結(jié)合具體函數(shù)【豆丁精選】

結(jié)合具體函數(shù)，了解函數(shù)奇偶性的含義.

1.奇偶性的定義[思考探究](1)奇偶函數(shù)的定義域有何特點？(2)是否存在既是奇函數(shù)又是偶函數(shù)的函數(shù)？提示：奇偶函數(shù)的定義域關(guān)于原點對稱.提示：存在.該函數(shù)的特點是定義域關(guān)于坐標原點對稱，且解析式化簡后等于0.2.奇偶函數(shù)的性質(zhì)(1)奇函數(shù)在關(guān)于原點對稱的區(qū)間上的單調(diào)性，偶函數(shù)在關(guān)于原點對稱的區(qū)間上的單調(diào)性(填“相同〞、“相反〞).(2)在公共定義域內(nèi)，①兩個奇函數(shù)的和函數(shù)是，兩個奇函數(shù)的積函數(shù)是；②兩個偶函數(shù)的和函數(shù)、積函數(shù)是.③一個奇函數(shù)，一個偶函數(shù)的積函數(shù)是.(3)假設(shè)f(x)是奇函數(shù)且在x＝0處有定義，那么f(0)＝.

相反奇函數(shù)偶函數(shù)偶函數(shù)奇函數(shù)0相同1.設(shè)f(x)是R上的任意函數(shù)，那么以下表達正確的選項是()A.f(x)f(－x)是奇函數(shù)B.f(x)|f(－x)|是奇函數(shù)C.f(x)－f(－x)是偶函數(shù)D.f(x)＋f(－x)是偶函數(shù)解析：令F(x)＝f(x)＋f(－x).F(－x)＝f(－x)＋f(x)為偶函數(shù)，故D正確.答案：D2.對任意實數(shù)x，以下函數(shù)中的奇函數(shù)是()A.y＝2x－3B.y＝－3x2C.y＝ln5xD.y＝－|x|cosx解析：假設(shè)f(x)＝ln5x，那么f(－x)＝ln5－x＝ln(5x)－1＝－ln5x＝－f(x).∴函數(shù)y＝ln5x為奇函數(shù).答案：C3.f(x)＝ax2＋bx是定義在[a－1,2a]上的偶函數(shù)，那么a＋b的值是()A.－B.C.D.－解析：∵函數(shù)f(x)＝ax2＋bx在x∈[a－1,2a]上為偶函數(shù)，∴b＝0，且a－1＋2a＝0，即b＝0，a＝.∴a＋b＝.答案：B4.函數(shù)y＝f(x)為奇函數(shù)，假設(shè)f(3)－f(2)＝1，那么f(－2)－f(－3)＝.解析：由題意得f(－2)－f(－3)＝－f(2)＋f(3)＝f(3)－f(2)＝1.答案：15.設(shè)函數(shù)f(x)＝為奇函數(shù)，那么a＝.解析：∵f(x)為奇函數(shù)，∴由f(－1)＝－f(1)得a＝－1.

