數(shù)學(xué)-江蘇省揚州市2024-2025學(xué)年高三上學(xué)期11月期中檢測試題和答案

2024—2025學(xué)年第一學(xué)期期中檢測高三數(shù)學(xué)參考答案123456789BDABCBCDABD15.【答案】(1)零假設(shè)為H0:該地居民喜歡喝茶與年齡沒有關(guān)系.即沒有90%的把握認為該地居民喜歡喝茶與年齡有關(guān).···········所以X的分布列為：X012P494919··············································································································12分所以X的期望為.················································13分16.【答案】(1)f(x)=sin2xcosφ+cos2xsinφ=sin(2x+φ)，又0<φ<，所以································所以f(x)的單調(diào)遞增區(qū)間為.·······························(2)因為f(x)=sin(2x+τ)圖象向右平移τ個單位得到y(tǒng)=sin[2(x?τ)+τ]=sin2x，再將y=sin2x圖象上各個點橫坐標變?yōu)樵瓉?倍得到y(tǒng)=sinx，所以g(x)=sinx；·10分所以不等式為sin(2x+)<sinx，不等式化為cos2x<sinx，結(jié)合函數(shù)y=sinx在(0,τ)上的圖象得τ<x<5τ,所以原不等式的解集為(,).···································································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，過點N作NT丄AM于T，連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為二面角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的正切值為.············································B1·B1NA1CDTHA1CDTHABzMD1MD1B1NQQAA1DDCyCyABx因為正方體棱長為2，M，N，Q分別為棱由（1）知A1N⊥平面AMQ.設(shè)m=(x222)是平面ANM的法向量，則,2z2所以cosA1N,mi==，所以二面角N－AM－Q的余弦值為，·····14分所以二面角N－AM－Q的正切值為.···········································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假設(shè)存在常數(shù)θ,k，對于所有滿足題意的α,β,都有sin2α?sin2β+k=6ksinαcosβ成立，則存在常數(shù)θ,k，對于所有滿足題意的α,β,利用參考公式，有2cos(α+β)sin(α?β)+k=6k.[(sin(α+β)+sin(α?β)].································14分由題意，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對于所有滿足題意的α,β成立， 故cosθ=cos[τ?(α+β)]=sin(α+β)=1，.············23思路二：也可以賦值：因為對于所有滿足題意的α,β,都有sin2α?sin2β+k=6ksinαcosβ,取α=β,則k=6ksinαcosα,則sin2α=sin(α+β)=，所以cosθ=cos[τ?(α+β)]=sin(α+β)=1，23 取?θ,β=0，則sinα=cosθ=，cosα=，則sin2α+k=6ksinα,即2××+k=6k×再證明等式恒成立.增，··························································所以當x=1時，f(x)取極小值0，無極大值.······················································4分又ex>x0+1>0，則f(ex)>f(x0+1)，+∞)，使得f(ex)<f(x0+1)成立，故不符合；············又ex>x0+1>1，則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)單調(diào)遞增，u(x)的圖象連續(xù)不間斷， 所以當x∈(0,x1)時，u(x)<0，即φ+∞)，使得f(ex)<f(x0+1)成立.4所以要比較f(4a+a2)與f(2a+a4)的大小，即比較4a+a2與2a+a4的大小，·········11分即比較4a?2a與a4?a2的大小.令F(x)=x2?x，則比較F(2a)與F(a2)的大小.易知F(x)在(1,+∞)上單調(diào)遞增，即比較2a與a2的大小，即比較aln2與2lna的大小，即比較與的大小.······································12分時，t(x)單調(diào)遞減，又t(2)=t(4