專題一 函數(shù)的圖象與性質(zhì)卷卷卷2018_函數(shù)圖象的辨識(shí)T3函數(shù)圖象的辨識(shí)T7抽象函數(shù)的奇偶性與周期性T112017利用函數(shù)的單調(diào)性、奇偶性解不等式T5_分段函數(shù)、解不等式T152016函數(shù)圖象辨識(shí)T7函數(shù)圖象的對(duì)稱性T12_縱向把握趨勢(shì)卷3年2考，涉及函數(shù)圖象的識(shí)別以及函數(shù)的單調(diào)性、奇偶性與不等式的綜合問題，試題均出現(xiàn)在選擇題上，難度適中，預(yù)計(jì)2019年會(huì)重點(diǎn)考查分段函數(shù)的有關(guān)性質(zhì)及應(yīng)用卷3年3考，涉及函數(shù)圖象的辨識(shí)以及抽象函數(shù)的性質(zhì)，其中函數(shù)圖象的識(shí)別難度較小，而函數(shù)性質(zhì)難度偏大，均出現(xiàn)在選擇題中，預(yù)計(jì)2019年會(huì)以選擇題的形式考查分段函數(shù)、函數(shù)的性質(zhì)等卷3年2考，涉及函數(shù)圖象的辨識(shí)、分段函數(shù)與不等式的綜合問題，既有選擇題，也有填空題，難度適中，預(yù)計(jì)2019年會(huì)以選擇題的形式考查函數(shù)的單調(diào)性、奇偶性等性質(zhì)橫向把握重點(diǎn)1.高考對(duì)此部分內(nèi)容的命題多集中于函數(shù)的概念、函數(shù)的性質(zhì)及分段函數(shù)等方面，多以選擇題、填空題形式考查，一般出現(xiàn)在第510或第1315題的位置上，難度一般主要考查函數(shù)的定義域、分段函數(shù)求值或分段函數(shù)中參數(shù)的求解及函數(shù)圖象的判斷2.此部分內(nèi)容有時(shí)也出現(xiàn)在選擇、填空中的壓軸題的位置，多與導(dǎo)數(shù)、不等式、創(chuàng)新性問題結(jié)合命題，難度較大.函數(shù)的概念及表示題組全練1(2018長春質(zhì)檢)函數(shù)y的定義域是()A1,0)(0,1)B1,0)(0,1C(1,0)(0,1 D(1,0)(0,1)解析：選D由題意得解得1<x<0或0<x<1.所以原函數(shù)的定義域?yàn)?1,0)(0,1)2已知函數(shù)f (x)則f (2 018)()A1 BeC. De2解析：選D由已知可得，當(dāng)x>2時(shí)，f (x)f (x4)，故f (x)在x>2時(shí)的周期為4，則f (2 018)f (2 018)f (2 0162)f (2)e2.3設(shè)f (x)若f (a)f (a1)，則f ()A2 B4C6 D8解析：選C當(dāng)0a1時(shí)，a11，f (a)，f (a1)2(a11)2a，f (a)f (a1)，2a，解得a或a0(舍去)f f (4)2(41)6.當(dāng)a1時(shí)，a12，f (a)2(a1)，f (a1)2(a11)2a，2(a1)2a，無解綜上，f 6.4已知函數(shù)f (x)則f (f (x)<2的解集為_解析：因?yàn)楫?dāng)x1時(shí)，f (x)x3x2，當(dāng)x<1時(shí)，f (x)2ex1<2，所以f (f (x)<2等價(jià)于f (x)<1，即2ex1<1，解得x<1ln 2，所以f (f (x)<2的解集為(，1ln 2)答案：(，1ln 2)5(2018成都模擬)設(shè)函數(shù)f ：RR滿足f (0)1，且對(duì)任意x，yR都有f (xy1)f (x)f (y)f (y)x2，則f (2 018)_.解析：令xy0，則f (1)f (0)f (0)f (0)02111022.令y0，則f (1)f (x)f (0)f (0)x2.將f (0)1，f (1)2代入，得f (x)1x，所以f (2 018)2 019.答案：2 019 系統(tǒng)方法1函數(shù)定義域的求法求函數(shù)的定義域，其實(shí)質(zhì)就是以函數(shù)解析式所含運(yùn)算有意義為準(zhǔn)則，列出不等式或不等式組，然后求出解集即可2分段函數(shù)問題的4種常見類型及解題策略常見類型解題策略求函數(shù)值弄清自變量所在區(qū)間，然后代入對(duì)應(yīng)的解析式，求“層層套”的函數(shù)值，要從最內(nèi)層逐層往外計(jì)算解不等式根據(jù)分段函數(shù)中自變量取值范圍的界定，代入相應(yīng)的解析式求解，但要注意取值范圍的大前提求參數(shù)“分段處理”，采用代入法列出各區(qū)間上的方程利用函數(shù)性質(zhì)求值必須依據(jù)條件找到函數(shù)滿足的性質(zhì)，利用該性質(zhì)求解函數(shù)的圖象及應(yīng)用由題知法(1)(2018全國卷)函數(shù)f (x)的圖象大致為()(2)如圖，已知l1l2，圓心在l1上、半徑為1 m 的圓O在t0時(shí)與l2相切于點(diǎn)A，圓O沿l1以1 m/s的速度勻速向上移動(dòng)，圓被直線l2所截上方圓弧長記為x，令ycos x，則y與時(shí)間t(0t1，單位：s)的函數(shù)yf (t)的圖象大致為()(3)已知函數(shù)f (x)若存在x1，x2，當(dāng)0x1<x2<2時(shí)，f (x1)f (x2)，則x1f (x2)的取值范圍是_解析(1)yexex是奇函數(shù)，yx2是偶函數(shù)，f (x)是奇函數(shù)，圖象關(guān)于原點(diǎn)對(duì)稱，排除A選項(xiàng)當(dāng)x1時(shí)，f (1)e>0，排除D選項(xiàng)又e>2，<，e>1，排除C選項(xiàng)故選B. (2)如圖，設(shè)MON，由弧長公式知x.