】方程f(x)=kx?的圖象有三個交點．由f(x)是R上的奇函數(shù)，則f(0)=0．當0<x≤1時，f(x)=xlnx，則f2222切線l2，設切點P(x0,y0)．其中y0=x0lnx0，f′(x)=lnx+1，則斜率kl=lnx0+1，切線x?x0)過點A0lnx0=0+10?x0kl=ln2+1kl=ln2+1=1?ln2，當直線l:y=kx?2繞點A|（0,?2)|在l1與l2之間旋轉時，直線l:y=kx?2與函a=an+=a+1n∈N，a+1a=an+=a+1n∈N，a+1nnnnn3nJannna+a+…+a00<1+題目要求．全部選對的得6分，部分選對的得部分分，有選錯的得0分.ffff(x)在(?∞,a)上遞增，f(x)在(a,b)上遞減，f(x)在(b,+∞)上遞增，故其大致圖象如下：則有m<a<k<b<t,故m+k<a+b<t+k，則A正確；又可得fff為1<a<k<b<t，則1<ab<kt，b<a2b<kta又導函數(shù)f′(x)在(b,+∞)單調遞增，故fa2b11.【答案】ABD【詳解】取BD中點O，連接OA、OC；因為AB=43OA=OC=ABsin60。=4×=2.因為AC=2，所以OA2+OC2=AC2，所以OA又因為AB=AD，O為BD的中點，則OA⊥BD，同理可得OC⊥BD，平面ABD，所以平面ABD⊥平面BDC，故A正確；因為OA⊥平面BDC，OC⊥BD，以O為原點，OB、OC、OA所在直線分別為x、y、z軸建立如下圖所示的空間直角坐標系，4所以點D到直線PQ的距離為d=?|---------PQ()=((2=7,故C錯誤；設P(a,0,0)、2+2=,當a=0a2PQ=λ?2+2a232+2=,當a=0a2PQ=λ?2+2a23PQ且------26則cosθ=26則cosθ==AD------ 與AD6,故6,故D正確；42?2531?2t=?max.因為f2?2lnx，所以f2?2=2xxxf所以m≤e2?2，故實數(shù)m的取值范圍是(?∞,e2?2.14．【答案】π【詳解】取BC中點O，連接AO,DO，因為AB=AC=BD=CD=DO⊥BC，設BO=m,AO=n，則m2+n2=1，則0<m<1且DO=AO=n，要想四面體體積最大，則AO為三棱錐A?BCD的高，即AO⊥平面BCD，1?m2，2 ?m3,5由余弦定理得由余弦定理得2323則外接球球心E在△BCD上的投影為點N，即EN⊥平面BCD，過點E作EF⊥AO于點F， 連接AE,DE，則AE,DE為外接球半徑，其中EF EN=OF，設EN=OF=h，AE=DE=R，則2|21513分）(1)f′(x)=?xex，·····································································································2分x)上為增函數(shù)····································5分x的解為x=1，則f(1)=?a，即切點為(1,?a)······································7分(2)由(1)知，f′(x)=?xex，當x>0時，f′(x)<0，f(x)在(0,+∞)上單調遞減·······················9分 ++…+ 23n6(1)因為AD∥BC，AD=2BC，又M為邊AD的中點，所以BC//AM，BC=AM，所以BCMA為平行四邊形，且AM=AB，所以BCMA為菱形，BM⊥AC 2分 4分又因為ACnCB1=C，AC?平面ACB1，CB1?平面ACB1，所以BM⊥平面ACB1由BM?平面BMB1，所以平面BMB1⊥平面ACB1····························································6分(2)由（1）知BM⊥AC，BC∥MD，BC=MD，故四邊形BCMD是平行四邊形，所以CD∥BM，以CD，CA，CB1分別為x，y，z軸建立空間直角坐標系················································7分設AD=2AB=2BC=2x，CB1=h，由∠BAD=60。，四棱柱ABCD?A1B1C1D1體積為9，解得x=2,h=3·······································································································10分=?3,1,3)，平面ABC的一個法向量為=(0,0,3)，設平面DCC1的法向量為=(x,y,z)，，A=?3,1,3)，平面ABC的一個法向量為=(0,0,3)，設平面DCC1的法向量為=(x,y,z)，,0,1×3得sin,=，即二面角A?DC?C1的正弦值為.···········································15分1715分）(1)f(x)的定義域為(0,+∞)，f′(x)=1+(2x?3)=2x2?3x+1 2分xxf所以f(x)的單調遞增區(qū)間為(|（0,,(1,+∞)，單調遞減區(qū)間為,1······································6分(2)f′(x)=，由f(x)在(0,+∞)上有兩個不同的極值點x1,x2，故2ax2?3ax+1=0有兩個不同的正根，則有〈12，解得a>8,9·····························8分因為f(x1)+f(x2)=ln(x1x2x+xx2+x22?2x1x2+x2)+4a=?ln(2a)+a?1，··································7設g(a)=?ln(2a)+a?1，a>，則g(a)=?=494a4a故g(a)在,+∞上單調遞增·······················································································13分又g(a)>g=?ln+=+ln，故f(x1)+f(x2)>+ln·········································15分1817分）1+m21+m22y1+y2?1+m21+m22y1+y2?4y1y21+m22?1又AB=y1?y2===2xy ?=2xy ?=4=12=1··············4分,所以C的標準方程為2=a2+b2(2)存在定圓滿足題意，方程為(x?6)2+y2=16，理由如下：因為過點F的直線與C的右支交于A,B兩點，所以直線AB斜率不為0，設直線AB方程為x=my+3m2?y1y2=,·····················································································7分3m2?1所以y0==，x0=my0+4=，······························································所以y0=由直線AB與C的右支交于A,B兩點可知y1y2由直線AB與C的右支交于A,B兩點可知y1y2=1+m2y+2=2·············································13分由對稱性可知，若存在定圓O1與圓M相內切，則定圓圓心O1一定在x軸上，不妨設定圓O1方程為(x?n)2+y2=r2(r>0)，則由圓O1與圓M相內切可知，MO1=?r，即?n2+2=?r2，2?所以存在定圓O1:(x?6)2+y2=16滿足題意.··································································17分1917分）?2,1上單調遞減···········································2分e?2=21?e?20e?2,1(x)=0,f(x)在e?2,x0上8 因為fe?2=4e?2?e?4=e?2(4?e?2)>0,f(1)=?1，所以f(x)在e?2,1上最小x+1?2x+1?2(lnx+x+1)，x1+1)·····························································6分2tte2tte2tt設h(t)=e2tt設h(t)=e上單調遞增，在(1,+∞)上單調遞減，·································