新高考數(shù)學(xué)二輪復(fù)習(xí)易錯(cuò)題專(zhuān)練易錯(cuò)點(diǎn)02 函數(shù)的性質(zhì)（含解析）

易錯(cuò)點(diǎn)02函數(shù)的性質(zhì)易錯(cuò)題【01】不理解函數(shù)概念在函數(shù)f：A→B中SKIPIF1<0都是非空數(shù)集,且滿(mǎn)足兩允許,兩不允許：允許B中有剩余元素,不允許A中有剩余元素；允許多對(duì)一,不允許一對(duì)多.易錯(cuò)題【02】研究函數(shù)單調(diào)性忽略定義域研究函數(shù)的單調(diào)性切記定義域優(yōu)先,分式形式的函數(shù)要保證分母不為零,對(duì)數(shù)型函數(shù)要保證底數(shù)大于零且不等于1,真數(shù)大于零.易錯(cuò)題【03】研究函數(shù)奇偶性與單調(diào)性忽略其等價(jià)形式的應(yīng)用1.注意與函數(shù)奇偶性有關(guān)的幾個(gè)結(jié)論：(1)SKIPIF1<0是偶函數(shù)SKIPIF1<0SKIPIF1<0；(2)SKIPIF1<0是奇函數(shù)SKIPIF1<0SKIPIF1<0；(3)若函數(shù)SKIPIF1<0在SKIPIF1<0處有意義,則SKIPIF1<0；(4)SKIPIF1<0是偶函數(shù),則SKIPIF1<0,SKIPIF1<0是偶函數(shù),則SKIPIF1<0.2.增函數(shù)與減函數(shù)的等價(jià)形式(1)若SKIPIF1<0,且SKIPIF1<0,SKIPIF1<0SKIPIF1<0SKIPIF1<0在SKIPIF1<0上是增函數(shù)；SKIPIF1<0SKIPIF1<0SKIPIF1<0在SKIPIF1<0上是減函數(shù).(2)若SKIPIF1<0,且SKIPIF1<0,則SKIPIF1<0是增函數(shù).易錯(cuò)題【04】不會(huì)利用對(duì)稱(chēng)性與奇偶性推導(dǎo)函數(shù)的周期性(1)函數(shù)SKIPIF1<0滿(mǎn)足SKIPIF1<0(SKIPIF1<0),若SKIPIF1<0為奇函數(shù),則其周期為SKIPIF1<0,若SKIPIF1<0為偶函數(shù),則其周期為SKIPIF1<0.(2)函數(shù)SKIPIF1<0SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0和SKIPIF1<0SKIPIF1<0都對(duì)稱(chēng),則函數(shù)SKIPIF1<0是以SKIPIF1<0為周期的周期函數(shù)；函數(shù)SKIPIF1<0SKIPIF1<0的圖象關(guān)于兩點(diǎn)SKIPIF1<0、SKIPIF1<0SKIPIF1<0都對(duì)稱(chēng),則函數(shù)SKIPIF1<0是以SKIPIF1<0為周期的周期函數(shù)；函數(shù)SKIPIF1<0SKIPIF1<0的圖象關(guān)于SKIPIF1<0和直線SKIPIF1<0SKIPIF1<0都對(duì)稱(chēng),則函數(shù)SKIPIF1<0是以SKIPIF1<0為周期的周期函數(shù). 01下列四個(gè)圖像中,是函數(shù)圖像的是()A．(1)(2) B．(1)(2)(3) C．(1)(3)(4) D．(1)(2)(3)(4)【警示】本題出錯(cuò)的主要原因不明白函數(shù)不允許一對(duì)多.【答案】C【問(wèn)診】根據(jù)函數(shù)的定義,一個(gè)自變量值對(duì)應(yīng)唯一一個(gè)函數(shù)值,或者多個(gè)自變量值對(duì)應(yīng)唯一一個(gè)函數(shù)值,顯然只有(2)不滿(mǎn)足.故選C.【叮囑】注意區(qū)分函數(shù)圖象與曲線的區(qū)別,曲線可以是一對(duì)多,但函數(shù)不允許一對(duì)多.1.函數(shù)y=f(x)的圖象與直線SKIPIF1<0的交點(diǎn)個(gè)數(shù)()A．至少1個(gè) B．至多1個(gè) C．僅有1個(gè)D．有0個(gè)、1個(gè)或多個(gè)【答案】B【解析】若1不在函數(shù)f(x)的定義域內(nèi),y=f(x)的圖象與直線SKIPIF1<0沒(méi)有交點(diǎn),若1在函數(shù)f(x)的定義域內(nèi),y=f(x)的圖象與直線SKIPIF1<0有1個(gè)交點(diǎn),故選B.2.