新高考數(shù)學(xué)一輪復(fù)習第1章 第04講 一元二次函數(shù)（方程不等式）精講+精練（學(xué)生版）

第04講一元二次函數(shù)（方程，不等式）(精講+精練）目錄第一部分：思維導(dǎo)圖（總覽全局）第二部分：知識點精準記憶第三部分：課前自我評估測試第四部分：典型例題剖析高頻考點一：一元二次（分式）不等式解法（不含參）高頻考點二：一元二次不等式解法（含參）高頻考點三：一元二次不等式與相應(yīng)的二次函數(shù)（方程）的關(guān)系高頻考點四：一元二次不等式恒成立問題①SKIPIF1<0上恒成立（優(yōu)選SKIPIF1<0法）②SKIPIF1<0上恒成立（優(yōu)選SKIPIF1<0法）③SKIPIF1<0上恒成立（優(yōu)選分離變量法）④SKIPIF1<0上恒成立（優(yōu)選分離變量法）⑤已知參數(shù)SKIPIF1<0，求SKIPIF1<0取值范圍（優(yōu)選變更主元法）高頻考點五：一元二次不等式的應(yīng)用第五部分：高考真題感悟第六部分：第04講一元二次函數(shù)（方程，不等式）（精練）第一部分：思維導(dǎo)圖總覽全局第一部分：思維導(dǎo)圖總覽全局第二部分：知識點精準記憶第二部分：知識點精準記憶1、二次函數(shù)（1）形式：形如SKIPIF1<0的函數(shù)叫做二次函數(shù).（2）特點：①函數(shù)SKIPIF1<0的圖象與SKIPIF1<0軸交點的橫坐標是方程SKIPIF1<0的實根.②當SKIPIF1<0且SKIPIF1<0(SKIPIF1<0)時，恒有SKIPIF1<0(SKIPIF1<0)；當SKIPIF1<0且SKIPIF1<0(SKIPIF1<0)時，恒有SKIPIF1<0(SKIPIF1<0).2、一元二次不等式只含有一個未知數(shù)，并且未知數(shù)的最高次數(shù)是2的不等式，稱為一元二次不等式.3.SKIPIF1<0或SKIPIF1<0型不等式的解集不等式解集SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<04、一元二次不等式與相應(yīng)的二次函數(shù)及一元二次方程的關(guān)系判別式SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0二次函數(shù)SKIPIF1<0的圖象一元二次方程SKIPIF1<0的根有兩相異實數(shù)根SKIPIF1<0，SKIPIF1<0(SKIPIF1<0)有兩相等實數(shù)根SKIPIF1<0沒有實數(shù)根一元二次不等式SKIPIF1<0的解集SKIPIF1<0SKIPIF1<0SKIPIF1<0一元二次不等式SKIPIF1<0的解集SKIPIF1<0SKIPIF1<0SKIPIF1<05、分式不等式解法（1）SKIPIF1<0（2）SKIPIF1<0（3）SKIPIF1<0（4）SKIPIF1<06、單絕對值不等式（1）SKIPIF1<0（2）SKIPIF1<0第三部分：課前自我評估測試第三部分：課前自我評估測試一、判斷題1．（2022·全國·高三專題練習）已知函數(shù)SKIPIF1<0，關(guān)于x的不等式SKIPIF1<0的解集為SKIPIF1<0，則SKIPIF1<0.___________（判斷對錯）二、單選題1．（2022·貴州畢節(jié)·高一期末）已知不等式SKIPIF1<0的解集為SKIPIF1<0，則a，b的值是（

）A．SKIPIF1<0，SKIPIF1<0 B．SKIPIF1<0，SKIPIF1<0 C．6，3 D．3，62．（2022·江西南昌·一模（理））已知集合SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．（2022·陜西西安·高二期末（文））若關(guān)于SKIPIF1<0的一元二次不等式SKIPIF1<0的解集為SKIPIF1<0，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<04．（2022·廣東珠海·高一期末）已知關(guān)于SKIPIF1<0的不等式SKIPIF1<0的解集是SKIPIF1<0，則SKIPIF1<0的值是（

