新高考數(shù)學(xué)二輪復(fù)習(xí)強(qiáng)化練習(xí)專題03 函數(shù)的圖象與應(yīng)用（講）（解析版）

第一篇熱點(diǎn)、難點(diǎn)突破篇專題03函數(shù)的圖象與應(yīng)用（講）真題體驗(yàn)感悟高考1．（2022·全國(guó)·高考真題（文））如圖是下列四個(gè)函數(shù)中的某個(gè)函數(shù)在區(qū)間SKIPIF1<0的大致圖像，則該函數(shù)是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】由函數(shù)圖像的特征結(jié)合函數(shù)的性質(zhì)逐項(xiàng)排除即可得解.【詳解】設(shè)SKIPIF1<0，則SKIPIF1<0，故排除B;設(shè)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，故排除C;設(shè)SKIPIF1<0，則SKIPIF1<0，故排除D.故選：A.2．（2022·全國(guó)·高考真題（理））函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0的圖象大致為（

）A． B．C． D．【答案】A【分析】由函數(shù)的奇偶性結(jié)合指數(shù)函數(shù)、三角函數(shù)的性質(zhì)逐項(xiàng)排除即可得解.【詳解】令SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0為奇函數(shù)，排除BD；又當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，排除C.故選：A.3．（2021·浙江·高考真題）已知函數(shù)SKIPIF1<0，則圖象為如圖的函數(shù)可能是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】由函數(shù)的奇偶性可排除A、B，結(jié)合導(dǎo)數(shù)判斷函數(shù)的單調(diào)性可判斷C，即可得解.【詳解】對(duì)于A，SKIPIF1<0，該函數(shù)為非奇非偶函數(shù)，與函數(shù)圖象不符，排除A；對(duì)于B，SKIPIF1<0，該函數(shù)為非奇非偶函數(shù)，與函數(shù)圖象不符，排除B；對(duì)于C，SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，與圖象不符，排除C.故選：D.4．（2020·天津·高考真題）已知函數(shù)SKIPIF1<0若函數(shù)SKIPIF1<0恰有4個(gè)零點(diǎn)，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】由SKIPIF1<0，結(jié)合已知，將問題轉(zhuǎn)化為SKIPIF1<0與SKIPIF1<0有SKIPIF1<0個(gè)不同交點(diǎn)，分SKIPIF1<0三種情況，數(shù)形結(jié)合討論即可得到答案.【詳解】注意到SKIPIF1<0，所以要使SKIPIF1<0恰有4個(gè)零點(diǎn)，只需方程SKIPIF1<0恰有3個(gè)實(shí)根即可，令SKIPIF1<0SKIPIF1<0，即SKIPIF1<0與SKIPIF1<0的圖象有SKIPIF1<0個(gè)不同交點(diǎn).因?yàn)镾KIPIF1<0，當(dāng)SKIPIF1<0時(shí)，此時(shí)SKIPIF1<0，如圖1，SKIPIF1<0與SKIPIF1<0有SKIPIF1<0個(gè)不同交點(diǎn)，不滿足題意；當(dāng)SKIPIF1<0時(shí)，如圖2，此時(shí)SKIPIF1<0與SKIPIF1<0恒有SKIPIF1<0個(gè)不同交點(diǎn)，滿足題意；當(dāng)SKIPIF1<0時(shí)，如圖3，當(dāng)SKIPIF1<0與SKIPIF1<0相切時(shí)，聯(lián)立方程得SKIPIF1<0，令SKIPIF1<0得SKIPIF1<0，解得SKIPIF1<0（負(fù)值舍去），所以SKIPIF1<0.綜上，SKIPIF1<0的取值范圍為SKIPIF1<0.故選：D.