下列四個(gè)圖像中,不是函數(shù)圖像的是()A． B．C． D．【答案】A【解析】對(duì)于A,由于一個(gè)自變量SKIPIF1<0對(duì)應(yīng)兩個(gè)SKIPIF1<0,不表示函數(shù),不是函數(shù)圖像,所以A符合題意,對(duì)于BCD,由圖像可知一個(gè)自變量SKIPIF1<0對(duì)應(yīng)唯一一個(gè)SKIPIF1<0,所以表示的是函數(shù)圖像,所以BCD不符合題意,故選A. 02(2022湖北省武漢市高三上學(xué)期月考)函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增,則實(shí)數(shù)SKIPIF1<0的取值范圍為()．A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【警示】本題出錯(cuò)注意原因是忽略在SKIPIF1<0上SKIPIF1<0.【答案】C【問(wèn)診】設(shè)SKIPIF1<0,可得SKIPIF1<0,則SKIPIF1<0是減函數(shù),要使得函數(shù)SKIPIF1<0為SKIPIF1<0上的增函數(shù),只需SKIPIF1<0為減函數(shù),且滿(mǎn)足SKIPIF1<0對(duì)于SKIPIF1<0恒成立,所以SKIPIF1<0,解得：SKIPIF1<0,所以實(shí)數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0,故選C.【叮囑】研究形如SKIPIF1<0的單調(diào)性一定要注意SKIPIF1<0.1.函數(shù)SKIPIF1<0的單調(diào)遞增區(qū)間為_(kāi)___________．【答案】SKIPIF1<0【解析】由SKIPIF1<0得SKIPIF1<0,所以SKIPIF1<0遞增區(qū)間為SKIPIF1<0．2.(2022屆安徽省安慶市重點(diǎn)高中高三上學(xué)期月考)函數(shù)SKIPIF1<0在SKIPIF1<0單調(diào)遞增,求a的取值范圍()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】令SKIPIF1<0,二次函數(shù)拋物線的對(duì)稱(chēng)軸方程為SKIPIF1<0,由復(fù)合函數(shù)的單調(diào)性可知,SKIPIF1<0.又SKIPIF1<0在SKIPIF1<0上恒成立,所以SKIPIF1<0,即SKIPIF1<0,所以SKIPIF1<0,解可得,SKIPIF1<0．故選C 03(2015新課標(biāo)全國(guó)卷1)若函數(shù)SKIPIF1<0為偶函數(shù),則SKIPIF1<0=【警示】本題失分注意原因是不知道用SKIPIF1<0求a的值,直接利用定義又不知道分子有理化.【答案】1【解析】因?yàn)镾KIPIF1<0是偶函數(shù),所以SKIPIF1<0SKIPIF1<0=SKIPIF1<0,所以SKIPIF1<0.【叮囑】研究函數(shù)的奇偶性與單調(diào)性要注意等價(jià)形式的應(yīng)用.1．(2022屆陜西省西安高三上學(xué)期期中)已知函數(shù)SKIPIF1<0(SKIPIF1<0,SKIPIF1<0),且SKIPIF1<0,則SKIPIF1<0()A．SKIPIF1<0 B．2 C．1 D．SKIPIF1<0【答案】C【解析】令SKIPIF1<0,因?yàn)镾KIPIF1<0,所以SKIPIF1<0為奇函數(shù),所以SKIPIF1<0,即SKIPIF1<0,又SKIPIF1<0,所以SKIPIF1<0,故選C.2．(2022屆北京市通州區(qū)高三上學(xué)期期中)已知函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0,SKIPIF1<0,SKIPIF1<0是偶函數(shù),SKIPIF1<0,有SKIPIF1<0,則()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】由SKIPIF1<0是偶函數(shù)可得SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱(chēng)因?yàn)镾KIPIF1<0,有SKIPIF1<0,所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞增,因?yàn)镾KIPIF1<0,所以SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,無(wú)法比較SKIPIF1<0與0的大小,故選B 