新高考數(shù)學(xué)一輪復(fù)習(xí)講義 第09講 二次函數(shù)與冪函數(shù)（含解析）

第09講二次函數(shù)與冪函數(shù)（精講）題型目錄一覽①冪函數(shù)的定義與圖像②冪函數(shù)的性質(zhì)和綜合應(yīng)用③二次函數(shù)單調(diào)性問題④二次函數(shù)最值問題⑤二次函數(shù)恒成立問題★【文末附錄-冪函數(shù)及冪函數(shù)解題思路思維導(dǎo)圖】一、知識(shí)點(diǎn)梳理一、知識(shí)點(diǎn)梳理1.冪函數(shù)的定義一般地，SKIPIF1<0(SKIPIF1<0為有理數(shù))的函數(shù)，即以\t"/item/%E5%B9%82%E5%87%BD%E6%95%B0/_blank"底數(shù)為\t"/item/%E5%B9%82%E5%87%BD%E6%95%B0/_blank"自變量，冪為\t"/item/%E5%B9%82%E5%87%BD%E6%95%B0/_blank"因變量，\t"/item/%E5%B9%82%E5%87%BD%E6%95%B0/_blank"指數(shù)為常數(shù)的函數(shù)稱為冪函數(shù).2.冪函數(shù)的特征：同時(shí)滿足一下三個(gè)條件才是冪函數(shù)①SKIPIF1<0的系數(shù)為1； ②SKIPIF1<0的底數(shù)是自變量； ③指數(shù)為常數(shù).(3)冪函數(shù)的圖象和性質(zhì)3.常見的冪函數(shù)圖像及性質(zhì)：函數(shù)SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0圖象定義域SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0值域SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0奇偶性奇偶奇非奇非偶奇單調(diào)性在SKIPIF1<0上單調(diào)遞增在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增在SKIPIF1<0上單調(diào)遞增在SKIPIF1<0上單調(diào)遞增在SKIPIF1<0和SKIPIF1<0上單調(diào)遞減公共點(diǎn)SKIPIF1<04．二次函數(shù)的圖像二次函數(shù)SKIPIF1<0的圖像是一條拋物線，二次項(xiàng)系數(shù)a的正負(fù)決定圖象的開口方向，對(duì)稱軸方程為SKIPIF1<0，頂點(diǎn)坐標(biāo)為SKIPIF1<0.【常用結(jié)論】1.冪函數(shù)SKIPIF1<0在第一象限內(nèi)圖象的畫法如下：①當(dāng)SKIPIF1<0時(shí)，其圖象可類似SKIPIF1<0畫出；②當(dāng)SKIPIF1<0時(shí)，其圖象可類似SKIPIF1<0畫出；③當(dāng)SKIPIF1<0時(shí)，其圖象可類似SKIPIF1<0畫出．2．實(shí)系數(shù)一元二次方程SKIPIF1<0的實(shí)根符號(hào)與系數(shù)之間的關(guān)系（1）方程有兩個(gè)不等正根SKIPIF1<0SKIPIF1<0SKIPIF1<0（2）方程有兩個(gè)不等負(fù)根SKIPIF1<0SKIPIF1<0SKIPIF1<0（3）方程有一正根和一負(fù)根，設(shè)兩根為SKIPIF1<0SKIPIF1<0SKIPIF1<0二、題型分類精講二、題型分類精講題型一冪函數(shù)的定義與圖像策略方法若冪函數(shù)y＝xα(α∈Z)是偶函數(shù)，則α必為偶數(shù)．當(dāng)α是分?jǐn)?shù)時(shí)，一般將其先化為根式，再判斷．【典例1】已知冪函數(shù)SKIPIF1<0滿足SKIPIF1<0，則SKIPIF1<0的值為（

）A．2 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】設(shè)出冪函數(shù)的解析式，根據(jù)已知，求出參數(shù)的關(guān)系式，即可計(jì)算作答.【詳解】依題意，設(shè)SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0.故選：B【題型訓(xùn)練】一、單選題1．現(xiàn)有下列函數(shù)：①SKIPIF1<0；②SKIPIF1<0；③SKIPIF1<0；④SKIPIF1<0；⑤SKIPIF1<0；⑥SKIPIF1<0；⑦SKIPIF1<0，其中冪函數(shù)的個(gè)數(shù)為（

