新高考數(shù)學(xué)二輪復(fù)習(xí)易錯(cuò)題專練易錯(cuò)點(diǎn)09 不等式（含解析）

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可能不會同時(shí)取到,會導(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)等號成立的條件,檢驗(yàn)等號能否同時(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參數(shù)進(jìn)行分類討論,分類要滿足互斥、無漏、最簡.(2)解形如SKIPIF1<0的不等式,首先要對SKIPIF1<0的符號進(jìn)行討論,當(dāng)a的符號確定后再根據(jù)判別式的符號或兩根的大小進(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________．【警示】本題常見的錯(cuò)誤解法是：由已知得eq\b\lc\{\rc\(\a\vs4\al\co1(1≤a－b≤2，①,2≤a＋b≤4，②))①＋②得3≤2a≤6,∴6≤4a≤12,又由①可得－2≤－a＋b≤－1,③②＋③得0≤2b≤3,∴－3≤－2b≤0,又f(－2)＝4a－2b,∴3≤4a－2b≤12,∴f(－2)的取值范圍是[3,12]．【答案】SKIPIF1<0【問診】正確解法是：由SKIPIF1<0得SKIPIF1<0∴f(－2)＝4a－2b＝3f(－1)＋f(1)．又∵1≤f(－1)≤2,2≤f(1)≤4,∴5≤3f(－1)＋f(1)≤10,故5≤f(－2)≤10.【叮囑】在求式子的范圍時(shí),如果多次使用不等式的可加性,式子中的等號不能同時(shí)取到,會導(dǎo)致范圍擴(kuò)大.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<0【答案】C【解析】∵SKIPIF1<0,SKIPIF1<0,∴兩式相加,得SKIPIF1<0,即1≤x≤4,故A錯(cuò)誤；∵SKIPIF1<0,∴SKIPIF1<0,解得SKIPIF1<0,故B錯(cuò)誤；∵SKIPIF1<0,又SKIPIF1<0,∴SKIPIF1<0,故C正確；∵SKIPIF1<0,又SKIPIF1<0且SKIPIF1<0,∴SKIPIF1<0,故D錯(cuò)誤.故選C.2.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【答案】C【解析】SKIPIF1<0.設(shè)SKIPIF1<0,所以SKIPIF1<0,解得：SKIPIF1<0,SKIPIF1<0,因?yàn)镾KIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,因?yàn)镾KIPIF1<0單調(diào)遞增,所以SKIPIF1<0.故選C. 02解不等式SKIPIF1<0.【警示】本題易錯(cuò)之處是誤以為SKIPIF1<0SKIPIF1<0.【問診】SKIPIF1<0SKIPIF1<0,所以SKIPIF1<0的解集為SKIPIF1<0.【叮囑】SKIPIF1<0,SKIPIF1<0且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<0【答案】D【解析】集合SKIPIF1<0,SKIPIF1<0SKIPIF1<0,故選D.2.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【答案】B【解析】由不等式SKIPIF1<0,可得SKIPIF1<0,解得SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0不一定成立,即充分性不成立；當(dāng)SKIPIF1<0時(shí),SKIPIF1<0成立,即必要性成立,所以“SKIPIF1<0”是“SKIPIF1<0”的必要不充分條件.故選B. 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________．【警示】本題錯(cuò)誤解法是：∵x>0,y>0,∴1＝eq\f(1,x)＋eq\f(2,y)≥2eq\r(\f(2,xy)),∴eq\r(xy)≥2eq\r(2),∴x＋y≥2eq\r(xy)＝4eq\r(2),∴x＋y的最小值為4eq\r(2).【答案】3＋2eq\r(2)【問診】eq\f(1,x)＋eq\f(2,y)≥2eq\r(\f(2,xy))取等號的條件是SKIPIF1<0,即SKIPIF1<0,x＋y≥2eq\r(xy)取等號的條件是SKIPIF1<0與SKIPIF1<0矛盾.正確解法為：∵x>0,y>0,∴x＋y＝(x＋y)(eq\f(1,x)＋eq\f(2,y))＝3＋eq\f(y,x)＋eq\f(2x,y)≥3＋2eq\r(2)(當(dāng)且僅當(dāng)y＝eq\r(2)x時(shí)取等號),∴當(dāng)x＝eq\r(2)＋1,y＝2＋eq\r(2)時(shí),(x＋y)min＝3＋2eq\r(2).【叮囑】多次使用基本不等式要驗(yàn)證等號成立的條件.