04(202全國(guó)卷甲卷理數(shù)12)設(shè)函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0,SKIPIF1<0為奇函數(shù),SKIPIF1<0為偶函數(shù),當(dāng)SKIPIF1<0,SKIPIF1<0時(shí),SKIPIF1<0．若SKIPIF1<0(3)SKIPIF1<0,則SKIPIF1<0SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【警示】不會(huì)利對(duì)稱(chēng)性推導(dǎo)周期性是本題失分的主要原因【答案】D【問(wèn)診】SKIPIF1<0為奇函數(shù),SKIPIF1<0SKIPIF1<0,且SKIPIF1<0,SKIPIF1<0偶函數(shù),SKIPIF1<0,SKIPIF1<0,即SKIPIF1<0,SKIPIF1<0．令SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0,SKIPIF1<0．當(dāng)SKIPIF1<0,SKIPIF1<0時(shí),SKIPIF1<0．SKIPIF1<0(2)SKIPIF1<0,SKIPIF1<0SKIPIF1<0SKIPIF1<0,又SKIPIF1<0(3)SKIPIF1<0,SKIPIF1<0,解得SKIPIF1<0,SKIPIF1<0SKIPIF1<0,SKIPIF1<0,SKIPIF1<0當(dāng)SKIPIF1<0,SKIPIF1<0時(shí),SKIPIF1<0,SKIPIF1<0．故選SKIPIF1<0．【叮囑】掌握由對(duì)稱(chēng)性推導(dǎo)周期性的結(jié)論1．(2022屆江西省贛州市十七校高三上學(xué)期期中聯(lián)考)已如SKIPIF1<0的圖像關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng),且對(duì)SKIPIF1<0,都有SKIPIF1<0成立,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,則SKIPIF1<0()A．SKIPIF1<0 B．2 C．0 D．SKIPIF1<0【答案】A【解析】SKIPIF1<0的圖像關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng),所以SKIPIF1<0關(guān)于原點(diǎn)對(duì)稱(chēng),為奇函數(shù)．由于SKIPIF1<0,所以SKIPIF1<0SKIPIF1<0,所以SKIPIF1<0是周期為4的周期函數(shù)．所以SKIPIF1<0．SKIPIF1<0,而SKIPIF1<0,∴SKIPIF1<0即SKIPIF1<0,SKIPIF1<0．2.(多選題)(2022屆山東師范大學(xué)附屬中學(xué)高三上學(xué)期月考)定義在SKIPIF1<0的奇函數(shù)SKIPIF1<0滿(mǎn)足SKIPIF1<0,當(dāng)SKIPIF1<0時(shí)SKIPIF1<0,則以下結(jié)論正確的有()A．SKIPIF1<0的周期為6 B．SKIPIF1<0的圖象關(guān)于SKIPIF1<0對(duì)稱(chēng)C．SKIPIF1<0 D．SKIPIF1<0的圖象關(guān)于SKIPIF1<0對(duì)稱(chēng)【答案】ACD【解析】因?yàn)镾KIPIF1<0滿(mǎn)足SKIPIF1<0,所以SKIPIF1<0,故函數(shù)SKIPIF1<0是周期為SKIPIF1<0的周期函數(shù),故A選項(xiàng)正確；、由于函數(shù)為SKIPIF1<0的奇函數(shù)SKIPIF1<0滿(mǎn)足SKIPIF1<0,所以SKIPIF1<0,所以根據(jù)周期性得SKIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0的圖象關(guān)于SKIPIF1<0對(duì)稱(chēng),故B錯(cuò)誤,D正確；對(duì)于C選項(xiàng),結(jié)合周期性得SKIPIF1<0,故正確.