PAGEPAGE1課時(shí)作業(yè)(十)第10講函數(shù)的圖像時(shí)間/30分鐘分值/80分基礎(chǔ)熱身1.函數(shù)y=x|x|的圖像大致是 ()ABCD圖K10-12.將函數(shù)f(x)=x2-2x的圖像向右平移1個(gè)單位長度得到函數(shù)g(x)的圖像,再將函數(shù)g(x)的圖像向上平移2個(gè)單位長度得到函數(shù)h(x)的圖像,則函數(shù)h(x)的最小值是 ()A.1 B.-2 C.2 D.-33.[2024·安徽皖江名校聯(lián)考]已知函數(shù)f(x)=ln|x|,g(x)=-x2+3,則y=f(x)·g(x)的圖像為 ()ABCD圖K10-24.若函數(shù)f(x)=ax+b,x<-1,ln(x+a),x≥-1的圖像如圖K10圖K10-3A.-1B.-5C.-1D.-25.設(shè)函數(shù)f(x)=12x-1,x<0,x12,x≥0,實(shí)力提升圖K10-46.[2024·廣東揭陽一模]函數(shù)f(x)的部分圖像如圖K10-4所示,則f(x)的解析式可能是 ()A.f(x)=x2(x2-π2)B.f(x)=xcosx+πC.f(x)=xsinxD.f(x)=x2+cosx-17.已知函數(shù)f(x)的圖像向左平移1個(gè)單位長度后得到的圖像關(guān)于y軸對稱,當(dāng)x2>x1>1時(shí),f(x2)-f(x1)x2-x1<0恒成立,設(shè)a=f-1A.c>a>b B.c>b>aC.a>c>b D.b>a>c8.[2024·青島5月模擬]已知函數(shù)f(x)=1x-lnx-1,則y=f(x)ABCD圖K10-59.函數(shù)f(x)=x2ln|x|的圖像大致是ABCD圖K10-610.[2024·保定一模]定義在R上的偶函數(shù)f(x)滿意f(x+1)=-f(x),當(dāng)x∈[0,1]時(shí),f(x)=-2x+1,設(shè)函數(shù)g(x)=12|x-1|,x∈(-1,3),則函數(shù)f(x)與g(x)的圖像全部A.2 B.4C.6 D.811.已知f(x)=ln(1-x),函數(shù)g(x)的圖像與f(x)的圖像關(guān)于點(diǎn)(1,0)對稱,則g(x)=.

12.若函數(shù)f(x)=ax-2x-1的圖像關(guān)于點(diǎn)(1,1)對稱,則實(shí)數(shù)13.方程x2-ax+1=0(a≠0)的根的個(gè)數(shù)是14.已知函數(shù)f(x)=|log2x|,x>0,-x2-2x,x≤0,關(guān)于x的方程f(x)=m(m∈R)有四個(gè)不同的實(shí)數(shù)解x1,x2,x3難點(diǎn)突破15.(5分)已知函數(shù)f(x)=-x2+2x,x≤0,ln(x+1),x>0,若|fA.(-∞,0] B.(-∞,1]C.[-2,1] D.[-2,0]16.(5分)[2024·廣東茂名3月聯(lián)考]已知函數(shù)f(x)=x2+x+2-x4-xA.函數(shù)f(x)在區(qū)間[-1,3]上單調(diào)遞增B.函數(shù)f(x)在區(qū)間[-1,3]上單調(diào)遞減C.函數(shù)f(x)的圖像關(guān)于直線x=1對稱D.函數(shù)f(x)的圖像關(guān)于點(diǎn)(1,0)對稱課時(shí)作業(yè)(十)1.D[解析]明顯y=x|x|是奇函數(shù),解除A,B,C,故選D.2.A[解析]依題意,得f(x)=x2-2x=(x-1)2-1,g(x)=(x-2)2-1,h(x)=(x-2)2+1,所以函數(shù)h(x)的最小值是1.故選A.3.C[解析]由已知得y=f(x)·g(x)為偶函數(shù),解除選項(xiàng)A,D,當(dāng)x∈(0,1)時(shí),f(x)<0,g(x)>0,y=f(x)·g(x)<0,解除B.故選C.4.C[解析]由圖像可得a·(-1)+b=3,ln(-1+a)=0,得a=2,b=5,所以f(x)=2x+5,x<-1,ln(x+2),x≥-1,故f(-3)=2×(5.