課時(shí)作業(yè)(四)1.C[解析]f(1)=0,f(2)=1,f(4)=2,∴B={0,1,2},∴A∩B={1,2}.2.D[解析]由函數(shù)的性質(zhì)可得1-2x>0,x+1≠0,解得x<12且x≠1,故f(x)的定義域?yàn)?∞,1)3.D[解析]函數(shù)y=x-1+1,定義域?yàn)閇1,+∞),根據(jù)冪函數(shù)性質(zhì)可知,該函數(shù)為增函數(shù),當(dāng)x=1時(shí),該函數(shù)取得最小值1,故函數(shù)y=x-1+1的值域?yàn)閇14.6[解析]f(4)=f(3×1+1)=1+3+2=6.5.32[解析]∵函數(shù)f(x)=15-x,x≤0,log4x,x>0,∴f(3)=15-(-3)=6.A[解析]f(4)=f(2)=f(0)=f(2)=f(4)=124=116,7.B[解析]由題意可知m≤1,∴f(m)=21|m|=14=22,∴1|m|=2,解得m=3(舍)或m=3.則f(1m)=f(4)=(42)2=48.B[解析]根據(jù)分段函數(shù)f(x)=x2-2,x<-1,2x-19.C[解析]由于f(x)=3+log2x,x>0,x2-x-1,x≤0,當(dāng)x>0時(shí),3+log2x≤5,即log2x≤2=log24,解得0<x≤4;當(dāng)x≤0時(shí),x2x1≤5,即(x3)(x+2)≤0,解得2≤x≤3,∴210.A[解析]由題意,f(x)+f(x)=lg(1+4x24x2)+4=4,∴f(ln2)+fln12=f(ln2)+f(ln2)=11.C[解析]由題意,得f(0)=30+1=2,f[f(0)]=f(2)=4a2=3a,解得a=2.故選C.12.2x+7[解析]設(shè)f(x)=ax+b(a≠0),則3f(x+1)2f(x1)=ax+5a+b,所以ax+5a+b=2x+17對任意實(shí)數(shù)x都成立,所以a=2,5a+b=17,解得a=2,13.(1,1][解析]∵-2≤2x≤2,x+1>0,∴1<x≤1,14.13[解析]∵f(3x)=x+2,設(shè)3x=t(t>0),則x=log3t,∴f(t)=log3t+2.∵f(a)=1,∴f(a)=log3a+2=1,解得a=115.-∞,-12∪12,+∞[解析]易知a=0不合題意.當(dāng)a>0時(shí),必有ax2+x+a>0在R上恒成立,即14a2<0,所以a>12;當(dāng)a<0時(shí),必有ax2+x+a<0在R上恒成立,即14a2<0,所以a<12.所以實(shí)數(shù)a的取值范圍是∞,1216.log373,1[解析]當(dāng)t=0時(shí)f[f(t)]=f(1)=3,不合題意;當(dāng)t∈(0,1]時(shí),f(t)=3t∈(1,3],又函數(shù)f(x)=3x,x∈[0,1],92-32x,x∈(1,3],所以f[f(t)]=9232×3t,又因?yàn)閒[f(t)]∈[0,1],所以0≤9232×3t課時(shí)作業(yè)(五)1.A[解析]依題意可得函數(shù)在(0,+∞)上單調(diào)遞減,故由選項(xiàng)可得A正確.2.B[解析]易知y=(x-1)2+2,因?yàn)?x1)2+2≥2,所以y3.B[解析]ln0.5<ln1=0,0<0.60.5<0.60=1,1=log0.60.6<log0.60.5,故a>c>b,故選B.4.D[解析]當(dāng)a=0時(shí),函數(shù)f(x)=12x+5在(∞,3)上是減函數(shù),符合題意;當(dāng)a≠0時(shí),則有a>0,-4(a-3)45.(3,+∞)[解析]令u(x)=|x3|,則在(∞,3)上u(x)為減函數(shù),在(3,+∞)上u(x)為增函數(shù).又∵0<12<1,∴在區(qū)間(3,+∞)上,函數(shù)y=log12|x36.D[解析]由x24>0得x<2或x>2,∴已知函數(shù)的定義域?yàn)?∞,2)∪(2,+∞),令u=x24,則y=log12u在(0,+∞)上是減函數(shù),又∵u=x24的圖像的對稱軸為直線x=0,且開口向上,∴u=x24在(∞,2)上是減函數(shù),由復(fù)合函數(shù)的單調(diào)性知,f(x)在(∞,2)上是增函數(shù).故選7.D[解析]a+b≤0可轉(zhuǎn)化為a≤b或b≤a,由于函數(shù)f(x)在R上是減函數(shù),所以f(a)≥f(b),f(b)≥f(a),兩式相加得f(a)+f(b)≥f(a)+f(b).8.D[解析]因?yàn)閒(x)=2-xx+1=1+3x+1在(1,+∞)上單調(diào)遞減,且f(2)=0,所以n=2,1≤9.C[解析]題中隱含a>0,∴2ax在區(qū)間[0,1]上是減函數(shù).∴y=logau應(yīng)為增函數(shù),且u=2ax在區(qū)間[0,1]上應(yīng)恒大于零,∴a>1,2-10.A[解析]當(dāng)x1≠x2時(shí),f(x1)-f(x2)x1-x2<0,∴f(x)是R上的減函數(shù).∵f(x11.6[解析]由題意知,f(x)=x+2,0≤x≤4,10-x,x>412.12<x<2[解析]易知函數(shù)在定義域內(nèi)為減函數(shù),所以由f(x+1)<f(2x1)及定義域?yàn)?0,+∞)得x+1>2x1>0,解得12<x<13.解:(1)∵f(1)=0,∴b=a+1.由f(x)≥0恒成立,知a>0且Δ=b24a=(a+1)24a=(a1)2≤0,∴a=1.從而f(x)=x2+2x+1.∴F(x)=((2)由(1)可知f(x)=x2+2x+1,∴g(x)=f(x)kx=x2+(2k)x+1,由g(x)在[2,2]上是單調(diào)函數(shù),知2-k2≤2或2-k2≥2,得k≤14.解:(1)設(shè)x1>x2>0,則x1x2∵當(dāng)x>1時(shí),f(x)>0,∴f(x1)f(x2)=fx1x2∴f(x1)>f(x2),∴函數(shù)f(x)為增函數(shù).(2)在f(x1)f(x2)=fx1x2中,令x1=9,x2∴f(9)f(3)=f(3).又f(3)=1,∴f(9)=2.∴不等式f(3x+6)+f1x>2,可轉(zhuǎn)化為f(3x+6)+f1x>f(∴f(3x+6)>f(9)f1x=f(9x由函數(shù)f(x)為(0,+∞)上的增函數(shù),可得3x+6>9x>0,∴0<x<1,∴原不等式的解集為(0,1).(3)∵函數(shù)f(x)在(0,3]上是增函數(shù),∴f(x)在(0,3]上的最大值為f(3)=1,∴不等式f(x)≤m22am+1對所有x∈(0,3],a∈[1,1]恒成立轉(zhuǎn)化為1≤m22am+1對所有a∈[1,1]恒成立,即m22am≥0對所有a∈[1,1]恒成立.設(shè)g(a)=2ma+m2,∴需滿足g(-1解該不等式組,得m≤2或m≥2或m=0,即實(shí)數(shù)m的取值范圍為(∞,2]∪{0}∪[2,+∞).15.A[解析]由題意知對任意x1<x2,f(x1)-f(x2)x1-x2>2,可得f(x1)+2x1<f(x2)+2x2,令F(x)=f(x)+2x,∴F(x)在定義域R內(nèi)單調(diào)遞增,由f(1)=1,得F(1)=f(1)+2=3.∵f(log2|3x1|)<3log2|3x1|等價(jià)于f(log2|3x1|)+2log2|3x1|<3,令t=log2|3x1|,有f(t)+2t<3,即有F(t)<F(1),∴t<1,即log2|3x1|<1,從而016.13,+∞[解析]由題意知,f(x)+f(x)=2,∴f(2x1)+f(x)>2可化為f(2x1)>f(x),又y=2017x,y=2017x,y=ln(x2+1+x)均為增函數(shù),∴函數(shù)f(x)在R上單調(diào)遞增,∴2x1>x,∴x>13,∴課時(shí)作業(yè)(六)1.A[解析]根據(jù)題意,函數(shù)y=3|x|為偶函數(shù),在(∞,0)上為增函數(shù).對于選項(xiàng)A,函數(shù)y=1x2為偶函數(shù),在(∞,0)上為增函數(shù),符合題意;對于選項(xiàng)B,函數(shù)y=log2|x|是偶函數(shù),在(∞,0)上為減函數(shù),不符合題意;對于選項(xiàng)C,函數(shù)y=1x為奇函數(shù),不符合題意;對于選項(xiàng)D,函數(shù)y=x31為非奇非偶函數(shù),不符合題意.故選A2.C[解析]f(2017)=f(1+672×3)=f(1)=1+1=2.3.D[解析]f(x)=(xa)(x+2)=x2+(2a)x2a為偶函數(shù),則2a=0,即a=2.4.D[解析]∵函數(shù)f(x)=x3+sinx+2(x∈R),∴f(a)=a3+sina+2=2,∴a3+sina=0,則f(a)=(a3sina)+2=2.5.3[解析]由題意得,函數(shù)y=f(x)為奇函數(shù),所以f(2)=f(2)=(221)=3.6.A[解析]f-52=f52+2=f-12=f12=2×12×112=17.D[解析]由題意,f(x)=ln(ex)+ln(e+x)=f(x),則函數(shù)f(x)是偶函數(shù).f(x)=ln(e2x2),在(0,e)上,y=e2x2單調(diào)遞減,所以f(x)=ln(e2x2)單調(diào)遞減.