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對稱,則SKIPIF1<0的最小值是()A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0【答案】C【解析】由圓SKIPIF1<0可得標(biāo)準(zhǔn)方程為SKIPIF1<0,因?yàn)閳ASKIPIF1<0關(guān)于直線SKIPIF1<0對稱,SKIPIF1<0該直線經(jīng)過圓心SKIPIF1<0,即SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0,即SKIPIF1<0時(shí)取等號,故選C.2.(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【答案】B【解析】由SKIPIF1<0,可得SKIPIF1<0,則SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0時(shí)等號成立；又由SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),等號成立,所以當(dāng)SKIPIF1<0與SKIPIF1<0同時(shí)取得最大值時(shí),則有SKIPIF1<0,解得SKIPIF1<0,此時(shí)SKIPIF1<0.故選B. 04已知SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0,(1)若對任意SKIPIF1<0,恒有SKIPIF1<0,求實(shí)數(shù)SKIPIF1<0的取值范圍；(2)若對任意SKIPIF1<0,恒有SKIPIF1<0,求實(shí)數(shù)SKIPIF1<0的取值范圍；【警示】本題易混淆單變量與雙變量【答案】(1)SKIPIF1<0；(2)SKIPIF1<0【問診】(1)設(shè)SKIPIF1<0SKIPIF1<0SKIPIF1<0,因?yàn)镾KIPIF1<0時(shí)SKIPIF1<0=SKIPIF1<0>0,所以SKIPIF1<0在SKIPIF1<0上是增函數(shù),由此可求得SKIPIF1<0的值域是[0,SKIPIF1<0],所以實(shí)數(shù)SKIPIF1<0的取值范圍是[0,SKIPIF1<0].對任意SKIPIF1<0,恒有SKIPIF1<0,即SKIPIF1<0時(shí)SKIPIF1<0SKIPIF1<0恒成立,即SKIPIF1<0SKIPIF1<0,由⑵可知SKIPIF1<0SKIPIF1<00.(2)由題中條件可得SKIPIF1<0的值域SKIPIF1<0SKIPIF1<0的值域SKIPIF1<0,若對任意SKIPIF1<0,恒有SKIPIF1<0,即SKIPIF1<0,即SKIPIF1<0,所以SKIPIF1<0.【叮囑】①若SKIPIF1<0值域?yàn)镾KIPIF1<0,則不等式SKIPIF1<0SKIPIF1<0恒成立SKIPIF1<0SKIPIF1<0SKIPIF1<0；不等式SKIPIF1<0SKIPIF1<0有解SKIPIF1<0SKIPIF1<0SKIPIF1<0；②若SKIPIF1<0值域?yàn)镾KIPIF1<0,則不等式SKIPIF1<0SKIPIF1<0恒成立SKIPIF1<0SKIPIF1<0SKIPIF1<0；若SKIPIF1<0值域?yàn)镾KIPIF1<0則不等式SKIPIF1<0SKIPIF1<0恒成立SKIPIF1<0SKIPIF1<0SKIPIF1<0.=3\*GB3③設(shè)SKIPIF1<0的最大值為SKIPIF1<0,對任意SKIPIF1<0,SKIPIF1<0的條件SKIPIF1<0,于是問題轉(zhuǎn)化為存在SKIPIF1<0,使得SKIPIF1<0,因此只需SKIPIF1<0的最小值大于SKIPIF1<0即SKIPIF1<0.1．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<0【答案】D【解析】因SKIPIF1<0,故當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,函數(shù)SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,函數(shù)SKIPIF1<0單調(diào)遞增,故SKIPIF1<0,又SKIPIF1<0,所以SKIPIF1<0,由題設(shè)可得SKIPIF1<0,解之得SKIPIF1<0,又由于SKIPIF1<0,所以SKIPIF1<0,故選D．2.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,若對任意SKIPIF1<0都有SKIPIF1<0,求實(shí)數(shù)SKIPIF1<0的取值范圍.【解析】(1)根據(jù)題意得到SKIPIF1<0,SKIPIF1<0,從而得到SKIPIF1<0,再解方程組即可；(2)根據(jù)題意得到SKIPIF1<0,設(shè)SKIPIF1<0,得到SKIPIF1<0,根據(jù)SKIPIF1<0,再利用二次函數(shù)的性質(zhì)得到SKIPIF1<0,SKIPIF1<0,從而得到SKIPIF1<0,解不等式即可.