高考數(shù)學(xué)復(fù)習(xí)拓展提升課七　雙變量問(wèn)題（導(dǎo)學(xué)案）

培優(yōu)增分拓展提升課七雙變量問(wèn)題考情分析:雙變量問(wèn)題,一直是高考命題的熱點(diǎn)問(wèn)題,試題常常出現(xiàn)在“壓軸題”的位置.這類試題綜合性強(qiáng),難度大,對(duì)學(xué)生的數(shù)學(xué)素養(yǎng)要求很高,因此具有很好的區(qū)分度,解題時(shí)往往顧此失彼,難以入手.解題關(guān)鍵:解決雙變量問(wèn)題的法寶是構(gòu)造一元變量函數(shù),借助導(dǎo)數(shù)知識(shí)獲得簡(jiǎn)捷、明了的解答.一、相互替代法統(tǒng)一變量[典例1]已知函數(shù)f(x)=1x-x+aln(1)討論函數(shù)f(x)的單調(diào)性;(2)若f(x)存在兩個(gè)極值點(diǎn)x1,x2,求證:f(x1)?【解題思路】(1)對(duì)f(x)求導(dǎo),確定函數(shù)f(x)的單調(diào)性;(2)由f(x)存在兩個(gè)極值點(diǎn)x1,x2,而f'(x)=-x2?ax+1x2,x2-ax+1=0,得到x1x2=1,利用x1x2=1將雙變量問(wèn)題轉(zhuǎn)化為關(guān)于x1或解析：(1)f(x)的定義域?yàn)?0,+∞),f'(x)=-1x2-1+ax①若a≤2,則f'(x)≤0,當(dāng)且僅當(dāng)a=2,x=1時(shí)f'(x)=0,所以f(x)在(0,+∞)上單調(diào)遞減.②若a>2,令f'(x)=0得,x=a?a2?42當(dāng)x∈(0,a?a2?42)∪(a+a當(dāng)x∈(a?a2?42,a+a所以f(x)在(0,a?a2?42),(a+a2?4(2)由題意f'(x)=-x2?ax+1x2(x>0)有零點(diǎn)x1,x2,即x1,x2滿足方程所以x1+x2=a,x1x2=1.所以x1=1x由(1)知a>2,不妨設(shè)x1<x2,則0<x1<1<x2.所以f(x1)?f(x2)x1?x2=-因此,要證f(x1)?2lnx2-x2+1x2<0(x設(shè)g(x)=2lnx-x+1x<0(x由(1)知g(x)在(0,+∞)上單調(diào)遞減,而g(1)=0,所以當(dāng)x∈(1,+∞)時(shí),g(x)<0.因此2lnx2-x2+1x2<0(x所以f(x1)?當(dāng)出現(xiàn)雙變量時(shí),若能將雙變量中所有變量統(tǒng)一用其中一個(gè)變量來(lái)替換,這樣就將雙變量問(wèn)題轉(zhuǎn)化為我們熟知的單變量問(wèn)題,從而為我們的解題帶來(lái)方便.二、齊次構(gòu)造法統(tǒng)一變量[典例2]已知函數(shù)f(x)=xlnx-12ax2-x,a∈R.若f(x)的兩個(gè)極值點(diǎn)為x1,x2,且x1<x2,求證x1x2>e2【解題思路】可以通過(guò)兩邊取對(duì)數(shù)將問(wèn)題轉(zhuǎn)化為lnx1+lnx2>2.由導(dǎo)數(shù)對(duì)應(yīng)方程的根x1,x2,可得lnx1?ax1=0,lnx2?ax2【證明】要證x1x2>e2,只需證lnx1+lnx2>2.若f(x)的兩個(gè)極值點(diǎn)為x1,x2,即函數(shù)f'(x)的兩個(gè)零點(diǎn).因?yàn)閒'(x)=lnx-ax,所以x1,x2是方程f'(x)=0的兩個(gè)不同實(shí)數(shù)根.于是ln兩式相加,整理得a=lnx兩式相減,整理得a=lnx因此lnx1+ln于是lnx1+lnx2=(lnx2?ln又0<x1<x2,可設(shè)t=x2x1,因此,lnx1+lnx2=(1+t)lnt要證lnx1+lnx2>2,即證(1+t)lntt設(shè)g(t)=lnt-2(t?1)tg'(t)=1t-2(t+1)?2(所以g(t)在(1,+∞)上為增函數(shù).g(t)>g(1)=0.于是,當(dāng)t>1時(shí),有l(wèi)nt>2(t所以lnx1+lnx2>2成立,即x1x2>e2.對(duì)于齊次構(gòu)造法解決極值點(diǎn)偏移的二元變量問(wèn)題,通常對(duì)導(dǎo)函數(shù)對(duì)應(yīng)的零點(diǎn)方程相加減,構(gòu)造出對(duì)應(yīng)零點(diǎn)的齊次方程,順其自然消去參數(shù),結(jié)合所證構(gòu)造對(duì)應(yīng)函數(shù)解決問(wèn)題.