在RtAOM中，|AO|1t，cos1t，ycos x2cos212(1t)21.又0t1，故選B.(3)畫出函數(shù)大致圖象如圖所示由圖象知，x1<，x2<1，x12x21，于是x1f (x2)x12x21x1，x1<，轉(zhuǎn)化為關(guān)于x1的二次函數(shù)在給定區(qū)間上的值域問題，易得x1f (x2)的取值范圍是.答案(1)B(2)B(3)類題通法1由函數(shù)解析式識(shí)別函數(shù)圖象的策略2根據(jù)動(dòng)點(diǎn)變化過程確定其函數(shù)圖象的策略(1)先根據(jù)已知條件求出函數(shù)解析式后再判斷其對(duì)應(yīng)的函數(shù)的圖象(2)采用“以靜觀動(dòng)”，即將動(dòng)點(diǎn)處于某些特殊的位置處考查圖象的變化特征，從而作出選擇(3)根據(jù)動(dòng)點(diǎn)中變量變化時(shí)，對(duì)因變量變化的影響，結(jié)合選項(xiàng)中圖象的變化趨勢(shì)作出判斷應(yīng)用通關(guān)1(2018全國卷)函數(shù)yx4x22的圖象大致為()解析：選D法一：令f (x)x4x22，則f (x)4x32x，令f (x)0，得x0或x，則f (x)>0的解集為，f (x)單調(diào)遞增；f (x)<0的解集為，f (x)單調(diào)遞減，結(jié)合圖象知選D.法二：當(dāng)x1時(shí)，y2，所以排除A、B選項(xiàng)當(dāng)x0時(shí)，y2，而當(dāng)x時(shí)，y22>2，所以排除C選項(xiàng)故選D.2.如圖，長方形ABCD的邊AB2，BC1，O是AB的中點(diǎn)，點(diǎn)P沿著邊BC，CD與DA運(yùn)動(dòng)，記BOPx.將動(dòng)點(diǎn)P到A，B兩點(diǎn)距離之和表示為x的函數(shù)f (x)，則yf (x)的圖象大致為()解析：選B當(dāng)x時(shí)，f (x)tan x，圖象不會(huì)是直線段，從而排除A、C.當(dāng)x時(shí)，f f 1，f 2.2<1，f <f f ，從而排除D，故選B.3已知f (x)2x1，g(x)1x2.規(guī)定：當(dāng)|f (x)|g(x)時(shí)，h(x)|f (x)|；當(dāng)|f (x)|<g(x)時(shí)，h(x)g(x)則h(x)()A有最小值1，最大值1 B有最大值1，無最小值C有最小值1，無最大值 D有最大值1，無最小值解析：選C作出函數(shù)g(x)1x2和函數(shù)|f (x)|2x1|的圖象如圖所示，得到函數(shù)h(x)的圖象如圖所示，由圖象得函數(shù)h(x)有最小值1，無最大值函數(shù)的性質(zhì)及應(yīng)用由題知法(1)(2018石家莊質(zhì)檢)已知函數(shù)f (x)為奇函數(shù)，當(dāng)x>0時(shí)，f (x)單調(diào)遞增，且f (1)0，若f (x1)>0，則x的取值范圍為()A(0,1)(2，) B(，0)(2，)C(，0)(3，) D(，1)(1，)(2)(2018益陽、湘潭調(diào)研)定義在R上的函數(shù)f (x)，滿足f (x5)f (x)，當(dāng)x(3,0時(shí)，f (x)x1，當(dāng)x(0,2時(shí)，f (x)log2x，則f (1)f (2)f (3)f (2 018)的值等于()A403 B405C806 D809(3)已知定義在R上的奇函數(shù)f (x)滿足f (x3)f (x)，且當(dāng)x時(shí)，f (x)x3，則f _.解析(1)由于函數(shù)f (x)是奇函數(shù)，且當(dāng)x>0時(shí)f (x)單調(diào)遞增，f (1)0，所以f (1)0，故由f (x1)>0，得1<x1<0或x1>1，所以0<x<1或x>2，故選A.(2)定義在R上的函數(shù)f (x)，滿足f (x5)f (x)，即函數(shù)f (x)的周期為5.又當(dāng)x(0,2時(shí)，f (x)log2x，所以f (1)log210，f (2)log221.當(dāng)x(3,0時(shí)，f (x)x1，所以f (3)f (2)1，f (4)f (1)0，f (5)f (0)1.所以f (1)f (2)f (3)f (2 018)403f (1)f (2)f (3)f (4)f (5)f (2 016)f (2 017)f (2 018)4031f (1)f (2)f (3)403011405.(3)由f (x3)f (x)知函數(shù)f (x)的周期為3，又函數(shù)f (x)為奇函數(shù)，所以f f f 3.答案(1)A(2)B(3)類題通法函數(shù)性質(zhì)的應(yīng)用技巧奇偶性具有奇偶性的函數(shù)在關(guān)于原點(diǎn)對(duì)稱的區(qū)間上其圖象、函數(shù)值、解析式和單調(diào)性聯(lián)系密切，研究問題時(shí)可轉(zhuǎn)化到只研究部分(一半)區(qū)間上尤其注意偶函數(shù)f (x)的性質(zhì)：f (|x|)f (x)單調(diào)性可以比較大小，求函數(shù)最值，解不等式，證明方程根的唯一性周期性利用周期性可以轉(zhuǎn)化函數(shù)的解析式、圖象和性質(zhì)，把不在已知區(qū)間上的問題，轉(zhuǎn)化到已知區(qū)間上求解對(duì)稱性利用其軸對(duì)稱或中心對(duì)稱可將研究的問題，轉(zhuǎn)化到另一對(duì)稱區(qū)間上研究 應(yīng)用通關(guān)1(2018貴陽模擬)已知函數(shù)f (x)，則下列結(jié)論正確的是()A函數(shù)f (x)的圖象關(guān)于點(diǎn)(1,2)中心對(duì)稱B函數(shù)f (x)在(，1)上是增函數(shù)C函數(shù)f (x)的圖象上至少存在兩點(diǎn)A，B，使得直線ABx軸D函數(shù)f (x)的圖象關(guān)于直線x1對(duì)稱解析：選A因?yàn)閥2，所以該函數(shù)圖象可以由y的圖象向右平移1個(gè)單位長度，再向上平移2個(gè)單位長度得到，所以函數(shù)f (x)的圖象關(guān)于點(diǎn)(1,2)中心對(duì)稱，A正確，D錯(cuò)誤易知函數(shù)f (x)在(，1)上單調(diào)遞減，故B錯(cuò)誤易知函數(shù)f (x)的圖象是由y的圖象平移得到的，所以不存在兩點(diǎn)A，B使得直線ABx軸，C錯(cuò)誤故選A.2(2019屆高三惠州調(diào)研)已知函數(shù)yf (x)的定義域?yàn)镽，且滿足下列三個(gè)條件：對(duì)任意的x1，x24,8，當(dāng)x1<x2時(shí)，都有>0恒成立；f (x4)f (x)；yf (x4)是偶函數(shù)若af (6)，bf (11)，cf (2 017)，則a，b，c的大小關(guān)系正確的是()Aa<b<c Bb<a<cCa<c<b Dc<b<a解析：選B由知函數(shù)f (x)在區(qū)間4,8上為單調(diào)遞增函數(shù)；由知f (x8)f (x4)f (x)，即函數(shù)f (x)的周期為8，所以cf (2 017)f (25281)f (1)，bf (11)f (3)；由可知函數(shù)f (x)的圖象關(guān)于直線x4對(duì)稱，所以bf (3)f (5)，cf (1)f (7)因?yàn)楹瘮?shù)f (x)在區(qū)間4,8上為單調(diào)遞增函數(shù)，所以f (5)<f (6)<f (7)，即b<a<c.3(2018全國卷)已知f (x)是定義域?yàn)?，)的奇函數(shù)，滿足f (1x)f (1x)若f (1)2，則f (1)f (2)f (3)f (50)()A50 B0C2 D50解析：選C法一：f (x)是奇函數(shù)，f (x)f (x)，f (1x)f (x1)由f (1x)f (1x)，得f (x1)f (x1)，f (x2)f (x)，f (x4)f (x2)f (x)，函數(shù)f (x)是周期為4的周期函數(shù)由f (x)為奇函數(shù)得f (0)0.又f (1x)f (1x)，f (x)的圖象關(guān)于直線x1對(duì)稱，f (2)f (0)0，f (2)0.又f (1)2，f (1)2，f (1)f (2)f (3)f (4)f (1)f (2)f (1)f (0)20200，f (1)f (2)f (3)f (4)f (49)f (50)012f (49)f (50)f (1)f (2)202.法二：由題意可設(shè)f (x)2sin，作出f (x)的部分圖象如圖所示由圖可知，f (x)的一個(gè)周期為4，所以f (1)f (2)f (3)f (50)12f (1)f (2)f (3)f (4)f (49)f (50)120f (1)f (2)2.重難增分(一)函數(shù)圖象與性質(zhì)的綜合應(yīng)用 典例細(xì)解(2016全國卷)已知函數(shù)f (x)(xR)滿足f (x)2f (x)，若函數(shù)y與yf (x)圖象的交點(diǎn)為(x1，y1)，(x2，y2)，(xm，ym)，則(xiyi)()A0BmC2m D4m學(xué)解題法一：利用函數(shù)的對(duì)稱性(學(xué)生用書不提供解題過程)因?yàn)閒 (x)2f (x)，所以f (x)f (x)2.因?yàn)?，1，所以函數(shù)yf (x)的圖象關(guān)于點(diǎn)(0,1)對(duì)稱函數(shù)y1，故其圖象也關(guān)于點(diǎn)(0,1)對(duì)稱所以函數(shù)y與yf (x)圖象的交點(diǎn)(x1，y1)，(x2，y2)，(xm，ym)成對(duì)出現(xiàn)，且每一對(duì)均關(guān)于點(diǎn)(0,1)對(duì)稱，所以i0，i2m，所以(xiyi)m.法二：構(gòu)造特殊函數(shù)(學(xué)生用書提供解題過程)因?yàn)閒 (x)2f (x)，所以f (x)f (x)2.因?yàn)?，1，所以函數(shù)yf (x)的圖象關(guān)于點(diǎn)(0,1)對(duì)稱可設(shè)yf (x)x1，由得交點(diǎn)(1,0)，(1,2)，則x1y1x2y22，結(jié)合選項(xiàng)，應(yīng)選B.答案B啟思維本題考查了抽象函數(shù)的性質(zhì)及圖象對(duì)稱性的應(yīng)用由于題目條件中的f (x)沒有具體的解析式，僅給出了它滿足的性質(zhì)f (x)2f (x)，即f (x)(xR)為抽象函數(shù)，顯然我們不可能求出這些交點(diǎn)的坐標(biāo)，這說明這些交點(diǎn)坐標(biāo)應(yīng)滿足某種規(guī)律，而這種規(guī)律必然和這兩個(gè)函數(shù)的性質(zhì)有關(guān)易知函數(shù)y關(guān)于點(diǎn)(0,1)成中心對(duì)稱，自然而然的讓我們有這樣的想法：函數(shù)f (x)(xR)的圖象是否也關(guān)于點(diǎn)(0,1)成中心對(duì)稱？