5.（2019·浙江·高考真題）已知SKIPIF1<0，函數(shù)SKIPIF1<0，若存在SKIPIF1<0，使得SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0的最大值是____.【答案】SKIPIF1<0【分析】本題主要考查含參絕對(duì)值不等式、函數(shù)方程思想及數(shù)形結(jié)合思想，屬于能力型考題.從研究SKIPIF1<0入手，令SKIPIF1<0，從而使問題加以轉(zhuǎn)化，通過繪制函數(shù)圖象，觀察得解.【詳解】使得SKIPIF1<0，使得令SKIPIF1<0，則原不等式轉(zhuǎn)化為存在SKIPIF1<0，由折線函數(shù)，如圖只需SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0的最大值是SKIPIF1<0總結(jié)規(guī)律預(yù)測(cè)考向（一）規(guī)律與預(yù)測(cè)高考對(duì)此部分內(nèi)容的命題多集中于函數(shù)圖象的辨識(shí)、函數(shù)圖象的變換、主要有由函數(shù)的性質(zhì)及解析式選圖；由函數(shù)的圖象來(lái)研究函數(shù)的性質(zhì)、圖象的變換、數(shù)形結(jié)合解決不等式、方程問題等．常常與導(dǎo)數(shù)結(jié)合考查.應(yīng)特別注意兩圖象交點(diǎn)、函數(shù)性質(zhì)、方程解的個(gè)數(shù)、不等式的解集等方面的應(yīng)用.關(guān)注抽象函數(shù)問題出現(xiàn).（二）本專題考向展示考點(diǎn)突破典例分析考向一做函數(shù)的圖象【核心知識(shí)】作函數(shù)圖象有兩種基本方法：一是描點(diǎn)法；二是圖象變換法，其中圖象變換有平移變換、伸縮變換、對(duì)稱變換．描點(diǎn)法步驟：(1)確定函數(shù)的定義域；(2)化簡(jiǎn)函數(shù)解析式；(3)討論函數(shù)的性質(zhì)(奇偶性、單調(diào)性、周期性、對(duì)稱性等)；(4)列表(尤其注意特殊點(diǎn)、零點(diǎn)、最大值點(diǎn)、最小值點(diǎn)、與坐標(biāo)軸的交點(diǎn)等)，描點(diǎn)，連線.【典例分析】典例1.（全國(guó)·高考真題（文））畫出函數(shù)SKIPIF1<0的圖象．【答案】見解析【分析】由SKIPIF1<0的圖象與函數(shù)圖象平移變換求解，【詳解】由SKIPIF1<0圖象向左平移一個(gè)單位即可，典例2.（2022·陜西·西安市鄠邑區(qū)第二中學(xué)高三階段練習(xí)）設(shè)函數(shù)SKIPIF1<0.(1)證明：函數(shù)SKIPIF1<0是偶函數(shù)；(2)畫出這個(gè)函數(shù)的圖象；【答案】(1)證明見解析(2)答案見解析【分析】（1）根據(jù)函數(shù)的奇偶性證得結(jié)論成立.（2）將SKIPIF1<0寫成分段函數(shù)的形式，從而畫出SKIPIF1<0的圖象.【詳解】（1）證明：因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0為偶函數(shù).（2）SKIPIF1<0，由此畫出SKIPIF1<0的圖象如下圖所示：典例3.（2021·全國(guó)·高考真題（文））已知函數(shù)SKIPIF1<0．（1）畫出SKIPIF1<0和SKIPIF1<0的圖像；（2）若SKIPIF1<0，求a的取值范圍．【答案】（1）圖像見解析；（2）SKIPIF1<0【分析】（1）分段去絕對(duì)值即可畫出圖像；（2）根據(jù)函數(shù)圖像數(shù)形結(jié)和可得需將SKIPIF1<0向左平移可滿足同角，求得SKIPIF1<0過SKIPIF1<0時(shí)SKIPIF1<0的值可求.【詳解】（1）可得SKIPIF1<0，畫出圖像如下：SKIPIF1<0，畫出函數(shù)圖像如下：（2）SKIPIF1<0，如圖，在同一個(gè)坐標(biāo)系里畫出SKIPIF1<0圖像，SKIPIF1<0是SKIPIF1<0平移了SKIPIF1<0個(gè)單位得到，則要使SKIPIF1<0，需將SKIPIF1<0向左平移，即SKIPIF1<0，當(dāng)SKIPIF1<0過SKIPIF1<0時(shí)，SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0（舍去），則數(shù)形結(jié)合可得需至少將SKIPIF1<0向左平移SKIPIF1<0個(gè)單位，SKIPIF1<0.