新高考數(shù)學(xué)一輪復(fù)習(xí)知識點總結(jié)與題型精練專題06 函數(shù)的概念及其性質(zhì)(原卷版)_第1頁
新高考數(shù)學(xué)一輪復(fù)習(xí)知識點總結(jié)與題型精練專題06 函數(shù)的概念及其性質(zhì)(原卷版)_第2頁
新高考數(shù)學(xué)一輪復(fù)習(xí)知識點總結(jié)與題型精練專題06 函數(shù)的概念及其性質(zhì)(原卷版)_第3頁
新高考數(shù)學(xué)一輪復(fù)習(xí)知識點總結(jié)與題型精練專題06 函數(shù)的概念及其性質(zhì)(原卷版)_第4頁
新高考數(shù)學(xué)一輪復(fù)習(xí)知識點總結(jié)與題型精練專題06 函數(shù)的概念及其性質(zhì)(原卷版)_第5頁
已閱讀5頁,還剩12頁未讀, 繼續(xù)免費閱讀

下載本文檔

版權(quán)說明:本文檔由用戶提供并上傳,收益歸屬內(nèi)容提供方,若內(nèi)容存在侵權(quán),請進(jìn)行舉報或認(rèn)領(lǐng)

文檔簡介

專題06函數(shù)的概念及其性質(zhì)【考綱要求】1.了解構(gòu)成函數(shù)的要素,會求一些簡單函數(shù)的定義域和值域;了解映射的概念.2.在實際情境中,會根據(jù)不同的需要選擇恰當(dāng)?shù)姆椒?如圖象法、列表法、解析法)表示函數(shù).3.了解簡單的分段函數(shù),并能簡單應(yīng)用(函數(shù)分段不超過三段).一、函數(shù)的概念及其表示【思維導(dǎo)圖】【考點總結(jié)】一、函數(shù)的概念(1)函數(shù)的概念:設(shè)A,B是非空的數(shù)集,如果按照某種確定的對應(yīng)關(guān)系f,使對于集合A中的任意一個數(shù)x,在集合B中都有唯一確定的數(shù)f(x)和它對應(yīng),那么就稱f:A→B為從集合A到集合B的一個函數(shù),記作y=f(x),xSKIPIF1<0A.其中,x叫做自變量,x的取值范圍A叫做函數(shù)的定義域;與x的值相對應(yīng)的y值叫做函數(shù)值,函數(shù)值的集合{f(x)|xSKIPIF1<0A}叫做函數(shù)的值域.顯然,值域是集合B的子集.二、具體函數(shù)定義域的求法函數(shù)的定義域是自變量x的取值范圍,如果未加特殊說明,函數(shù)的定義域就是指使函數(shù)關(guān)系式有意義的x的取值范圍,但在實際問題中,函數(shù)的定義域還要受到實際意義的制約.(1)求具體函數(shù)定義域的原則和方法主要有:①若f(x)為整式,則其定義域為實數(shù)集R.②若f(x)是分式,則其定義域是使分母不等于0的實數(shù)的集合.③若f(x)為偶次根式,則其定義域是使根號內(nèi)的式子大于或等于0的實數(shù)的集合.④若f(x)是由幾個部分的數(shù)學(xué)式子構(gòu)成的,那么函數(shù)的定義域是使各部分都有意義的實數(shù)的集合,即交集.⑤實際問題中,定義域要受到實際意義的制約.(2)求給出解析式的函數(shù)的定義域的步驟為:①列出使函數(shù)有意義的x所適合的式子(往往是一個不等式組);②解這個不等式組;③把不等式組的解表示成集合(或者區(qū)間)作為函數(shù)的定義域.三、抽象函數(shù)的定義域的求法求抽象函數(shù)的定義域是學(xué)習(xí)中的一個難點問題,常見的題型有如下兩種:①已知f(x)的定義域,求f(g(x))的定義域;②已知f(g(x))的定義域,求f(x)的定義域.下面介紹一下這兩種題型的解法.(1)已知f(x)的定義域,求f(g(x))的定義域.一般地,若f(x)的定義域為[a,b],則f(g(x))的定義域是指滿足不等式a≤g(x)≤b的x的取值范圍.其實質(zhì)是由g(x)的取值范圍,求x的取值范圍.(2)已知f(g(x))的定義域,求f(x)的定義域.函數(shù)f(g(x))的定義域為[a,b],指的是自變量xSKIPIF1<0[a,b].一般地,若f(g(x))的定義域為[a,b],則f(x)的定義域就是g(x)在區(qū)間[a,b]上的取值范圍(即g(x)的值域).其實質(zhì)是由x的取值范圍,求g(x)的取值范圍.四、函數(shù)值域的求法(1)常見函數(shù)的定義域和值域:①一次函數(shù)f(x)=kx+b(k≠0)的定義域是R,值域是R.②反比例函數(shù)f(x)=eq\f(k,x)(k≠0)的定義域是(-∞,0)SKIPIF1<0(0,+∞),值域是(-∞,0)SKIPIF1<0(0,+∞).