基于這個(gè)想法及選擇題的特點(diǎn)，那么解題方向不外乎兩個(gè)：一是判斷f (x)的對(duì)稱性，利用兩個(gè)函數(shù)的對(duì)稱性求解；二是構(gòu)造一個(gè)具體的函數(shù)f (x)來求解已知直線l與曲線yx3x2x1有三個(gè)不同的交點(diǎn)A(x1，y1)，B(x2，y2)，C(x3，y3)，且|AB|AC|，則(xiyi)()A4 B5C6 D7解析易知yx22x1(x1)20，所以函數(shù)yx3x2x1在R上單調(diào)遞增函數(shù)yx3x2x1的圖象如圖所示，yx3x2x1(x1)3，易知曲線關(guān)于點(diǎn)對(duì)稱，因?yàn)橹本€l與yx3x2x1的圖象交于不同的三個(gè)點(diǎn)，且滿足|AB|AC|，故B，C兩點(diǎn)一定關(guān)于點(diǎn)A對(duì)稱，故A，則有得故(xiyi)x1y1x2y2x3y3127，選D.答案D啟思維本題主要考查函數(shù)的圖象與性質(zhì)，解此類問題常利用函數(shù)的性質(zhì)作出函數(shù)圖象，數(shù)形結(jié)合法解題知能升級(jí)1解決抽象函數(shù)問題的2個(gè)常用方法性質(zhì)法先研究清楚函數(shù)的奇偶性、對(duì)稱性和周期性等性質(zhì)，這樣函數(shù)就不再抽象了，而是變得相對(duì)具體，我們就可以畫出符合性質(zhì)的草圖來解題特殊值法根據(jù)對(duì)題目給出的抽象的函數(shù)性質(zhì)的理解，我們找到一個(gè)符合題意的具體函數(shù)或給變量賦值，把抽象函數(shù)問題化為具體的數(shù)學(xué)問題，從而問題得解2.解決抽象函數(shù)問題常用的幾個(gè)結(jié)論(1)函數(shù)yf (x)關(guān)于xa對(duì)稱f (ax)f (ax)f (x)f (2ax)；(2)函數(shù)yf (x)關(guān)于點(diǎn)(a,0)對(duì)稱f (ax)f (ax)0f (2ax)f (x)0；(3)yf (xa)是偶函數(shù)函數(shù)yf (x)關(guān)于直線xa對(duì)稱；yf (xa)是奇函數(shù)函數(shù)yf (x)關(guān)于(a,0)對(duì)稱(4)對(duì)于函數(shù)f (x)定義域內(nèi)任一自變量的值x：若f (xa)f (x)，則T2a；若f (xa)，則T2a；若f (xa)，則T2a；(a>0)若f (xa)f (xb)(ab)，則T|ab|；若f (2ax)f (x)且f (2bx)f (x)(ab)，則T2|ba|.增分集訓(xùn)1定義在R上的函數(shù)yf (x)為減函數(shù)，且函數(shù)yf (x1)的圖象關(guān)于點(diǎn)(1,0)對(duì)稱若f (x22x)f (2bb2)0，且0x2，則xb的取值范圍是()A2,0 B2,2C0,2 D0,4解析：選B設(shè)P(x，y)為函數(shù)yf (x1)的圖象上的任意一點(diǎn)，P關(guān)于點(diǎn)(1,0)對(duì)稱的點(diǎn)為(2x，y)，f (2x1)f (x1)，即f (1x)f (x1)不等式f (x22x)f (2bb2)0可化為f (x22x)f (2bb2)f (112bb2)f (b22b)函數(shù)yf (x)為定義在R上的減函數(shù)，x22xb22b，即(x1)2(b1)2.0x2，或畫出可行域如圖中陰影部分所示設(shè)xbz，則bxz，由圖可知，當(dāng)直線bxz經(jīng)過點(diǎn)(0,2)時(shí)，z取得最小值2；當(dāng)直線bxz經(jīng)過點(diǎn)(2,0)時(shí)，z取得最大值2.綜上可得，xb的取值范圍是2,22(2018沈陽模擬)設(shè)f (x)是定義在R上的偶函數(shù)，F(xiàn)(x)(x2)3f (x2)17，G(x)，若F(x)的圖象與G(x)的圖象的交點(diǎn)分別為(x1，y1)，(x2，y2)，(xm，ym)，則(xiyi)_.解析：f (x)是定義在R上的偶函數(shù)，g(x)x3f (x)是定義在R上的奇函數(shù)，其圖象關(guān)于原點(diǎn)中心對(duì)稱，函數(shù)F(x)(x2)3f (x2)17g(x2)17的圖象關(guān)于點(diǎn)(2，17)中心對(duì)稱又函數(shù)G(x)17的圖象也關(guān)于點(diǎn)(2，17)中心對(duì)稱，F(xiàn)(x)和G(x)的圖象的交點(diǎn)也關(guān)于點(diǎn)(2，17)中心對(duì)稱，x1x2xm(2)22m，y1y2ym(17)217m，(xiyi)(x1x2xm)(y1y2ym)19m.答案：19m重難增分(二)新定義下的函數(shù)問題 典例細(xì)解我們將具有性質(zhì)f f (x)的函數(shù)，稱為滿足“倒負(fù)”變換的函數(shù)給出下列函數(shù)：f (x)ln；f (x)；f (x)其中滿足“倒負(fù)”變換的函數(shù)是()ABC D解析對(duì)于，因?yàn)閒 lnlnf (x)，所以不滿足“倒負(fù)”變換；對(duì)于，因?yàn)閒 f (x)，所以滿足“倒負(fù)”變換；對(duì)于，因?yàn)閒 即f 所以f f (x)，故滿足“倒負(fù)”變換綜上可知，選C.答案C啟思維本題是在現(xiàn)有函數(shù)的圖象與性質(zhì)的基礎(chǔ)上定義的一種新的函數(shù)性質(zhì)，考查在新情境下，靈活運(yùn)用有關(guān)函數(shù)知識(shí)求解“新定義”類數(shù)學(xué)問題的能力求解本題的關(guān)鍵是先準(zhǔn)確寫出f 的表達(dá)式，并加以整理，再具體考慮f 與f (x)是否相等設(shè)函數(shù)f (x)的定義域?yàn)镈，若f (x)滿足條件：存在a，bD(a<b)，使f (x)在a，b上的值域也是a，b，則稱函數(shù)f (x)為“優(yōu)美函數(shù)”若函數(shù)f (x)log2(4xt)為“優(yōu)美函數(shù)”，則t的取值范圍是()A. B.C. D.