【高考冲刺】2019届高考数学(理)倒计时模拟卷(5)(Word版,含答案)

20.1.依题意,直线l显然不平行于坐标轴,故y?k将x??y?1. ?x?1?可化为x??1k1, y?1代入x2?4y2?m2,消去x k2222得1?4ky?2ky?k1?m?0,①

????由直线l与椭圆相交于两个不同的点,

??4k2?4k2?1?m2??1?4k2??0,

4k2整理得m?.

1?4k222.设A?x1,y1?,B?x2,y2?

2k,

1?4k2uuur由①,得y1?y2?uuur因为AC?3CB,得y1??3y2,代入上式,得y2??k.

1?4k2于是,△OAB的面积S?2k2k11OC?y1?y2?2y2???, 21?4k24k21其中,上式取等号的条件是4k2?1,即k??.

2?k1由y2?,可得. y??21?4k241111将k?,y2??及k??,y2?

24245这两组值分别代入①,均可解出m2?.

228所以,△OAB的面积取得最大值时椭圆的方程是x2?y2?1.

5521.1.当a?11时, f(x)?x2?x?xex,f?(x)?x?1?e?xex?(x?1)(1?ex)

22x?(??,?1)时, f'(x)?0;x?(?1,0)时, f'(x)?0;x?(0,??)时, f'(x)?0 所以f(x)的递增区间是(?1,0),递减区间是(??,?1),(0,??)

2xxx2. g(x)?ax?ax?xe,g?(x)?2ax?a?e?xe

设h(x)?2ax?a?e?xe,则h?(x)?2a?2e?xe?2a?(x?2)e.

x因为x?0,所以x?2?2,e?1.又因为a?1,所以h?(x)?0,

xxxxx故h(x)?a(2x?1)?e(1?x)在(??,0)上为增函数.

x11?11又因h(0)?a?1?0,h(?)??e2?0,由零点存在性定理,存在唯一的x0?(?,0),

222有h(x0)?0.

当x????,x0?时, h(x)?g?(x)?0,即g(x)在(??,x0)上为减函数, 当x??x0,0?时, h(x)?g?(x)?0,即g(x)在?x0,0?上为增函数, 所以x0为函数g(x)的极小值点.

22.1.曲线C1的直角坐标方程为?x?2??y2?4,所以C1极坐标方程为??4cos?. 曲线C2的直角坐标方程为x2??y?2??4,所以C2极坐标方程为??4sin?; 2.设点P极点坐标??1,4cos??,即?1?4cos?. 点Q极坐标为??2,4sin???22???????6????,即?2?4sin????????, 6?则

???OP?OQ??1?2?4cos??4sin????6???3?1????16cos???sin??cos??8sin2??????4, ?2?26????∵???0,?????, 2???. ?∴2??当2?????5????,6?66?6??2,即???3时, OP?OQ取最大值4.

???3x?3,x??2?1?23.1.∵g(x)???5x?1,?2?x?,

4?1?3x?3,x???4当x??2?时, ?3x?3?6解得x??1,此时无解.

1771时, ?5x?1?6,解得x??,即??x?. 455411当?x时, 3x?3?6,解得x?3,即?x?3, 447综上, g(x)?6的解集为{x|??x?3}.

5当?2?x?2.因为存在x1,x2?R,使得f(x1)??g(x2)成立. 所以{y|y?f(x),x?R}I{y|y??g(x),x?R}??. 又f(x)?3|x?a|?|3x?1|?|(3x?3a)?(3x?1)|?|3a?1|,

99,??),则?g(x)?(??,]. 449135所以3a?1?,解得??a?.

41212135故a的取值范围为[?,].

1212由(1)可知g(x)?[?