南京市、盐城市2020届高三年级数学第二次模拟考试参考答案

高三数学答案第1页共11页南京市､盐城市2020届高三年级第二次模拟考试数学参考答案及评分标准说明：1．本解答给出的解法供参考．如果考生的解法与本解答不同，可根据试题的主要考查内容比照评分标准制订相应的评分细则．2．对计算题，当考生的解答在某一步出现错误时，如果后续部分的解答未改变该题的内容和难度，可视影响的程度决定给分，但不得超过该部分正确解答应得分数的一半；如果后续部分的解答有较严重的错误，就不再给分．3．解答右端所注分数，表示考生正确做到这一步应得的累加分数．4．只给整数分数，填空题不给中间分数．一、填空题（本大题共14小题，每小题5分，计70分.不需写出解答过程，请把答案写在答题纸的指定位置上）1．{1，3}2．53．－144．3255．126．07．π28．65π9．510．2211．812．±25513．214．(1，12＋ln2)二、解答题（本大题共6小题，计90分.解答应写出必要的文字说明，证明过程或演算步骤，请把答案写在答题纸的指定区域内）15．(本小题满分14分)证明：（1）因为点D，E分别为AB，BC的中点，所以DE∥AC．············································································2分因为AC平面PDE，DE平面PDE，所以AC∥平面PDE．···································································4分（2）因为点D，E分别为AB，BC的中点，所以DE＝12AC．又因为AC＝2，所以DE＝1，因为PD＝2，PE＝3，所以PD2＝PE2＋DE2，因此在△PDE中，PE⊥DE．··························································8分又平面PDE⊥平面ABC，且平面PDE∩平面ABC＝DE，PE平面PDE，所以PE⊥平面ABC，································································12分又因为PE平面PBC，所以平面PBC⊥平面ABC．·························································14分高三数学答案第2页共11页16．(本小题满分14分)解：（1）因为a＝bcosC＋csinB，由asinA＝bsinB＝csinC，得sinA＝sinBcosC＋sinCsinB．·····································2分又因为sinA＝sin[π－(B＋C)]＝sin(B＋C)＝sinBcosC＋cosBsinC，所以sinBcosC＋cosBsinC＝sinBcosC＋sinCsinB，即cosBsinC＝sinCsinB．··········································································4分因为0＜C＜π，所以sinC≠0，所以sinB＝cosB．又0＜B＜π，所以sinB≠0，从而cosB≠0，所以tanB＝1，所以B＝π4．·························································································6分（2）因为AD是∠BAC的平分线，设∠BAD＝θ，所以A＝2θ，因为cosA＝－725，所以cos2θ＝cosA＝－725，即2cos2θ－1＝－725，所以cos2θ＝925，因为0＜A＜π，所以0＜θ＜π2，所以cosθ＝35，所以sinθ＝1－cos2θ＝45．在△ABD中，sin∠ADB＝sin(B＋θ)＝sin(π4＋θ)＝sinπ4cosθ＋cosπ4sinθ＝22×(35＋45)＝7210．·············································8分由ADsinB＝ABsin∠ADB，所以AB＝AD·sin∠ADBsinB＝177×7210×2＝175．················10分在△ABC中，sinA＝1－cos2A＝2425，所以sinC＝sin(A＋B)＝sinAcosB＋cosAsinB＝22×(2425－725)＝17250．··············12分由bsinB＝csinC，所以b＝c·sinBsinC＝175×2217250＝5．··············································14分17．(本小题满分14分)解：（1）连接CD，因为BD与圆C相切，切点为D，所以△BCD为直角三角形．因为∠CBD＝θ，且圆形小岛的半径为1千米，所以DB＝1tanθ，BC＝1sinθ．因为岸边上的点A与小岛圆心C相距3千米，所以AB＝AC－BC＝3－1sinθ．·····2分又因为BE与圆C相切，所以BE＝DB＝1tanθ，优弧︵DE所对圆心角为2π－(π－2θ)＝π＋2θ，所以优弧︵DE长l为π＋2θ，·····························································4分高三数学答案第3页共11页所以f(θ)＝AB＋BD＋BE＋l＝3－1sinθ＋1tanθ＋1tanθ＋π＋2θ＝3＋π＋2θ＋2cosθ－1sinθ．······························································6分因为0＜AB＜2，所以0＜3－1sinθ＜2，解得13＜sinθ＜1，所以sinθ的取值范围为(13，1)．