1.B.应用三角代换和配方法,并注意正弦函数的有界性.
2.C.配方确定焦点的坐标.
3.A.首先由Δ≥0确定a的取值范围,再由韦达定理建立关于a的二次函数,通过配方法确定其最小值.
4.D.应用待定系数法.
5.(1/2)+file:///C:/DOCUME%7E1/leiwen/LOCALS%7E1/Temp/msoclip1/01/clip_image001.gif.令sinx=a+b,cosx=a-b,则由sin2x+cos2x=1,得a2+b2=(1/2),其中a∈[-(file:///C:/DOCUME%7E1/leiwen/LOCALS%7E1/Temp/msoclip1/01/clip_image001.gif/2),(file:///C:/DOCUME%7E1/leiwen/LOCALS%7E1/Temp/msoclip1/01/clip_image001.gif/2)].
∴ y=(a+b)(a-b)+2a=a2-b2+2a=a2-((1/2)-a2)+2a=2(a+(1/2))2-1.
∴ 当a=(file:///C:/DOCUME%7E1/leiwen/LOCALS%7E1/Temp/msoclip1/01/clip_image001.gif/2)时,ymax=(1/2)+file:///C:/DOCUME%7E1/leiwen/LOCALS%7E1/Temp/msoclip1/01/clip_image001.gif.
6.3条.用待定系数法设出直线的方程.
7.(a1+a2+…+an)/n.依题意,a是使得函数f(a)=(a-a1)2+(a-a2)2+…+(a-an)2取得最小值时所对应的值.显然,f(a)是关于a的二次函数,重新组合配方即可求出a的值.
8.令t=sinx+cosx(-file:///C:/DOCUME%7E1/leiwen/LOCALS%7E1/Temp/msoclip1/01/clip_image001.gif≤t≤file:///C:/DOCUME%7E1/leiwen/LOCALS%7E1/Temp/msoclip1/01/clip_image001.gif),则sinxcosx=(t2-1)/2.于是f(x)=g(t)=2at-(t2-1)/2-2a2=-(1/2)(t-2a)2+(1/2).
∵ -file:///C:/DOCUME%7E1/leiwen/LOCALS%7E1/Temp/msoclip1/01/clip_image001.gif≤t≤file:///C:/DOCUME%7E1/leiwen/LOCALS%7E1/Temp/msoclip1/01/clip_image001.gif,
∴ 当0<a≤(file:///C:/DOCUME%7E1/leiwen/LOCALS%7E1/Temp/msoclip1/01/clip_image001.gif/2)时,fmax(x)=(1/2);
当a>(file:///C:/DOCUME%7E1/leiwen/LOCALS%7E1/Temp/msoclip1/01/clip_image001.gif/2)时,
fmax(x)=-(1/2)(file:///C:/DOCUME%7E1/leiwen/LOCALS%7E1/Temp/msoclip1/01/clip_image001.gif-2a)2+(1/2).
9.依题意,设所求双曲线方程为(y2/a2)-(x2/b2)=1(a>0,b>0).由e=(file:///C:/DOCUME%7E1/leiwen/LOCALS%7E1/Temp/msoclip1/01/clip_image002.gif/2),得a=2b.从而双曲线方程为y2-4x2=4b2.设Q(x,y)为双曲线上任一点,则|PQ|2=x2+(y-5)2=(1/4)y2-b2+(y-5)2=(5/4)(y-4)2+5-b2.
∵ y≤-a=-2b或y≥a=2b,
∴ 当0<b≤2时,|PQ|2min=5-b2=4,得b=1;
当b>2时,|PQ|2min=(5/4)(2b-4)2+5-b2=4,得b=(3/2)(舍去)或b=(7/2).
故所求双曲线方程为y2-4x2=4或y2-4x2=49.
10.设存在实数a、b同时满足两个已知条件,由A∩B≠Φ知,方程组
有整数解x=n、y=m,代入消去m,得
3n2-an-(b-15)=0, ①
且Δ=a2+12b-180≥0. ②
由(a,b)∈C,得a2+b2≤144file:///C:/DOCUME%7E1/leiwen/LOCALS%7E1/Temp/msoclip1/01/clip_image004.gifa2≤144-b2.
结合②得180-12b≤144-b2.解得b=6,
从而Δ=0.这表明方程①有两个相等的整数根.由韦达定理,得
n2=n1·n2=(15-b)/3=3,即n=±file:///C:/DOCUME%7E1/leiwen/LOCALS%7E1/Temp/msoclip1/01/clip_image005.gif.
这与n为整数矛盾.
故不存在实数a、b同时满足条件.
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