2025版高考數(shù)學(xué)一輪復(fù)習(xí)易錯考點(diǎn)排查練函數(shù)與導(dǎo)數(shù)含解析新人教B版

PAGE易錯考點(diǎn)排查練函數(shù)與導(dǎo)數(shù)1.函數(shù)f(x)=QUOTE+lgQUOTE的定義域?yàn)?()A.(2,3) B.(2,4]C.(2,3)∪(3,4] D.(-1,3)∪(3,6]【解析】選C.由函數(shù)y=f(x)的表達(dá)式可知,函數(shù)f(x)的定義域應(yīng)滿意條件:QUOTE解之得QUOTE即函數(shù)f(x)的定義域?yàn)?2,3)∪(3,4].2.若函數(shù)f(x)=QUOTE(k為常數(shù))在定義域上為奇函數(shù),則k的值為 ()A.1 B.-1 C.±1 D.0【解析】選C.利用定義:f(-x)+f(x)=0,f(x)+f(-x)=QUOTE+QUOTE,化簡得f(x)+fQUOTE=QUOTEk2-22x-1=0,因?yàn)?+22x>0,所以k2-1=0,即k=±1.3.函數(shù)f(x)=2x-3QUOTE,x∈QUOTE的值域?yàn)?()A.[-2,0] B.(-3,0)C.QUOTE D.QUOTE【解析】選C.令QUOTE=t,因?yàn)閤∈QUOTE,所以t∈QUOTE,所以x=t2-1,所以y=2(t2-1)-3t=2QUOTE-QUOTE,所以t=QUOTE時,f(x)取最小值-QUOTE;t=2時,f(x)取最大值0,但是取不到,所以f(x)的值域?yàn)镼UOTE.4.已知函數(shù)f(x)=x3+ax2+bx+a2在x=1處取極值10,則a= ()A.4或-3 B.4或-11C.4 D.-3【解析】選C.因?yàn)閒(x)=x3+ax2+bx+a2,所以f′(x)=3x2+2ax+b.由題意得QUOTE即QUOTE,解得QUOTE,或QUOTE.當(dāng)QUOTE時,f′(x)=3x2-6x+3=3(x-1)2≥0,故函數(shù)f(x)單調(diào)遞增,無極值.不符合題意.所以a=4.5.已知函數(shù)y=QUOTE的定義域?yàn)镽,則實(shí)數(shù)k的取值范圍是 ()A.k≤0或k≥1 B.k≥1C.0≤k≤1 D.0<k≤1【解析】選C.因?yàn)楹瘮?shù)y=QUOTE的定義域?yàn)镽,則kx2-6kx+9≥0恒成立.①當(dāng)k<0時,函數(shù)f(x)=kx2-6kx+9是開口向下的拋物線,不符合題意;②當(dāng)k=0時,函數(shù)f(x)=9恒滿意kx2-6kx+9≥0,符合題意;③當(dāng)k>0時,函數(shù)f(x)=kx2-6kx+9滿意kx2-6kx+9≥0恒成立的條件是Δ=b2-4ac≤0,即36k2-4×9k≤0,解得0<k≤1.由①②③知實(shí)數(shù)k的取值范圍是0≤k≤1.6.已知函數(shù)f(x)=ax(a>0,且a≠1)在區(qū)間QUOTE上的值域?yàn)镼UOTE,則a= ()A.QUOTE B.QUOTE C.QUOTE或QUOTE D.QUOTE或4【解析】選C.分析知,m>0.探討:當(dāng)a>1時,QUOTE,所以am=2,m=2,所以a=QUOTE;當(dāng)0<a<1時,QUOTE,所以am=QUOTE,m=QUOTE,所以a=QUOTE.綜上,a=QUOTE或a=QUOTE.7.函數(shù)f(x)=x2-2lnx的單調(diào)遞減區(qū)間是 ()A.QUOTE B.QUOTEC.QUOTE D.QUOTE【解析】選A.由f(x)=x2-2lnx,得f′(x)=(x2-2lnx)′=2x-QUOTE.因?yàn)楹瘮?shù)f(x)=x2-2lnx的定義域?yàn)?0,+∞),由f′(x)<0,得2x-QUOTE<0,即(x+1)(x-1)<0,解得0<x<1.所以函數(shù)f(x)=x2-2lnx的單調(diào)遞減區(qū)間是(0,1).8.