新高考數(shù)學(xué)二輪復(fù)習(xí)培優(yōu)講義16 一元二次不等式和基本不等式問(wèn)題（含解析）

解密16一元二次不等式和基本不等式問(wèn)題【考點(diǎn)解密】一元二次不等式的解集判別式Δ＝b2－4acΔ>0Δ＝0Δ<0二次函數(shù)y＝ax2＋bx＋c(a>0)的圖象方程ax2＋bx＋c＝0(a>0)的根有兩相異實(shí)根x1，x2(x1<x2)有兩相等實(shí)根x1＝x2＝－eq\f(b,2a)沒(méi)有實(shí)數(shù)根ax2＋bx＋c>0(a>0)的解集{x|x<x1或x>x2}eq\b\lc\{\rc\}(\a\vs4\al\co1(x\b\lc\|\rc\(\a\vs4\al\co1(x≠－\f(b,2a))))){x|x∈R}ax2＋bx＋c<0(a>0)的解集{x|x1<x<x2}??1．基本不等式：eq\r(ab)≤eq\f(a＋b,2)(1)基本不等式成立的條件：a>0，b>0.(2)等號(hào)成立的條件：當(dāng)且僅當(dāng)a＝b時(shí)取等號(hào)．2．幾個(gè)重要的不等式(1)a2＋b2≥2ab(a，b∈R)．(2)eq\f(b,a)＋eq\f(a,b)≥2(a，b同號(hào))．(3)ab≤eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(a＋b,2)))2(a，b∈R)．(4)eq\f(a2＋b2,2)≥eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(a＋b,2)))2(a，b∈R)．以上不等式等號(hào)成立的條件均為a＝b.【方法技巧】1．利用基本不等式求最值問(wèn)題已知a>0，b>0，則(1)如果積ab是定值p，那么當(dāng)且僅當(dāng)a＝b時(shí)，a＋b有最小值2eq\r(p).(簡(jiǎn)記：積定和最小)(2)如果和a＋b是定值p，那么當(dāng)且僅當(dāng)a＝b時(shí)，ab有最大值eq\f(p2,4).(簡(jiǎn)記：和定積最大)2.利用基本不等式求最值時(shí)，要注意其必須滿(mǎn)足的三個(gè)條件：（1）“一正二定三相等”中的“一正”就是各項(xiàng)必須為正數(shù)；（2）“二定”就是要求和的最小值，必須把構(gòu)成和的二項(xiàng)之積轉(zhuǎn)化成定值；要求積的最大值，則必須把構(gòu)成積的因式的和轉(zhuǎn)化成定值；（3）“三相等”是利用基本不等式求最值時(shí)，必須驗(yàn)證等號(hào)成立的條件，若不能取等號(hào)則這個(gè)定值就不是所求的最值，這也是最容易發(fā)生錯(cuò)誤的地方，注意多次運(yùn)用不等式，等號(hào)成立條件是否一致.【核心題型】題型一：含參數(shù)的一元二次不等式問(wèn)題1．（2022·全國(guó)·高三專(zhuān)題練習(xí)）已知集合SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0，則實(shí)數(shù)a的取值范圍是A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】解分式不等式求得集合SKIPIF1<0，對(duì)SKIPIF1<0進(jìn)行分類(lèi)討論，結(jié)合SKIPIF1<0，求得實(shí)數(shù)SKIPIF1<0的取值范圍.【詳解】由SKIPIF1<0SKIPIF1<0SKIPIF1<0或SKIPIF1<0.所以SKIPIF1<0或SKIPIF1<0，所以SKIPIF1<0.由SKIPIF1<0，解得SKIPIF1<0或SKIPIF1<0.SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，此時(shí)SKIPIF1<0，滿(mǎn)足SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，由SKIPIF1<0得SKIPIF1<0，即SKIPIF1<0且SKIPIF1<0.綜上所述，實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0.故選：B【點(diǎn)睛】本小題主要考查分式不等式的解法，考查一元二次不等式的解法，考查根據(jù)交集、補(bǔ)集的運(yùn)算結(jié)果求參數(shù)的取值范圍，屬于中檔題.2．（2023·全國(guó)·高三專(zhuān)題練習(xí)）若關(guān)于x的不等式SKIPIF1<0的解集中恰有4個(gè)整數(shù)，則實(shí)數(shù)m的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】討論m與2的大小關(guān)系，求得不等式的解集，根據(jù)解集中恰有4個(gè)整數(shù)，確定m的取值范圍.【詳解】不等式SKIPIF1<0即SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，不等式解集為SKIPIF1<0，此時(shí)要使解集中恰有4個(gè)整數(shù)，這四個(gè)整數(shù)只能是3,4,5,6，故SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，不等式解集為SKIPIF1<0，此時(shí)不符合題意；當(dāng)SKIPIF1<0時(shí)，不等式解集為SKIPIF1<0，此時(shí)要使解集中恰有4個(gè)整數(shù)，這四個(gè)整數(shù)只能是SKIPIF1<0，故SKIPIF1<0，，故實(shí)數(shù)m的取值范圍為SKIPIF1<0，故選：C3．