2024-2025學年新教材高中數(shù)學第三章函數(shù)的概念與性質(zhì)3.2函數(shù)的基本性質(zhì)3.2.1單調(diào)性與最大小值第1課時函數(shù)的單調(diào)性分層演練新人教A版必修第一冊

3.2函數(shù)的基本性質(zhì)3.2.1單調(diào)性與最大(小)值第1課時函數(shù)的單調(diào)性A級基礎鞏固1.定義在區(qū)間[-5,5]上的函數(shù)y=f(x)的圖象如圖所示,則下列關于函數(shù)f(x)的說法錯誤的是 ()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]上沒有單調(diào)性答案:C2.若x1,x2∈(-∞,0),且x1<x2,函數(shù)f(x)=-1x,則f(x1)與f(x2)的大小關系是 (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“對隨意x1,x2∈(0,+∞)都有f(x1)-A.f(x)=2xB.f(x)=-3x+1C.f(x)=x2+4x+3 D.f(x)=x+1答案:C4.多選題若函數(shù)f(x)在R上為減函數(shù),則 ()A.函數(shù)f(-x)在R上為增函數(shù)B.函數(shù)|f(-x)|在R上為增函數(shù)C.f(a2-1)<f(a)D.f(a2+1)<f(a)解析:因為函數(shù)f(x)在R上為減函數(shù),所以f(-x)在R上為增函數(shù).因為a2+1-a=(a-12)2+34>0,所以a2+1>a.因為f(x)在R上為減函數(shù),所以f(a2+1)<f(a答案:AD5.多選題下列有關函數(shù)單調(diào)性的說法,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),當g(x)=-12x時,f(x)+g(x)=x2+2為增函數(shù);當g(x)=-3x時,f(x)+g(x)=-2x+2在R上為減函數(shù).所以不能確定f(x)+g(x答案:ABD6.函數(shù)f(x)=|x-1|+2的單調(diào)遞增區(qū)間為[1,+∞).解析:因為函數(shù)f(x)=|x-1|+2,所以當x≥1時,f(x)=|x-1|+2=x-1+2=x+1,單調(diào)遞增;當x<1時,f(x)=|x-1|+2=-x+1+2=-x+3,單調(diào)遞減.所以函數(shù)f(x)=|x-1|+2的單調(diào)遞增區(qū)間為[1,+∞).B級實力提升7.已知函數(shù)f(x)=(a-3)xA.(0,3) B.(0,3]C.(0,2) D.(0,2]解析:由題意,得實數(shù)a滿意a-3<0,答案:D8.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)的大小關系是f(a2-a+1)≤f(34解析:因為a2-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(C級挑戰(zhàn)創(chuàng)新9.探討函數(shù)f(x)=x+ax(a>0)的單調(diào)性解:f(x)=x+ax(a>0)因為定義域為{x|x∈R,且x≠0},所以可分開證明,設x1>x2>0,則f(x1)-f(x2)=x1+ax1-x2-ax2=(x1-x2)(1當0<x2<x1≤a時,恒有ax1x2>1,則1所以f(x1)-f(x2)<0,故f(x)在區(qū)間(0,a]上是減函數(shù);當x1>x2>a時,恒有0<ax1x2<1,則1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(-