7}，故選C.2.B3.D底面圓半徑為·、i2，所以圓錐SO的側(cè)面積為故選D.4.C解得m且m≠-2，即m的取值范圍為故選C.5.C726或63，故選C.6.A【解析】因?yàn)楹瘮?shù)f，則f又因?yàn)楹瘮?shù)f(x)是定義在(-∞,0)(0,+∞)上的奇函數(shù)，則f(x)+f(-x)=0，即解得a=1，則f所以f故選A.7.D8.C單調(diào)遞減，在(ln3,+∞)上單調(diào)遞增，所以函數(shù)g(t)在t=ln3處取得最小值3-3ln3，即函數(shù)f(x)在(0，+∞)上的最小值為3-3ln3，所以a≥3-3ln3，即a的最小值為3-3ln3，故選C.03-ex)+a≥0在(0，+∞)上有解，即a≥-lnx3-x(3-exx是增函數(shù).因?yàn)閑e-3<0，e所以由函數(shù)零點(diǎn)存在定理，設(shè)f,(x0)=0，則ex0- f(x)在(0，x0)上單調(diào)遞減，在(x0，+∞)上單調(diào)遞增，所以f(x)min=0f=-3lnx0-3x0+x0ex0=-3lnx0+xln3+3x0-數(shù)f(x)在(0，+∞)上的最小值為3-3ln3，所以a≥3-3ln3，即a的最小值為3-3ln3，故選C.9.AC【解析】對(duì)于A選項(xiàng)，因?yàn)閦4+1>0，x>y，所以成立，所以A選項(xiàng)正確；對(duì)于B選項(xiàng)，當(dāng)z2=0時(shí)，xz2=yz2，所以B選項(xiàng)不正確；對(duì)于C選項(xiàng)，因?yàn)閤>y，所以x-y>0，所以x-y，當(dāng)且僅當(dāng)x-y=1時(shí)，等號(hào)成立，所以C選項(xiàng)正確；對(duì)于故選AC.10.BCD【解析】因?yàn)閒(4x-1)為偶函數(shù)，則f(4x-1)=f(-4x-1)，則f(x-1)=f(-x-1)，所以函數(shù)f(x)的圖象關(guān)于直線x=-1對(duì)稱，所以A選項(xiàng)不正確；因?yàn)閒(x-1)=f(-x-1)，所以f,(x-1)=-f,(-x-1)，即f,(x-1)+f,(-x-1)=0，所以函數(shù)f,(x)的圖象關(guān)于點(diǎn)(-1,0)對(duì)稱，所以B選項(xiàng)正確；若f(x+4)+f(-x+4)=0，則函數(shù)f(x)的圖象關(guān)于點(diǎn)(4,0)對(duì)稱，又因?yàn)楹瘮?shù)f(x)的圖象關(guān)于直線x=-1對(duì)稱，所以函數(shù)f(x)的一個(gè)周期為20，所以f(-5)=f(15)，所以C選項(xiàng)正確；若f,(x+4)+f,(-x+4)=0，則函數(shù)f,(x)的圖象關(guān)于點(diǎn)(4,0)對(duì)稱，又因?yàn)楹瘮?shù)f,(x)的圖象關(guān)于點(diǎn)(-1,0)對(duì)稱，所以函數(shù)f,(x)的一個(gè)周期為10，所以f,(-5)=f,(5)，所以D選項(xiàng)正確，故選BCD. 所以A不正確；f cosα+sinα3cosα(π72scosα+sinα3cosα(π72s【解析】因?yàn)?-1)2+(-)2<4，所以點(diǎn)P(-1,-)在圓內(nèi).設(shè)O為坐標(biāo)原點(diǎn)，由題意得，是以2為首項(xiàng)，1為公差的等差數(shù)列，所以2nan=2+(n-1)=n+1，所以an14.-18【解析】在△ABC中，由cos上ACB可得sin上ACB又由正弦定理可得cos上ABC，整理得BC2-4BC-12=0，解得BC=6（負(fù)值舍去）.因?yàn)镺為△ABC所在平面內(nèi)一點(diǎn)，且OA=OB=OCcos上ABC，整理得BC2-4BC-12=0，解得BC=6（負(fù)值舍去）.因?yàn)镺為△ABC所在平15.（1）證明見(jiàn)解析（5分2【解析】（1）因?yàn)锳B=AC，上BAC=，D為BC的中點(diǎn)，所以AD丄BC，·······2分【另解】（1）因?yàn)锳B=AC，上BAC=，D為BC的中點(diǎn)，所以AD丄BC，·······2分因?yàn)镻A丄底面ABC，AB平面ABC，AC平面ABC，所以PA丄AB，PA丄AC，22+AC2，所以PB=PC.又因?yàn)镈為BC的中點(diǎn)，所以PD丄BC.···························································4分 因?yàn)镻A丄底面ABC，上BAC，所以AB，AC，AP兩兩垂直，故以A為坐標(biāo)原點(diǎn)，AB，AC，AP所在直線分別為x，y，z軸，建立如圖所示的空間直角坐標(biāo)系，00【另解1】（2）因?yàn)镻A丄底面ABC，上BAC，所以AB，AC，AP兩兩垂直，故以A為坐標(biāo)原點(diǎn)，AB，AC，AP所在直線分別為x，y，z軸，建立如圖所示的空間直角坐標(biāo)系，因?yàn)锽C=2，AB=AC，上BAC，D為BC的中點(diǎn)，所以AB=AC.00················································································································6分，令x=a，則y=a，z=，所以.