2014年高考真题理科数学

2014年普通高等学校招生全国统一考试理科（新课标卷二Ⅱ）第Ⅰ卷5.3xx|xMN01｝)2｝2｝12｝2izz（12z，zz）121-5|10|6ab=(12325-i-4-i)51)25215））1727510192732（）4=0123567xy03y0z2xy设x3xy0（）8323x设Fy2FC于O）339363329484BCNABAC，11111111则与）2110252fxfxxm3sinx0x0fxmm的2220）,66,,44,,22,,14,第Ⅱ卷第题第.第题第.a10x7x)sinx22sinxfxf20若fx10xfx在y1（x,1xx2200.aa，aa1.n1n1n1aa2nn1113a2n.aa12E为.；为.31234567tyyt.nttyyˆbiiˆ，ayi1ntt2ii1yxab21ab0M是CMF与x设F,F212222MF与C13C45FN在y.1e2xfx=exxfxgx0b求f2xxx0gx1.414221.4143）.1P是OA与O，D为O；PB2xC，20,.C3x2D在CC在Dl:yD.1axa(a0)fx=xfx≥2；f35.aDAD3A9B4BA6CDDCC12）1））11aa1得aan1nn12n21313又aa322221n1n31nanaa。22n2n12.a31nn1113123.nnn13123nn111111311-.n-12n31aa1an3122111aaaa213.123n交Q为.又EBAx31系xyzD0,3,0,E(0,,).AE2231(0,,).22C(,(,则设B(m(x,y,z)设n13ynAC0,即则131AE0.yn2213可取n(,1,3).m1又n为平面DAE的法向量。21由题设cos(n,n)，即212313，解得m.34m2221因为E为PD的中点，所以三棱锥E-ACD的高为。三棱锥E-ACD的体积211V332228313.27t7y77tt)21t17tt)(yy)11t1（3（21（17tt)(yy)14b110.5t1287tt)21，t1ay0.54.yt.将y0.592.36.8.）b2M(c,),2bac22cab2ac1c,2a2abacb将222212故C.FFMFMFyyDO为∥与1122b24MFa1b24a①5FN2FN由设得。111N(x,y)1y0113xc,2(cx)c2112y2y1119c2114ab2C2。9(a4a)121cab24a24a2a7,b4a282，ab27故.f'(x)eex020x）=xf(x)(,)在g(x)f(2x)bf(x)eeb(ee)(8b4)x2x2xxx=2ee2xb(ee)(4b2)g'(x)2xxx=2(ee2)(eeb2)xxxx=b2g'(x)0x0g(x)(,)在gxg(x)0；0xb1bb)时b22eeb22xxxxb1bb)g'(x)g0g(x)2而b3g2b2(2b22.3823g422212当；2＞；32b1b1bb)ln242当，32222g2=0，182228＜ln2）BEEC。PA2。。，ADDE2PB2.(x1)y1(0y1)22C.C1cos,xtsint,y0txt）D(1tt)CG1CDt与ttt3.33(,)22)33故D