解析f (x)log2(4xt)為增函數(shù)，且存在a，bD(a<b)，使f (x)在a，b上的值域也是a，b，則即所以a，b是方程4x2xt0的兩個(gè)不等的實(shí)根設(shè)2xm(m>0)，則方程m2mt0有兩個(gè)不等的實(shí)根，且兩根都大于0，所以解得0<t<.答案D啟思維(1)本題是一個(gè)新定義問題，讀懂題意后，即可由函數(shù)f (x)log2(4xt)為“優(yōu)美函數(shù)”，得到關(guān)于a，b的方程組，并構(gòu)造出以a，b為實(shí)數(shù)根的方程(2)在應(yīng)用換元法解題時(shí)，一定要注意挖掘隱含條件，確定新元的取值范圍，以防在解題過程中出現(xiàn)非等價(jià)轉(zhuǎn)化知能升級(jí)1函數(shù)新定義問題的常見形式(1)討論新函數(shù)的性質(zhì)；(2)利用新函數(shù)進(jìn)行運(yùn)算；(3)判斷新函數(shù)的圖象；(4)利用新概念判斷命題真假等2函數(shù)新定義問題的解題思路理解定義深刻理解題目中新函數(shù)的定義、新函數(shù)所具有的性質(zhì)或滿足的條件，將定義、性質(zhì)等與所求之間建立聯(lián)系合理轉(zhuǎn)化將題目中的新函數(shù)與已學(xué)函數(shù)聯(lián)系起來，仔細(xì)閱讀已知條件進(jìn)行分析，通過類比已學(xué)函數(shù)的性質(zhì)、圖象解決問題，或?qū)⑿潞瘮?shù)轉(zhuǎn)化為已學(xué)函數(shù)的復(fù)合函數(shù)等形式解決問題特值思想如果函數(shù)的某一性質(zhì)(一般是等式、不等式等)對(duì)某些數(shù)值恒成立，那么通過合理賦值可以得到特殊函數(shù)值甚至是函數(shù)解析式，進(jìn)而解決問題 增分集訓(xùn)1(2018武漢模擬)若存在正實(shí)數(shù)a，b，使得xR有f (xa)f (x)b恒成立，則稱f (x)為“限增函數(shù)”給出以下三個(gè)函數(shù)：f (x)x2x1；f (x)；f (x)sin(x2)，其中是“限增函數(shù)”的是()A BC D解析：選B對(duì)于，f (xa)f (x)b，即(xa)2(xa)1x2x1b，即2axa2ab，x對(duì)一切xR恒成立，顯然不存在這樣的正實(shí)數(shù)a，b.對(duì)于，f (x)，即b，|xa|x|b22b，而|xa|x|a，|x|a|x|b22b，則，顯然，當(dāng)ab2時(shí)式子恒成立，f (x)是“限增函數(shù)”對(duì)于，f (x)sin(x2)，1f (x)sin(x2)1，故f (xa)f (x)2，當(dāng)b2時(shí)，對(duì)于任意的正實(shí)數(shù)a，b都成立故選B.2對(duì)于函數(shù)f (x)和g(x)，設(shè)x|f (x)0，x|g(x)0，若存在，使得|1，則稱f (x)與g(x)互為“零點(diǎn)相鄰函數(shù)”若函數(shù)f (x)ex1x2與g(x)x2axa3互為“零點(diǎn)相鄰函數(shù)”，則實(shí)數(shù)a的取值范圍是()A2,4 B.C. D2,3解析：選Df (x)ex11>0，f (x)ex1x2是增函數(shù)又f (1)0，函數(shù)f (x)的零點(diǎn)為x1，1，|1|1，02，函數(shù)g(x)x2axa3在區(qū)間0,2上有零點(diǎn)由g(x)0，得a(0x2)，即a(x1)2(0x2)，設(shè)x1t(1t3)，則at2(1t3)，令h(t)t2(1t3)，易知h(t)在區(qū)間1,2)上是減函數(shù)，在區(qū)間(2,3上是增函數(shù)，2h(t)3，即2a3，故選D.3對(duì)任意實(shí)數(shù)a，b定義運(yùn)算“”：ab設(shè)f (x)(x21)(4x)，若函數(shù)yf (x)k的圖象與x軸恰有三個(gè)不同的交點(diǎn)，則實(shí)數(shù)k的取值范圍是()A(2,1) B0,1C2,0) D2,1)解析：選D當(dāng)x214x1，即x2或x3時(shí)，f (x)4x；當(dāng)x21<4x1，即2<x<3時(shí)，f (x)x21.作出f (x)的圖象如圖所示，由圖象可知，要使kf (x)有三個(gè)根，需滿足1<k2，即2k<1. 專題跟蹤檢測(cè)一、全練保分考法保大分1下列函數(shù)中，既是偶函數(shù)，又在區(qū)間(0，)上單調(diào)遞減的函數(shù)是()Ayx3 Byln |x|Cycos x Dy2|x|解析：選D顯然函數(shù)y2|x|是偶函數(shù)，當(dāng)x>0時(shí)，y2|x|x|x，函數(shù)yx在區(qū)間(0，)上是減函數(shù)故選D.2(2018貴陽模擬)若函數(shù)f (x)是定義在R上的奇函數(shù)，當(dāng)x0時(shí)，f (x)log2(x2)1，則f (6)()A2 B4C2 D4解析：選C根據(jù)題意得f (6)f (6)1log2(62)1log282.故選C.3(2018長春質(zhì)檢)已知函數(shù)f (x)則函數(shù)f (x)的值域?yàn)?)A1，) B(1，)C. DR解析：選B法一：當(dāng)x<1時(shí)，f (x)x22(1，)；當(dāng)x1時(shí)，f (x)2x1，綜上可知，函數(shù)f (x)的值域?yàn)?1，)故選B.