高考數學專題復習導練測 第三章 導數及其應用階段測試(四)課件 理 新人教A版.ppt
數學 A(理),第三章 導數及其應用,45分鐘階段測試(四),2,3,4,5,6,7,8,9,10,1,2,3,4,5,6,7,8,9,10,1,一、選擇題,2,3,4,5,6,7,8,9,10,1,又直線axy30的斜率為a,,答案 B,3,4,5,6,7,8,9,10,1,2,2.已知函數f(x)x3ax2x1在R上是單調減函數,則實數a的取值范圍是( ),解析 由題意,知f(x)3x22ax10在R上恒成立,,B,2,4,5,6,7,8,9,10,1,3,A.0 B.1 C.2 D.3,2,4,5,6,7,8,9,10,1,3,f(x)minf(1)0,a0, 即a的最大值為0. 答案 A,2,3,5,6,7,8,9,10,1,4,2,3,5,6,7,8,9,10,1,4,答案 D,2,3,4,6,7,8,9,10,1,5,5.已知函數f(x)對定義域R內的任意x都有f(x)f(4x),且當x2時,其導函數f(x)滿足xf(x)2f(x),若2a4,則( ) A.f(2a)f(3)f(log2a) B.f(3)f(log2a)f(2a) C.f(log2a)f(3)f(2a) D.f(log2a)f(2a)f(3),2,3,4,6,7,8,9,10,1,5,解析 由f(x)f(4x),可知函數圖象關于x2對稱. 由xf(x)2f(x),得(x2)f(x)0, 所以當20恒成立,函數f(x)單調遞增. 由2a4,得1log2a2,222a24,即42a16. 因為f(log2a)f(4log2a),所以24log2a3, 即24log2a32a,,2,3,4,6,7,8,9,10,1,5,所以f(4log2a)f(3)f(2a), 即f(log2a)f(3)f(2a). 答案 C,2,3,4,5,7,8,9,10,1,6,二、填空題,2,3,4,5,6,8,9,10,1,7,7.函數f(x)x33axb(a0)的極大值為6,極小值為2,則f(x)的單調遞減區(qū)間是_.,f(x),f(x)隨x的變化情況如下表:,2,3,4,5,6,8,9,10,1,7,答案 (1,1),2,3,4,5,6,9,10,1,7,8,8.已知f(x)2x36x23,對任意的x2,2都有f(x)a,則a的取值范圍為_.,解析 由f(x)6x212x0,得x0或x2. 又f(2)37,f(0)3,f(2)5, f(x)max3,又f(x)a,a3.,3,),三、解答題 9.已知函數f(x) x2aln x(aR). (1)若函數f(x)的圖象在x2處的切線方程為yxb,求a,b的值;,解 因為f(x)x (x0),,又f(x)在x2處的切線方程為yxb,,2,3,4,5,6,7,8,10,1,9,2,3,4,5,6,7,8,10,1,9,(2)若函數f(x)在(1,)上為增函數,求a的取值范圍.,解 若函數f(x)在(1,)上為增函數,,即ax2在(1,)上恒成立. 所以有a1.,2,3,4,5,6,7,8,9,1,10,10.(2014大綱全國)函數f(x)ln(x1) (a1). (1)討論f(x)的單調性;,當10,f(x)在(1,a22a)是增函數; 若x(a22a,0),則f(x)0,f(x)在(0,)是增函數.,2,3,4,5,6,7,8,9,1,10,當a2時,f(x)0,f(x)0成立當且僅當x0,f(x)在(1,)是增函數. 當a2時,若x(1,0),則f(x)0,f(x)在(1,0)是增函數; 若x(0,a22a),則f(x)0,f(x)在(a22a,)是增函數.,2,3,4,5,6,7,8,9,1,10,證明 由(1)知,當a2時,f(x)在(1,)是增函數. 當x(0,)時,f(x)f(0)0,,又由(1)知,當a3時,f(x)在0,3)是減函數.,2,3,4,5,6,7,8,9,1,10,當x(0,3)時,f(x)f(0)0,,2,3,4,5,6,7,8,9,1,10,2,3,4,5,6,7,8,9,1,10,根據、可知對任何nN*結論都成立.,