）A．1 B．2 C．3 D．4【答案】B【分析】根據(jù)冪函數(shù)的定義逐個(gè)辨析即可【詳解】?jī)绾瘮?shù)滿足SKIPIF1<0形式，故SKIPIF1<0，SKIPIF1<0滿足條件，共2個(gè)故選：B2．已知SKIPIF1<0為冪函數(shù)，且SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】根據(jù)冪函數(shù)及SKIPIF1<0求其解析式，進(jìn)而求SKIPIF1<0.【詳解】因?yàn)镾KIPIF1<0為冪函數(shù)，設(shè)SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，可得SKIPIF1<0，則SKIPIF1<0.故選：B3．下列冪函數(shù)中，定義域?yàn)镽的冪函數(shù)是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】利用分?jǐn)?shù)指數(shù)式與根式的互化，結(jié)合具體函數(shù)的定義域的求法逐項(xiàng)分析即可求出結(jié)果.【詳解】ASKIPIF1<0，則需要滿足SKIPIF1<0，即SKIPIF1<0，所以函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，故A不符合題意；BSKIPIF1<0，則需要滿足SKIPIF1<0，所以函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，故B不符合題意；CSKIPIF1<0，則需要滿足SKIPIF1<0，所以函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，故C不符合題意；DSKIPIF1<0，故函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0，故D正確；故選：D.4．已知冪函數(shù)SKIPIF1<0的圖像過點(diǎn)SKIPIF1<0，則SKIPIF1<0的值域是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】先求出冪函數(shù)解析式，根據(jù)解析式即可求出值域.【詳解】SKIPIF1<0冪函數(shù)SKIPIF1<0的圖像過點(diǎn)SKIPIF1<0，SKIPIF1<0，解得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0SKIPIF1<0的值域是SKIPIF1<0.故選：D.5．函數(shù)SKIPIF1<0的圖象大致為（

）A． B．C． D．【答案】C【分析】利用函數(shù)的奇偶性及冪函數(shù)的性質(zhì)進(jìn)行排除可得答案.【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0為偶函數(shù)，排除A,B選項(xiàng)；易知當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0為增函數(shù)，且增加幅度較為緩和，所以D不正確.故選：C.6．下列函數(shù)中，其圖像如圖所示的函數(shù)為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】根據(jù)函數(shù)的性質(zhì)逐項(xiàng)分析即得．【詳解】由圖象可知函數(shù)為奇函數(shù)，定義域?yàn)镾KIPIF1<0，且在SKIPIF1<0單調(diào)遞減，對(duì)于A，SKIPIF1<0，定義域?yàn)镾KIPIF1<0，SKIPIF1<0，所以函數(shù)為奇函數(shù)，在SKIPIF1<0單調(diào)遞減，故A正確；對(duì)于B，SKIPIF1<0，定義域?yàn)镾KIPIF1<0，故B錯(cuò)誤；對(duì)于C，SKIPIF1<0，定義域?yàn)镾KIPIF1<0，故C錯(cuò)誤；對(duì)于D，SKIPIF1<0，定義域?yàn)镾KIPIF1<0，SKIPIF1<0，函數(shù)為偶函數(shù)，故D錯(cuò)誤.故選：A．7．如圖所示是函數(shù)SKIPIF1<0(SKIPIF1<0且互質(zhì))的圖象，則（

）A．SKIPIF1<0是奇數(shù)且SKIPIF1<0 B．SKIPIF1<0是偶數(shù)，SKIPIF1<0是奇數(shù)，且SKIPIF1<0C．SKIPIF1<0是偶數(shù)，SKIPIF1<0是奇數(shù)，且SKIPIF1<0 D．SKIPIF1<0是偶數(shù)，且SKIPIF1<0【答案】C【分析】根據(jù)冪函數(shù)的性質(zhì)及圖象判斷即可；【詳解】解：SKIPIF1<0函數(shù)SKIPIF1<0的圖象關(guān)于SKIPIF1<0軸對(duì)稱，故SKIPIF1<0為奇數(shù)，SKIPIF1<0為偶數(shù)，在第一象限內(nèi)，函數(shù)是凸函數(shù)，故SKIPIF1<0，故選：C.二、填空題8．函數(shù)SKIPIF1<0的定義域?yàn)開______．【答案】SKIPIF1<0【解析】將函數(shù)解析式變形為SKIPIF1<0，即可求得原函數(shù)的定義域.【詳解】SKIPIF1<0，所以，SKIPIF1<0.因此，函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0.故答案為：SKIPIF1<0.9．設(shè)集合SKIPIF1<0，集合SKIPIF1<0，則SKIPIF1<0________．【答案】SKIPIF1<0/SKIPIF1<0【分析】根據(jù)指數(shù)函數(shù)與冪函數(shù)的性質(zhì)，先求出集合A、B，然后根據(jù)交集的定義即可求解.【詳解】解：因?yàn)榧蟂KIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，故答案為：SKIPIF1<0.10．若函數(shù)SKIPIF1<0的圖像經(jīng)過點(diǎn)SKIPIF1<0與SKIPIF1<0，則m的值為____________．【答案】81【分析】根據(jù)函數(shù)圖象過的點(diǎn)求得參數(shù)SKIPIF1<0，可得函數(shù)解析式，再代入求值即得答案.【詳解】由題意函數(shù)SKIPIF1<0的圖像經(jīng)過點(diǎn)SKIPIF1<0與SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0故SKIPIF1<0，故答案為：8111．冪函數(shù)SKIPIF1<0滿足：任意SKIPIF1<0有SKIPIF1<0，且SKIPIF1<0，請(qǐng)寫出符合上述條件的一個(gè)函數(shù)SKIPIF1<0___________．【答案】SKIPIF1<0（答案不唯一）【分析】取SKIPIF1<0，再驗(yàn)證奇偶性和函數(shù)值即可.【詳解】取SKIPIF1<0，則定義域?yàn)镽，且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，滿足SKIPIF1<0.故答案為：SKIPIF1<0.12．已知函數(shù)SKIPIF1<0若函數(shù)SKIPIF1<0在SKIPIF1<0上不是增函數(shù)，則a的一個(gè)取值為___________.【答案】-2(答案不唯一，滿足SKIPIF1<0或SKIPIF1<0即可)【分析】作出y=x和y=SKIPIF1<0的圖象，數(shù)形結(jié)合即可得a的范圍，從而得到a的可能取值．【詳解】y=x和y=SKIPIF1<0的圖象如圖所示：∴當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，y=SKIPIF1<0有部分函數(shù)值比y=x的函數(shù)值小，故當(dāng)SKIPIF1<0或SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上不是增函數(shù)．故答案為：-2．題型二冪函數(shù)的性質(zhì)和綜合應(yīng)用策略方法(1)緊扣冪函數(shù)SKIPIF1<0的定義、圖像、性質(zhì)，特別注意它的單調(diào)性在不等式中的作用，這里注意SKIPIF1<0為奇數(shù)時(shí)，SKIPIF1<0為奇函數(shù)，SKIPIF1<0為偶數(shù)時(shí)，SKIPIF1<0為偶函數(shù).(2)若冪函數(shù)y＝xα在(0，＋∞)上單調(diào)遞增，則α＞0；若在(0，＋∞)上單調(diào)遞減，則α＜0．(3)在比較冪值的大小時(shí)，必須結(jié)合冪值的特點(diǎn)，選擇適當(dāng)?shù)暮瘮?shù)，借助其單調(diào)性進(jìn)行比較．【典例1】函數(shù)SKIPIF1<0同時(shí)滿足①對(duì)于定義域內(nèi)的任意實(shí)數(shù)x，都有SKIPIF1<0；②在SKIPIF1<0上是減函數(shù)，則SKIPIF1<0的值為（