法二：作出分段函數(shù)f (x)的圖象(圖略)可知，該函數(shù)的值域?yàn)?1，)，故選B.4(2018陜西質(zhì)檢)設(shè)xR，定義符號(hào)函數(shù)sgn x則函數(shù)f (x)|x|sgn x的圖象大致是()解析：選C由符號(hào)函數(shù)解析式和絕對(duì)值運(yùn)算，可得f (x)x，選C.5(2018濮陽二模)若f (x)是奇函數(shù)，則f (g(2)的值為()A. BC1 D1解析：選Cf (x)是奇函數(shù)，x<0時(shí)，g(x)3，g(2)31，f (g(2)f (1)f (1)1.故選C.6(2018葫蘆島一模)設(shè)偶函數(shù)f (x)對(duì)任意xR，都有f (x3)，且當(dāng)x3，2時(shí)，f (x)4x，則f (107.5)()A10 B.C10 D解析：選B因?yàn)閒 (x3)，所以f (x6)f (x)，所以函數(shù)f (x)是以6為周期的函數(shù)，f (107.5)f (6175.5)f (5.5).故選B.7(2019屆高三合肥調(diào)研)函數(shù)f (x)(exex)的圖象大致是()解析：選D因?yàn)閒 (x)(exex)(x0)，所以f (x)(exex)(exex)f (x)，所以f (x)是偶函數(shù)，排除選項(xiàng)A、C；因?yàn)楹瘮?shù)f (x)在(0，)上是增函數(shù)，所以排除選項(xiàng)B，故選D.8.點(diǎn)P在邊長為1的正方形ABCD的邊上運(yùn)動(dòng)，M是CD的中點(diǎn)，則當(dāng)P沿ABCM運(yùn)動(dòng)時(shí)，點(diǎn)P經(jīng)過的路程x與APM的面積y的函數(shù)yf (x)的圖象的形狀大致是圖中的()解析：選A根據(jù)題意得f (x)畫出分段函數(shù)圖象可知A正確9(2018河北“五個(gè)一名校聯(lián)盟”模擬)已知奇函數(shù)f (x)滿足f (x1)f (1x)，若當(dāng)x(1,1)時(shí)，f (x)lg，且f (2 018a)1，則實(shí)數(shù)a的值可以是()A. B.C D解析：選Af (x1)f (1x)，f (x)f (2x)又函數(shù)f (x)為奇函數(shù)，f (x)f (x)，f (x)f (2x)，f (2x)f (x)，f (x4)f (x2)f (x)，函數(shù)f (x)為周期函數(shù)，且周期為4.當(dāng)x(1,1)時(shí)，令f (x)lg1，得x，又f (2 018a)f (2a)f (a)，a可以是.10已知函數(shù)f (x)則f (1)f (2)f (3)f (2 018)()A2 018 B1 513C1 009 D.解析：選D函數(shù)f (x)f (1)f (1)21，f (2)f (0)20，f (3)f (1)21，f (1)f (2)f (3)f (2 018)1 009f (1)1 009f (0)1 009211 00920.故選D.11(2018郴州二模)已知函數(shù)f (x)ex，其中e是自然對(duì)數(shù)的底數(shù)則關(guān)于x的不等式f (2x1)f (x1)>0的解集為()A.(2，) B(2，)C.(2，) D(，2)解析：選B函數(shù)f (x)exexex滿足f (x)f (x)，f (x)為奇函數(shù)且是單調(diào)遞增函數(shù)，關(guān)于x的不等式f (2x1)f (x1)>0，即為f (2x1)>f (x1)，2x1>x1，解得x>2，故選B.12(2018陜西二模)已知函數(shù)f (x)ex2(x<0)與g(x)ln(xa)2的圖象上存在關(guān)于y軸對(duì)稱的點(diǎn)，則a的取值范圍是()A. B(，e)C. D.解析：選B由題意知，方程f (x)g(x)0在(0，)上有解，即ex2ln(xa)20在(0，)上有解，即函數(shù)yex的圖象與yln(xa)的圖象在(0，)上有交點(diǎn)，函數(shù)yln(xa)的圖象是由函數(shù)yln x的圖象向左平移a個(gè)單位得到的，當(dāng)yln x向左平移且平移到過點(diǎn)(0,1)后開始，兩函數(shù)的圖象有交點(diǎn)，把點(diǎn)(0,1)代入yln(xa)得，1ln a，ae，a<e.故選B.13已知f (x)是定義在R上的偶函數(shù)，且f (x4)f (x2)若當(dāng)x3,0時(shí)，f (x)6x，則f (919)_.解析：f (x4)f (x2)，f (x6)f (x)，f (x)的周期為6，91915361，f (919)f (1)又f (x)為偶函數(shù)，f (919)f (1)f (1)6.答案：614(2018陜西質(zhì)檢)若函數(shù)f (x)axb，xa4，a的圖象關(guān)于原點(diǎn)對(duì)稱，則函數(shù)g(x)bx，x4，1的值域?yàn)開解析：由函數(shù)f (x)的圖象關(guān)于原點(diǎn)對(duì)稱，可得a4a0，即a2，則函數(shù)f (x)2xb，其定義域?yàn)?,2，所以f (0)0，所以b0，所以g(x)，易知g(x)在4，1上單調(diào)遞減，故值域?yàn)間(1)，g(4)，即.答案：15(2018青島一模)定義在R上的函數(shù)f (x)滿足f (x)則f (2 009)的值為_解析：f (2 009)f (2 008)f (2 007)f (2 007)f (2 006)f (2 007)f (2 006)，即當(dāng)x>3時(shí)滿足f (x)f (x3)f (x6)，函數(shù)f (x)的周期為6.f (2 009)f (33465)f (5)f (1)當(dāng)x0時(shí)f (x)log2(1x)，f (1)1，f (2 009)f (1)1.