2024数学新高考二卷
2024数学 新高考二卷一、选择题1. 已知集合A = {x | x^2 - 3x - 4 ≤ 0},B = {x | x^2 - 2x - 3 > 0},则A ∩ B = ( )A. {x | -1 ≤ x < -1 或 x > 3}B. {x | -1 < x ≤ 3}C. {x | -1 ≤ x < 3}D. {x | x < -1 或 x ≥ 3}2. 复数z满足(1 + i)z = 2i,则z = ( )A. 1 + iB. 1 - iC. -1 + iD. -1 - i3. 若直线l经过点P(1,2),且l与x轴、y轴的正半轴所围成的三角形面积为4,则l的方程为 ( )A. 2x + y - 4 = 0B. x + 2y - 4 = 0C. x + y - 3 = 0D. 2x - y = 0二、填空题1. 已知函数f(x) = 2x^2 - 4x + 1在[a, b]上的最小值为3,且b - a = 2,则a = _______。2. 在等差数列{an}中,若a2 + a3 + a4 = 21,则a1 + a5 = _______。3. 若向量a = (1, -1),b = (3, m),且a // b,则m = _______。三、解答题1.(本题20分)设f(x) = ln(x + 1) - x^2,求f(x)的单调区间。【解析】首先求f(x)的导数f'(x):f'(x) = \frac{1}{x + 1} - 2x= \frac{1 - 2x(x + 1)}{x + 1}= \frac{-2x^2 - 2x + 1}{x + 1}接下来分析f'(x)的符号:当f'(x) > 0时,解得x的取值范围,此时f(x)单调递增;当f'(x) < 0时,解得x的取值范围,此时f(x)单调递减。【答案】f(x)在(-1, \frac{1 - \sqrt{3}}{2})上单调递增,在(\frac