2025高考數(shù)學(xué)二輪復(fù)習(xí)-專題檢測(cè)1【課件】

專題檢測(cè)一202512345678910111213141516171819一、選擇題1.(2023·全國乙,理4)已知f(x)=是偶函數(shù),則a=(

)A.-2 B.-1 C.1 D.2D12345678910111213141516171819A12345678910111213141516171819解析

設(shè)f(x)=(3x-3-x)cos

x,則f(-x)=(3-x-3x)cos(-x)=-f(x),所以函數(shù)為奇函數(shù),排除B,D選項(xiàng).又f(1)=(3-3-1)cos

1>0,故選A.123456789101112131415161718193.(2020·新高考Ⅱ,8)若定義在R的奇函數(shù)f(x)在(-∞,0)單調(diào)遞減,且f(2)=0,則滿足xf(x-1)≥0的x的取值范圍是(

)A.[-1,1]∪[3,+∞) B.[-3,-1]∪[0,1]C.[-1,0]∪[1,+∞) D.[-1,0]∪[1,3]D解析

因?yàn)槎x在R的奇函數(shù)f(x)在(-∞,0)上單調(diào)遞減,且f(2)=0,所以f(x)在(0,+∞)上單調(diào)遞減,且f(-2)=0,f(0)=0,所以當(dāng)x∈(-∞,-2)∪(0,2)時(shí),f(x)>0,當(dāng)x∈(-2,0)∪(2,+∞)時(shí),f(x)<0,解得-1≤x≤0或1≤x≤3,所以滿足xf(x-1)≥0的x的取值范圍是[-1,0]∪[1,3].故選D.12345678910111213141516171819123456789101112131415161718194.生物豐富度指數(shù)

是河流水質(zhì)的一個(gè)評(píng)價(jià)指標(biāo),其中S,N分別表示河流中的生物種類數(shù)與生物個(gè)體總數(shù).生物豐富度指數(shù)d越大,水質(zhì)越好.如果某河流治理前后的生物種類數(shù)S沒有變化,生物個(gè)體總數(shù)由N1變?yōu)镹2,生物豐富度指數(shù)由2.1提高到3.15,則(

)D12345678910111213141516171819B12345678910111213141516171819A.b>c>a B.b>a>cC.c>b>a D.c>a>bA12345678910111213141516171819123456789101112131415161718197.(2024·新高考Ⅱ,8)設(shè)函數(shù)f(x)=(x+a)ln(x+b),若f(x)≥0,則a2+b2的最小值為(

)C解析

當(dāng)x≤-a時(shí),x+a≤0,當(dāng)x≥-a時(shí),x+a≥0,當(dāng)-b<x≤1-b時(shí),ln(x+b)≤0,當(dāng)x≥1-b時(shí),ln(x+b)≥0,要使f(x)≥0,必須-a=1-b,即b=a+1,123456789101112131415161718198.(2024·新高考Ⅰ,8)已知函數(shù)f(x)的定義域?yàn)镽,f(x)>f(x-1)+f(x-2),且當(dāng)x<3時(shí),f(x)=x,則下列結(jié)論中一定正確的是(

)A.f(10)>100 B.f(20)>1000C.f(10)<1000 D.f(20)<10000B12345678910111213141516171819解析

∵當(dāng)x<3時(shí),f(x)=x,∴f(1)=1,f(2)=2.∵f(x)的定義域?yàn)镽,且f(x)>f(x-1)+f(x-2),∴f(3)>f(2)+f(1)=3,f(4)>f(3)+f(2)>5,f(5)>f(4)+f(3)>8,f(6)>f(5)+f(4)>13,f(7)>f(6)+f(5)>21,f(8)>f(7)+f(6)>34,f(9)>f(8)+f(7)>55,f(10)>f(9)+f(8)>89,f(11)>f(10)+f(9)>144,f(12)>f(11)+f(10)>233,f(13)>f(12)+f(11)>377,f(14)>f(13)+f(12)>610,f(15)>f(14)+f(13)>987,f(16)>f(15)+f(14)>1

000.∴f(20)>1

000.結(jié)合各選項(xiàng)知,選項(xiàng)B一定正確.12345678910111213141516171819二、選擇題9.(2023·新高考Ⅰ,11)已知函數(shù)f(x)的定義域?yàn)镽,f(xy)=y2f(x)+x2f(y),則(