答案：116已知函數(shù)f (x)e|x|，函數(shù)g(x)對(duì)任意的x1，m(m>1)，都有f (x2)g(x)，則m的取值范圍是_解析：作出函數(shù)yh(x)e|x2|和yg(x)的圖象，如圖所示，由圖可知當(dāng)x1時(shí)，h(1)g(1)，又當(dāng)x4時(shí)，h(4)e2<g(4)4e，當(dāng)x>4時(shí)，由ex24e5x，得e2x74，即2x7ln 4，解得xln 2，又m>1，1<mln 2.答案：17設(shè)函數(shù)f (x)若函數(shù)f (x)在區(qū)間m,4上的值域?yàn)?,2，則實(shí)數(shù)m的取值范圍為_解析：畫出函數(shù)f (x)的圖象如圖所示，結(jié)合圖象易得，當(dāng)m8，1時(shí)，f (x)1,2，故實(shí)數(shù)m的取值范圍為8，1答案：8，118設(shè)函數(shù)f (x)1，g(x)ln(ax22x1)，若對(duì)任意的x1R，都存在實(shí)數(shù)x2，使得f (x1)g(x2)成立，則實(shí)數(shù)a的取值范圍為_解析：設(shè)g(x)ln(ax22x1)的值域?yàn)锳，f (x)1在R上的值域?yàn)?，0，(，0A，h(x)ax22x1至少要取遍(0,1中的每一個(gè)數(shù)，又h(0)1，實(shí)數(shù)a需要滿足a0或解得a1.實(shí)數(shù)a的取值范圍是(，1答案：(，119已知函數(shù)f (x)(p>1)，若對(duì)于任意a，b，cR，都有f (a)f (b)>f (c)成立，則實(shí)數(shù)m的取值范圍是_解析：因?yàn)閒 (x)1，所以當(dāng)m>1時(shí)，函數(shù)f (x)在R上是減函數(shù)，函數(shù)f (x)的值域?yàn)?1，m)，所以f (a)f (b)>2，f (c)<m.因?yàn)閒 (a)f (b)>f (c)對(duì)任意的a，b，cR恒成立，所以m2，所以1<m2.當(dāng)m1時(shí)，f (x)1，f (a)f (b)2>f (c)1，滿足題意當(dāng)m<1時(shí)，函數(shù)f (x)在R上是增函數(shù)，函數(shù)f (x)的值域?yàn)?m,1)，所以f (a)f (b)>2m，f (c)<1，所以2m1，所以m，所以m<1.綜上可知，m2，故所求實(shí)數(shù)m的取值范圍是.答案：20已知函數(shù)f (x)若f (x)的值域?yàn)镽，則實(shí)數(shù)a的取值范圍是_解析：依題意，當(dāng)x1時(shí)，f (x)1log2x單調(diào)遞增，f (x)1log2x在區(qū)間1，)上的值域是1，)因此，要使函數(shù)f (x)的值域是R，則需函數(shù)f (x)在(，1)上的值域M(，1)當(dāng)a1<0，即a<1時(shí)，函數(shù)f (x)在(，1)上單調(diào)遞減，函數(shù)f (x)在(，1)上的值域M(a3，)，顯然此時(shí)不能滿足M(，1)，因此a<1不滿足題意；當(dāng)a10，即a1時(shí)，函數(shù)f (x)在(，1)上的值域M2，此時(shí)不能滿足M(，1)，因此a1不滿足題意；當(dāng)a1>0，即a>1時(shí)，函數(shù)f (x)在(，1)上單調(diào)遞增，函數(shù)f (x)在(，1)上的值域M(，a3)，由M(，1)得解得1<a2.綜上所述，滿足題意的實(shí)數(shù)a的取值范圍是(1,2答案：(1,2二、強(qiáng)化壓軸考法拉開分1(2018惠州第一次調(diào)研)已知定義域?yàn)镽的偶函數(shù)f (x)在(，0上是減函數(shù)，且f (1)2，則不等式f (log2x)>2的解集為()A(2，) B.(2，)C.(，) D(，)解析：選B因?yàn)閒 (x)是R上的偶函數(shù)，且在(，0上是減函數(shù)，所以f (x)在0，)上是增函數(shù)因?yàn)閒 (1)2，所以f (1)2，所以f (log2x)>2f (|log2x|)>f (1)|log2x|>1log2x>1或log2x<1x>2或0<x<.故選B.2(2019屆高三太原模擬)已知函數(shù)f (x)是偶函數(shù)，f (x1)是奇函數(shù)，且對(duì)于任意x1，x20,1，且x1x2，都有(x1x2)f (x1)f (x2)<0，設(shè)af ，bf ，cf ，則下列結(jié)論正確的是()Aa>b>c Bb>a>cCb>c>a Dc>a>b解析：選B法一：因?yàn)楹瘮?shù)f (x)是偶函數(shù)，f (x1)是奇函數(shù)，所以f (x)f (x)，f (x1)f (x1)，所以f (x1)f (x1)，所以f (x)f (x2)，所以f (x)f (x4)，所以af f f ，bf f ，cf f ，又對(duì)于任意x1，x20,1，且x1x2，都有(x1x2)f (x1)f (x2)<0，所以f (x)在0,1上是減函數(shù)，因?yàn)?lt;<，所以b>a>c，故選B.