答案：－1判斷函數(shù)奇偶性的一般方法(1)首先確定函數(shù)的定義域，看其是否關(guān)于原點對稱的.否那么，既不是奇函數(shù)也不是偶函數(shù).(2)假設(shè)定義域關(guān)于原點對稱，那么可用下述方法進行判斷：①定義判斷：f(－x)＝f(x)?f(x)為偶函數(shù)，f(－x)＝－f(x)?f(x)為奇函數(shù).②等價形式判斷：f(－x)－f(x)＝0?f(x)為偶函數(shù)，f(－x)＋f(x)＝0?f(x)為奇函數(shù).或等價于：，那么f(x)為偶函數(shù)；＝－1，那么f(x)為奇函數(shù).(3)對于分段函數(shù)的奇偶性的判斷應(yīng)分段進行.[特別警示]分段函數(shù)的奇偶性判定，要注意定義域內(nèi)x取值的任意性，應(yīng)分段討論，討論時可依據(jù)x范圍取相應(yīng)的解析式化簡.此類問題也可利用圖象作判斷.判斷以下函數(shù)的奇偶性：[思路點撥](1)f(x)=x();(2)f(x)=log2(x+);(3)f(x)=;(4)f(x)=(5)f(x)=x2-|x-a|+2.[課堂筆記](1)函數(shù)定義域為(－∞，0)∪(0，＋∞).∴f(x)是偶函數(shù).∵f(-x)=-x()=f(x).(2)函數(shù)定義域為R.∴f(x)是奇函數(shù).∵f(-x)=log2(-x+)=log2=-log2(x+)=-f(x),(3)由得x＝－，或x＝.∴函數(shù)f(x)的定義域為{－，}.又∵對任意的x∈{－，}，－x∈{－，}且f(－x)＝－f(x)＝f(x)＝0，∴f(x)既是奇函數(shù)又是偶函數(shù).(4)函數(shù)定義域為(－∞，0)∪(0，＋∞).當x＜0時，－x＞0，那么f(－x)＝－(－x)2－x＝－(x2＋x)＝－f(x)；當x＞0時，－x＜0，那么f(－x)＝(－x)2－x＝x2－x＝－(－x2＋x)＝－f(x).∴對任意x∈(－∞，0)∪(0，＋∞)都有f(－x)＝－f(x).故f(x)為奇函數(shù).(5)函數(shù)f(x)的定義域為R.當a＝0時，f(x)＝f(－x)，∴f(x)是偶函數(shù)；當a≠0時，f(a)＝a2＋2，f(－a)＝a2－2|a|＋2.f(a)≠f(－a)，且f(a)＋f(－a)＝2(a2－|a|＋2)＝2(|a|－)2＋≠0，∴f(x)是非奇非偶函數(shù).判斷(或證明)抽象函數(shù)的奇偶性的步驟(1)利用函數(shù)奇偶性的定義，找準方向(想方法出現(xiàn)f(－x)，f(x))；(2)巧妙賦值，合理、靈活變形配湊；(3)找出f(－x)與f(x)的關(guān)系，得出結(jié)論.函數(shù)f(x)對一切x、y∈R，都有f(x＋y)＝f(x)＋f(y).(1)試判斷f(x)的奇偶性；(2)假設(shè)f(－3)＝a，用a表示f(12).[思路點撥][課堂筆記](1)顯然f(x)的定義域是R，關(guān)于原點對稱.又∵函數(shù)f(x)對一切x、y∈R都有f(x＋y)＝f(x)＋f(y).∴令x＝y(tǒng)＝0，得f(0)＝2f(0)，∴f(0)＝0.再令y＝－x，得f(0)＝f(x)＋f(－x)，∴f(－x)＝－f(x)，∴f(x)為奇函數(shù).(2)∵f(－3)＝a且f(x)為奇函數(shù)，∴f(3)＝－f(－3)＝－a.又∵f(x＋y)＝f(x)＋f(y)，x、y∈R，∴f(12)＝f(6＋6)＝f(6)＋f(6)＝2f(6)＝2f(3＋3)＝4f(3)＝－4a.(1)對抽象函數(shù)解不等式問題，應(yīng)充分利用函數(shù)的單調(diào)性，將“f〞脫掉，轉(zhuǎn)化為我們會求的不等式；(2)奇偶函數(shù)的不等式求解時，要注意到：奇函數(shù)在對稱的單調(diào)區(qū)間上有相同的單調(diào)性，偶函數(shù)在對稱的單調(diào)區(qū)間上有相反的單調(diào)性.函數(shù)f(x)的定義域為D＝{x|x∈R且x≠0}，且滿足對于任意x1，x2∈D，有f(x1·x2)＝f(x1)＋f(x2).(1)求f(1)的值；(2)判斷f(x)的奇偶性并證明你的結(jié)論；(3)如果f(4)＝1，f(3x＋1)＋f(2x－6)≤3，且f(x)在(0，＋∞)上是增函數(shù)，求x的取值范圍.[思路點撥][課堂筆記](1)∵對于任意x1，x2∈D，有f(x1·x2)＝f(x1)＋f(x2)，∴令x1＝x2＝1，得f(1)＝2f(1)，∴f(1)＝0.(2)令x1＝x2＝－1，有f(1)＝f(－1)＋f(－1)，∴f(－1)＝f(1)＝0.令x1＝－1，x2＝x有f(－x)＝f(－1)＋f(x)，∴f(－x)＝f(x)，∴f(x)為偶函數(shù).(3)依題設(shè)有f(4×4)=f(4)+f(4)=2.f(16×4)=f(16)+f(4)=3,∵f(3x+1)+f(2x-6)≤3,即f((3x+1)(2x-6))≤f(64).〔*〕法一：∵f(x)為偶函數(shù)，∴f(|(3x＋1)(2x－6)|)≤f(64).又∵f(x)在(0，＋∞)上是增函數(shù)，∴0＜|(3x＋1)(2x－6)|≤64.解上式，得3＜x≤5或－≤x＜－或－＜x＜3.∴x的取值范圍為{x|－≤x＜－或－＜x＜3或3＜x≤5}.法二：∵f(x)在(0，＋∞)上是增函數(shù)，∴(*)等價于不等式組或或∴3＜x≤5或≤x＜－或－＜x＜3.∴x的取值范圍為{x|－≤x＜－或－＜x＜3或3＜x≤5}.將本例中的條件f(x1·x2)＝f(x1)＋f(x2)改為f(x1＋x2)＝f(x1)＋f(x2)，定義域D＝{x|x≠0}改為D＝R，求解第(2)，(3)問.∴f(x)為奇函數(shù).解：(2)令x1＝x2＝0，得f(0)＝0；令x1＝x，x2＝－x，得f(0)＝f(x)＋f(－x)，即f(－x)＝－f(x)，

