2016屆高考數(shù)學(xué)一輪復(fù)習(xí)題組層級快練30含解析

1、PAGE PAGE 6題組層級快練(三十)1對于非零向量a，b，“ab0”是“ab”的()A充分不必要條件B必要不充分條件C充分必要條件 D既不充分也不必要條件答案A解析若ab0，則ab，所以ab；若ab，則ab，ab0不一定成立，故前者是后者的充分不必要條件2設(shè)a是任一向量，e是單位向量，且ae，則下列表示形式中正確的是()Aeeq f(a,|a|) Ba|a|eCa|a|e Da|a|e答案D解析對于A，當(dāng)a0時，eq f(a,|a|)沒有意義，錯誤；對于B，C，D當(dāng)a0時，選項B，C，D都對；當(dāng)a0時，由ae可知，a與e同向或反向，選D.3(2015北京東城期中)已知ABCD為平行四邊形

2、，若向量eq o(AB,sup10()a，eq o(AC,sup10()b，則向量eq o(BD,sup10()為()Aab BabCb2a Dab答案C4.如圖所示，在正六邊形ABCDEF中，eq o(BA,sup10()eq o(CD,sup10()eq o(EF,sup10()()A0 B.eq o(BE,sup10()C.eq o(AD,sup10() D.eq o(CF,sup10()答案D解析由于eq o(BA,sup10()eq o(DE,sup10()，故eq o(BA,sup10()eq o(CD,sup10()eq o(EF,sup10()eq o(CD,sup10()eq

3、 o(DE,sup10()eq o(EF,sup10()eq o(CF,sup10().5(2015廣東惠州二中模擬)已知點O，A，B不在同一條直線上，點P為該平面上一點，且eq o(OP,sup10()eq f(3o(OA,sup10()o(OB,sup10(),2)，則()A點P在線段AB上B點P在線段AB的反向延長線上C點P在線段AB的延長線上D點P不在直線AB上答案B解析eq o(OP,sup10()eq f(3o(OA,sup10()o(OB,sup10(),2)eq f(3,2)eq o(OA,sup10()eq f(1,2)eq o(OB,sup10()eq o(OA,sup10

4、()eq f(1,2)(eq o(OA,sup10()eq o(OB,sup10()eq o(OA,sup10()eq f(1,2)eq o(BA,sup10()，即eq o(OP,sup10()eq o(OA,sup10()eq o(AP,sup10()eq f(1,2)eq o(BA,sup10()，所以點P在線段AB的反向延長線上，故選B.6在ABC中，點D在邊AB上，CD平分ACB.若eq o(CB,sup10()a，eq o(CA,sup10()b，|a|1，|b|2，則eq o(CD,sup10()()A.eq f(1,3)aeq f(2,3)b B.eq f(2,3)aeq f(

5、1,3)bC.eq f(3,5)aeq f(4,5)b D.eq f(4,5)aeq f(3,5)b答案B解析由內(nèi)角平分線定理，得eq f(|CA|,|CB|)eq f(|AD|,|DB|)2.eq o(CD,sup10()eq o(CA,sup10()eq o(AD,sup10()eq o(CA,sup10()eq f(2,3)eq o(AB,sup10()eq o(CA,sup10()eq f(2,3)(eq o(CB,sup10()eq o(CA,sup10()eq f(2,3)eq o(CB,sup10()eq f(1,3)eq o(CA,sup10()eq f(2,3)aeq f(1

6、,3)b.故B正確7已知向量i與j不共線，且eq o(AB,sup10()imj，eq o(AD,sup10()nij，若A，B，D三點共線，則實數(shù)m，n應(yīng)該滿足的條件是()Amn1 Bmn1Cmn1 Dmn1答案C解析由A，B，D共線可設(shè)eq o(AB,sup10()eq o(AD,sup10()，于是有imj(nij)nij.又i，j不共線，因此eq blcrc (avs4alco1(n1，,m，)即有mn1.8O是平面上一定點，A，B，C是該平面上不共線的三個點，一動點P滿足：eq o(OP,sup10()eq o(OA,sup10()(eq o(AB,sup10()eq o(AC,su

7、p10()，(0，)，則直線AP一定通過ABC的()A外心 B內(nèi)心C重心 D垂心答案C解析取BC中點M.eq o(OP,sup10()eq o(OA,sup10()(eq o(AB,sup10()eq o(AC,sup10()，eq o(OP,sup10()eq o(OA,sup10()(eq o(AB,sup10()eq o(AC,sup10()，eq o(AP,sup10()2eq o(AD,sup10().A，P，D三點共線，AP一定通過ABC的重心，C正確9在四邊形ABCD中，eq o(AB,sup10()a2b，eq o(BC,sup10()4ab，eq o(CD,sup10()5a

8、3b，則四邊形ABCD的形狀是()A矩形 B平行四邊形C梯形 D以上都不對答案C解析由已知eq o(AD,sup10()eq o(AB,sup10()eq o(BC,sup10()eq o(CD,sup10()8a2b2(4ab)2eq o(BC,sup10().eq o(AD,sup10()eq o(BC,sup10().又eq o(AB,sup10()與eq o(CD,sup10()不平行，四邊形ABCD是梯形10已知四邊形ABCD是菱形，點P在對角線AC上(不包括端點A，C)的充要條件是eq o(AP,sup10()(eq o(AB,sup10()eq o(AD,sup10()，則的取值