）A．8 B．4 C．2 D．1【答案】B【分析】由SKIPIF1<0的值依次求出SKIPIF1<0的值，然后根據(jù)函數(shù)的性質(zhì)確定SKIPIF1<0，得函數(shù)解析式，計(jì)算函數(shù)值．【詳解】SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，代入SKIPIF1<0分別是SKIPIF1<0，在定義域內(nèi)SKIPIF1<0，即SKIPIF1<0是偶函數(shù)，因此SKIPIF1<0取值SKIPIF1<0或0，SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上不是減函數(shù)，只有SKIPIF1<0滿足，此時(shí)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．故選：B．【題型訓(xùn)練】一、單選題1．已知冪函數(shù)SKIPIF1<0為偶函數(shù)，則實(shí)數(shù)SKIPIF1<0的值為（

）A．3 B．2 C．1 D．1或2【答案】C【分析】由題意利用冪函數(shù)的定義和性質(zhì)，得出結(jié)論．【詳解】SKIPIF1<0冪函數(shù)SKIPIF1<0為偶函數(shù)，SKIPIF1<0，且SKIPIF1<0為偶數(shù)，則實(shí)數(shù)SKIPIF1<0，故選：C2．冪函數(shù)SKIPIF1<0SKIPIF1<0的圖象關(guān)于SKIPIF1<0軸對(duì)稱，且在SKIPIF1<0上是增函數(shù)，則SKIPIF1<0的值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0和SKIPIF1<0【答案】D【分析】分別代入SKIPIF1<0的值，由冪函數(shù)性質(zhì)判斷函數(shù)增減性即可.【詳解】因?yàn)镾KIPIF1<0，SKIPIF1<0，所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，由冪函數(shù)性質(zhì)得，在SKIPIF1<0上是減函數(shù)；所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，由冪函數(shù)性質(zhì)得，在SKIPIF1<0上是常函數(shù)；所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，由冪函數(shù)性質(zhì)得，圖象關(guān)于y軸對(duì)稱,在SKIPIF1<0上是增函數(shù)；所以當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，由冪函數(shù)性質(zhì)得，圖象關(guān)于y軸對(duì)稱,在SKIPIF1<0上是增函數(shù)；故選：D.3．已知SKIPIF1<0、SKIPIF1<0，則“SKIPIF1<0”是“SKIPIF1<0”的（

）條件A．充分不必要 B．必要不充分 C．充要 D．既不充分也不必要【答案】C【分析】利用函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增即可判斷出結(jié)論.【詳解】SKIPIF1<0是奇函數(shù)且為遞增函數(shù)，所以SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，同理，SKIPIF1<0，則SKIPIF1<0，函數(shù)單調(diào)遞增，得SKIPIF1<0；SKIPIF1<0“SKIPIF1<0”是“SKIPIF1<0”的充要條件.故選：C.4．已知冪函數(shù)SKIPIF1<0的圖象經(jīng)過點(diǎn)SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的取值范圍為(

)A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】首先根據(jù)已知條件求出SKIPIF1<0的解析式，再根據(jù)SKIPIF1<0的單調(diào)性和奇偶性求解即可.【詳解】由題意可知，SKIPIF1<0，解得，SKIPIF1<0，故SKIPIF1<0，易知，SKIPIF1<0為偶函數(shù)且在SKIPIF1<0上單調(diào)遞減，又因?yàn)镾KIPIF1<0，所以SKIPIF1<0，解得，SKIPIF1<0或SKIPIF1<0.故SKIPIF1<0的取值范圍為SKIPIF1<0.故選：C.5．已知SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】利用中間值SKIPIF1<0比較a,b的大小，再讓b，c與中間值SKIPIF1<0比較，判斷b,c的大小，即可得解.【詳解】SKIPIF1<0，又因?yàn)橥ㄟ^計(jì)算知SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0.故選：B二、多選題6．已知冪函數(shù)SKIPIF1<0的圖象經(jīng)過點(diǎn)SKIPIF1<0，則下列命題正確的有（

