新高考數(shù)學(xué)一輪復(fù)習(xí) 專項分層精練第10課 函數(shù)圖象（解析版）

第10課函數(shù)圖象（分層專項精練）【一層練基礎(chǔ)】一、單選題1．（2021·天津·統(tǒng)考高考真題）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，SKIPIF1<0，排除D，即可得解.【詳解】設(shè)SKIPIF1<0，則函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，關(guān)于原點對稱，又SKIPIF1<0，所以函數(shù)SKIPIF1<0為偶函數(shù)，排除AC；當(dāng)SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0，排除D.故選：B.2．（2023·四川瀘州·四川省瀘縣第一中學(xué)?？级＃┮阎瘮?shù)SKIPIF1<0，則（

）A．SKIPIF1<0在SKIPIF1<0上單調(diào)遞增 B．SKIPIF1<0的圖象關(guān)于點SKIPIF1<0對稱C．SKIPIF1<0為奇函數(shù) D．SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對稱【答案】D【分析】把SKIPIF1<0化簡成SKIPIF1<0，進而得到SKIPIF1<0是由SKIPIF1<0先向左平移一個單位，再向下平移一個單位得到的，然后根據(jù)SKIPIF1<0的圖象畫出SKIPIF1<0的圖象，即可判斷選項【詳解】化簡得SKIPIF1<0，SKIPIF1<0的可以看作是函數(shù)SKIPIF1<0先向左平移一個單位，再向下平移一個單位得到，先畫出SKIPIF1<0的圖象，再進行平移畫出SKIPIF1<0的圖象，明顯可見，對于原函數(shù)SKIPIF1<0，為奇函數(shù)，關(guān)于點SKIPIF1<0對稱，且在SKIPIF1<0和SKIPIF1<0上為單調(diào)減函數(shù)，所以，SKIPIF1<0經(jīng)過平移后變成的SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，關(guān)于SKIPIF1<0對稱，非奇函數(shù)也非偶函數(shù)，圖象關(guān)于直線SKIPIF1<0對稱，所以，D正確；A、B、C錯誤.故選：D3．（2022秋·黑龍江齊齊哈爾·高一齊齊哈爾市第一中學(xué)校?？计谥校┤鐖D所示，點SKIPIF1<0從點SKIPIF1<0出發(fā)，按逆時針方向沿邊長為SKIPIF1<0的正三角形SKIPIF1<0運動一周，SKIPIF1<0為△SKIPIF1<0的中心，設(shè)點SKIPIF1<0走過的路程為SKIPIF1<0，△SKIPIF1<0的面積為SKIPIF1<0（當(dāng)SKIPIF1<0、SKIPIF1<0、SKIPIF1<0三點共線時，記面積為0），則函數(shù)SKIPIF1<0的圖象大致為（）A．B．C． D．【答案】A【詳解】當(dāng)SKIPIF1<0時,SKIPIF1<0，為一次遞增函數(shù),去掉B；當(dāng)SKIPIF1<0(BC中點)時SKIPIF1<0為一次遞減函數(shù)，去掉C,D；所以選A.點睛：(1)運用函數(shù)性質(zhì)研究函數(shù)圖像時，先要正確理解和把握函數(shù)相關(guān)性質(zhì)本身的含義及其應(yīng)用方向.(2)在運用函數(shù)性質(zhì)特別是奇偶性、周期、對稱性、單調(diào)性、最值、零點時，要注意用好其與條件的相互關(guān)系，結(jié)合特征進行等價轉(zhuǎn)化研究.如奇偶性可實現(xiàn)自變量正負(fù)轉(zhuǎn)化，周期可實現(xiàn)自變量大小轉(zhuǎn)化，單調(diào)性可實現(xiàn)去SKIPIF1<0，即將函數(shù)值的大小轉(zhuǎn)化自變量大小關(guān)系4．（2023·廣西柳州·統(tǒng)考模擬預(yù)測）已知函數(shù)SKIPIF1<0，SKIPIF1<0，若函數(shù)SKIPIF1<0在SKIPIF1<0上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】D【分析】根據(jù)圖象判斷函數(shù)的奇偶性，結(jié)合特殊值，可得答案.【詳解】易知SKIPIF1<0為偶函數(shù)，由SKIPIF1<0，則SKIPIF1<0為奇函數(shù)，由圖象可知，該函數(shù)是奇函數(shù)，因為SKIPIF1<0是偶函數(shù)，SKIPIF1<0是奇函數(shù)，所以SKIPIF1<0是非奇非偶函數(shù)，A，B不符合題意.因為當(dāng)SKIPIF1<0時，SKIPIF1<0無意義，所以C不符合題意.故選：D.二、多選題5．（2023·全國·高三專題練習(xí)）關(guān)于函數(shù)SKIPIF1<0，下列描述正確的有（