································································8分（2）由f(θ)＝3＋π＋2θ＋2cosθ－1sinθ，得f'(θ)＝－2+cosθsin2θ＋2＝cosθ(1－2cosθ)sin2θ．·10分令f'(θ)＝0，解得cosθ＝12，因为θ为锐角，所以θ＝π3．····························12分设sinθ0＝13，θ0为锐角，则0＜θ0＜π3．当θ∈(θ0，π3)时，f'(θ)＜0，则f(θ)在(θ0，π3)单调递减；当θ∈(π3，π2)时，f'(θ)＞0，则f(θ)在(π3，π2)单调递增，所以f(θ)在θ＝π3时取得最小值．答：当θ＝π3时，栈道总长度最短．··························································14分18．(本小题满分16分)解：（1）记椭圆C的焦距为2c．因为椭圆C的离心率为12，所以ca＝12．因为椭圆C过点(0，3)，所以b＝3．因为a2－c2＝b2，解得c＝1，a＝2，故椭圆C的方程为x24＋y23＝1．··································································2分（2）①因为点B为椭圆C的上顶点，所心B点坐标为(0，3)．因为O为△BMN的垂心，所以BO⊥MN，即MN⊥y轴．由椭圆的对称性可知M，N两点关于y轴对称．··········································4分不妨设M(x0，y0)，则N(－x0，y0)，其中－3＜y0＜3．又因为MO⊥BN，所以→MO·→BN＝0，即(－x0，－y0)·(－x0，y0－3)＝0，得x20－y20＋3y0＝0．·············································································6分又点M(x0，y0)在椭圆上，则x024＋y023＝1．高三数学答案第4页共11页由x20－y20＋3y0＝0，x024＋y023＝1，解得y0＝－473或y0＝3(舍去)，此时|x0|＝2733．故MN＝2|x0|＝4733，即线段MN的长为4733．···········································8分②方法1设B(m，n)，记线段MN中点为D．因为O为△BMN的重心，所以→BO＝2→OD，则点D的坐标为(－m2，－n2)．······10分若n＝0，则|m|＝2，此时直线MN与x轴垂直，故原点O到直线MN的距离为|m2|，即为1．若n≠0，此时直线MN的斜率存在．设M(x1，y1)，N(x2，y2)，则x1＋x2＝－m，y1＋y2＝－n．又x124＋y123＝1，x224＋y223＝1，两式相减得(x1＋x2)(x1－x2)4＋(y1＋y2)(y1－y2)3＝0，可得kMN＝y1－y2x1－x2＝－3m4n．···································································12分故直线MN的方程为y＝－3m4n(x＋m2)－n2，即6mx＋8ny＋3m2＋4n2＝0，则点O到直线MN的距离为d＝|3m2＋4n2|36m2＋64n2．将m24＋n23＝1，代入得d＝3n2＋9．························································14分因为0＜n2≤3，所以dmin＝32．又32＜1，故原点O到直线MN距离的最小值为32．································16分方法2设M(x1，y1)，N(x2，y2)，B(x3，y3)．因为O为△BMN的重心，所以x1＋x2＋x3＝0，y1＋y2＋y3＝0，则x3＝－(x1＋x2)，y3＝－(y1＋y2)．··························································10分因为x234＋y233＝1，所以(x1＋x2)24＋(y1＋y2)23＝1．将x124＋y123＝1，x224＋y223＝1，代入得x1x24＋y1y23＝－12．··································12分若直线MN的斜率不存在，则线段MN的中点在x轴上，从而B点位于长轴的顶点处，由于OB＝2，所以此时原点O到直线MN的距离为1．若直线MN的斜率存在，设为k，则其方程为y＝kx＋n．高三数学答案第5页共11页由y＝kx＋n，x24＋y23＝1，消去y得，(3＋4k2)x2＋8knx＋4n2－12＝0．(*)则△＝(8kn)2－4(3＋4k2)(4n2－12)＞0，即3＋4k2＞n2，由根与系数关系可得x1＋x2＝－8kn3＋4k2，x1x2＝4n2－123＋4k2，则y1y2＝(kx1＋n)(kx2＋n)＝k2x1x2＋kn(x1＋x2)＋n2＝3n2－12k23＋4k2，代入x1x24＋y1y23＝－12，得14×4n2－123＋4k2＋13×3n2－12k23＋4k2＝－12，即n2＝k2＋34．······14分又3＋4k2＞n2，于是3＋4k2＞k2＋34，即3k2＋94＞0恒成立，因此k∈R．原点(0，0)到直线MN的距离为d＝|n|k2＋1＝k2＋34k2＋1＝1－14(k2＋1)．因为k2≥0，所以当k＝0时，dmin＝32，又32＜1，故原点O到直线MN距离的最小值为32．································16分19．(本小题满分16分)解：（1）因为h(x)＝f(x)x－g(x)＝x2－x－(a－16)－alnx，所以h'(x)＝2x－1－ax＝2x2－x－ax，令h'(x)＝0，得2x2－x－a＝0．因为函数h'(x)在[52，4]上存在零点，即y＝2x2