）．A．函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0B．函數(shù)SKIPIF1<0為非奇非偶函數(shù)C．過點(diǎn)SKIPIF1<0且與SKIPIF1<0圖象相切的直線方程為SKIPIF1<0D．若SKIPIF1<0，則SKIPIF1<0【答案】BC【分析】先利用待定系數(shù)法求出冪函數(shù)的解析式，寫出函數(shù)的定義域、判定奇偶性，即判定選項(xiàng)A錯(cuò)誤、選項(xiàng)B正確；設(shè)出切點(diǎn)坐標(biāo)，利用導(dǎo)數(shù)的幾何意義和過點(diǎn)SKIPIF1<0求出切線方程，進(jìn)而判定選項(xiàng)C正確；平方作差比較大小，進(jìn)而判定選項(xiàng)D錯(cuò)誤.【詳解】設(shè)SKIPIF1<0，將點(diǎn)SKIPIF1<0代入SKIPIF1<0，得SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，對(duì)于A：SKIPIF1<0的定義域?yàn)镾KIPIF1<0，即選項(xiàng)A錯(cuò)誤；對(duì)于B：因?yàn)镾KIPIF1<0的定義域?yàn)镾KIPIF1<0，所以SKIPIF1<0不具有奇偶性，即選項(xiàng)B正確；對(duì)于C：因?yàn)镾KIPIF1<0，所以SKIPIF1<0，設(shè)切點(diǎn)坐標(biāo)為SKIPIF1<0，則切線斜率為SKIPIF1<0，切線方程為SKIPIF1<0，又因?yàn)榍芯€過點(diǎn)SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，即切線方程為SKIPIF1<0，即SKIPIF1<0，即選項(xiàng)C正確；對(duì)于D：當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0，即SKIPIF1<0成立，即選項(xiàng)D錯(cuò)誤．故選：BC．7．已知函數(shù)SKIPIF1<0是冪函數(shù)，對(duì)任意SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，滿足SKIPIF1<0.若SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0的值為負(fù)值，則下列結(jié)論可能成立的有（