故選D.8.C[解析]由于f(x)在[3,6]上為增函數(shù),所以f(x)的最大值為f(6)=8,f(x)的最小值為f(3)=1,因?yàn)閒(x)為奇函數(shù),所以f(3)=f(3)=1,所以f(6)+f(3)=8+1=9.故選C.9.B[解析]因?yàn)閒(x)=x2+g(x),又函數(shù)f(x)為偶函數(shù),所以有(x)2+g(x)=x2+g(x),即g(x)=g(x),所以g(x)為偶函數(shù),由選項(xiàng)可知,只有選項(xiàng)B中的函數(shù)為偶函數(shù),故選B.10.B[解析]因?yàn)閒(x)=a-x-1a-x+1+loga1+x1-x=1-ax1+axloga1-x1+x=f(x),所以函數(shù)f(x)是奇函數(shù),11.B[解析]f(x)是R上的偶函數(shù),且在(∞,0]上是減函數(shù),所以f(x)在[0,+∞)上是增函數(shù),所以f(log2x)>2=f(1)?f(|log2x|)>f(1)?|log2x|>1?log2x>1或log2x<1?x>2或0<x<1212.B[解析]∵定義在R上的函數(shù)f(x)=2|xm|+1(m∈R)為偶函數(shù),∴m=0,∴f(x)=2|x|+1,∴當(dāng)x∈(∞,0)時(shí),f(x)是減函數(shù),當(dāng)x∈(0,+∞)時(shí),f(x)是增函數(shù).∵a=f(log22)=f(1),b=f(log24)=f(2),c=f(2m)=f(0),∴a,b,c的大小關(guān)系為c<a<b.13.4[解析]因?yàn)閥=f(x1)的圖像關(guān)于點(diǎn)(1,0)對稱,所以y=f(x)的圖像關(guān)于點(diǎn)(0,0)對稱,即函數(shù)f(x)為奇函數(shù),由f(x+6)+f(x)=2f(3)得f(x+12)+f(x+6)=2f(3),所以f(x+12)=f(x),T=12,因此f(22)=f(2)=f(2)=4.14.0[解析]因?yàn)閒(x)為奇函數(shù),f(x+1)為偶函數(shù),所以f(x+1)=f(x+1)=f(x1),所以f(x+2)=f(x),所以f(x+4)=f(x+2)=f(x),所以函數(shù)f(x)的周期為4,所以f(4)=f(0)=0,f(3)=f(1)=f(1).在f(x+1)=f(x+1)中,令x=1,可得f(2)=f(0)=0,所以f(1)+f(2)+f(3)+f(4)=0.15.A[解析]由題意知,f(0)=1+b=0,∴b=1,∴f(x)=log2(x+2)+x1,∴f(2)=3,且該函數(shù)在R上單調(diào)遞增.∵|f(x)|>3=f(2),∴f(x)>f(2)或f(x)<f(2)=f(2),∴x>2或x<2.16.C[解析]函數(shù)f(x)=e2x-exsinx+1e2x+1=1exsinx1+e2x,令g(x)=exsinx1+e2x,x∈R,則g(x)=e-xsin(-x)1+e-2x=ex(-sinx)e2x+1加練一課(一)1.D[解析]由題意知,f(x)在(0,+∞)上為減函數(shù),從而有a≤0,a≤-2或a>0,2.B[解析]由已知得f(1)=f(1),g(1)=g(1),則有-f(1)+g(13.D[解析]∵f(x)=x1exex=f(x),∴f(x)在R上為偶函數(shù).f'(x)=ex1ex+xex+1ex,∴當(dāng)x>0時(shí),f'(x)>0,∴f(x)在[0,+∞)上為增函數(shù).由f(x1)<f(x2),得f(|x1|)<f(|x2|),∴|x1|<|x2|,∴x14.B[解析]因?yàn)楹瘮?shù)f(x+1),f(x1)都是奇函數(shù),所以f(1)=f(1)=0,又因?yàn)槠婧瘮?shù)的圖像關(guān)于原點(diǎn)對稱,所以函數(shù)f(x)的圖像既關(guān)于點(diǎn)(1,0)對稱,又關(guān)于點(diǎn)(1,0)對稱,即f(2x)=f(x)和f(2x)=f(x),那么f(2x)=f(2x),所以函數(shù)的周期是4,則f(5)=f(1)=0.故選B.5.B[解析]∵f(x+2)=f(x)對x∈R恒成立,∴f(x)的周期為2,又∵f(x)是定義在R上的偶函,∴f-92=f-12=f12.∵當(dāng)x∈[0,1]時(shí),f(x)=2x,6.A[解析]∵f(x)是定義在R上的周期為3的偶函數(shù),∴f(5)=f(56)=f(1)=f(1).∵f(1)<1,f(5)=2a-3a+1,∴2a-3a+1<1,7.A[解析]由f(x+2)=1-f(x),得f(x+4)=1-f(x+2)=f(x),所以f(x)的周期為4,所以f(2018)=f(2).因?yàn)閒(2+2)=1-f(28.B[解析]由函數(shù)f(x)是偶函數(shù),f(x+1)是奇函數(shù),可知函數(shù)的周期為4,則a=f8211=f611,b=f509=f49,c=f247=f47.由(x1x2)[f(x1)f(x2)]<0,可知函數(shù)是區(qū)間[0,9.B[解析]因?yàn)楹瘮?shù)f(x)是R上的奇函數(shù),所以f(0)=0,若函數(shù)f(x)為R上的減函數(shù),則滿足當(dāng)x>0時(shí),函數(shù)為減函數(shù),且當(dāng)x=0時(shí),1a≤0,此時(shí)-a-2=a2≤0,-10.D[解析]∵y=f(x+4)為偶函數(shù),∴f(x+4)=f(x+4),令x=2,得f(2)=f(2+4)=f(2+4)=f(6),同理,f(3)=f(5),又知f(x)在(4,+∞)上為減函數(shù),∵5<6,∴f(5)>f(6),∴f(2)<f(3),∴f(2)=f(6)<f(5),f(3)=f(5)>f(6).故選D.11.B[解析]由于函數(shù)f(x)的周期為2,所以f-52=f-12=12+a,f92=f12=2512=110,所以12+a=110,即a=35,因此f(5a)=f(3)=f(112.C[解析]根據(jù)已知條件可知f(x)在(0,+∞)上是減函數(shù),且f(a)=f(|a|),f(b)=f(|b|),f(c)=f(|c|),|a|=lnπ>1,b=(lnπ)2>|a|,0<c=lnπ2<|a|,∴f(c)>f(a)>f(b13.C[解析]∵定義在R上的奇函數(shù)y=f(x),滿足對任意t∈R都有f(t)=f(1t),∴f(3)=f(13)=f(2)=f(2)=f(12)=f(1)=f(11)=f(0),f-32=f32=f1-32=f12.∵當(dāng)x∈0,12時(shí),f(x)=x2,∴f(0)=0,f12=122=14,∴f(14.(1,0)∪(0,1)[解析]∵f(x)為奇函數(shù),且在(0,+∞)上是增函數(shù),f(1)=0,∴f(1)=f(1)=0,且在(∞,0)上也是增函數(shù).∵f(x)-f(-x)x=2·f(x)x<0,即x>0,f(x)<0或x<0,f(x)>0,∴根據(jù)f(15.(0,1)∪(3,+∞)[解析]因?yàn)楹瘮?shù)f(x)=x3+2x是奇函數(shù),且在R上是增函數(shù),f(1)+f(log1a3)>0,所以f(log1a3)>f(1)=f(1),所以log1a3>1,所以1a>1,0<a<3或016.7[解析]因?yàn)楫?dāng)0≤x<2時(shí),f(x)=x3x,所以f(0)=0,f(1)=0.又f(x)是R上最小正周期為2的周期函數(shù),則f(6)=f(4)=f(2)=f(0)=0,f(5)=f(3)=f(1)=0,故函數(shù)y=f(x)的圖像在區(qū)間[0,6]上與x軸的交點(diǎn)有7個(gè).課時(shí)作業(yè)(七)1.C[解析]∵函數(shù)f(x)=k·xα是冪函數(shù),∴k=1.∵冪函數(shù)f(x)=xα的圖像過點(diǎn)12,22,∴12α=22,得α=12,則k+α=1+12=2.A[解析]∵函數(shù)y=x2+bx+c(x∈[0,+∞))是單調(diào)函數(shù),∴圖像的對稱軸x=b2在區(qū)間[0,+∞)的左邊,即b2≤0,得b≥3.A[解析]因?yàn)楹瘮?shù)f(x)=ax2+bx+c對任意實(shí)數(shù)x都有f(2+x)=f(2x)成立,所以該函數(shù)圖像關(guān)于直線x=2對稱,當(dāng)a<0時(shí)f(2)最大,四個(gè)選項(xiàng)均不滿足,當(dāng)a>0時(shí)f(2)最小,由21<42,得f(1)<f(4),故選A.4.C[解析]顯然y=|x|1n(n∈N,n>2)是偶函數(shù),故可排除A和B,又n∈N,n>2,所以應(yīng)選擇5.(4,+∞)[解析]由題意可知x23x4>0,得x<1或x>4,結(jié)合圖像易知該函數(shù)的單調(diào)遞增區(qū)間為(4,+∞).6.C[解析]由題知,對于任意實(shí)數(shù)x,x2+ax+1≥0恒成立,故只要Δ=a24≤0,即2≤a≤2即可,故實(shí)數(shù)a的取值范圍是[2,2].7.A[解析]設(shè)函數(shù)的解析式為f(x)=a(x+2)(x4),又其圖像過點(diǎn)1,92,所以92=a(1+2)×(14),得a=12,所以所求函數(shù)解析式為f(x)=12(x+2)(x4),即f(x)=12x28.D[解析]函數(shù)f(x)=ax2+(b2a)x2b為偶函數(shù),則b2a=0,故f(x)=ax24a=a(x2)(x+2),因?yàn)楹瘮?shù)f(x)在(0,+∞)上單調(diào)遞增,所以a>0.