故選ACD錯(cuò)1．(2022北京市朝陽(yáng)區(qū)高三上學(xué)期期中屆)若函數(shù)SKIPIF1<0為奇函數(shù),則實(shí)數(shù)SKIPIF1<0().A．SKIPIF1<0 B．SKIPIF1<0 C．0 D．1【答案】D【解析】因?yàn)楹瘮?shù)SKIPIF1<0為奇函數(shù),定義域?yàn)镾KIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0,解得SKIPIF1<0,經(jīng)檢驗(yàn)符合題意,故選D2．(2022屆廣東省深圳實(shí)驗(yàn)學(xué)校、湖南省長(zhǎng)沙市第一中學(xué)高三上學(xué)期聯(lián)考)已知函數(shù)SKIPIF1<0,若不等式SKIPIF1<0在SKIPIF1<0上有解,則實(shí)數(shù)a的取值范圍是()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,SKIPIF1<0,SKIPIF1<0.當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,SKIPIF1<0,SKIPIF1<0.所以SKIPIF1<0為偶函數(shù).又因?yàn)镾KIPIF1<0在SKIPIF1<0為減函數(shù),在SKIPIF1<0為增函數(shù).所以SKIPIF1<0.因?yàn)椴坏仁絊KIPIF1<0在SKIPIF1<0上有解,所以SKIPIF1<0,即SKIPIF1<0在SKIPIF1<0上有解,又因?yàn)镾KIPIF1<0在SKIPIF1<0為減函數(shù),SKIPIF1<0在SKIPIF1<0為增函數(shù),所以SKIPIF1<0.故選C3．已知偶函數(shù)SKIPIF1<0的圖象經(jīng)過(guò)點(diǎn)SKIPIF1<0,且當(dāng)SKIPIF1<0時(shí),不等式SKIPIF1<0恒成立,則使得SKIPIF1<0成立的SKIPIF1<0取值范圍為()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】由題意,SKIPIF1<0是偶函數(shù)且經(jīng)過(guò)點(diǎn)SKIPIF1<0所以點(diǎn)SKIPIF1<0也在圖象上,即SKIPIF1<0當(dāng)SKIPIF1<0時(shí),不等式SKIPIF1<0恒成立,所以函數(shù)SKIPIF1<0在SKIPIF1<0上為減函數(shù),因?yàn)镾KIPIF1<0,即SKIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0解得：SKIPIF1<0或SKIPIF1<0所以使得SKIPIF1<0成立的SKIPIF1<0取值范圍為：SKIPIF1<0.故選C.4．(2022屆新疆克拉瑪依市高三第三次模擬)已知定義在R上的奇函數(shù)SKIPIF1<0滿(mǎn)足SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,若函數(shù)SKIPIF1<0的所有零點(diǎn)為SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0()A．20 B．24 C．28 D．36【答案】C【解析】∵定義在SKIPIF1<0上的奇函數(shù)SKIPIF1<0滿(mǎn)足SKIPIF1<0,故圖象關(guān)于SKIPIF1<0對(duì)稱(chēng),∴SKIPIF1<0,故SKIPIF1<0,∴SKIPIF1<0,即周期為SKIPIF1<0,又SKIPIF1<0定義在SKIPIF1<0上的奇函數(shù),所以SKIPIF1<0是函數(shù)SKIPIF1<0一個(gè)對(duì)稱(chēng)中心,又因?yàn)楫?dāng)SKIPIF1<0時(shí),SKIPIF1<0,作出函數(shù)SKIPIF1<0的草圖,如下：函數(shù)SKIPIF1<0的所有零點(diǎn)即為SKIPIF1<0與SKIPIF1<0的交點(diǎn)的橫坐標(biāo),易知函數(shù)SKIPIF1<0經(jīng)過(guò)