③二次函數(shù)f(x)=ax2+bx+c(a≠0)的定義域是R.當(dāng)a>0時,值域是eq\b\lc\[\rc\)(\a\vs4\al\co1(f\b\lc\(\rc\)(\a\vs4\al\co1(-\f(b,2a))),+∞));當(dāng)a<0時,值域是eq\b\lc\(\rc\](\a\vs4\al\co1(-∞,f\b\lc\(\rc\)(\a\vs4\al\co1(-\f(b,2a))))).(2)求函數(shù)值域的常用方法.①觀察法:通過對解析式的簡單變形和觀察,利用熟知的基本函數(shù)的值域,求出函數(shù)的值域;如求函數(shù)y=eq\r(4-x2)的值域時,由x2≥0及4-x2≥0知eq\r(4-x2)SKIPIF1<0[0,2].故所求的值域為[0,2].②配方法:若函數(shù)是二次函數(shù)形式即可化為y=ax2+bx+c(a≠0)型的函數(shù),則可通過配方后再結(jié)合二次函數(shù)的性質(zhì)求值域,但要注意給定區(qū)間二次函數(shù)最值的求法.③換元法:對于一些無理函數(shù),可通過換元把它們轉(zhuǎn)化為有理函數(shù),然后利用有理函數(shù)求值域的方法,間接地求解原函數(shù)的值域.例如形如y=ax+b±eq\r(cx+d)的函數(shù),我們可令eq\r(cx+d)=t,將函數(shù)y轉(zhuǎn)化為關(guān)于自變量t的二次函數(shù),然后利用配方法求其值域.④分離常數(shù)法:將形如y=eq\f(cx+d,ax+b)(a≠0)的函數(shù),分離常數(shù),變形過程為eq\f(cx+d,ax+b)=eq\f(\f(c,a)(ax+b)+d-\f(bc,a),ax+b)=eq\f(c,a)+eq\f(d-\f(bc,a),ax+b),再結(jié)合x的范圍確定eq\f(d-\f(bc,a),ax+b)的取值范圍,從而確定函數(shù)的值域.(3)求函數(shù)的值域沒有通用的方法和固定的模式,要靠自己在解題過程中逐漸探索和積累.除了上述常用的方法外,還有最值法、數(shù)形結(jié)合法等,應(yīng)注意選擇最優(yōu)的解法.總之,求函數(shù)的值域關(guān)鍵是要重視對應(yīng)關(guān)系的作用,還要特別注意定義域?qū)χ涤虻闹萍s.五、分段函數(shù)(1)定義:有些函數(shù)在其定義域中,對于自變量x的不同取值范圍,對應(yīng)關(guān)系不同,這樣的函數(shù)通常稱為分段函數(shù).分段函數(shù)的表達(dá)式因其特點可以分成兩個或兩個以上的不同表達(dá)式,所以它的圖象也由幾部分構(gòu)成,有的可以是光滑的曲線段,有的也可以是一些孤立的點或幾條線段.六、函數(shù)解析式的求法求函數(shù)的解析式的常用方法有:(1)代入法:如已知f(x)=x2-1,求f(x+x2)時,有f(x+x2)=(x2+x)2-1.(2)待定系數(shù)法:已知f(x)的函數(shù)類型,要求f(x)的解析式時,可根據(jù)類型設(shè)其解析式,確定其系數(shù)即可.例如,一次函數(shù)可以設(shè)為f(x)=kx+b(k≠0);二次函數(shù)可以設(shè)為f(x)=ax2+bx+c(a≠0)等.(3)拼湊法:已知f(g(x))的解析式,要求f(x)時,可從f(g(x))的解析式中拼湊出“g(x)”,即用g(x)來表示,再將解析式兩邊的g(x)用x代替即可.(4)換元法:令t=g(x),再求出f(t)的解析式,然后用x代替f(g(x))解析式中所有的t即可.(5)方程組法:已知f(x)與f(g(x))滿足的關(guān)系式,要求f(x)時,可用g(x)代替兩邊的所有的x,得到關(guān)于f(x)及f(g(x))的方程組.解之即可得出f(x);(6)賦值法:給自變量賦予特殊值,觀察規(guī)律,從而求出函數(shù)的解析式.由具體的實際問題建立函數(shù)關(guān)系求解析式,一般是通過研究自變量、函數(shù)及其他量之間的等量關(guān)系,將函數(shù)用自變量和其他量的關(guān)系表示出來,但不要忘記確定自變量的取值范圍.