(完整word版)高一數(shù)學(必修1)專題復習一函數(shù)的單調(diào)性和奇偶性

1、高一數(shù)學(必修 1)專題復習一函數(shù)的單調(diào)性和奇偶性一基礎知識復習1 函數(shù)單調(diào)性的定義：如果函數(shù)f(x)對定義域內(nèi)的區(qū)間I內(nèi)的任意XX2，當為x2時都有f X!f X2,則f X在|內(nèi)是增函數(shù)；當X!X2時都有f捲f X2，則f X在I內(nèi)時減函數(shù).f X!f X2亠2單調(diào)性的定義的等價形式：設X!,X2a,b，那么1- 0 f X在XiX2a,b是增函數(shù)；f-Xi匚仝0 fX在a,b是減函數(shù)；X-iX2XiX2f Xif X20f (X)在a, b是減函數(shù).3函數(shù)單調(diào)性的應用：利用定義都是充要性命題.即若f (X)在區(qū)間I上遞增(遞減)且f(xj f(X2)X1X2(X1,X2I)；若f (x)

2、在區(qū)間|上遞遞減且f (x-i) f (x2)X-Ix2(x1,x2I). 比較函數(shù)值的大?。?可用來解不等式； 求函數(shù)的值域或最值等.4證明或判斷函數(shù)單調(diào)性的方法：討論函數(shù)單調(diào)性必須在其定義域內(nèi)進行，因此要研究 函數(shù)單調(diào)性必須先求函數(shù)的定義域，函數(shù)的單調(diào)區(qū)間是定義域的子集.(1)用定義.(2)用已知函數(shù)的單調(diào)性.(3)圖象法.(4)如果f (X)在區(qū)間I上是增(減)函數(shù)，那么f(x)在I的任一非空子區(qū)間上也是增 (減)函數(shù)(5)復合函數(shù)的單調(diào)性結論： “同增異減”.(6)奇函數(shù)在對稱的單調(diào)區(qū)間內(nèi)有相同的單調(diào)性，偶函數(shù)在對稱的單調(diào)區(qū)間內(nèi)具有相反 的單調(diào)性.(7) 在公共定義域內(nèi)，增函數(shù)f(X)

3、增函數(shù)g(x)是增函數(shù)；減函數(shù)f(x)減函數(shù)g(x)是減函數(shù)；增函數(shù)f(x)減函數(shù)g(x)是增函數(shù)；減函數(shù)f(x)增函數(shù)g(x)是減函數(shù).(8)函數(shù)y ax (a 0, b 0)在，Jb或Jb,上單調(diào)遞增；在x1a1a5 .函數(shù)的奇偶性的定義：設y f(x),x A,如果對于任意x A,都有f ( x) f (x),則稱函數(shù)y f (x)為奇函數(shù)；如果對于任意x A,都有f ( x) f (x)，則稱函數(shù)y f (x)為偶函數(shù).6 奇偶函數(shù)的性質：(1) 函數(shù)具有奇偶性的必要條件是其定義域關于原點對稱.(2)f (x)是偶函數(shù)f (x)的圖象關于y軸對稱；f (x)是奇函數(shù)f (x)的圖象關于

4、原點對稱.(3)f (x)為偶函數(shù)f(x) f( x) f(|x|)5.如果奇函數(shù)f(x)在區(qū)間3,7上是增函數(shù)，且最小值為5，那么在區(qū)間7, 3上是A.增函數(shù)且最小值為5B.增函數(shù)且最大值為5C.減函數(shù)且最小值為5D .減函數(shù)且最大值為56.右函數(shù)f (x)是疋義在R上的偶函數(shù)，在(,0上是減函數(shù)，且f(2)0，則使得f (x)0的x的取值范圍是()A.,2B.2,C,2 U 2,D .2,2A .x軸對稱B .y軸對稱C.原點對稱D.以上均不對27.若f (X)X2ax與g(x)在區(qū)間1, 2x 1上都是減函數(shù)，則a的取值范圍是A .1, 0 U0,1B .1,0 U 0, 1C.0, 1