【總結(jié)提升】函數(shù)圖象的畫法(1)直接法：當(dāng)函數(shù)表達(dá)式(或變形后的表達(dá)式)是熟悉的基本函數(shù)時(shí)，就可根據(jù)這些函數(shù)的特征描出圖象的關(guān)鍵點(diǎn)直接作出．(2)轉(zhuǎn)化法：含有絕對(duì)值符號(hào)的函數(shù)，可去掉絕對(duì)值符號(hào)，轉(zhuǎn)化為分段函數(shù)來(lái)畫圖象．.考向二基本初等函數(shù)的圖象【核心知識(shí)】1．指數(shù)函數(shù)y＝ax(a>0，a≠1)與對(duì)數(shù)函數(shù)y＝logax(a>0，a≠1)互為反函數(shù)，其圖象關(guān)于y＝x對(duì)稱，它們的圖象和性質(zhì)分0<a<1，a>1兩種情況，著重關(guān)注兩函數(shù)圖象的異同．2．冪函數(shù)y＝xα的圖象，主要掌握α＝1,2,3，SKIPIF1<0，－1五種情況．【典例分析】典例4.（2020·山東·高考真題）已知函數(shù)SKIPIF1<0是偶函數(shù)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則該函數(shù)在SKIPIF1<0上的圖像大致是（

）A． B．C． D．【答案】B【分析】根據(jù)偶函數(shù)，指數(shù)函數(shù)的知識(shí)確定正確選項(xiàng).【詳解】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上遞減，SKIPIF1<0是偶函數(shù)，所以SKIPIF1<0在SKIPIF1<0上遞增.注意到SKIPIF1<0，所以B選項(xiàng)符合.故選：B典例5.（2021·四川高三三模（理））函數(shù)SKIPIF1<0及SKIPIF1<0，則SKIPIF1<0及SKIPIF1<0的圖象可能為（）A． B．C． D．【答案】B【解析】討論SKIPIF1<0、SKIPIF1<0確定SKIPIF1<0的單調(diào)性和定義域、SKIPIF1<0在y軸上的截距，再討論SKIPIF1<0、SKIPIF1<0，結(jié)合SKIPIF1<0的單調(diào)性，即可確定函數(shù)的可能圖象.【詳解】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減，SKIPIF1<0單調(diào)遞減，所以SKIPIF1<0單調(diào)遞增且定義域?yàn)镾KIPIF1<0，此時(shí)SKIPIF1<0與y軸的截距在SKIPIF1<0上，排除C.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減，SKIPIF1<0單調(diào)遞增，所以SKIPIF1<0單調(diào)遞減且定義域?yàn)镾KIPIF1<0，此時(shí)SKIPIF1<0與y軸的截距在SKIPIF1<0上.∴當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞增；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0單調(diào)遞減，故只有B符合要求.故選：B.典例6.（2019·浙江·高考真題）在同一直角坐標(biāo)系中，函數(shù)SKIPIF1<0且SKIPIF1<0的圖象可能是A． B．C． D．【答案】D【解析】本題通過討論SKIPIF1<0的不同取值情況，分別討論本題指數(shù)函數(shù)、對(duì)數(shù)函數(shù)的圖象和，結(jié)合選項(xiàng)，判斷得出正確結(jié)論.題目不難，注重重要知識(shí)、基礎(chǔ)知識(shí)、邏輯推理能力的考查.【詳解】當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0過定點(diǎn)SKIPIF1<0且單調(diào)遞減，則函數(shù)SKIPIF1<0過定點(diǎn)SKIPIF1<0且單調(diào)遞增，函數(shù)SKIPIF1<0過定點(diǎn)SKIPIF1<0且單調(diào)遞減，D選項(xiàng)符合；當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0過定點(diǎn)SKIPIF1<0且單調(diào)遞增，則函數(shù)SKIPIF1<0過定點(diǎn)SKIPIF1<0且單調(diào)遞減，函數(shù)SKIPIF1<0過定點(diǎn)SKIPIF1<0且單調(diào)遞增，各選項(xiàng)均不符合.綜上，選D.