2011-2020年高考數(shù)學(xué)真題分類匯編 專題35 不等式選講（含解析）

專題35不等式選講十年大數(shù)據(jù)*全景展示年份題號(hào)考點(diǎn)考查內(nèi)容2011文理24不等式選講絕對(duì)值不等式的解法2012文理24不等式選講絕對(duì)值不等式的解法，不等式恒成立參數(shù)取值范圍問(wèn)題的解法2013卷1文理24不等式選講絕對(duì)值不等式的解法，不等式恒成立參數(shù)取值范圍問(wèn)題的解法卷2文理24不等式選講多元不等式的證明2014卷1文理24不等式選講基本不等式的應(yīng)用卷2文理24不等式選講絕對(duì)值不等式的解法2015卷1文理24不等式選講絕對(duì)值不等式的解法，不等式恒成立參數(shù)取值范圍問(wèn)題的解法卷2文理24不等式選講不等式的證明2016卷1文理24不等式選講分段函數(shù)的圖像，絕對(duì)值不等式的解法卷2文理24不等式選講絕對(duì)值不等式的解法，絕對(duì)值不等式的證明卷3文理24不等式選講絕對(duì)值不等式的解法，不等式恒成立參數(shù)取值范圍問(wèn)題的解法2017卷1文理23不等式選講絕對(duì)值不等式的解法，不等式恒成立參數(shù)取值范圍問(wèn)題的解法卷2文理23不等式選講不等式的證明卷3文理23不等式選講絕對(duì)值不等式的解法，絕對(duì)值不等式解集非空的參數(shù)取值范圍問(wèn)題2018卷1文理23不等式選講絕對(duì)值不等式的解法，不等式恒成立參數(shù)取值范圍問(wèn)題的解法卷2文理23不等式選講絕對(duì)值不等式的解法，不等式恒成立參數(shù)取值范圍問(wèn)題的解法卷3文理23不等式選講絕對(duì)值函數(shù)的圖象，不等式恒成立參數(shù)最值問(wèn)題的解法2019卷1文理23不等式選講三元條件不等式的證明卷2文理23不等式選講絕對(duì)值不等式的解法，不等式恒成立參數(shù)取值范圍問(wèn)題的解法卷3文理23不等式選講三元條件最值問(wèn)題的解法，三元條件不等式的證明2020卷1文理23不等式選講絕對(duì)值函數(shù)的圖像，絕對(duì)值不等式的解法卷2文理23不等式選講絕對(duì)值不等式的解法，不等式恒成立參數(shù)取值范圍問(wèn)題的解法卷3文理23不等式選講三元條件不等式的證明大數(shù)據(jù)分析*預(yù)測(cè)高考考點(diǎn)出現(xiàn)頻率2021年預(yù)測(cè)考點(diǎn)120絕對(duì)值不等式的求解23次考4次2021年主要考查絕對(duì)值不等式的解法、絕對(duì)值不等式的證明，不等式恒成立參數(shù)取值范圍問(wèn)題的解法等．考點(diǎn)121含絕對(duì)值不等式的恒成立問(wèn)題23次考12次考點(diǎn)122不等式的證明23次考7次十年試題分類*探求規(guī)律考點(diǎn)120絕對(duì)值不等式的求解1．(2020全國(guó)Ⅰ文理22)已知函數(shù)SKIPIF1<0．(1)畫出SKIPIF1<0的圖像；(2)求不等式SKIPIF1<0的解集．【解析】(1)∵SKIPIF1<0，作出圖像，如圖所示：(2)將函數(shù)SKIPIF1<0的圖像向左平移SKIPIF1<0個(gè)單位，可得函數(shù)SKIPIF1<0的圖像，如圖所示：由SKIPIF1<0，解得SKIPIF1<0，∴不等式的解集為SKIPIF1<0．2．(2020江蘇23)設(shè)SKIPIF1<0，解不等式SKIPIF1<0．【答案】SKIPIF1<0【思路導(dǎo)引】根據(jù)絕對(duì)值定義化為三個(gè)不等式組，解得結(jié)果．【解析】SKIPIF1<0或SKIPIF1<0或SKIPIF1<0，SKIPIF1<0或SKIPIF1<0或SKIPIF1<0，∴解集為SKIPIF1<0．3．(2016全國(guó)I文理)已知函數(shù)SKIPIF1<0．(I)在圖中畫出SKIPIF1<0的圖像；(II)求不等式SKIPIF1<0的解集．【解析】(1)如圖所示：(2)，．