江南十校2020高三第二次联考数学理科答案

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江南十校2020届高三第二次联考

数学(理科)试题参考答案

一㊁选择题:本大题共12小题,每小题5分,共60分.1.答案:D 解析:解得U =y y {}>0,A =x 1

解析:解得cos α=-5

13

,故f cos ()α=1,则f f cos ()[]α=f ()1=2㊂

3.答案:A

解析:如图所示,点P 在平面区域内任一点P ,点Q 在半圆x 2+y 2=10≤y ≤()1上,过点O 作直线x +y -5=0的垂线,垂足为P ,交半圆于Q ,此时PQ 取最小值,求得PQ

min

=

52

2

-1㊂

4.答案:B

解析:()f t =log b t 为增函数,0y ㊂当a <0时,()h t =t a 在第一象限单调递减,

a =log

b sin α且cos α>sin α,则x x >y ㊂5.答案:D

解析:由题得sin θ=x 2+12x ,由x 2+12x ≥1或x 2+1

2x

≤-1且-1≤sin θ≤1得:sin θ=±1,故x =±1㊂

6.答案:D

解析:y =sin 2x +cos 2x =2sin 2x +πæèçöø÷4=2sin2x +πæ

èçöø

÷8,y =-2cos 2x =2sin 2x -πæèçöø÷2=2sin 2x -πæèçöø

÷4,故向右移3π8个单位㊂　7.答案:C

解析:a 4=a 2+2d ,a 8=a 2+6d ,因为a 42=a 2㊃a 8且d ≠0,求得a 2=2d ,所以公比q =a 4

a 2

=2;或解:q =

a 8a 4=a 4a 2=a 8-a 4a 4-a 2=4d 2d

=2㊂

8.答案:C

解析:m ⊥α

α‖}

β⇒m ⊥β n ⊥ü

þ

ýïï

ï

ïβ⇒m ‖n.

9.答案:A

解析:()f ′x =e x +e -x +1>0x ()>0,故()f x 在0,+()¥上单调递增㊂b ∈0,(]1时,()[]f f b =b 成立,即()f b =b 有解,　则e -b +e b +b -a =b ,

故a =e -b +e b ,b ∈0,(]1㊂令e b =t ,则t ∈1,(]e ,e b +e -b =t +

1t ∈2,e +

1æ

è

çùûúúe ,即a ∈2,e +1æè

çùû

úúe ㊂

10.答案:C

解析:构造长方体ABCD -A 1B 1C 1D 1,使MN 与BD 1重合㊂设长方体长㊁宽㊁高分别为x ,y ,z ,则x 2+y 2+z 2=1㊂由题知x 2+z 2=a ,

y 2+z 2=b ,x 2+y 2=c ,a 2+b 2+c 2=2㊂

a +

b +()

c 2=a 2+b 2+c 2+2ab +2bc +2ac

≤3a 2+b 2+c ()2=6,故a +b +c ≤6㊂

11.答案:A

解析:连PI 延长x 轴于D ,连IF 1㊁IF 2㊂在△PF 1D 中有ID IP =DF 1PF 1,在△PF 2D 中有ID

IP =DF 2PF 2

,故ID IP =DF 1PF 1=DF 2

PF 2

=DF 1+DF 2PF 1+PF 2=2c 2a =e =1

2,故

S △IF 1F 2

S △PF 1F 2

=

ID PD =1

3

㊂

12.答案:B

解析:f (x +2)=f (2-x ),推得f (x +4)=f (-x )=f (x ),故f (x )最小正周期为4.f (x i )-f (x i +1)≤4-1=3,x n 取得最小值,则需尽可能多的x i 取到最高(低)点,

由2993=9923

以及2x =2得:x n (min)=99×2+1=199㊂

二㊁填空题:本大题共4小题,每小题5分,共20分.13.答案:

25

解析:sin(π+α)=2cos(π-α)可得tan α=2sin α(2cos 2

α2-1)=sin αcos α=sin αcos αsin 2α+cos 2α=tan αtan 2α+1=25

14.答案:2

解析:如图,作小圆的直径AE ,连DE ,则DE =4,AE =DE 2-DA 2=23=2r =BC AB 2+AC 2=BC 2=12≥2AB ㊃AC ,则AB ㊃AC ≤6,V =

13㊃12㊃AB ㊃AC ㊃AD =16×2AB ㊃AC =1

3

×AB ㊃AC ≤215.答案:{x |x >-

1

2

}解析:令=g (x )=xf (x ),由

x 1f (x 1)-x 2f (x 2)

x 1-x 2

<0,得g (x )在(-∞,0)为减函数,

且g (x )为偶函数,故g (x )在(0,+∞)上为增函数,g (x )-12

㊂16.答案:

33

解析:取F 1D 的中点Q ,连EQ ㊁PQ ㊂

PF →1㊃→PD =14(PF →1+→PD )2(PF →1-→PD )[]2=1

4(4→PQ 2-DF →12)=→PQ 2-14

DF →12,