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)*全景展示年份題號考點考查內(nèi)容2011文理24不等式選講絕對值不等式的解法2012文理24不等式選講絕對值不等式的解法，不等式恒成立參數(shù)取值范圍問題的解法2013卷1文理24不等式選講絕對值不等式的解法，不等式恒成立參數(shù)取值范圍問題的解法卷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不等式選講絕對值不等式的解法2015卷1文理24不等式選講絕對值不等式的解法，不等式恒成立參數(shù)取值范圍問題的解法卷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，絕對值不等式的解法卷2文理24不等式選講絕對值不等式的解法，絕對值不等式的證明卷3文理24不等式選講絕對值不等式的解法，不等式恒成立參數(shù)取值范圍問題的解法2017卷1文理23不等式選講絕對值不等式的解法，不等式恒成立參數(shù)取值范圍問題的解法卷2文理23不等式選講不等式的證明卷3文理23不等式選講絕對值不等式的解法，絕對值不等式解集非空的參數(shù)取值范圍問題2018卷1文理23不等式選講絕對值不等式的解法，不等式恒成立參數(shù)取值范圍問題的解法卷2文理23不等式選講絕對值不等式的解法，不等式恒成立參數(shù)取值范圍問題的解法卷3文理23不等式選講絕對值函數(shù)的圖象，不等式恒成立參數(shù)最值問題的解法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不等式選講絕對值不等式的解法，不等式恒成立參數(shù)取值范圍問題的解法卷3文理23不等式選講三元條件最值問題的解法，三元條件不等式的證明2020卷1文理23不等式選講絕對值函數(shù)的圖像，絕對值不等式的解法卷2文理23不等式選講絕對值不等式的解法，不等式恒成立參數(shù)取值范圍問題的解法卷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)測高考考點出現(xiàn)頻率2021年預(yù)測考點120絕對值不等式的求解23次考4次2021年主要考查絕對值不等式的解法、絕對值不等式的證明，不等式恒成立參數(shù)取值范圍問題的解法等．考點121含絕對值不等式的恒成立問題23次考12次考點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120絕對值不等式的求解1．(2020全國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)絕對值定義化為三個不等式組，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SKIPIF1<0＞3時，SKIPIF1<0=SKIPIF1<0，由SKIPIF1<0＜5得3＜SKIPIF1<0＜SKIPIF1<0；當(dāng)0＜SKIPIF1<0≤3時，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，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，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，因為，∴不等式組的解集為，由題設(shè)可得=，故．考點121含絕對值不等式的恒成立問題6．(2020全國Ⅱ文理22)已知函數(shù)SKIPIF1<0．(1)當(dāng)SKIPIF1<0時，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，SKIPIF1<0．當(dāng)SKIPIF1<0時，SKIPIF1<0，解得：SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1<0，無解；當(dāng)SKIPIF1<0時，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時取等號)，SKIPIF1<0，解得：SKIPIF1<0或SKIPIF1<0，SKIPIF1<0的取值范圍為SKIPIF1<0．7．(2019全國II文理23)[選修4-5：不等式選講](10分)已知(1)當(dāng)時，求不等式的解集；(2)若時，，求的取值范圍．【解析】(1)當(dāng)a=1時，．當(dāng)時，；當(dāng)時，，∴不等式的解集為．(2)因為，∴．當(dāng)，時，∴的取值范圍是．8．(2018全國Ⅰ文理)已知SKIPIF1<0．(1)當(dāng)SKIPIF1<0時，求不等式SKIPIF1<0的解集；(2)若SKIPIF1<0時不等式SKIPIF1<0成立，求SKIPIF1<0的取值范圍．【解析】(1)當(dāng)SKIPIF1<0時，SKIPIF1<0，即SKIPIF1<0故不等式SKIPIF1<0的解集為SKIPIF1<0．(2)當(dāng)SKIPIF1<0時SKIPIF1<0成立等價于當(dāng)SKIPIF1<0時SKIPIF1<0成立．若SKIPIF1<0，則當(dāng)SKIPIF1<0時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Ⅱ文理)設(shè)函數(shù)SKIPIF1<0．(1)當(dāng)SKIPIF1<0時，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，SKIPIF1<0可得SKIPIF1<0的解集為SKIPIF1<0．(2)SKIPIF1<0等價于SKIPIF1<0．而SKIPIF1<0，且當(dāng)SKIPIF1<0時等號成立．故SKIPIF1<0等價于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Ⅲ文理)設(shè)函數(shù)SKIPIF1<0．