當(dāng)，，解得或，；當(dāng)，，解得或，或；當(dāng)，，解得或，或．綜上，或或，，解集為．4．(2014全國(guó)II文理)設(shè)函數(shù)SKIPIF1<0=SKIPIF1<0(Ⅰ)證明：SKIPIF1<0SKIPIF1<02；(Ⅱ)若SKIPIF1<0，求SKIPIF1<0的取值范圍．【解析】(I)由SKIPIF1<0，有SKIPIF1<0SKIPIF1<0，∴SKIPIF1<0≥2．(Ⅱ)SKIPIF1<0．當(dāng)時(shí)SKIPIF1<0＞3時(shí)，SKIPIF1<0=SKIPIF1<0，由SKIPIF1<0＜5得3＜SKIPIF1<0＜SKIPIF1<0；當(dāng)0＜SKIPIF1<0≤3時(shí)，SKIPIF1<0=SKIPIF1<0，由SKIPIF1<0＜5得SKIPIF1<0＜SKIPIF1<0≤3．綜上：SKIPIF1<0的取值范圍是(SKIPIF1<0，SKIPIF1<0)．5．(2011新課標(biāo)文理)設(shè)函數(shù)，其中．(Ⅰ)當(dāng)時(shí)，求不等式的解集；(Ⅱ)若不等式的解集為，求a的值．【解析】(Ⅰ)當(dāng)時(shí)，可化為，由此可得或．故不等式的解集為或．(

Ⅱ)由得，此不等式化為不等式組或，即SKIPIF1<0或SKIPIF1<0，因?yàn)?，∴不等式組的解集為，由題設(shè)可得=，故．考點(diǎn)121含絕對(duì)值不等式的恒成立問(wèn)題6．(2020全國(guó)Ⅱ文理22)已知函數(shù)SKIPIF1<0．(1)當(dāng)SKIPIF1<0時(shí)，求不等式SKIPIF1<0的解集；(2)若SKIPIF1<0，求SKIPIF1<0的取值范圍．【答案】(1)SKIPIF1<0或SKIPIF1<0；(2)SKIPIF1<0．【思路導(dǎo)引】(1)分別在SKIPIF1<0、SKIPIF1<0和SKIPIF1<0三種情況下解不等式求得結(jié)果；(2)利用絕對(duì)值三角不等式可得到SKIPIF1<0，由此構(gòu)造不等式求得結(jié)果．【解析】(1)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，解得：SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，無(wú)解；當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，解得：SKIPIF1<0；綜上所述：SKIPIF1<0的解集為SKIPIF1<0或SKIPIF1<0．(2)SKIPIF1<0(當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào))，SKIPIF1<0，解得：SKIPIF1<0或SKIPIF1<0，SKIPIF1<0的取值范圍為SKIPIF1<0．7．(2019全國(guó)II文理23)[選修4-5：不等式選講](10分)已知(1)當(dāng)時(shí)，求不等式的解集；(2)若時(shí)，，求的取值范圍．【解析】(1)當(dāng)a=1時(shí)，．當(dāng)時(shí)，；當(dāng)時(shí)，，∴不等式的解集為．(2)因?yàn)椋啵?dāng)，時(shí)，∴的取值范圍是．8．(2018全國(guó)Ⅰ文理)已知SKIPIF1<0．(1)當(dāng)SKIPIF1<0時(shí)，求不等式SKIPIF1<0的解集；(2)若SKIPIF1<0時(shí)不等式SKIPIF1<0成立，求SKIPIF1<0的取值范圍．【解析】(1)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，即SKIPIF1<0故不等式SKIPIF1<0的解集為SKIPIF1<0．(2)當(dāng)SKIPIF1<0時(shí)SKIPIF1<0成立等價(jià)于當(dāng)SKIPIF1<0時(shí)SKIPIF1<0成立．若SKIPIF1<0，則當(dāng)SKIPIF1<0時(shí)SKIPIF1<0；若SKIPIF1<0，SKIPIF1<0的解集為SKIPIF1<0，∴SKIPIF1<0，故SKIPIF1<0．