根據(jù)二次函數(shù)的性質(zhì)可知,不等式f(2x)>0的解集為{x|2x>2或2x<2}={x|x<0或x>4},故選D.9.D[解析]函數(shù)f(x)=x2+2ax+1a=(xa)2+a2a+1,其圖像的對稱軸方程為x=a.當(dāng)a<0時(shí),f(x)max=f(0)=1a,所以1a=2,所以a=1;當(dāng)0≤a≤1時(shí),f(x)max=f(a)=a2a+1,所以a2a+1=2,所以a2a1=0,所以a=1±52(舍去);當(dāng)a>1時(shí),f(x)max=f(1)=a,所以a=2.綜上可知,a=1或a=10.B[解析]當(dāng)m=0時(shí),f(x)=8x+1>0不恒成立,g(x)=0,此時(shí)不符合條件;當(dāng)m<0時(shí),g(x)=mx在x>0時(shí)恒為負(fù),而f(x)=2mx22(4m)x+1的圖像開口向下,所以對任意x>0顯然不恒為正值;當(dāng)m>0時(shí),g(x)=mx在x>0時(shí)恒為正,所以只需f(x)=2mx22(4m)x+1在x≤0時(shí)恒為正即可,若b2a=4-m2m≥0,即0<m≤4,此時(shí)結(jié)論顯然成立,若b2a=4-m2m<0,即m>4,此時(shí)只要Δ=4(4m)28m<0即可,得4<m<11.12[解析]設(shè)冪函數(shù)f(x)=xa,把13,33代入函數(shù)方程f(x)=xa,得13a=33,解得a=12,則f(x)=x12,∴f(2)=212,∴l(xiāng)og2f(12.-22,0[解析]因?yàn)楹瘮?shù)圖像開口向上,所以據(jù)題意只需滿足f(13.解:(1)當(dāng)a=2時(shí),f(x)=x2+3x3,x∈[2,3],函數(shù)圖像的對稱軸為x=32∈[2,3∴f(x)min=f-32=9492f(x)max=f(3)=15,∴值域?yàn)?14,15.(2)函數(shù)圖像的對稱軸為直線x=2a①當(dāng)2a-12≤1,即af(x)max=f(3)=6a+3,∴6a+3=1,即a=13,滿足題意②當(dāng)2a-12>1,即a<f(x)max=f(1)=2a1,∴2a1=1,即a=1,滿足題意.綜上可知a=13或114.解:(1)由f(0)=1,得c=1,所以f(x)=ax2+bx+1.又f(x+1)f(x)=2x,所以a(x+1)2+b(x+1)+1(ax2+bx+1)=2x,即2ax+a+b=2x,所以2a=2,a+b=0,所以a=1,b=(2)f(x)>2x+m等價(jià)于x2x+1>2x+m,即x23x+1m>0,要使此不等式在區(qū)間[1,1]上恒成立,只需使函數(shù)g(x)=x23x+1m在區(qū)間[1,1]上的最小值大于0即可.因?yàn)間(x)=x23x+1m在區(qū)間[1,1]上單調(diào)遞減,所以g(x)min=g(1)=m1,由m1>0,得m<1.因此滿足條件的實(shí)數(shù)m的取值范圍是(∞,1).15.A[解析]若函數(shù)f(x)=ax2(1≤x≤2)與g(x)=x+2的圖像上存在關(guān)于x軸對稱的點(diǎn),則方程ax2=(x+2),即a=x2x2在區(qū)間[1,2]上有解.令h(x)=x2x2,1≤x≤2,由于h(x)=x2x2的圖像是開口朝上且以直線x=12為對稱軸的拋物線,故當(dāng)x=1時(shí),h(x)取得最小值2,當(dāng)x=2時(shí),h(x)取得最大值0,故a∈[2,0]16.-4,43[解析]f(x)=2x2+(x2a)|xa|可化為f(x)=3x2-3ax+2a2,x≥a,x2+3ax-2a2,x<a.若a>0,函數(shù)y=3x23ax+2a2(x≥a)單調(diào)遞增,此時(shí)函數(shù)y=x2+3ax2a2(x<a)的圖像的對稱軸為直線x=3a2,結(jié)合圖像可知要使函數(shù)f(x)在[2,1]上不單調(diào),則2<3a2<1,得0<a<43;若a=0,函數(shù)f(x)=3x2,x≥0,x2,x<0在[2,1]上不單調(diào),符合題意;若a<0,函數(shù)y=x2+3ax2a2(x<a)單調(diào)遞減,函數(shù)y=3x課時(shí)作業(yè)(八)1.D[解析]因?yàn)楹瘮?shù)y=0.86x在R上是減函數(shù),所以0<0.860.85<0.860.75<1,又1.30.86>1,所以c>a>b.2.A[解析]∵f(x)=5x,∴f(a+b)=5a+b=3,∴f(a)·f(b)=5a×5b=5a+b=3.故選A.3.B[解析]若ax<1且x>0,則必有0<a<1,所以C,D錯(cuò),又ax<bx,所以b>a,所以選B.4.B[解析]函數(shù)y=3x,y=5x是R上的增函數(shù),其圖像都是上升的,排除C和D;在第一象限內(nèi),底數(shù)越大,指數(shù)函數(shù)的圖像越靠近y軸,排除A.故選B.5.(3,1)[解析]令x3=0,得x=3,此時(shí)y=1,即函數(shù)f(x)=ax32的圖像必過定點(diǎn)(3,1).6.D[解析]由指數(shù)函數(shù)圖像的特征可知,當(dāng)0<a<1時(shí),函數(shù)y=ax+(b1)(a>0,a≠1)的圖像必經(jīng)過第二象限,故排除選項(xiàng)B,C.因?yàn)楹瘮?shù)y=ax+(b1)(a>0,a≠1)的圖像不經(jīng)過第二象限,所以其圖像與y軸的交點(diǎn)不在x軸上方,所以當(dāng)x=0時(shí),y=a0+(b1)≤0,即b≤0,故選項(xiàng)D正確.7.D[解析]∵(m2m)·4x2x<0在(∞,1]上恒成立,∴m2m<12x在(∞,1]上恒成立.∵f(x)=12x在(∞,1]上單調(diào)遞減,∴f(x)≥2,∴m2m<2,∴8.D[解析]由1ex≠0可得x≠0,排除A,C;當(dāng)x<0時(shí),0<ex<1,∴f(x)=1+ex1-ex>0,9.B[解析]因?yàn)閒(a)=g(b),所以ea1=b2+4b3,所以b2+4b2=ea>0,即b24b+2<0,所以22<b<2+2,故選B.10.A[解析]不等式可化為12x12-x>12y12-y,又f(x)=1211.(0,43][解析]令t=x23x+2=x32214,則t∈14,+∞,故13x2-3x+2≤13-112.解:(1)易知f(x)=(2x)24·2x6(0≤x≤3).令t=2x,∵0≤x≤3,∴1≤t≤8,∴h(t)=t24t6=(t2)210(1≤t≤8).∴當(dāng)t∈[1,2]時(shí),h(t)是減函數(shù);當(dāng)t∈(2,8]時(shí),h(t)是增函數(shù).∴f(x)min=h(2)=10,f(x)max=h(8)=26.(2)∵f(x)a≥0恒成立,即a≤f(x)恒成立,∴a≤f(x)min=10,∴a的取值范圍為(∞,10].13.D[解析]據(jù)題意可知2-a2>0,14.B[解析]f(x)=2x1+2x12=2x+1-11+2x12=111+2x12=1211+2x,因?yàn)?+2x>1,所以0<11+2x<1,所以課時(shí)作業(yè)(九)1.D[解析]由2x12>0,得x>1,故選D2.C[解析]∵0<a<b<1<c,∴ab<aa,ca<cb,logac>logbc,logbc<logba.故選C.3.A[解析]因?yàn)?x+1>1,所以f(x)=log2(3x+1)>log21=0,故選A.4.A[解析]∵2a=5b=m,∴a=log2m,b=log5m,∴1a=1log2m=logm2,1b=1log5m=logm5,∴1a+1b=logm2+logm5=logm10=2,5.1[解析]∵x=log210,∴xlog25=log22=1.6.D[解析]若a>1,則y=a-ax在[0,1]上單調(diào)遞減,則a-a=0,a-1=1,解得a=2,此時(shí),loga37+loga1123=log216=4;若0<a<1,則y=a-7.C[解析]當(dāng)x>0時(shí),f(x)=logax單調(diào)遞減,排除A,B;當(dāng)x<0時(shí),f(x)=loga(x)單調(diào)遞減,排除D.故選C.8.A[解析]f(x)=lgx232x+1=lgx342+716,令t=x342+716,當(dāng)x∈1,32時(shí),tmax=1,此時(shí)f(x)取到最大值0.9.A[解析]因?yàn)閒(x)是偶函數(shù),所以f(log2a)+f(log0.5a)≤2f(1)等價(jià)于2f(log2a)≤2f(1),即f(|log2a|)≤f(1).又因?yàn)閒(x)在[0,+∞)上是增函數(shù),所以|log2a|≤1,即1≤log2a≤1,解得12≤a≤2,所以實(shí)數(shù)a的最小值為110.22[解析]y=loga(2x3)+2的圖像恒過點(diǎn)P(2,2),設(shè)冪函數(shù)為f(x)=xa,則2a=2,∴a=12,故冪函數(shù)為f(x)=x12,∴f(8)=11.25[解析]由題意知函數(shù)f(x)=ax的反函數(shù)為g(x)=logax,又f(2)=9,∴a2=9,∴a=3,∴g(x)=log3x,∴g19+f(3)=log319+33=12.