(通用版)高考數(shù)學(xué)二輪復(fù)習(xí)選填題專項(xiàng)測(cè)試第8篇函數(shù)圖像02（含解析）

高考數(shù)學(xué)選填題專項(xiàng)測(cè)試02（函數(shù)圖像）第I卷（選擇題)一、單選題：本大題共12小題，每小題5分，共60分。在每小題給出的四個(gè)選項(xiàng)中，只有一項(xiàng)是符合題目要求的。1．（·山西高三月考）函數(shù)SKIPIF1<0，的大致圖象是（）A． B．C． D．【答案】C【解析】【分析】先確定函數(shù)SKIPIF1<0為偶函數(shù),排除B,D選項(xiàng),再取特值即可判斷最終結(jié)論.【詳解】因?yàn)閒(﹣x)=(﹣x)2ecos(﹣x)=x2ecosx=f(x),所以函數(shù)f(x)為偶函數(shù),排除B?D選項(xiàng),因?yàn)閒(π)=π2ecosπ=π2e﹣1>0,所以排除A選項(xiàng),故選:C.【點(diǎn)睛】本題考查函數(shù)圖象的識(shí)別,難度不大.對(duì)于判斷函數(shù)圖象的試題,排除法是十分常用的方法,一般通過函數(shù)的奇偶性?單調(diào)性和特殊值即可判斷.2．（·洪洞縣第一中學(xué)高三期中）函數(shù)SKIPIF1<0的導(dǎo)函數(shù)SKIPIF1<0的圖象如圖所示，則（）A．SKIPIF1<0為SKIPIF1<0的極大值點(diǎn) B．SKIPIF1<0為SKIPIF1<0的極大值點(diǎn)C．SKIPIF1<0為SKIPIF1<0的極大值點(diǎn) D．SKIPIF1<0為SKIPIF1<0的極小值點(diǎn)【答案】A【解析】【分析】觀察各極值點(diǎn)附近左右的導(dǎo)數(shù)符號(hào)，可得出正確選項(xiàng).【詳解】對(duì)于A選項(xiàng)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0為SKIPIF1<0的極大值點(diǎn)，A選項(xiàng)正確；對(duì)于B選項(xiàng)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0為SKIPIF1<0的極小值點(diǎn)，B選項(xiàng)錯(cuò)誤；對(duì)于C選項(xiàng)，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0為SKIPIF1<0的極小值點(diǎn)，C選項(xiàng)錯(cuò)誤；對(duì)于D選項(xiàng)，由于函數(shù)SKIPIF1<0為可導(dǎo)函數(shù)，且SKIPIF1<0，SKIPIF1<0不是SKIPIF1<0的極值點(diǎn)，D選項(xiàng)錯(cuò)誤.故選：A.【點(diǎn)睛】本題考查利用導(dǎo)數(shù)的圖象判斷極值點(diǎn)，解題時(shí)要充分利用極大值點(diǎn)和極小值點(diǎn)的概念加以理解，考查分析問題與解決問題的能力，屬于中等題.3．（·廣西師大附屬外國(guó)語學(xué)校高三（理）已知函數(shù)SKIPIF1<0的導(dǎo)數(shù)為SKIPIF1<0，且SKIPIF1<0，則函數(shù)SKIPIF1<0圖象的大致形狀是（）A． B．C． D．【答案】A【解析】【分析】根據(jù)導(dǎo)函數(shù)求出SKIPIF1<0，討論SKIPIF1<0的函數(shù)圖象，結(jié)合奇偶性和特殊值即可得解.【詳解】SKIPIF1<0，SKIPIF1<0，SKIPIF1<0,SKIPIF1<0SKIPIF1<0所以SKIPIF1<0為奇函數(shù)，且當(dāng)SKIPIF1<0時(shí)，有g(shù)(x)<0.結(jié)合選項(xiàng)，只有A符合題意.【點(diǎn)睛】此題考查根據(jù)導(dǎo)數(shù)值求參數(shù)的取值，根據(jù)函數(shù)的性質(zhì)確定函數(shù)圖象，關(guān)鍵在于根據(jù)導(dǎo)函數(shù)準(zhǔn)確求解.4．（·四川成都七中高三開學(xué)考試）若函數(shù)SKIPIF1<0的圖象關(guān)于原點(diǎn)對(duì)稱，則實(shí)數(shù)SKIPIF1<0等于（）A．SKIPIF1<0 B．SKIPIF1<0 C．1 D．2【答案】A【解析】【分析】由題意知，函數(shù)SKIPIF1<0為奇函數(shù),利用SKIPIF1<0,化簡(jiǎn)整理即可求出實(shí)數(shù)SKIPIF1<0.【詳解】因?yàn)楹瘮?shù)SKIPIF1<0的圖象關(guān)于原點(diǎn)對(duì)稱，所以函數(shù)SKIPIF1<0為奇函數(shù)，則有SKIPIF1<0，即SKIPIF1<0，化簡(jiǎn)可得，SKIPIF1<0，解可得SKIPIF1<0.