二、函數(shù)的性質(zhì)【考點總結(jié)】一、函數(shù)的單調(diào)性(1)增函數(shù)和減函數(shù)名稱[定義幾何意義圖形表示增函數(shù)對于定義域I內(nèi)某個區(qū)間D上的任意兩個自變量的值x1,x2,當(dāng)x1<x2時,都有f(x1)<f(x2),那么就說f(x)在區(qū)間D上是增函數(shù),區(qū)間D稱為f(x)的單調(diào)遞增區(qū)間f(x)的圖象在區(qū)間D上是“上升”的減函數(shù)對于定義域I內(nèi)某個區(qū)間D上的任意兩個自變量的值x1,x2,當(dāng)x1<x2時,都有f(x1)>f(x2),那么就說f(x)在區(qū)間D上是減函數(shù),區(qū)間D稱為f(x)的單調(diào)遞減區(qū)間f(x)的圖象在區(qū)間D上是“下降”的(2)單調(diào)性如果函數(shù)y=f(x)在區(qū)間D上是增函數(shù)或減函數(shù),那么就說函數(shù)y=f(x)在這一區(qū)間具有(嚴(yán)格的)單調(diào)性.區(qū)間D叫做函數(shù)y=f(x)的單調(diào)區(qū)間.二、函數(shù)的最值(1)最大值和最小值定義:一般地,設(shè)函數(shù)y=f(x)的定義域為I,如果存在實數(shù)M滿足:①對于任意的xSKIPIF1<0I,都有f(x)≤M(f(x)≥M);②存在x0SKIPIF1<0I,使得f(x0)=M.那么,我們稱M是函數(shù)y=f(x)的最大(小)值.幾何意義:函數(shù)y=f(x)的最大(小)值是其圖象上最高(低)點的縱坐標(biāo).(2)最值函數(shù)的最大值和最小值統(tǒng)稱為函數(shù)的最值,則函數(shù)y=f(x)的最值是圖象上最高點或最低點的縱坐標(biāo).三、函數(shù)的奇偶性函數(shù)奇偶性的定義及圖象特點奇偶性定義圖象特點偶函數(shù)如果對于函數(shù)SKIPIF1<0的定義域內(nèi)任意一個SKIPIF1<0,都有SKIPIF1<0,那么函數(shù)SKIPIF1<0就叫做偶函數(shù)關(guān)于SKIPIF1<0軸對稱奇函數(shù)如果對于函數(shù)SKIPIF1<0的定義域內(nèi)任意一個SKIPIF1<0,都有SKIPIF1<0,那么函數(shù)SKIPIF1<0就叫做奇函數(shù)關(guān)于原點對稱判斷SKIPIF1<0與SKIPIF1<0的關(guān)系時,也可以使用如下結(jié)論:如果SKIPIF1<0或SKIPIF1<0,則函數(shù)SKIPIF1<0為偶函數(shù);如果SKIPIF1<0或SKIPIF1<0,則函數(shù)SKIPIF1<0為奇函數(shù).注意:由函數(shù)奇偶性的定義可知,函數(shù)具有奇偶性的一個前提條件是:對于定義域內(nèi)的任意一個SKIPIF1<0,SKIPIF1<0也在定義域內(nèi)(即定義域關(guān)于原點對稱).五、函數(shù)的對稱性(1)若函數(shù)SKIPIF1<0為偶函數(shù),則函數(shù)SKIPIF1<0關(guān)于SKIPIF1<0對稱.(2)若函數(shù)SKIPIF1<0為奇函數(shù),則函數(shù)SKIPIF1<0關(guān)于點SKIPIF1<0對稱.(3)若SKIPIF1<0,則函數(shù)SKIPIF1<0關(guān)于SKIPIF1<0對稱.(4)若SKIPIF1<0,則函數(shù)SKIPIF1<0關(guān)于點SKIPIF1<0對稱.六、函數(shù)的周期性(1)周期函數(shù):對于函數(shù)SKIPIF1<0,如果存在一個非零常數(shù)SKIPIF1<0,使得當(dāng)SKIPIF1<0取定義域內(nèi)的任何值時,都有SKIPIF1<0,那么就稱函數(shù)SKIPIF1<0為周期函數(shù),稱SKIPIF1<0為這個函數(shù)的周期.(2)最小正周期:如果在周期函數(shù)SKIPIF1<0的所有周期中存在一個最小的正數(shù),那么稱這個最小整數(shù)叫做SKIPIF1<0的最小正周期.【題型匯編】題型一:函數(shù)的定義題型二:函數(shù)的定義域題型三:函數(shù)的值域題型四:函數(shù)的解析式題型五:分段函數(shù)題型六:函數(shù)的單調(diào)性題型七:函數(shù)的最值題型八:函數(shù)的奇偶性題型九:函數(shù)的周期性題型十:函數(shù)的對稱性【題型講解】題型一:函數(shù)的定義一、單選題1.(2022·安徽省舒城中學(xué)三模(理))已知下表為函數(shù)SKIPIF1<0部分自變量取值及其對應(yīng)函數(shù)值,為便于研究,相關(guān)函數(shù)值非整數(shù)值時,取值精確到0.01.SKIPIF1<03.271.57SKIPIF1<0SKIPIF1<00.260.42SKIPIF1<0SKIPIF1<00SKIPIF1<0SKIPIF1<0SKIPIF1<00.270.260.210.20SKIPIF1<0SKIPIF1<00下列關(guān)于函數(shù)SKIPIF1<0的敘述不正確的是(

