20.(本小题满分8分)
解:(Ⅰ)以DA,DC,DP所在直线分别为x轴,y轴,z轴建立空间直角坐标系,如图示:则A(2,0,0),B(2,2,0),C(0,2,0). 设P(0,0,2m)(m?0),则E(1,1,m)
AE?(?1,1,m),DP?(0,0,2m)
?cos
2m21?1?m2?2m?3?m?1 3 ?点E坐标是(1,1,1)
(Ⅱ)?F?平面PAD,?可设F(x,0,z)?EF?(x?1,?1,z?1) 要使EF?平面PCB,则须EF?CB,且EF?PC 即(x?1,?1,z?1)?(2,0,0)?0,且(x?1,?1,z?1)?(0,2,?2)?0 解得x?1,z?0
?存在点F(1,0,0),即点F是AD的中点时,EF?平面PCB
21.(本小题满分8分)
解:(Ⅰ)设函数y?f(x)的图像上任一点Q(x0,y0)关于原点的对称点为P(x,y)
?x0?x?2?0?x0??x则?y?y 即? ?点Q(x0,y0)在函数y?f(x)的图像上.
y??y0?0??0?2??y?x2?2x,即y??x2?2x,故g(x)??x2?2x
(Ⅱ)由g(x)?f(x)?|x?1|可得2x2?|x?1|?0 当x?1时,2x2?x?1?0 此时不等式无解 当x<1时,2x2?x?1?0 ??1?x?
12因此,原不等式的解集为??1,?.
2??22.(本小题满分8分)
解:(Ⅰ)a?3b?(x,0)?3(1,y)?(x?3,3y)
a?3b?(x,0)?3(1,y)?(x?3,?3y) ?(a?3b)?(a?3b)
x2?(a?3b)(a?3b)?0 ?(x?3)(x?3)?3y?(?3y)?0 得?y2?1
3x2?P点的轨迹方程为:?y2?1
3?1??y?kx?m2(Ⅱ)由方程组?消去y,得(1?3k2)x2?6kmx?3m2?3?0(*) ?x2?y?1??3依题意知1?3k2?0,??(?6km)2?4(1?3k2)(?3m2?3)=12(m2?1?3k2)>0 设x1,x2为方程(*)的两根,AB中点为(x0,y0) 则x1?x2??x0?x1?x23kmm?y?kx?m? 0021?3k21?3k23kmm,)22 1?3k1?3km13km?(?)(x?)22 1?3kk1?3k6km 1?3k2故AB中点M的坐标为(?线段AB的垂直平分线方程为:y?将D(0,?1)坐标代入,化简得:4m?3k2?1
?m2?1?3k2?022m?4m?0,m?4或m?0, k故m,k满足?,消去得:2?4m?3k?1又?4m?3k2?1??1,?m??,故m?(?,0)?(4,??)
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中职数学单独招生考试模拟测试4
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