新高考數(shù)學(xué)二輪復(fù)習(xí)易錯(cuò)點(diǎn)09 不等式（原卷版）

易錯(cuò)點(diǎn)09不等式易錯(cuò)題【01】利用同向相加求范圍出錯(cuò)利用同向相加求變量或式子的取值范圍,是最常用的方法,但如果多次使用不等式的可加性,變量或式子中的等號(hào)可能不會(huì)同時(shí)取到,會(huì)導(dǎo)致范圍擴(kuò)大.易錯(cuò)題【02】解分?jǐn)?shù)不等式忽略分母不為零解含有分?jǐn)?shù)的不等式,在去分母時(shí)要注意分母不為零的限制條件,防止出現(xiàn)增解,如SKIPIF1<0.易錯(cuò)題【03】連續(xù)使用均值不等式忽略等號(hào)能否同時(shí)成立連續(xù)使用均值不等式求最值或范圍,要注意判斷每個(gè)等號(hào)成立的條件,檢驗(yàn)等號(hào)能否同時(shí)成立.易錯(cuò)題【04】混淆單變量與雙變量(1)SKIPIF1<0恒成立SKIPIF1<0的最小值大于零；(2)SKIPIF1<0恒成立SKIPIF1<0；(3)SKIPIF1<0SKIPIF1<0使得SKIPIF1<0成立SKIPIF1<0的最大值大于零；(4)SKIPIF1<0使得SKIPIF1<0恒成立SKIPIF1<0；易錯(cuò)題【05】解含有參數(shù)的不等式分類(lèi)不當(dāng)致誤(1)解含有參數(shù)的不等式要注意判斷是否需要對(duì)參數(shù)進(jìn)行分類(lèi)討論,分類(lèi)要滿足互斥、無(wú)漏、最簡(jiǎn).(2)解形如SKIPIF1<0的不等式,首先要對(duì)SKIPIF1<0的符號(hào)進(jìn)行討論,當(dāng)a的符號(hào)確定后再根據(jù)判別式的符號(hào)或兩根的大小進(jìn)行討論. 01設(shè)f(x)＝ax2＋bx,若1≤f(－1)≤2,2≤f(1)≤4,則f(－2)的取值范圍是________．1.已知實(shí)數(shù)x,y滿足SKIPIF1<0,SKIPIF1<0,則()A．1≤x≤3 B．SKIPIF1<02≤y≤1 C．2≤4x+y≤15 D．SKIPIF1<0xSKIPIF1<0ySKIPIF1<02.已知SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0的取值范圍是()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0 02解不等式SKIPIF1<0..1．設(shè)集合SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02.設(shè)SKIPIF1<0,那么“SKIPIF1<0”是“SKIPIF1<0”的()A．充分不必要條件 B．必要不充分條件C．充分必要條件 D．既不充分也不必要條件 03已知x>0,y>0,且eq\f(1,x)＋eq\f(2,y)＝1,則x＋y的最小值是________．1.(2022屆遼寧省東北育才學(xué)校高三上學(xué)期模擬)圓SKIPIF1<0關(guān)于直線SKIPIF1<0對(duì)稱(chēng),則SKIPIF1<0的最小值是()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02.(2022屆河南省名校大聯(lián)考高三上學(xué)期期中)已知正實(shí)數(shù)SKIPIF1<0,SKIPIF1<0,SKIPIF1<0滿足SKIPIF1<0,則當(dāng)SKIPIF1<0與SKIPIF1<0同時(shí)取得最大值時(shí),SKIPIF1<0()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0 04已知SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0,(1)若對(duì)任意SKIPIF1<0,恒有SKIPIF1<0,求實(shí)數(shù)SKIPIF1<0的取值范圍；(2)若對(duì)任意SKIPIF1<0,恒有SKIPIF1<0,求實(shí)數(shù)SKIPIF1<0的取值范圍；1．已知SKIPIF1<0SKIPIF1<0,在區(qū)間SKIPIF1<0上存在三個(gè)不同的實(shí)數(shù)SKIPIF1<0,使得以SKIPIF1<0為邊長(zhǎng)的三角形是直角三角形,則SKIPIF1<0的取值范圍是A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<02.