考向三函數(shù)圖象的變換及應(yīng)用【核心知識(shí)】利用圖象變換法作函數(shù)的圖象(1)平移變換(2)對(duì)稱變換y＝f(x)的圖象eq\o(→,\s\up7(關(guān)于x軸對(duì)稱))y＝－f(x)的圖象；y＝f(x)的圖象eq\o(→,\s\up7(關(guān)于y軸對(duì)稱))y＝f(－x)的圖象；y＝f(x)的圖象eq\o(→,\s\up7(關(guān)于原點(diǎn)對(duì)稱))y＝－f(－x)的圖象；(3)伸縮變換y＝f(x)eq\o(→,\s\up7(縱坐標(biāo)不變),\s\do5(各點(diǎn)橫坐標(biāo)變?yōu)樵瓉?lái)的\f(1,a)（a>0）倍))y＝f(ax).y＝f(x)eq\o(→,\s\up7(橫坐標(biāo)不變),\s\do5(各點(diǎn)縱坐標(biāo)變?yōu)樵瓉?lái)的A（A>0）倍))y＝Af(x).(4)翻轉(zhuǎn)變換y＝f(x)的圖象eq\o(→,\s\up7(x軸下方部分翻折到上方),\s\do5(x軸及上方部分不變))y＝|f(x)|的圖象；y＝f(x)的圖象eq\o(→,\s\up7(y軸右側(cè)部分翻折到左側(cè)),\s\do5(原y軸左側(cè)部分去掉，右側(cè)不變))y＝f(|x|)的圖象.典例7.（全國(guó)·高考真題（文））若函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，則函數(shù)SKIPIF1<0與SKIPIF1<0的圖象關(guān)于（

）A．直線SKIPIF1<0對(duì)稱 B．直線SKIPIF1<0對(duì)稱C．直線SKIPIF1<0對(duì)稱 D．直線SKIPIF1<0對(duì)稱【答案】C【分析】根據(jù)函數(shù)圖象的變換規(guī)律，結(jié)合SKIPIF1<0與SKIPIF1<0的圖象的關(guān)系即得.【詳解】因?yàn)楹瘮?shù)SKIPIF1<0的圖象是SKIPIF1<0的圖象向右平移1個(gè)單位得到的，SKIPIF1<0的圖象是SKIPIF1<0的圖象也向右平移1個(gè)單位得到的；又因?yàn)镾KIPIF1<0與SKIPIF1<0的圖象是關(guān)于SKIPIF1<0軸（直線SKIPIF1<0）對(duì)稱，所以函數(shù)SKIPIF1<0與SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱.故選：SKIPIF1<0.典例8.（2021·北京高三二模）已知指數(shù)函數(shù)SKIPIF1<0，將函數(shù)SKIPIF1<0的圖象上的每個(gè)點(diǎn)的橫坐標(biāo)不變，縱坐標(biāo)擴(kuò)大為原來(lái)的SKIPIF1<0倍，得到函數(shù)SKIPIF1<0的圖象，再將SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度，所得圖象恰好與函數(shù)SKIPIF1<0的圖象重合，則a的值是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】根據(jù)函數(shù)圖象變換求出變換后的函數(shù)解析式，結(jié)合已知條件可得出關(guān)于實(shí)數(shù)SKIPIF1<0的等式，進(jìn)而可求得實(shí)數(shù)SKIPIF1<0的值.【詳解】由題意可得SKIPIF1<0，再將SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度，得到函數(shù)SKIPIF1<0，又因?yàn)镾KIPIF1<0，所以，SKIPIF1<0，整理可得SKIPIF1<0，因?yàn)镾KIPIF1<0且SKIPIF1<0，解得SKIPIF1<0.故選：D.典例9.（2022·河南·高三階段練習(xí)（文））設(shè)函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，且滿足SKIPIF1<0是偶函數(shù)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則下列說法不正確的是（

