数学-江苏省扬州市2024-2025学年高三上学期11月期中检测试题和答案

2024—2025学年第一学期期中检测高三数学参考答案123456789BDABCBCDABD15.【答案】(1)零假设为H0:该地居民喜欢喝茶与年龄没有关系.即没有90%的把握认为该地居民喜欢喝茶与年龄有关.···········所以X的分布列为：X012P494919··············································································································12分所以X的期望为.················································13分16.【答案】(1)f(x)=sin2xcosφ+cos2xsinφ=sin(2x+φ)，又0<φ<，所以································所以f(x)的单调递增区间为.·······························(2)因为f(x)=sin(2x+τ)图象向右平移τ个单位得到y=sin[2(x−τ)+τ]=sin2x，再将y=sin2x图象上各个点横坐标变为原来2倍得到y=sinx，所以g(x)=sinx；·10分所以不等式为sin(2x+)<sinx，不等式化为cos2x<sinx，结合函数y=sinx在(0,τ)上的图象得τ<x<5τ,所以原不等式的解集为(,).···································································15分A1D1⊥平面ABB1A1，A1N平面ABB1A1，所以A1D1⊥A1N，所以QM⊥A1N，··································································2分BBQH=900,所以上A1HQ=900,即A1N⊥AQ；············································又AQ、AM平面AMQ，AQAM=A，所以A1N⊥平面AMQ.····················································································7分(2)方法一：在ΔANM中，过点N作NT丄AM于T，连HT.由(1)知NH⊥平面AMQ，故NH⊥AM，又NH、NT平面NHT，所以AM⊥平面NHT，所以AM⊥HT，所以上NTH为二面角N－AM－Q的平面角.······················································10分在Rt△A1B1N中，A1N=BH，Q所以A1H=所以NH=.···································································12分 55在Rt△AHT中，sin上QAM=所以，·············所以二面角N－AM－Q的正切值为.············································B1·B1NA1CDTHA1CDTHABzMD1MD1B1NQQAA1DDCyCyABx因为正方体棱长为2，M，N，Q分别为棱由（1）知A1N⊥平面AMQ.设m=(x222)是平面ANM的法向量，则,2z2所以cosA1N,mi==，所以二面角N－AM－Q的余弦值为，·····14分所以二面角N－AM－Q的正切值为.···········································所以cosB=0或者sinC=sinB.·········方法一：在△CBQ中，由正弦定理，得在△BPQ中，由正弦定理，得APQBC 2sinαcosα+ 2sinαcosα+2方法二：在△ABP中，由正弦定理，得所以 .方法三：在△CBQ中，由正弦定理，得所以 4(sinα+cosα)(cosα+sinα)−23sinα(cosα+sinα)−2(cosα−sinα)(sinα+cosα) =②假设存在常数θ,k，对于所有满足题意的α,β,都有sin2α−sin2β+k=6ksinαcosβ成立，则存在常数θ,k，对于所有满足题意的α,β,利用参考公式，有2cos(α+β)sin(α−β)+k=6k.[(sin(α+β)+sin(α−β)].································14分由题意，α+β=是定值，所以sin(α+β)，co(α+β)是定值，[2cos(α+β)−3k]sin(α−β)+k[1−3sin(α+β)]=0对于所有满足题意的α,β成立， 故cosθ=cos[τ−(α+β)]=sin(α+β)=1，.············23思路二：也可以赋值：因为对于所有满足题意的α,β,都有sin2α−sin2β+k=6ksinαcosβ,取α=β,则k=6ksinαcosα,则sin2α=sin(α+β)=，所以cosθ=cos[τ−(α+β)]=sin(α+β)=1，23 取−θ,β=0，则sinα=cosθ=，cosα=，则sin2α+k=6ksinα,即2××+k=6k×再证明等式恒成立.增，··························································所以当x=1时，f(x)取极小值0，无极大值.······················································4分又ex>x0+1>0，则f(ex)>f(x0+1)，+∞)，使得f(ex)<f(x0+1)成立，故不符合；············又ex>x0+1>1，则f(ex)>f(x0+1)，+∞)，使得f(ex)<f(x0+1)成立，故不符合；············+∞)，使得f(ex)<f(x0+1)成立，故符合.xxxx又g(0)=0，所以g(x)>0，故不存在x0∈(0,+∞)，使得f(ex)<f(x0+1)成立.··········5分x又g(0)=0，所以g(x)>0，故不存在x0∈(0,+∞)，使得f(ex)<f(x0+1)成立；·······7分又u(x)单调递增，u(x)的图象连续不间断， 所以当x∈(0,x1)时，u(x)<0，即φ+∞)，使得f(ex)<f(x0+1)成立.4所以要比较f(4a+a2)与f(2a+a4)的大小，即比较4a+a2与2a+a4的大小，·········11分即比较4a−2a与a4−a2的大小.令F(x)=x2−x，则比较F(2a)与F(a2)的大小.易知F(x)在(1,+∞)上单调递增，即比较2a与a2的大小，即比较aln2与2lna的大小，即比较与的大小.······································12分时，t(x)单调递减，又t(2)=t(4