[-1,1][解析]在同始終角坐標(biāo)系中,作出函數(shù)y=f(x)的圖像和直線y=1,如圖所示,它們相交于(-1,1)和(1,1)兩點(diǎn),由f(a)≤1,得-1≤a≤1.6.C[解析]當(dāng)x∈(0,π)時(shí),f(x)>0,解除選項(xiàng)A;由圖知f(x)是偶函數(shù),而f(x)=xcosx+π是非奇非偶函數(shù),解除選項(xiàng)B;又f(π)=0,而選項(xiàng)D中f(π)>0,解除選項(xiàng)D.故選C.7.D[解析]因?yàn)楹瘮?shù)f(x)的圖像向左平移1個(gè)單位長度后得到的圖像關(guān)于y軸對稱,所以f(x)的圖像關(guān)于直線x=1對稱,所以a=f-12=f52,又f(x)在(1,+∞)上單調(diào)遞減,所以f(2)>f52>f(3),即8.B[解析]由f1e=11e+1-1=e>0,解除選項(xiàng)A;由f(e)=1e-2>0解除選項(xiàng)C;又f(e2)=1e2-3>0,e2-3>e-2,所以f(e)>f(e9.D[解析]函數(shù)f(x)是奇函數(shù),解除選項(xiàng)B.當(dāng)0<x<1時(shí),f(x)<0,解除選項(xiàng)A.當(dāng)x>1時(shí),f(x)=x2lnx,f'(x)=lnx-12(lnx)2,令f'(x)>0,得x>e,令f'(x)<0,得1<x<e,所以f(x)在(1,e)上單調(diào)遞減,在(e,+∞)上單調(diào)遞增10.B[解析]由f(x+1)=-f(x),得f(x+2)=-f(x+1)=f(x),所以f(x)是周期為2的周期函數(shù).函數(shù)g(x)=12|x-1|,x∈(-1,3)的圖像關(guān)于直線x=1對稱,作出y=f(x)與y=g(x)的圖像,由圖可得兩圖像的四個(gè)交點(diǎn)的橫坐標(biāo)關(guān)于直線x=1對稱,其和為2×2=11.-ln(x-1)[解析]設(shè)P(x,y)為函數(shù)y=g(x)的圖像上隨意一點(diǎn),則點(diǎn)P(x,y)關(guān)于點(diǎn)(1,0)的對稱點(diǎn)Q(2-x,-y)在函數(shù)y=f(x)的圖像上,即-y=f(2-x)=ln(x-1),所以y=-ln(x-1),即g(x)=-ln(x-1).12.1[解析]函數(shù)f(x)=ax-2x-1=a+a-2x-1,當(dāng)a=2時(shí),f(x)=2(x≠1),函數(shù)f(x)的圖像不關(guān)于點(diǎn)(1,1)對稱,故a≠2,f(x)的圖像的對稱中心為點(diǎn)(1,13.1[解析]令f(x)=x2+1,g(x)=ax,作出函數(shù)f(x)和g(x)的圖像(圖略),易知不論a>0還是a<0,兩個(gè)函數(shù)的圖像都有1個(gè)交點(diǎn),所以方程x2-ax+1=0(a≠0)有114.(0,1)[解析]作出函數(shù)f(x)的圖像和直線y=m,如圖所示.當(dāng)函數(shù)f(x)的圖像與直線y=m有四個(gè)不同交點(diǎn)時(shí),0<m<1,不妨設(shè)交點(diǎn)的橫坐標(biāo)從左到右依次為x1,x2,x3,x4,則-2<x1<-1<x2<0<x3<1<x4<2.又0<x1x2<1,|log2x3|=|log2x4|,即log21x3=log2x4,所以x3x4=1,所以0<x1x2x3x4<15.D[解析]作出y=|f(x)|的圖像(如圖所示).①當(dāng)x>0時(shí),只有a≤0才能滿意|f(x)|≥ax,可解除選項(xiàng)B,C.②當(dāng)x≤0時(shí),y=|f(x)|=|-x2+2x|=x2-2x,故由|f(x)|≥ax得x2-2x≥ax.當(dāng)x=0時(shí),不等式為0≥0成立;當(dāng)x<0時(shí),不等式等價(jià)為x-2≤a,因?yàn)閤-2<-2,所以a≥-2.綜上可知,a∈[-2,0].故選D.16.C[解析]由f(x)=x2+x+2-x4-x得f