高考數(shù)學(xué)大一輪復(fù)習(xí) 課時(shí)作業(yè)8 指數(shù)與指數(shù)函數(shù) 理 新人教A版-新人教A版高三數(shù)學(xué)試題

課時(shí)作業(yè)(八)第8講指數(shù)與指數(shù)函數(shù)時(shí)間/30分鐘分值/75分基礎(chǔ)熱身 1.[2018·青島二模]已知方程x2-3x+1=0的兩個(gè)根為x1,x2,則2x1·2x2A.3 B.6C.8 D.22.已知函數(shù)f(x)=ax-1+4的圖像恒過(guò)定點(diǎn)P,則點(diǎn)P的坐標(biāo)是 ()A.(1,5) B.(1,4)C.(0,4) D.(4,0)3.已知a=0.80.7,b=0.80.9,c=1.20.8,則a,b,c的大小關(guān)系是 ()A.a>b>c B.c>a>bC.b>a>c D.c>b>a4.函數(shù)f(x)=ax與g(x)=-x+a在同一坐標(biāo)系中的圖像可能是()ABCD圖K8-15.不等式3-x2+2x>能力提升6.函數(shù)y=ax在[0,1]上的最大值與最小值的和為54,則函數(shù)y=3·a2x-1在[0,1]上的最大值為 (A.16 B.15C.12 D.37.[2018·三明5月質(zhì)檢]若a=π-2,b=aa,c=aaa,則a,b,c的大小關(guān)系為 (A.c>b>a B.b>c>aC.b>a>c D.a>b>c8.若對(duì)于任意x∈(-∞,-1],都有(3m-1)2x<1成立,則m的取值范圍是 ()A.-∞,1C.(-∞,1) D.(-∞,1]9.已知函數(shù)y=f(x)的圖像關(guān)于直線x=1對(duì)稱(chēng),當(dāng)x<1時(shí),f(x)=12x-1,那么當(dāng)x>1時(shí),函數(shù)f(xA.(-∞,0) B.(1,2)C.(2,+∞) D.(2,5)10.已知實(shí)數(shù)a≠1,函數(shù)f(x)=4x,x≥0,2a-x,x<011.[2018·湖南八校聯(lián)考]對(duì)于給定的函數(shù)f(x)=ax-a-x(x∈R,a>0,且a≠1),下面五個(gè)結(jié)論中正確的是.(填序號(hào))

①函數(shù)f(x)的圖像關(guān)于原點(diǎn)對(duì)稱(chēng);②函數(shù)f(x)在R上不具有單調(diào)性;③函數(shù)f(|x|)的圖像關(guān)于y軸對(duì)稱(chēng);④當(dāng)0<a<1時(shí),函數(shù)f(|x|)的最大值是0;⑤當(dāng)a>1時(shí),函數(shù)f(|x|)的最大值是0.12.(10分)已知函數(shù)f(x)=a·4x-a·2x+1+1-b(a>0)在區(qū)間[1,2]上有最大值9和最小值1.(1)求a,b的值;(2)若不等式f(x)-k·4x≥0在x∈[-1,1]時(shí)有解,求實(shí)數(shù)k的取值范圍.難點(diǎn)突破13.(5分)已知函數(shù)f(x)=ex-e-x2,x1,x2,x3∈R,且x1+x2>0,x2+x3>0,x3+x1>0,則f(x1)+f(x2)+f(A.一定等于零 B.一定大于零C.一定小于零 D.正負(fù)都有可能14.(5分)已知函數(shù)f(x)=2-x,給出下列結(jié)論:①若x>0,則f(x)>1;②對(duì)于任意的x1,x2∈R,x1-x2≠0,必有(x1-x2)·[f(x1)-f(x2)]<0;③若0<x1<x2,則x2f(x1)<x1f(x2);④對(duì)于任意的x1,x2∈R,x1-x2≠0,必有f(x1其中所有正確結(jié)論的序號(hào)是.