··············································7分【另解2】（2）取AB的中點(diǎn)M，連接PM和DM，由（1）知，上PDA即為二面角P-BC-A的平面角，所以上PDA·················7分因?yàn)锽C=2，AB=AC，上BACD為BC的中點(diǎn)，所以AB=ACAD=1，所因?yàn)镈為BC的中點(diǎn)，M為AB的中點(diǎn)，所以DM為△ABC的中位線，DM∥AC，DM=AC所以上PDM為異面直線PD與AC所成的角.····································9分在△PMD中，由余弦定理可得cos上PDM=PD2+DM2-PM22PD.DM2=評(píng)分細(xì)則： 16.（1）開始充電時(shí)的電量為27.5%（3分）；充滿電需min t-1【解析】（1）由題意知，當(dāng)t=0時(shí)電池開始充電，此時(shí)Q(t)=55.210，將t=0代入得-1=27.5，····································································2分當(dāng)Q(t)=100，即電量達(dá)到100%時(shí)，電池充滿電，已知Q(60)=80，且Q(t)單調(diào)遞增，故充滿電所需的時(shí)間必大于60min，·····································································4分故充滿電需min.·····················································································7分t2b，因?yàn)楫?dāng)t=10時(shí)，電量不產(chǎn)生躍變，故在t=10附近，第一段和第二段函數(shù)連續(xù).················································································································9分評(píng)分細(xì)則： 17.（14分25分3）0（6分）所以sinAcosC+sinAsinC=sin(A+C)，整理得sinC(sinAcosA)=0，··································································2分226222，·············································6分6解得AC=2，即AC=2，所以AB=2.·······················································7分△ABC2（3）由正弦定理得 =sin2C-cosC-sinC+cos-cosC-sinC+cosCsinC 442=sin2CcosCπ(5π)π(π)(π5π)cosCπ(5π)π(π)(π5π)).令f則易知fmax=f【另解】（3）由余弦定理得a2=b2+c2-2bccosA，22bπ(5π)π(π)(π5π)bπ(5π)π(π)(π5π),令f=-3t，則易知fmax=f3，AB23，AB2評(píng)分細(xì)則：導(dǎo)出導(dǎo)出A=給1分.f,(x)的極小值為lna+2，無(wú)極大值（7分3）證明見(jiàn)解析（6分） 則曲線y=f(x)在(1,f(1))處的切線方程為y-0=3(x-1)，即3x-y-3=0.··········4分················································································································6分當(dāng)a>0時(shí)，令g,(x)==0，解得x=a，···················································8分(0,a)時(shí)，g,(x)<0，所以函數(shù)g(x)在(0,a)上單調(diào)遞減；（3）不等式f(x)-2x>-2，即(x+a)lnx-2x+2>0，即(x+a)lnx-2(x-1)>0，令由（2）可知，a>0時(shí)，函數(shù)y=lnx+0,a)上單調(diào)遞減，在(a,+∞)上單調(diào)遞增，即當(dāng)a>1時(shí)，f(x)-2x>-2在(1,+∞)上恒成立.···············································17分評(píng)分細(xì)則：f,(x)的極小因?yàn)閍n+1，an+2，an成等差數(shù)列，所以2an+2=an+1+an，即an·········2分所以a234a34所以數(shù)列{bn}的前4項(xiàng)和為.··················································4分所以cnn，所以cnn.n所以m<p，·······························································································6分*，則→m.