）A．SKIPIF1<0在區(qū)間SKIPIF1<0上單調(diào)遞增 B．SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對稱C．若SKIPIF1<0則SKIPIF1<0 D．SKIPIF1<0有且僅有兩個零點【答案】ABD【分析】作出函數(shù)SKIPIF1<0的圖象，由圖象觀察性質(zhì)判斷各選項．【詳解】根據(jù)圖象變換作出函數(shù)SKIPIF1<0的圖象（SKIPIF1<0，作出SKIPIF1<0的圖象，再作出其關(guān)于SKIPIF1<0軸對稱的圖象，然后向右平移2個單位，最后把SKIPIF1<0軸下方的部分關(guān)于SKIPIF1<0軸翻折上去即可得），如圖，由圖象知SKIPIF1<0在SKIPIF1<0是單調(diào)遞增，A正確，函數(shù)圖象關(guān)于直線SKIPIF1<0對稱，B正確；SKIPIF1<0，直線SKIPIF1<0與函數(shù)SKIPIF1<0圖象相交可能是4個交點，如圖，如果最左邊兩個交點橫坐標(biāo)分別是SKIPIF1<0，則SKIPIF1<0不成立，C錯誤，SKIPIF1<0與SKIPIF1<0軸僅有兩個公共點，即函數(shù)僅有兩個零點，D正確．故選：ABD．6．（2021·全國·高三專題練習(xí)）如圖所示的函數(shù)圖象，對應(yīng)的函數(shù)解析式不可能是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】ABC【解析】根據(jù)圖象用特殊值驗證、排除可得答案.【詳解】由圖象可知當(dāng)SKIPIF1<0時，SKIPIF1<0，而A中函數(shù)當(dāng)SKIPIF1<0時，SKIPIF1<0，B中函數(shù)當(dāng)SKIPIF1<0時，SKIPIF1<0，故A和B不可能；C中函數(shù)的定義域是SKIPIF1<0，與圖象不符，故C不可能.對于SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，所以D符合，故選：ABC.【點睛】本題考查了函數(shù)圖象的性質(zhì)，屬于基礎(chǔ)題.【二層練綜合】一、單選題1．（2023·江西南昌·南昌市八一中學(xué)?？既＃┖瘮?shù)SKIPIF1<0在SKIPIF1<0上的圖像大致為（

）A． B．C． D．【答案】B【分析】根據(jù)給定的函數(shù)，由奇偶性排除兩個選項，再取特值即可判斷作答.【詳解】函數(shù)SKIPIF1<0定義域為SKIPIF1<0，而SKIPIF1<0，且SKIPIF1<0，即函數(shù)SKIPIF1<0既不是奇函數(shù)也不是偶函數(shù)，其圖象關(guān)于原點不對稱，排除選項CD；而當(dāng)SKIPIF1<0時，SKIPIF1<0，排除選項A，選項B符合要求.故選：B2．（2022秋·寧夏·高三六盤山高級中學(xué)?？计谥校┒x在SKIPIF1<0上的函數(shù)SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，則方程SKIPIF1<0在SKIPIF1<0上解的個數(shù)為（

