2024-2025學(xué)年新教材高中數(shù)學(xué)課時(shí)素養(yǎng)檢測二十第三章函數(shù)的概念與性質(zhì)3.2.1.1函數(shù)的單調(diào)性含解析新人教A版必修第一冊

PAGE課時(shí)素養(yǎng)檢測二十函數(shù)的單調(diào)性(30分鐘60分)一、選擇題(每小題5分,共30分,多選題全部選對得5分,選對但不全對的得3分,有選錯(cuò)的得0分)1.下列四個(gè)函數(shù)中在(0,+∞)上單調(diào)遞增的是 ()A.f(x)=8-x B.f(x)=(x-2)2C.f(x)=QUOTE+1 D.f(x)=x2+2x【解析】選D.A在R上為減函數(shù),B在(0,2)上單調(diào)遞減,C在(0,+∞)上單調(diào)遞減.2.函數(shù)y=|x+2|在區(qū)間[-3,0]上 ()A.遞減 B.遞增C.先減后增 D.先增后減【解析】選C.因?yàn)閥=|x+2|=QUOTE作出y=|x+2|的圖象,如圖所示,易知在[-3,-2)上單調(diào)遞減,在[-2,0]上單調(diào)遞增.【補(bǔ)償訓(xùn)練】函數(shù)f(x)=QUOTE的單調(diào)減區(qū)間是 ()A.(-∞,+∞) B.QUOTEC.QUOTE D.QUOTE【解析】選C.由-2x+1≥0,得x≤QUOTE,又一次函數(shù)y=-2x+1為R上的減函數(shù),故f(x)=QUOTE的單調(diào)減區(qū)間為QUOTE.3.(2024·宣城高一檢測)已知函數(shù)y=ax和y=-QUOTE在(0,+∞)上都是單調(diào)遞減的,則函數(shù)f(x)=bx+a在R上是 ()A.減函數(shù)且f(0)<0 B.增函數(shù)且f(0)<0C.減函數(shù)且f(0)>0 D.增函數(shù)且f(0)>0【解析】選A.因?yàn)閥=ax和y=-QUOTE在(0,+∞)上都是單調(diào)遞減的,所以a<0,b<0,f(x)=bx+a在R上為減函數(shù)且f(0)=a<0.4.函數(shù)y=f(x)在R上為增函數(shù),且f(2m)>f(-m+12),則實(shí)數(shù)m的取值范圍是()A.(-∞,-4)B.(0,+∞)C.(4,+∞)D.(-∞,-4)∪(4,+∞)【解析】選C.因?yàn)楹瘮?shù)y=f(x)在R上為增函數(shù),且f(2m)>f(-m+12),所以2m>-m+12,即m>4.5.若f(x)=-x2+2ax與g(x)=QUOTE在區(qū)間[1,2]上都是單調(diào)遞減的,則a的取值范圍是 ()A.(-1,0)∪(0,1] B.(-1,0)∪(0,1)C.(0,1) D.(0,1]【解析】選D.f(x)=-(x-a)2+a2,當(dāng)a≤1時(shí),f(x)在[1,2]上是單調(diào)遞減的;g(x)=QUOTE,當(dāng)a>0時(shí),g(x)在[1,2]上是單調(diào)遞減的,則a的取值范圍是0<a≤1.6.(多選題)函數(shù)f(x)=4x2-mx+5在區(qū)間[-2,+∞)上是單調(diào)遞增的,則下列選項(xiàng)正確的是 ()A.f(1)≥25 B.f(-1)≤-7C.f(1)≤25 D.f(-1)≥-7【解析】選A、B.因?yàn)楹瘮?shù)f(x)的對稱軸為x=QUOTE,所以f(x)在QUOTE上是單調(diào)遞增的.所以QUOTE≤-2,所以m≤-16.則f(1)=4-m+5=9-m≥25.f(-1)=4+m+5=9+m≤-7.【補(bǔ)償訓(xùn)練】假如f(x)=ax2-(2-a)x+1在區(qū)間QUOTE上是單調(diào)遞減的,則a的取值范圍是()A.(0,1] B.[0,1)C.[0,1] D.(0,1)【解析】選C.a=0時(shí),f(x)=-2x+1,在區(qū)間QUOTE上是單調(diào)遞減的,符合題意;當(dāng)a≠0時(shí),假如f(x)=ax2-(2-a)x+1在區(qū)間QUOTE上是單調(diào)遞減的,必有QUOTE解得0<a≤1,綜上所述,a的取值范圍是[0,1].二、填空題(每小題5分,共10分)7.假如二次函數(shù)f(x)=x2-(a-1)x+5在區(qū)間QUOTE上是單調(diào)遞增的,則實(shí)數(shù)a的取值范圍為________.

