2013年高考理科数学新课标1卷解析版

2013年高考理科数学新课标1卷解析版55)ABUxx或x2Azz()4545b4a045zabi(3i)(abi)abi4aib4ib.ab5)x2y25－a2b2C()21x41x1x32cb2a25b2a214b1ba121a2xeyx.a2s)1St3,3S3,4tt3Stt3,4，2.)33333333BE2，AECE4DExADx24AEDEx3AD5，所以V为AD2R3.2233nS()nnmm(mm1a2,a3d1Sd0aaa5，0mm1m1212mm1a2a(2md(m1)2m15m5.故amm11)A、16+8C、16+161V16.V22416V24822122m()aC，bCC7Cmm2m1mm2m12m2mx2y2＋a2b2FE()x2y2x2y2＋＋x2y2＋x2y2＋9x2y211ab1b222A(x,y)、B(x,y)，运用点差法，所以直线ABka21122xy2212ab222bb2ab22y(x(ab)xbxba0xx2222224a2122b9b9,a18.a2222x22xx0a)ln(x1)x0f(x)x0时，g(x)f(x)x2x，21g(x)2x2，g(0)2a2x0g(x)f(x)ln(x1)，g(x)'g(x)上''x1aa02a0BC的三边长分别为aBCSnnnnnnnnnncanban若b＋c＝2a，a＝n＝n)n1111122nn4a2a132c11，p(abc)bb,ca；3321111aaa5aggga；21故S111226611243aaaa57aa2aaggg631111aa，baca，，Sa；2111122331262662121212SS；215434aaaa97aaa5a35ggg同理，aab11ac，11a，S1S.aS1112288228283121313132ta(1t)b，60，cbabc0torrrrtb(1t)bb0t)0t2.221nS＝a＋ann3n3nn(a；n1n2323a1a1n2aSS2ananan1(2)n1.n1nnn1ann1n25；52255cos.5222)f(x(x4x3)x(a4)x42abf(x22aa8x28b11ab15，令f(x)0，得展开式中x的系数为8，故(xf(x)，的系数为，故3'x36x7x202)(x4x1)05252和2231212CBP60PBA30，由余弦定理得：【答案1）因为，所以，所以007PAPBBA2PBBAcosPBA；2223sinPBAPBsin3cos4sin，sin1500sin(30)03故tan.43sinPBAsin1500sin(30)0BCAA111111BACCC11111OOAOABOCABAA，0OC、、11111OAABABOACACOACOxOAyA(1,0,0)，A(0,3,0)，1111111(3,0)AC(0,3,3)rB(1,0,0)，BBn(x,y,z)BBCC，111100ngrn(BBCCACCC111n11g1105sin.441X1111344C()ggP()；3442222644111P(X1，P(X，P(XX4XP114E(x)506.25X为=1，圆N：(x－1)=9，动圆PMN2222PCCMCP的圆心，圆NN(1,0)PMPN42x2y21(xN；43CP(x,y)PMPN2R2PP2)y4的半径最长时，其方程为(x；2223lxAB；QPRQMr，解得Q(4,0)yk(x4)llxl与x1k2221kkyx2M1k24442k.7472)y4P(x22线l2点Pxk2xaf(0)a4；f(x)''g(x)e(cxdc)g(0)2c4c2f(x)x4x2，g(x)e(2x2)；'x'2xF(x)k1F(x)0得kg(x)f(x)F(x)(ke1)(2x4)F(0)0'x'xlnk,x2，122k0x(2,x)F(x)0x(x,)时F(x)0ke2（1）若1F(x)''11在(2,)F(x)2x2x4x2x(x02111111eF(x)(e1)(2x4)0F(x)(2,)在F(2)0k2'x2eF(2)2ke220k2k1,e2.kg(x)f(x)F(x)kC3交ABEBCEBECE，DBBE为3CDEBDEDBDC，DG是BCBG23BOG60ABEBCECBE30，CFBF.00245cosxtC55sint1yC2C1C与C1245cosxt4)(y5)25xy8x10y160，(x222255sinty8cos10sin160故C1；210xxy2y0（2）Cx22C、C或2y1y212().42C1a1，a22125x,x122