）A．3 B．4 C．5 D．6【答案】B【分析】首先將問題轉(zhuǎn)化為SKIPIF1<0與SKIPIF1<0在SKIPIF1<0上的交點個數(shù)，然后根據(jù)SKIPIF1<0的對稱性和周期性以及已知條件作出SKIPIF1<0的圖像，再利用導(dǎo)函數(shù)作出SKIPIF1<0的大致圖像，結(jié)合圖像即可求解.【詳解】由題意可知，方程SKIPIF1<0在SKIPIF1<0上解的個數(shù)可轉(zhuǎn)化為SKIPIF1<0與SKIPIF1<0在SKIPIF1<0上的交點個數(shù)，因為SKIPIF1<0，所以SKIPIF1<0的圖像關(guān)于SKIPIF1<0對稱；又由SKIPIF1<0，故SKIPIF1<0，從而SKIPIF1<0是周期為2的周期函數(shù)，又由SKIPIF1<0可得，SKIPIF1<0，從而SKIPIF1<0；SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，在SKIPIF1<0單調(diào)遞減，且SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，故SKIPIF1<0與SKIPIF1<0在SKIPIF1<0上的圖像如下：從而SKIPIF1<0與SKIPIF1<0在SKIPIF1<0上的交點個數(shù)為4，故方程SKIPIF1<0在SKIPIF1<0上解的個數(shù)為4.故選：B.3．（2023·全國·高三專題練習(xí)）已知函數(shù)SKIPIF1<0滿足SKIPIF1<0對任意SKIPIF1<0恒成立，又函數(shù)SKIPIF1<0的圖象關(guān)于點SKIPIF1<0對稱，且SKIPIF1<0則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】首先利用賦值法求出SKIPIF1<0，代入等式賦值得到SKIPIF1<0，即對稱軸為SKIPIF1<0，再根據(jù)函數(shù)圖象的平移規(guī)律判斷函數(shù)為奇函數(shù)，進一步求得函數(shù)周期，進而得到SKIPIF1<0，則可求出結(jié)果.【詳解】因為對任意SKIPIF1<0，都有SKIPIF1<0令SKIPIF1<0得SKIPIF1<0解得SKIPIF1<0則SKIPIF1<0即SKIPIF1<0所以函數(shù)SKIPIF1<0的圖象關(guān)于直線SKIPIF1<0對稱.又函數(shù)SKIPIF1<0的圖象關(guān)于點SKIPIF1<0對稱，則函數(shù)SKIPIF1<0的圖象關(guān)于點SKIPIF1<0對稱，即函數(shù)SKIPIF1<0為奇函數(shù)，所以SKIPIF1<0所以SKIPIF1<0所以8是函數(shù)SKIPIF1<0的一個周期，所以SKIPIF1<0故選：D.4．（2022秋·河南·高一校聯(lián)考階段練習(xí)）若函數(shù)SKIPIF1<0的大致圖象如圖，則SKIPIF1<0的解析式可能是(

