解答題專項(xiàng)訓(xùn)練(函數(shù)與導(dǎo)數(shù))

高考數(shù)學(xué)精品復(fù)習(xí)資料 2019.5專題升級訓(xùn)練 解答題專項(xiàng)訓(xùn)練(函數(shù)與導(dǎo)數(shù))1.已知函數(shù)f(x)=x2+(x0,aR).(1)討論函數(shù)f(x)的奇偶性,并說明理由;(2)若函數(shù)f(x)在2,+)上為增函數(shù),求a的取值范圍.2.設(shè)定義在(0,+)上的函數(shù)f(x)=ax+b(a>0).(1)求f(x)的最小值;(2)若曲線y=f(x)在點(diǎn)(1,f(1)處的切線方程為y=x,求a,b的值.3.已知定義在實(shí)數(shù)集R上的奇函數(shù)f(x)有最小正周期2,且當(dāng)x(0,1)時,f(x)=.(1)求函數(shù)f(x)在(-1,1)上的解析式;(2)判斷f(x)在(0,1)上的單調(diào)性;(3)當(dāng)取何值時,方程f(x)=在(-1,1)上有實(shí)數(shù)解?4.(20xx·山東濟(jì)寧模擬,21)設(shè)函數(shù)f(x)=ln x,g(x)=ax+,函數(shù)f(x)的圖象與x軸的交點(diǎn)也在函數(shù)g(x)的圖象上,且在此點(diǎn)有公切線.(1)求a,b的值;(2)試比較f(x)與g(x)的大小.5.已知函數(shù)f(x)=ex-ax-1(aR).(1)討論f(x)=ex-ax-1(aR)的單調(diào)性;(2)若a=1,求證:當(dāng)x0時,f(x)f(-x).6.已知函數(shù)f(x)滿足f(x)=f'(1)ex-1-f(0)x+x2.(1)求f(x)的解析式及單調(diào)區(qū)間;(2)若f(x)x2+ax+b,求(a+1)b的最大值.7.已知函數(shù)f(x)=在x=1處取得極值2,設(shè)函數(shù)y=f(x)圖象上任意一點(diǎn)(x0,f(x0)處的切線斜率為k.(1)求k的取值范圍;來源:(2)若對于任意0<x1<x2<1,存在k,使得k=,求證:x1<|x0|<x2.來源:8.(20xx·山西太原模擬,21)設(shè)函數(shù)f(x)=x2+ax-ln x(aR).(1)當(dāng)a=1時,求函數(shù)f(x)的極值;(2)當(dāng)a>1時,討論函數(shù)f(x)的單調(diào)性;(3)若對任意a(3,4)及任意x1,x21,2,恒有m+ln 2>|f(x1)-f(x2)|成立,求實(shí)數(shù)m的取值范圍.#1.解:(1)當(dāng)a=0時,f(x)=x2,對任意x(-,0)(0,+),f(-x)=(-x)2=x2=f(x),f(x)為偶函數(shù).當(dāng)a0時,f(x)=x2+(a0,x0),取x=±1,得f(-1)+f(1)=20,f(-1)-f(1)=-2a0,f(-1)-f(1),f(-1)f(1).函數(shù)f(x)既不是奇函數(shù),也不是偶函數(shù).(2)若函數(shù)f(x)在2,+)上為增函數(shù),則f'(x)0在2,+)上恒成立,來源:即2x-0在2,+)上恒成立,即a2x3在2,+)上恒成立,只需a(2x3)min,x2,+),a16.a的取值范圍是(-,16.2.解:(1)f(x)=ax+b2+b=b+2,來源:當(dāng)且僅當(dāng)ax=1時,f(x)取得最小值為b+2.(2)由題意得:f(1)=a+b=,f'(x)=a-f'(1)=a-,由得:a=2,b=-1.3.解:(1)f(x)是xR上的奇函數(shù),f(0)=0.設(shè)x(-1,0),則-x(0,1),f(-x)=-f(x),f(x)=-,f(x)=(2)設(shè)0<x1<x2<1,f(x1)-f(x2)=來源:=,0<x1<x2<1,>20=1,f(x1)-f(x2)>0,f(x)在(0,1)上為減函數(shù).(3)f(x)在(0,1)上為減函數(shù),<f(x)<,即f(x).同理,f(x)在(-1,0)上的值域?yàn)?又f(0)=0,當(dāng),或=0時,方程f(x)=在x(-1,1)上有實(shí)數(shù)解.4.解:(1)f(x)=ln x的圖象與x軸的交點(diǎn)坐標(biāo)是(1,0),依題意,得g(1)=a+b=0,又f'(x)=,g'(x)=a-,f(x)與g(x)在點(diǎn)(1,0)處有公切線,g'(1)=f'(1)=1,即a-b=1.由得a=,b=-.(2)令F(x)=f(x)-g(x),則F(x)=ln x-=ln x-x+.F'(x)=-0.F(x)在(0,+)上為減函數(shù),當(dāng)0<x<1時,F(x)>F(1)=0,即f(x)>g(x);當(dāng)x=1時,F(x)=F(1)=0,即f(x)=g(x);當(dāng)x>1時,F(x)<F(1)=0,即f(x)<g(x).