（2021·全國(guó)·高三專(zhuān)題練習(xí)）已知定義在SKIPIF1<0上的函數(shù)SKIPIF1<0滿(mǎn)足SKIPIF1<0，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，則關(guān)于SKIPIF1<0的不等式SKIPIF1<0（其中SKIPIF1<0）的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0或SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0或SKIPIF1<0【答案】A【分析】先判斷函數(shù)SKIPIF1<0單調(diào)遞減，再利用已知條件和函數(shù)的單調(diào)性得SKIPIF1<0，解不等式即得解.【詳解】任取SKIPIF1<0，由已知得SKIPIF1<0，即SKIPIF1<0，所以函數(shù)SKIPIF1<0單調(diào)遞減．由SKIPIF1<0可得SKIPIF1<0，即SKIPIF1<0SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，即SKIPIF1<0，又因?yàn)镾KIPIF1<0，所以SKIPIF1<0，此時(shí)原不等式解集為SKIPIF1<0．故選：A【點(diǎn)睛】方法點(diǎn)睛：解抽象函數(shù)不等式一般先要判斷函數(shù)的單調(diào)性，再利用單調(diào)性化抽象函數(shù)不等式為具體的函數(shù)不等式解答.題型二：一元二次不等式根分布問(wèn)題4．（2021·全國(guó)·高三專(zhuān)題練習(xí)）已知SKIPIF1<0若函數(shù)SKIPIF1<0恰有5個(gè)零點(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】B【解析】先作出函數(shù)SKIPIF1<0的圖象，然后結(jié)合函數(shù)的零點(diǎn)與方程的根的關(guān)系，得到方程SKIPIF1<0的一個(gè)根在SKIPIF1<0，一個(gè)根在SKIPIF1<0，結(jié)合一元二次方程的根的分布問(wèn)題即可求解．【詳解】解：作出函數(shù)SKIPIF1<0的圖象如圖所示，令SKIPIF1<0，則由圖可知，當(dāng)SKIPIF1<0時(shí)，方程SKIPIF1<0只有一個(gè)根；當(dāng)SKIPIF1<0時(shí)，方程SKIPIF1<0有兩個(gè)根；當(dāng)SKIPIF1<0時(shí)，方程SKIPIF1<0只有一個(gè)根；顯然SKIPIF1<0不是方程SKIPIF1<0的根；若SKIPIF1<0是方程SKIPIF1<0的根，則SKIPIF1<0，此時(shí)SKIPIF1<0，結(jié)合圖象可知，此時(shí)方程SKIPIF1<0和方程SKIPIF1<0共有4個(gè)根，則函數(shù)SKIPIF1<0有4個(gè)零點(diǎn)，不滿(mǎn)足題意；∴SKIPIF1<0恰有5個(gè)零點(diǎn)等價(jià)于方程SKIPIF1<0恰有5個(gè)實(shí)根，等價(jià)于方程SKIPIF1<0的一個(gè)根在SKIPIF1<0，一個(gè)根在SKIPIF1<0，令SKIPIF1<0，則SKIPIF1<0，∴SKIPIF1<0，故選：B．【點(diǎn)睛】本題主要考查由函數(shù)的零點(diǎn)求解參數(shù)范圍問(wèn)題，體現(xiàn)了轉(zhuǎn)化思想及數(shù)形結(jié)合思想的應(yīng)用，屬于難題．5．（2022·安徽·南陵中學(xué)校聯(lián)考模擬預(yù)測(cè)）在區(qū)間SKIPIF1<0上任取兩個(gè)實(shí)數(shù)a，b，則方程SKIPIF1<0有兩個(gè)不同的非負(fù)根的概率為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】根據(jù)方程SKIPIF1<0有兩個(gè)不同的非負(fù)根，可得SKIPIF1<0，在平面直角坐標(biāo)系作出可行域，結(jié)合圖象，根據(jù)幾何概型即可得解.【詳解】解：因?yàn)榉匠蘏KIPIF1<0有兩個(gè)不同的非負(fù)根，所以SKIPIF1<0，則SKIPIF1<0，如圖，作出不等式組所表示得平面區(qū)域?yàn)镾KIPIF1<0，在區(qū)間SKIPIF1<0上任取兩個(gè)實(shí)數(shù)a，b，所表示得平面區(qū)域?yàn)檎叫蜸KIPIF1<0，SKIPIF1<0，所以方程SKIPIF1<0有兩個(gè)不同的非負(fù)根的概率為SKIPIF1<0.故選：B.6．（2023·四川綿陽(yáng)·四川省綿陽(yáng)南山中學(xué)?？家荒＃┮阎瘮?shù)SKIPIF1<0有兩個(gè)不同的極值點(diǎn)SKIPIF1<0，且不等式SKIPIF1<0恒成立，則實(shí)數(shù)t的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】把函數(shù)SKIPIF1<0有兩個(gè)不同的極值點(diǎn)SKIPIF1<0轉(zhuǎn)化為根的分布求出a的范圍，利用分離參數(shù)法得到SKIPIF1<0.把SKIPIF1<0轉(zhuǎn)化為SKIPIF1<0，令SKIPIF1<0，利用導(dǎo)數(shù)求出SKIPIF1<0的值域，即可得到答案.【詳解】SKIPIF1<0，因?yàn)楹瘮?shù)SKIPIF1<0有兩個(gè)不同的極值點(diǎn)SKIPIF1<0，SKIPIF1<0，所以方程SKIPIF1<0有兩個(gè)不相等的正實(shí)數(shù)根，于是有SKIPIF1<0，解得SKIPIF1<0.因?yàn)椴坏仁絊KIPIF1<0恒成立，所以SKIPIF1<0恒成立.SKIPIF1<0SKIPIF1<0SKIPIF1<0，設(shè)SKIPIF1<0，SKIPIF1<0，故SKIPIF1<0在SKIPIF1<0上單調(diào)遞增，故SKIPIF1<0，所以SKIPIF1<0.因此實(shí)數(shù)t的取值范圍是SKIPIF1<0.