新高考數(shù)學(xué)一輪復(fù)習(xí)知識點(diǎn)總結(jié)與題型精練專題07 基本初等函數(shù)（含解析）

專題07基本初等函數(shù)【考綱要求】1、掌握二次函數(shù)的圖象與性質(zhì)，會求二次函數(shù)的最值(值域)、單調(diào)區(qū)間．2、了解指數(shù)函數(shù)模型的實(shí)際背景，理解有理數(shù)指數(shù)冪的含義，了解實(shí)數(shù)指數(shù)冪的意義，掌握冪的運(yùn)算．3、理解指數(shù)函數(shù)的概念及其單調(diào)性，掌握指數(shù)函數(shù)圖象通過的特殊點(diǎn)，知道指數(shù)函數(shù)是重要的函數(shù)模型．4、理解對數(shù)的概念及其運(yùn)算性質(zhì)，知道用換底公式將一般對數(shù)轉(zhuǎn)化成自然對數(shù)或常用對數(shù)；了解對數(shù)在簡化運(yùn)算中的作用．5、理解對數(shù)函數(shù)的概念及其單調(diào)性，掌握對數(shù)函數(shù)圖象通過的特殊點(diǎn)，知道對數(shù)函數(shù)是重要的函數(shù)模型．一、二次函數(shù)【考點(diǎn)總結(jié)】1．二次函數(shù)(1)二次函數(shù)解析式的三種形式①一般式：f(x)＝ax2＋bx＋c(a≠0)．②頂點(diǎn)式：f(x)＝a(x－m)2＋n(a≠0)．③零點(diǎn)式：f(x)＝a(x－x1)(x－x2)(a≠0)．(2)二次函數(shù)的圖象和性質(zhì)解析式f(x)＝ax2＋bx＋c(a>0)f(x)＝ax2＋bx＋c(a<0)圖象定義域(－∞，＋∞)(－∞，＋∞)值域eq\b\lc\[\rc\)(\a\vs4\al\co1(\f(4ac－b2,4a)，＋∞))eq\b\lc\(\rc\](\a\vs4\al\co1(－∞，\f(4ac－b2,4a)))單調(diào)性在eq\b\lc\(\rc\](\a\vs4\al\co1(－∞，－\f(b,2a)))上單調(diào)遞減；在eq\b\lc\(\rc\)(\a\vs4\al\co1(－\f(b,2a)，＋∞))上單調(diào)遞增在eq\b\lc\(\rc\](\a\vs4\al\co1(－∞，－\f(b,2a)))上單調(diào)遞增；在eq\b\lc\(\rc\)(\a\vs4\al\co1(－\f(b,2a)，＋∞))上單調(diào)遞減對稱性函數(shù)的圖象關(guān)于x＝－eq\f(b,2a)對稱二、冪函數(shù)【思維導(dǎo)圖】【考點(diǎn)總結(jié)】1．冪函數(shù)(1)定義：形如y＝xα(α∈R)的函數(shù)稱為冪函數(shù)，其中底數(shù)x是自變量，α為常數(shù)．常見的五類冪函數(shù)為y＝x，y＝x2，y＝x3，y＝xeq\s\up6(\f(1,2))，y＝x－1.(2)五種冪函數(shù)的圖象(3)性質(zhì)①冪函數(shù)在(0，＋∞)上都有定義；②當(dāng)α>0時，冪函數(shù)的圖象都過點(diǎn)(1，1)和(0，0)，且在(0，＋∞)上單調(diào)遞增；③當(dāng)α<0時，冪函數(shù)的圖象都過點(diǎn)(1，1)，且在(0，＋∞)上單調(diào)遞減．三、指數(shù)與指數(shù)函數(shù)【思維導(dǎo)圖】【考點(diǎn)總結(jié)】1．根式(1)根式的概念①若xn＝a，則x叫做a的n次方根，其中n>1且n∈N*.式子eq\r(n,a)叫做根式，這里n叫做根指數(shù)，a叫做被開方數(shù)．②a的n次方根的表示：xn＝a?eq\b\lc\{(\a\vs4\al\co1(x＝\r(n,a)，當(dāng)n為奇數(shù)且n∈N*，n>1時，,x＝±\r(n,a)，當(dāng)n為偶數(shù)且n∈N*時.))(2)根式的性質(zhì)①(eq\r(n,a))n＝a(n∈N*，且n>1)；②eq\r(n,an)＝eq\b\lc\{(\a\vs4\al\co1(a，n為奇數(shù)，,|a|＝\b\lc\{(\a\vs4\al\co1(a，a≥0，,－a，a<0，))n為偶數(shù).))2．有理數(shù)指數(shù)冪(1)冪的有關(guān)概念①正分?jǐn)?shù)指數(shù)冪：aeq\s\up6(\f(m,n))＝eq\r(n,am)(a>0，m，n∈N*，且n>1)；②負(fù)分?jǐn)?shù)指數(shù)冪：a－eq\s\up6(\f(m,n))＝eq\f(1,a\s\up6(\f(m,n)))＝eq\f(1,\r(n,am))(a>0，m，n∈N*，且n>1)；③0的正分?jǐn)?shù)指數(shù)冪等于0，0的負(fù)分?jǐn)?shù)指數(shù)冪無意義．(2)有理數(shù)指數(shù)冪的運(yùn)算性質(zhì)①aras＝ar＋s(a>0，r，s∈Q)；②(ar)s＝ars(a>0，r，s∈Q)；③(ab)r＝arbr(a>0，b>0，r∈Q)．3．指數(shù)函數(shù)的圖象與性質(zhì)y＝ax(a>0且a≠1)a>10<a<1圖象定義域R值域(0，＋∞)性質(zhì)過定點(diǎn)(0，1)當(dāng)x>0時，y>1；當(dāng)x<0時，0<y<1當(dāng)x>0時，0<y<1；當(dāng)x<0時，y>1在R上是增函數(shù)在R上是減函數(shù)四、對數(shù)與對數(shù)函數(shù)【考點(diǎn)總結(jié)】1．