广东省东莞市2026届高三下学期数学模拟考试试题含答案

这是一份广东省东莞市2026届高三下学期数学模拟考试试题含答案，共10页。试卷主要包含了单项选择题，多项选择题，填空题，解答题等内容，欢迎下载使用。
1. 已知集合 A={−3,−1,1,3},B=x∣x3=x ，则 A∩B= ()
A. {−1,0,1} B. {−1,1} C. {1} D. {−1}
2. 在复平面内,复数 z=1−ii 对应的点位于( )
A. 第一象限 B. 第二象限 C. 第三象限 D. 第四象限
3. 样本数据2,3,5,8,9,10的 25% 分位数为( )
A. 1.5 B. 2 C. 2.5 D. 3
4. 设 a>0 且 a≠1,b>0 ,下列各项中,能推出 lgab>0 的一项是( )
A. a>1 且 b>1 B. a>1 或 01 时, lgab>0 , A 正确;
对于 B ,若 a=2,b=12,lg212=−10,yn>0 在直线 ln 上,所以 yn+yn−1=p+1xn−xn−1 ,①
又点 Pn−1,Pn 也在抛物线 y2=2px 上,
所以 xn−1=yn−122p,xn=yn22p ,代入①得:
所以 yn+yn−1=p+1yn22p−yn−122p=p+1yn+yn−1yn−yn−12p ,
因为 yn+yn−1>0 ,所以 yn−yn−1=2pp+1 ,
所以数列 yn 为首项为 y1=2pp+1 ,公差为 2pp+1 的等差数列.
(3)由题意如图所示:

由题意可知: Pnxn,yn,Pn+1xn+1,yn+1,Pn+2xn+2,yn+2yn>0,yn+1>0,yn+2>0 , 且 yn≠yn+1≠yn+2,xn=yn22p,xn+1=yn+122p,xn+2=yn+222p ,
则直线 lPnPn+2 的斜率为:
k=yn+2−ynxn+2−xn=yn+2−ynyn+222p−yn22p=2pyn+2−ynyn+22−yn2=2pyn+2−ynyn+2+ynyn+2−yn=2pyn+2+yn,
所以直线 lPnPn+2 的方程为: y−yn=2pyn+2+ynx−xn ,
即 2pyn+2+ynx−y−2pxnyn+2+yn+yn=0 ,
所以点 Pn+1 到直线 lPnPn+2 的距离为:
d=2pxn+1yn+2+yn−yn+1−2pxnyn+2+yn+yn2pyn+2+ynxn+1−xn−yn+1−yn2pyn+2+yn2+1
将 xn=yn22p,xn+1=yn+122p 代入上式可得:
d=yn+1+ynyn+2+yn−1yn+1−yn4p2+yn+2+yn2yn+2+yn2=yn+1+yn−yn+2+ynyn+2+ynyn+1−yn4p2+yn+2+yn2yn+2+yn2
=yn+1−yn+2yn+2+yn×yn+1−yn4p2+yn+2+yn2yn+2+yn2=yn+2−yn+1yn+2+yn×yn+1−yn4p2+yn+2+yn2yn+2+yn2
由 (2) 得: yn+1−yn=2pp+1,yn+2−yn+1=2pp+1 ,
所以 d=2pp+1×2pyn+2+yn4p2+yn+2+yn2yn+2+yn2=4p2p+124p2+yn+2+yn2 ,
又 PnPn+2=xn+2−xn2+yn+2−yn2=yn+222p−yn22p2+yn+2−yn2
=yn+2+yn2yn+2−yn24p2+yn+2−yn2=yn+2+yn24pp+124p2+4pp+12
=2p+14p2+yn+2+yn2,
所以 ΔPnPn+1Pn+2 的面积为:
S=12PnPn+2d=12×2p+14p2+yn+2+yn2×4p2p+124p2+yn+2+yn2=4p2p+13,
设 fp=4p2p+13p>0 ,则 f′p=p8−4pp+14 ,
令 f′p=0⇒p8−4pp+14=0⇒8−4p=0⇒p=2 ,
当 02 时, f′p