法二：因?yàn)楹瘮?shù)f (x)是偶函數(shù)，f (x1)是奇函數(shù)，且對(duì)于任意x1，x20,1，且x1x2，都有(x1x2)f (x1)f (x2)<0，即f (x)在0,1上是減函數(shù)，不妨取f (x)cosx，則af coscos，bf coscos，cf coscos，因?yàn)楹瘮?shù)ycos x在0,1上是減函數(shù)，且<<<1，所以b>a>c，故選B.3(2018全國卷)設(shè)函數(shù)f (x)則滿足f (x1)<f (2x)的x的取值范圍是()A(，1 B(0，)C(1,0) D(，0)解析：選D法一：當(dāng)即x1時(shí)，f (x1)<f (2x)，即為2(x1)<22x，即(x1)<2x，解得x<1.因此不等式的解集為(，1當(dāng)時(shí)，不等式組無解當(dāng)即1<x0時(shí)，f (x1)<f (2x)，即為1<22x，解得x<0.因此不等式的解集為(1,0)當(dāng)即x>0時(shí)，f (x1)1，f (2x)1，不合題意綜上，不等式f (x1)<f (2x)的解集為(，0)法二：f (x)函數(shù)f (x)的圖象如圖所示結(jié)合圖象知，要使f (x1)<f (2x)，則需或x<0，故選D.4已知函數(shù)yf (x)是R上的偶函數(shù)，對(duì)于xR都有f (x4)f (x)f (2)成立，且f (3)1，當(dāng)x1，x20,2，且x1x2時(shí)，都有>0.給出下列命題：f (221)1；函數(shù)yf (x)圖象的一條對(duì)稱軸方程為x4；函數(shù)yf (x)在6，4上為減函數(shù)；方程f (x)0在6,6上有4個(gè)根其中正確的命題個(gè)數(shù)為()A1 B2C3 D4解析：選D令x2，由f (x4)f (x)f (2)得f (2)0.因?yàn)楹瘮?shù)yf (x)是R上的偶函數(shù)，所以f (2)f (2)0，所以f (x4)f (x)，即函數(shù)yf (x)是以4為周期的周期函數(shù)，所以f (221)f (5541)f (1)因?yàn)閒 (3)1，所以f (3)f (1)1，從而f (221)1，正確因?yàn)楹瘮?shù)圖象關(guān)于y軸對(duì)稱，函數(shù)的周期為4，所以函數(shù)yf (x)圖象的一條對(duì)稱軸方程為x4，正確因?yàn)楫?dāng)x1，x20,2，且x1x2時(shí)，都有>0，設(shè)x1<x2，則f (x1)<f (x2)，易知函數(shù)yf (x)在0,2上是增函數(shù)根據(jù)圖象的對(duì)稱性，易知函數(shù)yf (x)在2,0上是減函數(shù)，又根據(jù)周期性，易知函數(shù)yf (x)在6，4上為減函數(shù)，正確因?yàn)閒 (2)f (2)0，由函數(shù)f (x)的單調(diào)性及周期性，可知在6,6上有且僅有f (2)f (2)f (6)f (6)0，即方程f (x)0在6,6上有4個(gè)根綜上所述，四個(gè)命題都正確故選D.5(2018長沙模擬)定義運(yùn)算：xy例如：343，(2)44，則函數(shù)f (x)x2(2xx2)的最大值為_解析：由已知得f (x)x2(2xx2)畫出函數(shù)f (x)的大致圖象(圖略)可知，函數(shù)f (x)的最大值為4.答案：46(2019屆高三石家莊檢測(cè))已知定義域?yàn)镽的函數(shù)f (x)是奇函數(shù)，當(dāng)x0時(shí)，f (x)|xa2|a2，且對(duì)xR，恒有f (x1)f (x)，則實(shí)數(shù)a的取值范圍為_解析：定義域?yàn)镽的函數(shù)f (x)是奇函數(shù)，當(dāng)x0時(shí)，f (x)|xa2|a2作出函數(shù)f (x)的圖象如圖所示當(dāng)x<0時(shí)，函數(shù)的最大值為a2，對(duì)xR，恒有f (x1)f (x)，要滿足f (x1)f (x)，1要大于等于a2,3a2的區(qū)間長度3a2(a2)，13a2(a2)，解得a.答案：7已知函數(shù)yf (x)與yF(x)的圖象關(guān)于y軸對(duì)稱，當(dāng)函數(shù)yf (x)和yF(x)在區(qū)間a，b同時(shí)遞增或同時(shí)遞減時(shí)，把區(qū)間a，b叫作函數(shù)yf (x)的“不動(dòng)區(qū)間”若區(qū)間1,2為函數(shù)f (x)|2xt|的“不動(dòng)區(qū)間”，則實(shí)數(shù)t的取值范圍是_解析：函數(shù)yf (x)與yF(x)的圖象關(guān)于y軸對(duì)稱，F(xiàn)(x)f (x)|2xt|.區(qū)間1,2為函數(shù)f (x)|2xt|的“不動(dòng)區(qū)間”，函數(shù)f (x)|2xt|和函數(shù)F(x)|2xt|在1,2上單調(diào)性相同y2xt和函數(shù)y2xt的單調(diào)性相反(2xt)(2xt)0在1,2上恒成立，即2xt2x在1,2上恒成立，即t2.答案：8定義在R上的函數(shù)f (x)在(，2)上是增函數(shù)，且f (x2)是偶函數(shù)，若對(duì)一切實(shí)數(shù)x，不等式f (2sin x2)>f (sin x1m)恒成立，則實(shí)數(shù)m的取值范圍為_解析：因?yàn)閒 (x2)是偶函數(shù)，所以函數(shù)f (x)的圖象關(guān)于x2對(duì)稱又f (x)在(，2)上為增函數(shù)，則f (x)在(2，)上為減函數(shù)，所以不等式f (2sin x2)>f (sin x1m)恒成立等價(jià)于|2sin x22|<|sin x1m2|，即|2sin x|<|sin x1m|，兩邊同時(shí)平方，得3sin2x2(1m)sin x(1m)2<0，即(3sin x1m)(sin x1m)<0，即或即或即或即m<2或m>4，故實(shí)數(shù)m的取值范圍為(，2)(4，)答案：(，2)(4，)