故選：A【點(diǎn)睛】導(dǎo)數(shù)的應(yīng)用主要有：（1）利用導(dǎo)函數(shù)幾何意義求切線(xiàn)方程；（2）利用導(dǎo)數(shù)研究原函數(shù)的單調(diào)性，求極值（最值）；（3）利用導(dǎo)數(shù)求參數(shù)的取值范圍.題型三：一元二次不等式恒成立問(wèn)題、7．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知實(shí)數(shù)SKIPIF1<0，SKIPIF1<0滿(mǎn)足如下兩個(gè)條件：（1）關(guān)于SKIPIF1<0的方程SKIPIF1<0有兩個(gè)異號(hào)的實(shí)根；（2）SKIPIF1<0，若對(duì)于上述的一切實(shí)數(shù)SKIPIF1<0，SKIPIF1<0，不等式SKIPIF1<0恒成立，則實(shí)數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】首先判斷SKIPIF1<0，再化簡(jiǎn)SKIPIF1<0，利用基本不等式求解.【詳解】解：設(shè)方程SKIPIF1<0的兩個(gè)異號(hào)的實(shí)根分別為SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0．又SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（當(dāng)且僅當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)取“SKIPIF1<0”），由不等式SKIPIF1<0恒成立，得SKIPIF1<0，解得SKIPIF1<0．SKIPIF1<0實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0．故選：A．8．（2022·浙江·模擬預(yù)測(cè)）已知函數(shù)SKIPIF1<0，若對(duì)任意的實(shí)數(shù)x，恒有SKIPIF1<0成立，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】C【分析】首先令SKIPIF1<0，然后判斷SKIPIF1<0的奇偶性和單調(diào)性，然后將原不等式轉(zhuǎn)化為SKIPIF1<0，再利用SKIPIF1<0的奇偶性和單調(diào)性得SKIPIF1<0對(duì)于任意的實(shí)數(shù)SKIPIF1<0恒成立，最后解二次函數(shù)恒成立問(wèn)題即可.【詳解】令SKIPIF1<0，由于SKIPIF1<0，所以得SKIPIF1<0為奇函數(shù).又因?yàn)镾KIPIF1<0在SKIPIF1<0上單調(diào)遞減，所以SKIPIF1<0在SKIPIF1<0上單調(diào)遞減.已知對(duì)于任意的實(shí)數(shù)SKIPIF1<0，恒有SKIPIF1<0，整理得：SKIPIF1<0，即SKIPIF1<0，由于SKIPIF1<0為奇函數(shù)，得SKIPIF1<0，由于SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，得SKIPIF1<0對(duì)于任意的實(shí)數(shù)SKIPIF1<0恒成立，即SKIPIF1<0對(duì)于任意的實(shí)數(shù)SKIPIF1<0恒成立.當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0不恒成立，故SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，有SKIPIF1<0，解得SKIPIF1<0.故選：C9．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知函數(shù)SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0恒成立，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】C【分析】根據(jù)函數(shù)解析式畫(huà)出函數(shù)圖象，即可判斷函數(shù)為奇函數(shù)且在定義域上單調(diào)遞減，則不等式等價(jià)于SKIPIF1<0，即SKIPIF1<0恒成立，再分SKIPIF1<0和SKIPIF1<0兩種情況討論，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，即可求出參數(shù)SKIPIF1<0的取值范圍；【詳解】解：因?yàn)镾KIPIF1<0，所以函數(shù)圖象如下所示：由函數(shù)圖象可知函數(shù)為定義域SKIPIF1<0上單調(diào)遞減的奇函數(shù)，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0，因?yàn)镾KIPIF1<0，SKIPIF1<0恒成立，即SKIPIF1<0，SKIPIF1<0恒成立，所以SKIPIF1<0恒成立，即SKIPIF1<0恒成立，當(dāng)SKIPIF1<0，顯然不成立，當(dāng)SKIPIF1<0時(shí)，則m>0Δ=81?48m≤0，解得SKIPIF1<0，即SKIPIF1<0；故選：C題型四：一元二次不等式在某區(qū)間成立問(wèn)題10．（2017·天津·高考真題）已知函數(shù)SKIPIF1<0設(shè)SKIPIF1<0，若關(guān)于x的不等式SKIPIF1<0在R上恒成立，則a的取值范圍是A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【詳解】不等式SKIPIF1<0為SKIPIF1<0(*)，當(dāng)SKIPIF1<0時(shí)，(*)式即為SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0（SKIPIF1<0時(shí)取等號(hào)），SKIPIF1<0（SKIPIF1<0時(shí)取等號(hào)），所以SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，(*)式為SKIPIF1<0，SKIPIF1<0，又SKIPIF1<0（當(dāng)SKIPIF1<0時(shí)取等號(hào)），SKIPIF1<0（當(dāng)SKIPIF1<0時(shí)取等號(hào)），所以SKIPIF1<0，綜上SKIPIF1<0．