）A．SKIPIF1<0 B．2 C．22 D．SKIPIF1<05．（2022·寧夏·高三階段練習（文））已知集合SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0第四部分：典型例題剖析第四部分：典型例題剖析高頻考點一：一元二次（分式）不等式解法（不含參）1．（2022·河北·模擬預(yù)測）已知集合SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（2022·湖南·高一課時練習）下面四個不等式中解集為空集的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<03．（2022·河南·信陽高中高一期末（理））設(shè)集合SKIPIF1<0，N=x∈Zx2?12x?5≤0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<04．（2022·河南南陽·高二期末（文））不等式SKIPIF1<0的一個必要不充分條件是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05．（2022·河南洛陽·高二期末（文））不等式SKIPIF1<0的解集為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<06．（2022·全國·高三專題練習）設(shè)集合SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0高頻考點二：一元二次不等式解法（含參）一元二次不等式解法（含參問題）談?wù)撊瓌t：①最高項系數(shù)含參，從參數(shù)等于0開始討論；如：SKIPIF1<0，最高項系數(shù)為SKIPIF1<0討論時，從SKIPIF1<0開始討論.②兩根大小不確定，從兩根相等開始討論；如SKIPIF1<0兩根分別為：SKIPIF1<0，SKIPIF1<0，討論時從SKIPIF1<0開始討論③根是否在定義域內(nèi)：如SKIPIF1<0此時兩根SKIPIF1<0，SKIPIF1<0，討論時注意SKIPIF1<0（舍去）1．（2022·北京·清華附中高一期末）求下列關(guān)于SKIPIF1<0的不等式的解集：SKIPIF1<02．（2022·河北唐山·高一期末）已知關(guān)于x的不等式：SKIPIF1<0．(1)當SKIPIF1<0時，解此不等式；(2)當SKIPIF1<0時，解此不等式．

3．（2022·福建·莆田第二十五中學(xué)高一期末）解關(guān)于SKIPIF1<0的不等式SKIPIF1<0.4．（2022·全國·高三專題練習）解關(guān)于SKIPIF1<0的不等式：SKIPIF1<0.高頻考點三：一元二次不等式與相應(yīng)的二次函數(shù)（方程）的關(guān)系1．（2021·甘肅·甘南藏族自治州合作第一中學(xué)高一期中）若不等式SKIPIF1<0的解集為[-1，2]，則SKIPIF1<0=（

）A．-2 B．-1 C．1 D．22．（2021·四川省南充高級中學(xué)高二開學(xué)考試（理））已知不等式SKIPIF1<0的解集為SKIPIF1<0，則SKIPIF1<0___________.3．（2022·天津市紅橋區(qū)教師發(fā)展中心高一期末）若函數(shù)SKIPIF1<0的兩個零點是2和3，則不等式SKIPIF1<0的解集為________．4．（2022·上海閔行·高一期末）已知SKIPIF1<0、SKIPIF1<0，關(guān)于SKIPIF1<0的不等式SKIPIF1<0的解集為SKIPIF1<0，則SKIPIF1<0___________.5．（2022·上海·曹楊二中高一期末）已知a為常數(shù)，若關(guān)于x的不等式SKIPIF1<0的解集為SKIPIF1<0，則SKIPIF1<0______．高頻考點四：一元二次不等式恒成立問題①SKIPIF1<0上恒成立二次型+SKIPIF1<0（范圍）優(yōu)選SKIPIF1<0法（注意最高項系數(shù)含參數(shù)，從0開始討論）1．（2022·福建寧德·高一期末）SKIPIF1<0不等式SKIPIF1<0恒成立，則SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0或SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<02．（2022·全國·高三專題練習）已知SKIPIF1<0，“SKIPIF1<0對SKIPIF1<0恒成立”的一個充要條件是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．（2021·吉林·汪清縣汪清第四中學(xué)高一階段練習）若不等式SKIPIF1<0對任意SKIPIF1<0恒成立，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0或SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．（2021·全國·高一課時練習）若不等式SKIPIF1<0對任意SKIPIF1<0均成立，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05．（2020·河北省尚義縣第一中學(xué)高一期中）若命題SKIPIF1<0為真命題，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0或SKIPIF1<0②SKIPIF1<0上恒成立二次型+SKIPIF1<0（范圍）優(yōu)選SKIPIF1<0法（注意最高項系數(shù)含參數(shù)，從0開始討論）1．（2022·河南·新蔡縣第一高級中學(xué)高二階段練習（文））如果“SKIPIF1<0，使SKIPIF1<0.”是真命題，那么實數(shù)a的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（2020·寧夏·隆德縣中學(xué)高三階段練習（理））已知命題“SKIPIF1<0，SKIPIF1<0”是真命題，則實數(shù)SKIPIF1<0的取值范圍（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0）D．SKIPIF1<0SKIPIF1<03．（2022·江蘇南通·高一期末）若命題“SKIPIF1<0”是真命題，則實數(shù)SKIPIF1<0的取值范圍是（