)A.f(0)=0B.f(1)=0C.f(x)是偶函數(shù)D.x=0為f(x)的極小值點(diǎn)ABC解析

對(duì)于選項(xiàng)A,令x=0,y=0,f(0)=0,所以A正確;對(duì)于選項(xiàng)B,令x=1,y=1,f(1×1)=12×f(1)+12×f(1)=2f(1),解得f(1)=0,所以B正確;對(duì)于選項(xiàng)C,令x=-1,y=-1,f[(-1)×(-1)]=(-1)2×f(-1)+(-1)2×f(-1)=2f(-1),解得f(-1)=0;再令x=-1,y=x,f[(-1)×x]=x2×f(-1)+(-1)2×f(x),f(-x)=f(x),所以C正確;對(duì)于選項(xiàng)D,用特值法,函數(shù)f(x)=0,為常數(shù)函數(shù),且滿足f(xy)=y2f(x)+x2f(y),而常數(shù)函數(shù)沒有極值點(diǎn),所以D錯(cuò)誤.故選ABC.123456789101112131415161718191234567891011121314151617181910.(2020·新高考Ⅰ,11)已知a>0,b>0,且a+b=1,則(

)ABD123456789101112131415161718191234567891011121314151617181911.(2024·新高考Ⅰ,10)設(shè)函數(shù)f(x)=(x-1)2(x-4),則(

)A.x=3是函數(shù)f(x)的極小值點(diǎn)B.當(dāng)0<x<1時(shí),f(x)<f(x2)C.當(dāng)1<x<2時(shí),-4<f(2x-1)<0D.當(dāng)-1<x<0時(shí),f(2-x)>f(x)AC12345678910111213141516171819解析

∵f(x)=(x-1)2(x-4),∴函數(shù)f(x)的定義域?yàn)镽,且f'(x)=3(x-1)(x-3).令f'(x)=3(x-1)(x-3)=0,得x=1或x=3.當(dāng)x<1或x>3時(shí),f'(x)>0,f(x)單調(diào)遞增,當(dāng)1<x<3時(shí),f'(x)<0,f(x)單調(diào)遞減,∴x=3是函數(shù)f(x)的極小值點(diǎn),∴A正確.當(dāng)0<x<1時(shí),0<x2<x<1,又由上可知當(dāng)0<x<1時(shí),f(x)單調(diào)遞增,∴f(x2)<f(x),∴B錯(cuò)誤.當(dāng)1<x<2時(shí),f(2x-1)=(2x-1-1)2(2x-1-4)=4(x-1)2(2x-5)<0,f(2x-1)+4=4(x-2)2(2x-1)>0,即f(2x-1)>-4,∴C正確.f(2-x)-f(x)=(2-x-1)2(2-x-4)-(x-1)2(x-4)=-2(x-1)3.當(dāng)-1<x<0時(shí),f(2-x)-f(x)=-2(x-1)3的值有正有負(fù),∴D錯(cuò)誤.故選AC.12345678910111213141516171819三、填空題12.(2024·廣東廣州三模)函數(shù)

其中a>0且a≠1,若函數(shù)是單調(diào)函數(shù),則a的一個(gè)可能取值為

.

4(答案不唯一)解析

因?yàn)閍>0且a≠1,若函數(shù)是單調(diào)函數(shù),結(jié)合二次函數(shù)性質(zhì)可知f(x)在R上單調(diào)遞增,故答案為4(答案不唯一).1234567891011121314151617181913.函數(shù)f(x)=|2x-1|-2lnx的最小值為

.