故選:A【點(diǎn)睛】本題考查奇函數(shù)的定義和性質(zhì);根據(jù)題意,挖掘題中隱含的條件:函數(shù)SKIPIF1<0為奇函數(shù)是求解本題的關(guān)鍵;屬于中檔題.5．（·廣東高三期末）函數(shù)SKIPIF1<0的部分圖象大致是（）A． B．C． D．SKIPIF1<0【答案】B【解析】【分析】分析函數(shù)的定義域、奇偶性以及函數(shù)值的正負(fù)變化，排除錯(cuò)誤選項(xiàng)可得答案.【詳解】由SKIPIF1<0，可得SKIPIF1<0，故SKIPIF1<0是奇函數(shù)，圖象關(guān)于原點(diǎn)對(duì)稱，排除A.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，排除C,D.故選：B.【點(diǎn)睛】本題考查函數(shù)圖象的識(shí)別，一般利用函數(shù)的定義域、值域、奇偶性、單調(diào)性等性質(zhì)分析函數(shù)圖象的特征，排除錯(cuò)誤選項(xiàng)得到答案.6．（·山西大同一中高三月考）函數(shù)SKIPIF1<0的圖象大致為（）A． B．C． D．【答案】D【解析】【分析】先確定函數(shù)的定義域，再判斷函數(shù)的奇偶性和值域，由此確定正確選項(xiàng)。【詳解】函數(shù)的定義域?yàn)镾KIPIF1<0，SKIPIF1<0，則函數(shù)SKIPIF1<0為偶函數(shù)，圖象關(guān)于y軸對(duì)稱，排除B，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，排除A，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，排除C，故選：D.【點(diǎn)睛】本題通過判斷函數(shù)圖像考查函數(shù)的基本性質(zhì)，屬于基礎(chǔ)題。7．（·廣西高三月考（理））已知函數(shù)SKIPIF1<0的大致圖象如圖所示，則函數(shù)SKIPIF1<0的解析式可能為()A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】【分析】結(jié)合圖像，判斷函數(shù)的性質(zhì)即可求解.【詳解】從圖像可知，SKIPIF1<0為偶函數(shù)，對(duì)于A，SKIPIF1<0SKIPIF1<0，排除A；對(duì)于B，SKIPIF1<0SKIPIF1<0，排除B；SKIPIF1<0和SKIPIF1<0其定義域均為SKIPIF1<0，當(dāng)SKIPIF1<0從SKIPIF1<0的右側(cè)趨近SKIPIF1<0時(shí)，SKIPIF1<0，SKIPIF1<0，即SKIPIF1<0，結(jié)合圖像排除D項(xiàng)，【點(diǎn)睛】本題考查了函數(shù)圖像的識(shí)別，注意從函數(shù)的性質(zhì)進(jìn)行深入分析，考查了函數(shù)的性質(zhì)，屬于基礎(chǔ)題.8．（·安慶市第二中學(xué)高三期末）已知函數(shù)SKIPIF1<0，則SKIPIF1<0的圖象大致為（）A． B．C． D．【答案】A 【解析】【分析】利用特殊值，對(duì)函數(shù)圖象進(jìn)行排除，由此得出正確選項(xiàng).【詳解】由于SKIPIF1<0，排除B選項(xiàng).由于SKIPIF1<0，SKIPIF1<0，函數(shù)單調(diào)遞減，排除C選項(xiàng).由于SKIPIF1<0，排除D選項(xiàng).故選A.【點(diǎn)睛】本小題主要考查已知具體函數(shù)的解析式，判斷函數(shù)的圖象，屬于基礎(chǔ)題.9．（·洪洞縣第一中學(xué)高三期中）函數(shù)SKIPIF1<0（SKIPIF1<0）的圖象如圖所示，為了得到SKIPIF1<0的圖象，可以將SKIPIF1<0的圖象（）A．向左平移SKIPIF1<0個(gè)單位長(zhǎng)度 B．向左平移SKIPIF1<0個(gè)單位長(zhǎng)度C．向右平移SKIPIF1<0個(gè)單位長(zhǎng)度 D．向右平移SKIPIF1<0個(gè)單位長(zhǎng)度【答案】B【解析】【分析】先根據(jù)函數(shù)圖象求得函數(shù)SKIPIF1<0的解析式，結(jié)合誘導(dǎo)公式將正弦函數(shù)化為余弦函數(shù)，即可確定經(jīng)過怎樣平移得到SKIPIF1<0的圖象.