在區(qū)間QUOTE中任取一個實(shí)數(shù)a,使函數(shù)f(x)=QUOTE在R上是增函數(shù)的概率為 ()A.QUOTE B.QUOTE C.QUOTE D.QUOTE【解析】選A.因?yàn)楹瘮?shù)f(x)=QUOTE是增函數(shù)所以QUOTE解得1<a≤2.所以從區(qū)間(0,6)中任取一個值a,則函數(shù)f(x)=QUOTE是增函數(shù)的概率為P=QUOTE.9.已知點(diǎn)A(1,2)在函數(shù)f(x)=ax3的圖象上,則過點(diǎn)A的曲線C:y=f(x)的切線方程是 ()A.6x-y-4=0B.x-4y+7=0C.6x-y-4=0或x-4y+7=0D.6x-y-4=0或3x-2y+1=0【解析】選D.由于點(diǎn)A(1,2)在函數(shù)f(x)=ax3的圖象上,則a=2,即y=2x3,y′=6x2,設(shè)切點(diǎn)為(m,2m3),則切線的斜率為k=6m2,由點(diǎn)斜式得:y-2m3=6m2(x-m).代入點(diǎn)A(1,2)得,2-2m3(m-1)2(2m+1)=0.解得m=1或-QUOTE,即斜率為6或QUOTE,則過點(diǎn)A的曲線C:y=f(x)的切線方程是:y-2=6(x-1)或y-2=QUOTE(x-1),即6x-y-4=0或3x-2y+1=0.10.已知函數(shù)f(x)=2x3-4x+2(ex-e-x),若f(5a-2)+f(3a2)≤0,則實(shí)數(shù)a的取值范圍是()A.QUOTE B.QUOTEC.QUOTE D.QUOTE【解析】選D.由函數(shù)f(x)=2x3-4x+2(ex-e-x),可得f(-x)=2(-x)3-4(-x)+2(e-x-ex)=-[2x3-4x+2(ex-e-x)]=-f(x),所以函數(shù)f(x)為奇函數(shù),又f′(x)=6x2-4+2(ex+QUOTE),因?yàn)閑x+QUOTE≥2QUOTE=2,所以f′(x)≥0,所以函數(shù)f(x)為單調(diào)遞增函數(shù),因?yàn)閒(5a-2)+f(3a2)≤0,即f(3a2)≤-f(5a-2)=f(2-5a),所以3a2≤2-5a?3a2+5a-2≤0,解得-2≤a≤QUOTE.11.(2024·沈陽模擬)已知函數(shù)f(x)=3x3-ax2+x-5在區(qū)間[1,2]上單調(diào)遞增,則a的取值范圍是 世紀(jì)金榜導(dǎo)學(xué)號()A.QUOTE B.QUOTEC.QUOTE D.QUOTE【解析】選B.f′(x)=9x2-2ax+1,因?yàn)閒(x)=3x3-ax2+x-5在區(qū)間[1,2]上單調(diào)遞增,所以f′(x)=9x2-2ax+1≥0在區(qū)間[1,2]上恒成立.即a≤QUOTE=QUOTE+QUOTE,當(dāng)x=1時,QUOTE+QUOTE有最小值5,即a≤5.12.已知函數(shù)f(x)=xlnx-QUOTEax2+(a-1)x(a∈R)在x=1處取得極大值,則實(shí)數(shù)a的取值范圍是 世紀(jì)金榜導(dǎo)學(xué)號()A.QUOTE B.(-∞,1)C.QUOTE D.(1,+∞)【解析】選D.當(dāng)a=0時,f(x)=xlnx-x,f′(x)=lnx,由f′(x)>0,得x>1,由f′(x)<0,得0<x<1,所以f(x)在(0,1)上遞減,在(1,+∞)上遞增,所以f(x)在x=1處有微小值,即a=0不合題意,解除A,B;當(dāng)a=1時,f(x)=xlnx-QUOTEx2,f′(x)=lnx-x+1=g(x),g′(x)=QUOTE,由g′(x)>0得0<x<1,由g′(x)<0得x>1,所以g(x)有最大值g(1)=0,所以f′(x)≤0,所以f(x)在(0,+∞)上遞減,在x=1處無極值,解除C.13.若函數(shù)f(x)=lg(x2+ax-a-1)在區(qū)間[2,+∞)上單調(diào)遞增,則實(shí)數(shù)a的取值范圍是__________.

