内容发布更新时间 : 2024/11/16 3:30:00星期一 下面是文章的全部内容请认真阅读。

解:(1)证明:PD?平面ABCD,PD?PCD,?平面PCD?平面ABCD,平面PCD平面ABCD?CD,MD?平面ABCD,MD?CD,?MD?平面PCD,CF?平面PCD,?CF?MD,又CF?MF,MD,MF?平面MDF,MD?CF?平面MDF.11

(2)CF?平面MDF,?CF?DF,又易知?PCD?600,??CDF?300,从而CF=CD=,221DECFDE233313EF∥DC,??,即=,?DE?,?PE?,S?CDE?CD?DE?,DPCP442832MD?ME2?DE2?PE2?DE2?(33236)?()2?,442MF?M,11362?VM?CDE?S?CDE?MD????.338216219.设各项均为正数的数列?an?的前n项和为Sn,且Sn满足Sn?(n2?n?3)Sn?3(n2?n)?0,n?N?.(1)求a1的值;(2)求数列?an?的通项公式;(3)证明：对一切正整数n，有

1111????.a1?a1?1?a2?a2?1?an?an?1?3

解:(1)令n?1得:S12?(?1)S1?3?2?0,即S12?S1?6?0,?(S1?3)(S1?2)?0,S1?0,?S1?2,即a1?2.22(2)由Sn?(n2?n?3)Sn?3(n2?n)?0,得:(Sn?3)?S?(n?n)?n???0,an?0(n?N?),?Sn?0,从而Sn?3?0,?Sn?n2?n,2?当n?2时,an?Sn?Sn?1?n2?n??(n?1)?(n?1)????2n,又a1?2?2?1,?an?2n(n?N?).(3)当k?N?时,k2?kk313?k2???(k?)(k?),221644111111??????ak(ak?1)2k(2k?1)4k(k?1)4(k?1)(k?3)244???111?11?????11?1?1?4?4?k?(k?1)??(k?)??(k?1)???44?4?4??11??a1(a1?1)a2(a2?1)?1an(an?1)

???1?111111??(?)?(?)???1111?4?1?12?12?3?n?(n?1)???444444?111111?(?)???.41?1(n?1)?134n?3344x2y2520.已知椭圆C:2?2?1(a?b?0)的一个焦点为(5,0),离心率为.ab3(1)求椭圆C的标准方程;(2)若动点P(x0,y0)为椭圆C外一点,且点P到椭圆C的两条切线相互垂直,求点P的轨迹方程.解:(1)c?5,e?c55??,?a?3,b2?a2?c2?9?5?4,aa3x2y2?椭圆C的标准方程为:??1.94(2)若一切线垂直x轴,则另一切线垂直于y轴,则这样的点P共4个,它们的坐标分别为(?3,?2),(3,?2).若两切线不垂直于坐标轴,设切线方程为y?y0?k(x?x0),x2y2即y?k(x?x0)?y0,将之代入椭圆方程??1中并整理得:942(9k2?4)x2?18k(y0?kx0)x?9?(y?kx)?4?0?0??0,依题意,??0,2222?即:(18k)2(y0?kx0)2?36?(y?kx)?4(9k?4)?0,即4(y?kx)?4(9k?4)?0,0000??

y02?4?(x0?9)k?2x0y0k?y0?4?0,两切线相互垂直,?k1k2??1,即:2??1,x0?9222?x02?y02?13,显然(?3,?2),(3,?2)这四点也满足以上方程,?点P的轨迹方程为x2?y2?13.

121.已知函数f(x)?x3?x2?ax?1(a?R).3(1)求函数f(x)的单调区间;111(2)当a?0时,试讨论是否存在x0?(0,)(,1),使得f(x0)=f().222 解:(1)f'(x)?x2?2x?a,方程x2?2x?a?0的判别式:??4?4a,当a?1时,方程x2?2x?a?0的两根为?1?1?a,当x?(??,?1?1?a)时,f'(x)?0,?此时f(x)为增函数,?当a?1时,??0,?f'(x)?0,此时f(x)在(??,??)上为增函数.当x?(?1?1?a,?1?1?a)时,f'(x)?0,此时f(x)为减函数,当x?(?1?1?a,??)时,f'(x)?0,此时f(x)为增函数,综上,a?1时,f(x)在(??,??)上为增函数,当a?1时,f(x)的单调递增区间为(??,?1?1?a),(?1?1?a,??),f(x)的单调递减区间为(?1?1?a,?1?1?a). 1111?11?(2)f(x0)?f()?x03?x02?ax0?1??()3?()2?a()?1?2322?32?1?1??1?1??x03?()3???x02?()2??a(x0?)3?2??2?2x1?1?1111??(x0?)(x02?0?)??(x0?)(x0?)?a(x0?)3?224?2221x02x011?(x0?)(???x0??a)23612211?(x0?)(4x02?14x0?7?12a)122111?若存在x0?(0,)(,1),使得f(x0)?f(),22211必须4x02?14x0?7?12a?0在(0,)(,1)上有解.22a?0,???142?16(7?12a)?4(21?48a)?0,方程的两根为:依题意,0??14?221?48a?7?21?48a?7?21?48a?,x0?0,?x0只能是,844

?7+21?48a257?1,即7?21?48a?11,?49?21?48a?121,即??a??,41212?7+21?48a155又由=,得a??,故欲使满足题意的x0存在,则a??,424425557111?当a?(?,?)(?,?)时,存在唯一的x0?(0,)(,1)满足f(x0)?f().124412222257111?5?当a?(??,?][?,0)???时,不存在x0?(0,)(,1)使f(x0)?f().1212222?4?