2024-2025學(xué)年高中數(shù)學(xué)第一章集合與函數(shù)概念1.3.1單調(diào)性與最大小值第一課時函數(shù)的單調(diào)性課時作業(yè)新人教A版必修1

PAGE1-第一課時函數(shù)的單調(diào)性選題明細表學(xué)問點、方法題號函數(shù)單調(diào)性概念1,2,9函數(shù)單調(diào)性的判定、證明3函數(shù)單調(diào)性的應(yīng)用4,5,6,7,8,10,11,12基礎(chǔ)鞏固1.下列說法中正確的有(A)①若x1,x2∈I,當(dāng)x1<x2時,f(x1)<f(x2),則y=f(x)在I上是增函數(shù);②函數(shù)y=x2在R上是增函數(shù);③函數(shù)y=-1x④y=1x的單調(diào)遞減區(qū)間是(-∞,0)∪(0,+∞(A)0個 (B)1個 (C)2個 (D)3個解析:由于①中的x1,x2不是隨意的,因此①不正確;②③④明顯不正確.2.下列函數(shù)f(x)中,滿意“對隨意x1,x2∈(0,+∞),當(dāng)x1<x2時,都有f(x1)<f(x2)”的是(C)(A)f(x)=x2-2x+3 (B)f(x)=1(C)f(x)=x+1 (D)f(x)=|x-1|解析:因為對隨意x1,x2∈(0,+∞),當(dāng)x1<x2時,都有f(x1)<f(x2),所以f(x)在(0,+∞)上為增函數(shù),只有C選項符合題意.3.已知函數(shù)y=-mx和y=nx在(0,+∞(A)減函數(shù)且f(0)<0 (B)增函數(shù)且f(0)<0(C)減函數(shù)且f(0)>0 (D)增函數(shù)且f(0)>0解析:因為y=-mx和y=nx在(0,+∞所以m<0,n<0,f(x)=mx+n為減函數(shù)且f(0)=n<0.故選A.4.(2024·山東濰坊市高一上期中聯(lián)考)設(shè)函數(shù)f(x)在(-∞,+∞)上是減函數(shù),則(D)(A)f(a)>f(2a) (B)f(a2)<f(a)(C)f(a2+a)<f(a) (D)f(a2+1)<f(a)解析:因為a2+1-a=(a-12)2+3所以a2+1>a,又f(x)在(-∞,+∞)上是減函數(shù),所以f(a2+1)<f(a).5.若定義在R上的函數(shù)f(x)滿意對隨意的x1,x2∈[0,+∞)(x1≠x2),有f((A)f(3)<f(2)<f(4) (B)f(1)<f(2)<f(3)(C)f(2)<f(1)<f(3) (D)f(3)<f(1)<f(0)解析:若對隨意的x1,x2∈[0,+∞)(x1≠x2),有f(x26.函數(shù)y=|x|(1-x)在區(qū)間A上是增函數(shù),那么區(qū)間A是(B)(A)(-∞,0) (B)[0,12(C)[0,+∞) (D)(12,+∞解析:y=|x|(1-x)=x=-=-畫出函數(shù)的大致圖象如圖所示,由圖易知原函數(shù)在[0,12]上單調(diào)遞增.故A=[0,127.(2024·山東濟寧高一期末)已知函數(shù)f(x)=4x2-kx-8,x∈[5,20]的圖象上隨意兩點連線不平行于x軸,則k的取值范圍是.

解析:由題意,函數(shù)f(x)=4x2-kx-8在區(qū)間[5,20]上為單調(diào)函數(shù),則由f(x)圖象對稱軸方程為x=k8知k8≤5或k8≥20,即k≤答案:(-∞,40]∪[160,+∞)8.已知函數(shù)f(x)=x2+3,x解析:由題意知a>0,且02+3≥a×0+b,即b≤3.答案:a>0,b≤39.(2024·山東煙臺市高一上期中)已知函數(shù)f(x)的定義域為[a,b],對隨意x1,x2∈[a,b],且x1≠x2,下列條件中能推出f(x)在定義域內(nèi)為增函數(shù)的有(寫出全部正確的序號).