）A．SKIPIF1<0B．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的取值范圍為SKIPIF1<0C．SKIPIF1<0為奇函數(shù)D．方程SKIPIF1<0僅有5個(gè)不同實(shí)數(shù)解【答案】D【分析】由已知條件可得函數(shù)的對(duì)稱中心及對(duì)稱軸，利用對(duì)稱中心和對(duì)稱軸將已知區(qū)間圖象進(jìn)行多次對(duì)稱變換，可得函數(shù)SKIPIF1<0的圖象，依據(jù)圖象對(duì)各個(gè)選項(xiàng)進(jìn)行判斷即可.【詳解】∵SKIPIF1<0，∴SKIPIF1<0，∴SKIPIF1<0當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∴函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0的圖象如圖：∵SKIPIF1<0是偶函數(shù)，∴SKIPIF1<0，即SKIPIF1<0∴SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對(duì)稱，SKIPIF1<0在區(qū)間SKIPIF1<0的圖象如圖：∵SKIPIF1<0，∴將SKIPIF1<0中的SKIPIF1<0替換為SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0∴SKIPIF1<0的圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱，SKIPIF1<0在區(qū)間SKIPIF1<0的圖象如圖：由函數(shù)圖象的對(duì)稱軸直線SKIPIF1<0和對(duì)稱中心SKIPIF1<0進(jìn)行多次對(duì)稱變換，可得函數(shù)圖象如圖：由函數(shù)圖象可知，SKIPIF1<0是周期為SKIPIF1<0的周期函數(shù)，函數(shù)SKIPIF1<0的對(duì)稱軸為直線SKIPIF1<0（SKIPIF1<0Z），對(duì)稱中心為點(diǎn)SKIPIF1<0（SKIPIF1<0Z），另外，函數(shù)的周期性還可以通過以下方法進(jìn)行證明：將SKIPIF1<0中的SKIPIF1<0替換為SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0，由已知有SKIPIF1<0，∴SKIPIF1<0將SKIPIF1<0中SKIPIF1<0分別替換為SKIPIF1<0和SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0和SKIPIF1<0，即SKIPIF1<0SKIPIF1<0SKIPIF1<0∴SKIPIF1<0將SKIPIF1<0中SKIPIF1<0替換為SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0，∴SKIPIF1<0是周期為SKIPIF1<0的周期函數(shù).對(duì)于A，SKIPIF1<0，故A正確；對(duì)于B，當(dāng)SKIPIF1<0時(shí)，由圖象可知其值域?yàn)镾KIPIF1<0，故B正確；對(duì)于C，由圖象知，其圖象的對(duì)稱中心為點(diǎn)SKIPIF1<0（SKIPIF1<0Z），當(dāng)SKIPIF1<0時(shí)，點(diǎn)SKIPIF1<0為SKIPIF1<0圖象的對(duì)稱中心，因此將SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度，所得函數(shù)SKIPIF1<0為奇函數(shù)，故C正確；對(duì)于D，將函數(shù)SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度，再將SKIPIF1<0軸下方的圖象翻折至SKIPIF1<0軸上方，得到函數(shù)SKIPIF1<0的圖象，易知SKIPIF1<0的圖象過點(diǎn)SKIPIF1<0如圖，SKIPIF1<0的圖象與SKIPIF1<0的圖象有6個(gè)交點(diǎn)，所以方程SKIPIF1<0有6個(gè)不同實(shí)數(shù)解，故D錯(cuò)誤．故選：D.【規(guī)律方法】圖象變換法常用的有平移變換、伸縮變換和對(duì)稱變換．尤其注意y＝f(x)與y＝f(－x)，y＝－f(x)，y＝－f(－x)，y＝f(|x|)，y＝|f(x)|及y＝af(x)＋b的相互關(guān)系．考向四函數(shù)圖象的識(shí)別【核心知識(shí)】識(shí)別函數(shù)圖象的方法基本方法有：(1)直接法(直接求出函數(shù)的解析式并作出其圖象)；(2)特例排除法(其中用特殊點(diǎn)法破解函數(shù)圖象問題需尋找特殊的點(diǎn)，即根據(jù)已知函數(shù)的圖象或已知函數(shù)的解析式，取特殊點(diǎn)，判斷各選項(xiàng)的圖象是否經(jīng)過該特殊點(diǎn))；(3)性質(zhì)驗(yàn)證法．【典例分析】典例10.（2022·天津·高考真題）函數(shù)SKIPIF1<0的圖像為（