(1)SKIPIF1<0,又SKIPIF1<0是奇函數(shù),SKIPIF1<0,SKIPIF1<0,解得SKIPIF1<0,此時(shí),經(jīng)檢驗(yàn)SKIPIF1<0滿足題意,SKIPIF1<0(2)由題意知SKIPIF1<0,令SKIPIF1<0,SKIPIF1<0,由可知函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞減,在SKIPIF1<0上單調(diào)遞增,SKIPIF1<0,函數(shù)SKIPIF1<0的對稱軸方程為SKIPIF1<0,函數(shù)SKIPIF1<0在SKIPIF1<0上單調(diào)遞增,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0；當(dāng)SKIPIF1<0時(shí),SKIPIF1<0；即SKIPIF1<0,SKIPIF1<0,又對SKIPIF1<0,都有SKIPIF1<0恒成立,SKIPIF1<0即SKIPIF1<0解得,SKIPIF1<0又SKIPIF1<0,SKIPIF1<0的取值范圍是SKIPIF1<0. 05解關(guān)于x的不等式ax2－(a＋1)x＋1<0.【警示】本題錯(cuò)誤解法為：原不等式化為a(x－eq\f(1,a))(x－1)<0.∴當(dāng)a>1時(shí),不等式的解集為eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(1,a)，1)).當(dāng)a<1時(shí),不等式的解集為eq\b\lc\(\rc\)(\a\vs4\al\co1(1，\f(1,a))).【答案】當(dāng)a<0時(shí),不等式的解集為eq\b\lc\(\rc\)(\a\vs4\al\co1(－∞，\f(1,a)))∪(1,＋∞)；當(dāng)a＝0時(shí),不等式的解集為(1,＋∞)；當(dāng)0<a<1時(shí),不等式的解集為eq\b\lc\(\rc\)(\a\vs4\al\co1(1，\f(1,a)))；當(dāng)a＝1時(shí),不等式的解集為SKIPIF1<0；當(dāng)a>1時(shí),不等式的解集為eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(1,a)，1)).【問診】解本題容易出現(xiàn)的錯(cuò)誤是：(1)認(rèn)定這個(gè)不等式就是一元二次不等式,忽視了對a＝0時(shí)的討論；(2)在不等式兩端約掉系數(shù)a時(shí),若a<0,忘記改變不等號的方向；(3)忽視了對根的大小的討論,特別是等根的討論；(4)分類討論后,最后對結(jié)論不進(jìn)行整合．正確解法：當(dāng)a＝0時(shí),不等式的解集為{x|x>1}．當(dāng)a≠0時(shí),不等式化為aeq\b\lc\(\rc\)(\a\vs4\al\co1(x－\f(1,a)))(x－1)<0.當(dāng)a<0時(shí),原不等式等價(jià)于eq\b\lc\(\rc\)(\a\vs4\al\co1(x－\f(1,a)))(x－1)>0,不等式的解集為{x|x>1或x<eq\f(1,a)}；當(dāng)0<a<1時(shí),1<eq\f(1,a),不等式的解集為{x|1<x<eq\f(1,a)}；當(dāng)a>1時(shí),eq\f(1,a)<1,不等式的解集為{x|eq\f(1,a)<x<1}；當(dāng)a＝1時(shí),不等式的解集為?.綜上所述,當(dāng)a<0時(shí),不等式的解集為eq\b\lc\(\rc\)(\a\vs4\al\co1(－∞，\f(1,a)))∪(1,＋∞)；當(dāng)a＝0時(shí),不等式的解集為(1,＋∞)；當(dāng)0<a<1時(shí),不等式的解集為eq\b\lc\(\rc\)(\a\vs4\al\co1(1，\f(1,a)))；當(dāng)a＝1時(shí),不等式的解集為SKIPIF1<0；當(dāng)a>1時(shí),不等式的解集為eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(1,a)，1)).【糾錯(cuò)筆記】解形如ax2＋bx＋c>0的不等式,應(yīng)對系數(shù)a分a>0,a＝0,a<0進(jìn)行討論,還要討論各根的大小,最后根據(jù)不同情況分別寫出不等式的解集．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．【解析】(1)當(dāng)SKIPIF1<0時(shí),SKIPIF1<0是開口向上,對稱軸為SKIPIF1<0的二次函數(shù),又SKIPIF1<0,所以當(dāng)SKIPIF1<0時(shí),函數(shù)SKIPIF1<0單調(diào)遞減；當(dāng)SKIPIF1<0時(shí),函數(shù)SKIPIF1<0單調(diào)遞增；所以SKIPIF1<0,又SKIPIF1<0,SKIPIF1<0,因此SKIPIF1<0在SKIPIF1<0上的值域?yàn)镾KIPIF1<0.