1、1,2,3A=|33,By yxxA=AB=31,33,60,3,61yx=ee2xxy+=tanyx=ee2xxy=p12x q0mxpqm3m 3m 5m 5m()f x0 x 2()32f xxx=+1,3x()f x214214山东省枣庄2023-2024高三上学期期中考试数学试题(4,3)Atan()2+=tan(3)=12125252nannS0na 610S=1870S=24S=90135150180sin 2()sin2xf xx=+323 345 3854abcd2e111aacbd=22()()acbd+89101121(sin15cos15)2=222sin 22.5co
2、s 22.52=1cos28 cos32cos62 cos582=3(tan103)cos502=xyR，()()()1f xyf xf y+=+0 x()1f x (1)1f=(0)0f=()f x(1,1)()f xR2(1)(2)(2023)2023fff+=x，y223xyxy+=2xy+2 3xy+226xy+224xy+a()(ln2)f xxxax=1x2x12xx104a122xx+1()0f x21()2f x 2()1 log(2)f xx=+ABCABCabc，23A=7a=1b=ABC1234(0,)+()f x()e1xfx(ln)efxx+1()2 e2f=x|12
3、1Ax axa=+|3Bx x=5x 4a=ABABa 1 2 3 4 5 6 2()(3)3f xxaxa=+，aRx()0f x 4,)x+()9f x a()3sin2cos21f xxx=+()f xABCABCabc()2f A=2 3a=ABC()f xR0 x 2()exf xx=()f x()f x22()31(5)(3)40axxfaxfaxa x+xRanannS222nnna San=+na*nN2 2namm()ln1f xxmx=+()(e1)xg xx=()f x0mx()()f xg xm(2,032369e4a=|57Axx=|3Bx x=5x|57ABxx=2
4、a 211aa+A=AB2a 211aa+A AB2132aa 152aa+2a=4a 高三数学试题参考答案a|2a a 4a()(3)()f xxxa=3a()0f x(,3)a3a=()0f x 3a()0f x(3,)a()(3)()f xxxa=4,)x+()9f x 93axx+999332(3)39333xxxxxx+=+=933xx=6x=9a a|9a a()3sin2cos21f xxx=+312(sin2cos2)122xx=+2sin(2)16x=+32 22 262kxk+k Z263kxk+k Z2,63kk+k Z()2f A=，1sin(2)62A+=(0,)A5
5、266A+=3A=2 3a=2222cosabcbcA=+2212bcbc+=2212212bcbcbc=+12bc 2 3bc=13sin3 324SbcAbc=ABC3 3()f xR0 x 2()exf xx=0 x 时0 x 22()()e e)xxf xfxxx=22e,0()e,0 xxxxf xxx=()f xR22(31)31(5)5f axxaxxfaxax+xR22(31)31(5)5f axxaxxf axax+xR()()h xf xx=+()h xR2315axxax xR2(3)40axa x+xR0a 0a 20(1)36aa+19aa(1,9)=1n221122
6、aa=+12a=2n 222nnna San=+，可得2112()()2nnnnnSSSSSn=+2212nnSSn=222222222121321()()()nnnSSSSSSSSnn+=+=+2(1)nSnnn n=+=+1(1)(1)nnnaSSn nnn=+.5 分12a=(1)(1)nan nnn=+1(1)(2)(1)nannn n+=+故1212211nnnnaannnn+=+21(11)2(2)(2)(11)nnnnnnnnnn+=+222(1)21222(2)(11)nnnnnnnnn+=+222(11)22(2)(11)nnnnnnn+=+.8 分2222222(11)21
7、2(2)nnnnnnn+=+=+22112nnn+10nnaa+na2 2nam12 2am2m()f x(0)+，1()fxmx=+0m()0f x()f x0m 1(0,)xm()0f x()f x1(,)xm+()0f x()f x1xm=()f x1ln()0m=1m=x()()f xg xln11exxmx+(0)+，ln1()exxxx+=22eln()xxxxx+=2()elnxq xxx=+21()(+2)e0 xq xxxx=+()q x1()02q(1)0q()q x0 x0200eln0 xxx+=00ln0000lnelnexxxxxx=()exh xx=00()(ln)h xhx=()exh xx=(0)+，00lnxx=()x0(0,)x0(,)x+0000000ln111()()e1xxxxxxxx+=+=m(,0