對數(shù)概念如果ax＝N(a>0，且a≠1)，那么數(shù)x叫做以a為底數(shù)N的對數(shù)，記作x＝logaN，其中a叫做對數(shù)的底數(shù)，N叫做真數(shù)，logaN叫做對數(shù)式性質(zhì)對數(shù)式與指數(shù)式的互化：ax＝N?x＝logaN(a>0，且a≠1)loga1＝0，logaa＝1，alogaN＝N(a>0，且a≠1)運(yùn)算法則loga(M·N)＝logaM＋logaNa>0，且a≠1，M>0，N>0logaeq\f(M,N)＝logaM－logaNlogaMn＝nlogaM(n∈R)換底公式logab＝eq\f(logcb,logca)(a>0，且a≠1，c>0，且c≠1，b>0)2.對數(shù)函數(shù)的圖象與性質(zhì)a>10<a<1圖象續(xù)表a>10<a<1性質(zhì)定義域：(0，＋∞)值域：R過定點(diǎn)(1，0)當(dāng)x>1時，y>0當(dāng)0<x<1時，y<0當(dāng)x>1時，y<0當(dāng)0<x<1時，y>0在(0，＋∞)上是增函數(shù)在(0，＋∞)上是減函數(shù)3.反函數(shù)指數(shù)函數(shù)y＝ax與對數(shù)函數(shù)y＝logax互為反函數(shù)，它們的圖象關(guān)于直線y＝x對稱．【題型匯編】題型一：二次函數(shù)的概念題型二：二次函數(shù)的圖象與性質(zhì)題型三：冪函數(shù)的圖象與性質(zhì)題型四：指數(shù)函數(shù)的圖象與性質(zhì)題型五：對數(shù)函數(shù)的圖象與性質(zhì)【題型講解】題型一：二次函數(shù)的概念一、單選題1．（2022·上海松江·二模）已知正方形SKIPIF1<0的邊長為4，點(diǎn)SKIPIF1<0、SKIPIF1<0分別在邊SKIPIF1<0、SKIPIF1<0上，且SKIPIF1<0，SKIPIF1<0，若點(diǎn)SKIPIF1<0在正方形SKIPIF1<0的邊上，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】【分析】建立平面直角坐標(biāo)系，利用向量的數(shù)量積運(yùn)算及二次函數(shù)求值域即可得解.【詳解】如圖，建立平面直角坐標(biāo)系，則SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0在SKIPIF1<0上時，設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，即SKIPIF1<0，當(dāng)SKIPIF1<0在SKIPIF1<0上時，設(shè)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，知SKIPIF1<0，當(dāng)SKIPIF1<0在SKIPIF1<0上時，設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，即SKIPIF1<0，當(dāng)SKIPIF1<0在SKIPIF1<0上時，設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，即SKIPIF1<0.綜上可得，SKIPIF1<0，故選：C2．（2022·北京·北大附中三模）已知半徑為SKIPIF1<0的圓SKIPIF1<0經(jīng)過點(diǎn)SKIPIF1<0，且與直線SKIPIF1<0相切，則其圓心到直線SKIPIF1<0距離的最小值為（

）A．1 B．SKIPIF1<0 C．2 D．SKIPIF1<0【答案】B【解析】【分析】先求出得圓心SKIPIF1<0的軌跡方程，再利用點(diǎn)到直線的距離公式表示出距離，最后根據(jù)二次函數(shù)的最值求解方法可求得答案.【詳解】依題意，設(shè)圓SKIPIF1<0的圓心SKIPIF1<0，動點(diǎn)SKIPIF1<0到點(diǎn)SKIPIF1<0的距離等于到直線SKIPIF1<0的距離，根據(jù)拋物線的定義可得圓心SKIPIF1<0的軌跡方程為SKIPIF1<0，設(shè)圓心SKIPIF1<0到直線SKIPIF1<0距離為SKIPIF1<0，SKIPIF1<0當(dāng)SKIPIF1<0時,SKIPIF1<0故選：B方法二：可以設(shè)與直線SKIPIF1<0平行的拋物線的切線方程，聯(lián)立方程，利用判別式等于零，得到切線方程，再利用平行線的距離公式得解；方法三：在第一象限分析問題，轉(zhuǎn)化為求函數(shù)SKIPIF1<0的切線與直線SKIPIF1<0平行，再利用平行線的距離公式得解.3．（2022·江西南昌·三模（理））已知SKIPIF1<0的內(nèi)角SKIPIF1<0，SKIPIF1<0，SKIPIF1<0所對的邊分別為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0.SKIPIF1<0，SKIPIF1<0分別為線段SKIPIF1<0，SKIPIF1<0上的動點(diǎn)，SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】【分析】根據(jù)已知條件運(yùn)用正弦定理求出SKIPIF1<0的長，根據(jù)SKIPIF1<0設(shè)出邊的關(guān)系，再利用余弦定理表示出SKIPIF1<0，從函數(shù)的角度求其最值.【詳解】依題意，如圖所示，在SKIPIF1<0中，SKIPIF1<0，SKIPIF1<0，由正弦定理得，SKIPIF1<0，又SKIPIF1<0，解得：SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，在SKIPIF1<0中，由余弦定理得，SKIPIF1<0SKIPIF1<0SKIPIF1<0，對于二次函數(shù)SKIPIF1<0開口向上，對稱軸SKIPIF1<0SKIPIF1<0，SKIPIF1<0的最小值為SKIPIF1<0.4．（2022·北京·二模）如圖，已知正方體SKIPIF1<0的棱長為1，則線段SKIPIF1<0上的動點(diǎn)P到直線SKIPIF1<0的距離的最小值為（