1123456789101112131415161718191234567891011121314151617181991234567891011121314151617181912345678910111213141516171819四、解答題15.(13分)(2024·上海,18)若f(x)=logax(a>0,a≠1).(1)曲線y=f(x)過(4,2),求f(2x-2)<f(x)的解集;(2)存在x使得f(x+1),f(ax),f(x+2)成等差數(shù)列,求實(shí)數(shù)a的取值范圍.解

(1)因?yàn)閥=f(x)的圖象過點(diǎn)(4,2),故loga4=2,故a2=4,即a=2(負(fù)值舍去).而f(x)=log2x在(0,+∞)上單調(diào)遞增,又f(2x-2)<f(x),所以0<2x-2<x,即1<x<2,故f(2x-2)<f(x)的解集為{x|1<x<2}.12345678910111213141516171819(2)因?yàn)榇嬖趚使得f(x+1),f(ax),f(x+2)成等差數(shù)列,所以2f(ax)=f(x+1)+f(x+2),即2loga(ax)=loga(x+1)+loga(x+2)有解.因?yàn)閍>0,a≠1,故x>0,故a2x2=(x+1)(x+2)在(0,+∞)上有解,12345678910111213141516171819(1)若a=0,求曲線y=f(x)在點(diǎn)(1,f(1))處的切線方程;(2)若f(x)在x=-1處取得極值,求f(x)的單調(diào)區(qū)間,以及其最大值與最小值.12345678910111213141516171819x(-∞,-1)-1(-1,4)4(4,+∞)f'(x)+0-0+f(x)單調(diào)遞增極大值單調(diào)遞減極小值單調(diào)遞增123456789101112131415161718191234567891011121314151617181917.(15分)(2020·全國Ⅰ,文20)已知函數(shù)f(x)=ex-a(x+2).(1)當(dāng)a=1時(shí),討論f(x)的單調(diào)性;(2)若f(x)有兩個(gè)零點(diǎn),求實(shí)數(shù)a的取值范圍.解

(1)當(dāng)a=1時(shí),f(x)=ex-x-2,則f'(x)=ex-1.當(dāng)x<0時(shí),f'(x)<0;當(dāng)x>0時(shí),f'(x)>0.所以f(x)在(-∞,0)上單調(diào)遞減,在(0,+∞)上單調(diào)遞增.12345678910111213141516171819(2)f'(x)=ex-a.當(dāng)a≤0時(shí),f'(x)>0,所以f(x)在(-∞,+∞)單調(diào)遞增,故f(x)至多存在1個(gè)零點(diǎn),不合題意.當(dāng)a>0時(shí),由f'(x)=0可得x=ln

a.當(dāng)x∈(-∞,ln

a)時(shí),f'(x)<0;當(dāng)x∈(ln

a,+∞)時(shí)f'(x)>0.所以f(x)在(-∞,ln

a)單調(diào)遞減,在(ln

a,+∞)單調(diào)遞增,故當(dāng)x=ln

a時(shí),f(x)取得最小值,最小值為f(ln

a)=-a(1+ln

a).12345678910111213141516171819由于f(-2)=e-2>0,所以f(x)在(-∞,ln

a)存在唯一零點(diǎn).由(1)知,當(dāng)x>2時(shí),ex-x-2>0,所以當(dāng)x>4且x>2ln(2a)時(shí),故f(x)在(ln

a,+∞)存在唯一零點(diǎn).從而f(x)在(-∞,+∞)有兩個(gè)零點(diǎn).1234567891011121314151617181918.(17分)(2022·北京,20)已知函數(shù)f(x)=exln(1+x).(1)求曲線y=f(x)在點(diǎn)(0,f(0))處的切線方程;(2)設(shè)g(x)=f'(x),討論函數(shù)g(x)在[0,+∞)上的單調(diào)性;(3)證明:對(duì)任意的s,t∈(0,+∞),有f(s+t)>f(s)+f(t).因此曲線y=f(x)在點(diǎn)(0,f(0))處的切線方程為y=x.12345678910111213141516171819所以對(duì)任意x∈[0,+∞),g'(x)>0恒成立,故g(x)在[0,+∞)上單調(diào)遞增.12345678910111213141516171819(3)證明

設(shè)函數(shù)F(x)=f(x+t)-f(x)(x>0),F'(x)=f'(x+t)-f'(x)=g(x+t)-g(x).因?yàn)閠>0,所以x+t>x>0.因?yàn)間(x)在[0,+∞)上單調(diào)遞增,所以g(x+t)>g(x),即g(x+t)-g(x)>0,F'(x)>0,所以F(x)在(0,+∞)上單調(diào)遞增.又因?yàn)閟>0,所以F(s)>F(0),即f(s+t)-f