(3)∵f(4)＝1，∴f(8)＝f(4)＋f(4)＝2，f(12)＝f(4＋8)＝f(4)＋f(8)＝3.又∵f(3x＋1)＋f(2x－6)≤3，∴f(3x＋1＋2x－6)≤f(12)，即f(5x－5)≤f(12).又∵f(x)在(0，＋∞)上為增函數(shù)，f(x)為奇函數(shù)，∴f(x)在R上是增函數(shù)，∴5x－5≤12，∴x≤.函數(shù)奇偶性的判定以及利用函數(shù)的奇偶性求參數(shù)是高考對函數(shù)奇偶性的常規(guī)考法，09年山東、陜西等省將函數(shù)的奇偶性、單調(diào)性以及比較大小等問題綜合出現(xiàn)在高考試題中，這是高考新的一個考查方向.[考題印證](2021·山東高考)定義在R上的奇函數(shù)f(x)滿足f(x－4)＝－f(x)，且在區(qū)間[0,2]上是增函數(shù)，那么()A.f(－25)＜f(11)＜f(80)B.f(80)＜f(11)＜f(－25)C.f(11)＜f(80)＜f(－25)D.f(－25)＜f(80)＜f(11)【解析】∵f(x－4)＝－f(x)，∴T＝8.又f(x)是奇函數(shù)，∴f(0)＝0.∵f(x)在[0,2]上是增函數(shù)，且f(x)＞0，∴f(x)在[－2,0]上也是增函數(shù)，且f(x)＜0.又x∈[2,4]時，f(x)＝－f(x－4)＞0，且f(x)為減函數(shù).同理f(x)在[4,6]為減函數(shù)且f(x)＜0.如圖.∵f(－25)＝f(－1)＜0，f(11)＝f(3)＞0，f(80)＝f(0)＝0，∴f(－25)＜f(80)＜f(11).【答案】D[自主體驗](2021·陜西高考)定義在R上的偶函數(shù)f(x)滿足：對任意的x1，x2∈(－∞，0](x1≠x2)，有(x2－x1)(f(x2)－f(x1))＞0.那么當n∈N*時，有()A.f(－n)＜f(n－1)＜f(n＋1)B.f(n－1)＜f(－n)＜f(n＋1)C.f(n＋1)＜f(－n)＜f(n－1)D.f(n＋1)＜f(n－1)＜f(－n)解析：由(x2－x1)(f(x2)－f(x1))＞0得f(x)在x∈(－∞，0]為增函數(shù).又f(x)為偶函數(shù)，所以f(x)在x∈(0，＋∞)為減函數(shù).又f(－n)＝f(n)且0≤n－1＜n＜n＋1，∴f(n＋1)＜f(n)＜f(n－1)，即f(n＋1)＜f(－n)＜f(n－1).答案：C