綜上可知,當(dāng)0<x1時,f(x)g(x);當(dāng)x>1時,f(x)<g(x).5. 解: (1) f'(x)=ex-a.當(dāng)a0時,f'(x)0恒成立,當(dāng)a>0時,令f'(x)>0,得x>ln a;令f'(x)<0,得x<ln a.綜上,當(dāng)a0時,f(x)在(-,+)上單調(diào)遞增;當(dāng)a>0時,增區(qū)間是(ln a,+),減區(qū)間是(-,ln a).(2)證明:令g(x)=f(x)-f(-x)=ex-2x,g'(x)=ex+e-x-20,g(x)在0,+)上是增函數(shù),g(x)g(0)=0,f(x)f(-x).6.解:(1)f(x)=f'(1)ex-1-f(0)x+x2=ex-f(0)x+x2f'(x)=f'(1)ex-1-f(0)+x,令x=1得f(0)=1.f(x)=f'(1)ex-1-x+x2f(0)=f'(1)e-1=1f'(1)=e,得:f(x)=ex-x+x2.令g(x)=f'(x)=ex-1+x,則g'(x)=ex+1>0y=g(x)在xR上單調(diào)遞增,f'(x)在R上單調(diào)遞增,f'(x)>0=f'(0)x>0,f'(x)<0=f'(0)x<0,得f(x)的解析式為f(x)=ex-x+x2,且單調(diào)遞增區(qū)間為(0,+),單調(diào)遞減區(qū)間為(-,0).(2)令h(x)=f(x)-x2-ax-b,則h(x)=ex-(a+1)x-b0,h'(x)=ex-(a+1).當(dāng)a+10時,h'(x)>0y=h(x)在xR上單調(diào)遞增,x-時,h(x)-與h(x)0矛盾.當(dāng)a+1>0時,h'(x)>0x>ln(a+1),h'(x)<0x<ln(a+1),得:當(dāng)x=ln(a+1)時,h(x)min =(a+1)-(a+1)ln(a+1)-b0,(a+1)b(a+1)2-(a+1)2ln(a+1)(a+1>0).令F(x)=x2-x2ln x(x>0),則F'(x)=x(1-2ln x),F'(x)>00<x<,F'(x)<0x>.當(dāng)x=時,F(x)max =.當(dāng)a=-1,b=時,(a+1)b的最大值為.7. 解: (1) f'(x)=.由f'(1)=0及f(1)=2,得a=4,b=1.k=f'(x0)=4,設(shè)=t,t(0,1,得k.(2)證明:f'(x)=,令f'(x)>0x(-1,1).f(x)的增區(qū)間為(-1,1),故當(dāng)0<x1<x2<1時,>0,即k>0,故x0(-1,1).由于f'(x0)=f'(-x0),故只需要證明x0(0,1)時結(jié)論成立.由k=,得f(x2)-kx2=f(x1)-kx1,記h(x)=f(x)-kx,則h(x2)=h(x1).h'(x)=f'(x)-k,則h'(x0)=0,設(shè)g(x)=,x(0,1),g'(x)=<0,g(x)為減函數(shù),故f'(x)為減函數(shù).故當(dāng)x>x0時,有f'(x)<f'(x0)=k,此時h'(x)<0,h(x)為減函數(shù).當(dāng)x<x0時,h'(x)>0,h(x)為增函數(shù).所以h(x0)為h(x)的唯一的極大值,因此要使h(x2)=h(x1),必有x1<x0<x2.綜上,有x1<|x0|<x2成立.8.解:(1)函數(shù)的定義域?yàn)?0,+).當(dāng)a=1時,f(x)=x-ln x,f'(x)=1-,當(dāng)0<x<1時,f'(x)<0;當(dāng)x>1時,f'(x)>0.f(x)極小值=f(1)=1,無極大值.(2)f'(x)=(1-a)x+a-.當(dāng)=1,即a=2時,f'(x)=-0,f(x)在定義域上單調(diào)遞減;當(dāng)<1,即a>2時,令f'(x)<0,得0<x<或x>1.令f'(x)>0,得<x<1.當(dāng)>1,即1<a<2時,令f'(x)<0,得0<x<1或x>令f'(x)>0,得1<x<.綜上知,當(dāng)a=2時,f(x)在(0,+)上單調(diào)遞減;當(dāng)a>2時,f(x)在和(1,+)上單調(diào)遞減,在上單調(diào)遞增;當(dāng)1<a<2時,f(x)在(0,1)和上單調(diào)遞減,在上單調(diào)遞增.(3)由(2)知,當(dāng)a(3,4)時,f(x)在1,2上單調(diào)遞減,f(1)是最大值,f(2)是最小值.|f(x1)-f(x2)|f(1)-f(2)=+ln 2,m+ln 2>+ln 2.而a(3,4),經(jīng)整理得m>,由3<a<4得0<,m.