浙江省部分学校2025-2026学第二学期高二下学期期中联考数学试题（含答案）

这是一份浙江省部分学校2025-2026学第二学期高二下学期期中联考数学试题（含答案），共21页。试卷主要包含了8 13等内容，欢迎下载使用。

12.8 13.( __ ∞, 0] U { } 14.
15 ．（1）因为an+1 = 3an __ 2 ，所以an+1 __ 1 = 3 (an __ 1)，·························2 分又a1 = 2 ，所以a1 __ 1 = 1 ≠ 0，···············································4 分所以 即{an __ 1}是首项为 1 ，公比为 3 的等比数列．·····················6 分（2）由（1）得an __ 1 = 3n__1 ，即an = 3n__1 + 1，································8 分
所以Sn n．
(写出求和公式得 2 分，化简正确得 1 分）······································11 分设函数f n, n ∈ N* , 易知f(n)递增，
检验：n = 4 时，f (n) = 44 < 50,
\l "bkmark1" n = 5 时，f (n) = 126 > 50 ，所以满足条件的最小值n = 5 13 分
\l "bkmark2" 16 ．（1）f' (x) = x2 __ 4x + 3，1 分
\l "bkmark3" 因式分得f' (x) = (x __ 1)(x __ 3)，2 分
令f' (x) = 0 ，得x = 1 或x = 3
（列表或文字说明均可）······················································4 分所以：单调递增区间为( __ ∞, 1)和(3, + ∞) ，单调递减区间为(1,3) ·················5 分（2） 由（1）知，在区间[ __ 1.4]上的极值点为x = 1 和x = 3 ，区间端点为x =__ 1 和x = 4。计算函数值：
题号
1
2
3
4
5
6
7
8
答案
C
B
C
D
A
B
B
D
题号
9
10
11
答案
AD
ABC
BCD
x
(-∞, -1)
1
(1, 3)
3
(3, +∞ )
f, (x )
+
0
-
0
+
f (x)
单调递增
极大
单调递减
极小
单调递增
f (3) = 9 __ 18 + 9 + 1 = 1
（每个 1 分）
\l "bkmark4" 9 分
\l "bkmark5" 比较得：最大值为7 ，最小值为__ 13 10 分
\l "bkmark6" 3 3
（3）由 (1) 知，f (x)在x = 1 处取得极大值 ，在x = 3 处取得极小值 1 ············12 分方程f(x) = k有三个不同的实数根H直线y = k与曲线y = f (x)有三个不同交点 ······14 分所以k的取值范围为k ∈ (1, ) ·················································15 分
17 ．（1）需抽取 4 次，按顺序可看作为 4 个位置，
两张中奖奖券置于第二，四位，有放法数AEQ \* jc3 \* hps11 \\al(\s\up 5(2),2) = 2 ；································3 分其余二个位置放两张不中奖奖券，有放法数AEQ \* jc3 \* hps10 \\al(\s\up 4(2),4) = 12
由乘法原理方法数为：2 × 12 = 24 种不同的抽取顺序；·····························6 分
（2）至多 4 次可分为恰好 2 次，恰好 3 次，恰好 4 次找到所有中奖奖券，
恰好 2 次，即前 2 次抽取都是不中奖奖券，方法数为AEQ \* jc3 \* hps11 \\al(\s\up 5(2),2) = 2 ；·····················8 分
恰好 3 次，即第 3 次是中奖奖券，前 2 次中有 1 次是中奖奖券，方法数为CEQ \* jc3 \* hps11 \\al(\s\up 5(1),2)A EQ \* jc3 \* hps11 \\al(\s\up 5(2),2)CEQ \* jc3 \* hps11 \\al(\s\up 5(1),4) = 16 ···
··········································································10 分
恰好 4 次，即第 4 次是中奖奖券，前 3 次中有 1 次是中奖奖券，方法数CEQ \* jc3 \* hps11 \\al(\s\up 5(1),3)A EQ \* jc3 \* hps11 \\al(\s\up 5(2),2)A EQ \* jc3 \* hps11 \\al(\s\up 5(2),4) = 72 ；·····
··········································································12 分
也可以是前四次全是不中奖奖券，方法数AEQ \* jc3 \* hps10 \\al(\s\up 4(4),4) = 24 ································14 分
故共有2 +16 + 72 + 24 = 114 种不同的抽取情况 ··································15 分18 ．（1）设数列{an } 的公比为q ，由4a2 , 2a3 , a4 成等差数列可得4a2 + a4 = 4a3 ，·······2 分故4 + q2 = 4q ，解得q = 2 ，···················································4 分由s3 = 14 可得 解得a1 = 2 ，
故an = 2n ，即数列{an } 的通项公式为an = 2n , n ∈ N* . ·······························7 分
（2） 由（1）可得bn 分故 ················ 13 分易知单调递减，故 单调递增，
即{Tn }为递增数列，则Tn ≥ T ····································15 分又当n → +∞ 时， → 0 且 > 0 ，所以Tn
故 Tn 所以m ≥ ···················································17 分19 ．（1）当 a = 1 时， f (x ) = (2 _ x )lnx ，函数定义域为x ∈(0, +∞ )
故f, = _lnx f, ···············································2 分又f(1) = 0 ，所以切线方程为y = x _1 . ··········································3 分
（2） 由题意得f, = _lnx
若f(x)不存在单调增区间，则__lnx + 2a _1 ≤ 0 恒成立，即2a ≤ xlnx + x 恒成立，······5 分
x
令g (x) = xlnx + x，g, (x) = 2 + lnx ，
当x ∈(e_2 , +∞ ) 时g,(x) > 0 ，当x ∈(0, e_2) 时g, (x) < 0
所以g(x)在(0, e_2)单调递减，在(e_2 , +∞ )单调递增，·····························7 分所以g min = g 所以2a ≤ _ 即a ≤ _
因此所求实数a 的取值范围为 ······································8 分（3）由（2）知f, = _lnx ························10 分
所以f, (x ) 在(0, +∞ )单调递减，又f, (e_2) = 1+ 2ae2 > 0 ， f,
所以必存在正数x0 ，使得f, = _lnx 即2a = x0 + x0lnx0，············12 分由（2）知当x 时， 2a < 0 即a < 0 ，当x 时， 2a = 0 即a = 0 ，
当x 时， 2a > 0 即a > 0 ，
由上可知f(x)在(0, x0)单调递增，在(x0 , +∞)单调递减，
所以f(x)max = f (x0) = (2a - x0)lnx0 ，··········································13 分所以a +b ≥ (2a - x0)lnx0 ，即b-5a ≥ (2a - x0)lnx0 -6a ，
b - 5a ≥ x0 (lnx0)2 - 3x0lnx0 - 3x0
令h = x 2 - 3x ln x - 3x
因为h, (x ) = (lnx )2 - lnx - 6 = (lnx - 3)(lnx + 2) ···································15 分当x 时， h (x) 单调递减，当x ∈(e3 , +∞ ) 时， h (x) 单调递增，
所以h(x )min = h (e3) = -3e3 ，
所以b-5a 的最小值为-3e3 ··················································17 分