2025高考數(shù)學(xué)一輪復(fù)習(xí)-第7講-函數(shù)的單調(diào)性與最值-專項(xiàng)訓(xùn)練【含答案】

2025高考數(shù)學(xué)一輪復(fù)習(xí)-第7講-函數(shù)的單調(diào)性與最值-專項(xiàng)訓(xùn)練1.設(shè)a∈R,則“a≥1”是“函數(shù)f(x)=ax-1x-1在(1,+∞)上單調(diào)遞減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2.已知函數(shù)f(x)是定義域?yàn)?0,+∞)的減函數(shù),若f(2-2m)>f(1+m),則實(shí)數(shù)m的取值范圍是()A.13,+∞C.13,1 3.已知函數(shù)f(x)=x+2x-3,則函數(shù)f(x)有(A.最小值1,無最大值B.最大值32C.最小值32D.無最大值,無最小值4.已知函數(shù)f(x)=-x|x|+2x,則下列結(jié)論正確的是()A.增區(qū)間是(0,+∞)B.減區(qū)間是(-∞,-1)C.增區(qū)間是(-∞,-1)D.增區(qū)間是(-1,1)5.函數(shù)f(x)=x2-x-1A.53 B.23 C.1 D.6.(多選題)已知函數(shù)f(x)同時(shí)滿足性質(zhì):①f(-x)=f(x);②?x1,x2∈(0,1),f(x1)-f(x2)x1-A.f(x)=x2 B.f(x)=1C.f(x)=cos4x D.f(x)=ln(1-|x|)7.(多選題)已知函數(shù)f(x)=2-x-21-2x+a.關(guān)于函數(shù)f(x)的單調(diào)性,下列判斷錯(cuò)誤的是()A.f(x)在(-∞,2)上單調(diào)遞增B.f(x)在(-∞,2)上單調(diào)遞減C.f(x)在14D.f(x)在148.已知函數(shù)f(x)=x+lnx-1,則不等式f(x)<0的解集是.

9.已知函數(shù)f(x)=log2(x2-ax+3a)在[2,+∞)上單調(diào)遞增,則實(shí)數(shù)a的取值范圍是.

10.寫出一個(gè)同時(shí)具有下列性質(zhì)①②③的函數(shù):f(x)=.

①定義域?yàn)镽;②值域?yàn)?-∞,1);③對(duì)任意x1,x2∈(0,+∞),且x1≠x2,均有f(x111.已知函數(shù)f(x)=x2-2x+3在區(qū)間[t,t+1]上是單調(diào)函數(shù),則實(shí)數(shù)t的取值范圍是()A.[1,+∞) B.[0,1]C.(-∞,0] D.(-∞,0]∪[1,+∞)12.設(shè)函數(shù)f(x)=1,x>0,0,x=0,-1,x<0,gA.(-∞,0] B.[0,1)C.[1,+∞) D.[-1,0]13.定義在(0,+∞)上的函數(shù)y=f(x)滿足:?x1,x2∈(0,+∞),且x1≠x2,x2f(x1)-x1f(x2)x1-x2A.(9,+∞) B.(0,9)C.(0,3) D.(3,+∞)14.(多選題)若函數(shù)f(x)的定義域與值域的交集為[a,b],則稱f(x)為“[a,b]交匯函數(shù)”,下列函數(shù)是[0,2]交匯函數(shù)的是()A.f(x)=x2-4x+4,x∈(-∞,2]B.f(x)=-x+2C.f(x)=-2x+2D.f(x)=215.若函數(shù)f(x)和g(x)的值域相同,但定義域不同,則稱f(x)和g(x)是“同象函數(shù)”.已知函數(shù)f(x)=x2+2,寫出一個(gè)與f(x)是“同象函數(shù)”的函數(shù)g(x)的解析式:g(x)=.

16.已知函數(shù)f(x)=x2+a|x-1|-4在區(qū)間[0,+∞)上單調(diào)遞增,則實(shí)數(shù)a的取值范圍為.

17.(多選題)已知函數(shù)f(x)的定義域?yàn)镽,且f(x+y)=f(x)f(y)+f(x)+f(y),當(dāng)x>0時(shí),f(x)>0,f(2)=3,則()A.f(1)=1B.函數(shù)f(x)在區(qū)間(0,+∞)上單調(diào)遞增C.函數(shù)f(x)是奇函數(shù)D.函數(shù)f(x)的一個(gè)解析式為f(x)=2x-118.函數(shù)f(x)的定義域?yàn)镽,并滿足以下條件:①對(duì)任意x∈R,都有f(x)>0;②對(duì)任意x,y∈R,都有f(xy)=[f(x)]y;③f13>1(1)求f(0)的值.(2)求證:f(x)在R上單調(diào)遞增.(3)若a>b>c>0,且b2=ac,求證:f(a)+f(c)>2f(b).參考答案1.B2.C3.C4.D5.B6.ABC7.BCD8.(0,1)9.(-4,4]10.f(x)=1-1211.D12.B13.D14.ABD15.x+2(或x+m+2,m∈R或4x+m16.[-2,0]17.ABD18.解解法一:(1)令x=0,y=2,得f(0)=[f(0)]2.因?yàn)閒(0)>0,所以f(0)=1.(2)任取x1,x2∈(-∞,+∞),且x1<x2.設(shè)x1=13p1,x2=13p2,則p1<pf(x1)-f(x2)=f13p因?yàn)閒13>1,p1<p2,所以f(x1)<f(x2所以f(x)在R上單調(diào)遞增.(3)由(1)(2)知f(b)>f(0)=1.因?yàn)閒(b)>1,且f(a)=fb·ab=[f(b)]ab,f(c)=fb·cb所以f(a)+f(c)=[f(b)]ab+[f(b)]cb>2[f(b)所以f(a)+f(c)>2f(b).解法二:(1)因?yàn)閷?duì)任意x,y∈R,有f(xy)=[f(x)]y,且對(duì)任意x∈R,f(x)>0,所以f(x)=f(x·1)=[f(1)]x,當(dāng)x=0時(shí),f(0)=[f(1)]0=1.故f(0)=1.(2)因?yàn)閒13>1,所以f(1)=f3×13=f133>1,所