綜上，SKIPIF1<0的取值范圍為SKIPIF1<0．9．(2018全國(guó)Ⅱ文理)設(shè)函數(shù)SKIPIF1<0．(1)當(dāng)SKIPIF1<0時(shí)，求不等式SKIPIF1<0的解集；(2)若SKIPIF1<0，求SKIPIF1<0的取值范圍．【解析】(1)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0可得SKIPIF1<0的解集為SKIPIF1<0．(2)SKIPIF1<0等價(jià)于SKIPIF1<0．而SKIPIF1<0，且當(dāng)SKIPIF1<0時(shí)等號(hào)成立．故SKIPIF1<0等價(jià)于SKIPIF1<0．由SKIPIF1<0可得SKIPIF1<0或SKIPIF1<0，∴SKIPIF1<0的取值范圍是SKIPIF1<0．10．(2018全國(guó)Ⅲ文理)設(shè)函數(shù)SKIPIF1<0．(1)畫出SKIPIF1<0的圖像；(2)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，求SKIPIF1<0的最小值．【解析】(1)SKIPIF1<0SKIPIF1<0的圖像如圖所示．(2)由(1)知，SKIPIF1<0的圖像與SKIPIF1<0軸交點(diǎn)的縱坐標(biāo)為2，且各部分所在直線斜率的最大值為3，故當(dāng)且僅當(dāng)SKIPIF1<0且SKIPIF1<0時(shí)，SKIPIF1<0在SKIPIF1<0成立，因此SKIPIF1<0的最小值為5．11．(2018江蘇)若SKIPIF1<0，SKIPIF1<0，SKIPIF1<0為實(shí)數(shù)，且SKIPIF1<0，求SKIPIF1<0的最小值．【解析】由柯西不等式，得SKIPIF1<0．因?yàn)镾KIPIF1<0，∴SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，不等式取等號(hào)，此時(shí)SKIPIF1<0，∴SKIPIF1<0的最小值為4．12．(2017全國(guó)Ⅰ文理)已知函數(shù)SKIPIF1<0，SKIPIF1<0．(1)當(dāng)SKIPIF1<0時(shí)，求不等式SKIPIF1<0的解集；(2)若不等式SKIPIF1<0的解集包含SKIPIF1<0，求SKIPIF1<0的取值范圍．【解析】(1)當(dāng)SKIPIF1<0時(shí)，不等式SKIPIF1<0等價(jià)于SKIPIF1<0．①當(dāng)SKIPIF1<0時(shí)，①式化為SKIPIF1<0，無(wú)解；當(dāng)SKIPIF1<0時(shí)，①式化為SKIPIF1<0，從而SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，①式化為SKIPIF1<0，從而SKIPIF1<0，∴SKIPIF1<0的解集為SKIPIF1<0．(2)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，∴SKIPIF1<0的解集包含SKIPIF1<0，等價(jià)于當(dāng)SKIPIF1<0時(shí)SKIPIF1<0．又SKIPIF1<0在SKIPIF1<0的最小值必為SKIPIF1<0與SKIPIF1<0之一，∴SKIPIF1<0且SKIPIF1<0，得SKIPIF1<0，∴SKIPIF1<0的取值范圍為SKIPIF1<0．13．(2017全國(guó)Ⅲ文理)已知函數(shù)SKIPIF1<0．(1)求不等式SKIPIF1<0的解集；(2)若不等式SKIPIF1<0的解集非空，求SKIPIF1<0的取值范圍．【解析】(1)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0無(wú)解；當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0得，SKIPIF1<0，解得SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，由SKIPIF1<0解得SKIPIF1<0．