）A．1 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】【分析】利用坐標(biāo)法，設(shè)SKIPIF1<0，可得動點(diǎn)P到直線SKIPIF1<0的距離為SKIPIF1<0SKIPIF1<0，然后利用二次函數(shù)的性質(zhì)即得.【詳解】如圖建立空間直角坐標(biāo)系，則SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，∴動點(diǎn)P到直線SKIPIF1<0的距離為SKIPIF1<0SKIPIF1<0，當(dāng)SKIPIF1<0時取等號，即線段SKIPIF1<0上的動點(diǎn)P到直線SKIPIF1<0的距離的最小值為SKIPIF1<0.故選：D.5．（2022·江西·上饒市第一中學(xué)二模（文））已知集合SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】【分析】解不等式得集合A，求二次函數(shù)值域得集合B，然后由集合的交集運(yùn)算可得.【詳解】由SKIPIF1<0解得SKIPIF1<0，即SKIPIF1<0，易知SKIPIF1<0，即SKIPIF1<0則SKIPIF1<0.故選：A6．（2022·北京市第十二中學(xué)三模）若函數(shù)SKIPIF1<0的值域?yàn)镽，則a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】【分析】由SKIPIF1<0時，SKIPIF1<0，由題意，當(dāng)SKIPIF1<0時，SKIPIF1<0，對SKIPIF1<0分SKIPIF1<0和SKIPIF1<0兩種情況討論即可求解.【詳解】解：由SKIPIF1<0時，SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0的值域?yàn)镽，所以當(dāng)SKIPIF1<0時，SKIPIF1<0，分兩種情況討論：①當(dāng)SKIPIF1<0時，SKIPIF1<0SKIPIF1<0，所以只需SKIPIF1<0，解得SKIPIF1<0，所以SKIPIF1<0；②當(dāng)SKIPIF1<0時，SKIPIF1<0，所以只需SKIPIF1<0，顯然成立，所以SKIPIF1<0.綜上，SKIPIF1<0的取值范圍是SKIPIF1<0.故選：D.7．（2022·四川·三模（理））設(shè)函數(shù)SKIPIF1<0的定義城為R，且SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，若存在SKIPIF1<0時，使SKIPIF1<0，則k的最大值為（

）．A．1 B．2 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】【分析】先根據(jù)SKIPIF1<0得到從SKIPIF1<0開始，SKIPIF1<0每右移1個單位，圖像就會向上移1個單位，然后確定函數(shù)SKIPIF1<0的SKIPIF1<0由小到大，SKIPIF1<0第一次取到SKIPIF1<0時，SKIPIF1<0的范圍，進(jìn)而可得該范圍內(nèi)函數(shù)SKIPIF1<0的解析式，令SKIPIF1<0，求出SKIPIF1<0，進(jìn)而可得k的最大值.【詳解】當(dāng)SKIPIF1<0時，SKIPIF1<0由SKIPIF1<0得SKIPIF1<0，即從SKIPIF1<0開始，SKIPIF1<0每右移1個單位，圖像就會向上移1個單位，當(dāng)SKIPIF1<0時，SKIPIF1<0，又SKIPIF1<0，故當(dāng)函數(shù)SKIPIF1<0的SKIPIF1<0由小到大，SKIPIF1<0第一次取到SKIPIF1<0時，SKIPIF1<0又當(dāng)SKIPIF1<0時，SKIPIF1<0，令SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0，若存在SKIPIF1<0時，使SKIPIF1<0，則必有SKIPIF1<0，所以k的最大值為SKIPIF1<0.故選：D.8．（2022·安徽·淮南第一中學(xué)一模（理））已知雙曲線SKIPIF1<0（SKIPIF1<0，SKIPIF1<0）的左、右焦點(diǎn)分別是SKIPIF1<0、SKIPIF1<0，且SKIPIF1<0，若P是該雙曲線右支上一點(diǎn)，且滿足SKIPIF1<0，則SKIPIF1<0面積的最大值是（