1.以下函數(shù)中，在其定義域內(nèi)既是奇函數(shù)又是減函數(shù)的是()A.y＝－x3，x∈RB.y＝sinx，x∈RC.y＝x，x∈RD.y＝()x，x∈R解析：y＝－x3為奇函數(shù)且為減函數(shù)；y＝sinx為奇函數(shù)，但不是單調(diào)函數(shù)；y＝x為增函數(shù)；y＝()x不是奇函數(shù).答案：A2.(2021·泉州模擬)假設(shè)x∈R、n∈N*，定義：＝x(x＋1)(x＋2)…(x＋n－1)，例如＝(－5)(－4)(－3)(－2)(－1)＝－120，那么函數(shù)f(x)＝的奇偶性為()A.是奇函數(shù)而不是偶函數(shù)B.是偶函數(shù)而不是奇函數(shù)C.既是奇函數(shù)又是偶函數(shù)D.既不是奇函數(shù)又不是偶函數(shù)解析：∵＝x(x＋1)(x＋2)…(x＋n－1)，∴＝(x－9)(x－8)(x－7)…(x＋9)＝(x2－92)(x2－82)…(x2－12)x.∴＝(x2－92)(x2－82)…(x2－12)x2，∴f(x)＝是偶函數(shù).答案：B3.函數(shù)f(x)＝x3＋sinx＋1(x∈R)，假設(shè)f(a)＝2，那么f(－a)的值為()A.3B.0C.－1D.－2解析：f(a)＝a3＋sina＋1，①f(－a)＝(－a)3＋sin(－a)＋1＝－a3－sina＋1，②①＋②得f(a)＋f(－a)＝2，∴f(－a)＝2－f(a)＝2－2＝0.答案：B4.f(x)，g(x)分別是定義在R上的奇函數(shù)和偶函數(shù)，且f(x)－g(x)＝()x，那么f(1)，g(0)，g(－1)之間的大小關(guān)系是.解析：∵f(x)和g(x)分別為奇函數(shù)和偶函數(shù)，且f(x)－g(x)＝()x，①∴f(－x)－g(－x)＝()－x，即－f(x)－g(x)＝2x，∴f(x)＋g(x)＝－2x，②由①②得答案：f(1)>g(0)>g(－1)5.設(shè)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，假設(shè)f(1－m)＜f(m)，那么實數(shù)m的取值范圍是.解析：∵f(x)是偶函數(shù)，∴f(－x)＝f(x)＝f(|x|).∴不等式f(1－m)＜f(m)?f(|1－m|)＜f(|m|).又當x∈[0,2]時，f(x)是減函數(shù).答案：解得6.函數(shù)f(x)＝(a、b、c∈N)是奇函數(shù)，又f(1)＝2，f(2)＜3，求a、b、c的值.解：∵f(x)為奇函數(shù)∴f(－x)＝－f(x)，即.∴c＝0，即f(x)＝.又∵f(1)＝2，f(2)＜3，即∴4a＋1＜3a＋3，∴a＜2.又∵a∈N，∴a＝0或a＝1.當a＝0時，b＝，舍去.當a＝1時，b＝1，∴a＝1，b＝1，c＝0.

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