三、分離構(gòu)造法統(tǒng)一變量角度1兩個(gè)變量若能分離[典例3]已知函數(shù)f(x)=(a+1)lnx+ax2+1,若a<-1,對(duì)?x1,x2∈(0,+∞),|f(x1)-f(x2)|≥4|x1-x2|,求a的取值范圍.【解題思路】由|f(x1)-f(x2)|≥4|x1-x2|想到利用單調(diào)性去掉絕對(duì)值符號(hào),分離兩個(gè)變量,得到f(x2)+4x2≥f(x1)+4x1,根據(jù)特征構(gòu)造函數(shù)g(x)=f(x)+4x,再利用導(dǎo)數(shù)求解.解析：f(x)的定義域?yàn)?0,+∞),f'(x)=a+1x+2ax.當(dāng)a<-1時(shí),f'(所以函數(shù)f(x)在(0,+∞)上單調(diào)遞減.不妨設(shè)x1≥x2,從而對(duì)?x1,x2∈(0,+∞),|f(x1)-f(x2)|≥4|x1-x2|等價(jià)于f(x2)+4x2≥f(x1)+4x1①.令g(x)=f(x)+4x,則g'(x)=a+1x+2①等價(jià)于g(x)在(0,+∞)上單調(diào)遞減,即a+1x+2從而a≤?4x?12x2故a的取值范圍為(-∞,-2].角度2若兩個(gè)變量不能分離[典例4]已知函數(shù)f(x)=lnx,(1)求函數(shù)g(x)=(x2+1)f(x)-2x+2(x≥1)的最小值;(2)當(dāng)0<a<b時(shí),求證:f(b)-f(a)>2a【解題思路】對(duì)于第二問(wèn),由于兩個(gè)變量a,b的獨(dú)立性不強(qiáng),不能直接構(gòu)造.根據(jù)(1)中的結(jié)論可得lnx>2(x?1)x2+1,將兩個(gè)變量合二為一,令x解析：(1)g'(x)=2xlnx+x2+1x-2=2xlnx+x+因?yàn)閤≥1,所以g'(x)≥0,所以g(x)在[1,+∞)上是增函數(shù),所以g(x)min=g(1)=0.(2)由(1)知,當(dāng)x>1時(shí),g(x)>0,所以(x2+1)lnx>2(x-1),即lnx>2(x令x=ba,則lnba>2(b所以f(b)-f(a)>2a對(duì)于雙變量問(wèn)題,若兩個(gè)變量地位均等,相互獨(dú)立,能分離,則分離構(gòu)造一元函數(shù);如不能分離,則合二為一,構(gòu)造一元函數(shù).四、主元法構(gòu)造統(tǒng)一變量[典例5]已知函數(shù)f(x)=aex-lnx-1,求證:當(dāng)a≥1e時(shí),f(x)≥0【解題思路】本題先將f(x)視為關(guān)于參數(shù)a的一次函數(shù),利用函數(shù)單調(diào)性將參數(shù)消掉,進(jìn)而轉(zhuǎn)化為關(guān)于x的單變量問(wèn)題求解.【證明】令h(a)=aex-lnx-1,則h(a)在1e,+∞上單調(diào)遞增,所以h(a)min=h(1要證f(x)≥0,只需證h(a)≥0,即證h(a)min=h(1e)=ex-1-lnx-1≥0構(gòu)造函數(shù)g(x)=ex-1-lnx-1,則g'(x)=ex-1-1x在(0,+∞)上單調(diào)遞增注意到g'(1)=0,所以當(dāng)0<x<1時(shí),g'(x)<0;當(dāng)x>1時(shí),g'(x)>0.故g(x)min=g(1)=0,g(x)≥0,即ex-1-lnx-1≥0,所以當(dāng)a≥1e時(shí),f(x)≥0當(dāng)問(wèn)題中含有變量較多且變量之間互不影響時(shí),可以選擇一個(gè)作為主元,并以此為線索解決問(wèn)題,這樣的方法叫做主元法.對(duì)于含所有參數(shù)的函數(shù)不等式證明,巧妙設(shè)置主元,將參數(shù)消掉,進(jìn)而轉(zhuǎn)化為單變量問(wèn)題來(lái)處理,可以使問(wèn)題化繁為簡(jiǎn),事半功倍.五、切線法統(tǒng)一變量[典例6]已知函數(shù)f(x)=lnx+(e-a)x-2b,若不等式f(x)≤0,對(duì)x∈(0,+∞)恒成立,則ba的最小值為 (A.-12e B.-2e C.1e D.【解題思路】問(wèn)題轉(zhuǎn)化為lnx≤(a-e)x+2b恒成立.設(shè)(x0,y0)為曲線y=lnx上的任意點(diǎn),求切線方程得到lnx≤1x0x+lnx0-1,利用系數(shù)相等得到a=e+1x0,b=解析：選A.由題意知lnx≤(a-e)x+2b對(duì)x∈(0,+∞)恒成立,設(shè)y=lnx上一點(diǎn)為(x0,y0),則lnx≤1x0(x-x0)+lnx0=1x0x+ln故a則