）A．SKIPIF1<0 B．1 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】【分析】根據(jù)已知條件，結(jié)合雙曲線的定義求出SKIPIF1<0與SKIPIF1<0，然后在SKIPIF1<0中，利用余弦定理求出SKIPIF1<0，再根據(jù)面積公式及二次函數(shù)的知識即可求解.【詳解】解：因?yàn)镻是該雙曲線右支上一點(diǎn)，所以由雙曲線的定義有SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，設(shè)SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時等號成立，所以SKIPIF1<0面積的最大值是SKIPIF1<0，故選：A.9．（2022·安徽淮北·一模（理））已知SKIPIF1<0是橢圓SKIPIF1<0的右焦點(diǎn)，點(diǎn)SKIPIF1<0在SKIPIF1<0上，直線SKIPIF1<0與SKIPIF1<0軸交于點(diǎn)SKIPIF1<0，點(diǎn)SKIPIF1<0為C上的動點(diǎn)，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】【分析】由題可得橢圓SKIPIF1<0，進(jìn)而可得SKIPIF1<0，利用向量數(shù)量積的坐標(biāo)表示可得SKIPIF1<0SKIPIF1<0，再結(jié)合條件及二次函數(shù)的性質(zhì)即求.【詳解】由題可得SKIPIF1<0，∴SKIPIF1<0，即橢圓SKIPIF1<0，∴SKIPIF1<0，直線SKIPIF1<0方程為SKIPIF1<0，∴SKIPIF1<0，又SKIPIF1<0，設(shè)SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0，又SKIPIF1<0，∴當(dāng)SKIPIF1<0時，SKIPIF1<0有最小值為SKIPIF1<0.故選：C.10．（2022·四川巴中·一模（理））已知集合SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】【分析】求出集合SKIPIF1<0中SKIPIF1<0的范圍，與集合SKIPIF1<0取交集【詳解】集合SKIPIF1<0中，根據(jù)SKIPIF1<0得：SKIPIF1<0，所以集合SKIPIF1<0，集合SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0故選：B二、多選題1．（2022·重慶·一模）已知SKIPIF1<0，SKIPIF1<0且SKIPIF1<0，則下列結(jié)論正確的是（

）A．SKIPIF1<0的最大值為SKIPIF1<0 B．SKIPIF1<0的最大值為SKIPIF1<0C．SKIPIF1<0的最小值為SKIPIF1<0 D．SKIPIF1<0的最大值為SKIPIF1<0【答案】BC【解析】【分析】利用基本不等式直接判斷A；利用基本不等式求得SKIPIF1<0的最大值可判斷B；利用基本不等式“1”的代換可判斷C；利用二次函數(shù)的性質(zhì)可判斷D；【詳解】SKIPIF1<0，SKIPIF1<0且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0對于A，利用基本不等式得SKIPIF1<0，化簡得SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時，等號成立，所以SKIPIF1<0的最大值為SKIPIF1<0，故A錯誤；對于B，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時，等號成立，所以SKIPIF1<0的最大值為SKIPIF1<0，故B正確；對于C，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時，等號成立，所以SKIPIF1<0的最小值為SKIPIF1<0，故C正確；對于D，SKIPIF1<0SKIPIF1<0利用二次函數(shù)的性質(zhì)知，當(dāng)SKIPIF1<0時，函數(shù)單調(diào)遞減；當(dāng)SKIPIF1<0時，函數(shù)單調(diào)遞增，SKIPIF1<0，SKIPIF1<0，故D錯誤；故選：BC題型二：二次函數(shù)的圖象與性質(zhì)一、單選題1．（2022·上海浦東新·二模）已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，實(shí)數(shù)SKIPIF1<0滿足SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，現(xiàn)有如下兩個結(jié)論：①對于任意的實(shí)數(shù)SKIPIF1<0，存在實(shí)數(shù)SKIPIF1<0，使得SKIPIF1<0；②存在實(shí)數(shù)SKIPIF1<0，對于任意的SKIPIF1<0，都有SKIPIF1<0；則（