5、D .0, 1&若函數(shù)f (x)是定義在R上的奇函數(shù)，則函數(shù)F(x)(4)若奇函數(shù)二.訓練題目(一)選擇題1.下列函數(shù)中,f(x)的定義域包含0，則f(0)在區(qū)間(,0上是增函數(shù)的是4x 822.若函數(shù)f (x) x 2(a1)xlog*22在區(qū)間x)C.yD .y ,1 xx 1,4上是減函數(shù)，則實數(shù)a的取值范圍是3函數(shù)3,C.,3D.3f (x)在遞增區(qū)間是2,3IB.4,7，則y1,10f (x 3)的遞增區(qū)間是()1,7C.D.4,104已知函數(shù)f x為R上的減函數(shù)，則滿足f 1的實數(shù)x的范圍是(A.1,B.0,x1,00,11,f (x)f (x)的圖象關于(9.設f (x)

6、是R上的任意函數(shù)，下列敘述正確的是()2.已知奇函數(shù)f(x)在0,單調(diào)遞增，且f(3)0,則不等式xf(x) 0的解集A .f (x) f ( x)是奇函數(shù)C .f (x) f ( x)是偶函數(shù)10 .已知f (x)是偶函數(shù)，x R，當x|為| | X2|,則()A .f( Xj f( X2)C .f(X1)f ( X2)(二)填空題1 .已知f (x)是R上的奇函數(shù)，且在(0,為_ .-B .f (X) f ( X)是奇函數(shù)D .f (X) f ( X)是偶函數(shù)0時，f (x)為增函數(shù)，若X10, X20,且B .f ( X1)f ( X2)D .f (X1)f ( X2)上是增函數(shù)，則f

7、(X)在(，0)上的單調(diào)性3已知偶函數(shù)f(x)在0,2內(nèi)單調(diào)遞減，若a f( 1),bf(logj),c f(lg 0.5),2 4則a、b、4. 若函數(shù)5. 已知yc之間的大小關系是_ .f (x) a x b 2在0,上為增函數(shù)，則實數(shù)f (x)為奇函數(shù)，若f (3) f (2)1，則f( 2)f (x)(x 1)(x a)為奇函數(shù)，則a _x26設函數(shù)a、b的范圍是f(3)7.已知函數(shù)f (x) ax bx c,x 2a 3,1是偶函數(shù)，則a3cx dx 5，其中a, b, c, d為常數(shù)，若75&已知f (x) ax bxf(7)_9.已知函數(shù)f( 7)7，則則當x 0,f (

8、x)是定義在時，f (x)上的偶函數(shù)，當x,0時,f(x)10.定義在(1,1)上的函數(shù)f(x)x m2xnx是可函數(shù)，則常數(shù)m(三)解答題1.寫出下列函數(shù)的單調(diào)區(qū)間(1)y2x 13x(3)y2.判斷下列各函數(shù)的奇偶性：(1)f(x) 2(x 1)36x(x2x2(3)f (x)1x 23 利用單調(diào)性的定義:(1)證明函數(shù)f(x)(2)討論函數(shù)f(x)2)(2)(4)f(x)f(x)(x 0)(x 0)+g)上是減函數(shù).ax廠(a 0)在(1,1)上的單調(diào)性.4. (1)已知奇函數(shù)f(x)在定義域(1,1)內(nèi)單調(diào)遞減，且f(1 m) f (1 m2) 0，求m的取值范圍.(2)設定義在2,2上的偶函數(shù)f (x)在區(qū)間0,2上單調(diào)遞減，若f(1 m) f (m)，求實數(shù)m的取值范圍.5設函數(shù)f (x) x21 ax，其中a 0求證：當a1時，函數(shù)f(x)在區(qū)間0,上是單調(diào)函數(shù).xe a6設a 0,f(x)?是R上的偶函數(shù).(1)求a的值；(2)證明f(x)在(0,)a e上為增函數(shù)