【數(shù)學(xué)試卷+答案】2025福州高二上期末質(zhì)檢

2024-2025學(xué)年第一學(xué)期福州市高二年級期末質(zhì)量檢測數(shù)學(xué)【解析】設(shè)P(x,y)，由PAPO(x5)2+y2=r2上存在點P使得PA=2PO，則圓C1與圓(x5)2+y2=r2有公共r∈[2,6].二、選擇題：本題共3小題，每小題6分，共18分22 象知，當(dāng)m=4時直線2x2y+m=0與曲線G有且只有一個公共點，所以當(dāng)直線 yO xO三、填空題：本題共3小題，每小題5分，共15分.依題意得O,M分別為F/F,AF的中點，所以O(shè)MⅡBF，由AF.BF=0得，AF丄BF，AF/=2a，解：解法一1）設(shè)等差數(shù)列{an}的公差為d，··········································1分解方程組，得{ld=-2,所以an=5+(n-1)×(-2)=7-2n；····················································（2）Sn=na1+d················=-n2+6n····························································.························*，所以當(dāng)n=3時，································································12分解法二1）同解法一；········································································6分當(dāng)n≥4時，an<0；··················即當(dāng)n=3時，Sn最大，········································································12分解：解法一1）設(shè)圓C的方程為(x-a)2+(y-b)2=r2，·····························1分所以圓的方程為(x-2)2+(y-1)2=5；······················································10分（2）圓心C(2,1)到直線x+y-2=0的距離d=22 22········································所以.·······················解法二1）設(shè)線段OP的中點為M，由O(0,0)，P(4,0)，可得點M的坐標(biāo)為(2,0)，線段OP的垂直平分線的方程是x=2，······························ 所以圓的方程為(x-2)2+(y-1)2=5；······················································10分2=5,得，x2-3x=0，······························設(shè)A(x1,y1)，B(x2,y2)，x1<x2，所以x1=0，x2=3，························································所以A(0,2)，B(3,-1)，所以.·······················17.本小題考查直線與平面的位置關(guān)系、平面與平面的夾角、空間向量在立體幾何中的應(yīng)用等基礎(chǔ)知識；考查直觀想象能力、運算求解能力、邏輯思維能力；考查數(shù)形結(jié)合思想、化歸與轉(zhuǎn)化思想；考查直觀想象、數(shù)學(xué)運算、邏輯推理素養(yǎng)，體現(xiàn)基礎(chǔ)性和解：解法一1）設(shè)棱SC的中點為N，連接MN，DN，如圖所示，····························································2分因為M為棱SB的中點，所以MN//BC，MN=BC.······3分又因為AD//BC，AD=，所以MN//AD，MN=AD，·······································5分SMNMADBC所以四邊形MNDA是平行四邊形，所以AM//DN.······································6分又AM丈平面SCD，DN平面SCD，所以AM//平面SCD.···········································································7分（2）以A為坐標(biāo)原點，AB,AD,AS所在直線分別為x,y,z軸，建立如圖所示的空間直角坐標(biāo)系A(chǔ)xyz.·······················································8分則B(2,0,0)，C(2,4,0)，D(0,2,0)，S(0,0,4)，·········································9分所以AB=(2,0,0)，SB=(2,0,-4),DC=(2,2,0)，SD=(0,2,-4)，··················10分取平面SAB的一個法向量為m=(0,1,0)，···············設(shè)平面SCD的法向量為n=(x,y,z)，令x=2，得n=(2,-2,-1).·························zSASADMMyyB/Bx所以平面SAB與平面SCD夾角的余弦值為2.··········································15分M解法二1）取BC的中點N，連接AN，MN，··············································2分MAD因為M是SB的中點，所以MN∥SC，AD又因為MN丈平面SCD，SC平面SCD，CBNC所以MN∥平面SCD，·················································3分所以AN∥CD，又因為AN丈平面SCD，CD平面SCD，所以AN∥平面SCD，············································································5分又因為ANMN=N，所以平面AMN∥平面SCD，··································································6分又AM平面AMN，所以AM//平面SDC.··············································7分（2）同解法一.·····································解法三1）以A為坐標(biāo)原點，AB,AD,AS所在直線分別為x,y,z軸，建立如圖所示的空間直角坐標(biāo)系A(chǔ)xyz.·······················································1分則B(2,0,0)，C(2,4,0)，D(0,2,0)，S(0,0,4)，M(1,0,2)，··························3分z所以DC=(2,2,0)，SD=(0,2,-4)，AM=(1,0,2)，··············4分zS設(shè)平面SCD的法向量為n=(x,y,z)，所以AM.n=2—2=0，·························又AM丈平面SCD，所以AM//平面SCD.············································································10分（2）依題意得，AB=(2,0,0)，SB=取平面SAB的一個法向量為m=(0,1,0)，·················································1所以平面SAB與平面SCD夾角的余弦值為.············································15分18.本小題主要考查數(shù)列的前n項和與通項的關(guān)系、數(shù)列的求和以及數(shù)列的性質(zhì)等基礎(chǔ)知識；考查運算求解能力、邏輯思維能力；考查函數(shù)查數(shù)學(xué)運算、邏輯推理、數(shù)學(xué)抽象素養(yǎng)，體，································兩式相減得：an=2an2an1， n n;··························即對任意的n∈N*····················設(shè)，·················································所以{cn}是單調(diào)遞減數(shù)列，則cnc1=1，所以k>1，考查直觀想象、數(shù)學(xué)運算、邏輯推理、數(shù)學(xué)抽象素養(yǎng)，體分.解方程組，得,······················································（2i）設(shè)G(x,y)，解方程組，得{·2或{·2ly=2,ly=-2.()()所以有兩個點滿足“共軛點對”[A,G]，且點G的坐標(biāo)為|(2,2,,(|-2,-()()·································································································10分（ii）由（i）得，直線G1G2的方程為y=x.假設(shè)存在定點Q(2m,m)，依題意可知直線l斜率存在，設(shè)直線l:y-1=k(x-2)，即y=kx+1-2k，由{22消去y得，(4k2+1)x2+8k設(shè)C(x1,y1),D(x2,y2)，······························································································y1-my2-mkx1+1-2k-mkx2+1-2k-m所以kQC.kQD=y1-my2-mkx1+1-2k-mkx2+1-2k-m2·····················13分k2x1x2+(1-m-2k)k(x1+x2)+(1-m-22·····················13分.+(1-m-2k