）A．SKIPIF1<0，SKIPIF1<0 B．SKIPIF1<0，SKIPIF1<0C．SKIPIF1<0，SKIPIF1<0 D．SKIPIF1<0，SKIPIF1<0【答案】BC【解析】首先根據(jù)函數(shù)是冪函數(shù)，求得SKIPIF1<0的兩個(gè)值，然后根據(jù)題意判斷函數(shù)在SKIPIF1<0上是增函數(shù)，確定SKIPIF1<0的具體值，再結(jié)合函數(shù)的奇偶性可判斷得正確選項(xiàng).【詳解】由于函數(shù)SKIPIF1<0為冪函數(shù)，故SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0.由于“對(duì)任意SKIPIF1<0，且SKIPIF1<0，滿足SKIPIF1<0”知，函數(shù)在SKIPIF1<0上為增函數(shù)，故SKIPIF1<0.易見SKIPIF1<0，故函數(shù)SKIPIF1<0是單調(diào)遞增的奇函數(shù).由于SKIPIF1<0，即SKIPIF1<0，得SKIPIF1<0，所以SKIPIF1<0，此時(shí)，若當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故SKIPIF1<0，故SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0知，SKIPIF1<0，故SKIPIF1<0或SKIPIF1<0或SKIPIF1<0，即SKIPIF1<0或SKIPIF1<0或SKIPIF1<0.綜上可知，SKIPIF1<0，且SKIPIF1<0或SKIPIF1<0或SKIPIF1<0.故選：BC.【點(diǎn)睛】本題解題關(guān)鍵是熟知冪函數(shù)定義和性質(zhì)突破參數(shù)m，再綜合應(yīng)用奇偶性和單調(diào)性的性質(zhì)確定SKIPIF1<0和SKIPIF1<0的符號(hào)情況.三、填空題8．已知冪函數(shù)SKIPIF1<0是偶函數(shù)，在SKIPIF1<0上遞增的，且滿足SKIPIF1<0.請(qǐng)寫出一個(gè)滿足條件的SKIPIF1<0的值，SKIPIF1<0__________.【答案】SKIPIF1<0【分析】結(jié)合偶函數(shù)和單調(diào)性及SKIPIF1<0可得，答案不是唯一的.【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0；因?yàn)镾KIPIF1<0在SKIPIF1<0上遞增的，所以SKIPIF1<0；因?yàn)閮绾瘮?shù)SKIPIF1<0是偶函數(shù)，所以SKIPIF1<0的值可以為SKIPIF1<0.故答案為：SKIPIF1<0.【點(diǎn)睛】本題主要考查冪函數(shù)的性質(zhì)，冪函數(shù)的單調(diào)性和奇偶性取決于SKIPIF1<0，側(cè)重考查數(shù)學(xué)抽象的核心素養(yǎng).9．已知y=f(x)是奇函數(shù)，當(dāng)x≥0時(shí)，SKIPIF1<0，則f(-8)的值是____.【答案】SKIPIF1<0【分析】先求SKIPIF1<0，再根據(jù)奇函數(shù)求SKIPIF1<0【詳解】SKIPIF1<0，因?yàn)镾KIPIF1<0為奇函數(shù)，所以SKIPIF1<0故答案為：SKIPIF1<0【點(diǎn)睛】本題考查根據(jù)奇函數(shù)性質(zhì)求函數(shù)值，考查基本分析求解能力，屬基礎(chǔ)題.10．已知函數(shù)SKIPIF1<0，則關(guān)于SKIPIF1<0的表達(dá)式SKIPIF1<0的解集為__________.【答案】SKIPIF1<0【分析】利用冪函數(shù)的性質(zhì)及函數(shù)的奇偶性和單調(diào)性即可求解.【詳解】由題意可知，SKIPIF1<0的定義域?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以函數(shù)SKIPIF1<0是奇函數(shù)，由冪函數(shù)的性質(zhì)知，函數(shù)SKIPIF1<0在函數(shù)SKIPIF1<0上單調(diào)遞增，由SKIPIF1<0，得SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，所以關(guān)于SKIPIF1<0的表達(dá)式SKIPIF1<0的解集為SKIPIF1<0.故答案為：SKIPIF1<0.四、解答題11．已知冪函數(shù)SKIPIF1<0的圖像關(guān)于y軸對(duì)稱．(1)求SKIPIF1<0的解析式；(2)求函數(shù)SKIPIF1<0在SKIPIF1<0上的值域．【答案】(1)SKIPIF1<0(2)SKIPIF1<0【分析】(1)根據(jù)冪函數(shù)的定義和性質(zhì)求出m的值即可；(2)由(1)求出函數(shù)SKIPIF1<0的解析式，結(jié)合二次函數(shù)的性質(zhì)即可得出結(jié)果.【詳解】（1）因?yàn)镾KIPIF1<0是冪函數(shù)，所以SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0．又SKIPIF1<0的圖像關(guān)于y軸對(duì)稱，所以SKIPIF1<0，故SKIPIF1<0．（2）由（1）可知，SKIPIF1<0．因?yàn)镾KIPIF1<0，所以SKIPIF1<0，又函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0．故SKIPIF1<0在SKIPIF1<0上的值域?yàn)镾KIPIF1<0．12．已知冪函數(shù)SKIPIF1<0的定義域?yàn)镽.(1)求實(shí)數(shù)SKIPIF1<0的值；(2)若函數(shù)SKIPIF1<0在SKIPIF1<0上不單調(diào)，求實(shí)數(shù)SKIPIF1<0的取值范圍.【答案】(1)SKIPIF1<0；(2)SKIPIF1<0．【分析】（1）由冪函數(shù)定義求得參數(shù)SKIPIF1<0值；（2）由二次函數(shù)的單調(diào)性知對(duì)稱軸在開區(qū)間SKIPIF1<0上，再由指數(shù)函數(shù)性質(zhì)，對(duì)數(shù)的定義得結(jié)論．【詳解】（1）由題意SKIPIF1<0且SKIPIF1<0，解得SKIPIF1<0；（2）由（1）SKIPIF1<0，SKIPIF1<0的對(duì)稱軸SKIPIF1<0，因?yàn)镾KIPIF1<0在SKIPIF1<0上不單調(diào)，所以SKIPIF1<0，解得SKIPIF1<0．題型三二次函數(shù)單調(diào)性問題策略方法二次函數(shù)單調(diào)性問題的求解策略(1)對(duì)于二次函數(shù)的單調(diào)性，關(guān)鍵是開口方向與對(duì)稱軸的位置，若開口方向或?qū)ΨQ軸的位置不確定，則需要分類討論求解．(2)利用二次函數(shù)的單調(diào)性比較大小，一定要將待比較的兩數(shù)通過二次函數(shù)的對(duì)稱性轉(zhuǎn)化到同一單調(diào)區(qū)間上比較．【典例1】“函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上不單調(diào)”是“SKIPIF1<0”的（

