5.4函數(shù)的奇偶性第1課時奇偶性的概念一、選擇題1.函數(shù)y=f(x),x∈[-1,a](a>-1)是奇函數(shù),則a等于 ()A.-1 B.0C.1 D.無法確定2.下列函數(shù)中,既是偶函數(shù),又在區(qū)間(0,+∞)上單調(diào)遞減的為 ()A.y=1x2 BC.y=x2 D.y=x3.下列說法中錯誤的是 ()A.奇函數(shù)的圖象關(guān)于坐標(biāo)原點對稱B.圖象關(guān)于y軸對稱的函數(shù)是偶函數(shù)C.奇函數(shù)一定滿足f(0)=0D.偶函數(shù)的圖象不一定與y軸相交4.已知f(x)=x5+ax3+bx-8,且f(-2)=10,則f(2)等于 ()A.-26 B.-18C.-10 D.105.若函數(shù)f(x)=(m-1)x2+(m-2)x+(m2-7m+12)為偶函數(shù),則m的值是 ()A.1 B.2C.3 D.46.已知函數(shù)f(x)=1,x≥0,-1,x<0,g(x)=x3,則f(A.是奇函數(shù),但不是偶函數(shù)B.是偶函數(shù),但不是奇函數(shù)C.既是奇函數(shù)也是偶函數(shù)D.既不是奇函數(shù)也不是偶函數(shù)7.如圖為奇函數(shù)y=f(x)的部分圖象,則f(-2)+f(-1)的值為 ()A.-2 B.2C.1 D.08.下列函數(shù)為奇函數(shù)的是 ()A.f(x)=x3 B.f(x)=x5C.f(x)=x+1xD.f(x)=19.(多選題)設(shè)函數(shù)f(x),g(x)的定義域都為R,且f(x)是奇函數(shù),g(x)是偶函數(shù),則下列結(jié)論中正確的是 ()A.y=f(x)g(x)為奇函數(shù)B.y=|f(x)|g(x)為偶函數(shù)C.y=f(x)|g(x)|為偶函數(shù)D.y=|f(x)g(x)|為偶函數(shù)二、填空題10.若函數(shù)y=(x-1)(x+a)為偶函數(shù),則a=.

11.已知函數(shù)f(x)為奇函數(shù),且當(dāng)x>0時,f(x)=x2+1x,則f(-1)=12.設(shè)f(x)=ax2+bx+2是定義在[1+a,2]上的偶函數(shù),則f(x)的值域是.

三、解答題13.判斷下列函數(shù)的奇偶性:(1)f(x)=1x2+x2,x∈(-1,0)(2)f(x)=1-14.已知函數(shù)f(x)=x+mx,且此函數(shù)的圖象過點(1,5)(1)求實數(shù)m的值;(2)判斷f(x)的奇偶性;(3)判斷函數(shù)f(x)在[2,+∞)上的單調(diào)性,證明你的結(jié)論.15.已知y=f(x)是定義在R上的奇函數(shù),則下列函數(shù)中為奇函數(shù)的是 ()①y=f(|x|);②y=f(-x);③y=xf(x);④y=f(x)+x.A.①③ B.②③C.①④ D.②④16.已知函數(shù)f(x)對一切x,y都有f(x+y)=f(x)+f(y).(1)求證:f(x)是奇函數(shù);(2)若f(-3)=a,試用a表示f(12).5.4函數(shù)的奇偶性第1課時奇偶性的概念1.C[解析]由-1+a=0,得a=1.故選C.2.A[解析]易判斷A,C為偶函數(shù),B,D為奇函數(shù),但函數(shù)y=x2在(0,+∞)上單調(diào)遞增,所以選A.3.C[解析]根據(jù)奇偶函數(shù)的性質(zhì)知A,B中說法正確;對于C,如f(x)=1x,x∈(-∞,0)∪(0,+∞),易得函數(shù)f(x)是奇函數(shù),它的圖象不過原點,故C中說法錯誤;對于D,如g(x)=1x2,x∈(-∞,0)∪(0,+∞),易得函數(shù)g(x)是偶函數(shù),它的圖象不與y軸相交,故D中說法正確4.A[解析]令g(x)=x5+ax3+bx,則g(-x)=-g(x),∴g(x)為奇函數(shù).