）A．（﹣∞，1） B．（﹣∞，1] C．（1，+∞） D．[1，+∞）4．（2021·山西·朔州市平魯區(qū)李林中學(xué)高一階段練習）若命題“SKIPIF1<0”是真命題，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05．（2021·天津·耀華中學(xué)高一期中）若命題“SKIPIF1<0，使得不等式SKIPIF1<0”成立，則實數(shù)SKIPIF1<0的取值集合是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0③SKIPIF1<0上恒成立（優(yōu)選分離變量法）SKIPIF1<0SKIPIF1<01．（2022·海南·嘉積中學(xué)高一階段練習）對任意的SKIPIF1<0，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<02．（2021·河北·石家莊市藁城區(qū)第一中學(xué)高三開學(xué)考試）若關(guān)于SKIPIF1<0的不等式SKIPIF1<0在SKIPIF1<0上恒成立，則實數(shù)SKIPIF1<0的取值范圍為SKIPIF1<0SKIPIF1<0A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．（2021·福建·泉州市第六中學(xué)高一期中）已知關(guān)于SKIPIF1<0的不等式SKIPIF1<0對任意SKIPIF1<0恒成立，則有（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．（2021·黑龍江·雞西市第一中學(xué)校高一期中）已知函數(shù)SKIPIF1<0，若SKIPIF1<0在SKIPIF1<0上恒成立，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05．（2022·全國·高三專題練習）若對任意的SKIPIF1<0恒成立，則m的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<06．（2022·甘肅張掖·高一期末）設(shè)函數(shù)SKIPIF1<0．(1)若不等式SKIPIF1<0的解集是SKIPIF1<0，求不等式SKIPIF1<0的解集；(2)當SKIPIF1<0時，SKIPIF1<0在SKIPIF1<0上恒成立，求實數(shù)SKIPIF1<0的取值范圍．7．（2021·山東·棗莊市第三中學(xué)高一階段練習）已知函數(shù)SKIPIF1<0（a∈R）.(1)若關(guān)于x的不等式SKIPIF1<0<0的解集為（1，b），求a和b的值;(2)若對任意x∈SKIPIF1<0，SKIPIF1<0恒成立，求實數(shù)a的取值范圍.④SKIPIF1<0上恒成立（優(yōu)選分離變量法）SKIPIF1<0SKIPIF1<01．（2021·河南信陽·高二期中（理））若關(guān)于SKIPIF1<0的不等式SKIPIF1<0在SKIPIF1<0有解，則SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<02．（2021·安徽·池州市第一中學(xué)高一期中）若關(guān)于x的不等式SKIPIF1<0在SKIPIF1<0上有解則實數(shù)m的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．（2021·河南·高二期中（理））已知關(guān)于SKIPIF1<0的不等式SKIPIF1<0在SKIPIF1<0上有解，則實數(shù)SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．（2021·山西·大同一中高一期中）若關(guān)于SKIPIF1<0的不等式SKIPIF1<0在SKIPIF1<0內(nèi)有解，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05．（2021·河北·石家莊市第四十四中學(xué)高一期中）若關(guān)于x的不等式SKIPIF1<0在區(qū)間SKIPIF1<0上有解，則實數(shù)m的取值范圍是__________．6．（2021·福建省龍巖第一中學(xué)高一期中）已知不等式SKIPIF1<0有解，則實數(shù)SKIPIF1<0的取值范圍為__________.7．（2021·河北·石家莊市藁城區(qū)第一中學(xué)高一階段練習）已知函數(shù)SKIPIF1<0；(1)若關(guān)于SKIPIF1<0的不等式SKIPIF1<0的解集為SKIPIF1<0，求實數(shù)SKIPIF1<0的值；(2)存在SKIPIF1<0使得SKIPIF1<0成立，求實數(shù)SKIPIF1<0的取值范圍.⑤已知參數(shù)SKIPIF1<0，求SKIPIF1<0取值范圍（變更主元法）1．（2022·全國·高三專題練習）已知SKIPIF1<0，SKIPIF1<0，不等式SKIPIF1<0恒成立，則SKIPIF1<0的取值范圍為SKIPIF1<0SKIPIF1<0A．SKIPIF1<0，SKIPIF1<0，SKIPIF1<0 B．SKIPIF1<0，SKIPIF1<0，SKIPIF1<0C．SKIPIF1<0，SKIPIF1<0，SKIPIF1<0 D．SKIPIF1<02．（2022·全國·高三專題練習）不等式SKIPIF1<0對一切SKIPIF1<0恒成立，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<03．（2021·全國·高一課時練習）對任意的SKIPIF1<0，函數(shù)SKIPIF1<0的值總大于0，則SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．（2021·江西吉安·高一期中）若不等式SKIPIF1<0對任意SKIPIF1<0成立，則SKIPIF1<0的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0高頻考點五：一元二次不等式的應(yīng)用1．（2021·全國·高一課時練習）某文具店購進一批新型臺燈，每盞的最低售價為15元，若每盞按最低售價銷售，每天能賣出45盞，每盞售價每提高1元，日銷售量將減少3盞，為了使這批臺燈每天獲得600元以上的銷售收入，則這批臺燈的銷售單價x（單位：元）的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（2021·河北·石家莊一中高一階段練習）某城市對一種每件售價為160元的商品征收附加稅，稅率為SKIPIF1<0（即每銷售100元征稅SKIPIF1<0元），若年銷售量為SKIPIF1<0萬件，要使附加稅不少于128萬元，則SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．（2021·全國·高一專題練習）某文具店購進一批新型臺燈，若按每盞臺燈15元的價格銷售，每天能賣出30盞；若售價每提高1元，日銷售量將減少2盞，現(xiàn)決定提價銷售，為了使這批臺燈每天獲得400元以上（不含400元）的銷售收入．則這批臺燈的銷售單價SKIPIF1<0（單位：元）的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<04．（2021·全國·高一課時練習）一服裝廠生產(chǎn)某種風衣，日產(chǎn)量為SKIPIF1<0件時，售價為SKIPIF1<0元/件，每天的總成本為SKIPIF1<0元，且SKIPIF1<0，SKIPIF1<0，要使獲得的日利潤不少于1300元，則SKIPIF1<0的取值范圍為A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<05．（2021·全國·高一專題練習）某小型服裝廠生產(chǎn)一種風衣，日銷售量SKIPIF1<0（件）與單價SKIPIF1<0（元）之間的關(guān)系為SKIPIF1<0，生產(chǎn)SKIPIF1<0件所需成本為SKIPIF1<0（元），其中SKIPIF1<0元，若要求每天獲利不少于1300元，則日銷售量SKIPIF1<0的取值范圍是（