)A.SKIPIF1<0為奇函數(shù) B.SKIPIF1<0在SKIPIF1<0上沒有零點C.SKIPIF1<0在SKIPIF1<0上單調(diào)遞減 D.SKIPIF1<02.(2022·江西萍鄉(xiāng)·三模(理))已知定義域為SKIPIF1<0的函數(shù)SKIPIF1<0的圖象關(guān)于點SKIPIF1<0成中心對稱,且當(dāng)SKIPIF1<0時,SKIPIF1<0,若SKIPIF1<0,則SKIPIF1<0(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<03.(2022·江西九江·三模(理))已知函數(shù)SKIPIF1<0是定義在SKIPIF1<0的奇函數(shù),且當(dāng)SKIPIF1<0時,SKIPIF1<0.若SKIPIF1<0,則SKIPIF1<0(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<04.(2022·廣西桂林·二模(文))已知函數(shù)SKIPIF1<0,且SKIPIF1<0.給出如下結(jié)論:①SKIPIF1<0;②SKIPIF1<0;③SKIPIF1<0;④SKIPIF1<0.其中正確結(jié)論是(

)A.①③ B.②③ C.①④ D.②④5.(2022·山東濟(jì)南·二模)已知函數(shù)SKIPIF1<0若SKIPIF1<0,則m的值為(

)A.SKIPIF1<0 B.2 C.9 D.2或9二、多選題1.(2022·湖北十堰·三模)已知函數(shù)SKIPIF1<0,則(

)A.SKIPIF1<0,SKIPIF1<0,SKIPIF1<0成等差數(shù)列 B.SKIPIF1<0,SKIPIF1<0,SKIPIF1<0成等差數(shù)列題型二:函數(shù)的定義域一、單選題1.(2022·河南鄭州·三模(理))設(shè)全集SKIPIF1<0,集合SKIPIF1<0,SKIPIF1<0,則下面Venn圖中陰影部分表示的集合是(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<02.(2022·遼寧大連·二模)設(shè)集合SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<03.(2022·江西贛州·二模(文))下列四個命題中正確的是(