解:(1)由題意知當(dāng)a=2時(shí),f(x)=log2(32x),令t=32x,則t∈(1,3],∴函數(shù)f(x)在[0,1)上的值域?yàn)?0,log23].(2)令u=3ax,則u=3ax在[1,2]上恒為正.∵a>0,a≠1,∴u=3ax在[1,2]上單調(diào)遞減,∴32a>0,即a∈(0,1)∪1,32.又函數(shù)f(x)在[1,2]上單調(diào)遞減,u=3ax在[1,2]上單調(diào)遞減,∴a>1,即a∈1,32.又函數(shù)f(x)在[1,2]上的最大值為1,∴f(1)=loga(3a)=1,∴a=32,與a∈1,32矛盾,∴a不存在.13.解:(1)f(x)=logax-5x+5由x-5x+5>0,可得f(x)的定義域?yàn)閧x|x<5或x>∵f(x)=logax+5x-5=logax-∴函數(shù)f(x)為奇函數(shù).(2)假設(shè)存在滿足條件的實(shí)數(shù)m.∵f(x+2)+f(mx)=logax-3x+7·∴-x2+(m-2)x-3(m-5)-x2+(m-2)x+7(m+5)為常數(shù),設(shè)其為k,則(∴k-1=0∴存在實(shí)數(shù)m=2滿足條件.14.C[解析]令f(x)=2x3+x2,則f(x)在R上單調(diào)遞增,且f(0)·f(1)=2×1=2<0,即a∈(0,1).在同一坐標(biāo)系中作出y=1x,y=log2x,y=log5x的圖像,由圖像得1<b<c,故c>b>a.故選C15.A[解析]當(dāng)0<a<1時(shí),函數(shù)f(x)在區(qū)間12,23上是減函數(shù),所以loga43a>0,即0<43a<1,解得13<a<43,故13<a<1;當(dāng)a>1時(shí),函數(shù)f(x)在區(qū)間12,23上是增函數(shù),所以loga(1a)>0,即1a>1,解得a<0,此時(shí)無解.綜上所述,實(shí)數(shù)a的取值范圍是13,課時(shí)作業(yè)(十)1.D[解析]由點(diǎn)(x,y)關(guān)于原點(diǎn)的對稱點(diǎn)是(x,y),可知D正確.2.A[解析]將函數(shù)y=2x的圖像向右平移3個(gè)單位長度得到y(tǒng)=2x3的圖像,再向下平移1個(gè)單位長度得到y(tǒng)=2x31的圖像.3.C[解析]函數(shù)的定義域?yàn)閧x|x≠0},排除A.當(dāng)x<0時(shí),y>0,排除B.當(dāng)x→+∞時(shí),有x3<3x1,此時(shí)y→0,排除D.故選C.4.B[解析]當(dāng)x=1時(shí),y=1+1e<0,排除A,C;當(dāng)x=2時(shí),y=322e2>3218>0,排除D.故選B5.1[解析]由圖像可知不等式2<f(x+t)<4即為f(3)<f(x+t)<f(0),故x+t∈(0,3),即不等式的解集為(t,3t),依題意可得t=1.6.D[解析]作出函數(shù)f(x)=|x+a|與g(x)=x1的圖像,如圖所示,觀察圖像可知,當(dāng)a≤1,即a≥1時(shí),不等式f(x)≥g(x)恒成立,因此a的取值范圍是[1,+∞).7.D[解析]∵f(x)=12|x|x2+2為偶函數(shù),∴f(x)的圖像關(guān)于y軸對稱,排除A;又f(0)=3>0,排除C;當(dāng)x>0時(shí),y=12|x|單調(diào)遞減,y=x2單調(diào)遞減,∴f(x)=12|x|x2+2在(8.B[解析]當(dāng)h=H時(shí),對應(yīng)陰影部分的面積為0,∴排除C與D;當(dāng)h=H2時(shí),對應(yīng)陰影部分的面積小于整個(gè)圖形面積的一半,∴排除A.故選B9.B[解析]函數(shù)y=f(x)是偶函數(shù),根據(jù)已知得,函數(shù)f(x)在[0,+∞)上單調(diào)遞增,又函數(shù)y=f(x-1)1的圖像是把函數(shù)y=f(x)的圖像先向右平移1個(gè)單位長度,再向下平移1個(gè)單位長度得到的,故只可能是選項(xiàng)B10.B[解析]由題意可得,函數(shù)f(x)的圖像和函數(shù)g(x)的圖像有兩個(gè)交點(diǎn),如圖所示,易得A(2,1),kOA=12,數(shù)形結(jié)合可得1<k<111.A[解析]方程f(x)+lgx=0的實(shí)根的個(gè)數(shù)就是函數(shù)y=x[x]與y=lgx圖像交點(diǎn)的個(gè)數(shù),函數(shù)y=x[x]是周期為1的周期函數(shù),在同一坐標(biāo)系中作出這兩個(gè)函數(shù)的圖像,如圖所示,可知它們共有8個(gè)交點(diǎn),所以方程f(x)+lgx=0有8個(gè)實(shí)根.12.k≥3[解析]由題意知函數(shù)f(x)的圖像與恒過定點(diǎn)(0,2)的直線y=kx2有兩個(gè)交點(diǎn),由圖像可知,實(shí)數(shù)k的取值范圍是k≥3.13.2[解析]由函數(shù)y=f(x)的圖像與y=2xa的圖像關(guān)于直線y=x對稱,可得f(x)=alog2(x),由f(2)+f(4)=1,可得alog22alog24=1,解得a=2.14.3[解析]畫出函數(shù)f(x)的圖像如圖所示.∵m>3,∴m>4mm2,故方程f(x)=b的實(shí)根個(gè)數(shù)最多為3.15.B[解析]f(x)=1+1x-1,x>1,-1+11-x,x<1,g(x)=1+x,x≥0,1,x<0,作出兩函數(shù)的圖像如圖所示.當(dāng)0≤x<1時(shí),由1+11-x=x+1,解得x=5-12;當(dāng)x>1時(shí),16.D[解析]由題意可得函數(shù)f(x)是以4為周期的周期函數(shù),作出函數(shù)y=f(x)與函數(shù)y=ax的圖像.由圖像可得,方程(x4)2+1=ax,即x2+(a8)x+15=0在(3,5)上有2個(gè)不同的實(shí)數(shù)根,則Δ=(a-8)2-60>0,9+3(a-8)+15>0,25+5(a-8)+15>0,3<8-a2<5,a加練一課(二)1.D[解析]將y=lgx的圖像關(guān)于y軸對稱得到y(tǒng)=lg(x)的圖像,再向右平移2個(gè)單位長度,得到y(tǒng)=lg[(x2)]的圖像,將得到的圖像在x軸下方的部分翻折上來,即得到f(x)=|lg(2x)|的圖像.由圖像知,只有D選項(xiàng)滿足題意.2.A[解析]由函數(shù)圖像可知,函數(shù)f(x)為奇函數(shù),應(yīng)排除B,C.若函數(shù)為f(x)=x1x,則x→+∞時(shí),f(x)→+∞,排除D.故選A3.D[解析]作出函數(shù)y=f(x)與y=k的圖像,如圖所示.由圖可知k∈(0,1],故選D.4.C[解析]令g(x)=log2(x+1),可知g(x)的定義域?yàn)?1,+∞),作出函數(shù)g(x)的圖像如圖.結(jié)合圖像知,不等式f(x)≥log2(x+1)的解集為{x|1<x≤1}.5.A[解析]在同一直角坐標(biāo)系中,分別作出y=f(x)和y=|lgx|的圖像,如圖,結(jié)合圖像知,共有10個(gè)交點(diǎn).6.C[解析]畫出y=max{2x,2x3,6x}的示意圖,如圖.由圖可知,y的最小值為22=62=4,故選C.7.D[解析]函數(shù)f(x)的圖像如圖所示.易得f(x)=f(x),從而函數(shù)f(x)是偶函數(shù),且在[0,+∞)上是增函數(shù).又0<|x1|<|x2|,∴f(x2)>f(x1),即f(x1)f(x2)<0.8.B[解析]依題意知,f(x)=(x22)?(x1)=x2-2,-1≤x≤2,x-1,x<-1或x>2,結(jié)合圖像(圖略)可知,當(dāng)c∈(2,1]∪(1,2]時(shí),函數(shù)y=f(9.C[解析]由題意可得f(x)在(3,1)上的圖像如圖,由圖可知不等式xf(x)>0在(3,1)上的解集為(3,1)∪(0,1),故選C.10.B[解析]根據(jù)題意可知,作出函數(shù)y=ln(x)(x<0)關(guān)于原點(diǎn)對稱的函數(shù)y=lnx(x>0)的圖像,使它與直線y=kx1(x>0)的交點(diǎn)個(gè)數(shù)為2即可.當(dāng)直線y=kx1與y=lnx的圖像相切時(shí),設(shè)切點(diǎn)為(m,lnm),又y=lnx的導(dǎo)數(shù)為y'=1x,則km1=lnm,k=1m,解得m=1,k=1,可得函數(shù)y=lnx(x>0)的圖像的過點(diǎn)(0,1)的切線的斜率為1.結(jié)合圖像可知,k∈(0,1)時(shí)兩函數(shù)圖像有兩個(gè)交點(diǎn).故選11.C[解析]設(shè)y=h(x)與y=f(x)的圖像關(guān)于y軸對稱,則h(x)=f(x)=ln(-x),x<0,-x,x≥0∵f(x)與g(x)的圖像上存在關(guān)于y軸對稱的點(diǎn),∴y=h(x)與y=g(x)的圖像有交點(diǎn),∴a≤e,即a≥e.12.