）.A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0第五部分：高考真題感悟第五部分：高考真題感悟1．（2020·山東·高考真題）已知二次函數(shù)SKIPIF1<0的圖像如圖所示，則不等式SKIPIF1<0的解集是（

）SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02．（2019·全國·高考真題（理））設(shè)集合A={x|x2-5x+6>0}，B={x|x-1<0}，則A∩B=A．(-∞，1) B．(-2，1)C．(-3，-1) D．(3，+∞)3．（2017·天津·高考真題（理））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<04．（2019·天津·高考真題（文））設(shè)SKIPIF1<0，使不等式SKIPIF1<0成立的SKIPIF1<0的取值范圍為__________.5．（2018·天津·高考真題（文））已知SKIPIF1<0，函數(shù)SKIPIF1<0若對任意x∈［–3，+SKIPIF1<0），f(x)≤SKIPIF1<0恒成立，則a的取值范圍是__________．第六部分：第六部分：第04講一元二次函數(shù)（方程，不等式）（精練）一、單選題1．（2022·河南濮陽·高二開學(xué)考試（理））若不等式SKIPIF1<0的解集為SKIPIF1<0，則SKIPIF1<0的值分別為（

）A．SKIPIF1<0，SKIPIF1<0 B．SKIPIF1<0，SKIPIF1<0 C．SKIPIF1<0，SKIPIF1<0 D．SKIPIF1<0，SKIPIF1<02．（2022·江蘇南通·高三階段練習）當SKIPIF1<0時，不等式SKIPIF1<0恒成立，則實數(shù)SKIPIF1<0的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<03．（2022·吉林·農(nóng)安縣教師進修學(xué)校高一期末）不等式SKIPIF1<0對一切SKIPIF1<0恒成立，則實數(shù)a的取值范圍是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<04．（2022·河南焦作·高二期末（理））若存在SKIPIF1<0，使得不等式SKIPIF1<0成立，則實數(shù)k的取值范圍為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05．（2022·河南·高一階段練習）已知關(guān)于SKIPIF1<0的不等式SKIPIF1<0的解集為SKIPIF1<0，則下列結(jié)論錯誤的是（

）A．SKIPIF1<0 B．a(chǎn)b的最大值為SKIPIF1<0C．SKIPIF1<0的最小值為4 D．SKIPIF1<0的最小值為SKIPIF1<06．（2022·浙江·安吉縣高級中學(xué)高一開學(xué)考試）已知關(guān)于SKIPIF1<0的不等式SKIPIF1<0的解集為SKIPIF1<0或SKIPIF1<0，則下列說法正確的是（

）A．SKIPIF1<0 B．不等式SKIPIF1<0的解集為SKIPIF1<0C．SKIPIF1<0 D．不等式SKIPIF1<0的解集為SKIPIF1<07．（2022·江蘇南京·高一期末）已知SKIPIF1<0