)A.若函數(shù)SKIPIF1<0的定義域為SKIPIF1<0,則SKIPIF1<0的定義域為SKIPIF1<0B.若正三角形SKIPIF1<0的邊長為SKIPIF1<0,則SKIPIF1<0C.已知函數(shù)SKIPIF1<0,則函數(shù)SKIPIF1<0的零點為SKIPIF1<0D.“SKIPIF1<0”是“SKIPIF1<0”的既不充分也不必要條件4.(2022·四川·內(nèi)江市教育科學(xué)研究所三模(理))已知集合SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0二、多選題1.(2022·山東威?!と#┮阎瘮?shù)SKIPIF1<0,則(

)A.當(dāng)SKIPIF1<0時,函數(shù)SKIPIF1<0的定義域為SKIPIF1<0B.當(dāng)SKIPIF1<0時,函數(shù)SKIPIF1<0的值域為SKIPIF1<0C.當(dāng)SKIPIF1<0時,函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減D.當(dāng)SKIPIF1<0時,關(guān)于x的方程SKIPIF1<0有兩個解2.(2022·江蘇江蘇·一模)下列函數(shù)中,最大值是1的函數(shù)有(

)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0題型三:函數(shù)的值域一、單選題1.(2022·廣東佛山·三模)箕舌線因意大利著名的女?dāng)?shù)學(xué)家瑪麗亞·阿涅西的深入研究而聞名于世.如圖所示,過原點的動直線交定圓SKIPIF1<0于點SKIPIF1<0,交直線SKIPIF1<0于點SKIPIF1<0,過SKIPIF1<0和SKIPIF1<0分別作SKIPIF1<0軸和SKIPIF1<0軸的平行線交于點SKIPIF1<0,則點SKIPIF1<0的軌跡叫做箕舌線.記箕舌線函數(shù)為SKIPIF1<0,設(shè)SKIPIF1<0,下列說法正確的是(

)A.SKIPIF1<0是奇函數(shù) B.點SKIPIF1<0的橫坐標(biāo)為SKIPIF1<0C.點SKIPIF1<0的縱坐標(biāo)為SKIPIF1<0 D.SKIPIF1<0的值域是SKIPIF1<02.(2022·陜西西安·三模(理))已知集合SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0(

)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<03.(2022·安徽·蕪湖一中一模(文))已知集合A={1,2,3},B={y|y=2x-1,x∈A},則A∩B=(

)A.{1} B.{1,2,3}C.{1,3} D.{1,3,5}4.(2022·四川瀘州·模擬預(yù)測(文))設(shè)集合SKIPIF1<0,則SKIPIF1<0(

)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0二、多選題1.(2022·湖北武漢·模擬預(yù)測)高斯是德國著名數(shù)學(xué)家,近代數(shù)學(xué)奠基者之一,享有“數(shù)學(xué)王子”的稱號,他和阿基米德,牛頓并列為世界三大數(shù)學(xué)家,用SKIPIF1<0表示不超過x的最大整數(shù),則SKIPIF1<0稱為高斯函數(shù),例如SKIPIF1<0,SKIPIF1<0.則下列說法正確的是(

)A.函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0(SKIPIF1<0)上單調(diào)遞增B.若函數(shù)SKIPIF1<0,則SKIPIF1<0的值域為SKIPIF1<0C.若函數(shù)SKIPIF1<0,則SKIPIF1<0的值域為SKIPIF1<0D.SKIPIF1<0,SKIPIF1<0題型四:函數(shù)的解析式一、單選題1.(2022·陜西·西北工業(yè)大學(xué)附屬中學(xué)二模(理))已知SKIPIF1<0,則SKIPIF1<0(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<02.(2022·河北保定·二模)若函數(shù)SKIPIF1<0,則函數(shù)SKIPIF1<0的最小值為(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<03.(2022·陜西·略陽縣天津高級中學(xué)二模(理))若SKIPIF1<0,則SKIPIF1<0等于(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0二、多選題1.(2022·江蘇·華羅庚中學(xué)三模)SKIPIF1<0是定義在SKIPIF1<0上的函數(shù),若SKIPIF1<0是奇函數(shù),SKIPIF1<0是偶函數(shù),函數(shù)SKIPIF1<0,則(