）A．①②均正確 B．①②均不正確C．①正確，②不正確 D．①不正確，②正確【答案】C【解析】【分析】對①，根據(jù)SKIPIF1<0，SKIPIF1<0的幾何意義，判斷得出SKIPIF1<0與SKIPIF1<0一定有兩個交點(diǎn)分析即可對②，通過化簡SKIPIF1<0，將題意轉(zhuǎn)換為：存在實(shí)數(shù)SKIPIF1<0，使得SKIPIF1<0在SKIPIF1<0上為減函數(shù)，再分析出當(dāng)SKIPIF1<0時函數(shù)有增區(qū)間，推出矛盾即可【詳解】對①，SKIPIF1<0的幾何意義為SKIPIF1<0與SKIPIF1<0兩點(diǎn)間的斜率，同理SKIPIF1<0的幾何意義為SKIPIF1<0與SKIPIF1<0兩點(diǎn)間的斜率.數(shù)形結(jié)合可得，當(dāng)SKIPIF1<0時，存在SKIPIF1<0；當(dāng)SKIPIF1<0時，存在SKIPIF1<0，使得SKIPIF1<0，即SKIPIF1<0成立.即對于任意的實(shí)數(shù)SKIPIF1<0，存在實(shí)數(shù)SKIPIF1<0，使得SKIPIF1<0，故①正確；對②，若存在實(shí)數(shù)SKIPIF1<0，對于任意的SKIPIF1<0，都有SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0.即存在實(shí)數(shù)SKIPIF1<0，對于任意的SKIPIF1<0，SKIPIF1<0恒成立.設(shè)SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0為減函數(shù).故原題意可轉(zhuǎn)化為：存在實(shí)數(shù)SKIPIF1<0，使得SKIPIF1<0在SKIPIF1<0上為減函數(shù).因?yàn)楫?dāng)SKIPIF1<0時，SKIPIF1<0，因?yàn)镾KIPIF1<0對稱軸為SKIPIF1<0，故當(dāng)SKIPIF1<0時SKIPIF1<0一定為增函數(shù)，故不存在實(shí)數(shù)SKIPIF1<0，使得SKIPIF1<0在SKIPIF1<0上為減函數(shù).故②錯誤故選：C2．（2022·遼寧·三模）函數(shù)SKIPIF1<0的最大值為（

）A．2 B．3 C．4 D．5【答案】B【解析】【分析】利用三角函數(shù)的平方關(guān)系將SKIPIF1<0化為SKIPIF1<0，配方后結(jié)合二次函數(shù)知識，求得答案.【詳解】SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0取得最大值，且最大值為3,故選：B3．（2022·江西鷹潭·二模（理））已知函數(shù)SKIPIF1<0的極大值點(diǎn)SKIPIF1<0，極小值點(diǎn)SKIPIF1<0，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】【分析】求出SKIPIF1<0的導(dǎo)函數(shù)SKIPIF1<0，由當(dāng)SKIPIF1<0時取得極大值，當(dāng)SKIPIF1<0時取得極小值，可得SKIPIF1<0、SKIPIF1<0是方程SKIPIF1<0的兩個根，根據(jù)一元二次方程根的分布可以得到參數(shù)SKIPIF1<0、SKIPIF1<0滿足的不等式組，畫出其表示的平面區(qū)域，根據(jù)SKIPIF1<0的幾何意義即可求解【詳解】SKIPIF1<0又因?yàn)楫?dāng)SKIPIF1<0時取得極大值，當(dāng)SKIPIF1<0時取得極小值，可得SKIPIF1<0、SKIPIF1<0是方程SKIPIF1<0的兩個根，根據(jù)一元二次方程根的分布可得SKIPIF1<0即：SKIPIF1<0作出該不等式組表示的平面區(qū)域如圖中陰影部分所示（不包括邊界），可求出邊界交點(diǎn)坐標(biāo)分別為SKIPIF1<0、SKIPIF1<0、SKIPIF1<0，SKIPIF1<0表示平面區(qū)域內(nèi)的點(diǎn)SKIPIF1<0與點(diǎn)SKIPIF1<0連線的斜率，由圖可知SKIPIF1<0SKIPIF1<0，根據(jù)傾斜角的變化，可得SKIPIF1<0故選:B4．（2022·北京昌平·二模）已知函數(shù)SKIPIF1<0，則關(guān)于SKIPIF1<0的不等式SKIPIF1<0的解集是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】【分析】由二次函數(shù)的性質(zhì)判斷SKIPIF1<0區(qū)間單調(diào)性，根據(jù)解析式知SKIPIF1<0恒過SKIPIF1<0且SKIPIF1<0，進(jìn)而確定區(qū)間值域，再由對數(shù)函數(shù)性質(zhì)求SKIPIF1<0的對應(yīng)區(qū)間值域，即可得不等式解集.【詳解】由題設(shè)，SKIPIF1<0對稱軸為SKIPIF1<0且圖象開口向下，則SKIPIF1<0在SKIPIF1<0上遞增，SKIPIF1<0上遞減，由SKIPIF1<0，即SKIPIF1<0恒過SKIPIF1<0且SKIPIF1<0，所以SKIPIF1<0上SKIPIF1<0，SKIPIF1<0上SKIPIF1<0，而SKIPIF1<0在SKIPIF1<0上遞增，且SKIPIF1<0上SKIPIF1<0，SKIPIF1<0上SKIPIF1<0，所以SKIPIF1<0的解集為SKIPIF1<0.故選：C5．