故選A．【考點(diǎn)】不等式、恒成立問(wèn)題【名師點(diǎn)睛】首先滿(mǎn)足SKIPIF1<0轉(zhuǎn)化為SKIPIF1<0去解決，由于涉及分段函數(shù)問(wèn)題要遵循分段處理原則，分別對(duì)SKIPIF1<0的兩種不同情況進(jìn)行討論，針對(duì)每種情況根據(jù)SKIPIF1<0的范圍，利用極端原理，求出對(duì)應(yīng)的SKIPIF1<0的范圍.11．（2022秋·湖北襄陽(yáng)·高三校考階段練習(xí)）若命題“SKIPIF1<0”為假命題，則實(shí)數(shù)x的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】等價(jià)于“SKIPIF1<0”為真命題．令SKIPIF1<0，解不等式SKIPIF1<0即得解.【詳解】解：命題“SKIPIF1<0”為假命題，其否定為真命題，即“SKIPIF1<0”為真命題．令SKIPIF1<0，則SKIPIF1<0，即SKIPIF1<0，解得SKIPIF1<0，所以實(shí)數(shù)x的取值范圍為SKIPIF1<0.故選：C12．（2022·四川攀枝花·統(tǒng)考二模）已知函數(shù)SKIPIF1<0，若關(guān)于SKIPIF1<0的不等式SKIPIF1<0恒成立，則實(shí)數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】先判斷SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上恒成立；若SKIPIF1<0在SKIPIF1<0上恒成立，轉(zhuǎn)化為SKIPIF1<0在SKIPIF1<0上恒成立．【詳解】當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0恒成立，二次函數(shù)的對(duì)稱(chēng)軸為SKIPIF1<0，（1）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上單調(diào)遞減，則SKIPIF1<0恒成立，（2）當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，所以SKIPIF1<0綜上可知，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0上恒成立；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0恒成立，即SKIPIF1<0在SKIPIF1<0上恒成立，令SKIPIF1<0，則SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，函數(shù)單增，又SKIPIF1<0，所以SKIPIF1<0；綜上可知，SKIPIF1<0的取值范圍是SKIPIF1<0，故選：D題型五：基本不等式求積最大值問(wèn)題13．（2021·全國(guó)·統(tǒng)考高考真題）已知SKIPIF1<0，SKIPIF1<0是橢圓SKIPIF1<0：SKIPIF1<0的兩個(gè)焦點(diǎn)，點(diǎn)SKIPIF1<0在SKIPIF1<0上，則SKIPIF1<0的最大值為（

）A．13 B．12 C．9 D．6【答案】C【分析】本題通過(guò)利用橢圓定義得到SKIPIF1<0，借助基本不等式SKIPIF1<0即可得到答案．【詳解】由題，SKIPIF1<0，則SKIPIF1<0，所以SKIPIF1<0（當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，等號(hào)成立）．故選：C．【點(diǎn)睛】14．（2022·全國(guó)·高三專(zhuān)題練習(xí)）已知SKIPIF1<0的外心為點(diǎn)O，M為邊SKIPIF1<0上的一點(diǎn)，且SKIPIF1<0，則SKIPIF1<0的面積的最大值等于（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】首先用SKIPIF1<0、SKIPIF1<0表示SKIPIF1<0，再根據(jù)向量數(shù)量積的運(yùn)算律及基本不等式求出SKIPIF1<0的最大值，最后根據(jù)三角形面積公式計(jì)算可得；【詳解】解：因?yàn)镾KIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0SKIPIF1<0所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，取等號(hào)；所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，取等號(hào)；故選：C15．（2022·全國(guó)·高三專(zhuān)題練習(xí)）已知△ABC的三邊分別為a，b，c，若滿(mǎn)足a2+b2+2c2＝8，則△ABC面積的最大值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【解析】根據(jù)a2+b2+2c2＝8，得到SKIPIF1<0，由余弦定理得到SKIPIF1<0，由正弦定理得到SKIPIF1<0，兩式平方相加得SKIPIF1<0，而SKIPIF1<0，兩式結(jié)合有SKIPIF1<0，再用基本不等式求解.