(1)畫出SKIPIF1<0的圖像；(2)當(dāng)SKIPIF1<0時，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，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為實數(shù)，且SKIPIF1<0，求SKIPIF1<0的最小值．【解析】由柯西不等式，得SKIPIF1<0．因為SKIPIF1<0，∴SKIPIF1<0，當(dāng)且僅當(dāng)SKIPIF1<0時，不等式取等號，此時SKIPIF1<0，∴SKIPIF1<0的最小值為4．12．(2017全國Ⅰ文理)已知函數(shù)SKIPIF1<0，SKIPIF1<0．(1)當(dāng)SKIPIF1<0時，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，不等式SKIPIF1<0等價于SKIPIF1<0．①當(dāng)SKIPIF1<0時，①式化為SKIPIF1<0，無解；當(dāng)SKIPIF1<0時，①式化為SKIPIF1<0，從而SKIPIF1<0；當(dāng)SKIPIF1<0時，①式化為SKIPIF1<0，從而SKIPIF1<0，∴SKIPIF1<0的解集為SKIPIF1<0．(2)當(dāng)SKIPIF1<0時，SKIPIF1<0，∴SKIPIF1<0的解集包含SKIPIF1<0，等價于當(dāng)SKIPIF1<0時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，SKIPIF1<0無解；當(dāng)SKIPIF1<0時，由SKIPIF1<0得，SKIPIF1<0，解得SKIPIF1<0；當(dāng)SKIPIF1<0時，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，SKIPIF1<0，故m的取值范圍為SKIPIF1<0．14．(2016全國III文理)已知函數(shù)SKIPIF1<0(Ⅰ)當(dāng)a=2時，求不等式SKIPIF1<0的解集；(Ⅱ)設(shè)函數(shù)SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，求a的取值范圍．【解析】(Ⅰ)當(dāng)SKIPIF1<0時，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，SKIPIF1<0SKIPIF1<0SKIPIF1<0，當(dāng)SKIPIF1<0時等號成立，∴當(dāng)SKIPIF1<0時，SKIPIF1<0等價于SKIPIF1<0．①當(dāng)SKIPIF1<0時，①等價于SKIPIF1<0，無解．當(dāng)SKIPIF1<0時，①等價于SKIPIF1<0，解得SKIPIF1<0．∴SKIPIF1<0的取值范圍是SKIPIF1<0．15．(2015全國I文理)已知函數(shù)SKIPIF1<0，SKIPIF1<0．(Ⅰ)當(dāng)SKIPIF1<0時，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，不等式SKIPIF1<0化為SKIPIF1<0，當(dāng)SKIPIF1<0時，不等式化為SKIPIF1<0，無解；當(dāng)SKIPIF1<0時，不等式化為SKIPIF1<0，解得SKIPIF1<0；當(dāng)SKIPIF1<0時，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頂點分別為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？并說明理由．【解析】(=1\*ROMANI)由SKIPIF1<0，得SKIPIF1<0，且當(dāng)SKIPIF1<0時取等號．故SKIPIF1<0SKIPIF1<0，且當(dāng)SKIPIF1<0時取等號．∴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I文理)已知函數(shù)=，=．(Ⅰ)當(dāng)=-2時，求不等式＜的解集；(Ⅱ)設(shè)＞-1，且當(dāng)∈[，)時，≤，求的取值范圍．【解析】(Ⅰ)當(dāng)=SKIPIF1<02時，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，＜0，∴原不等式解集是．(Ⅱ)當(dāng)∈[，)時，=，不等式≤化為SKIPIF1<0，∴SKIPIF1<0對∈[，)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，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，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122不等式的證明18．(2020全國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)證明見解析(2)證明見解析．【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，取等號，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)因為，又，故有，∴．(2)因為為正數(shù)且，故有=24．∴．20．(2019全國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=–，時等號成立．∴的最小值為．(2)由于，故由已知，當(dāng)且僅當(dāng)，，時等號成立，因此的最小值為．由題設(shè)知，解得或．21．(2017全國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，因為SKIPIF1<0∴SKIPIF1<0，因此SKIPIF1<0．23．(2016全國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，SKIPIF1<0．【解析】(I)當(dāng)SKIPIF1<0時，SKIPIF1<0，若SKIPIF1<0；當(dāng)SKIPIF1<0時，SKIPIF1