）A． B．C． D．【答案】D【分析】分析函數(shù)SKIPIF1<0的定義域、奇偶性、單調(diào)性及其在SKIPIF1<0上的函數(shù)值符號(hào)，結(jié)合排除法可得出合適的選項(xiàng).【詳解】函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，且SKIPIF1<0，函數(shù)SKIPIF1<0為奇函數(shù)，A選項(xiàng)錯(cuò)誤；又當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，C選項(xiàng)錯(cuò)誤；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0函數(shù)單調(diào)遞增，故B選項(xiàng)錯(cuò)誤；故選：D.典例11.（2021·天津·高考真題）函數(shù)SKIPIF1<0的圖像大致為（

）A． B．C． D．【答案】B【分析】由函數(shù)為偶函數(shù)可排除AC，再由當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，排除D，即可得解.【詳解】設(shè)SKIPIF1<0，則函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，關(guān)于原點(diǎn)對(duì)稱，又SKIPIF1<0，所以函數(shù)SKIPIF1<0為偶函數(shù)，排除AC；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0，排除D.故選：B.典例12.（2022·四川綿陽(yáng)·一模（理））函數(shù)SKIPIF1<0的圖象大致為（

）A． B．C． D．【答案】D【分析】先利用導(dǎo)函數(shù)研究SKIPIF1<0上的單調(diào)性，得到SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0，進(jìn)而研究SKIPIF1<0上的單調(diào)性，得到在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0，從而選出正確答案.【詳解】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0在SKIPIF1<0處取得極小值，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，且SKIPIF1<0，顯然SKIPIF1<0，綜上：只有D選項(xiàng)滿足要求.故選：D【總結(jié)提升】識(shí)圖的三種常用方法1．抓住函數(shù)的性質(zhì)，定性分析：（1）由函數(shù)的定義域，判斷圖象的左、右位置，由函數(shù)的值域，判斷圖象的上、下位置；（2）由函數(shù)的單調(diào)性，判斷圖象的變化趨勢(shì)；（3）由函數(shù)的奇偶性，判斷圖象的對(duì)稱性；（4）由函數(shù)的周期性，判斷圖象的循環(huán)往復(fù)．2．抓住函數(shù)的特征，定量計(jì)算：從函數(shù)的特征點(diǎn)，利用特征點(diǎn)、特殊值的計(jì)算分析解決問題．3．根據(jù)實(shí)際背景、圖形判斷函數(shù)圖象的方法：(1)根據(jù)題目所給條件確定函數(shù)解析式，從而判斷函數(shù)圖象(定量分析)；(2)根據(jù)自變量取不同值時(shí)函數(shù)值的變化、增減速度等判斷函數(shù)圖象(定性分析)．考向五由函數(shù)圖象確定解析式【核心知識(shí)】從圖象與軸的交點(diǎn)及左、右、上、下分布范圍、變化趨勢(shì)、對(duì)稱性等方面找準(zhǔn)解析式與圖象的對(duì)應(yīng)關(guān)系．【典例分析】典例13.（2022·重慶·高三階段練習(xí)）已知函數(shù)SKIPIF1<0的圖象如圖1所示，則圖2所表示的函數(shù)是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】根函數(shù)圖象判斷兩個(gè)函數(shù)見的位置關(guān)系，進(jìn)而可得解.【詳解】由圖知，將SKIPIF1<0的圖象關(guān)于SKIPIF1<0軸對(duì)稱后再向下平移SKIPIF1<0個(gè)單位即得圖2，又將SKIPIF1<0的圖象關(guān)于SKIPIF1<0軸對(duì)稱后可得函數(shù)SKIPIF1<0，再向下平移SKIPIF1<0個(gè)單位，可得SKIPIF1<0所以解析式為SKIPIF1<0，故選：C.典例14.