【詳解】因?yàn)閍2+b2+2c2＝8，所以SKIPIF1<0，由余弦定理得SKIPIF1<0，即SKIPIF1<0①由正弦定理得SKIPIF1<0，即SKIPIF1<0②由①，②平方相加得SKIPIF1<0，所以SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0且SKIPIF1<0即SKIPIF1<0時(shí)，取等號(hào).故選：B【點(diǎn)睛】本題主要考查了正弦定理和余弦定理及基本不等式的應(yīng)用，還考查了運(yùn)算求解的能力，屬于中檔題.題型六：基本不等式求和最小值問(wèn)題16．（2021秋·江蘇蘇州·高三張家港高級(jí)中學(xué)校考期中）在SKIPIF1<0中，SKIPIF1<0為SKIPIF1<0上一點(diǎn)，SKIPIF1<0，SKIPIF1<0為SKIPIF1<0上任一點(diǎn)，若SKIPIF1<0，則SKIPIF1<0的最小值是A．9 B．10C．11 D．12【答案】D【分析】由題意結(jié)合向量共線(xiàn)的充分必要條件首先確定SKIPIF1<0的關(guān)系，然后結(jié)合均值不等式的結(jié)論整理計(jì)算即可求得最終結(jié)果.【詳解】由題意可知：SKIPIF1<0，SKIPIF1<0三點(diǎn)共線(xiàn)，則：SKIPIF1<0，據(jù)此有：SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立.綜上可得：SKIPIF1<0的最小值是12.本題選擇D選項(xiàng).【點(diǎn)睛】本題主要考查三點(diǎn)共線(xiàn)的充分必要條件，均值不等式求最值的方法等知識(shí)，意在考查學(xué)生的轉(zhuǎn)化能力和計(jì)算求解能力.17．（2023·全國(guó)·高三專(zhuān)題練習(xí)）在平面四邊形SKIPIF1<0中，已知SKIPIF1<0的面積是SKIPIF1<0的面積的2倍.若存在正實(shí)數(shù)SKIPIF1<0使得SKIPIF1<0成立，則SKIPIF1<0的最小值為（

）A．1 B．2 C．3 D．4【答案】A【分析】由面積比得SKIPIF1<0，再利用SKIPIF1<0三點(diǎn)共線(xiàn)可得出SKIPIF1<0的關(guān)系，從而利用基本不等式可求得SKIPIF1<0的最小值．【詳解】如圖，設(shè)SKIPIF1<0與SKIPIF1<0交于點(diǎn)SKIPIF1<0，由SKIPIF1<0的面積是SKIPIF1<0的面積的2倍，可得SKIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0三點(diǎn)共線(xiàn)，即SKIPIF1<0共線(xiàn)，所以存在實(shí)數(shù)SKIPIF1<0使得SKIPIF1<0，因?yàn)镾KIPIF1<0，所以SKIPIF1<0，消去k，可得SKIPIF1<0，又因?yàn)镾KIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)等號(hào)成立．所以SKIPIF1<0的最小值為1．故選：A．18．（2023秋·天津?yàn)I海新·高三大港一中校考階段練習(xí)）已知SKIPIF1<0是橢圓與雙曲線(xiàn)的公共焦點(diǎn)，P是它們的一個(gè)公共點(diǎn)，且SKIPIF1<0，線(xiàn)段SKIPIF1<0的垂直平分線(xiàn)過(guò)SKIPIF1<0，若橢圓的離心率為SKIPIF1<0，雙曲線(xiàn)的離心率為SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．3 C．6 D．SKIPIF1<0【答案】C【分析】利用橢圓和雙曲線(xiàn)的性質(zhì)，用橢圓雙曲線(xiàn)的焦距長(zhǎng)軸長(zhǎng)表示SKIPIF1<0，再利用均值不等式得到答案．【詳解】設(shè)橢圓長(zhǎng)軸SKIPIF1<0，雙曲線(xiàn)實(shí)軸SKIPIF1<0，由題意可知：SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0，兩式相減，可得：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào)，SKIPIF1<0的最小值為6，故選：C．【點(diǎn)睛】本題考查了橢圓雙曲線(xiàn)的性質(zhì)，用橢圓雙曲線(xiàn)的焦距長(zhǎng)軸長(zhǎng)表示SKIPIF1<0是解題的關(guān)鍵，意在考查學(xué)生的計(jì)算能力．題型七：二次或二次商式的最值問(wèn)題19．（2023·全國(guó)·高三專(zhuān)題練習(xí)）若a，b，c均為正實(shí)數(shù)，則SKIPIF1<0的最大值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】對(duì)原式變形，兩次利用基本不等式，求解即可.【詳解】因?yàn)閍，b均為正實(shí)數(shù)，則SKIPIF1<0SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，且SKIPIF1<0，即SKIPIF1<0時(shí)取等號(hào)，則SKIPIF1<0的最大值為SKIPIF1<0．故選：A．20．（2022秋·吉林四平·高三四平市第一高級(jí)中學(xué)?？