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11导数一、选择题1.(福建11)如果函数y=f(x)的图象如右图,那么导函数/()yfx的图象可能是(A)2.(辽宁6)设P为曲线C:223yxx上的点,且曲线C在点P处切线倾斜角的取值范围为04,,则点P横坐标的取值范围为(A)A.112,B.10,C.01,D.112,3.(全国Ⅰ4)曲线324yxx在点(13),处的切线的倾斜角为(B)A.30°B.45°C.60°D.120°4.(全国Ⅱ7)设曲线2axy在点(1,a)处的切线与直线062yx平行,则a(A)A.1B.12C.12D.1二、填空题1.(北京13)如图,函数()fx的图象是折线段ABC,其中ABC,,的坐标分别为(04)(20)(64),,,,,,则((0))ff_________;2函数()fx在1x处的导数(1)f_________.22BCAyx1O345612342.(江苏14)13)(3xaxxf对于1,1x总有0)(xf成立,则a=4三、解答题1.(安徽20)(本小题满分12分)设函数323()(1)1,32afxxxaxa其中为实数。(Ⅰ)已知函数()fx在1x处取得极值,求a的值;(Ⅱ)已知不等式'2()1fxxxa对任意(0,)a都成立,求实数x的取值范围。解:(1)'2()3(1)fxaxxa,由于函数()fx在1x时取得极值,所以'(1)0f即310,1aaa∴(2)方法一由题设知:223(1)1axxaxxa对任意(0,)a都成立即22(2)20axxx对任意(0,)a都成立设22()(2)2()gaaxxxaR,则对任意xR,()ga为单调递增函数()aR所以对任意(0,)a,()0ga恒成立的充分必要条件是(0)0g即220xx,20x∴于是x的取值范围是|20xx方法二由题设知:223(1)1axxaxxa对任意(0,)a都成立即22(2)20axxx对任意(0,)a都成立于是2222xxax对任意(0,)a都成立,即22202xxx20x∴于是x的取值范围是|20xx2.(北京17)(本小题共13分)已知函数32()3(0)fxxaxbxcb,且()()2gxfx是奇函数.(Ⅰ)求a,c的值;(Ⅱ)求函数()fx的单调区间.解:(Ⅰ)因为函数()()2gxfx为奇函数,所以,对任意的xR,()()gxgx,即()2()2fxfx.又32()3fxxaxbxc所以32323232xaxbxcxaxbxc.所以22aacc,.解得02ac,.(Ⅱ)由(Ⅰ)得3()32fxxbx.所以2()33(0)fxxbb.当0b时,由()0fx得xb.x变化时,()fx的变化情况如下表:x()b,b()bb,bb(,)()fx00所以,当0b时,函数()fx在()b,上单调递增,在()bb,上单调递减,在()b,上单调递增.当0b时,()0fx,所以函数()fx在(),上单调递增.3.(福建21)(本小题满分12分)已知函数32()2fxxmxnx的图象过点(-1,-6),且函数()()6gxfxx的图象关于y轴对称.(Ⅰ)求m、n的值及函数y=f(x)的单调区间;(Ⅱ)若a>0,求函数y=f(x)在区间(a-1,a+1)内的极值.解:(1)由函数f(x)图象过点(-1,-6),得m-n=-3,……①由f(x)=x3+mx2+nx-2,得f′(x)=3x2+2mx+n,则g(x)=f′(x)+6x=3x2+(2m+6)x+n;而g(x)图象关于y轴对称,所以-3262m=0,所以m=-3,代入①得n=0.于是f′(x)=3x2-6x=3x(x-2).由f′(x)得x2或x0,故f(x)的单调递增区间是(-∞,0),(2,+∞);由f′(x)0得0x2,故f(x)的单调递减区间是(0,2).(Ⅱ)由(Ⅰ)得f′(x)=3x(x-2),令f′(x)=0得x=0或x=2.当x变化时,f′(x)、f(x)的变化情况如下表:X(-∞.0)0(0,2)2(2,+∞)f′(x)+0-0+f(x)极大值极小值由此可得:当0a1时,f(x)在(a-1,a+1)内有极大值f(O)=-2,无极小值;当a=1时,f(x)在(a-1,a+1)内无极值;当1a3时,f(x)在(a-1,a+1)内有极小值f(2)=-6,无极大值;当a≥3时,f(x)在(a-1,a+1)内无极值.综上得:当0a1时,f(x)有极大值-2,无极小值,当1a3时,f(x)有极小值-6,无极大值;当a=1或a≥3时,f(x)无极值.4.(宁夏21)(本小题满分12分)设函数()bfxaxx,曲线()yfx在点(2(2))f,处的切线方程为74120xy.(Ⅰ)求()fx的解析式;(Ⅱ)证明:曲线()yfx上任一点处的切线与直线0x和直线yx所围成的三角形面积为定值,并求此定值.21.解:(Ⅰ)方程74120xy可化为734yx.当2x时,12y.······················································································2分又2()bfxax,于是1222744baba,,解得13.ab,故3()fxxx.