（2022·浙江·模擬預(yù)測(cè)）已知SKIPIF1<0，若SKIPIF1<0的圖像如圖所示，SKIPIF1<0的解析式可能是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】根據(jù)函數(shù)定義域和奇偶性分析判斷.【詳解】對(duì)A：SKIPIF1<0的定義域?yàn)镾KIPIF1<0，且為奇函數(shù)，與圖像符合；對(duì)B：SKIPIF1<0的定義域?yàn)镾KIPIF1<0，且為偶函數(shù)，與圖像不符合，B錯(cuò)誤；對(duì)C：SKIPIF1<0的定義域?yàn)镾KIPIF1<0，且為奇函數(shù)，與圖象不符合，C錯(cuò)誤；對(duì)D：SKIPIF1<0的定義域?yàn)镾KIPIF1<0，且為偶函數(shù)，與圖像不符合，D錯(cuò)誤；故選：A.典例15.（2021·福建高三三模）若函數(shù)SKIPIF1<0的大致圖象如圖所示，則SKIPIF1<0的解析式可能是（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】利用排除法，取特殊值分析判斷即可得答案【詳解】解：由圖可知，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，取SKIPIF1<0，則對(duì)于B，SKIPIF1<0，所以排除B，對(duì)于D，SKIPIF1<0，所以排除D，當(dāng)SKIPIF1<0時(shí)，對(duì)于A，SKIPIF1<0，此函數(shù)是由SKIPIF1<0向右平移1個(gè)單位，再向上平移1個(gè)單位，所以SKIPIF1<0時(shí)，SKIPIF1<0恒成立，而圖中，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0可以小于1，所以排除A,故選：C【總結(jié)提升】1.根據(jù)已知或作出的函數(shù)圖象，從最高點(diǎn)、最低點(diǎn)，分析函數(shù)的最值；2.從圖象的對(duì)稱性，分析函數(shù)的奇偶性；3.從圖象的走向趨勢(shì)，分析函數(shù)的單調(diào)性、周期性；4.從圖象與x軸的交點(diǎn)情況，分析函數(shù)的零點(diǎn).考向六函數(shù)圖象與函數(shù)的零點(diǎn)【核心知識(shí)】在研究函數(shù)性質(zhì)特別是單調(diào)性、最值、零點(diǎn)時(shí)，要注意用好其與圖象的關(guān)系，結(jié)合圖象研究.函數(shù)圖象的應(yīng)用主要體現(xiàn)為數(shù)形結(jié)合思想，借助于函數(shù)圖象的特點(diǎn)和變化規(guī)律，求解有關(guān)不等式恒成立、最值、交點(diǎn)、方程的根等問題．求解兩個(gè)函數(shù)圖象在給定區(qū)間上的交點(diǎn)個(gè)數(shù)問題時(shí)，可以先畫出已知函數(shù)完整的圖象，再觀察．【典例分析】典例16.（2022·湖北·高三期中）己知函數(shù)SKIPIF1<0，則函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】確定函數(shù)SKIPIF1<0的值域，利用換元法令SKIPIF1<0，則SKIPIF1<0，則將函數(shù)SKIPIF1<0的零點(diǎn)問題轉(zhuǎn)化為函數(shù)SKIPIF1<0的圖象的交點(diǎn)問題，作函數(shù)SKIPIF1<0圖象，確定其交點(diǎn)以及其橫坐標(biāo)范圍，再結(jié)合SKIPIF1<0的圖象，即可確定SKIPIF1<0的零點(diǎn)個(gè)數(shù).【詳解】已知SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，作出其圖象如圖示：可知SKIPIF1<0值域?yàn)镾KIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，則函數(shù)SKIPIF1<0的零點(diǎn)問題即為函數(shù)SKIPIF1<0的圖象的交點(diǎn)問題，而SKIPIF1<0，作出函數(shù)SKIPIF1<0的圖象如圖示：可知：SKIPIF1<0的圖象有兩個(gè)交點(diǎn)，橫坐標(biāo)分別在SKIPIF1<0之間，不妨設(shè)交點(diǎn)橫坐標(biāo)為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0圖象和直線SKIPIF1<0可知，二者有兩個(gè)交點(diǎn)，即此時(shí)SKIPIF1<0有兩個(gè)零點(diǎn)；當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0圖象和直線SKIPIF1<0可知，二者有3個(gè)交點(diǎn)，即此時(shí)SKIPIF1<0有3個(gè)零點(diǎn)，故函數(shù)SKIPIF1<0的零點(diǎn)個(gè)數(shù)是5，故選：B.典例17.【多選題】（2022·湖北·丹江口市第一中學(xué)模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，設(shè)函數(shù)SKIPIF1<0，則下列說法正確的是（