计谀┮阎獢?shù)列SKIPIF1<0的首項(xiàng)是SKIPIF1<0，前SKIPIF1<0項(xiàng)和為SKIPIF1<0，且SKIPIF1<0，設(shè)SKIPIF1<0，若存在常數(shù)SKIPIF1<0，使不等式SKIPIF1<0恒成立，則SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】首先由數(shù)列通項(xiàng)與前SKIPIF1<0項(xiàng)和的關(guān)系得到數(shù)列SKIPIF1<0的遞推關(guān)系SKIPIF1<0，再構(gòu)造等比數(shù)列SKIPIF1<0，求數(shù)列SKIPIF1<0的通項(xiàng)公式，進(jìn)一步求出數(shù)列SKIPIF1<0的通項(xiàng)公式，從而可求數(shù)列SKIPIF1<0通項(xiàng)公式，代入所求式子SKIPIF1<0，分子、分母同除以SKIPIF1<0構(gòu)造基本不等式即可求出SKIPIF1<0的最大值，從而求出SKIPIF1<0的范圍.【詳解】由SKIPIF1<0，則當(dāng)SKIPIF1<0時(shí)，得SKIPIF1<0，兩式相減得SKIPIF1<0，變形可得：SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0，所以SKIPIF1<0，SKIPIF1<0，∴數(shù)列SKIPIF1<0是以SKIPIF1<0為首項(xiàng)、SKIPIF1<0為公比的等比數(shù)列，故SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立，故SKIPIF1<0.故選：C.【點(diǎn)睛】關(guān)鍵點(diǎn)點(diǎn)睛：構(gòu)造等比數(shù)列SKIPIF1<0求SKIPIF1<0的通項(xiàng)公式，即可得SKIPIF1<0通項(xiàng)公式，再由不等式恒成立，結(jié)合基本不等式求SKIPIF1<0的最值，即可求參數(shù)范圍.21．（2023·全國(guó)·高三專(zhuān)題練習(xí)）設(shè)正實(shí)數(shù)SKIPIF1<0滿(mǎn)足SKIPIF1<0，不等式SKIPIF1<0恒成立，則SKIPIF1<0的最大值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】設(shè)SKIPIF1<0，求出SKIPIF1<0的值，代入SKIPIF1<0中化簡(jiǎn)，利用基本不等式求出結(jié)果.【詳解】設(shè)SKIPIF1<0，則SKIPIF1<0所以SKIPIF1<0SKIPIF1<0當(dāng)且僅當(dāng)SKIPIF1<0即SKIPIF1<0時(shí)取等號(hào)所以SKIPIF1<0的最小值是SKIPIF1<0，則SKIPIF1<0的最大值為SKIPIF1<0.故選A【點(diǎn)睛】本題考查基本不等式，解題的關(guān)鍵是設(shè)SKIPIF1<0，得出SKIPIF1<0進(jìn)行代換，屬于偏難題目.題型八：基本不等式中1的秒用22．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．13 B．19 C．21 D．27【答案】D【分析】利用基本不等式“1”的妙用求最小值.【詳解】SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0，b＝6時(shí)，等號(hào)成立，故SKIPIF1<0的最小值為27故選：D23．（2022秋·廣東深圳·高三深圳市南山區(qū)華僑城中學(xué)?？茧A段練習(xí)）已知a，b為正實(shí)數(shù)，直線(xiàn)SKIPIF1<0與曲線(xiàn)SKIPIF1<0相切，則SKIPIF1<0的最小值為（

）A．8 B．9 C．10 D．13【答案】B【分析】設(shè)切點(diǎn)為SKIPIF1<0,求函數(shù)的導(dǎo)數(shù)，由已知切線(xiàn)的方程，可得切線(xiàn)的斜率，求得切點(diǎn)的坐標(biāo)，可得SKIPIF1<0,再由乘1法結(jié)合基本不等式，即可得到所求最小值．【詳解】設(shè)切點(diǎn)為SKIPIF1<0，SKIPIF1<0的導(dǎo)數(shù)為SKIPIF1<0，由切線(xiàn)的方程SKIPIF1<0可得切線(xiàn)的斜率為1，令SKIPIF1<0，則SKIPIF1<0，故切點(diǎn)為SKIPIF1<0，代入SKIPIF1<0，得SKIPIF1<0，SKIPIF1<0、SKIPIF1<0為正實(shí)數(shù)，則SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，SKIPIF1<0時(shí)，SKIPIF1<0取得最小值9，故選：B24．（2022秋·福建泉州·高三福建省南安國(guó)光中學(xué)校考階段練習(xí)）在SKIPIF1<0中，點(diǎn)SKIPIF1<0滿(mǎn)足SKIPIF1<0，過(guò)點(diǎn)SKIPIF1<0的直線(xiàn)與SKIPIF1<0，SKIPIF1<0所在的直線(xiàn)分別交于點(diǎn)SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．3 B．SKIPIF1<0 C．1 D．SKIPIF1<0【答案】A【分析】由向量加減的幾何意義可得SKIPIF1<0，結(jié)合已知有SKIPIF1<0，根據(jù)三點(diǎn)共線(xiàn)知SKIPIF1<0，應(yīng)用基本不等式“1”的代換即可求最值，注意等號(hào)成立的條件.【詳解】由題設(shè)，如下圖示：SKIPIF1<0，又SKIPIF1<0，SKIPIF1<0，∴SKIPIF1<0，由SKIPIF1<0三點(diǎn)共線(xiàn)，有SKIPIF1<0，∴SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號(hào)成立.故選：A【點(diǎn)睛】關(guān)鍵點(diǎn)點(diǎn)睛：利用向量線(xiàn)性運(yùn)算的幾何表示，得到SKIPIF1<0、SKIPIF1<0、SKIPIF1<0的線(xiàn)性關(guān)系，根據(jù)三點(diǎn)共線(xiàn)有SKIPIF1<0，再結(jié)合基本不等式求最值.題型九：條件等式求最值25．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】A【分析】由已知可得SKIPIF1<0，將SKIPIF1<0展開(kāi)利用基本不等式即可求得最小值.【詳解】由SKIPIF1<0可得SKIPIF1<0，所以SKIPIF1<0，因?yàn)镾KIPIF1<0，SKIPIF1<0，則SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0即SKIPIF1<0時(shí)等號(hào)成立，所以SKIPIF1<0的最小值為SKIPIF1<0，故選：A.26．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知a，SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的最大值為（