∴SKIPIF1<0的解集為SKIPIF1<0．(2)由SKIPIF1<0得SKIPIF1<0，而SKIPIF1<0SKIPIF1<0，且當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，故m的取值范圍為SKIPIF1<0．14．(2016全國(guó)III文理)已知函數(shù)SKIPIF1<0(Ⅰ)當(dāng)a=2時(shí)，求不等式SKIPIF1<0的解集；(Ⅱ)設(shè)函數(shù)SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，求a的取值范圍．【解析】(Ⅰ)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0．解不等式SKIPIF1<0，得SKIPIF1<0，因此SKIPIF1<0的解集為SKIPIF1<0．(Ⅱ)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)等號(hào)成立，∴當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0等價(jià)于SKIPIF1<0．①當(dāng)SKIPIF1<0時(shí)，①等價(jià)于SKIPIF1<0，無(wú)解．當(dāng)SKIPIF1<0時(shí)，①等價(jià)于SKIPIF1<0，解得SKIPIF1<0．∴SKIPIF1<0的取值范圍是SKIPIF1<0．15．(2015全國(guó)I文理)已知函數(shù)SKIPIF1<0，SKIPIF1<0．(Ⅰ)當(dāng)SKIPIF1<0時(shí)，求不等式SKIPIF1<0的解集；(Ⅱ)若SKIPIF1<0的圖像與SKIPIF1<0軸圍成的三角形面積大于6，求SKIPIF1<0的取值范圍．【解析】(Ⅰ)當(dāng)SKIPIF1<0時(shí)，不等式SKIPIF1<0化為SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，不等式化為SKIPIF1<0，無(wú)解；當(dāng)SKIPIF1<0時(shí)，不等式化為SKIPIF1<0，解得SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，不等式化為SKIPIF1<0，解得SKIPIF1<0．∴SKIPIF1<0的解集為SKIPIF1<0．(Ⅱ)有題設(shè)可得，SKIPIF1<0，∴函數(shù)SKIPIF1<0圖象與SKIPIF1<0軸圍成的三角形的三個(gè)頂點(diǎn)分別為SKIPIF1<0，SKIPIF1<0的面積為SKIPIF1<0．有題設(shè)得SKIPIF1<0，故SKIPIF1<0．∴SKIPIF1<0的取值范圍為SKIPIF1<0．16．(2014全國(guó)I文理)若SKIPIF1<0，且SKIPIF1<0．(Ⅰ)求SKIPIF1<0的最小值；(Ⅱ)是否存在SKIPIF1<0，使得SKIPIF1<0？并說(shuō)明理由．【解析】(=1\*ROMANI)由SKIPIF1<0，得SKIPIF1<0，且當(dāng)SKIPIF1<0時(shí)取等號(hào)．故SKIPIF1<0SKIPIF1<0，且當(dāng)SKIPIF1<0時(shí)取等號(hào)．∴SKIPIF1<0的最小值為SKIPIF1<0．(=2\*ROMANII)由(=1\*ROMANI)知，SKIPIF1<0．由于SKIPIF1<0，從而不存在SKIPIF1<0，使得SKIPIF1<0．16．(2013全國(guó)I文理)已知函數(shù)=，=．(Ⅰ)當(dāng)=-2時(shí)，求不等式＜的解集；(Ⅱ)設(shè)＞-1，且當(dāng)∈[，)時(shí)，≤，求的取值范圍．【解析】(Ⅰ)當(dāng)=SKIPIF1<02時(shí)，不等式＜化為，設(shè)函數(shù)=，=，其圖像如圖所示，從圖像可知，當(dāng)且僅當(dāng)時(shí)，＜0，∴原不等式解集是．