【詳解】根據(jù)函數(shù)SKIPIF1<0圖象可知，SKIPIF1<0，則SKIPIF1<0，由周期公式SKIPIF1<0，可得SKIPIF1<0，所以SKIPIF1<0，將最低點(diǎn)坐標(biāo)SKIPIF1<0代入即可得SKIPIF1<0，由正弦函數(shù)的圖象與性質(zhì)可得SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0，由誘導(dǎo)公式可知SKIPIF1<0，即SKIPIF1<0，所以為得到SKIPIF1<0，需將SKIPIF1<0向左平移SKIPIF1<0個(gè)單位長(zhǎng)度，故選：B.【點(diǎn)睛】本題考查了根據(jù)部分函數(shù)圖象求三角函數(shù)解析式，利用誘導(dǎo)公式變換正弦函數(shù)與余弦函數(shù)，三角函數(shù)圖像平移變換的應(yīng)用，屬于中檔題.10．（·安慶市第二中學(xué)高三期末（理））函數(shù)f（x）=（x2+2x）e2x的圖象大致是()A． B．C． D．【答案】A【解析】【分析】利用導(dǎo)數(shù)判斷出SKIPIF1<0的單調(diào)區(qū)間，結(jié)合函數(shù)值的符號(hào)，選出正確選項(xiàng).【詳解】由于SKIPIF1<0，而SKIPIF1<0的判別式SKIPIF1<0，所以SKIPIF1<0開口向上且有兩個(gè)根SKIPIF1<0，不妨設(shè)SKIPIF1<0，所以SKIPIF1<0在SKIPIF1<0上遞增，在SKIPIF1<0上遞減.所以C,D選項(xiàng)不正確.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以B選項(xiàng)不正確.由此得出A選項(xiàng)正確.故選：A【點(diǎn)睛】本小題主要考查利用導(dǎo)數(shù)判斷函數(shù)的圖像，屬于基礎(chǔ)題.11．（·河北高三月考）若函數(shù)SKIPIF1<0（其中SKIPIF1<0，SKIPIF1<0圖象的一個(gè)對(duì)稱中心為SKIPIF1<0，SKIPIF1<0，其相鄰一條對(duì)稱軸方程為SKIPIF1<0，該對(duì)稱軸處所對(duì)應(yīng)的函數(shù)值為SKIPIF1<0，為了得到SKIPIF1<0的圖象，則只要將SKIPIF1<0的圖象()A．向右平移SKIPIF1<0個(gè)單位長(zhǎng)度 B．向左平移SKIPIF1<0個(gè)單位長(zhǎng)度C．向左平移SKIPIF1<0個(gè)單位長(zhǎng)度 D．向右平移SKIPIF1<0個(gè)單位長(zhǎng)度【答案】B【解析】【分析】由函數(shù)的圖象的頂點(diǎn)坐標(biāo)求出A，由周期求出SKIPIF1<0，由五點(diǎn)法作圖求出SKIPIF1<0的值，可得SKIPIF1<0的解析式，再根據(jù)函數(shù)SKIPIF1<0的圖象變換規(guī)律，誘導(dǎo)公式，得出結(jié)論．【詳解】根據(jù)已知函數(shù)SKIPIF1<0SKIPIF1<0其中SKIPIF1<0，SKIPIF1<0的圖象過點(diǎn)SKIPIF1<0，SKIPIF1<0，可得SKIPIF1<0，SKIPIF1<0，解得：SKIPIF1<0．再根據(jù)五點(diǎn)法作圖可得SKIPIF1<0，可得：SKIPIF1<0，可得函數(shù)解析式為：SKIPIF1<0故把SKIPIF1<0的圖象向左平移SKIPIF1<0個(gè)單位長(zhǎng)度，可得SKIPIF1<0的圖象，故選B．【點(diǎn)睛】本題主要考查由函數(shù)SKIPIF1<0的部分圖象求解析式，由函數(shù)的圖象的頂點(diǎn)坐標(biāo)求出A，由周期求出SKIPIF1<0，由五點(diǎn)法作圖求出SKIPIF1<0的值，函數(shù)SKIPIF1<0的圖象變換規(guī)律，誘導(dǎo)公式的應(yīng)用，屬于中檔題．12．（·山西高三期末）已知SKIPIF1<0若函數(shù)SKIPIF1<0的圖象與SKIPIF1<0的圖象有3個(gè)不同的交點(diǎn)，則實(shí)數(shù)SKIPIF1<0的取值范圍是（）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】【分析】采用數(shù)形結(jié)合的方法，作出SKIPIF1<0圖像，根據(jù)直線SKIPIF1<0過定點(diǎn)SKIPIF1<0以及兩函數(shù)圖像有3個(gè)交點(diǎn)，可得結(jié)果.