）A．2 B．3 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】由題知SKIPIF1<0，進(jìn)而得SKIPIF1<0，再結(jié)合已知得SKIPIF1<0，即可得答案.【詳解】解：SKIPIF1<0，則SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，“=”成立，又a，SKIPIF1<0，所以SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，“=”成立，所以SKIPIF1<0的最大值為SKIPIF1<0.故選：C27．（2023·全國(guó)·高三專(zhuān)題練習(xí)）設(shè)SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．2 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】依題意可得SKIPIF1<0，利用基本不等式計(jì)算可得；【詳解】解：因?yàn)镾KIPIF1<0，SKIPIF1<0，且SKIPIF1<0，所以SKIPIF1<0，所以SKIPIF1<0當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0，SKIPIF1<0或SKIPIF1<0時(shí)取等號(hào)；故選：D題型十：對(duì)勾函數(shù)求最值28．（2023·全國(guó)·高三專(zhuān)題練習(xí)）在銳角SKIPIF1<0中，角A，B，C的對(duì)邊分別為a，b，c，SKIPIF1<0的面積為S，若SKIPIF1<0，則SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】由面積公式與正余弦定理化簡(jiǎn)后得出SKIPIF1<0關(guān)系后求解【詳解】在SKIPIF1<0中，SKIPIF1<0，故題干條件可化為SKIPIF1<0，由余弦定理得SKIPIF1<0，故SKIPIF1<0，又由正弦定理化簡(jiǎn)得：SKIPIF1<0，整理得SKIPIF1<0，故SKIPIF1<0或SKIPIF1<0（舍去），得SKIPIF1<0SKIPIF1<0為銳角三角形，故SKIPIF1<0，解得SKIPIF1<0，故SKIPIF1<0SKIPIF1<0故選：C29．（2022·天津?qū)氎妗ぬ旖蚴袑氎鎱^(qū)第一中學(xué)?？级＃┫铝薪Y(jié)論正確的是（

）A．當(dāng)SKIPIF1<0且SKIPIF1<0時(shí)，SKIPIF1<0 B．SKIPIF1<0的最大值是2C．SKIPIF1<0的最小值是2 D．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0【答案】D【分析】A、B選項(xiàng)取特殊值判斷即可；C選項(xiàng)基本不等式取等的件不成立；D選項(xiàng)由雙勾函數(shù)的單調(diào)性即可判斷.【詳解】A選項(xiàng)：令SKIPIF1<0，顯然SKIPIF1<0，故A錯(cuò)誤；B選項(xiàng)：令SKIPIF1<0，顯然SKIPIF1<0，故B錯(cuò)誤；C選項(xiàng)：SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0取等，顯然SKIPIF1<0無(wú)解，即SKIPIF1<0不能等于2，故C錯(cuò)誤；D選項(xiàng)：令SKIPIF1<0，由雙勾函數(shù)知SKIPIF1<0在SKIPIF1<0單減，即SKIPIF1<0時(shí)取得最小值5，即SKIPIF1<0，故D正確.故選：D.題型十一：基本不等式恒成立問(wèn)題30．（2022·四川綿陽(yáng)·四川省綿陽(yáng)江油中學(xué)?？寄M預(yù)測(cè)）已知圓SKIPIF1<0與圓SKIPIF1<0(SKIPIF1<0是正實(shí)數(shù))相交于SKIPIF1<0兩點(diǎn)，SKIPIF1<0為坐標(biāo)原點(diǎn).當(dāng)SKIPIF1<0的面積最大時(shí)，則SKIPIF1<0的最小值是（

）A．SKIPIF1<0 B．8 C．7 D．SKIPIF1<0【答案】B【分析】由相交兩圓的方程，求出直線(xiàn)AB方程，SKIPIF1<0最大時(shí)SKIPIF1<0為直角，由點(diǎn)直線(xiàn)距離求出m，n的關(guān)系，利用函數(shù)單調(diào)性即可得解.【詳解】因圓SKIPIF1<0與圓SKIPIF1<0相交，則直線(xiàn)AB方程為：SKIPIF1<0，又|OA|=|OB|=1，則SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0取“=”，即SKIPIF1<0為等腰直角三角形，點(diǎn)O到直線(xiàn)AB的距離為SKIPIF1<0，則SKIPIF1<0，SKIPIF1<0，而SKIPIF1<0是正實(shí)數(shù)，則SKIPIF1<0，即SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取“=”，SKIPIF1<0令函數(shù)SKIPIF1<0，則SKIPIF1<0，f(x)在SKIPIF1<0上遞減，SKIPIF1<0，所以SKIPIF1<0的最小值是8.