{#{QQABIQIAogCIQAAAAQgCUQVwCAMQkgGCAQgOBAAEoAAAyRNABAA=}#}{#{QQABIQIAogCIQAAAAQgCUQVwCAMQkgGCAQgOBAAEoAAAyRNABAA=}#}{#{QQABIQIAogCIQAAAAQgCUQVwCAMQkgGCAQgOBAAEoAAAyRNABAA=}#}{#{QQABIQIAogCIQAAAAQgCUQVwCAMQkgGCAQgOBAAEoAAAyRNABAA=}#}【高三数学试题参考答案第1页(共6页)】2025届高三10月联考数学参考答案、提示及评分细则1.【答案】A【解析】因为A={-1,1},B⊆A,所以m2=1m=-1{,即得m=-1.2.【答案】D【解析】当a=-2,b=1时,a2>b2,但是ea<eb,故a2>b2⇒/ea>eb当a=1,b=-2时,ea>eb,但是a2<b2,故ea>eb⇒/a2>b2故“a2>b2”是“ea>eb”的既不充分也不必要条件.3.【答案】C【解析】由已知可得-2=a+bi+a-bi,解得a=-1.4.【答案】C【解析】依题意有tanθ>0,且tanθ=cos(θ+π4)cos(θ-π4)=22cosθ-22sinθ22cosθ+22sinθ=1-tanθ1+tanθ,故tan2θ+2tanθ-1=0,结合tanθ>0,解得tanθ=2-1.5.【答案】B【解析】因为f(x)在R上单调递增,所以当x<0,y=-x2+ax单调递增,所以a≥0,当x≥0时,f′(x)=ex-a≥0,由f(x)单调递增可知a≤1,且当x=0,0≤f(0)=1,所以a的取值范围是[0,1].故选:B.6.【答案】C【解析】设P(x,y),则PA2+PB2+PC2=(x+1)2+(y-1)2+(x-1)2+(y+1)2+(x-3)2+(y-3)2=3(x2+y2-2x-2y)+22=70,故P(x,y)的轨迹方程为(x-1)2+(y-1)2=18.PA→PB→=PO→2-OA→2=PO→2-2,而PO≤32+2=42,故PA→PB→≤30,选C.7.【答案】B【解析】对于A,令x=±1,则f(e)=±1,有2个函数值对应,故A错误;对于C,取x=0,可知f(x2+2x)=f(0)=0,再取x=-2,可知f(x2+2x)=f(0)=2,故C错误;
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