一元二次函數(shù)、方程和不等式練習(xí)題- 高三數(shù)學(xué)一輪復(fù)習(xí)

高考數(shù)學(xué)一輪復(fù)習(xí)專題二一元二次函數(shù)、方程和不等式一、選擇題1.若集合A=xx+2x?1≤0,B={x|x2-x-2<0},則A∩B= ()A.[-2,2) B.(-1,1]C.(-1,1) D.(-1,2)2.已知集合U={-2,-1,0,1,2},A={0},B={x|x2+x-2<0},則(?UA)∩B= ()A.{-1} B.{1}C.{-1,1,2} D.{-2,-1,1}3.設(shè)x∈R,則“x2-3x<0”是“1<x<2”的 ()A.充要條件 B.充分不必要條件C.必要不充分條件 D.既不充分也不必要條件4.設(shè)a,b是實(shí)數(shù),則“a>b”是“a2>b2”的 ()A.充分不必要條件 B.必要不充分條件C.充要條件 D.既不充分也不必要條件5.設(shè)x∈R,則“2x<4”是“x2-x-2<0”的 ()A.充分不必要條件 B.必要不充分條件C.充要條件 D.既不充分也不必要條件6.若x>1,則4x+1x?1的最小值等于 (A.4 B.6 C.8 D.97.已知x∈[0,2],則x(2?x)的最大值是 A.8 B.2 C.1 D.08.若log4(3a+4b)=log2ab,則a+b的最小值是 ()A.6+23 B.7+23 C.6+43 D.7+439.設(shè)a>0,b>1,若a+b=2,則4a+1b?1的最小值為 A.6 B.9 C.32 D.1810.已知實(shí)數(shù)a,b,且ab>0,則aba2+b2A.16 B.14 C.1二、填空題11.若不等式2kx2+kx-38<0對(duì)一切實(shí)數(shù)x都成立,則k的取值范圍為12.若實(shí)數(shù)x,y滿足xy=1,則x2+2y2的最小值為.

13.設(shè)a>0,b>0,若a與b2的等差中項(xiàng)是2,則log2a+2log2b的最大值是.

14.函數(shù)y=a1-x(a>0,且a≠1)的圖象恒過定點(diǎn)A,若點(diǎn)A在直線mx+ny-1=0(mn>0)上,則1m+1n的最小值為15.已知a>0,b>0,且ab=1,則12a+12b+16.已知x>0,y>0,且2x+8y-xy=0,則x+y的最小值為.

17.若4x+4y=1,則x+y的取值范圍是.

18.已知正實(shí)數(shù)a,b滿足lg(a+b)=lg2ba+lgab,則12a+119.若a>0,b>0,當(dāng)(a+4b)2+1ab取得最小值為時(shí),a+b=20.已知a>0,b>0,且a+4b=2,則a2+2b21.若a>0,b>0,則1a+ab2+b22.已知a>1,b>1,若loga2+logb16=3,則log2(ab)的最小值為.

B能力提升一、選擇題1.已知實(shí)數(shù)a>b>0,c∈R,則下列不等式恒成立的是 ()A.ac<bc B.b+1aC.b+1a+1>ba 2.若x>0,則下列說法正確的是 ()A.x+1x的最小值為2 B.x+1xC.2x+12x的最小值為2 D.lgx+13.已知a>0,b>0,a-1b2=ba,則當(dāng)a+1b取最小值時(shí),a2+1b2的值為 A.2 B.22 C.3 D.44.(2022·河?xùn)|區(qū)期中)若x>0,y>0,且2x+1y=1,x+2y>m2+7m恒成立,則實(shí)數(shù)m的取值范圍是 (A.(-8,1) B.(-∞,-8)∪(1,+∞)C.(-∞,-1)∪(8,+∞) D.(-1,8)二、填空題5.已知正實(shí)數(shù)a,b滿足a+b=1,則a2+4a+b6.二次函數(shù)y=ax2+4x+c的最小值為0,則1a+1c的最小值為7.函數(shù)y=(x+5)(x+2)x+1(8.若a,b是正實(shí)數(shù),且a+b=1,則1a+1ab的最小值為9.已知首項(xiàng)與公比相等且不為1的等比數(shù)列{an}中,若m,n∈N*,滿足aman2=a62,則2m10.若a,b均為正實(shí)數(shù),且a+2b=1,則a+b+111.已知正實(shí)數(shù)x,y,z滿足x2+y2+z2=22,則xy+yz的最大值為.