9、范圍是()A(0,1) B(1,0)C(0，eq f(r(2),2) D(eq f(r(2),2)，0)答案A解析如圖所示，點P在對角線AC上(不包括端點A，C)，eq o(AP,sup10()eq o(AC,sup10()(eq o(AB,sup10()eq o(AD,sup10()由eq o(AP,sup10()與eq o(AC,sup10()同向知，0.又|eq o(AP,sup10()|eq o(AC,sup10()|，eq f(|o(AP,sup10()|,|o(AC,sup10()|)1，(0,1)反之亦然11設(shè)A1，A2，A3，A4是平面直角坐標(biāo)系中兩兩不同的四點，若eq o(A

10、1A3,sup10()eq o(A1A2,sup10()(R)，eq o(A1A4,sup10()eq o(A1A2,sup10()(R)，且eq f(1,)eq f(1,)2，則稱A3，A4調(diào)和分割A(yù)1，A2.已知平面上的點C，D調(diào)和分割點A，B，則下列說法正確的是()AC可能是線段AB的中點BD可能是線段AB的中點CC，D可能同時在線段AB上DC，D不可能同時在線段AB的延長線上答案D解析若A成立，則eq f(1,2)，而eq f(1,)0，不可能；同理B也不可能；若C成立，則01，且02，與已知矛盾；若C，D同時在線段AB的延長線上時，1，且1，eq f(1,)eq f(1,)2，與已知

11、矛盾，故C，D不可能同時在線段AB的延長線上，故D正確12.如圖所示，下列結(jié)論不正確的是_eq o(PQ,sup10()eq f(3,2)aeq f(3,2)b；eq o(PT,sup10()eq f(3,2)aeq f(3,2)b；eq o(PS,sup10()eq f(3,2)aeq f(1,2)b；eq o(PR,sup10()eq f(3,2)ab.答案解析由abeq f(2,3)eq o(PQ,sup10()，知eq o(PQ,sup10()eq f(3,2)aeq f(3,2)b，正確；由eq o(PT,sup10()eq f(3,2)aeq f(3,2)b，從而錯誤；eq o(P

12、S,sup10()eq o(PT,sup10()b，故eq o(PS,sup10()eq f(3,2)aeq f(1,2)b，正確；eq o(PR,sup10()eq o(PT,sup10()2beq f(3,2)aeq f(1,2)b，錯誤故正確的為.13.如圖所示，已知B30，AOB90，點C在AB上，OCAB，用eq o(OA,sup10()和eq o(OB,sup10()來表示向量eq o(OC,sup10()，則eq o(OC,sup10()等于_答案eq f(3,4)eq o(OA,sup10()eq f(1,4)eq o(OB,sup10()解析eq o(OC,sup10()eq

13、 o(OA,sup10()eq o(AC,sup10()eq o(OA,sup10()eq f(1,4)eq o(AB,sup10()eq o(OA,sup10()eq f(1,4)(eq o(OB,sup10()eq o(OA,sup10()eq f(3,4)eq o(OA,sup10()eq f(1,4)eq o(OB,sup10().14設(shè)a和b是兩個不共線的向量，若eq o(AB,sup10()2akb，eq o(CB,sup10()ab，eq o(CD,sup10()2ab，且A，B，D三點共線，則實數(shù)k的值等于_答案4解析A，B，D三點共線，eq o(AB,sup10()eq o(

14、BD,sup10().eq o(AB,sup10()2akb，eq o(BD,sup10()eq o(BC,sup10()eq o(CD,sup10()a2b，k4.故填4.15已知O為ABC內(nèi)一點，且eq o(OA,sup10()eq o(OC,sup10()2eq o(OB,sup10()0，則AOC與ABC的面積之比是_答案12解析如圖所示，取AC中點D.eq o(OA,sup10()eq o(OC,sup10()2eq o(OD,sup10().eq o(OD,sup10()eq o(BO,sup10().O為BD中點，面積比為高之比16已知向量a2e13e2，b2e13e2，其中e1

15、，e2不共線，向量c2e19e2.問是否存在這樣的實數(shù)，使向量dab與c共線？答案當(dāng)2時共線解析d(2e13e2)(2e13e2)(22)e1(33)e2.要使d與c共線，則應(yīng)有實數(shù)k，使dkc.即(22)e1(33)e22ke19ke2.即eq blcrc (avs4alco1(222k，,339k，)得2.故存在這樣的實數(shù)，只要2，就能使d與c共線17如圖所示，已知點G是ABO的重心(1)求eq o(GA,sup10()eq o(GB,sup10()eq o(GO,sup10()；(2)若PQ過ABO的重心G，且eq o(OA,sup10()a，eq o(OB,sup10()b，eq o(

16、OP,sup10()ma，eq o(OQ,sup10()nb，求證：eq f(1,m)eq f(1,n)3.答案(1)eq o(GA,sup10()eq o(GB,sup10()eq o(GO,sup10()0(2)略解析(1)如圖所示，延長OG交AB于M點，則M是AB的中點eq o(GA,sup10()eq o(GB,sup10()2eq o(GM,sup10().G是ABO的重心，eq o(GO,sup10()2eq o(GM,sup10().eq o(GA,sup10()eq o(GB,sup10()eq o(GO,sup10()0.(2)M是AB邊的中點，eq o(OM,sup10()eq f(1,2)(eq o(OA,sup10()eq o(OB,sup10()eq f(1,2)(ab)又G是ABO的重心，eq o(OG,sup10()eq f(2,3)eq o(OM,sup10()eq f(1,3)(ab)eq o(PG,sup10()eq o(OG,sup10()eq o(OP,sup10()eq f(1,3)(