【解析】因?yàn)槎魏瘮?shù)f(x)=x2-(a-1)x+5的圖象的對稱軸為直線x=QUOTE,又函數(shù)f(x)在區(qū)間QUOTE上是單調(diào)遞增的,所以QUOTE≤QUOTE,解得a≤2.答案:(-∞,2]8.已知函數(shù)f(x)在R上是減函數(shù),A(0,-2),B(-3,2)是其圖象上的兩點(diǎn),那么不等式-2<f(x)<2的解集為________.

【解析】因?yàn)锳(0,-2),B(-3,2)在函數(shù)y=f(x)的圖象上,所以f(0)=-2,f(-3)=2,故-2<f(x)<2可化為f(0)<f(x)<f(-3),又f(x)在R上是減函數(shù),因此-3<x<0.答案:(-3,0)三、解答題(每小題10分,共20分)9.作出函數(shù)f(x)=QUOTE的圖象,并指出函數(shù)的單調(diào)區(qū)間.【解析】f(x)=QUOTE的圖象如圖所示,由圖象可知,函數(shù)的單調(diào)遞減區(qū)間為(-∞,1]和(1,2];單調(diào)遞增區(qū)間為(2,+∞).10.已知函數(shù)f(x)=QUOTE.(1)求f(x)的定義域.(2)證明函數(shù)f(x)=QUOTE在[1,+∞)上是單調(diào)遞增的.【解析】(1)由題意知x+1≠0,即x≠-1.所以f(x)的定義域?yàn)?-∞,-1)∪(-1,+∞).(2)?x1,x2∈[1,+∞),且x1<x2,則f(x2)-f(x1)=QUOTE-QUOTE=QUOTE=QUOTE.因?yàn)閤1<x2,所以x2-x1>0.又因?yàn)閤1,x2∈[1,+∞),所以x2+1>0,x1+1>0.所以f(x2)-f(x1)>0,所以f(x2)>f(x1).所以函數(shù)f(x)=QUOTE在[1,+∞)上是單調(diào)遞增的.(35分鐘70分)一、選擇題(每小題5分,共20分,多選題全部選對得5分,選對但不全對的得3分,有選錯(cuò)的得0分)1.(多選題)關(guān)于函數(shù)y=-QUOTE的單調(diào)性的敘述正確的是 ()A.函數(shù)在(-∞,0)上是單調(diào)遞增的B.函數(shù)在(-∞,0)∪(0,+∞)上是單調(diào)遞增的C.函數(shù)在[0,+∞)上是單調(diào)遞增的D.函數(shù)的單調(diào)遞增區(qū)間是(-∞,0)和(0,+∞)【解析】選A、D.結(jié)合函數(shù)y=-QUOTE的圖象可知,其在(-∞,0)和(0,+∞)上都是單調(diào)遞增的.2.函數(shù)y=QUOTE的單調(diào)遞增區(qū)間是 ()A.(-∞,-3] B.QUOTEC.(-∞,1) D.[-1,+∞)【解析】選B.由2x-3≥0,得x≥QUOTE.又因?yàn)閠=2x-3在(-∞,+∞)上單調(diào)遞增,y=QUOTE在定義域上是增函數(shù),所以y=QUOTE的單調(diào)遞增區(qū)間是QUOTE.3.已知f(x)為R上的減函數(shù),則滿意fQUOTE<f(1)的實(shí)數(shù)x的取值范圍是()A.(-1,1) B.(0,1)C.(-1,0)∪(0,1) D.(-∞,-1)∪(1,+∞)【解析】選C.由已知條件:QUOTE>1,不等式等價(jià)于QUOTE解得-1<x<1,且x≠0.4.已知函數(shù)f(x)=QUOTE是(-∞,+∞)上的減函數(shù),則實(shí)數(shù)a的取值范圍是 ()A.QUOTE B.QUOTEC.QUOTE D.QUOTE【解析】選A.當(dāng)x<0時(shí),函數(shù)f(x)=x2-ax+1是減函數(shù),解得a≥0,當(dāng)x≥0時(shí),函數(shù)f(x)=-x+3a是減函數(shù),分段點(diǎn)0處的值應(yīng)滿意1≥3a,解得a≤QUOTE,所以0≤a≤QUOTE.二、填空題(每小題5分,共20分)5.已知函數(shù)f(x)=2ax2+4(a-3)x+1在區(qū)間(-∞,3)上是單調(diào)遞減的,則a的取值范圍是________.若函數(shù)f(x)=2ax2+4(a-3)x+1的單調(diào)減區(qū)間是(-∞,3),則a為________.