12.設(shè)m>n>0,則m2+1mn+1m(13.已知a>0,b>0,且a+2b=2,則2a+1b+ab的最小值為14.已知a>0,b>0,a+b=1,則1a+3b+115.已知a>0,b>0,c>0,a+b+c=2,則4a+b+a16.已知x,y為正實(shí)數(shù),則2yx+9x17.已知a>0,b>0,且a+b=1,則1a2-11b2-1的最小值是.

18.已知正實(shí)數(shù)x,y滿足x+y=1x+9y+6,則x+y的最小值是19.已知實(shí)數(shù)x,y滿足x>1,y>0且x+4y+1x?1+1y=11,則1x?120.已知x,y為正實(shí)數(shù),且xy+2x+4y=41,則x+y的最小值為.

21.已知正實(shí)數(shù)a,b滿足a+b=1,則1a+2ab22.設(shè)正實(shí)數(shù)x,y,z滿足x2-3xy+4y2-z=0,則當(dāng)xyz取得最大值時(shí),x+2y-z的最大值為23.(2021·塘沽一中三模)已知正實(shí)數(shù)a,b滿足a+b=1,則2aa2+b24.已知ab>0,則(a2+425.已知a>0,b>0,且3a+2+3b+2=1,則a+2答案A基礎(chǔ)夯實(shí)1.CA=xx+2x?1≤0={x|-2≤x<1},B={x|x2-x-2<0}={x|-1<x<2},A∩B={x|-1<x<1}.故選C2.A?UA={-2,-1,1,2},B={x|-2<x<1},(?UA)∩B={-1}.故選A.3.C∵x2-3x<0,∴0<x<3.∵{x|1<x<2}?{x|0<x<3},∴“x2-3x<0”是“1<x<2”的必要不充分條件.故選C.4.D由a>b不一定能推出a2>b2,如1>-2;反之也不成立,如(-2)2>12,故“a>b”是“a2>b2”的既不充分也不必要條件.故選D.5.B由2x<4可得x<2,由x2-x-2<0可得-1<x<2,所以“2x<4”是“x2-x-2<0”的必要不充分條件.故選B.6.C由x>1可得x-1>0,則4x+1x?1=4(x-1)+1x?1+4≥24(x?1)·1x?1+4=8,當(dāng)且僅當(dāng)4(x-1)=1x?17.C∵x(2?x)=?x2+2x=?(x?1)2+1,∵x∈[0,2],∴x-1∈[-1,1],∴(x-1)2∈[0,1],∴-(x-1)2+1∈[0,1],故若x∈8.D∵log4(3a+4b)=log2ab,∴3a+4b=ab>0,∴3b+4a=1且a>0,b>0,∴a+b=(a+b)3b+4a=3+4+3ab+4ba≥7+212=7+43,當(dāng)且僅當(dāng)3ab=4ba時(shí)等號(hào)成立,∴9.B∵a>0,b>1,a+b=2,∴a+b-1=1,∴4a+1b?1=[a+(b-1)]4a+1b?1=4+1+4(b?1)a+ab?1≥5+24(b?1)a·ab?1=9,當(dāng)且僅當(dāng)4(b?1)a=ab?1,10.A∵ab>0,∴ab>0,ba>0.∴aba2+b2+a2b2+4=1ab+ba+ab+4ab11.(-3,0]解析:當(dāng)k=0時(shí),-38<0,滿足題意;當(dāng)k≠0時(shí),則k<0,Δ<0,即k<0,k2?4·212.22解析:∵xy=1,∴x2+2y2≥22xy=22,當(dāng)且僅當(dāng)x2=2y2,即x=±42時(shí)取等號(hào),∴x2+2y2的最小值為2213.2解析:由題意知a+b2=4,又∵a>0,b>0,∴l(xiāng)og2a+2log2b=log2ab2≤log2a+b222=2,當(dāng)且僅當(dāng)a=b2=2時(shí)等號(hào)成立,∴l(xiāng)og2a+2log2b的最大值為14.4解析:y=a1-x的圖象恒過定點(diǎn)A(1,1),∴m+n=1.又∵mn>0,∴1m+1n=1m+1n(m+n)=2+nm+mn≥4.當(dāng)且僅當(dāng)m=n=12時(shí)等號(hào)成立,∴115.4解析:∵a>0,b>0,∴a+b>0,ab=1,∴12a+12b+8a+b=ab2a+ab2b+8a+b=a+b2+8a+b≥2a+b2·8a+b=4.