高考数学解答题规范专题练 （含答案）

这是一份高考数学解答题规范专题练 （含答案），共12页。
(1)满足有解三角形的序号组合有哪些？
(2)请在(1)所有组合中任选一组，求对应△ABC的面积．
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2．(2023·河北衡水中学模拟)黄河鲤是我国华北地区的主要淡水养殖品种之一，其鳞片金黄、体型梭长，尤以色泽鲜丽、肉质细嫩、气味清香而著称．为研究黄河鲤早期生长发育的规律，丰富黄河鲤早期养殖经验，某院校研究小组以当地某水产养殖基地的黄河鲤仔鱼为研究对象，从出卵开始持续观察20天，试验期间，每天固定时段从试验水体中随机取出同批次9尾黄河鲤仔鱼测量体长，取其均值作为第ti天的观测值yi(单位：mm)，其中ti＝i，i＝1,2,3，…，20.根据以往的统计资料，该组数据(ti，yi)可以用Lgistic曲线拟合模型y＝eq \f(1,\f(1,u)＋abt)或Lgistic非线性回归模型y＝eq \f(u,1＋ea－bt)进行统计分析，其中a，b，u为参数．基于这两个模型，绘制得到如图所示的散点图和残差图．
(1)你认为哪个模型的拟合效果更好？分别结合散点图和残差图进行说明；
(2)假定u＝12.5，且黄河鲤仔鱼的体长y与天数t具有很强的相关关系．现对数据进行初步处理，得到如下统计量的值：eq \x\t(t)＝eq \f(1,20)eq \i\su(i＝1,20,t)i＝10.5，eq \x\t(z)＝eq \f(1,20)eq \i\su(i＝1,20,t)i＝－3.83，eq \x\t(w)＝eq \f(1,20)eq \i\su(i＝1,20,t)i＝－1.608，eq \i\su(i＝1,20,t)(ti－eq \x\t(t))2＝665，eq \i\su(i＝1,20,t)(zi－eq \x\t(z))(ti－eq \x\t(t))＝－109.06，eq \i\su(i＝1,20,t)(wi－eq \x\t(w))(ti－eq \x\t(t))＝－138.32，其中zi＝lneq \b\lc\(\rc\)(\a\vs4\al\c1(\f(1,yi)－\f(1,u)))，wi＝lneq \b\lc\(\rc\)(\a\vs4\al\c1(\f(u,yi)－1))，根据(1)的判断结果及给定数据，求y关于t的经验回归方程，并预测第22天时仔鱼的体长(结果精确到小数点后2位)．
附：对于一组数据(x1，y1)，(x2，y2)，…，(xn，yn)，其经验回归直线eq \(y,\s\up6(^))＝eq \(a,\s\up6(^))＋eq \(b,\s\up6(^))x的斜率和截距的最小二乘估计分别为eq \(b,\s\up6(^))＝eq \f(\i\su(i＝1,n, )xi－\x\t(x)yi－\x\t(y),\i\su(i＝1,n, )xi－\x\t(x)2)，eq \(a,\s\up6(^))＝eq \x\t(y)－eq \(b,\s\up6(^))eq \x\t(x)；参考数据：e－4≈0.018 3.
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3．(2023·锦州模拟)记Sn为各项均为正数的等比数列{an}的前n项和，S3＝14，且a3,3a2，a4成等差数列．
(1)求{an}的通项公式；
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(2)在an和an＋1之间插入n个数，使得这n＋2个数依次组成公差为dn的等差数列，求数列eq \b\lc\{\rc\}(\a\vs4\al\c1(\f(1,dn)))的前n项和Tn.
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4．(2023·镇江模拟)在直角梯形AA1B1B中，A1B1∥AB，AA1⊥AB，AB＝AA1＝2A1B1＝6，直角梯形AA1B1B绕直角边AA1旋转一周得到如图所示的圆台A1A，已知点P，Q分别在线段CC1，BC上，二面角B1－AA1－C1的大小为θ.
(1)若θ＝120°，eq \(CP,\s\up6(→))＝eq \f(2,3)eq \(CC1,\s\up6(—→))，AQ⊥AB，证明：PQ∥平面AA1B1B；
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(2)若θ＝90°，点P为CC1上的动点，点Q为BC的中点，求PQ与平面AA1C1C所成最大角的正切值，并求此时平面APQ与平面APC夹角的余弦值．
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5．(2023·哈尔滨三中模拟)已知平面内动点M到定点F(0,1)的距离和到定直线y＝4的距离的比为定值eq \f(1,2).