··························································································6分(Ⅱ)设00()Pxy,为曲线上任一点,由231yx知曲线在点00()Pxy,处的切线方程为002031()yyxxx,即00200331()yxxxxx.令0x得06yx,从而得切线与直线0x的交点坐标为060x,.令yx得02yxx,从而得切线与直线yx的交点坐标为00(22)xx,.············10分所以点00()Pxy,处的切线与直线0x,yx所围成的三角形面积为016262xx.故曲线()yfx上任一点处的切线与直线0x,yx所围成的三角形的面积为定值,此定值为6.··································································································12分5.(江西21)已知函数4322411()(0)43fxxaxaxaa(1)求函数()yfx的单调区间;(2)若函数()yfx的图像与直线1y恰有两个交点,求a的取值范围.解:(1)因为322()2(2)()fxxaxaxxxaxa令()0fx得1232,0,xaxxa由0a时,()fx在()0fx根的左右的符号如下表所示x(,2)a2a(2,0)a0(0,)aa(,)a()fx000()fx极小值极大值极小值所以()fx的递增区间为(2,0)(,)aa与()fx的递减区间为(2)(0)aa,与,(2)由(1)得到45()(2)3fxfaa极小值,47()()12fxfaa极小值4()(0)fxfa极大值要使()fx的图像与直线1y恰有两个交点,只要44571312aa或41a,即4127a或01a.6.(湖南21)已知函数43219()42fxxxxcx有三个极值点。(I)证明:275c;(II)若存在实数c,使函数)(xf在区间,2aa上单调递减,求a的取值范围。解:(I)因为函数43219()42fxxxxcx有三个极值点,所以32()390fxxxxc有三个互异的实根.设32()39,gxxxxc则2()3693(3)(1),gxxxxx当3x时,()0,gx()gx在(,3)上为增函数;当31x时,()0,gx()gx在(3,1)上为减函数;当1x时,()0,gx()gx在(1,)上为增函数;所以函数()gx在3x时取极大值,在1x时取极小值.当(3)0g或(1)0g时,()0gx最多只有两个不同实根.因为()0gx有三个不同实根,所以(3)0g且(1)0g.即2727270c,且1390c,解得27,c且5,c故275c.(II)由(I)的证明可知,当275c时,()fx有三个极值点.不妨设为123xxx,,(123xxx),则123()()()().fxxxxxxx所以()fx的单调递减区间是1(]x,,23[,]xx若)(xf在区间,2aa上单调递减,则,2aa1(]x,,或,2aa23[,]xx,若,2aa1(]x,,则12ax.由(I)知,13x,于是5.a若,2aa23[,]xx,则2ax且32ax.由(I)知,231.x又32()39,fxxxxc当27c时,2()(3)(3)fxxx;当5c时,2()(5)(1)fxxx.因此,当275c时,313.x所以3,a且23.a即31.a故5,a或31.a反之,当5,a或31a时,总可找到(27,5),c使函数)(xf在区间,2aa上单调递减.综上所述,a的取值范围是(5)(3,1),.7.(辽宁22)(本小题满分14分)设函数322()31()fxaxbxaxabR,在1xx,2xx处取得极值,且122xx.(Ⅰ)若1a,求b的值,并求()fx的单调区间;(Ⅱ)若0a,求b的取值范围.解:22()323fxaxbxa.①·····································································2分(Ⅰ)当1a时,2()323fxxbx;由题意知12xx,为方程23230xbx的两根,所以2124363bxx.由122xx,得0b.···············································································4分从而2()31fxxx,2()333(1)(1)fxxxx.当(11)x,时,()0fx;当(1)(1)x∞,,∞时,()0fx.故()fx在(11),单调递减,在(1)∞,,(1),∞单调递增.·······························6分(Ⅱ)由①式及题意知12xx,为方程223230xbxa的两根,所以23124363baxxa.从而221229(1)xxbaa,由上式及题设知01a≤.·············································································8分考虑23()99gaaa,22()1827273gaaaa
本文标题:08年高文科数学分类导数
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