當(dāng)且僅當(dāng)a+b=4時(shí)取等號(hào),結(jié)合ab=1,解得16.18解析:∵2x+8y-xy=0,∴2y+8x=1,又x>0,y>0,∴x+y=(x+y)8x+2y=8+2+8yx+2xy≥10+28yx·2xy=18.當(dāng)且僅當(dāng)17.(-∞,-1]解析:∵4x+4y=1≥24x·4y=2x+y+1,∴x+y+1≤0,∴x+y≤-1,∴x+y的取值范圍是(-∞18.1+52解析:因?yàn)閘g(a+b)=lg2ba+lgab,所以lg(a+b)=lg2,所以a+b=2,所以12a+12b+ab=141a+1b(a+b)+ab=12+b4a+5a4b≥12+2b4a·5a19.854解析:∵a>0,b>0,∴a+4b≥4ab,當(dāng)且僅當(dāng)a=4b時(shí)等號(hào)成立,∴(a+4b)2+1ab≥16ab+1ab≥8.當(dāng)且僅當(dāng)16ab=1ab時(shí)等號(hào)成立.由a=4b,16ab=1ab,解得a=1,b=14,∴20.5解析:因?yàn)閍>0,b>0,且a+4b=2,則a2+2bab=ab+2a=ab+a+4ba=1+ab+4ba≥1+2ab·4ba=5.當(dāng)且僅當(dāng)ab=4ba且21.22解析:∵a>0,b>0,∴1a+ab2+b≥21a·ab2+b=2b+b≥22b·b=22,當(dāng)且僅當(dāng)1a=ab2且2b=b,即a=22.3解析:令x=loga2,y=logb16,則a=21x,b=24y,x+y=3,所以log2(ab)=log2a+log2b=1x+4y,所以1x+4y=131x+4y(x+y)=131+4+yx+4xy≥135+2yx·4xy=3,當(dāng)且僅當(dāng)B能力提升1.C當(dāng)c≥0時(shí),不等式ac<bc不成立,A錯(cuò)誤;b+1a+1-ba=ab+a?ab?ba(a+1)=a?ba(a+1)>0,故B錯(cuò)誤,2.A對(duì)于A,由于x>0,故x+1x≥2,當(dāng)且僅當(dāng)x=1時(shí),等號(hào)成立,A正確;對(duì)于B,x+1+1x+1-1≥2-1=1,當(dāng)且僅當(dāng)x=0時(shí),等號(hào)成立,由于x>0,與題意矛盾,B錯(cuò)誤;對(duì)于C,2x+12x≥22x·12x=2,當(dāng)且僅當(dāng)x=0時(shí),等號(hào)成立,由于x>0,與題意矛盾,C錯(cuò)誤;對(duì)于D,當(dāng)x=1時(shí),lgx3.C∵a-1b2=ba,∴a2+1b2=ba+2ab.又∵a>0,b>0,∴a+1b2=a2+1b2+2ab=ba+4ab≥4,當(dāng)且僅當(dāng)b=2a=2±2時(shí)等號(hào)成立,∴a+1b≥2,當(dāng)且僅當(dāng)b=2a時(shí)等號(hào)成立.∴a+1b取最小值時(shí),4.A∵x>0,y>0,且2x+1y=1,∴x+2y=(x+2y)·2x+1y=4+4yx+xy≥4+24yx·xy=8,當(dāng)且僅當(dāng)4yx=xy且2x+1y=1,即x=4,y=2時(shí)取等號(hào),∵x+2y>m2+7m恒成立,∴8>m2+7m,解得-8<5.10解析:已知正實(shí)數(shù)a,b滿足a+b=1,則a2+4a+b2+1b=a+4a+b+1b=a+b+4a+1b=1+4a+1b=1+(a+b)4a+1b=1+5+ab+4ba≥6+2ab·4ba=10,當(dāng)且僅當(dāng)ab=4ba且6.1解析:因?yàn)槎魏瘮?shù)有最小值,所以a>0,且4ac?424a=0,則ac=4,且c>0,所以1a+1c=14ac1a+1c=14(a+c)≥14×2ac=1,當(dāng)且僅當(dāng)a=c=27.9解析:∵x>-1,∴x+1>0.