(1,0)[解析]在同一坐標(biāo)系內(nèi)作出y=log2(x),y=x+1的圖像,可知滿足條件的x∈(1,0).13.(0,+∞)[解析]在同一坐標(biāo)系中畫出函數(shù)y=|x|與y=ax的圖像,如圖所示.由圖像知,當(dāng)a>0時(shí),方程|x|=ax只有一個(gè)解.14.116,1[解析]由x2logax<0,得x2<logax.設(shè)f(x)=x2,g(x)=logax.由題意知,當(dāng)x∈0,12時(shí),函數(shù)f(x)的圖像在函數(shù)g(x)的圖像下方,由圖可知0<a<1,f12≤g12,即015.-∞,34∪54,+∞[解析]對任意x∈R,都有f(x)≤|k1|成立,即f(x)max≤|k1|.作出函數(shù)f(x)的圖像如圖所示,由圖像可知,當(dāng)x=12時(shí),f(x)取得最大值14,所以|k1|≥14,16.①②④[解析]畫出函數(shù)y=|log2x|和y=m(m>0)的圖像,由圖可知,0<x1<1<x2.因?yàn)閥1=y2,所以log2x1=log2x2,解得x1x2=1,所以x1+x2>2,所以2x1+2x2>22x1+x2>222=4.綜上可知①正確課時(shí)作業(yè)(十一)1.A[解析]∵函數(shù)f(x)=m+log2x(x≥1)存在零點(diǎn),∴方程m+log2x=0在x≥1時(shí)有解,∴m=log2x≤log21=0.2.B[解析]顯然函數(shù)f(x)=log2x+x2在(0,+∞)上連續(xù),又f(1)=0+12<0,f(2)=1+22>0,故函數(shù)f(x)=log2x+x2的零點(diǎn)所在的區(qū)間是(1,2).3.B[解析]由題意知,當(dāng)x=1時(shí),f[g(1)]=f(1)=0,所以零點(diǎn)為1.4.B[解析]方程x2+(m1)x+m22=0對應(yīng)的二次函數(shù)為f(x)=x2+(m1)x+m22,由題意知,只需f(0)<0且f(1)<0,則m2-2<0,1+(m-5.1[解析]由于f(0)=1,f(x)圖像的對稱軸方程x=a2>1,f(2)=52a<0,結(jié)合圖像可知函數(shù)f(x)在(0,2)上恰有1個(gè)零點(diǎn)6.C[解析]由f(x)f(ex)=0,得|lnx|=|ln(ex)|,x∈(0,e),故x=ex或ex=1x,共有3個(gè)解,故選C7.B[解析]∵f(2)=ln21<0,f(3)=ln323>0,∴x0∈(2,3),∴g(x0)=[x0]=2,故選B8.A[解析]令f(x)=2x+x+1=0,可知x<0,即a<0;令g(x)=log2x+x+1=0,則0<x<1,即0<b<1;令h(x)=log2x1=0,可知x=2,即c=2.顯然a<b<c.故選A.9.B[解析]當(dāng)x<0時(shí),必存在x0=ea<0,使得f(x0)=0,因此對任意實(shí)數(shù)a,f(x)在(∞,0)內(nèi)必有一個(gè)零點(diǎn);當(dāng)x≥0時(shí),f(x)是周期為1的周期函數(shù),且0≤x<1時(shí),f(x)=1x.因此可畫出函數(shù)的大致圖像,如圖所示,可知函數(shù)f(x)的零點(diǎn)個(gè)數(shù)為1,故選B.10.D[解析]∵f(x)=x+2,x>a,x2+5x+2,x≤a,∴g(x)=f(x)2x=-x+2,x>a,x2+3x+2,x≤a,而方程x+2=0的根為2,方程x2+3x+2=0的根為111.C[解析]f(x)滿足f(x)=f(x)且f(x)=f(2x),可得f(2+x)=f(x)=f(x),進(jìn)而有f(x+4)=f(x),即f(x)是以4為周期的周期函數(shù).根據(jù)函數(shù)是奇函數(shù)且圖像關(guān)于直線x=1對稱,可以畫出函數(shù)在[2,2]上的圖像,并根據(jù)周期性擴(kuò)展到區(qū)間(2,6)上(圖略),可知在區(qū)間(0,2]上方程f(x)=12的兩個(gè)實(shí)根之和為2.在區(qū)間(4,6)上方程f(x)=12的兩個(gè)實(shí)根之和為10.綜上,在(0,6)上方程f(x)12=012.a<0[解析]令f(x)=x1x,顯然f(x)在區(qū)間(0,1]上單調(diào)遞增,則f(x)≤0,且x→0時(shí)f(x)→∞,要使原方程在(0,1]內(nèi)無實(shí)數(shù)根,則必有a>0,得a<013.12,1[解析]由題意,作出函數(shù)f(x)的圖像如圖所示,方程f(x)=g(x)有兩個(gè)不等實(shí)數(shù)根等價(jià)于函數(shù)f(x)=|x2|+1與g(x)=kx的圖像有兩個(gè)不同的交點(diǎn).g(x)=kx表示過原點(diǎn)的直線,斜率為k,如圖,當(dāng)直線過點(diǎn)(2,1)時(shí),k=12,有一個(gè)交點(diǎn),當(dāng)k=1時(shí),有一個(gè)交點(diǎn),結(jié)合圖像可得,14.(0,1)[解析]由g(x)=f(x)k=0,得f(x)=k.作出函數(shù)y=k與y=f(x)的圖像如圖所示.由圖可知,若函數(shù)g(x)=f(x)k有兩個(gè)零點(diǎn),則實(shí)數(shù)k的取值范圍是(0,1).15.C[解析]f(x)=lg|x-3|,x≠3,3,x=3的圖像關(guān)于直線x=3對稱,且直線y=3與y=f(x若函數(shù)F(x)=[f(x)]2+bf(x)+c有且只有3個(gè)不同的零點(diǎn)x1,x2,x3,則關(guān)于f(x)的方程[f(x)]2+bf(x)+c=0只能有一個(gè)解,從而只能有f(x)=3,所以x1,x2,x3就是方程f(x)=3的解.根據(jù)圖像的對稱性知,x1+x2+x3=3×3=9,則ln(x1+x2+x3)=ln9=2ln3,故選C.16.C[解析]在區(qū)間[2,3]上方程axf(x)+2a=0恰有四個(gè)不相等的實(shí)數(shù)根,等價(jià)于函數(shù)f(x)和g(x)=a(x+2)的圖像有四個(gè)不同的交點(diǎn).∵f(x+2)=f(x),∴函數(shù)f(x)的周期是2.當(dāng)1≤x≤0時(shí),0≤x≤1,此時(shí)f(x)=2x,又∵f(x)是定義在R上的偶函數(shù),∴f(x)=2x=f(x),即f(x)=2x,1≤x≤0.作出函數(shù)f(x)和g(x)的圖像,當(dāng)g(x)經(jīng)過點(diǎn)A(1,2)時(shí),兩個(gè)圖像有3個(gè)交點(diǎn),此時(shí)g(1)=3a=2,解得a=23,當(dāng)g(x)經(jīng)過點(diǎn)B(3,2)時(shí),兩個(gè)圖像有5個(gè)交點(diǎn),此時(shí)g(3)=5a=2,解得a=25,要使方程axf(x)+2a=0在區(qū)間[2,3]上恰有四個(gè)不相等的實(shí)數(shù)根,則25<a<23課時(shí)作業(yè)(十二)1.B[解析]由題意得h=205t(0≤t≤4),故選B.2.C[解析]當(dāng)1≤x≤10時(shí),y≤40;當(dāng)x>100時(shí),y>150.因此所求人數(shù)x∈(10,100],由2x+10=60,得x=25,故選C.3.C[解析]當(dāng)x=1時(shí),由3000=alog3(1+2),得a=3000,到2018年即第7年,可得y=3000log3(7+2)=6000,故選C.4.C[解析]對于C,當(dāng)x=1時(shí),y=100;當(dāng)x=2時(shí),y=200;當(dāng)x=3時(shí),y=400;當(dāng)x=4時(shí),y=800,與第4個(gè)月銷售臺數(shù)790比較接近.故選C.5.a142[解析]由題意得D=aAA=Aa22+a24,且A≥0,∴當(dāng)A=a2,即A=a24時(shí),D6.D[解析]買小包裝時(shí)每克費(fèi)用為3100元,買大包裝時(shí)每克費(fèi)用為8.4300=2.8100元,而3100>2.8100,所以買大包裝實(shí)惠.賣3小包的利潤為3×(31.80.5)=2.1(元),賣1大包的利潤是8.41.8×30.7=2.3(元),而2.3>7.D[解析]從圖像可以看出,首次服用該藥物1單位約10分鐘后,該藥物的血藥濃度大于最低有效濃度,藥物發(fā)揮治療作用,A正確;第一次服藥4小時(shí)后與第2次服藥1小時(shí)后,血藥濃度之和大于最低中毒濃度,因此一定會產(chǎn)生藥物中毒,B正確,D錯(cuò)誤;服藥5.5小時(shí)后,血藥濃度小于最低有效濃度,此時(shí)再服藥,血藥濃度增加,正好能發(fā)揮作用,C正確.故選D.8.D[解析]銷售額先下降后上升,很明顯只有選項(xiàng)C和D符合,又因?yàn)閒(1)=8,f(3)=2,所以只有選項(xiàng)D符合.9.B[解析]由題意可得a+b+c=2.4,a=0.3b2-1.2b+1.5,x=3a+5b+8c,整理得x=1.5(b1)2+13.2,當(dāng)b=1時(shí),10.148.4[解析]據(jù)題意有0.568×50+0.598×150+0.288×50+0.318×50=148.4(元).11.y=2x,0≤x≤2,-12(x-4)2+6,2<x≤4[解析]易知0≤x≤4,當(dāng)0≤x≤2時(shí),y=212.