2023年高考诊断性测试数学参考答案及评分标准一、选择题ACBBADBC二、选择题9.AC10.ACD11.ABD12.ABD三、填空题4314811.6012.13.2xy1014.5，755四、解答题17.解：（1）设数列an的公比为q，因为3a1,a3,5a2成等差数列，2所以3a15a1q2a1q，······································································1分1即35qq22，解得q3或q，2因为an各项均为正数，所以q0，所以.·····································2分4a131由Sa55，得5532a，解得a1.···························4分433111nn11所以aan133.·········································································5分n1（2）由（1）知，bnn3.··································································6分012n1则Tnn1323333123n所以3Tnn1323333，···········································7分01nn1两式相减可得2Tnn3333，·······································8分13nn3n，132n11整理可得T3n.·······························································10分n4418.解：（1）因为c2bcosAb，由正弦定理得sinCBAB2sincossin.···2分又ABCπ，所以高三数学答案（第1页，共7页）sin(ABBAABABABB)2sincossincoscossinsin()sin.····4分ππππ因为ABC为锐角三角形，所以AB(0,),(0,)，AB(,)，2222ππ又yxsin在(,)上单调递增，所以ABB，即AB2.···············6分22（2）由（1）可知，，所以在ABD中，ABCBAD，ADAB21由正弦定理得：，所以ADBD，sinBBBsin(π2)sin2cosB1sinB所以SABADsinBtanB.····································9分ABD2cosBπππ又因为ABC为锐角三角形，所以0B，02B，0π3B，222ππ解得B，···············································································11分6433所以tanB(,1)，即ABD面积的取值范围为(,1).·······················12分33u19.解：（1）Logistic非线性回归模型y拟合效果更好.·····················1分1eabt从散点图看，散点更均匀地分布在该模型拟合曲线附近；从残差图看，该模型下的残差更均匀地集中在以残差为0的直线为对称轴的水平带状区域内.····················3分uu（2）将y两边取对数得ln(1)abt,···································5分1eabty20(ww)(tt)ii138.32则ˆi1,ˆ,7分b200.208b0.208·····················2665()ttii1aˆw(bˆ)t1.6080.20810.50.576.····································9分高三数学答案（第2页，共7页）12.5所以y关于t的经验回归方程为y.··································10分1e0.5760.208t12.512.5当t22时，体长y12.28mm.·························12分1e0.5760.208221e4z20.解：（1）证明：取BC中点E，连接BD,,DEVE，V因为ABCD为菱形，且BAD60，所以BCD为等边三角形，故DEBC.····1分DC又在等边三角形中，，2分VBCVEBC·······OyDEVEE，所以BC面DEV.··········4分EVD面，所以BCVD；·············5分AxB（2）由，，可得DEV就是二面角ABCV的平面角，所以DEV60，··········································································6分在DEV中，VEDE3，所以为边长为3的等边三角形，由（1）可知，面底面ABCD，取DE中点O，以O为坐标原点，以DA,,OEOV所在的方向为x,,yz轴的正向，建立空间直角坐标系Oxyz，····7分33333在VOE中，OE,OV，可得A(2,,0)，B(1,,0)，C(1,,0)，2222233333V(0,0,)，故CB(2,0,0)，CV(1,,)，AV(2,,).·········8分22222设n(,,)xyz为平面VBC的一个法向量，则有20x33，令y3，则z1，得n(0,3,1)，·················10分xyz022设直线VA与平面VBC所成角为，|nAV|337则有sin|cosn,AV|，|n||AV|271437故直线与平面所成角的正弦值为.······································12分14高三数学答案（第3页，共7页）|PF|221.解：（1）设P(,)xy，由题意，|x22|2(xy2)222因为|PF|(x2)22y，所以，···················2分|x22|22xy22即(x2)22y|x22|，两边平方并整理得1.242故点P的轨迹C的方程为.··················································4分1（2）设直线l方程为ykx1(≤k≤2)，2xy22122联立42，消y并整理得，(2k1)x4kx20，ykx14k2设A(x,y),B(x,y)，则xx，xx，·············5分11221221k21221k2221k又yyk(xx)2，可得线段AB中点坐标为(,)，121221k22kk22121112k所以线段中垂线的方程为yx()，2k221k2k1k令y0，可得N(,0)，··························································6分21k21对于直线ykx1，令y0，可得M(,0)，k11kk2所以|MN|||.··················································7分k2k221k(2k1)高三数学答案（第4页，共7页）4kk8212又|AB|1k2|xx|1k2()28k22，122k212k212k21······································································································9分|AB|2k8k226所以2，10分28(k1)214··························|MN|1k2k1566令tk21[,5]，则y8(k21)148t14，4kt2165因为yt814在[,5]上单调递增，t422|AB|44所以y[5,170]，故[5,170].····························12分55|MN|55122.解：（1）f(x)acosx1，················································1分(x1)2因为0为fx()的一个极值点，所以fa(0)20，所以a2.·············2分（2）①当10x时，fx()2110，所以单减，所以对x(1,0]，有f(x)f(0)1，此时函数fx()无零点；··········································3分2当0x时，f(x)2sinx，fx()在(0,)上单调递减，又2(x1)322f(0)20，f()20，由零点存在定理，存在x(0,)，使得02(1)322fx()0，且当xx(0,)时，fx()0，即fx()单调递增，当xx(,)时，002fx()0，即单调递减.又因为f(0)0，所以xx(0,0]，fx()0，在(0,x0)单增；因为高三数学答案（第5页，共7页）1fx()0，f()10,所以存在xx(,)，当x(,)xx时，010012(1)222fx()0，fx()单增，当xx(,)时，fx()0，单减.12所以，当xx(0,)时，单增，f(x)f(0)1；当xx(,)时，单减，1121f(x)f()20，此时fx()在(0,)上无零点；···················5分221221当x(,)时，f(x)2cosx10，所以fx()在(,)单减，又2(x1)221f()0，f()00，由零点存在定理，函数fx()在(,)上存在唯212一零点；································································································6分1当x时，f(x)2sinxx210，此时函数无零点；x1综上，fx()在区间(1,)上存在唯一零点.···········································7分1②因为f()210，由（1）中fx()在(0,)上的单调性分析，知4(1)224x，所以fx()在(0,)单增，所以对x(0,)，有f(x)f(0)1，即14441112sinxx1，所以sinxx(1).··································8分x121x111111111令xk(≥2)，则sin()···9分k2k22k2k21k21k2kkk1n111111111所以sin()()()···················10分2k2k2334nn12n1高三数学答案（第6页，共7页）111111因为sinxx，x(0,