2024秋新教材高中數(shù)學(xué)第三章函數(shù)的概念與性質(zhì)3.2函數(shù)的基本性質(zhì)3.2.1第1課時(shí)函數(shù)的單調(diào)性分層演練含解析新人教A版必修第一冊(cè)

第1課時(shí)函數(shù)的單調(diào)性分層演練綜合提升A級(jí)基礎(chǔ)鞏固1.定義在區(qū)間[-5,5]上的函數(shù)y=f(x)的圖象如圖所示,則下列關(guān)于函數(shù)f(x)的說(shuō)法錯(cuò)誤的是 ()A.函數(shù)在區(qū)間[-5,-3]上單調(diào)遞增B.函數(shù)在區(qū)間[1,4]上單調(diào)遞增C.函數(shù)在區(qū)間[-3,1]∪[4,5]上單調(diào)遞減D.函數(shù)在區(qū)間[-5,5]上沒(méi)有單調(diào)性答案:C2.若x1,x2∈(-∞,0),且x1<x2,函數(shù)f(x)=-1xf(x1)與f(x2)的大小關(guān)系是 ()A.f(x1)>f(x2) B.f(x1)<f(x2)C.f(x1)=f(x2) D.以上都有可能答案:B3.下列函數(shù)中,滿意“對(duì)隨意x1,x2∈(0,+∞)都有f(x1)-A.f(x)=2x B.f(x)=-3xC.f(x)=x2+4x+3 D.f(x)=x+1答案:C4.函數(shù)f(x)=|x-1|+2的單調(diào)遞增區(qū)間為[1,+∞).5.已知函數(shù)f(x)=1x(1)設(shè)f(x)的定義域?yàn)锳,求集合A;(2)推斷函數(shù)f(x)在區(qū)間(1,+∞)上的單調(diào)性,并用定義加以證明.解:(1)由x2-1≠0,得x≠±1,所以函數(shù)f(x)=1x2-1的定義域?yàn)锳={x|x∈(2)函數(shù)f(x)=1x2證明:任取x1,x2∈(1,+∞),且x1<x2,f(x2)-f(x1)=1x22-1因?yàn)閤1>1,x2>1,所以x12-1>0,x22-1>0,x1+又因?yàn)閤1<x2,所以x1-x2<0,故f(x2)-f(x1)<0.因此,函數(shù)f(x)=1x2B級(jí)實(shí)力提升6.已知函數(shù)f(x)=(a-3)x+5A.(0,3) B.(0,3]C.(0,2) D.(0,2]解析:由題意,得實(shí)數(shù)a滿意a解得0<a≤2.答案:D7.已知函數(shù)f(x)是區(qū)間(0,+∞)上的減函數(shù),則f(a2-a+1)與f(34)的大小關(guān)系是f(a2-a+1)≤f(34解析:因?yàn)閍2-a+1=(a-12)2+34≥34>0,且f(x)是區(qū)間(0,+∞)上的減函數(shù),所以f(a2-a+1)≤f(8.探討函數(shù)f(x)=x+ax(a>0)的單調(diào)性解:f(x)=x+ax(a>0)因?yàn)槎x域?yàn)閧x|x∈R,且x≠0},所以可分開(kāi)證明,設(shè)x1>x2>0,則f(x1)-f(x2)=x1+ax1-x2-ax2=(x1(1-ax1當(dāng)0<x2<x1≤a時(shí),恒有ax1x所以f(x1)-f(x2)<0,故f(x)在區(qū)間(0,a]上是減函數(shù);當(dāng)x1>x2>a時(shí),恒有0<ax1x所以f(x1)-f(x2)>0,故f(x)在區(qū)間(a,+∞)上是增函數(shù).同理可證f(x)在區(qū)間(-∞,-a)上是增函數(shù),在區(qū)間[-a,0)上是減函數(shù).綜上所述,f(x)在區(qū)間(-∞,-a),(a,+∞)上是增函數(shù),在區(qū)間[-a,0),(0,a]上是減函數(shù).C級(jí)挑戰(zhàn)創(chuàng)新9.多選題已知函數(shù)f(x)=8+2x-x2,則下列結(jié)論不正確的是 ()A.f(x)在區(qū)間(-∞,1]上是減函數(shù)B.f(x)在區(qū)間(-∞,1]上是增函數(shù)C.f(x)在區(qū)間[-1,+∞)上是減函數(shù)D.f(x)在區(qū)間[-1,+∞)上是增函數(shù)解析:f(x)=8+2x-x2=-(x-1)2+9,結(jié)合它的圖象(圖略)知B項(xiàng)正確,A,C,D項(xiàng)錯(cuò)誤,故選A、C、D.答案:ACD10.多選題下列有關(guān)函數(shù)單調(diào)性的說(shuō)法,正確的是 ()A.若f(x)為增函數(shù),g(x)為增函數(shù),則f(x)+g(x)為增函數(shù)B.若f(x)為減函數(shù),g(x)為減函數(shù),則f(x)+g(x)為減函數(shù)C.若f(x)為增函數(shù),g(x)為減函數(shù),則f(x)+g(x)為增函數(shù)D.若f(x)為減函數(shù),g(x)為增函數(shù),則f(x)-g(x)為減函數(shù)解析:若f(x)為增函數(shù),g(x)為減函數(shù),則f(x)+g(x)的增減性不確定.例如:f(x)=x+2為R上的增函數(shù),