又f(x)=g(x)-8,∴f(-2)=g(-2)-8=10,∴g(-2)=18,∴g(2)=-18,∴f(2)=g(2)-8=-18-8=-26.5.B[解析]由f(-x)=f(x),得(m-1)x2-(m-2)x+(m2-7m+12)=(m-1)x2+(m-2)x+(m2-7m+12),解得m=2.6.B[解析]由題知f(x)·g(x)=x3,x≥0,-x3,x<0.若x>0,則-x<0,f(-x)=-(-x)3=x3=f(x);若x<0,則-x>0,f(-x)=(-x)3=-x3=f(x7.A[解析]由圖知f(1)=12,f(2)=32,又f(x)為奇函數(shù),所以f(-2)+f(-1)=-f(2)-f(1)=-32-12=-8.ABC[解析]由奇函數(shù)的定義可知A,B,C是奇函數(shù).故選ABC.9.ABD[解析]∵f(x)是奇函數(shù),g(x)是偶函數(shù),∴y=|f(x)|為偶函數(shù),y=|g(x)|為偶函數(shù),∴y=f(x)g(x)為奇函數(shù),y=|f(x)|g(x)為偶函數(shù),y=f(x)|g(x)|為奇函數(shù),y=|f(x)g(x)|為偶函數(shù).故選ABD.10.1[解析]∵函數(shù)y=(x-1)(x+a)=x2+(a-1)x-a為偶函數(shù),∴x2-(a-1)x-a=x2+(a-1)x-a恒成立,∴a-1=0,∴a=1.11.-2[解析]∵f(x)為奇函數(shù),且當(dāng)x>0時,f(x)=x2+1x,∴f(-1)=-f(1)=-12+12.[-10,2][解析]因為f(x)是定義在[1+a,2]上的偶函數(shù),所以1+a=-2,可得a=-3.易知b=0,所以f(x)=-3x2+2,當(dāng)x∈[-2,2]時,f(x)∈[-10,2],即函數(shù)f(x)的值域為[-10,2].13.解:(1)函數(shù)f(x)的定義域為(-1,0)∪(0,1],不關(guān)于原點對稱,故此函數(shù)為非奇非偶函數(shù).(2)由1-x2≥0,得-1≤x≤1.又|x+2|-2≠0,∴x≠0,∴-1≤x≤1且x≠0,x+2>0,∴f(x)=1-x2設(shè)函數(shù)f(x)的定義域為A,∵對于任意的x∈A,都有-x∈A,且f(-x)=1-(-x)2-x=-1-x2x14.解:(1)∵f(x)的圖象過點(1,5),∴1+m=5,解得m=4.(2)對于f(x)=x+4x,∵x∴f(x)的定義域為(-∞,0)∪(0,+∞),關(guān)于原點對稱.∵f(-x)=-x+4-x=-f(x),∴f(x(3)函數(shù)f(x)在[2,+∞)上單調(diào)遞增.證明如下:設(shè)x1,x2∈[2,+∞),且x1<x2,則f(x1)-f(x2)=x1+4x1-x2-4x2=(x1-x2)+∵x1,x2∈[2,+∞),且x1<x2,∴x1-x2<0,x1x2>4,∴f(x1)-f(x2)<0,∴f(x)在[2,+∞)上單調(diào)遞增.15.D[解析]①f(|-x|)=f(|x|),故y=f(|x|)為偶函數(shù);②f[-(-x)]=f(x)=-f(-x),故y=f(-x)為奇函數(shù);③-xf(-x)=-x·[-f(x)]=xf(x),故y=xf(x)為偶函數(shù);④f(-x)+(-x)=-[f(x)+x],故y=f(x)+x為奇函數(shù).可知②④正確,故選D.16.解:(1)證明:由已知f(x+y)=f(x)+f(y),令y=-x得f(0)=f(x)+f(-x),令x=