)A.當(dāng)SKIPIF1<0時,SKIPIF1<0 B.當(dāng)SKIPIF1<0時,SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0題型五:分段函數(shù)一、單選題1.(2022·天津·耀華中學(xué)二模)已知函數(shù)SKIPIF1<0SKIPIF1<0,若函數(shù)SKIPIF1<0在SKIPIF1<0內(nèi)恰有5個零點,則a的取值范圍是(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<02.(2022·內(nèi)蒙古赤峰·三模(文))已知函數(shù)SKIPIF1<0若SKIPIF1<0,且SKIPIF1<0,則SKIPIF1<0的最大值是(

)A.SKIPIF1<0 B.1 C.2 D.SKIPIF1<03.(2022·天津市武清區(qū)楊村第一中學(xué)二模)設(shè)SKIPIF1<0,函數(shù)SKIPIF1<0.若SKIPIF1<0在SKIPIF1<0上單調(diào)遞增,且函數(shù)SKIPIF1<0與SKIPIF1<0的圖象有三個交點,則SKIPIF1<0的取值范圍是(

)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<0二、多選題1.(2022·山東泰安·一模)已知函數(shù)SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,則下列結(jié)論正確的是(

)A.SKIPIF1<0在SKIPIF1<0上單調(diào)遞增B.當(dāng)SKIPIF1<0時,方程SKIPIF1<0有且只有3個不同實根C.SKIPIF1<0的值域為SKIPIF1<0D.若對于任意的SKIPIF1<0,都有SKIPIF1<0成立,則SKIPIF1<0題型六:函數(shù)的單調(diào)性一、單選題1.(2022·湖南師大附中三模)下列兩數(shù)的大小關(guān)系中正確的是(

)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<02.(2022·北京·首都師范大學(xué)附屬中學(xué)三模)下列函數(shù)中,既是偶函數(shù)又在SKIPIF1<0上單調(diào)遞減的是(

)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<03.(2022·江西師大附中三模(理))下列函數(shù)中既是奇函數(shù)又是增函數(shù)的是(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0二、多選題1.(2022·遼寧·育明高中一模)下列說法中正確的是(

)A.若SKIPIF1<0,則SKIPIF1<0B.若SKIPIF1<0,則SKIPIF1<0C.若定義域為SKIPIF1<0的奇函數(shù)SKIPIF1<0在SKIPIF1<0單調(diào)遞減,且SKIPIF1<0,則滿足SKIPIF1<0的SKIPIF1<0的取值范圍為SKIPIF1<0D.若SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0題型七:函數(shù)的最值一、單選題1.(2022·上海普陀·二模)已知定義在SKIPIF1<0上的偶函數(shù)SKIPIF1<0,滿足SKIPIF1<0對任意的實數(shù)SKIPIF1<0都成立,且值域為SKIPIF1<0.設(shè)函數(shù)SKIPIF1<0,(SKIPIF1<0),若對任意的SKIPIF1<0,存在SKIPIF1<0,使得SKIPIF1<0成立,則實數(shù)SKIPIF1<0的取值范圍為(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<02.(2022·上海長寧·二模)若函數(shù)SKIPIF1<0存在反函數(shù),則常數(shù)SKIPIF1<0的取值范圍為(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<03.(2022·湖北·荊門市龍泉中學(xué)二模)設(shè)SKIPIF1<0且SKIPIF1<0,若SKIPIF1<0對SKIPIF1<0恒成立,則a的取值范圍是(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<04.(2022·山西晉城·三模(理))已知函數(shù)SKIPIF1<0,若對任意SKIPIF1<0,SKIPIF1<0,SKIPIF1<0恒成立,則m的最大值為(

)A.-1 B.0 C.1 D.e二、多選題1.(2022·江蘇·二模)已知定義在SKIPIF1<0上的函數(shù)SKIPIF1<0,則(

)A.任意SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0均能作為一個三角形的三條邊長B.存在SKIPIF1<0,使得SKIPIF1<0,SKIPIF1<0,SKIPIF1<0不能作為一個三角形的三條邊長C.任意SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0均不能成為一個直角三角形的三條邊長D.存在SKIPIF1<0,使得SKIPIF1<0,SKIPIF1<0,SKIPIF1<0能成為一個直角三角形的三條邊長題型八:函數(shù)的奇偶性一、單選題1.(2022·上海金山·二模)對于定義在SKIPIF1<0上的函數(shù)SKIPIF1<0,若同時滿足:(1)對任意的SKIPIF1<0,均有SKIPIF1<0;(2)對任意的SKIPIF1<0,存在SKIPIF1<0,且SKIPIF1<0,使得SKIPIF1<0成立,則稱函數(shù)SKIPIF1<0為“等均”函數(shù).下列函數(shù)中:①SKIPIF1<0;②SKIPIF1<0;③SKIPIF1<0;④SKIPIF1<0,“等均”函數(shù)的個數(shù)是(

