2021_2022學(xué)年新教材高中數(shù)學(xué)第2章函數(shù)習(xí)題課_函數(shù)性質(zhì)的綜合應(yīng)用鞏固練習(xí)含解析北師大版必修第一冊

1、習(xí)題課函數(shù)性質(zhì)的綜合應(yīng)用課后訓(xùn)練鞏固提升一、A組1.若函數(shù)f(x)是R上的偶函數(shù),且在區(qū)間0,+)上單調(diào)遞增,則下列關(guān)系成立的是()A.f(-3)f(0)f(1)B.f(-3)f(1)f(0)C.f(1)f(0)f(-3)D.f(1)f(-3)f(0)解析:f(-3)=f(3),且f(x)在區(qū)間0,+)上單調(diào)遞增,f(-3)f(1)f(0).答案:B2.已知函數(shù)f(x)=3x-3x(x0),則函數(shù)()A.是奇函數(shù),且在區(qū)間(0,+)上單調(diào)遞減B.是偶函數(shù),且在區(qū)間(0,+)上單調(diào)遞減C.是奇函數(shù),且在區(qū)間(0,+)上單調(diào)遞增D.是偶函數(shù),且在區(qū)間(0,+)上單調(diào)遞增解析:f(x)=3x-3x(

2、x0),f(-x)=-3x+3x=-f(x),f(x)是奇函數(shù).函數(shù)y=3x和y=-3x(x0)在區(qū)間(0,+)上均單調(diào)遞增,函數(shù)f(x)=3x-3x(x0)在區(qū)間(0,+)上單調(diào)遞增,故選C.答案:C3.若對任意x,yR,有f(x+y)=f(x)+f(y),則函數(shù)f(x)滿足()A.f(0)=0且f(x)為奇函數(shù)B.f(0)=0且f(x)為偶函數(shù)C.f(x)為增函數(shù)且為奇函數(shù)D.f(x)為增函數(shù)且為偶函數(shù)解析:對任意的x,yR,都有f(x+y)=f(x)+f(y),令x=y=0,得f(0)=f(0)+f(0)=2f(0),f(0)=0.令y=-x,得f(x-x)=f(x)+f(-x)=f(0

3、)=0,f(-x)=-f(x),函數(shù)f(x)為奇函數(shù).答案:A4.已知偶函數(shù)f(x)在區(qū)間0,+)上單調(diào)遞增,則滿足f(2x)f13的x的取值范圍是.解析:函數(shù)f(x)為偶函數(shù),則f(x)=f(|x|),于是f(2x)f13可化為f(|2x|)f13.又f(x)在區(qū)間0,+)上單調(diào)遞增,|2x|13,解得-16x0,求實數(shù)a的取值范圍.解:由題意,f(a-1)+f(4a-5)0,即f(a-1)-f(4a-5).又因為函數(shù)y=f(x)為奇函數(shù),所以f(a-1)f(5-4a).又函數(shù)y=f(x)在區(qū)間-1,1上是增函數(shù),所以-1a-11,-15-4a1,a-15-4a,解得0a2,1a32,a65

4、,所以65f(1)B.f(0)f(2)C.f(1)f(2)D.f(1)f(3)解析:函數(shù)y=f(x+2)為偶函數(shù),f(-x+2)=f(x+2),即f(2+x)=f(2-x),函數(shù)f(x)的圖象關(guān)于直線x=2對稱.又函數(shù)f(x)在區(qū)間(2,+)上單調(diào)遞增,函數(shù)f(x)在區(qū)間(-,2)上單調(diào)遞減,畫出函數(shù)f(x)的圖象(圖略),可得f(0)f(1),f(0)f(2),f(1)f(2),f(1)=f(3).答案:D2.已知f(x)是定義在區(qū)間a,b上的奇函數(shù),且f(x)在區(qū)間a,b上的最大值為m,則函數(shù)F(x)=f(x)+3在區(qū)間a,b上的最大值與最小值之和為()A.2m+3B.2m+6C.6-2m

5、D.6解析:因為奇函數(shù)f(x)在區(qū)間a,b上的最大值為m,所以它在區(qū)間a,b上的最小值為-m,所以函數(shù)F(x)=f(x)+3在區(qū)間a,b上的最大值與最小值之和為m+3+(-m+3)=6.答案:D3.函數(shù)f(x)是定義在R上的奇函數(shù),則下列說法:f(0)=0;若f(x)在區(qū)間0,+)上有最小值-1,則f(x)在區(qū)間(-,0上有最大值1;若f(x)在區(qū)間1,+)上單調(diào)遞增,則f(x)在區(qū)間(-,-1上單調(diào)遞減;若x0時,f(x)=x2-2x,則x0時,f(x)=-x2-2x.其中正確的個數(shù)是()A.1B.2C.3D.4解析:f(x)是R上的奇函數(shù),則f(0)=0,正確;其圖象關(guān)于原點對稱,且在對稱

6、區(qū)間上具有相同的單調(diào)性,最值互為相反數(shù),所以正確,不正確;對于,設(shè)x0,所以f(-x)=(-x)2-2(-x)=x2+2x.又f(-x)=-f(x),所以f(x)=-x2-2x,正確.答案:C4.已知函數(shù)y=f(x)滿足:y=f(x+1)是偶函數(shù);在區(qū)間1,+)上單調(diào)遞增,若x10,且x1+x2f(-x2)B.f(-x1)f(-x2)C.f(-x1)=f(-x2)D.無法確定解析:因為f(x+1)是偶函數(shù),所以f(-x+1)=f(x+1),即f(-x)=f(x+2),由x10,且x1+x22+x22.又y=f(x)在區(qū)間1,+)上單調(diào)遞增,所以f(-x1)f(2+x2)=f(-x2).答案:A

7、5.若函數(shù)f(x)=(k-2)x2+(k-1)x+3是偶函數(shù),則函數(shù)g(x)=kx2+2x-3的單調(diào)遞減區(qū)間是.解析:由函數(shù)f(x)=(k-2)x2+(k-1)x+3是偶函數(shù),可得k=1,所以函數(shù)g(x)=x2+2x-3=(x+1)2-4.故函數(shù)圖象開口向上,對稱軸為直線x=-1.因而函數(shù)g(x)=kx2+2x-3的單調(diào)遞減區(qū)間是(-,-1.答案:(-,-16.定義在R上的函數(shù)y=f(x)在區(qū)間(-,2)上單調(diào)遞增,且函數(shù)y=f(x+2)為偶函數(shù),則f(-1),f(4),f512的大小關(guān)系是.解析:因為y=f(x+2)為偶函數(shù),所以函數(shù)y=f(x)的圖象關(guān)于直線x=2對稱.又y=f(x)在區(qū)間(-,2)上單調(diào)遞增,所以y=f(x)在區(qū)間(2,+)上單調(diào)遞減,畫出函數(shù)y=f(x)的草圖(圖略)可知f512f(-1)f(4).答案:f512f(-1)f13的x的取值范圍.解:因為f(x)為偶函數(shù),所以f(x)=f(|x|)