【詳解】函數(shù)SKIPIF1<0，SKIPIF1<0圖像有3個(gè)交點(diǎn)，且直線SKIPIF1<0過定點(diǎn)SKIPIF1<0，如圖根據(jù)圖形可知：SKIPIF1<0，當(dāng)直線SKIPIF1<0與SKIPIF1<0相切時(shí)，設(shè)切點(diǎn)SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，在點(diǎn)SKIPIF1<0處的切線方程：SKIPIF1<0又過定點(diǎn)SKIPIF1<0，代入上式，可得SKIPIF1<0，所以SKIPIF1<0，當(dāng)直線SKIPIF1<0過點(diǎn)SKIPIF1<0時(shí)則SKIPIF1<0，所以可知SKIPIF1<0，故選：D【點(diǎn)睛】本題考根據(jù)方程根的個(gè)數(shù)求參數(shù)，熟練使用等價(jià)轉(zhuǎn)化的思想以及數(shù)形結(jié)合的方法，使問題化繁為簡(jiǎn)，考驗(yàn)對(duì)問題的分析能力，屬中檔題.第II卷（非選擇題)二、填空題：本大題共4小題，每小題5分，共20分。把答案填在題中的橫線上。13．（·山西高三期末）已知函數(shù)SKIPIF1<0的圖象過點(diǎn)SKIPIF1<0，則實(shí)數(shù)SKIPIF1<0_________.【答案】SKIPIF1<0【解析】【分析】根據(jù)對(duì)數(shù)的運(yùn)算，直接代值計(jì)算即可..【詳解】由題可知：SKIPIF1<0，則SKIPIF1<0，故答案為：SKIPIF1<0【點(diǎn)睛】本題考查對(duì)數(shù)式的運(yùn)算，屬基礎(chǔ)題.14．（·北京101中學(xué)高三）把下面不完整的命題補(bǔ)充完整，并使之成為真命題．若函數(shù)SKIPIF1<0的圖象與SKIPIF1<0的圖象關(guān)于對(duì)稱，則函數(shù)SKIPIF1<0．（注：填上你認(rèn)為可以成為真命題的一種情形即可，不必考慮所有可能的情形）【答案】SKIPIF1<0軸，SKIPIF1<0；或：SKIPIF1<0軸，SKIPIF1<0；或：原點(diǎn)，SKIPIF1<0；或：直線SKIPIF1<0，SKIPIF1<0【解析】基于對(duì)對(duì)數(shù)函數(shù)圖象、指數(shù)函數(shù)圖象的認(rèn)識(shí)，從多角度考慮．SKIPIF1<0軸，SKIPIF1<0；或：SKIPIF1<0軸，SKIPIF1<0；或：原點(diǎn)，SKIPIF1<0；或：直線SKIPIF1<0，SKIPIF1<0均可．15．（·安慶市第二中學(xué)高三期末）將函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度得到SKIPIF1<0圖像，則下列說法中正確的是（填序號(hào)）。①函數(shù)SKIPIF1<0的最小正周期是SKIPIF1<0；②SKIPIF1<0圖像關(guān)于直線SKIPIF1<0對(duì)稱；③函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞減；④SKIPIF1<0圖像關(guān)于點(diǎn)SKIPIF1<0對(duì)稱；【答案】①②④【解析】【分析】根據(jù)三角函數(shù)的圖象平移關(guān)系求出SKIPIF1<0的解析式，結(jié)合函數(shù)的單調(diào)性，對(duì)稱性分別進(jìn)行判斷即可．【詳解】由題意，將函數(shù)SKIPIF1<0的圖象向右平移SKIPIF1<0個(gè)單位長(zhǎng)度，可得SKIPIF1<0，對(duì)于SKIPIF1<0,函數(shù)的最小正周期為SKIPIF1<0，所以該選項(xiàng)是正確的；對(duì)于SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0為最大值，SKIPIF1<0函數(shù)SKIPIF1<0圖象關(guān)于直線SKIPIF1<0，對(duì)稱是正確的；對(duì)于SKIPIF1<0中，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，則函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上先減后增，SKIPIF1<0不正確；對(duì)于SKIPIF1<0中，令SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0圖象關(guān)于點(diǎn)SKIPIF1<0對(duì)稱是正確的．【點(diǎn)睛】本題主要考查命題的真假判斷，涉及三角函數(shù)的單調(diào)性，對(duì)稱性，求出解析式是解決本題的關(guān)鍵．16.（·四川成都七中高三開學(xué)考試（理））有如下結(jié)論：若無窮等比數(shù)列SKIPIF1<0的公比SKIPIF1<0滿足SKIPIF1<0，則它的各項(xiàng)和SKIPIF1<0．已知函數(shù)SKIPIF1<0，則SKIPIF1<0的