{#{QQABRYaQogioAAJAAQgCEQGwCEEQkgGAAQgGgBAAIAABCQFABAA=}#}{#{QQABRYaQogioAAJAAQgCEQGwCEEQkgGAAQgGgBAAIAABCQFABAA=}#}{#{QQABRYaQogioAAJAAQgCEQGwCEEQkgGAAQgGgBAAIAABCQFABAA=}#}{#{QQABRYaQogioAAJAAQgCEQGwCEEQkgGAAQgGgBAAIAABCQFABAA=}#}盐城市2025届高三年级第一学期期中考试数学参考答案1.C2.C3.B4.A5.D6.D7.B8.B9.ACD10.AB11.ABD12.0,213.1214.8tan2n15.解：⑴因为函数f(x)为偶函数，所以xR，都有f(x)f(x)，················2分即exkexsinxexkexsinx，即xR，都有k1exexsinx0，···············································4分所以k1.··························································································6分⑵当k0时，f(x)exsinx，f(x)f(x)exsinxexsin(x)(exex)sinx，所以(exex)sinx0.············································································9分当x0时，显然不合题意；当x0,2时，exex0，则sinx0，此时，x0,；当x2，0时，exex0，则sinx0，此时，x,0，·············12分综上，不等式f(x)f(x)0的解集为,00,.·····························13分16.解：⑴因为f(x)sinxcosx2sin(x)，所以f(A)2sin(A)2，44所以sin(A)1，·············································································4分4又因为A0,，所以A.································································6分41212⑵因为ABBCBABCBC，所以BABCBC，221法一：所以BA在BC方向上的投影向量为BC，2过点A作边BC的垂线，垂足为H，则H是边BC的中点，所以ABAC，即bc，······································································9分在ABC中，因为a2b2c22bccosA，即12b22b2，第1页共6页{#{QQABRYaQogioAAJAAQgCEQGwCEEQkgGAAQgGgBAAIAABCQFABAA=}#}22所以b2，··············································································12分212221所以SbcsinAb.················································15分ABC2441a2c2b2a2c2b21法二：所以cacosB即ca，22ac22即a2c2b21即bc，····································································9分在ABC中，因为a2b2c22bccosA，即12b22b2，22所以b2··················································································12分212221所以SbcsinAb.················································15分ABC24417．解：⑴设ADx，119则SABACsinBAC63sin3，BAC2232113xSABADsinBAD6xsin，BAD2262113xSACADsinCAD3xsin，CAD2264由SBACSBADSCAD············································································3分933得3xx，224得x23，即AD的长为23.································································5分⑵在VABC中，由AB6，AC3，BAC及余弦定理可得3BC2AB2AC22ABACcosBAC6232263cos27，3得BC33，·······················································································7分又AB6，AC3，得AB2AC2BC2，得ACB，21而CADBAC，∴ADCACDCAD，263又VPCD为等腰三角形，∴VPCD为等边三角形，·······································9分第2页共6页{#{QQABRYaQogioAAJAAQgCEQGwCEEQkgGAAQgGgBAAIAABCQFABAA=}#}在VACD中，AD23，AC3，CAD，由余弦定理可得6CD2AD2AC22ADACcosDAC(23)2322233cos3，6得CD3，······················································································11分则PD3，BD23，设ABP，则DBP，BPD，66PDBD在△DBP中，由正弦定理得，sinDBPsinBPD323则，···································································13分sin()sin()661313得2sin()sin()，则2(cossin)cossin，662222sin3得cos33sin，得tan，cos93即tanABP的值为．·······································································15分92218.⑴证：由(an1)4Sn，得(an11)4Sn1(n2)，2222所以(an1)(an11)4an，即(an1)(an11)，得(anan12)(anan1)0，··································································2分因数列an各项为正，所以anan10，于是anan12(n2)，所以数列an是公差为2的等差数列，·······················································3分2又(a11)4S14a1，所以a11，所以an12(n1)2n1.·····································································4分⑵解：由⑴知，an2n1，1所以T(12n1)nnnn2，n22n21由T2n2(n1)2，即nn22n2(n1)2，所以2，nn2n21令c2，················································································6分nn2n112n21n(2n11)(n1)(2n21)2n2(n1)1所以cc，n1nn1nn(n1)n(n1)当n1时，cn1cn，即c2c1；第3页共6页{#{QQABRYaQogioAAJAAQgCEQGwCEEQkgGAAQgGgBAAIAABCQFABAA=}#}当n2时，cn1cn，即c2c3；当n3时，cn1cn；············································································8分201所以c21，故的最大值为1.··········································9分222n12n1⑶解：由⑴得bnbn1bn22，所以bn1bn2bn32，1于是bn34bn，又b1b21，b1b2b32，所以b322b1，·······················10分由P3k(b1b4Lb3k2)(b2b5Lb3k1)(b3b6Lb3k)1(14k)4所以P4(bbLb)4[](4k1)，3k143k21434n所以当n3k时，P(431)；···························································12分n3由P3k1(b4b7Lb3k1)(b2b5Lb3k1)(b3b6Lb3k)b1，k1(14)7k4因b44b14，所以P7(bbLb)1714，3k1253k11433P11也适合上式，7n14所以当n3k1时，P43；·····················································14分n33由P3k2(b4b7Lb3k1)(b5b8Lb3k2)(b3b6Lb3k)b1b2，k1(14)10k4又b54b24，所以P10(bbLb)210[]24，3k2143k21433P22也适合上式，10n24所以当n3k2时，P43；···················································16分n334n(431),n3k,kN*3n1734综上所述，Pn4,n3k1,kN.···········································17分3310n2443,n3k2,kN3319.解：⑴f(x)(x1)ex，················································