2021雅安高三下学期5月第三次诊断考试数学（文）试题PDF版含答案

这是一份2021雅安高三下学期5月第三次诊断考试数学（文）试题PDF版含答案，文件包含2021届高三四川省雅安市高三第三次诊断考试数学文答案docx、2021届高三四川省雅安市高三第三次诊断考试数学文试题pdf版pdf等2份试卷配套教学资源，其中试卷共13页， 欢迎下载使用。
雅安市高中2018级第三次诊断性考试数学（文科）参考答案及评分标准一.选择题1.C  2.A  3.D   4.B   5.C   6.D   7.C  8.D   9.B  10.C  11.B  12.A 二.填空题13.1   14.   15. 5   16. ②③④三、解答题 17解：∵是与的等差中项∴=∴ ∴........................................................3分∵∴∴  ∴    ....................6分（2）  ∵∴........................................9分 ..................................11分∴.............................................................................................................................12分 18解：（1）由已知可得，40岁及以下采用乘坐成雅高铁出行的有人··························································1分 40岁及以下40岁上合计乘成雅高铁401050不乘成雅高铁203050合计6040100列联表如表:    ········································································4分由列联表中的数据计算可得的观测值····································6分由于，故有的把握认为“采用乘坐成雅高铁出行与年龄有关”.7分（2）采用分层抽样的方法，从“岁(含)以下”的人中抽取人，记为1.2.3，从“岁以上”的人中抽取人，记为a.b.         ··············8分则基本事件为（1,2），（1,3），（1,a），（1,b），（2,3)，（2,a），（2,b），（3,a），（3,b），（a,b）,共10个.··································································10分符合条件的共6种，故抽到2人中恰有一人为40岁以上的概率为6/10=0.6.·········12分  解（1）由题意，可知在等腰梯形中，，∵，分别为，的中点，∴，.                                                        ∴折叠后，，.                                                ∵，∴平面.  ···································4分                                  又平面，∴.  ······································6分                          （2）易知，.∵，∴.                                                又，∴四边形为平行四边形.∴，故.                                                    ∵平面平面，平面平面，且，∴平面.  ·······························································9分 ∴                                            .即三棱锥的体积为. ···········································12分 20.解：（1）①，且过点，②③由①②③解得：，椭圆的标准方程，..........................................4分（2）（i）若的斜率不存在，则此时.........................................5分（ii）若的斜率存在，设，设的方程为：，，.........................6分由韦达定理得：...........................7分则：，............................................8分         =...............................................11分所以：=1........................................................12分另解：（2）当直线AB的斜率为0时，，..........5分当直线AB的斜率不为0时，设直线AB为：，设则：，....................................6分，..........................................7分则：，..............................................8分，...................................................11分所以：.........................................................12分解：（1）由题意，，可得a＝（x>0），...............1分转化为函数T(x)＝与直线y＝a在(0，＋∞)上有两个不同交点...........2分T′(x)＝(x>0)，故当x∈(0,1)时，T′(x)>0；当x∈(1，＋∞)时，T′(x)<0，故T(x)在(0,1)上单调递增，在(1，＋∞)上单调递减，························4分所以T(x)max＝T(1)＝1.又T＝0，故当x∈时，T(x)<0，当x∈时，T(x)>0. ·········5分可得a∈(0,1)．··························································6分另解：则：，...........1分①当时，恒成立，不满足题意；······.3分 ②当时，单调递减，则，当.................................5分综上：...........................................................6分(2)证明：　h′(x)＝－a，   因为x1，x2是ln x－ax＋1＝0的两个根，故ln x1－ax1＋1＝0，  ln x2－ax2＋1＝0⇒   a＝，··················8分要证h′(x1x2)<1－a，只需证x1x2>1，  即证ln x1＋ln x2>0，  即证(ax1－1)＋(ax2－1)>0，即只需证明 a>成立，即证>.··························9分不妨设0<x1<x2，故ln <＝.( * )···························10分令t＝∈(0,1)，φ(t)＝ln t－，······································11分φ′(t)＝－＝>0，则h(t)在(0,1)上单调递增，则φ(t)< φ(1)＝0，故（*）式成立，即要证不等式得证．······································12分22. 解 ：（1）直线l的直角坐标方程为：························2分曲线C的极坐标方程为：，即，化为直角坐标方程：.将曲线C上所有点的横坐标缩短为原来的一半，纵坐标不变，得到曲线：. ···············································5分（2）直线的极坐标方程为，展开可得：.可得直角坐标方程：.可得参数方程：（为参数）.  ····························7分代入曲线的直角坐标方程可得：.解得，.∴. ··········································10分23.解：（1）或解出或无解,  所以，原不等式的解集为[0，1]··························5分另解：（1）当时，等价于，则∴或无解综上，原不等式的解集为[0，1]·················································.5分（2）当时，，因为，所以恒成立，即恒成立，所以满足的解集为；而，当时，，当时，，作出的图像如下图所示，要使的解集为，则需或，解得或；综上可得：a的取值范围是.  ····10 分