）A．充分不必要條件 B．必要不充分條件C．充分且必要條件 D．既不充分也不必要條件【答案】C【分析】根據(jù)二次函數(shù)的單調(diào)性以及充分且必要條件的概念可得答案.【詳解】由函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上不單調(diào)，可得SKIPIF1<0，即SKIPIF1<0；由SKIPIF1<0，得SKIPIF1<0，得函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上不單調(diào)，所以“函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上不單調(diào)”是“SKIPIF1<0”的充分且必要條件.故選：C【題型訓(xùn)練】一、單選題1．若二次函數(shù)SKIPIF1<0，滿足SKIPIF1<0，則下列不等式成立的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】首先根據(jù)SKIPIF1<0，判斷出二次函數(shù)的對(duì)稱軸，然后再根據(jù)二次函數(shù)的單調(diào)性即可得出答案.【詳解】因?yàn)镾KIPIF1<0，所以二次函數(shù)SKIPIF1<0的對(duì)稱軸為SKIPIF1<0，又因?yàn)镾KIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0.故選：B.2．已知SKIPIF1<0在SKIPIF1<0為單調(diào)函數(shù)，則a的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】求出SKIPIF1<0的單調(diào)性，從而得到SKIPIF1<0.【詳解】SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，故要想在SKIPIF1<0為單調(diào)函數(shù)，需滿足SKIPIF1<0，故選：D3．已知二次函數(shù)SKIPIF1<0的兩個(gè)零點(diǎn)都在區(qū)間SKIPIF1<0內(nèi)，則a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】根據(jù)二次函數(shù)的對(duì)稱軸與單調(diào)區(qū)間，結(jié)合已知可得到關(guān)于a的不等式，進(jìn)而求解.【詳解】二次函數(shù)SKIPIF1<0，對(duì)稱軸為SKIPIF1<0，開口向上，在SKIPIF1<0上單調(diào)遞減，在SKIPIF1<0上單調(diào)遞增，要使二次函數(shù)SKIPIF1<0的兩個(gè)零點(diǎn)都在區(qū)間SKIPIF1<0內(nèi)，需SKIPIF1<0，解得SKIPIF1<0故實(shí)數(shù)a的取值范圍是SKIPIF1<0故選：C4．已知函數(shù)SKIPIF1<0，若函數(shù)SKIPIF1<0在R上為減函數(shù)，則實(shí)數(shù)a的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】利用二次函數(shù)、指數(shù)函數(shù)的單調(diào)性以及函數(shù)單調(diào)性的定義，建立關(guān)于a的不等式組，解不等式組即可得答案.【詳解】解：因?yàn)楹瘮?shù)SKIPIF1<0在R上為減函數(shù)，所以SKIPIF1<0，解得SKIPIF1<0，所以實(shí)數(shù)a的取值范圍為SKIPIF1<0，故選：B.二、填空題5．若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0單調(diào)遞減，則實(shí)數(shù)SKIPIF1<0的取值范圍為__．【答案】SKIPIF1<0【分析】根據(jù)一元二次函數(shù)單調(diào)性，結(jié)合條件，可知SKIPIF1<0，然后求出SKIPIF1<0的取值范圍即可.【詳解】易知二次函數(shù)SKIPIF1<0的單調(diào)遞減區(qū)間為SKIPIF1<0，又因?yàn)楹瘮?shù)SKIPIF1<0在區(qū)間SKIPIF1<0單調(diào)遞減，所以SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0.故答案為：SKIPIF1<0.6．若函數(shù)SKIPIF1<0滿足下列性質(zhì)：（1）定義域?yàn)镾KIPIF1<0，值域?yàn)镾KIPIF1<0；（2）圖象關(guān)于直線SKIPIF1<0對(duì)稱；（3）對(duì)任意的SKIPIF1<0，且SKIPIF1<0，都有SKIPIF1<0．寫出函數(shù)SKIPIF1<0的一個(gè)解析式：_______．【答案】SKIPIF1<0（不唯一）【分析】根據(jù)二次函數(shù)的對(duì)稱性、值域及單調(diào)性可得一個(gè)符合條件的函數(shù)式.【詳解】由二次函數(shù)的對(duì)稱性、值域及單調(diào)性可得解析式SKIPIF1<0，此時(shí)SKIPIF1<0對(duì)稱軸為SKIPIF1<0，開口向上，滿足（SKIPIF1<0），因?yàn)閷?duì)任意SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，都有SKIPIF1<0，等價(jià)于SKIPIF1<0在SKIPIF1<0上單調(diào)減，∴SKIPIF1<0，滿足（SKIPIF1<0），又SKIPIF1<0，滿足（SKIPIF1<0），故答案為：SKIPIF1<0（不唯一）．7．若定義在R上的二次函數(shù)f(x)=ax2-4ax+b在區(qū)間[0,2]上是增函數(shù),且f(m)≥f(0),則實(shí)數(shù)m的取值范圍是_____________.【答案】[0,4]【分析】可先求出二次函數(shù)的對(duì)稱軸SKIPIF1<0，再根據(jù)函數(shù)的增減性及對(duì)稱性可求得m的取值范圍．【詳解】二次函數(shù)的對(duì)稱軸SKIPIF1<0，SKIPIF1<0時(shí)函數(shù)單調(diào)遞增，SKIPIF1<0，二次函數(shù)開口向下，SKIPIF1<0函數(shù)單調(diào)遞減，根據(jù)二次函數(shù)的對(duì)稱性，SKIPIF1<0，f(m)≥f(0)，SKIPIF1<0【點(diǎn)睛】二次函數(shù)是對(duì)稱函數(shù)，解題時(shí)，一定要根據(jù)對(duì)稱性來解題，防止漏解錯(cuò)解．題型四二次函數(shù)最值問題策略方法二次函數(shù)最值問題的類型及解題思路(1)類型：①對(duì)稱軸、區(qū)間都是給定的；②對(duì)稱軸動(dòng)、區(qū)間固定；③對(duì)稱軸定、區(qū)間變動(dòng)．(2)解決這類問題的思路：抓住“三點(diǎn)一軸”數(shù)形結(jié)合，“三點(diǎn)”是指區(qū)間兩個(gè)端點(diǎn)和中點(diǎn)，“一軸”指的是對(duì)稱軸，結(jié)合配方法，根據(jù)函數(shù)的單調(diào)性及分類討論的思想解決問題．【典例1】若函數(shù)f(x)＝ax2＋2ax＋1在[－1，2]上有最大值4，則a的值為(