【解析】①當(dāng)a=0時(shí),f(x)=-12x+1在(-∞,3)上是單調(diào)遞減的;②當(dāng)a>0時(shí),要使f(x)=2ax2+4(a-3)x+1在區(qū)間(-∞,3)上是單調(diào)遞減的,則對稱軸x=QUOTE必在x=3的右邊,即QUOTE≥3,故0<a≤QUOTE;③當(dāng)a<0時(shí),不行能在區(qū)間(-∞,3)上恒為單調(diào)遞減的.綜合知:a的取值范圍是QUOTE.若函數(shù)f(x)=2ax2+4(a-3)x+1的單調(diào)減區(qū)間是(-∞,3),則對稱軸x=QUOTE=3,則a=QUOTE.答案:QUOTEQUOTE6.已知函數(shù)f(x)=|x+a|在區(qū)間(-∞,1]上單調(diào)遞減,則a的取值范圍是______.

【解析】f(x)=|x+a|=QUOTE所以f(x)在(-∞,-a]上單調(diào)遞減;又f(x)在(-∞,1]上單調(diào)遞減;所以-a≥1,所以a≤-1.答案:(-∞,-1]7.已知定義在[1,4]上的函數(shù)f(x)是減函數(shù),則滿意不等式f(1-2a)-f(3-a)>0的實(shí)數(shù)a的取值范圍為________.

【解析】由題意,可得f(1-2a)>f(3-a).因?yàn)閒(x)在定義域[1,4]上單調(diào)遞減,所以QUOTE解得-1≤a≤0,所以實(shí)數(shù)a的取值范圍為[-1,0].答案:[-1,0]8.已知函數(shù)f(x)=QUOTE則滿意不等式f(1-x2)>f(2x)的x的范圍是________.

【解析】f(x)=QUOTE的圖象如圖所示,不等式f(1-x2)>f(2x)等價(jià)于QUOTE或QUOTE解得-1<x<QUOTE-1.答案:(-1,QUOTE-1)三、解答題(每小題10分,共30分)9.畫出函數(shù)y=-x2+2|x|+3的圖象,并指出該函數(shù)的單調(diào)區(qū)間.【解析】x≥0時(shí),y=-x2+2x+3;x<0時(shí),y=-x2-2x+3.所以y=QUOTE畫出該函數(shù)的圖象如圖所示,由圖象知,該函數(shù)的單調(diào)遞增區(qū)間是(-∞,-1],(0,1];單調(diào)遞減區(qū)間是(-1,0],(1,+∞).10.已知函數(shù)f(x)=x-QUOTE+QUOTE在(1,+∞)上是單調(diào)遞增的,求實(shí)數(shù)a的取值范圍.【解析】設(shè)1<x1<x2,所以x1x2>1.因?yàn)楹瘮?shù)f(x)在(1,+∞)上是單調(diào)遞增的,所以f(x1)-f(x2)=x1-QUOTE+QUOTE-QUOTE=(x1-x2)·QUOTE<0.因?yàn)閤1-x2<0,所以1+QUOTE>0,即a>-x1x2.因?yàn)?<x1<x2,x1x2>1,所以-x1x2<-1,所以a≥-1.所以a的取值范圍是[-1,+∞).11.設(shè)f(x)=x2+1,g(x)=f(f(x)),F(x)=g(x)-λf(x).問是否存在實(shí)數(shù)λ,使F(x)在區(qū)間QUOTE上是單調(diào)遞減的且在區(qū)間QUOTE上是單調(diào)遞增的?【解析】假設(shè)存在這樣的實(shí)數(shù)λ,則由f(x)=x2+1,g(x)=f(f(x)),得g(x)=(x2+1)2+1,所以F(x)=g(x)-λf(x)=x4+(2-λ)·x2+2-λ.令t=x2,則t=x2在(-∞,0)上單調(diào)遞減,且當(dāng)x∈QUOTE時(shí),t>QUOTE