已知函數(shù)SKIPIF1<0是定義域上的奇函數(shù),且SKIPIF1<0.(1)求函數(shù)SKIPIF1<0的解析式,判斷函數(shù)SKIPIF1<0在SKIPIF1<0上的單調(diào)性并證明；(2)令SKIPIF1<0,若對(duì)任意SKIPIF1<0都有SKIPIF1<0,求實(shí)數(shù)SKIPIF1<0的取值范圍. 05解關(guān)于x的不等式ax2－(a＋1)x＋1<0.1．已知函數(shù)SKIPIF1<0．(1)當(dāng)SKIPIF1<0時(shí),求SKIPIF1<0在SKIPIF1<0上的值域；(2)當(dāng)SKIPIF1<0時(shí),解關(guān)于SKIPIF1<0的不等式SKIPIF1<0．2.設(shè)函數(shù)SKIPIF1<0.(1)若關(guān)于SKIPIF1<0的不等式SKIPIF1<0有實(shí)數(shù)解,求實(shí)數(shù)SKIPIF1<0的取值范圍；(2)若不等式SKIPIF1<0對(duì)于實(shí)數(shù)SKIPIF1<0時(shí)恒成立,求實(shí)數(shù)SKIPIF1<0的取值范圍；(3)解關(guān)于SKIPIF1<0的不等式：SKIPIF1<0.錯(cuò)1.已知SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0是SKIPIF1<0的什么條件()A．既不充分又不必要條件 B．充要條件C．必要不充分條件 D．充分不必要條件2．(2021屆海南熱帶海洋學(xué)院附屬中學(xué)高三月考)關(guān)于SKIPIF1<0的不等式SKIPIF1<0對(duì)SKIPIF1<0恒成立的一個(gè)必要不充分條件是()A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0或SKIPIF1<0 D．SKIPIF1<0或SKIPIF1<03．(2022屆重慶市第一中學(xué)高三上學(xué)期期中)若SKIPIF1<0,則SKIPIF1<0的最小值為()A．SKIPIF1<0 B．1 C．2 D．44．已知實(shí)數(shù)SKIPIF1<0滿足SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0的取值范圍是()A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<05．(2021屆浙江省紹興市高三上學(xué)期測(cè)試)已知SKIPIF1<0,不等式SKIPIF1<0在SKIPIF1<0上恒成立,則()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<06．已知函數(shù)SKIPIF1<0,若對(duì)任意SKIPIF1<0,都有SKIPIF1<0,則實(shí)數(shù)m的取值范圍是()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<07．(多選題)當(dāng)SKIPIF1<0,SKIPIF1<0,SKIPIF1<0時(shí),SKIPIF1<0恒成立,則SKIPIF1<0的取值可能是()A．SKIPIF1<0 B．SKIPIF1<0 C．1 D．28．已知SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0的最大值為_(kāi)__________.9．函數(shù)SKIPIF1<0,a為參數(shù),(1)解關(guān)于x的不等式SKIPIF1<0；(2)當(dāng)SKIPIF1<0,SKIPIF1<0最大值為M,最小值為m,若SKIPIF1<0,求參數(shù)a的取值范圍；(3)若SKIPIF1<0在區(qū)間SKIPIF1<0上滿足SKIPIF1<0有兩解,求a的取值范圍10．已知關(guān)于x的不等式(kx－k2－4)(x－4)＞0,其中k∈R.(1)當(dāng)k變化時(shí),試求不等式的解集A；(2)對(duì)于不等式的解集A,若滿足A∩Z＝B(其中Z為整數(shù)集)．試探究集合B能否為有限集？若能,求出使得集合B中元素個(gè)數(shù)最少的k的所有取值,并用列舉法表示集合B；若不能,請(qǐng)說(shuō)明理由．11．已知函數(shù)SKIPIF1<0(SK