【解析】因?yàn)楹瘮?shù)f(x)=lg(x2+ax-a-1)在區(qū)間[2,+∞)上單調(diào)遞增,所以QUOTE,解得a>-3.答案:(-3,+∞)14.函數(shù)f(x)=QUOTE+log2QUOTE為奇函數(shù),則實(shí)數(shù)a=________.

【解析】因?yàn)楹瘮?shù)f(x)=QUOTE+log2QUOTE為奇函數(shù),所以fQUOTE=-f(x)即fQUOTE+f(x)=0,則-QUOTE+log2QUOTE+QUOTE+log2QUOTE=0,即log2QUOTE=0,所以QUOTE·QUOTE=QUOTE=1,則1-a2x2=1-x2,所以a2=1,則a=±1,當(dāng)a=-1時,f(x)=QUOTE+log2QUOTE,則f(x)定義域?yàn)镼UOTE且QUOTE,此時定義域不關(guān)于原點(diǎn)對稱,為非奇非偶函數(shù),不滿意題意.當(dāng)a=1時,f(x)=QUOTE+log2QUOTE,滿意題意,所以a=1.答案:115.已知f(x)=4x-m·2x+1,設(shè)g(x)=QUOTE,若存在不相等的實(shí)數(shù)a,b同時滿意方程g(a)+g(b)=0和f(a)+f(b)=0,則實(shí)數(shù)m的取值范圍為________. 世紀(jì)金榜導(dǎo)學(xué)號

【解析】因?yàn)間QUOTE=QUOTE=QUOTE=-g(x),所以g(x)為R上的奇函數(shù),又gQUOTE+gQUOTE=0且a≠b,所以b=-a且a≠0,所以fQUOTE+fQUOTE=fQUOTE+fQUOTE=4a+4-a-mQUOTE=0,即m=QUOTE=QUOTE=QUOTE-QUOTE.令h(x)=QUOTE-QUOTE,則h′(x)=QUOTE+QUOTE>0,所以h(x)在QUOTE上單調(diào)遞增,所以h(x)>h(2)=1-QUOTE=QUOTE,又2a+2-a>2,所以hQUOTE=QUOTE-QUOTE>QUOTE,所以m∈QUOTE.答案:QUOTE16.已知f(x)是定義在R上的奇函數(shù),且當(dāng)x>0時,f(x)=-x2+2x.若函數(shù)f(x)在區(qū)間[-1,a-2]上單調(diào)遞增,則實(shí)數(shù)a的取值范圍為________. 世紀(jì)金榜導(dǎo)學(xué)號

【解析】因?yàn)閒(x)是定義在R上的奇函數(shù),所以fQUOTE=-f(x)且fQUOTE=0.當(dāng)x<0時,-x>0,所以f(x)=-fQUOTE=-QUOTE=x2+2x,又fQUOTE滿意f(x)=x2+2x,所以f(x)=QUOTE所以f(x)圖象如圖所示因?yàn)閒(x)在區(qū)間[-1,a-2]上單調(diào)遞增,所以-1<a-2≤1,解得a∈QUOTE,所以a的取值范圍為QUOTE.答案:QUOTE給易錯點(diǎn)找題號序號易錯點(diǎn)題號練后感悟1忽視真數(shù)大于零和分母不為零.12忽視對底數(shù)a進(jìn)行分類探討.63奇函數(shù)用結(jié)論f(0)=0錯選A.24利用導(dǎo)數(shù)的幾何意義求切線問題中忽視