（2022·江蘇·華羅庚中學(xué)三模）若函數(shù)SKIPIF1<0的定義域和值域的交集為空集，則正數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】【分析】首先得到函數(shù)的定義域，再分析當(dāng)SKIPIF1<0時SKIPIF1<0的取值，即可得到SKIPIF1<0，再對SKIPIF1<0時分SKIPIF1<0和SKIPIF1<0兩種情況討論，求出此時SKIPIF1<0的取值，即可得到SKIPIF1<0的值域，從而得到不等式，解得即可；【詳解】解：因?yàn)镾KIPIF1<0，所以SKIPIF1<0的定義域?yàn)镾KIPIF1<0，SKIPIF1<0，當(dāng)SKIPIF1<0時SKIPIF1<0，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以SKIPIF1<0；要使定義域和值域的交集為空集，顯然SKIPIF1<0，當(dāng)SKIPIF1<0時SKIPIF1<0，若SKIPIF1<0則SKIPIF1<0，此時顯然不滿足定義域和值域的交集為空集，若SKIPIF1<0時SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，此時SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0，即SKIPIF1<0故選：B6．（2022·寧夏·銀川一中三模（文））已知SKIPIF1<0的最小值為2，則SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】【分析】注意觀察SKIPIF1<0時，SKIPIF1<0，所以讓SKIPIF1<0時，SKIPIF1<0恒成立即可，根據(jù)參變分離和換元方法即可得解.【詳解】當(dāng)SKIPIF1<0時，SKIPIF1<0，又因?yàn)镾KIPIF1<0的最小值為2，，所以需要當(dāng)SKIPIF1<0時，SKIPIF1<0恒成立，所以SKIPIF1<0在SKIPIF1<0恒成立，所以SKIPIF1<0在SKIPIF1<0恒成立，即SKIPIF1<0在SKIPIF1<0恒成立，令SKIPIF1<0，則SKIPIF1<0，原式轉(zhuǎn)化為SKIPIF1<0在SKIPIF1<0恒成立，SKIPIF1<0是二次函數(shù)，開口向下，對稱軸為直線SKIPIF1<0，所以在SKIPIF1<0上SKIPIF1<0最大值為SKIPIF1<0，所以SKIPIF1<0，故選：D.7．（2022·北京·一模）已知直線SKIPIF1<0是圓SKIPIF1<0的一條對稱軸，則SKIPIF1<0的最大值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．1 D．SKIPIF1<0【答案】A【解析】【分析】圓心必然在直線l上，得到SKIPIF1<0的關(guān)系式，再考慮求最大值.【詳解】由于直線l是圓的對稱軸，所以圓的圓心必定在直線l上，將圓的一般方程轉(zhuǎn)變?yōu)闃?biāo)準(zhǔn)方程：SKIPIF1<0，圓心為SKIPIF1<0，將圓心坐標(biāo)代入直線l的方程得SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，函數(shù)SKIPIF1<0是開口向下，以SKIPIF1<0

為對稱軸的拋物線，所以SKIPIF1<0，故選：A.8．（2022·山東濟(jì)南·二模）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【解析】【分析】首先根據(jù)SKIPIF1<0，判斷出二次函數(shù)的對稱軸，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.二、多選題1．（2022·福建莆田·三模）已知函數(shù)SKIPIF1<0，函數(shù)SKIPIF1<0，則下列結(jié)論正確的是（

）A．若SKIPIF1<0有3個不同的零點(diǎn)，則a的取值范圍是SKIPIF1<0B．若SKIPIF1<0有4個不同的零點(diǎn)，則a的取值范圍是SKIPIF1<0C．若SKIPIF1<0有4個不同的零點(diǎn)SKIPIF1<0，則SKIPIF1<0D．若SKIPIF1<0有4個不同的零點(diǎn)SKIPIF1<0，則SKIPIF1<0的取值范圍是SKIPIF1<0【答案】BCD【解析】【分析】根據(jù)題意，將問題轉(zhuǎn)化為函數(shù)SKIPIF1<0與SKIPIF1<0圖像交點(diǎn)個數(shù)問題，進(jìn)而數(shù)形結(jié)合求解即可得答案.【詳解】解：令SKIPIF1<0得SKIPIF1<0，即所以SKIPIF1<0零點(diǎn)個數(shù)為函數(shù)SKIPIF1<0與SKIPIF1<0圖像交點(diǎn)個數(shù)，故，作出函數(shù)SKIPIF1<0圖像如圖，由圖可知，SKIPIF1<0有3個不同的零點(diǎn)，則a的取值范圍是SKIPIF1<0，故A選項(xiàng)錯誤；SKIPIF1<0有4個不同的零點(diǎn)，則a的取值范圍是SKIPIF1<0，故B選項(xiàng)正確；SKIPIF1<0有4個不同的零點(diǎn)SKIPIF1<0，此時SKIPIF1<0關(guān)于直線SKIPIF1<0對稱，所以SKIPIF1<0，故C選項(xiàng)正確；由C選項(xiàng)可知SKIPIF1<0，所以SKIPIF1<0，由于SKIPIF1<0有4個不同的零點(diǎn)，a的取值范圍是SKIPIF1<0，故SKIPIF1<0，所以SKIPIF1<0，故D選項(xiàng)正確.故選：BCD題型三：冪函數(shù)的圖象與性質(zhì)一、單選題1．