2022屆高考數(shù)學(xué)4-5不等式選講配套練習(xí)

選修4—5不等式選講隨堂操練穩(wěn)固x,不等式|10|-|-2|8的解集為【答案】[0)12x10【解析】令=|10|-|-2|=2x810x212x2則可畫出其函數(shù)圖象如下圖:由圖象能夠察看出使y8的的范圍為[0)∴|10|-|-2|8的解集為[0)2（2022江西高考，理15）關(guān)于實(shí)數(shù),,若|-1|1|-2|1,則|-21|的最大值為【答案】5【解析】|-21|=|-1-2-2-2||-1|2|-2|2122=53解不等式|x29|3x【解法一】原不等式等價(jià)于x290①x293xx290②或3x9x2由①得=-3或3x4由②得2x3所以原不等式的解集為{|=-3或2x4}【解法二】原不等式可化為|x29|x3兩邊平方得29)2(x3)2移項(xiàng)并分解因式得3)2(x2)(x4)0且x3所以原不等式的解集為{|=-3或2x4}=|2-4|11畫出函數(shù)=f的圖象;2若不等式f(x)ax的解集非空,求a的取值范圍2x5x2【解】1由于f=則函數(shù)=f的圖象如下圖2x3x22由函數(shù)=f與函數(shù)=a的圖象可知,當(dāng)且僅當(dāng)a1或2af(x)ax13322(2)[)abab(ab)2a3b3ab(a2b2)a2a(ab)b2b(ba)(ab)[(a)5(b)5]abab(a)5(b)5(ab)[(a)5(b)5]0ab(a)5(b)5(ab)[(a)5(b)5]0a3b3ab(a2b2)2x10x11x2x1-1,所以2221x1;2所以不等式的解集為{|1112xx1x2}{|1x}={|-1ax(3a33a331x2222x1x2f(x)3a3xx22x23x4(3)xx22x2x2x2x20x2x2x23x4-6,>-3∴原不等式的解集為(3)4（2022天津高考，理13）已知會(huì)合A={xR||3||-4|9},B={xR|x4t16t(0)},則會(huì)合tAB2x5}【答案】{|【解析】解不等式|3||-4|91當(dāng)xx44x33x4x3x44時(shí),|3||-4|x3x49∴x5即4x5綜上所述,A={xR|4x5}∵t(0)∴x4t1624t162當(dāng)且僅當(dāng)t1時(shí)等號(hào)建立tt2∴B={xR|x2}∴AB{xR|4x5}{xR|x2}={xR|2x5}的不等式|-3|-|-4|-1【解析】a>|-3|-|-4|)min令=|-3|-|-4|,由幾何意義得1y1故a>-16若不等式|x1|>|a-2|1關(guān)于一切非零實(shí)數(shù)均建立,則實(shí)數(shù)a的取值范圍是x【答案】1,3【解析】∵|x12∴ab且ab證明b2ac3x|:a|-2|1>c,c=0,【證明】b2ac3b2ac3aab2ac3a23a2b2ac03a2(ac)2ac2a2acc20(ac)(2ac>0a>b>c,abc=0,a-c>0,2ac=aac=a-b>0,即知a-c2ac>0故b2ac3a8解不等式2(x2)(x3)0或|5xx2|5xx2x25x6x25x6x5x60x2x3x25x60(x6)(x1)01x6yx25xx25xx25x6x1x26x25x61x32x43x7-4,∴-4x77x070時(shí),275x2舍去22223∴原不等式的解集為-4,-2|-1|a有解,求實(shí)數(shù)a的取值范圍【解】設(shè)f=|-1|,則f=2x1x11x1所以f的最小值為1所以當(dāng)a1時(shí)f(x)a有解,即實(shí)數(shù)a的取值范圍是[1)|2-1|2x102x101x42x12x4ababx(2x1)3x(2x1)323231ab1a1bxx1x111xx1(1)(1)0)f(ababf(x)11x1x1ab1ababab1ab1ab1a10b1當(dāng)a=1時(shí),求不等式f(x)3x2的解集;2若不等式f(x)0的解集為{|x1},求a的值【解】1當(dāng)a=1時(shí)f(x)3x2可化為|-1|2由此可得x3或x1故不等式f(x)3x2的解集為{|x3或x-1}2由f(x)0得|-a|3x0xa此不等式化為不等式組xa3x0或xax3x0axaxa即a或axx422}因?yàn)閍>0,所以不等式組的解集為{|xa由題設(shè)可得拓展延長(zhǎng)

1故a=22=|-a|,1若不等式f(x)3的解集為{|1x5},求實(shí)數(shù)a的值;2在1的條件下,若f(x)f(x5)m對(duì)一確實(shí)數(shù)恒建立,求實(shí)數(shù)的取值范圍m【解法一】1由f(x)3得|-a|3解得a3xa3又已知不等式f(x)3的解集為{|1x5},a31所以35a解得a=22當(dāng)a=2時(shí),f=|-2|設(shè)g=ff5,于是2x1x3g=|-2||3|=53x22x1x2所以當(dāng)5;當(dāng)3x2時(shí),g=5;當(dāng)>2時(shí),g>5綜上可得,g的最小值為5進(jìn)而,若f(x)f(x5)m即g(x)m對(duì)一確實(shí)數(shù)恒建立,則m的取值范圍為(5]【解法二