令x+1=t,則x=t-1,∴y=(x+5)(x+2)x+1=(t+4)(t+1)t=t+4t+5≥9,當(dāng)且僅當(dāng)t=4t,8.3+22解析:∵a>0,b>0,a+b=1,∴1a+1ab=1a+1ab·(a+b)=1+ba+1b+1a=1+ba+1b+1a·(a+b)=1+ba+ab+1+1+ba=3+2ba+ab≥3+22.當(dāng)且僅當(dāng)29.23解析:設(shè)等比數(shù)列{an}的公比為q,則首項(xiàng)a1=q,由aman2=a62得:a1qm-1·(a1qn-1)2=(a1q5)2,則:qm+2n=q12,∴m+2n=12.∴2m+1n=112·2m+1n·(m+2n)=112·2+4nm+mn+2=112·4+4nm+mn.∵m,n∈N*,∴4nm>0,mn>0,則4nm+mn≥24nm·mn=4(當(dāng)且僅當(dāng)4nm=mn,即2n=m=6時(shí)取等號(hào)10.43+7解析:∵a>0,b>0,a+2b=1,∴a+b+1ab=2a+3bab=2b+3a(a+2b)=2ab+6ba+3+4≥7+43.當(dāng)且僅當(dāng)11.2解析:由于22=x2+y2+z2=x2+12y2+12y2+z2=x2+22y2+22y2+z2≥2xy+2yz,當(dāng)且僅當(dāng)x=z=22y時(shí)取等號(hào),∴xy+yz≤2,可得xy+yz的最大值為2.12.4解析:因?yàn)閙>n>0,所以m2+1mn+1m(m?n)=m2-mn+mn+1mn+1m(m?n)=(m2-mn)+1m(m?n)+mn+13.92解析:∵a>0,b>0,∴2=a+2b≥22ab,即0<ab≤12,當(dāng)且僅當(dāng)a=2b,即a=1,b=12時(shí)等號(hào)成立.∵2a+1b+ab=a+2bab+ab=2ab+ab,令y=2t+t,0<t≤12,由對(duì)勾函數(shù)性質(zhì)知y=2t+t在0,12上單調(diào)遞減,∴當(dāng)t=12時(shí),函數(shù)取得最小值,且ymin=212+12=92,即14.22+35解析:由a+b=1可得(a+3b)+(4a+2b)=5,所以1a+3b+12a+b=1a+3b+24a+2b=151a+3b+24a+2b[(a+3b)+(4a+2b)]=154a+2ba+3b+2(a+3b)15.2+22解析:∵a+b+c=2,∴a+b=2-c.∵a>0,b>0,c>0,∴2-c>0,∴0<c<2,∴4a+b+a+bc=42?c+2?cc=42?c+2c-1=12[(2-c)+c]42?c+2c-1=124+2+4c2?c+2(2?c)c-1=2+2c2?c16.62-4解析:令yx=t>0,則2yx+9x2x+y=2t+92+t=2(t+2)+92+t-4≥22(t+2)·92+t-4=62-4,當(dāng)且僅當(dāng)2(t+2)=92+t,17.9解析:1a2-11b2-1=1a+11a-11b+11b-1=1a+11?aa1b+11?bb=1a+1ba1b+1ab=1a+11b+1=1ab+b+aab+1=2ab+1,∵a+b=1,∴a+b≥2ab,即1≥2ab,∴ab≤14,∴2ab+1≥9,當(dāng)且僅當(dāng)a=b=12時(shí),取得等號(hào),即1a2-11b218.8解析:因?yàn)檎龑?shí)數(shù)x,y滿足x+y=1x+9y+6,由等式性質(zhì)得(x+y)2=(1x+9y)(x+y)+6(x+y)=10+yx+9xy+6(x+y),所以(x+y)2-6(x+y)=10+yx+9xy≥10+2yx·9xy=16,當(dāng)且僅當(dāng)yx=9xy時(shí)取等號(hào),設(shè)x+y=t,則t2-6t≥16,解得t≥8或t≤-2,所以x+19.9解析:令1x?1+1y=t,∴x-1+4y=10-t,(x-1+4y)(1x?1+1y)=(10-t)t.∵5+4yx?1+x?1y≥5+24yx?1·x?1y=9,當(dāng)且僅當(dāng)4yx?1=x?1y時(shí)等號(hào)成立.∴(10-t20.8