定點(diǎn)SKIPIF1<0,且關(guān)于SKIPIF1<0中心對(duì)稱(chēng),又SKIPIF1<0,分別作出函數(shù)SKIPIF1<0和SKIPIF1<0的圖象,則函數(shù)SKIPIF1<0的圖象在函數(shù)SKIPIF1<0和SKIPIF1<0的圖象之間,如下圖所示：則SKIPIF1<0與SKIPIF1<0交點(diǎn)關(guān)于SKIPIF1<0中心對(duì)稱(chēng),由圖像可知關(guān)于SKIPIF1<0對(duì)稱(chēng)的點(diǎn)共有3對(duì),同時(shí)還經(jīng)過(guò)點(diǎn)SKIPIF1<0,所以SKIPIF1<0.故選C.5．(2022屆重慶市涪陵實(shí)驗(yàn)中學(xué)校高三上學(xué)期期中)已知SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng),且對(duì)SKIPIF1<0,都有SKIPIF1<0成立,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,則SKIPIF1<0()A．SKIPIF1<0 B．SKIPIF1<0 C．0 D．2【答案】B【解析】因?yàn)镾KIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng),所以函數(shù)SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱(chēng),即函數(shù)SKIPIF1<0為奇函數(shù),所以SKIPIF1<0,又對(duì)SKIPIF1<0,都有SKIPIF1<0成立,所以SKIPIF1<0SKIPIF1<0,所以SKIPIF1<0,所以函數(shù)SKIPIF1<0是周期為4的周期函數(shù),因?yàn)镾KIPIF1<0時(shí),SKIPIF1<0,所以SKIPIF1<0,故選B.6．(2022屆吉林省長(zhǎng)春市重點(diǎn)高中高三上學(xué)期月考)已知函數(shù)f(x)＝lg(x2－2x－3)在(－∞,a)單調(diào)遞減,則a的取值范圍是()A．(－∞,－1] B．(－∞,2] C．[5,＋∞) D．[3,＋∞)【答案】A【解析】SKIPIF1<0是增函數(shù),SKIPIF1<0在SKIPIF1<0上遞減,在SKIPIF1<0遞增,因此SKIPIF1<0在SKIPIF1<0上遞減,則有SKIPIF1<0,解得SKIPIF1<0．故選A．7．(多選題)(2022屆決勝新高考名校交流高三聯(lián)考卷)SKIPIF1<0是定義在SKIPIF1<0上的偶函數(shù),對(duì)SKIPIF1<0,均有SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,則下列結(jié)論正確的是()A．函數(shù)SKIPIF1<0的一個(gè)周期為SKIPIF1<0 B．SKIPIF1<0C．當(dāng)SKIPIF1<0時(shí),SKIPIF1<0 D．函數(shù)SKIPIF1<0在SKIPIF1<0內(nèi)有SKIPIF1<0個(gè)零點(diǎn)【答案】AC【解析】SKIPIF1<0是定義在SKIPIF1<0上的偶函數(shù),對(duì)SKIPIF1<0,均有SKIPIF1<0,SKIPIF1<0故函數(shù)的周期為SKIPIF1<0,故選項(xiàng)A正確；SKIPIF1<0,故選項(xiàng)B錯(cuò)誤；當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,則SKIPIF1<0,故選項(xiàng)C正確；易知SKIPIF1<0,于是函數(shù)SKIPIF1<0在SKIPIF1<0內(nèi)有SKIPIF1<0個(gè)零點(diǎn),故選項(xiàng)D錯(cuò)誤,故選AC．8．(2022屆廣東省普通高中高三上學(xué)期11月階段性檢測(cè))已知定義在SKIPIF1<0上的函數(shù)SKIPIF1<0滿(mǎn)足SKIPIF1<0,且SKIPIF1<0,則()A．SKIPIF1<0為奇函數(shù) B．SKIPIF1<0的圖象關(guān)于SKIPIF1<0對(duì)稱(chēng)C．SKIPIF1<0為偶函數(shù) D．SKIPIF1<0