解:(1)由題意可列方程組64=c兩式相除,解得c(2)由題意可列不等式1281214t≤所以1214t≤128,即14t≥故至少排氣32分鐘,這個(gè)地下車庫中的一氧化碳含量才能達(dá)到正常狀態(tài).13.解:(1)當(dāng)0<x≤40時(shí),W=xR(x)(16x+40)=6x2+384x40;當(dāng)x>40時(shí),W=xR(x)(16x+40)=40000x16所以W=-(2)①當(dāng)0<x≤40時(shí),W=6(x32)2+6104,所以當(dāng)x=32時(shí),Wmax=6104.②當(dāng)x>40時(shí),W=40000x16由于40000x+16x≥240當(dāng)且僅當(dāng)40000x=16x,即x=50時(shí),W綜合①②知,當(dāng)x=32時(shí),W取得最大值6104,即當(dāng)年產(chǎn)量為32萬部時(shí),該公司在該款手機(jī)的生產(chǎn)中所獲得的利潤最大,最大利潤為6104萬美元.14.5[解析]由題知f(n)=26n8n+n(n-1)2×260=n2+19n60.令f(n)>0,即n2+19n60>0,解得4<n<15,15.5[解析]∵5min后甲桶和乙桶中的水量相等,∴函數(shù)y=f(t)=aent滿足f(5)=ae5n=12a,可得n=15ln12.令f(k)=14a,則15ln12·k=ln14,即為15ln12·k=2ln12,解得課時(shí)作業(yè)(十三)1.C[解析]因?yàn)閒'(x)=1x2cosx+1x(sinx),所以f(π)+f'π2=1π+2π×2.A[解析]因?yàn)閒'(x)=sinx+xcosx+a,且f'π2=1,所以sinπ2+π2cosπ2+a=1,3.C[解析]y'=cosx+ex,故切線斜率k=2,切線方程為y=2x+1,即2xy+1=0.4.A[解析]求導(dǎo)可得y'=-4ex+e-x+2,∵ex+ex+2≥2ex·e-x+2=4,∴y'∈[1,0),即tanα∈[1,0),又5.1[解析]∵f(x)=ex·sinx,∴f'(x)=(ex)'sinx+ex(sinx)'=ex·sinx+ex·cosx,∴f'(0)=0+1=1.6.B[解析]設(shè)切點(diǎn)坐標(biāo)為(m,n)(m>0),對y=14x23lnx求導(dǎo)得y'=12x3x,可得切線的斜率為12m3m=12,解方程可得m=2(7.B[解析]g'(x)=3x2+5x+3x,則g'(1)=11,又g(1)=72+b,故曲線y=g(x)在x=1處的切線方程為y72+b=11(x1),由該切線過點(diǎn)(0,5),得b=58.C[解析]∵f(x)=x32x2+x+6,∴f'(x)=3x24x+1,∴f'(1)=8,故切線方程為y2=8(x+1),即8xy+10=0.令x=0,得y=10,令y=0,得x=54,∴所求面積S=12×54×109.B[解析]由題意知,方程f'(x)=1e有解,即exm=1e有解,即ex=m1e有解,故只要m1e>0,即m>1e10.A[解析]設(shè)與直線2xy+3=0平行且與曲線y=2lnx相切的直線方程為2xy+m=0.設(shè)切點(diǎn)為P(x0,y0),∵y'=2x,∴斜率k=2x0=2,解得x0=1,因此y0=2ln1=0,∴切點(diǎn)為P(1,0),則點(diǎn)P到直線2xy+3=0的距離d=|2-0+3|22+(-1)2=5,∴曲線11.1[解析]當(dāng)x>0時(shí),f'(x)=2lnx+2x-1x,則f'(1)=1,∵函數(shù)f(x)是偶函數(shù),∴f'(112.1+ln2[解析]由y=lnx,可得y'=1x,設(shè)切點(diǎn)坐標(biāo)為(x0,y0),由曲線y=lnx的一條切線是直線y=12x+b,可得1x0=12,解得x0=2,則切點(diǎn)坐標(biāo)為(2,ln2),所以ln2=1+b,b=113.解:(1)由題意得f'(x)=x24x+3=(x2)21≥1,即曲線C上任意一點(diǎn)處的切線斜率的取值范圍是[1,+∞).(2)設(shè)曲線C的其中一條切線的斜率為k(k≠0),則由(2)中條件并結(jié)合(1)中結(jié)論可知k≥解得1≤k<0或k≥1,故由1≤x24x+3<0或x24x+3≥1,得x∈(∞,22]∪(1,3)∪[2+2,+∞).14.B[解析]f(x)=e2x2ex+ax1的導(dǎo)函數(shù)為f'(x)=2e2x2ex+a,由題意可得2e2x2ex+a=3的解有兩個(gè),即有ex122=7-2a4,即為ex=12+7-2a2或ex=127-2a2,即有7215.0或1[解析]設(shè)公共切點(diǎn)的橫坐標(biāo)為x0,函數(shù)y=2x3+1的導(dǎo)函數(shù)為y'=6x2,y=3x2b的導(dǎo)函數(shù)為y'=6x.由圖像在一個(gè)公共點(diǎn)處的切線相同,可得6x02=6x0且1+2x03=3x02b,解得x0=0,b=1或x0=1,b=0.課時(shí)作業(yè)(十四)第1課時(shí)1.A[解析]函數(shù)f(x)的定義域是(0,+∞),且f'(x)=11x=x-1x,令f'(x)<0,解得0<x<1,所以函數(shù)f(x)的單調(diào)遞減區(qū)間是(02.D[解析]因?yàn)閒'(x)=sinx1<0,所以f(x)在(0,π)上是減函數(shù),故選D.3.A[解析]f'(x)=32x2+a,當(dāng)a≥0時(shí),f'(x)≥0恒成立,故“a>0”是“f(x)在R上單調(diào)遞增”的充分不必要條件4.A[解析]因?yàn)閒(x)=xsinx,所以f(x)=(x)·sin(x)=xsinx=f(x),所以函數(shù)f(x)是偶函數(shù),所以f-π3=fπ3.又當(dāng)x∈0,π2時(shí),f'(x)=sinx+xcosx>0,所以函數(shù)f(x)在0,π2上是增函數(shù),所以fπ5<f(1)<fπ3,即f-π3>f(1)>f5.k≥1[解析]因?yàn)閒'(x)=cosx+k≥0,所以k≥cosx,x∈(0,π)恒成立.當(dāng)x∈(0,π)時(shí),1<cosx<1,所以k≥1.6.B[解析]若函數(shù)f(x)=x33a2x2+x在區(qū)間[1,2]上單調(diào)遞減,則f'(x)=x2ax+1≤0在[1,2]上恒成立,即a≥x+1x在[1,2]上恒成立,又當(dāng)x∈[1,2]時(shí),x+1xmax=2+12=52,所以a≥7.C[解析]由題意知x>0,f'(x)=1+ax,若函數(shù)f(x)=x+alnx不是單調(diào)函數(shù),則方程1+ax=0在(0,+∞)上有解,即x=a>0,所以a<8.B[解析]依題意得,當(dāng)x>0時(shí),f'(x)>0,f(x)是增函數(shù);當(dāng)x<0時(shí),f'(x)<0,f(x)是減函數(shù).又f(3)=f(5)=1,因此不等式f(x)<1的解集是(3,5).9.C[解析]令函數(shù)F(x)=f(x)x22017,則F'(x)=f'(x)2x<0,則函數(shù)F(x)是減函數(shù),且滿足F(2)=f(2)42017=0,故不等式f(x)>x2+2017可化為F(x)>F(2),即原不等式f(x)>x2+2017的解集為{x|x<2}.應(yīng)選答案C.10.C[解析]由題意可得f'(x)=lnx+1bx+2x,滿足題意時(shí)f'(x)=lnx+1bx+2x≥0恒成立,即b≤x(lnx+2x+1).令g(x)=x(lnx+2x+1),則g'(x)=lnx+4x+2,很明顯g'(x)是定義域內(nèi)的增函數(shù),則g'(x)≥g'(1)=6>0,則函數(shù)g(x)在定義域內(nèi)單調(diào)遞增,在[1,e]上,g(x)min=g(1)=3,所以實(shí)數(shù)b的取值范圍是(∞,3].11.(0,e)[解析]由f'(x)=lnxx'=1-lnxx2>0(x>0),可得1-ln12.(∞,1)∪(1,+∞)[解析]設(shè)F(x)=f(x)12x,則F'(x)=f'(x)12,因?yàn)閒'(x)<12,所以F'(x)=f'(x)12<0,即函數(shù)F(x)在R上為減函數(shù).因?yàn)閒(x2)<x22+12,所以f(x2)x22<f(1)12,所以F(x2)<F(1),又函數(shù)F(x)在R上為減函數(shù),所以x2>1,即x∈(∞13.解:(1)f(x)的定義域?yàn)镽,且f'(x)=(ax+a1)ex.①當(dāng)a=0時(shí),f'(x)=ex<0,此時(shí)f(x)的單調(diào)遞減區(qū)間為(∞,+∞).