广东省中山市2026届下学期高三高考一模 数学试题

·PAGE1/2026届高三模拟测试一数学试题2B非选择题必须用黑色字迹的钢笔或签字笔作答考生必须保证答题卡的整洁．考试结束后一选择题本题共 小题每小题5 z=1-i()A.1- B.1+ C.2- D.2+【详解因为复数z=2= 2(1+i =2+2i=2+2i=1+i，

1-

(1-

(1

1- 篮球和乒乓球的人数为( A. B. C. D.20-x25-45 若m⎳α,n⊂α则m⎳ B.若m⎳α,α⎳β则m⎳C.若m⊥α,m⎳β则α⊥ D.若m⊥α,m⊥n则n⎳Am⎳α,n⊂αm⎳nm,nABm⎳α,α⎳βm⎳βm⊂βBCm⎳βm1⊂βm⎳m1m⊥αm1⊥αm1⊂βα⊥βCDm⊥α,m⊥nn⎳αn⊂αD错误 关于实数x有下列甲乙丙三个陈述分别为甲：x>2乙：x<4丙：x>3．如果甲乙丙三个陈述中有且仅有一个正确则x的取值范围可以为( (3, B.(- C.

D.[4,x>2x≥4x≤3xx4x2x3x2x的范围为(-∞,2]；x3则甲：x2.x的取值范围为(-∞,2].y2xy

ylog1xA.x轴对 B.y轴对 C.原点对 D.直线y=x对

=2-x=

.y=axy=logax(a>0a≠1)(x-4y=x(x-4

ylog1xyx对称

25-

=1表示的圆锥曲线的离心率e=(

25-

(x-4(x-4

(x-

2+y2=25-4x

25(x-42+y2252200xx2

= = a5，c4e=

=4.C.- B.C.

= D.sinα=

所以 αsinα<0,cosα<sinα=-2，cosα=-tanβ=17

sin2β+cos2ββsinβ0,cosβsinβ=-34，cosβ=-2

=sinαcosβ+cosαsinβ=-

×-

+-

×- =2 任意一点则PB⋅PC的最小值为 B.-

-

-3ACDCDAC2BDE为△ABD所在平面内

λ)ADAE 由λAB+(2-2λAC23AEmin23AEBD时

sin∠ABD=2

=3又∠ABD∠ABD=πADAB4△ABD为正三角形，于是∠BAC=

4xAB

PBPCxABACAPxABACx(1x)AB16x2(4x329≥-9x3时取等号 PBPC的最小值为-4二、择题本 题每 6分共 部分选对的得部分分有选错的得 ≠±1Sn S53，S109S15AB21S10

a1(1- 1-

a1(1- 1-

a1(1-1- q10a1q92a1q3na1qna1q2n2q3nqnq2n，2(1-

=(1-

+(1-

a1(1- 1-

a1(1- 1-

a1(1-1- S2nR上的奇函数f

f

+

=

=

则

是奇函 B.

值域为(-

h(x-

=h(x)-1-Af

+

2ex

-f

=2e-xf

=ex+e-

=

=ex-e-xh(-ex+e-

=e-x-exe-x

=-ex-e-ex+e-

=-h(xh(xAB

不可能为单调增函数BC

ex+e-

=e2x-

=1- ，e2x11+∞)h(x(-1,1)CDhx=ex-e-xh

=ex-y-e-(x-y) ex+e-

ey+e-

ex-y+e-(x-

-

=ex-e-x-ey-e-yex+e- ey+e-

(ex-e-x)(ey+e-y)-(ey-e-y)(ex+e-x)(ex+e-x)(ey+e-y)=exey+exe-y-e-xey-e-xe-y-(eyex-eye-x-e-yex+e-ye-(ex+e-x)(ey+e-

2ex-y-2e-(x- (ex+e-x)(ey+e-(ex-e-x)(ey-e-1h(x)h(y1-(ex+e-x)(ey+e-y)

(ex+e-x)(ey+e-y)-(ex-e-x)(ey-e-y)(ex+e-x)(ey+e-y)=exey+exe-y+e-xey+e-xe-y-(exey-exe-y-e-xey+e-xe-(ex+e-x)(ey+e-

2ex-y+2e-(x- (ex+e-x)(ey+e-h(x)-所 1-

2ex-y-2e-(x-(ex+e-x)(ey+e-2ex-y+2e-(x-(ex+e-x)(ey+e-

=2ex-y-2e-(x-2ex-y+2e-(x-

=ex-y-e-(x-ex-y+e-(x-

=h(x-D正确. Q1Q2Q3Q1Q2Q3(如图①)．某老师绘制了一次数学小测验中甲(如图②)根据该图判断下列说法正确的是()方差最小，AB都对，C错，D对.三填空题本题 小题每小题 分已知抛物线C:y=4x的焦点为F，M是C上一点，△MOF的面积为2则|MF|= S△MOF1OFyM2yMx

=由抛物线的定义知，|MF|

+

=x∈0,

cos(ωx+

≤1(ω>0)恒成立则ω的取值范围为 【详解因为cos(ωx+ ≤1所以-1≤cos(ωx+ ≤1 π2kπωx+

≤

2kπ,kZ

+2kπ≤ωx+

≤

+2kπ,k∈Zx∈

≤ωx+

≤ωπ+ππωπ+π≤2π0ω1 ω的取值范围为0,1 空间中有四个半径为2的小球每个球都与其它三个球外切．现另有一小球与这四个球均外切则 6PABCPABCOR，POABCD，PDPABC的高，PD⊥ABCPO=4AD=

