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。试卷主要包含了O是坐标原点，椭圆C等内容，欢迎下载使用。
问题：已知锐角△ABC的内角A，B，C的对边分别为a，b，c，且满足________．
(1)求A；
(2)若a＝4，求b＋c的取值范围．
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2．(2023·襄阳四中模拟)设正项数列{an}的前n项和为Sn，已知a3＝5，且aeq \\al(2,n＋1)＝4Sn＋4n＋1.
(1)求{an}的通项公式；
(2)若bn＝eq \f(－1n·2n,anan＋1)，求数列{bn}的前n项和Tn.
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3．(2023·天津模拟)O是坐标原点，椭圆C：eq \f(x2,a2)＋eq \f(y2,b2)＝1(a>b>0)的左、右焦点分别为F1，F2，点M在椭圆上，当△MF1F2的面积最大时，∠F1MF2＝120°且最大面积为2eq \r(3).
(1)求椭圆C的标准方程；
(2)直线l：x＝2与椭圆C在第一象限交于点N，点A是第四象限内的点且在椭圆C上，线段AB被直线l垂直平分，直线NB与椭圆交于另一点D，求证：ON∥AD.
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4．(2023·浙江联考)如图1，在矩形ABCD中，AB＝2，BC＝1，E是DC的中点，将△DAE沿AE折起，使得点D到达点P的位置，且PB＝PC，如图2所示．F是棱PB上的一点．
(1)若F是棱PB的中点，求证：CF∥平面PAE；
(2)是否存在点F，使得平面AEF与平面AEC的夹角的余弦值为eq \f(4\r(17),17)？若存在，则求出eq \f(PF,FB)的值；若不存在，请说明理由．
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5．(2023·长沙模拟)盲盒是指消费者不能提前得知具体产品款式的玩具盒子，具有随机属性．某品牌推出2款盲盒套餐，A款盲盒套餐包含4款不同单品，且必包含隐藏款X；B款盲盒套餐包含2款不同单品，有50%的可能性出现隐藏款X.为避免盲目购买与黄牛囤积，每人每天只能购买1件盲盒套餐，开售第二日，销售门店对80名购买了套餐的消费者进行了问卷调查，得到如下数据：
(1)依据小概率值α＝0.01的独立性检验，能否认为A，B款盲盒套餐的选择与年龄有关联？
(2)甲、乙、丙三人每人购买1件B款盲盒套餐，记随机变量ξ为其中隐藏款X的个数，求ξ的分布列和均值；
(3)某消费者在开售首日与次日分别购买了A款盲盒套餐与B款盲盒套餐各1件，并将6件单品全部打乱放在一起，从中随机抽取1件打开后发现为隐藏款X，求该隐藏款来自于B款盲盒套餐的概率．
附：χ2＝eq \f(nad－bc2,a＋bc＋da＋cb＋d)，其中n＝a＋b＋c＋d.
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6．已知函数f(x)＝ex－ax＋sin x－1.
(1)当a＝2时，求函数f(x)的单调区间；
(2)当1≤af′(0)＝2－a>0，
∴f(x)在(0，＋∞)上单调递增，
∴f(x)>f(0)＝0，此时f(x)在(0，＋∞)上无零点．
②当x∈(－∞，－π]时，－ax≥π，
有f(x)＝ex－ax＋sin x－1≥ex＋π＋sin x－1>0，
此时f(x)在(－∞，－π]上无零点．
③当x∈(－π，0)时，sin x0，
∴f′(x)在(－π，0)上单调递增，
又f′(0)＝2－a>0，f′(－π)＝e－π－1－a0，f(x0)