)A.1 B.2 C.3 D.42.(2022·上海虹口·二模)函數(shù)SKIPIF1<0是定義域為SKIPIF1<0的奇函數(shù),且對于任意的SKIPIF1<0,都有SKIPIF1<0成立.如果SKIPIF1<0,則實數(shù)SKIPIF1<0的取值集合是(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<03.(2022·海南??凇ざ#┮阎瘮?shù)SKIPIF1<0是定義在SKIPIF1<0上的奇函數(shù),函數(shù)SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對稱,若SKIPIF1<0,則SKIPIF1<0(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<04.(2022·湖南·雅禮中學(xué)二模)函數(shù)SKIPIF1<0的定義域為SKIPIF1<0,若SKIPIF1<0是奇函數(shù),SKIPIF1<0是偶函數(shù),則(

)A.SKIPIF1<0是奇函數(shù) B.SKIPIF1<0是偶函數(shù)C.SKIPIF1<0 D.SKIPIF1<0二、多選題1.(2022·全國·高考真題)已知函數(shù)SKIPIF1<0及其導(dǎo)函數(shù)SKIPIF1<0的定義域均為SKIPIF1<0,記SKIPIF1<0,若SKIPIF1<0,SKIPIF1<0均為偶函數(shù),則(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0題型九:函數(shù)的周期性一、單選題1.(2022·江西師大附中三模(文))定義在R上的函數(shù)SKIPIF1<0滿足SKIPIF1<0,且當(dāng)SKIPIF1<0時,SKIPIF1<0.則函數(shù)SKIPIF1<0的所有零點之和為(

)A.7 B.14 C.21 D.282.(2022·四川·內(nèi)江市教育科學(xué)研究所三模(理))已知函數(shù)SKIPIF1<0滿足:對任意SKIPIF1<0,SKIPIF1<0.當(dāng)SKIPIF1<0時,SKIPIF1<0,則SKIPIF1<0(

)A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<03.(2022·湖北武漢·二模)定義在SKIPIF1<0上的函數(shù)SKIPIF1<0滿足SKIPIF1<0,則下列是周期函數(shù)的是(

)A.SKIPIF1<0 B.SKIPIF1<0C.SKIPIF1<0 D.SKIPIF1<04.(2022·新疆阿勒泰·三模(理))已知定義域為R的奇函數(shù)SKIPIF1<0滿足SKIPIF1<0,且當(dāng)SKIPIF1<0

溫馨提示

  • 1. 本站所有資源如無特殊說明,都需要本地電腦安裝OFFICE2007和PDF閱讀器。圖紙軟件為CAD,CAXA,PROE,UG,SolidWorks等.壓縮文件請下載最新的WinRAR軟件解壓。
  • 2. 本站的文檔不包含任何第三方提供的附件圖紙等,如果需要附件,請聯(lián)系上傳者。文件的所有權(quán)益歸上傳用戶所有。
  • 3. 本站RAR壓縮包中若帶圖紙,網(wǎng)頁內(nèi)容里面會有圖紙預(yù)覽,若沒有圖紙預(yù)覽就沒有圖紙。
  • 4. 未經(jīng)權(quán)益所有人同意不得將文件中的內(nèi)容挪作商業(yè)或盈利用途。
  • 5. 人人文庫網(wǎng)僅提供信息存儲空間,僅對用戶上傳內(nèi)容的表現(xiàn)方式做保護(hù)處理,對用戶上傳分享的文檔內(nèi)容本身不做任何修改或編輯,并不能對任何下載內(nèi)容負(fù)責(zé)。
  • 6. 下載文件中如有侵權(quán)或不適當(dāng)內(nèi)容,請與我們聯(lián)系,我們立即糾正。
  • 7. 本站不保證下載資源的準(zhǔn)確性、安全性和完整性, 同時也不承擔(dān)用戶因使用這些下載資源對自己和他人造成任何形式的傷害或損失。

評論

0/150

提交評論