故選：B【點(diǎn)睛】方法點(diǎn)睛：圓的弦長(zhǎng)的常用求法：(1)幾何法：求圓的半徑為r，弦心距為d，弦長(zhǎng)為l，則SKIPIF1<0；(2)代數(shù)方法：運(yùn)用根與系數(shù)的關(guān)系及弦長(zhǎng)公式：SKIPIF1<0.31．（2023·上?！じ呷龑?zhuān)題練習(xí)）已知P是曲線(xiàn)SKIPIF1<0上的一動(dòng)點(diǎn)，曲線(xiàn)C在P點(diǎn)處的切線(xiàn)的傾斜角為SKIPIF1<0，若SKIPIF1<0，則實(shí)數(shù)a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】D【分析】對(duì)函數(shù)求導(dǎo)，利用導(dǎo)數(shù)的幾何意義以及給定傾斜角的范圍，轉(zhuǎn)化為恒成立問(wèn)題求解a的范圍即可.【詳解】因?yàn)镾KIPIF1<0，所以SKIPIF1<0，因?yàn)榍€(xiàn)在M處的切線(xiàn)的傾斜角SKIPIF1<0，所以SKIPIF1<0對(duì)于任意的SKIPIF1<0恒成立，即SKIPIF1<0對(duì)任意SKIPIF1<0恒成立，即SKIPIF1<0，又SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0，即SKIPIF1<0時(shí)，等號(hào)成立，故SKIPIF1<0，所以a的取值范圍是SKIPIF1<0．故選：D．32．（2021秋·河南濮陽(yáng)·高三濮陽(yáng)外國(guó)語(yǔ)學(xué)校?？茧A段練習(xí)）若對(duì)任意正數(shù)SKIPIF1<0，不等式SKIPIF1<0恒成立，則實(shí)數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】原不等式即SKIPIF1<0，再利用基本不等式求得SKIPIF1<0的最大值，可得SKIPIF1<0的范圍．【詳解】依題意得，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0恒成立，又因?yàn)镾KIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào)，所以，SKIPIF1<0的最大值為SKIPIF1<0，所以SKIPIF1<0，解得SKIPIF1<0的取值范圍為SKIPIF1<0．故選：B33．（2022·山西朔州·統(tǒng)考三模）若存在實(shí)數(shù)x，y，使得SKIPIF1<0成立，且對(duì)任意a，SKIPIF1<0，SKIPIF1<0，則實(shí)數(shù)t的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】B【分析】根據(jù)不等式組有解，得出SKIPIF1<0的一個(gè)范圍，利用基本不等式得出SKIPIF1<0的又一個(gè)范圍，兩者的公共部分即為所求．【詳解】SKIPIF1<0的解為SKIPIF1<0，若存在實(shí)數(shù)x，y，使得SKIPIF1<0成立，則SKIPIF1<0應(yīng)滿(mǎn)足SKIPIF1<0，即SKIPIF1<0，所以SKIPIF1<0，又SKIPIF1<0，所以SKIPIF1<0，所以t的取值范圍是SKIPIF1<0，故選：B．【高考必刷】一、單選題34．（2023·云南曲靖·統(tǒng)考一模）若SKIPIF1<0,則在“函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0”的條件下,“函數(shù)SKIPIF1<0為奇函數(shù)”的概率為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【分析】先列出所有的結(jié)果數(shù),由于函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0,則SKIPIF1<0,SKIPIF1<0恒成立,可得SKIPIF1<0,在所有結(jié)果數(shù)中選出滿(mǎn)足的情況,求出概率,根據(jù)SKIPIF1<0為奇函數(shù)可得SKIPIF1<0或SKIPIF1<0,在所有結(jié)果數(shù)中選出同時(shí)滿(mǎn)足兩個(gè)事件情況,求出其概率,再根據(jù)條件概率的計(jì)算公式即可計(jì)算出結(jié)果.【詳解】解:用所有的有序數(shù)對(duì)SKIPIF1<0表示滿(mǎn)足SKIPIF1<0的結(jié)果，則所有的情況為:SKIPIF1<0,共9種,記“函數(shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0”為事件A,因?yàn)楹瘮?shù)SKIPIF1<0的定義域?yàn)镾KIPIF1<0,所以SKIPIF1<0,SKIPIF1<0恒成立,即SKIPIF1<0,即SKIPIF1<0,其中滿(mǎn)足SKIPIF1<0的基本事件有:SKIPIF1<0共6種,故SKIPIF1<0．記“函數(shù)SKIPIF1<0為奇函數(shù)”為事件B．已知SKIPIF1<0是奇函數(shù),且定義域?yàn)镾KIPIF1<0,則SKIPIF1<0,即SKIPIF1<0,即SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0．滿(mǎn)足SKIPIF1<0或SKIPIF1<0的情況有SKIPIF1<0共3種,所以,即同時(shí)滿(mǎn)足事件A和事件B的情況有SKIPIF1<0共3種,故SKIPIF1<0,所以SKIPIF1<0.故選:C3