高一函數(shù)經(jīng)典難題講解

1、高一經(jīng)典難題講解高一經(jīng)典難題講解11.1.已知函數(shù)已知函數(shù)f(x)=(x1a)(ax)x∈Rf(x)=(x1a)(ax)x∈R且x≠ax≠a當(dāng)f(x)f(x)的定義域?yàn)榈亩x域?yàn)閇a[a1a12]1a12]時(shí)求f(x)f(x)值解:由題知，已知函數(shù)f(x)=(x1a)(ax)所以，f(x)=11(ax)當(dāng)f(x)的定義域?yàn)閇a1a12]時(shí)x∈[a1a12](ax)∈[121]1(ax)∈[12]f(x)=11(ax)∈[01]2.2.設(shè)

2、a為非負(fù)數(shù)為非負(fù)數(shù)函數(shù)函數(shù)f(x)=x|xa|a.f(x)=x|xa|a.(1)(1)當(dāng)a=2a=2時(shí)求函數(shù)的求函數(shù)的單調(diào)區(qū)間單調(diào)區(qū)間(2)(2)討論函數(shù)討論函數(shù)y=f(x)y=f(x)的零點(diǎn)個(gè)數(shù)的零點(diǎn)個(gè)數(shù)解析：(1)∵函數(shù)f(x)=x|x2|2當(dāng)x=2時(shí)，f(x)=x^22x2，為開口向上拋物線，對(duì)稱軸為x=1∴當(dāng)x∈(∞1)時(shí)，f(x)單調(diào)增；當(dāng)x∈[12]時(shí)，f(x)單調(diào)減；當(dāng)x∈(2∞)時(shí)，f(x)單調(diào)增；(2).f(x)=x|

3、xa|a=0x|xa|=a①a=0時(shí)x=0零點(diǎn)個(gè)數(shù)為1；a0時(shí)x0由①，x=ax^2axa=0x1=[a√(a^24a)]204時(shí)，②無實(shí)根，零點(diǎn)個(gè)數(shù)為1。a=a4x^2axa=0③x12=[a土√(a^24a)]2；x4時(shí)零點(diǎn)個(gè)數(shù)為1；a=土4時(shí)，零點(diǎn)個(gè)數(shù)為2；4a0或0a4時(shí)，零點(diǎn)個(gè)數(shù)為3.3.3.已知函數(shù)已知函數(shù)f(x)=log3f(x)=log3為底為底1m(x2)x31m(x2)x3的圖像關(guān)于原點(diǎn)對(duì)稱的圖像關(guān)于原點(diǎn)對(duì)稱（1）求

4、常數(shù)）求常數(shù)m的值的值（2）當(dāng)）當(dāng)x∈x∈（3434）時(shí)，求）時(shí)，求f(x)f(x)的值域；的值域；（3）判斷）判斷f(x)f(x)的單調(diào)性并證明。的單調(diào)性并證明。解：1、函數(shù)f(x)=log3[1m(x2)[(x3)圖象關(guān)于原點(diǎn)對(duì)稱，高一經(jīng)典難題講解高一經(jīng)典難題講解3聯(lián)立log4[(4^x1)2^x]=log4(a2^x43a)∴(4^x1)2^x=a2^x43a不妨設(shè)t=2^xt＞0t^21t=at43at^21=at^243at(

5、a1)t^243at1=0設(shè)u(t)=(a1)t^243at1∵兩函數(shù)圖像只有1個(gè)公共點(diǎn)，在這里就變成了有且只有一個(gè)正根1.當(dāng)a=1時(shí)t=34不滿足(舍)2.當(dāng)△=0時(shí)a=34或a=3a=34時(shí)t=12＜0（舍）a=3時(shí)t=12滿足3.當(dāng)一正根一負(fù)根時(shí)(a1)u(0)＜0(根據(jù)根的分布)∴a＞1綜上所述，得a=3或a＞15.5.這個(gè)是概念的問題:1.對(duì)于f(x)取值范圍（0，無窮），f(x)bf(x)c=0最多有兩個(gè)不同的f(x)。2.

6、對(duì)f(x)的圖像進(jìn)行分析，知道f(x)=1對(duì)應(yīng)的x值有三個(gè)，即除x=2外另有兩個(gè)關(guān)于x=2對(duì)稱的x。f(x)不等于1時(shí)對(duì)應(yīng)的x值有兩個(gè)，即兩個(gè)關(guān)于x=2對(duì)稱的兩個(gè)x。3.題意說f(x)bf(x)c=0對(duì)應(yīng)的x根有5個(gè)，顯然滿足f(x)bf(x)c=0的f(x)有兩個(gè)，一個(gè)f(x)對(duì)應(yīng)三個(gè)x值，設(shè)為x1x2x3另一個(gè)f（x）對(duì)應(yīng)兩個(gè)x設(shè)為x4x5根據(jù)以上分析，應(yīng)有x1x3=22x2=2x4x5=22=4則f（x1x2x3x4x5）=f(1