)A．SKIPIF1<0 B．－3 C．SKIPIF1<0或－3 D．4【答案】C【分析】按SKIPIF1<0分類討論求SKIPIF1<0的最大值，然后由最大值為4得參數(shù)值．【詳解】由題意得f(x)＝a(x＋1)2＋1－a．①當(dāng)a＝0時(shí)，函數(shù)f(x)在區(qū)間[－1，2]上的值為常數(shù)1，不符合題意，舍去；②當(dāng)a>0時(shí)，函數(shù)f(x)在區(qū)間[－1，2]上是增函數(shù)，最大值為f(2)＝8a＋1＝4，解得SKIPIF1<0；③當(dāng)a<0時(shí)，函數(shù)f(x)在區(qū)間[－1，2]上是減函數(shù)，最大值為f(－1)＝1－a＝4，解得a＝－3．綜上可知，a的值為SKIPIF1<0或－3．故選：C．【題型訓(xùn)練】一、單選題1．已知函數(shù)SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0的最小值為SKIPIF1<0，則SKIPIF1<0的最大值為（

）A．1 B．0 C．SKIPIF1<0 D．2【答案】A【分析】根據(jù)二次函數(shù)性質(zhì)求得最小值，由最小值得SKIPIF1<0值，從而再求得最大值．【詳解】∵SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，∴其最小值為SKIPIF1<0，∴其最大值為SKIPIF1<0.故選：A．2．已知函數(shù)SKIPIF1<0在SKIPIF1<0上為單調(diào)遞增函數(shù)，則實(shí)數(shù)m的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】求導(dǎo)，由單調(diào)性得到SKIPIF1<0在SKIPIF1<0上恒成立，由二次函數(shù)數(shù)形結(jié)合得到不等關(guān)系，求出m的取值范圍.【詳解】SKIPIF1<0，因?yàn)镾KIPIF1<0在SKIPIF1<0上為單調(diào)遞增函數(shù)，所以SKIPIF1<0在SKIPIF1<0上恒成立，令SKIPIF1<0，要滿足SKIPIF1<0①，或SKIPIF1<0②，由①得：SKIPIF1<0，由②得：SKIPIF1<0，綜上：實(shí)數(shù)m的取值范圍是SKIPIF1<0.故選：D3．若函數(shù)SKIPIF1<0在SKIPIF1<0上的最小值為-1，則SKIPIF1<0（

）A．2或SKIPIF1<0 B．1或SKIPIF1<0 C．2 D．1【答案】D【分析】先求出二次函數(shù)的對(duì)稱軸，然后討論對(duì)稱軸與區(qū)間SKIPIF1<0的關(guān)系，求出其最小值，列方程可求出SKIPIF1<0的值【詳解】函數(shù)SKIPIF1<0圖象的對(duì)稱軸為SKIPIF1<0，圖象開口向上，（1）當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增．則SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，不符合SKIPIF1<0；（2）當(dāng)SKIPIF1<0時(shí)．則SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0或SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0符合；（3）當(dāng)SKIPIF1<0時(shí)，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0SKIPIF1<0不符合，綜上可得SKIPIF1<0．故選：D4．已知二次函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】由二次函數(shù)的值域可得出SKIPIF1<0，可得出SKIPIF1<0，則有SKIPIF1<0，利用基本不等式可求得結(jié)果.【詳解】若SKIPIF1<0，則函數(shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0，不合乎題意，因?yàn)槎魏瘮?shù)SKIPIF1<0的值域?yàn)镾KIPIF1<0，則SKIPIF1<0，且SKIPIF1<0，所以，SKIPIF1<0，可得SKIPIF1<0，則SKIPIF1<0，所以，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立，因此，SKIPIF1<0的最小值為SKIPIF1<0.故選：B.5．設(shè)二次函數(shù)SKIPIF1<0在SKIPIF1<0上有最大值，最大值為SKIPIF1<0，當(dāng)SKIPIF1<0取最小值時(shí)，SKIPIF1<0（