②當(dāng)a>0時(shí),由f'(x)>0,得x>a-由f'(x)<0,得x<a-此時(shí)f(x)的單調(diào)遞減區(qū)間為∞,a-1a,單調(diào)遞增區(qū)間為a-1a,③當(dāng)a<0時(shí),由f'(x)>0,得x<a-由f'(x)<0,得x>a-此時(shí)f(x)的單調(diào)遞減區(qū)間為a-1a,+∞,單調(diào)遞增區(qū)間為∞,a-1(2)證明:當(dāng)m>n>0時(shí),要證men+n<nem+m,只要證m(en1)<n(em1),即證em-1m>e設(shè)g(x)=ex-1x,x>0,則g'(x)=(設(shè)h(x)=(x1)ex+1,由(1)知h(x)在[0,+∞)上單調(diào)遞增,所以當(dāng)x>0時(shí),h(x)>h(0)=0,于是g'(x)>0,所以g(x)在(0,+∞)上單調(diào)遞增,所以當(dāng)m>n>0時(shí),(*)式成立,故當(dāng)m>n>0時(shí),men+n<nem+m.14.解:(1)函數(shù)f(x)的定義域?yàn)镽,f'(x)=exa.當(dāng)a≤0時(shí),f'(x)>0,∴f(x)在R上為增函數(shù);當(dāng)a>0時(shí),由f'(x)=0得x=lna,則當(dāng)x∈(∞,lna)時(shí),f'(x)<0,∴函數(shù)f(x)在(∞,lna)上為減函數(shù),當(dāng)x∈(lna,+∞)時(shí),f'(x)>0,∴函數(shù)f(x)在(lna,+∞)上為增函數(shù).(2)當(dāng)a=1時(shí),g(x)=(xm)(exx)ex+x2+x.∵g(x)在(2,+∞)上為增函數(shù),∴g'(x)=xexmex+m+1≥0在(2,+∞)上恒成立,即m≤xex+1ex-1在令h(x)=xex+1ex-1,則h'(x)=(ex)令L(x)=exx2,L'(x)=ex1>0在(2,+∞)上恒成立,即L(x)=exx2在(2,+∞)上為增函數(shù),即L(x)>L(2)=e24>0,∴h'(x)>0在(2,+∞)上成立,即h(x)=xex+1ex-1在(2,+∞)上為增函數(shù),∴h(x)∴m≤2e∴實(shí)數(shù)m的取值范圍是∞,2e2+115.C[解析]由f(x)=f(2x)可得函數(shù)f(x)的圖像關(guān)于直線x=1對稱,∵當(dāng)x∈(∞,1)時(shí),(x1)·f'(x)<0,∴函數(shù)f(x)在(∞,1)上是增函數(shù),在(1,+∞)上是減函數(shù).由于a=f(0),b=f12,c=f(3)=f(23)=f(1),12>0>116.12,+∞[解析]f'(x)=lnx2ax+1,若f(x)在(0,+∞)上單調(diào)遞減,則lnx2ax+1≤0在(0,+∞)上恒成立,即a≥lnx+12x在(0,+∞)上恒成立.令g(x)=lnx+12x,x∈(0,+∞),則g'(x)=lnx2x2,令g'(x)>0,解得0<x<1,令g'(x)<0,解得x>1,故g(x)在(0,1)上單調(diào)遞增,在(1,+∞)上單調(diào)遞減,故g第2課時(shí)1.C[解析]因?yàn)閥=x·ex,所以y'=ex+xex=(1+x)ex,當(dāng)x∈(∞,1)時(shí),y'<0,當(dāng)x∈(1,+∞)時(shí),y'>0,所以當(dāng)x=1時(shí),ymin=(1)×e1=1e2.D[解析]x=2是函數(shù)f(x)=x33ax+2的極小值點(diǎn),即x=2是f'(x)=3x23a=0的根,將x=2代入得a=4,所以函數(shù)解析式為f(x)=x312x+2.由3x212=0,得x=±2,故函數(shù)在(2,2)上是減函數(shù),在(∞,2)和(2,+∞)上是增函數(shù),由此可知當(dāng)x=2時(shí)函數(shù)f(x)取得極大值,極大值為f(2)=18.3.B[解析]由函數(shù)極值的定義和導(dǎo)函數(shù)的圖像可知,f'(x)在(a,b)上與x軸的交點(diǎn)個(gè)數(shù)為4,但是在原點(diǎn)附近的導(dǎo)數(shù)值恒大于零,故x=0不是函數(shù)f(x)的極值點(diǎn),其余的3個(gè)交點(diǎn)都是極值點(diǎn),其中有2個(gè)點(diǎn)附近的導(dǎo)數(shù)值左正右負(fù),故極大值點(diǎn)有2個(gè).4.8[解析]f'(x)=6x24x,令f'(x)=0,得x=0或x=23.∵f(1)=4,f(0)=0,f23=827,f(2)=8,∴5.40[解析]由y'=x239x40=0,得x=1或x=40.當(dāng)0<x<40時(shí),y'<0;當(dāng)x>40時(shí),y'>0.所以當(dāng)x=40時(shí),y有最小值.6.B[解析]求得f(x)的導(dǎo)函數(shù)f'(x)=3x2+2x+5a,三次函數(shù)f(x)有極值,則f'(x)=0有兩個(gè)不相等的解,∴Δ=460a>0,∴a<1157.B[解析]因?yàn)閒'(x)=1x1=1-xx,當(dāng)x∈(0,1)時(shí),f'(x)>0,當(dāng)x∈(1,e]時(shí),f'(x)<0,所以f(x)的單調(diào)遞增區(qū)間是(0,1),單調(diào)遞減區(qū)間是(1,e],所以當(dāng)x=1時(shí),f(x)取得最大值ln18.D[解析]由題意可得f(m)=m3+am2+bm=0,m≠0,則m2+am+b=0①,且f'(m)=3m2+2am+b=0②,①②化簡得m=a2,∴f'(x)=3x2+2ax+b=0的兩根分別為a2和a6,∴b=a24.易知f-a6=12,解得a=9.B[解析]因?yàn)閒(x)=x(lnxax),所以f'(x)=lnx2ax+1.由題可知f'(x)在(0,+∞)上有兩個(gè)不同的零點(diǎn),令f'(x)=0,則2a=lnx+1x.令g(x)=lnx+1x,則g'(x)=-lnxx2,所以g(x)在(0,1)上單調(diào)遞增,在(1,+∞)上單調(diào)遞減,又因?yàn)楫?dāng)x從右邊趨近于0時(shí),g(x)→∞,當(dāng)x→+∞時(shí),g(x)→0,而g(x)max=g(1)=1,所以只需010.C[解析]由題意,f'(x)=x2+2x=x(x+2),故f(x)在(∞,2),(0,+∞)上是單調(diào)遞增的,在(2,0)上是單調(diào)遞減的,作出其大致圖像如圖所示,令13x3+x223=23,得x=0或x=3,則結(jié)合圖像可知,-3≤a<0,a11.4[解析]由f(x)=x3+ax2+bx+a2,得f'(x)=3x2+2ax+b,由題知f'(1)=0,f(1)=10,即2a+b+3=0,a2+a+b+1=10,解得a=4,b=-11或a=-12.1e[解析]當(dāng)x>0時(shí),f(x)=exx1,f'(x)=ex(x-1)x2,∴當(dāng)x∈(0,1)時(shí),f'(x)<0,函數(shù)單調(diào)遞減,當(dāng)x∈(1,+∞)時(shí),f'(x)>0,函數(shù)單調(diào)遞增,∴當(dāng)x=1時(shí),函數(shù)取得極小值也即最小值e1.∵函數(shù)f(x)13.解:(1)由題意可得函數(shù)f(x)=12ax2+lnx的定義域?yàn)?0,+∞由求導(dǎo)公式可得f'(x)=ax+1x=a當(dāng)a≥0時(shí),f'(x)=ax2+1x>0,f(x)在(0,當(dāng)a<0時(shí),令ax2+1x>0,得x<-1a,即f(x)在0同理由ax2+1x<0,得x>-1a,即f(x)在-綜上,當(dāng)a≥0時(shí),f(x)的單調(diào)遞增區(qū)間為(0,+∞);當(dāng)a<0時(shí),f(x)的單調(diào)遞增區(qū)間為0,-1a,單調(diào)遞減區(qū)間為-1a,+∞.(2)由(1)可知,若a≥0,則f(x)在(0,1]上單調(diào)遞增,故函數(shù)在x=1處取到最大值,f(1)=12a=1,解得a=2,與a≥0矛盾,應(yīng)舍去若0<-1a≤1,即a≤1,則函數(shù)f(x)在0,-1a上單調(diào)遞增,在-1a,+∞上單調(diào)遞減,故函數(shù)在x=-1a處取得最大值,f-1a=若-1a>1,即1<a<0,則f(x)在(0,1]故函數(shù)在x=1處取到最大值,f(1)=12a=1,解得a=2,應(yīng)舍去綜上可得所求a的值為e.14.解:(1)當(dāng)m=1時(shí),f'(x)=1x(2x1)=2x2-x-1x=(2當(dāng)x∈(0,1)時(shí),f'(x)>0,f(x)單調(diào)遞增;當(dāng)x∈(1,+∞)時(shí),f'(x)<0,f(x)單調(diào)遞減.所以函數(shù)f(x)在x=1處取得極大值,也是最大值,且f(x)max=f(1)=0,f(x)無最小值.(2)f'(x)=2mx2-mx+1x,x當(dāng)m=0時(shí),f'(x)=1x>0,函數(shù)f(x)在(0,+∞)上單調(diào)遞增,無極值點(diǎn)當(dāng)m>0時(shí),設(shè)g(x)=2mx2mx+1,Δ=m28m.①若0<m≤8,則Δ≤0,f'(x)≥0,函數(shù)f(x)在(0,+∞)上單調(diào)遞增,無極值點(diǎn).