小学、初中、高中各种试卷真题知识归纳文案合同PPT等免费下载www.doc985.com无锡市第六高级中学2025年10月高三教学质量调研数学试卷答案和解析一、单选题1.B2.D3.4.D5.6.C7.8.A二、多选题9.ACD10.AC11.三、填空题12．❑√213．❑√314．[e241,+∞\)−四、解答题15.【答案】解：\(1\)由题意，设b⃗=(x,y)，因为|⃗b|=2，所以√x2+y2=2，即x2+y2=4，①......................................2分又因为向量⃗a,⃗b的夹角为60°，所以cos⟨a⃗,b⃗⟩=a⃗·b⃗|a⃗||b⃗|=y2=12，解得y=1，......................................4分将y=1代入①，解得x=±√3，所以b⃗=(√3,1)或b⃗=(−√3,1)．......................................6分\(2\)因为\(a→+b→\)⊥\(a→−b→\),λ∈R，所以\(a⃗+b⃗\)·\(a⃗−b⃗\)=0，即⃗a2=⃗b2，所以|⃗a|=|⃗b|=1，......................................8分小学、初中、高中各种试卷真题知识归纳文案合同PPT等免费下载www.doc985.com所以|⃗a+λ⃗b|=√(⃗a+λ⃗b)2=√⃗a2+2λ⃗a·⃗b+λ2⃗b2=√λ2+λ+1=√(λ+12)2+34，......................................10分所以当λ=−12时，|a⃗+λb⃗|有最小值√32．......................................13分16.【答案】（1）在△ABC中，由asinB=❑√3bsinA2及正弦定理，得sinAsinB=❑√3sinBsinA2，···········2分则2sinA2cosA2sinB=❑√3sinBsinA2，而A,B∈(0,π)，sinBsinA2>0，因此cosA2=❑√32，解得A2=π6，所以A=π3.·········································································5分（2）由（1）知A=π3，由cosB=2❑√77，得sinB=❑√1−cos2B=❑√217，·································7分sinC=sin(A+B)=sinAcosB+cosAsinB=❑√32×2❑√77+12×❑√217=3❑√2114，····························11分由正弦定理得c=bsinCsinB=4×3❑√2114❑√217=6，而⃗AD=12(⃗AB+⃗AC)，所以¿⃗AD∨¿12❑√⃗AB2+⃗AC2+2⃗AB⋅⃗AC=12❑√62+42+2×6×4×12=❑√19.····························15分17.【答案】解：(1)由于g(x)为奇函数，且定义域为R，∴g(0)=0，即40−n20=0，n=1．······························2分当n=1时，g(x)=4x−12x=2x−2−x,g(−x)=2−x−2x=−g(x)，∴n=1时g(x)为奇函数．......................................4分小学、初中、高中各种试卷真题知识归纳文案合同PPT等免费下载www.doc985.com f(x)=log4(4x+1)+mx，∴f(−x)=log4(4−x+1)−mx=log4(4x+1)−(m+1)x， f(x)是偶函数，∴f(−x)=f(x)，m=−m−1，得到m=−12，......................................6分由此可得：m+n的值为12．·································7分(2) ℎ(x)=f(x)+12x=log4(4x+1)，∴ℎ[log4(2a+1)]=log4(2a+2)，·································9分又 g(x)=2x−2−x在区间¿上是增函数，∴当x≥1时，g(x)min=g(1)=32，·································11分由题意得{2a+2<4❑322a+1>02a+2>0,∴−12<a<3综上，a的取值范围{a∨−12<a<3}.·································15分18.【答案】解：(1)当a＝0时，f(x)＝x3＋3x2＋1，所以f′(x)＝3x2＋6x＝3x(x＋2)，令f′(x)＝0，得x＝0或x＝－2.………………………(1分)列表如下：x(－∞，－2)－2(－2，0)0(0，＋∞)f′(x)＋0－0＋f(x)极大值极小值所以f(x)在x＝－2处取极大值，即x1＝－2，且f(x1)＝5.…………………………(3分)由f(x1)＝f(x3)＝5，所以x＋3x＋1＝5，即x＋3x－4＝0，所以(x3－1)(x3＋2)2＝0.小学、初中、高中各种试卷真题知识归纳文案合同PPT等免费下载www.doc985.com因为x1≠x3，所以x3＝1，…………………………(5分)所以2x1＋x3＝...