(2)∵SKIPIF1<0．①當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,即解集為SKIPIF1<0；②當(dāng)SKIPIF1<0時(shí),SKIPIF1<0且SKIPIF1<0開口方向向下,所以SKIPIF1<0的解集為SKIPIF1<0③當(dāng)SKIPIF1<0時(shí),若SKIPIF1<0,即SKIPIF1<0時(shí),原不等式的解集為SKIPIF1<0；若SKIPIF1<0,即SKIPIF1<0,原不等式的解集為SKIPIF1<0若SKIPIF1<0,即SKIPIF1<0,原不等式的解集為SKIPIF1<0綜上,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0的解集為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí),SKIPIF1<0的解集為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí),SKIPIF1<0的解集為SKIPIF1<0當(dāng)SKIPIF1<0時(shí),SKIPIF1<0的解集為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí),SKIPIF1<0的解集為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.【解析】(1)依題意,SKIPIF1<0有實(shí)數(shù)解,即不等式SKIPIF1<0有實(shí)數(shù)解,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0有實(shí)數(shù)解,則SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),取SKIPIF1<0,則SKIPIF1<0成立,即SKIPIF1<0有實(shí)數(shù)解,于是得SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),二次函數(shù)SKIPIF1<0的圖象開口向下,要SKIPIF1<0有解,當(dāng)且僅當(dāng)SKIPIF1<0,從而得SKIPIF1<0,綜上,SKIPIF1<0,所以實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0；(2)不等式SKIPIF1<0對于實(shí)數(shù)SKIPIF1<0時(shí)恒成立,即SKIPIF1<0,顯然SKIPIF1<0,函數(shù)SKIPIF1<0在SKIPIF1<0上遞增,從而得SKIPIF1<0,即SKIPIF1<0,解得SKIPIF1<0,所以實(shí)數(shù)SKIPIF1<0的取值范圍是SKIPIF1<0；(3)不等式SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),不等式可化為SKIPIF1<0,而SKIPIF1<0,解得SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),不等式可化為SKIPIF1<0,當(dāng)SKIPIF1<0,即SKIPIF1<0時(shí),SKIPIF1<0,當(dāng)SKIPIF1<0,即SKIPIF1<0時(shí),SKIPIF1<0或SKIPIF1<0,當(dāng)SKIPIF1<0,即SKIPIF1<0時(shí),SKIPIF1<0或SKIPIF1<0,所以,當(dāng)SKIPIF1<0時(shí),原不等式的解集為SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),原不等式的解集為SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),原不等式的解集為SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),原不等式的解集為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【答案】D【解析】SKIPIF1<0,SKIPIF1<0,由于SKIPIF1<0,解得：SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0是SKIPIF1<0的充分不必要條件.故選D.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<0【答案】A【解析】關(guān)于SKIPIF1<0的不等式SKIPIF1<0對SKIPIF1<0恒成立,則SKIPIF1<0,根據(jù)題意知,選項(xiàng)能推出題干,題干推不出選項(xiàng),故題干的范圍是選項(xiàng)范圍的子集,只有A選項(xiàng)符合題意.故選A.3．(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．