福建省福州市2024-2025学年高三年级上学期第一次质量检测数学试卷答案

2024-2025学年福州市高三年级第一次质量检测数学答案一、单项选择题：本题共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。题目12345678答案DCCBADBC二、多项选择题：本题共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。全部选对的得6分，部分选对的得部分分，有选错的得0分。题目91011答案ADABDBCD三、填空题：本大题共3小题，每小题5分，共15分。题目121314答案35(24，25)四、解答题：本题共5小题，共77分。解答应写出文字说明、证明过程或演算步骤。15.（13分）已知数列an满足a1=2，aann+1=+32．（1）证明：数列an+1是等比数列；（2）求an的前n项和Sn．【解法一】（1）证明：因为aann+1=+32，且，所以an+10，···················································································1分aa+13+2+1所以nn+1=········································································3分aann++113(a+1)==n3，····································································5分an+1又a1+=13，所以数列an+1是以3为首项，为公比的等比数列．·······························6分nn（2）由（1）得an+=13，所以an=−31，·············································8分2n所以Sn=(3−1)+(3−1)++(3−1)数学试题第1页（共14页）{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}=(3+323+3++3n)−n····························································10分33−n+1=−n············································································12分13−33n+1−=−n.············································································13分2【解法二】（1）证明：因为aann+1=+32，所以an+1+1=3an+3=3(an+1)，·····························································2分因为a1=2，所以a1+1=30，所以an+10，········································4分a+1所以n+1=3,an+1所以数列an+1是以3为首项，3为公比的等比数列．·······························6分（2）略，同解法一．16.（15分）已知△ABC的内角ABC,,的对边分别为abc,,，且2acosC=3bcosC+3ccosB．（1）求角C；（2）若a=4，b=3，D为AB中点，求CD的长．【解法一】（1）因为，由正弦定理，得2sinACBCBCcos=+3sincos3cossin··············································2分=+3sin(BC)··································································4分=−3sin(πA)=3sinA，······································································6分3因为0Aπ，则sinA0，所以cosC=，··········································7分2π由于0Cπ，则C=；····································································8分61（2）因为D为AB中点，故CD=+CACB，······································10分2()数学试题第2页（共14页）{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}212所以CD=+CACB······································································11分4()12211π=CA+CB+CACBcos············································13分44261113=3+16+34442231=，·················································································14分431所以CD的长为．······································································15分2【解法二】（1）因为2acosC=3bcosC+3ccosB，a2+b2−c2a2+c2−b2由余弦定理，得2acosC=+3b3c···························2分22abac=3a，····························································4分3所以cosC=，················································································6分2π由于0Cπ，则C=；····································································8分6π（2）由（1）知，ACB=，6在△ABC中，由余弦定理，得c2=a2+b2−2abcosACB···································································10分3=422+(3)−2432=7，···························································································11分故c=7，·······················································································12分因为D为AB中点，所以cosADC+cosBDC=0，AD2+CD2−AC2BD2+CD2−BC2故+=0，··········································13分22ADCDBDCD22772+CD2−34+CD2−222()所以+=0，7722CDCD22数学试题第3页（共14页）{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}31解得CD2=，················································································14分431故CD的长为．···········································································15分2【解法三】（1）略，同解法一或解法二；π（2）由（1）知，ACB=，6在△ABC中，由余弦定理，得c2=a2+b2−2abcosACB···································································10分23=42+3−243()2=7，···························································································11分故c=7，·······················································································12分b2+−c2a2所以cosA=2bc22(3)+−(7)42=2373=−，·············································································13分7在△ACD中，由余弦定理，得CD2=AC2+AD2−2ACADcosA22773=3+−23−()22731=，·······················································································14分4故的长为．···········································································15分17.（15分）如图，在四棱锥S−ABCD中，BC⊥平面SAB，AD∥BC，SA==BC1，SB=2，=SBA45o．（1）求证：SA⊥平面ABCD；数学试题第4页（共14页）{#{QQABBQwEggAAAIAAARgCAwmKCkIQkBGCCQgOgFAIMAAAQRFABAA=}#}1（2）若AD=，求平面SCD与平面SAB的夹角的余弦值．2【解法一】（1）在△SAB中，因为SA=1，=SBA45o，SB=2，由正弦定理，得SASB=，·········································································1分sinSBAsinSAB12所以=，······································································2分sin45sinSAB所以sin=SAB1，因为0SAB180，所以SAB=90，所以SA⊥AB．···················································································4分因为BC⊥平面SAB，SA平面SAB，所以BC⊥SA，···················································································5分又BCAB=B，所以SA⊥平面ABCD；·········································································6分（2）解：由（1）知平面，又AB,AD平面ABCD，所以，SA⊥AD，