①f(x1)-f(x2)x1③若x1<x2時,都有f(x1)-f(x2)<0;④若x1<x2時,都有f(解析:①中,f(x1)-f(x2)x1-x2>1,則肯定有f(x1)-f(x2)x1-x2>0,所以f(x)為增函數(shù);②中,當(dāng)x1<x2時可得f(x1)<f(x2),當(dāng)x1>x2時可得f(x答案:①②③實力提升10.假如函數(shù)f(x)=ax2+2x-3在區(qū)間(-∞,4)上是單調(diào)遞增的,則實數(shù)a的取值范圍是(D)(A)(-14,+∞) (B)[-14,+(C)[-14,0) (D)[-14解析:當(dāng)a=0時,f(x)=2x-3在定義域R上是單調(diào)遞增的,故在(-∞,4)上單調(diào)遞增;當(dāng)a≠0時,二次函數(shù)f(x)的對稱軸為x=-1a因為f(x)在(-∞,4)上單調(diào)遞增,所以a<0,且-1a≥4,解得-14綜上,實數(shù)a的取值范圍是[-14,0]11.(2024·四川西昌市高一上期中)已知函數(shù)f(x)的圖象如圖所示.(1)依據(jù)函數(shù)圖象,寫出f(x)的單調(diào)區(qū)間;(2)若f(x)在[a-1,a+1]上單調(diào)遞增,求a的取值范圍.解:(1)由函數(shù)圖象得f(x)在(-∞,-1]和[2,+∞)上單調(diào)遞增,f(x)在(-1,2)上單調(diào)遞減.(2)因為f(x)在[a-1,a+1]上單調(diào)遞增,所以a+1≤-1或a-1≥2,解得a≤-2或a≥3,故a的取值范圍為(-∞,-2]∪[3,+∞).探究創(chuàng)新12.(2024·河南信陽中學(xué)高一期中)已知函數(shù)f(x)=ax2-2x-5a+6對隨意兩個不相等的實數(shù)x(A)(0,+∞) (B)[12,+∞(C)(0,12] (D)[12解析:由題意知函數(shù)f(x)在[2,+∞)上是增函數(shù),令g(x)=ax2-2x-5a+6,則a≠0時a即12≤a≤[老師備用1]設(shè)f(x),g(x)都是單調(diào)函數(shù),有如下四個命題,其中正確的命題是(C)①若f(x)單調(diào)遞增,g(x)單調(diào)遞增,則f(x)-g(x)單調(diào)遞增;②若f(x)單調(diào)遞增,g(x)單調(diào)遞減,則f(x)-g(x)單調(diào)遞增;③若f(x)單調(diào)遞減,g(x)單調(diào)遞增,則f(x)-g(x)單調(diào)遞減;④若f(x)單調(diào)遞減,g(x)單調(diào)遞減,則f(x)-g(x)單調(diào)遞減.(A)①② (B)①④ (C)②③ (D)②④解析:若函數(shù)f(x),g(x)單調(diào)性相同,則函數(shù)f(x)-g(x)的單調(diào)性不確定,故①④不正確.由-g(x)與g(x)的單調(diào)性相反知②③正確.故選C.[老師備用2](2024·唐山市縣中11校聯(lián)盟高一第一學(xué)期期中)已知f(11+x2)=x2(1)求函數(shù)f(x)的解析式與定義域;(2)推斷函數(shù)f(x)在(0,1]上的單調(diào)性,并用定義法加以證明.解:(1)令11+x2=t,則x2因為f(11+x2)=x2所以f(t)=t-1+1t因為1+x2≥1,所以0<t≤1,則f(x)的定義域為(0,1],所以f(x)=x-1+1x(0<x≤(2)f(