(Ⅱ)當(dāng)∈[，)時(shí)，=，不等式≤化為SKIPIF1<0，∴SKIPIF1<0對(duì)∈[，)都成立，故SKIPIF1<0，即≤，∴的取值范圍為(SKIPIF1<01，]．17．(2012新課標(biāo)文理)已知函數(shù)SKIPIF1<0．(Ⅰ)當(dāng)SKIPIF1<0時(shí)，求不等式SKIPIF1<0的解集；(Ⅱ)若SKIPIF1<0的解集包含SKIPIF1<0，求SKIPIF1<0的取值范圍．【解析】(1)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0SKIPIF1<0或SKIPIF1<0或SKIPIF1<0SKIPIF1<0或SKIPIF1<0．(2)原命題SKIPIF1<0在SKIPIF1<0上恒成立SKIPIF1<0在SKIPIF1<0上恒成立SKIPIF1<0在SKIPIF1<0上恒成立SKIPIF1<0．考點(diǎn)122不等式的證明18．(2020全國(guó)Ⅲ文理23)設(shè)SKIPIF1<0．(1)證明：SKIPIF1<0；(2)用SKIPIF1<0表示SKIPIF1<0的最大值，證明：SKIPIF1<0．【答案】(1)證明見(jiàn)解析(2)證明見(jiàn)解析．【思路導(dǎo)引】(1)根據(jù)題設(shè)條件SKIPIF1<0兩邊平方，再利用均值不等式證明即可；(2)思路一：不妨設(shè)SKIPIF1<0，由題意得出SKIPIF1<0，由SKIPIF1<0，結(jié)合基本不等式，即可得出證明．思路二：假設(shè)出SKIPIF1<0中最大值，根據(jù)反證法與基本不等式推出矛盾，即可得出結(jié)論．【解析】(1)證明：SKIPIF1<0SKIPIF1<0即SKIPIF1<0SKIPIF1<0(2)證法一：不妨設(shè)SKIPIF1<0，由SKIPIF1<0可知，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時(shí)，取等號(hào)，SKIPIF1<0，即SKIPIF1<0．證法二：不妨設(shè)SKIPIF1<0，則SKIPIF1<0而SKIPIF1<0矛盾，∴命題得證．19．(2019全國(guó)I文理23)已知a，b，c為正數(shù)，且滿足abc=1．證明：(1)；(2)．【解析】(1)因?yàn)?，又，故有，∴?2)因?yàn)闉檎龜?shù)且，故有=24．∴．20．(2019全國(guó)III文理23)設(shè)，且．(1)求的最小值；(2)若成立，證明：或．【解析】(1)由于，故由已知得，當(dāng)且僅當(dāng)x=，y=–，時(shí)等號(hào)成立．∴的最小值為．(2)由于，故由已知，當(dāng)且僅當(dāng)，，時(shí)等號(hào)成立，因此的最小值為．由題設(shè)知，解得或．21．(2017全國(guó)Ⅱ文理)已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，證明：(1)SKIPIF1<0；(2)SKIPIF1<0．【解析】(1)SKIPIF1<0SKIPIF1<0SKIPIF1<0．(2)∵SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0，∴SKIPIF1<0，因此SKIPIF1<0．22．(2017江蘇)已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0為實(shí)數(shù)，且SKIPIF1<0，SKIPIF1<0，證明SKIPIF1<0．【解析】證明：由柯西不等式可得：SKIPIF1<0，因?yàn)镾KIPIF1<0∴SKIPIF1<0，因此SKIPIF1<0．23．(2016全國(guó)II文理)已知函數(shù)SKIPIF1<0，M為不等式SKIPIF1<0的解集．(I)求M；(II)證明：當(dāng)a，SKIPIF1<0時(shí)，SKIPIF1<0．【解析】(I)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，若SKIPIF1<0；當(dāng)SKIPIF1<0時(shí)，SKIPIF1