）A．0 B．1 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】根據(jù)二次函數(shù)的性質(zhì)求出SKIPIF1<0，然后利用基本不等式即得.【詳解】SKIPIF1<0在SKIPIF1<0上有最大值SKIPIF1<0，SKIPIF1<0且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0的最大值為SKIPIF1<0，即SKIPIF1<0且SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，即SKIPIF1<0時(shí)，SKIPIF1<0有最小值2，故選：A．二、填空題6．若函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0內(nèi)存在最小值，則SKIPIF1<0的取值范圍是___________．【答案】SKIPIF1<0【分析】根據(jù)二次函數(shù)的性質(zhì)確定在開區(qū)間SKIPIF1<0內(nèi)存在最小值的情況列不等式，即可得SKIPIF1<0的取值范圍是.【詳解】解：二次函數(shù)SKIPIF1<0的對(duì)稱軸為SKIPIF1<0，且二次函數(shù)開口向上若函數(shù)在開區(qū)間SKIPIF1<0內(nèi)存在最小值，則SKIPIF1<0，即SKIPIF1<0，此時(shí)函數(shù)在SKIPIF1<0處能取到最小值，故SKIPIF1<0的取值范圍是SKIPIF1<0.故答案為：SKIPIF1<0.7．若函數(shù)SKIPIF1<0的定義域和值域均為SKIPIF1<0，則SKIPIF1<0的值為__________.【答案】SKIPIF1<0【分析】由二次函數(shù)的解析式，可知二次函數(shù)關(guān)于SKIPIF1<0成軸對(duì)稱，即可得到SKIPIF1<0，從而得到方程組，解得即可.【詳解】解：因?yàn)镾KIPIF1<0，對(duì)稱軸為SKIPIF1<0，開口向上，所以函數(shù)在SKIPIF1<0上單調(diào)遞增，又因?yàn)槎x域和值域均為SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0（舍去）或SKIPIF1<0，所以SKIPIF1<0.故答案為：SKIPIF1<08．函數(shù)SKIPIF1<0（SKIPIF1<0，且SKIPIF1<0）在SKIPIF1<0上的最大值為13，則實(shí)數(shù)SKIPIF1<0的值為___________.【答案】SKIPIF1<0或SKIPIF1<0【分析】令SKIPIF1<0，討論SKIPIF1<0或SKIPIF1<0，求出SKIPIF1<0的取值范圍，再利用二次函數(shù)的單調(diào)性即可求解.【詳解】∵SKIPIF1<0令SKIPIF1<0，則SKIPIF1<0，則SKIPIF1<0，其對(duì)稱軸為SKIPIF1<0.該二次函數(shù)在SKIPIF1<0上是增函數(shù).①若SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，故當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0，解得SKIPIF1<0(SKIPIF1<0舍去).②若SKIPIF1<0，由SKIPIF1<0，可得SKIPIF1<0，故當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，SKIPIF1<0.∴SKIPIF1<0或SKIPIF1<0(舍去).綜上可得SKIPIF1<0或SKIPIF1<0.故答案為：SKIPIF1<0或SKIPIF1<0.9．設(shè)SKIPIF1<0，SKIPIF1<0，若函數(shù)SKIPIF1<0在定義域上滿足：①是非奇非偶函數(shù)；②既不是增函數(shù)也不是減函數(shù)；③有最大值，則實(shí)數(shù)a的取值范圍是__________．【答案】SKIPIF1<0【分析】對(duì)①：根據(jù)奇偶函數(shù)的定義可得SKIPIF1<0；對(duì)②：分類討論可得二次項(xiàng)系數(shù)小于零，且對(duì)稱軸為SKIPIF1<0，求出a的取值范圍；對(duì)③：結(jié)合②中所求的范圍驗(yàn)證即可.【詳解】對(duì)①：∵SKIPIF1<0，即SKIPIF1<0，故SKIPIF1<0不是奇函數(shù)；若SKIPIF1<0是偶函數(shù)，則SKIPIF1<0，可得SKIPIF1<0，即SKIPIF1<0；故若SKIPIF1<0是非奇非偶函數(shù)，則SKIPIF1<0；對(duì)③：若SKIPIF1<0在SKIPIF1<0上有最大值，則有：當(dāng)SKIPIF1<0時(shí)，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，無最值，不合題意；當(dāng)SKIPIF1<0時(shí)，則SKIPIF1<0為二次函數(shù)且對(duì)稱軸為SKIPIF1<0，由題意可得SKIPIF1<0，解得SKIPIF1<0，故若SKIPIF1<0在SKIPIF1<0上有最大值，則SKIPIF1<0；對(duì)②：若SKIPIF1<0，則SKIPIF1<0開口向下，且對(duì)稱軸為SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上既不是增函數(shù)也不是減函數(shù)；綜上所述：實(shí)數(shù)a的取值范圍為SKIPIF1<0.故答案為：SKIPIF1<0.題型五二次函數(shù)恒成立問題策略方法由不等式恒成立求參數(shù)取值范圍的思路及關(guān)鍵(1)一般有兩個(gè)解題思路：一是分離參數(shù)；二是不分離參數(shù)．(2)兩種思路都是將問題歸結(jié)為求函數(shù)的最值，至于用哪種方法，關(guān)鍵是看參數(shù)是否已分離．這兩個(gè)思路的依據(jù)是：a≥f(x)恒成立?a≥f(x)max，a≤f(x)恒成立?a≤f(x)min．2．a(chǎn)x2＋bx＋c＜0(a＞0)在區(qū)間[m，n]上恒成立的條件．設(shè)f(x)＝ax2＋bx＋c，則eq\b\lc\{\rc\(\a\vs4\al\co1(fm＜0，,fn＜0.))【典例1】設(shè)函數(shù)SKIPIF1<0，若對(duì)于SKIPIF1<0，SKIPIF1<0恒成立，則實(shí)數(shù)SKIPIF1<0的取值范圍為___________.【答案】SKIPIF1<0【分析】整理可得SKIPIF1<0在SKIPIF1<0上恒成立，根據(jù)x的范圍，可求得SKIPIF1<0的范圍，分析即可得答案.【詳解】由題意，SKIPIF1<0可得SKIPIF1<0，即SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0在