課時(shí)作業(yè)(八)1.C[解析]由題得x1+x2=3,∴2x1·2x2=2x1+x22.A[解析]令x-1=0?x=1,又f(1)=5,故圖像恒過(guò)定點(diǎn)P(1,5).3.B[解析]易知b=0.80.9<0.80.7=a<1<1.20.8=c,故選B.4.A[解析]因?yàn)楹瘮?shù)g(x)單調(diào)遞減,所以排除選項(xiàng)C,D,又因?yàn)楹瘮?shù)f(x)=ax單調(diào)遞增時(shí),a>1,所以當(dāng)x=0時(shí),g(0)=a>1=f(0),所以排除選項(xiàng)B,故選A.5.(-1,4)[解析]由3-x2+2x>13x∴-x2+2x>-x-4,即x2-3x-4<0,∴不等式3-x2+2x>16.C[解析]∵函數(shù)y=ax在定義域上是單調(diào)函數(shù),且y=ax在[0,1]上的最大值與最小值的和為54,∴1+a=54,解得a=14,∴函數(shù)y=3·a2x-1=3·142x-1=12·116x.∵函數(shù)y=12·116x在定義域上為減函數(shù),∴當(dāng)x=0時(shí),函數(shù)y=3·a7.B[解析]由題意可知a=π-2=1π2∈(0,1),即則函數(shù)f(x)=ax單調(diào)遞減,則aa>a1,即aa>a.由于aa>a,所以結(jié)合函數(shù)的單調(diào)性可得aaa<aa,即由于0<a<1,故aa<1,結(jié)合函數(shù)的單調(diào)性可得aaa>a1,綜上可得,a,b,c的大小關(guān)系為b>c>a.8.C[解析]∵2x>0,∴不等式(3m-1)2x<1對(duì)于任意x∈(-∞,-1]恒成立等價(jià)于3m-1<12x=12x對(duì)于任意x∈(-∞,-1]恒成立.∵x≤-1,∴12x≥12-1=2,∴3m-1<2,解得m<1,9.C[解析]如圖,畫(huà)出函數(shù)y=f(x)的圖像,可知當(dāng)x>1時(shí),函數(shù)f(x)的單調(diào)遞增區(qū)間為(2,+∞),故選C.10.12[解析]當(dāng)a<1時(shí),41-a=21,所以a=1當(dāng)a>1時(shí),22a-1=4a-1,無(wú)解.所以a的值為1211.①③④[解析]∵f(-x)=-f(x),x∈R,∴f(x)為奇函數(shù),∴f(x)的圖像關(guān)于原點(diǎn)對(duì)稱(chēng),①正確;當(dāng)a>1時(shí),f(x)在R上為增函數(shù),當(dāng)0<a<1時(shí),f(x)在R上為減函數(shù),②錯(cuò)誤;y=f(|x|)是偶函數(shù),其圖像關(guān)于y軸對(duì)稱(chēng),③正確;當(dāng)0<a<1時(shí),y=f(|x|)在(-∞,0)上為增函數(shù),在[0,+∞)上為減函數(shù),∴當(dāng)x=0時(shí),y=f(|x|)取得最大值,為0,④正確;當(dāng)a>1時(shí),y=f(|x|)在(-∞,0)上為減函數(shù),在[0,+∞)上為增函數(shù),∴當(dāng)x=0時(shí),y=f(|x|)取得最小值,為0,⑤錯(cuò)誤.綜上,正確結(jié)論是①③④.12.解:(1)令n=2x∈[2,4],則y=an2-2an+1-b(a>0),n∈[2,4]有最大值9和最小值1,易知函數(shù)y=an2-2an+1-b的圖像的對(duì)稱(chēng)軸為直線n=1,∴當(dāng)n=2時(shí),ymin=4a-4a+1-b=1,當(dāng)n=4時(shí),ymax=16a-8a+1-b=9,∴a=1,b=0.(2)由(1)知,4x-2·2x+1-k·4x≥0在x∈[-1,1]時(shí)有解.設(shè)2x=t,∵x∈[-1,1],∴t∈12∴t2-2t+1-kt2≥0在t∈12,∴k≤t2-2t+1t2=1-2再令1t=m,則m∈1∴k≤m2-2m+1=(m-1)2≤1,即k≤1,故實(shí)數(shù)k的取值范圍是(-∞,1].13.B[解析]由已知可得f(x)為奇函數(shù),且f(x)在R上是增函數(shù).由x1+x2>0?x1>-x2?f(x1)>f(-x2)=-f(x2),同理可得f(x2)>-f(x3),f(x3)>-f(x1),故f(x1)+f(x2)+f(x3)>-[f(x2)+f(x3)+f(x1)]?f(x1)+f(x2)+f(x3)>0.14.②④[解析]f(x)=2-x=12對(duì)于①,當(dāng)x>0時(shí),12x∈(0,1),故①對(duì)于②,f(x)=12x在