)A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】根據(jù)定義域排除A，根據(jù)奇偶性排除B，根據(jù)值域或單調(diào)性排除C.【詳解】由圖可知函數(shù)定義域為{x|x≠0}，由此排除A；該函數(shù)圖像關(guān)于原點對稱，則該函數(shù)為奇函數(shù)，需滿足f(x)＋f(－x)＝0，對于B項：f(x)＋f(－x)≠0，故排除B；C和D均滿足f(x)＋f(－x)＝0，對于C：SKIPIF1<0，當(dāng)x→＋∞時，SKIPIF1<0→0，故SKIPIF1<0，∵y＝SKIPIF1<0增長的速率比y＝SKIPIF1<0增長的速率慢，∴SKIPIF1<0，即圖像在x軸上方無限接近于x軸正半軸，與題意不符，故排除C.綜上，D選項正確.故選：D.二、填空題5．（2011·江蘇南京·統(tǒng)考一模）如圖放置的等腰直角SKIPIF1<0薄片，SKIPIF1<0沿x軸滾動，設(shè)頂點SKIPIF1<0的軌跡方程是SKIPIF1<0，則SKIPIF1<0在其相鄰兩個零點間的圖象與x軸所圍區(qū)域的面積為．【答案】SKIPIF1<0【分析】由題意確定軌跡的形狀，作出圖象，進而求解其與x軸所圍區(qū)域的面積，即可得到答案.【詳解】作出點A的軌跡中相鄰兩個零點間的圖象，如圖所示，其軌跡為兩段圓弧，一段是以C為圓心，CA為半徑的四分之一圓弧；一段是以B為圓心，BA為半徑，圓心角為SKIPIF1<0的圓弧，其與x軸圍成的圖形的面積為SKIPIF1<0.故答案為：SKIPIF1<0.6．（2023·全國·高三專題練習(xí)）定義在SKIPIF1<0上函數(shù)SKIPIF1<0滿足SKIPIF1<0，且當(dāng)SKIPIF1<0時，SKIPIF1<0．若當(dāng)SKIPIF1<0時，SKIPIF1<0，則SKIPIF1<0的最小值等于．【答案】SKIPIF1<0【分析】轉(zhuǎn)化條件為在區(qū)間SKIPIF1<0上，SKIPIF1<0，作出函數(shù)的圖象，數(shù)形結(jié)合即可得解．【詳解】當(dāng)SKIPIF1<0時，故SKIPIF1<0，當(dāng)SKIPIF1<0時，故SKIPIF1<0…，可得在區(qū)間SKIPIF1<0上，SKIPIF1<0，所以當(dāng)SKIPIF1<0時，SKIPIF1<0，作函數(shù)SKIPIF1<0的圖象，如圖所示，當(dāng)SKIPIF1<0時，由SKIPIF1<0得SKIPIF1<0，由圖象可知當(dāng)SKIPIF1<0時，SKIPIF1<0，所以SKIPIF1<0的最小值為SKIPIF1<0．故答案為：SKIPIF1<0．【三層練能力】一、單選題1．（2023·全國·高三專題練習(xí)）已知函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0與SKIPIF1<0的圖象恰有5個不同公共點，則實數(shù)a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】利用導(dǎo)數(shù)分段畫出函數(shù)SKIPIF1<0的大致圖象，將函數(shù)SKIPIF1<0與SKIPIF1<0的圖象恰有5個不同公共點的問題轉(zhuǎn)化為方程SKIPIF1<0有5個不同的根的問題，然后采用換元法將問題變?yōu)橛懻揝KIPIF1<0在給定區(qū)間上有解的問題.【詳解】當(dāng)SKIPIF1<0時，SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，故SKIPIF1<0時，SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0有極大值SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，作出SKIPIF1<0的大致圖象如圖：函數(shù)SKIPIF1<0與SKIPIF1<0的圖象恰有5個不同公共點，即方程SKIPIF1<0有5個不同的根，令SKIPIF1<0，根據(jù)其圖象，討論SKIPIF1<0有解情況如下：令SKIPIF1<0,(1當(dāng)SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上各有一個解時，即SKIPIF1<0，解得SKIPIF1<0，（2）當(dāng)SKIPIF1<0在SKIPIF1<0和SKIPIF1<0上各有一個解時，SKIPIF1<0，解得SKIPIF1<0，（3）當(dāng)SKIPIF1<0有一個根為6時，解得SKIPIF1<0，此時另一個根為SKIPIF1<0，不合題意；（4）當(dāng)SKIPIF1<0有一個根為1時，解得SKIPIF1<0，此時另一個根也為1，不合題意，綜上可知：SKIPIF1<0，故選：A二、多選題2．（2023春·云南紅河·高二開遠市第一中學(xué)校校考階段練習(xí)）已知SKIPIF1<0，若存在SKIPIF1<0，使得SKIPIF1<0，則下列結(jié)論正確的有（

）A．實數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0的最大值為SKIPIF1<0【答案】AC【分析】畫出SKIPIF1<0的圖象，數(shù)形結(jié)合得到SKIPIF1<0，且SKIPIF1<0