4【答案】D【解析】由題意得SKIPIF1<0,則SKIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0,所以SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0,即SKIPIF1<0時(shí)等號成立,所以SKIPIF1<0的最小值為4.故選D4．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<0【答案】B【解析】令SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,又SKIPIF1<0,SKIPIF1<0,∴SKIPIF1<0,故選B．5．(2021屆浙江省紹興市高三上學(xué)期測試)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<0【答案】D【解析】∵SKIPIF1<0,且SKIPIF1<0,∴SKIPIF1<0,∴SKIPIF1<0,∴SKIPIF1<0SKIPIF1<0SKIPIF1<0,∵上述不等式恒成立,∴SKIPIF1<0,即SKIPIF1<0(否則取SKIPIF1<0,則左邊SKIPIF1<0,矛盾),此時(shí)不等式轉(zhuǎn)化為SKIPIF1<0,∴SKIPIF1<0,解得SKIPIF1<0,∴SKIPIF1<0,故選D．6．已知函數(shù)SKIPIF1<0,若對任意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<0【答案】B【解析】由已知得SKIPIF1<0,令SKIPIF1<0,因?yàn)镾KIPIF1<0,所以SKIPIF1<0,所以SKIPIF1<0,所以,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,即SKIPIF1<0,所以對任意SKIPIF1<0,SKIPIF1<0,所以對任意SKIPIF1<0,都有SKIPIF1<0,等價(jià)于SKIPIF1<0,即SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0,所以實(shí)數(shù)m的取值范圍是SKIPIF1<0,故選B.7．(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．2【答案】AB【解析】因?yàn)镾KIPIF1<0,SKIPIF1<0,所以SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0時(shí),等號成立．因?yàn)镾KIPIF1<0．若SKIPIF1<0恒成立,則SKIPIF1<0,解得SKIPIF1<0．故選AB.8．已知SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0的最大值為___________.【答案】SKIPIF1<0.【解析】設(shè)SKIPIF1<0,則SKIPIF1<0SKIPIF1<0,設(shè)SKIPIF1<0,則原式SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取“=”.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【解析】(1)由題意可得：SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,不等式的解為SKIPIF1<0或SKIPIF1<0；當(dāng)SKIPIF1<0時(shí),不等式的解為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí),SKIPIF1<0不等式的解為SKIPIF1<0或SKIPIF1<0；綜上：當(dāng)SKIPIF1<0時(shí),不等式的解集為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí),不等式的解集為SKIPIF1<0；當(dāng)SKIPIF1<0時(shí),不等式的解集為SKIPIF1<0；(2)由題意：SKIPIF1<0,即SKIPIF1<0是開口向上,以SKIPIF1<0為對稱軸的二次函數(shù),當(dāng)SKIPIF1<0時(shí),即SKIPIF1<0時(shí),滿足SKIPIF1<0,即SKIPIF1<0,解得SKIPIF1<0；當(dāng)SKIPIF1<0時(shí),即SKIPIF1<0時(shí),有SKIPIF1<0,可得SKIPIF1<0,故a不存在；綜上可得參數(shù)a的取值范圍SKIPIF1<0；(3)由題意：SKIPIF1<0,可得SKIPIF1<0且SKIPIF1<0,且SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0,由因?yàn)镾KIPIF1<0的對稱軸為SKIPIF1<0,故可得SKIPIF1<0在SKIPIF1<0上單調(diào)遞減,在SKIP