）A．若SKIPIF1<0有4個(gè)零點(diǎn)，則SKIPIF1<0B．存在實(shí)數(shù)t，使得SKIPIF1<0有5個(gè)零點(diǎn)C．當(dāng)SKIPIF1<0有6個(gè)零點(diǎn)時(shí)．記零點(diǎn)分別為SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0D．對(duì)任意SKIPIF1<0恒有2個(gè)零點(diǎn)【答案】BC【分析】由SKIPIF1<0可得SKIPIF1<0或SKIPIF1<0，作函數(shù)SKIPIF1<0的圖象，觀察圖象判斷A，B，D，由條件觀察圖象確定SKIPIF1<0的關(guān)系，由此判斷C，【詳解】SKIPIF1<0的大致圖像如圖所示，令SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0．若SKIPIF1<0有4個(gè)零點(diǎn)，則實(shí)數(shù)t的取值范圍為SKIPIF1<0或SKIPIF1<0或SKIPIF1<0，故A項(xiàng)錯(cuò)誤；由圖可知，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有5個(gè)零點(diǎn)，故B項(xiàng)正確；當(dāng)SKIPIF1<0有6個(gè)零點(diǎn)時(shí)，則SKIPIF1<0，所以SKIPIF1<0，即有SKIPIF1<0，故C項(xiàng)正確；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0有4個(gè)零點(diǎn)，故D項(xiàng)錯(cuò)誤，故選：BC．典例18.（2019·江蘇·高考真題）設(shè)SKIPIF1<0是定義在SKIPIF1<0上的兩個(gè)周期函數(shù)，SKIPIF1<0的周期為4，SKIPIF1<0的周期為2，且SKIPIF1<0是奇函數(shù).當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，其中SKIPIF1<0.若在區(qū)間SKIPIF1<0上，關(guān)于SKIPIF1<0的方程SKIPIF1<0有8個(gè)不同的實(shí)數(shù)根，則SKIPIF1<0的取值范圍是_____.【答案】SKIPIF1<0.【分析】分別考查函數(shù)SKIPIF1<0和函數(shù)SKIPIF1<0圖像的性質(zhì)，考查臨界條件確定k的取值范圍即可.【詳解】當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0即SKIPIF1<0又SKIPIF1<0為奇函數(shù)，其圖象關(guān)于原點(diǎn)對(duì)稱，其周期為SKIPIF1<0，如圖，函數(shù)SKIPIF1<0與SKIPIF1<0的圖象，要使SKIPIF1<0在SKIPIF1<0上有SKIPIF1<0個(gè)實(shí)根，只需二者圖象有SKIPIF1<0個(gè)交點(diǎn)即可.

當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0與SKIPIF1<0的圖象有SKIPIF1<0個(gè)交點(diǎn)；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的圖象為恒過點(diǎn)SKIPIF1<0的直線，只需函數(shù)SKIPIF1<0與SKIPIF1<0的圖象有SKIPIF1<0個(gè)交點(diǎn).當(dāng)SKIPIF1<0與SKIPIF1<0圖象相切時(shí)，圓心SKIPIF1<0到直線SKIPIF1<0的距離為SKIPIF1<0，即SKIPIF1<0，得SKIPIF1<0，函數(shù)SKIPIF1<0與SKIPIF1<0的圖象有SKIPIF1<0個(gè)交點(diǎn)；當(dāng)SKIPIF1<0過點(diǎn)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0