②若m>8,則Δ>0,設(shè)方程2mx2mx+1=0的兩個(gè)根分別為x1,x2(不妨設(shè)x1<x2),因?yàn)閤1+x2=12,g(0)=1>0,所以0<x1<14,x2>所以當(dāng)x∈(0,x1)時(shí),f'(x)>0,函數(shù)f(x)單調(diào)遞增;當(dāng)x∈(x1,x2)時(shí),f'(x)<0,函數(shù)f(x)單調(diào)遞減;當(dāng)x∈(x2,+∞)時(shí),f'(x)>0,函數(shù)f(x)單調(diào)遞增.因此函數(shù)有兩個(gè)極值點(diǎn),滿足題意.當(dāng)m<0時(shí),Δ>0,由g(0)=1>0,可得x1<0.所以當(dāng)x∈(0,x2)時(shí),f'(x)>0,函數(shù)f(x)單調(diào)遞增;當(dāng)x∈(x2,+∞)時(shí),f'(x)<0,函數(shù)f(x)單調(diào)遞減.因此函數(shù)有一個(gè)極值點(diǎn),滿足題意.綜上,函數(shù)f(x)有一個(gè)極值點(diǎn)時(shí)m<0,函數(shù)f(x)有兩個(gè)極值點(diǎn)時(shí)m>8.15.B[解析]因?yàn)閒(x)+xf'(x)=lnx,即[xf(x)]'=lnx,所以xf(x)=xlnxx+c,其中c為常數(shù).又因?yàn)閒(1)=0,所以xf(x)=xlnxx+1,f(x)=lnx1+1x,所以f'(x)=1x1x2=x-1x2(x>0).當(dāng)0<x<1時(shí),f'(x)<0,當(dāng)x>1時(shí),f'(x)>0,所以函數(shù)f(x16.C[解析]f'(x)=ax(1+2a)+2x=ax2-(2a+1)x+2x(a>0,x>0),若f(x)在12,1內(nèi)有極大值,則f'(x)在12,1內(nèi)先大于0,再小于第3課時(shí)1.解:(1)∵f'(x)=2x6x11,f'(1)=15,f(1)=14∴曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程為y(14)=15(x1),即15x+y1=0.(2)關(guān)于x的不等式f(x)≤(a3)x2+(2a13)x2恒成立?2lnxax22ax+2x+2≤0恒成立.令h(x)=2lnxax22ax+2x+2(x>0),則h'(x)=2x2ax2a+2=-2(ax-當(dāng)a≤0時(shí),h'(x)>0恒成立,h(x)在(0,+∞)上單調(diào)遞增,且當(dāng)x→+∞時(shí),h(x)→+∞,不符合題意.當(dāng)a>0時(shí),若x∈0,1a,則h'(x)>0,若x∈1a,+∞,則h'(x)<0,故h(x)在0,1a上單調(diào)遞增,在1a,+∞上單調(diào)遞減,h(x)max=h1a=2lna+1a≤0.令g(a)=2lna+1a,則g(a)在(0,+∞)上單調(diào)遞減又g(1)=1>0,g(2)=122ln2<0,∴當(dāng)a≥2時(shí),g(a)<0恒成立∴整數(shù)a的最小值為2.2.解:(1)當(dāng)a=0時(shí),f(x)=xlnx+x1,由f'(x)=lnx,得f'(e)=1,f(e)=1,∴曲線y=f(x)在點(diǎn)P(e,f(e))處的切線方程為y+1=(xe),即x+y+1e=0.(2)f'(x)=2ax1lnx(2a1)=2a(x1)lnx,易知當(dāng)x∈[1,+∞)時(shí),lnx≤x1,則f'(x)≥2a(x1)(x1)=(2a1)(x1).當(dāng)2a1≥0,即a≥12時(shí),由x∈[1,+∞)得f'(x)≥0恒成立f(x)在[1,+∞)上單調(diào)遞增,f(x)≥f(1)=0,符合題意.當(dāng)a≤0時(shí),由x∈[1,+∞)得f'(x)≤0恒成立,f(x)在[1,+∞)上單調(diào)遞減,f(x)≤f(1)=0,顯然不合題意,a≤0舍去.當(dāng)0<a<12時(shí),由lnx≤x1,得ln1x≤1x1,即lnx≥則f'(x)≤2a(x1)11x=x-1x(2ax∵0<a<12,∴12a當(dāng)x∈1,12a時(shí),f'(x)≤0恒成立,∴f(x)在1,12a上單調(diào)遞減,∴當(dāng)x∈1,12a時(shí),f(x)≤f(1)=0,顯然不合題意,0<a<12舍去綜上可得a∈12,+∞.3.證明:(1)f'(x)=1xa設(shè)f(x)的圖像與x軸相切于點(diǎn)(x0,0),則f(x0)=0,f'(所以f(x)=lnx+1x1f(x)≤(x-1)2x等價(jià)于ln設(shè)h(x)=lnxx+1,則h'(x)=1x1當(dāng)0<x<1時(shí),h'(x)>0,h(x)單調(diào)遞增;當(dāng)x>1時(shí),h'(x)<0,h(x)單調(diào)遞減.所以h(x)≤h(1)=0,即lnx≤x1,(*)所以f(x)≤(x(2)設(shè)g(x)=(b1)logbxx2-12,則g'(x)=由g'(x)=0,得x=b-1ln由(*)式可得,當(dāng)x>1時(shí),lnx<x1,即x-1ln以1x代換x可得ln1x<1x1,有l(wèi)nx>x-所以當(dāng)b>1時(shí),有1<t<b.當(dāng)1<x<t時(shí),g'(x)>0,g(x)單調(diào)遞增;當(dāng)t<x<b時(shí),g'(x)<0,g(x)單調(diào)遞減.又因?yàn)間(1)=g(b)=0,所以g(x)>0,即(b1)logbx>x24.解:(1)∵f(x)=ln(x1)k(x1)+1,∴f'(x)=1x-1k,∴當(dāng)k≤0時(shí),f'(x)=1x-1k>0,f(x)在(1,+∞當(dāng)k>0時(shí),令f'(x)>0,得1<x<1+1k,令f(x)<0,得x>1+1∴f(x)在1,1+1k上是增函數(shù),在1+1k,+∞上是減函數(shù).(2)∵f(x)≤0恒成立,∴?x>1,ln(x1)k(x1)+1≤0,∴?x>1,ln(x1)≤k(x1)1,∴k>0.由(1)知,當(dāng)k>0時(shí),f(x)max=f1+1k=lnk≤0,解得k≥1.故實(shí)數(shù)k的取值范圍是[1,+∞).(3)證明:令k=1,則由(2)知,ln(x1)≤x2對任意x∈(1,+∞)恒成立,即lnx≤x1對任意x∈(0,+∞)恒成立.取x=n2,則2lnn≤n21,即lnnn+1<n-1∴l(xiāng)n23+ln34+ln45+…+lnnn+1<n(n-15.解:(1)f'(x)=2a(x+1)ln(x+1)+a(x+1)+b,∵f'(0)=a+b=0,f(e1)=ae2+b(e1)=a(e2e+1)=e2e+1,∴a=1,b=1.(2)證明:f(x)=(x+1)2ln(x+1)x,設(shè)g(x)=(x+1)2ln(x+1)xx2(x≥0),則g'(x)=2(x+1)ln(x+1)x.令φ(x)=2(x+1)ln(x+1)x,則φ'(x)=2ln(x+1)+1>0,∴φ(x)在[0,+∞)上單調(diào)遞增,∴φ(x)≥φ(0)=0,∴g(x)在[0,+∞)上單調(diào)遞增,∴g(x)≥g(0)=0,∴f(x)≥x2.(3)設(shè)h(x)=(x+1)2ln(x+1)xmx2,h'(x)=2(x+1)ln(x+1)+x2mx,由(2)可知(x+1)2ln(x+1)≥x2+x=x(x+1),∴(x+1)ln(x+1)≥x,∴h'(x)≥3x2mx.①當(dāng)32m≥0,即m≤32時(shí),h'(x)≥0∴h(x)在[0,+∞)上單調(diào)遞增,∴h(x)≥h(0)=0,滿足題意.②當(dāng)32m<0,即m>32時(shí),令u(x)=h'(x)=2(x+1)ln(x+1)+(12m)x,則u'(x)=2ln(x+1)+32m令u'(x0)=0,得x0=e2m-3當(dāng)x∈[0,x0)時(shí),u(x)≤u(0)=0,∴h(x)在[0,x0)上單調(diào)遞減,∴h(x)≤h(0)=0,不滿足題意.綜上,m≤326.解:(1)當(dāng)a=1時(shí),f(x)=e2x+ln(x+1),f'(x)=2e2x+1x①f(0)=1,f'(0)=2+11=3,∴曲線y=f(x)在(0,f(0))處的切線方程為3xy+1=0②證明:設(shè)F(x)=e2x+ln(x+1)(x+1)2x(x≥0),則F'(x)=2e2x+1x+12(x+1)令G(x)=F'(x),則G'(x)=4e2x1(x+1)22=e2x1(x+1)2+2(e2x1∴G(x)在[0,+∞)上單調(diào)遞增,∴G(x)≥G(0)=0,∴F(x)在[0,+∞)上單調(diào)遞增,∴F(x)≥F(0)=0,故得證.(2)原問題等價(jià)于存在x0≥0,使得e2x0ln(x0+a)x設(shè)u(x)=e2xln(x+a)x2,則u'(x)=2e2x1x+令φ(x)=2e2x1x+a2x,則φ'(x)=4e2x+1(∴φ(x)在[0,+∞)上單調(diào)遞增,∴φ(x)≥φ(0)=21a當(dāng)a≥12時(shí),φ(0)=21a≥∴u(x)在[0,+∞)上單調(diào)遞增,∴u(x)min=u(0)=1lna<0,∴a>e.當(dāng)a<12時(shí),ln(x+a)<lnx+12,設(shè)h(x)=x12lnx+12(x≥0),則h'(x)=11x+1令h'(x)>0得x>12,h'(x)<0得0<x<1∴h(x)在0,12上單調(diào)遞減,在12,+∞上單調(diào)遞增,∴h