AB2-(

2=

16-4=434所以PD AP2-AD2 42-(4332 4Rt△AOD中，ODPDOP=4

OD2

2+4

-R,R=64OrR2=6四解答题本题共5小题 分．解答应写出文字说明证明过程或演算步骤0.80.7)(2X为三次射击中命中的次数X(2)分布列答案见解析E(XAi(i=1,2,3

==0.8，P(A2A1=

=

=

⋅

+

⋅=0.8×0.7+0.2×0.9=P(X=0

=0.2×0.1×0.1=2P(X=

=0.8×0.7×0.7=P(X=1P(X=2

=0.8×0.3×0.1+0.2×0.9×0.3+0.2×0.1×0.9==0.8×0.7×0.3+0.8×0.3×0.9+0.2×0.9×0.7=X所以E(X00.00210.09620.5130.392∠BAC=π求平面四边形ABCD(1)7(2)0,33+6由∠ABD=∠CBD=π得∠ABC=π而∠BAC=π,AC=3则AB= =1， BC= 2CD AC2+CD2=7在△ABD中由正弦定理得 = ，sin∠ADB

1= 2 71

由(1知∠ABC=π,∠BCD=2π∠ACBθ(0<θ2π 在△ABC中由正弦定理得 = = =2， AB2sinθ,BC2sinθ+

ABCDSS△ABCS△ACDS△ABCS△BCD1ABBCsinπ1BC2sin =3sinθsin(θ+ +3sin2(θ+ =3(1sinθ+3cosθ(3sinθ+3 =3(3+3sinθcosθ=333sin2θ02θ<4π-3sin2θ 0<333sin2θ≤33+60S≤33+6 ABCD面积的取值范围是0,33+6P-ABCDABCDAB=AD=CD=1，BC=PC(1PD的长；(2PB①求平面PAB与平面PCD (2)①21② ，ABCDAAEBCEDDFBCF2BD=3BC2，CD1BD2CD2BC2，CDBPBDPBDPC2PD=4-1=3BDPcos∠BDP=3+3-9=-12×3 所以∠BDP120°CD DDB，DCx,y轴的正方向，DDCzA(3,-1,0，B(3,0,0，C(0,1,0，D(0,0,0，P(-3,0,3 PAB=(x,y,z 31 13AB=(2,2,0，AP=(-3,2,2_

3x+1y=0AB⋅m=0所以

x1y=-3，z=3_

AP⋅m 所以=(1,-3 PCD=(a,b,c

3

2,0,2_

b=0DC⋅n=0所以

a=3c1，b_

a c=0 DP⋅n (3,0,1PABPCD7×2 则 (3 3PB

,0,-

PB的中点为E(3,0, (3,1,3 4，AE=- _

-9-所以点A到平面α的距离为AE⋅PB= 8=3所以 3

27+27+ AQ4已知函数y=f(x的定义域为D对于给定实数t定义集合Vf(t=xf(x+t≥f 13x若D=R求证y=f(xt使得Vf(t=Vf(-t=若f(x=1ax-e2-xlnx-e，D=(0,+∞且对于任 t∈D都有 = e-

e-

(1)Vf(2(-∞,-31,+∞(2)证明见解析 (3)a∈1,1

≥f f(x+2f(x(x+2313(x+2x36x212x186(x+3(x-1Vf(2(-∞,-3充分性因Vf(t=Vf(-t=R，∀x∈R,f(x+t≥f(x,f(x-t≥f(xf(xf(x+tf(xf(x+tt0f(xf(xTf(xf(x=f(x±T所以Vf(T=xf(x+T≥f =R，Vf(-T=xf(x-T≥f =tT0Vf(tVf(-t

在

fr

=ax-e-

-lnx+1≥e-a0x→+∞,fr(x→-∞fr(xa0fr

=ax-e-

-lnx+1=h(xe-

=ax-1x∈0,

<

x∈1

>

h

=fr

0lnae(a-e-

≥

=lnx-e(x-e-

gr

=

-ee-x∈0e-

>

x∈e-1

<

g

=

=

0

≤a≤a1,1

C:

+

=1a>b>0

l:2x-y=

的离心率为2

x2y21P2

(n

在椭圆C(1C(2)Tl:x=4TCMNMN②是否存在点T∠MPF=∠NPFT

y2=

T427F(c,0lOlOFOAOx2y21Ax2y21OFOA1cCe=

=1可得a=2故b a2-c2 22-12=3C的标准方程为4+3M(x,yN(x,yCMx1x+y1y1 2 x1x+y1y=1 联立 y 可得(3x2+4y2x2-24xx+16(3-4y2= 3

MC

+

13x24y2 所以方程(*)12x224x1x12x20x22xxx2Δ=4x2-4x2=

CM

+

=1TM

+

=TN

+

=x1+ty1=1T

x2+ty2=1 MNx

31MNx

3=MNy=0x=1MNF(1,02②将点P的坐标代入椭圆方程得( +n2

=22n>0n

故点P(2,2 PMPNαβPF2因为∠MPF∠NPFαγγββγγααβ2-且有tanγ=kPF 2- =-26PMPNPMPNk1tan(α

=tan2γ=2tanγ

=46 1-

1-(- tan(α

=tanα+tanβ=k1+k1=461- 1- MNpx

+qy-2

=12

+

+2

= 3(x-

2+4(y-2

2+4(x-

+166(y-2

= 3(x-

2+4(y-2

+166(y-2

+q(y-2

= 整理可得4166

y-2

2+4q+166

x-

y-2

+(4p

x-

=x

k

y-2y-2则方程(*)可化为4+166

k2+166p

k+(4p

=4(46qk1k