（2022·山東·德州市教育科學(xué)研究院三模）已知對數(shù)函數(shù)SKIPIF1<0的圖像經(jīng)過點(diǎn)SKIPIF1<0與點(diǎn)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】【分析】根據(jù)對數(shù)函數(shù)可以解得SKIPIF1<0，SKIPIF1<0，再結(jié)合中間值法比較大?。驹斀狻吭O(shè)SKIPIF1<0，由題意可得：SKIPIF1<0，則SKIPIF1<0∴SKIPIF1<0SKIPIF1<0，SKIPIF1<0，SKIPIF1<0∴SKIPIF1<0故選：C．2．（2022·江西·二模（文））已知SKIPIF1<0，則a，b，c的大小關(guān)系是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】【分析】利用對數(shù)函數(shù)、三角函數(shù)、冪函數(shù)的單調(diào)性比較大小即可.【詳解】SKIPIF1<0，因?yàn)镾KIPIF1<0在SKIPIF1<0是單調(diào)遞增函數(shù)，所以SKIPIF1<0，因?yàn)镾KIPIF1<0在SKIPIF1<0是單調(diào)遞增函數(shù)，所以SKIPIF1<0所以SKIPIF1<0，故選：C.3．（2022·四川眉山·三模（文））下列結(jié)論正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】【分析】對于A、B：作出SKIPIF1<0和SKIPIF1<0在第一象限的圖像判斷出：在SKIPIF1<0上，有SKIPIF1<0，在SKIPIF1<0上，有SKIPIF1<0，在SKIPIF1<0上，有SKIPIF1<0.即可判斷A、B；對于C：判斷出SKIPIF1<0,SKIPIF1<0，即可判斷；對于D：判斷出SKIPIF1<0，SKIPIF1<0，即可判斷.【詳解】對于A、B：作出SKIPIF1<0和SKIPIF1<0在第一象限的圖像如圖所示：其中SKIPIF1<0的圖像用虛線表示，SKIPIF1<0的圖像用虛線表示.可得，在SKIPIF1<0上，有SKIPIF1<0，在SKIPIF1<0上，有SKIPIF1<0，在SKIPIF1<0上，有SKIPIF1<0.因?yàn)镾KIPIF1<0，所以SKIPIF1<0，故A正確；因?yàn)镾KIPIF1<0，所以SKIPIF1<0，故B錯誤；對于C：SKIPIF1<0,而SKIPIF1<0，所以SKIPIF1<0.故C錯誤；對于D：SKIPIF1<0，而SKIPIF1<0，所以SKIPIF1<0.故D錯誤.故選：A4．（2022·北京·二模）下列函數(shù)中，與函數(shù)SKIPIF1<0的奇偶性相同，且在SKIPIF1<0上有相同單調(diào)性的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】【分析】根據(jù)指對函數(shù)的性質(zhì)判斷A、B，由正弦函數(shù)性質(zhì)判斷C，對于D有SKIPIF1<0，即可判斷奇偶性和SKIPIF1<0單調(diào)性.【詳解】由SKIPIF1<0為奇函數(shù)且在SKIPIF1<0上遞增，A、B：SKIPIF1<0、SKIPIF1<0非奇非偶函數(shù)，排除；C：SKIPIF1<0為奇函數(shù)，但在SKIPIF1<0上不單調(diào)，排除；D：SKIPIF1<0，顯然SKIPIF1<0且定義域關(guān)于原點(diǎn)對稱，在SKIPIF1<0上遞增，滿足.故選：D5．（2022·江西·南昌市八一中學(xué)三模（文））已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則a，b，c的大小關(guān)系為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】【分析】結(jié)合對數(shù)函數(shù)、指數(shù)函數(shù)和冪函數(shù)的單調(diào)性直接比較大小即可.【詳解】依題意，SKIPIF1<0，SKIPIF1<0，而SKIPIF1<0，即SKIPIF1<0，故SKIPIF1<0.故選：C．6．（2022·廣東·二模）定義在SKIPIF1<0上的下列函數(shù)中，既是奇函數(shù)，又是增函數(shù)的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】【分析】由正弦函數(shù)，指數(shù)函數(shù)和冪函數(shù)的性質(zhì)對各個選項(xiàng)進(jìn)行分析判斷即可得到答案.【詳解】A.SKIPIF1<0,由正弦函數(shù)的性質(zhì)可知SKIPIF1<0在SKIPIF1<0上不為增函數(shù)，故排除；B.SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，故排除；C.SKIPIF1<0，故函數(shù)在SKIPIF1<0上為偶函數(shù)，故排除；D.SKIPIF1<0，SKIPIF1<0，故函數(shù)在SKIPIF1<0上為奇函數(shù)，且由冪函數(shù)的性質(zhì)知SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，滿足題意；故選：D7．