(1)求动点M的轨迹方程；
(2)设动点M的轨迹为曲线C，过点(1,0)的直线交曲线C于不同的两点A，B，过点A，B分别作直线x＝t的垂线，垂足分别为A1，B1，判断是否存在常数t，使得四边形AA1B1B的对角线交于一定点．若存在，求出常数t的值和该定点坐标；若不存在，说明理由．
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6．(2023·佛山模拟)已知函数f(x)＝ex－aln(ax＋1)－1，其中a>0，x≥0.
(1)当a＝1时，求函数f(x)的零点；
(2)若函数f(x)≥0恒成立，求a的取值范围．
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1．(2023·杭州第二中学模拟)已知a，b，c分别为△ABC内角A，B，C的对边，若△ABC同时满足下列四个条件中的三个：①a＝eq \r(3)；②b＝2；③eq \f(sin B＋sin C,sin A)＝eq \f(a＋c,b－c)；④cs2eq \b\lc\(\rc\)(\a\vs4\al\c1(\f(B－C,2)))－sin Bsin C＝eq \f(1,4).
(1)满足有解三角形的序号组合有哪些？
(2)请在(1)所有组合中任选一组，求对应△ABC的面积．
解 (1)对于③，eq \f(b＋c,a)＝eq \f(a＋c,b－c)⇒eq \f(a2＋c2－b2,2ac)＝－eq \f(1,2)，
∵B∈(0，π)，
∴B＝eq \f(2π,3)；
对于④，eq \f(1＋csB－C,2)－sin Bsin C＝eq \f(1,4)⇒cs(B－C)－2sin Bsin C＝－eq \f(1,2)，
即cs(B＋C)＝－eq \f(1,2)，且A＋B＋C＝π，00，设A(x1，y1)，B(x2，y2)，则A1(t，y1)，B1(t，y2)．
则由根与系数的关系得y1＋y2＝eq \f(－8m,4m2＋3)，y1y2＝eq \f(－8,4m2＋3)，
则y1＋y2＝my1y2.
若存在常数t，使得四边形AA1B1B的对角线交于一定点，
由对称性知，该定点一定在x轴上，设该定点为D(s,0)，则A1，B，D三点共线．
又eq \(A1B,\s\up6(—→))＝(x2－t，y2－y1)，eq \(A1D,\s\up6(—→))＝(s－t，－y1)，则－y1(x2－t)＝(y2－y1)(s－t)
⇒－y1x2＋y2t＝(y2－y1)s
⇒s＝eq \f(－y1x2＋y2t,y2－y1)
＝eq \f(－y1my2＋1＋y2t,y2－y1)
＝eq \f(－my1y2－y1＋ty2,y2－y1)
＝eq \f(－y1＋y2－y1＋ty2,y2－y1)
＝eq \f(t－1y2－2y1,y2－y1).
由s为定值，则eq \f(t－1,2)＝1⇒t＝3，s＝2.
同理，若A，B1，D三点共线，可得t＝3，s＝2.
故存在常数t＝3，使得四边形AA1B1B的对角线交于一定点，该定点为(2,0)．
6．(2023·佛山模拟)已知函数f(x)＝ex－aln(ax＋1)－1，其中a>0，x≥0.
(1)当a＝1时，求函数f(x)的零点；
(2)若函数f(x)≥0恒成立，求a的取值范围．
解 (1)当a＝1时，f(x)＝ex－ln(x＋1)－1，
f′(x)＝ex－eq \f(1,x＋1)，
当x≥0时，ex≥1≥eq \f(1,x＋1)，得f′(x)≥0恒成立．
即可得f(x)在[0，＋∞)上单调递增．
而此时f(0)＝0，
即可得f(x)在[0，＋∞)上仅有1个零点，且该零点为0.
(2)函数f(x)≥0等价于ex－aln(ax＋1)－1≥0，
因为a>0，所以eq \f(ex,a)≥ln(ax＋1)＋eq \f(1,a)，
得ex－ln a≥ln(ax＋1)＋eq \f(1,a)，
所以ex－ln a＋x－ln a≥ln(ax＋1)＋eq \f(1,a)＋x－ln a，
所以ex－ln a＋x－ln a≥ln eq \f(ax＋1,a)＋eq \f(ax＋1,a)＝＋ln eq \f(ax＋1,a)，
构造函数g(x)＝ex＋x，上式等价于g(x－ln a)≥geq \b\lc\(\rc\)(\a\vs4\al\c1(ln \f(ax＋1,a)))，
函数g(x)在定义域内为增函数，
从而可得x－ln a≥ln eq \f(ax＋1,a)成立．
化简可得x≥ln(ax＋1)，等价于ex－ax－1≥0恒成立．
设函数h(x)＝ex－ax－1，易知h(0)＝0，
h′(x)＝ex－a，
当01时，x∈[0，ln a)时，h′(x)＝ex－a