1
2018~2019 学年第一学期期末考试试卷
高二数学参考答案与评分标准(文)
说明:
一、本解答给出了一种或几种解法供参考,如果考生的解法与本解答不同,可根据试题的
主要考查内容比照评分标准制订相应的评分细则.
二、对解答题,当考生的解答在某一步出现错误时,如果后继部分的解答未改变该题的内
容和难度,可视影响的程度决定后继部分的给分,但不得超过该部分正确解答应得分数的一半;
如果后继部分的解答有较严重的错误,就不再给分.
三、解答右端所注分数,表示考生正确做到这一步应得的累加分数.
四、只给整数分数,选择题和填空题不给中间分.
一.选择题:
(1)(B);( 2)(A);( 3)( D);(4)( B);(5)( A);(6)( C);(7)( B);(8)(C);(9)(C);
(10)( B);(11)( A);(12)( D).
二.填空题:
(13)(0,2) ;( 14)81;( 15)①④;(16)5.
三.解答题:
(17) (本小题满分 12 分)
(I) 动点 ),( yxP 满足到点 )0,2(F 的距离比到直线 1x 距离多 1.
动点 满足到点 的距离与到直线 2x 的距离相等。
动点 P 是以 )0,2(F 为焦点, 为准线的抛物线,方程为 xy 82 ………………4 分
(II)当直线斜率不存在时, )1,1( Q 显然不为中点,
当直线斜率存在时,设为 k ,设 ),( 11 yxA , ),( 22 yxB
2
2
2
1
2
1
8
8
xy
xy ,得 )(8 21
2
2
2
1 xxyy , )(8))(( 212121 xxyyyy
又 是线段 AB 的中点, 221 yy , 4
21
21
xx
yyk
故直线的方程为 )1(41 xy ,化为一般形式即: 034 yx . ………………10 分
2
(18) (本小题满分 12 分)
解:设正四棱柱底面边长为 xcm,高为 hcm,
则有8 4 72xh, 18 2 ,(0 9)h x x
2 2 3 2( ) (18 2 ) 2 18V x x h x x x x ···························2 分
2( ) 6 36 6 ( 6)V ' x x x x x ······················································4 分
由 ( ) 0V ' x 得:06x,由 ( ) 0V ' x 得:69x··················6 分
()Vx 在 (0,6) 上是增函数,在(6,9) 上是减函数···········································8 分
6x时, ()Vx取最大值 (6) 216V , 6h ···········································10 分
∴底面边长与高均为 6cm 时,容积最大·····················································12 分
(19)(本小题满分 12 分)
解:(Ⅰ)由题意得 )2()1(23)( 2 aaxaxxf
又
3)2()0(
0)0(
aaf
bf
,解得 0b , 3a 或 1a ····················4 分
(Ⅱ)法一:∵函数 )(xf 在区间 )1,1( 单调递减,∴ ( 1,1)x 时 ( ) 0f ' x
2( ) 3 2(1 ) ( 2) ( )(3 2)f ' x x a x a a x a x a 零点为 12
2, 3
ax a x ,·····6 分
故有
1
2 13
a
a
,或
2 13
1
a
a
························································10 分
解得 51aa 或
故 a 的取值范围为( , 5] [1, ) U ······················································12 分
法二:∵函数 在区间 单调递减,∴ 时 ,
只需 (1) 0
( 1) 0
f'
f'
························································8 分
即 3 2(1 ) ( 2) 0
3 2(1 ) ( 2) 0
a a a
a a a
,解得
故 a 的取值范围为 ······················································12 分
3
(20) (本小题满分 12 分)
解:(I)∵ 2 2 2 1c a b ,∴ 2
2
ce a ·············································4 分
(II)设直线 l 的方程为:y=k(x )A(x1,y1) ,B(x2,y2),M(5
4,0)
联立
2
2
2 2 2 21 (1 2 ) 4 2 2 02
1
x y y k x k x k
y k x
消去 得:
则
2
12 2
2
12 2
4
12
22
12
0
kxx k
kxx k
…………………………………………………………………………8 分
∵ 1 1 2 2
55( , ) ( , ) 44MA x y MB x y
1 2 1 2 1 2 1 2 1 2
5 5 5 25 7( ( ) = ( ) + + = 4 4 4 16 16MA MB x x y y x x x x y y )
7,=16x R MA MB 对任意 有 为定值.···············································12 分
(21)(本小题满分 12 分)
解:(Ⅰ)椭圆离心率
221
2
c a be aa
,又 3bc , 2 2 2a b c,
解得 2, 3ab, 椭圆方程:
22
1.43
xy···········································4 分
(Ⅱ)显然,直线 l 的斜率不能为 0,
设直线 l 的方程为 1x my, 11( , )A x y , 22( , )B x y ,
4
联立
22
143
1
xy
x my
消去 x 得 22(3 4) 6 9 0m y my .
则
12 2
12 2
6
34
9
34
0
myy m
yy m
…………………8 分
2
2
2
1 2 1 2 1 2 1 2 1 2 22
11| || | | | ( ) 4 122 (3 4)F AB
mS F F y y y y y y y y m
V .
设 2 1tm,则 1t , 22=1mt ,
2 2
12 12
13( 1) 4 3
F AB
tS t t t
,············10 分
设函数 1( ) 3f t t t, [1, )t 时为增函数, ( ) (1) 4f t f ,
2
12 313
F ABS
t t
,
即 0m 时, 2F AB 面积的最大值为 3.···············································12 分
(22)( 本小题满分 12 分)
解:(I)由题意知,函数的定义域为(0,+ ∞),当 a=-2 时, ()f ' x =2x- 2
x
= 2( 1)( 1)xx
x
,
由 <0 得 0<x<1,故 f(x)的单调递减区间是(0,1).·····················4 分
(II)设 g(x)= f(x)-(3- )=x2+aln x+ -3,则 g(1)=0,
()g' x =2x+ a
x
- 2
2
x
=
3
2
22x ax
x
,
当 0a 时,x∈(1,+∞)时,
33
22
2 2 2( 1)( ) 0x ax x axg' x xx
,g(x)在[1,+∞)
单调递增,g(x)≥g(1)=0 恒成立;·····························································8 分
当 0a 时,设 3( ) 2 2h x x ax ,∵ (1) 0ha, 0 1x ,使得 0(1, )xx 时,
5
( ) 0hx ,∴ 2
()( ) 0hxg' x x,g(x)在 0[1, ]x 单调递减, 0( ) (1) 0g x g ,与条件矛盾,
故 a 的取值范围为[0,+ ∞) ·····························································12 分