（2022·內(nèi)蒙古包頭·二模（文））下列函數(shù)中是減函數(shù)的為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【解析】【分析】根據(jù)二次函數(shù)、正比例函數(shù)、指數(shù)函數(shù)、冪函數(shù)的單調(diào)性逐一判斷即可.【詳解】A：因?yàn)楹瘮?shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，所以該函數(shù)不是減函數(shù)，不符合題意；B：因?yàn)楹瘮?shù)SKIPIF1<0是增函數(shù)，所以不符合題意；C：因?yàn)楹瘮?shù)SKIPIF1<0是增函數(shù)，所以不符合題意；D：因?yàn)楹瘮?shù)SKIPIF1<0是減函數(shù)，所以符合題意，故選：D8．（2022·安徽·蕪湖一中三模（文））設(shè)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則a，b，c的大小關(guān)系為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】【分析】根據(jù)指數(shù)函數(shù)和對數(shù)函數(shù)單調(diào)性，即可求解.【詳解】解：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，故選：A.二、多選題1．（2022·山東威?！と＃┤鬝KIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】BC【解析】【分析】根據(jù)冪函數(shù)、指數(shù)函數(shù)、對數(shù)函數(shù)的單調(diào)性分別可判斷A、B、C，結(jié)合C和對數(shù)換底公式即可判斷D.【詳解】對于A，∵冪函數(shù)y=SKIPIF1<0在SKIPIF1<0單調(diào)遞增，∴根據(jù)SKIPIF1<0可知SKIPIF1<0，故A錯誤；對于B，∵指數(shù)函數(shù)y=SKIPIF1<0在R上單調(diào)遞減，∴根據(jù)SKIPIF1<0可知SKIPIF1<0，故B正確；對于C，∵對數(shù)函數(shù)y=SKIPIF1<0(SKIPIF1<0)在SKIPIF1<0上單調(diào)遞減，∴根據(jù)SKIPIF1<0可知SKIPIF1<0，故C正確；對于D，由C可知SKIPIF1<0，∴SKIPIF1<0，即SKIPIF1<0，故D錯誤．故選：BC．2．（2022·山東濱州·二模）若實(shí)數(shù)a，b滿足SKIPIF1<0，則下列結(jié)論中正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】BCD【解析】【分析】根據(jù)給定條件，求出a，b的關(guān)系，再利用不等式性質(zhì)判斷A，B；指對數(shù)函數(shù)、冪函數(shù)單調(diào)性分析判斷C，D作答.【詳解】因SKIPIF1<0，則SKIPIF1<0，于是有SKIPIF1<0，A不正確；SKIPIF1<0，即SKIPIF1<0，B正確；由SKIPIF1<0得：SKIPIF1<0，因此，SKIPIF1<0，C正確；因SKIPIF1<0，函數(shù)SKIPIF1<0在R上單調(diào)遞減，函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，則SKIPIF1<0，D正確.故選：BCD題型四：指數(shù)的圖象與性質(zhì)一、單選題1．（2022·青?！ご笸ɑ刈逋磷遄灾慰h教學(xué)研究室三模（文））已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則正數(shù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的大小關(guān)系為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【解析】【分析】由已知求出m，n，p，再借助商值比較法及“媒介”數(shù)推理判斷作答.【詳解】由SKIPIF1<0，得SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，因此，SKIPIF1<0，即SKIPIF1<0，由SKIPIF1<0，得SKIPIF1<0，于是得SKIPIF1<0，所以正數(shù)SKIPIF1<0，SKIPIF1<0，SKIPIF1<0的大小關(guān)系為SKIPIF1<0.故選：A2．（2022·青?！ご笸ɑ刈逋磷遄灾慰h教學(xué)研究室三模（文））若函數(shù)SKIPIF1<0滿足SKIPIF1<0，且當(dāng)SKIPIF1<0時，SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．10 C．4 D．2【答案】B【解析】【分析】首先得到SKIPIF1<0的周期，再根據(jù)函數(shù)的周期性計(jì)算可得；【詳解】解：由SKIPIF1<0，得SKIPIF1<0，∴函數(shù)SKIPIF1<0是周期函數(shù)，且4是它的一個周期，又當(dāng)SKIPIF1<0時，SKIPIF1<0，∴SKIPIF1<0；故選：B.3．（2022·山東臨沂·三模）已知SKIPIF1<0，則a，b，c的大小關(guān)系是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】【分析】分別化簡SKIPIF1