高中數(shù)學(xué)人教A版選修11練習(xí)：第3章 導(dǎo)數(shù)及其應(yīng)用3.3.1 Word版含解析

1、高中數(shù)學(xué)選修精品教學(xué)資料第三章 3.3 3.3.1a級(jí)基礎(chǔ)鞏固一、選擇題1函數(shù)f(x)x33x21的遞減區(qū)間是(b)a(,0)b(0,2)c(,2)d(2,)解析f (x)3x26x,令f (x)3x26x<0,解得0<x<2,所以函數(shù)f(x)x33x21的遞減區(qū)間是(0,2)2函數(shù)f(x)2xsin x在(,)上(a)a是增函數(shù)b是減函數(shù)c在(0,)上增,在(,0)上減d在(0,)上減,在(,0)上增解析f (x)2cosx>0在(,)上恒成立3(2016·江西撫州高二檢測(cè))函數(shù)yx3x2mx1是r上的單調(diào)函數(shù),則實(shí)數(shù)m的取值范圍是(c)a(,)b(,)c,

2、)d(,)解析y3x22xm,由題意知3x22xm0在r上恒成立,412m0,m.4設(shè)f (x)是函數(shù)f(x)的導(dǎo)函數(shù),yf (x)的圖象如圖所示,則yf(x)的圖象最有可能的是(c)思路分析由導(dǎo)函數(shù)f (x)的圖象位于x軸上方(下方),確定f(x)的單調(diào)性,對(duì)比f(wàn)(x)的圖象,用排除法求解解析由f (x)的圖象知,x(,0)時(shí),f (x)>0,f(x)為增函數(shù),x(0,2)時(shí),f (x)<0,f(x)為減函數(shù),x(2,)時(shí),f (x)>0,f(x)為增函數(shù)只有c符合題意,故選c5(2016·貴州貴陽(yáng)一中月考)函數(shù)yxln x在(0,5)上的單調(diào)性是(c)a單調(diào)遞增

3、b單調(diào)遞減c在(0,)上單調(diào)遞減,在(,5)上單調(diào)遞增d在(0,)上單調(diào)遞增,在(,5)上單調(diào)遞減解析函數(shù)的定義域?yàn)?0,)yln x1,令y>0,得x>.令y<0,得0<x<.函數(shù)yxln x在(0,)上單調(diào)遞減,在(,5)上單調(diào)遞增6若函數(shù)f(x)kxln x在區(qū)間(1,)上單調(diào)遞增,則k的取值范圍是(d)a(,2b(,1c2,)d1,)解析由條件知f (x)k0在(1,)上恒成立,k1.把函數(shù)的單調(diào)性轉(zhuǎn)化為恒成立問(wèn)題是解決問(wèn)題的關(guān)鍵二、填空題7函數(shù)yx3x2x的單調(diào)遞增區(qū)間為(,),(1,).解析y3x22x1(3x1)(x1),由y>0得,x>

4、1或x<.8若函數(shù)f(x)x3bx2cxd的單調(diào)減區(qū)間為(1,3),則b_3_,c_9_.解析f (x)3x22bxc,由條件知,即,解得b3,c9.三、解答題9(2016·北京昌平區(qū)高二檢測(cè))設(shè)函數(shù)f(x)x3mx21的導(dǎo)函數(shù)f(x),且f(1)3.(1)求函數(shù)f(x)在點(diǎn)(1,f(1)處的切線方程；(2)求函數(shù)f(x)的單調(diào)區(qū)間.解析(1)f(x)x22mx,f(x)12m3,m1.f(x)x3x21,f(1).切線方程為y3(x1),即3x3y40.(2)f(x)x22xx(x2),令f(x)>0,得x>0或x<2,令f(x)<0,得2<x&

5、lt;0,函數(shù)f(x)的單調(diào)遞增區(qū)間為(,2),(0,),遞減區(qū)間為(2,0)b級(jí)素養(yǎng)提升一、選擇題1函數(shù)yf(x)的圖象如圖所示,則yf (x)的圖象可能是(d)解析由f(x)的圖象知,f(x)在(,0)上單調(diào)遞增,在(0,)上單調(diào)遞減,在(0,)上f (x)0,在(,0)上f (x)0,故選d2下列函數(shù)中,在區(qū)間(1,1)上是減函數(shù)的是(c)ay23x2byln xcydysin x解析a中,y6x,當(dāng)1<x<0時(shí),y>0,當(dāng)0<x<1時(shí),y<0,故函數(shù)y23x2在區(qū)間(1,1)上不是減函數(shù),b中,yln x在x0處無(wú)意義；c中,y<0對(duì)x(1,1

6、)恒成立,函數(shù)y在區(qū)間(1,1)上是減函數(shù)；d中,ycos x>0對(duì)x(1,1)恒成立,函數(shù)ysin x在(1,1)上是增函數(shù)3定義在r上的函數(shù)f(x),若(x1)·f (x)<0,則下列各項(xiàng)正確的是(c)af(0)f(2)>2f(1)bf(0)f(2)2f(1)cf(0)f(2)<2f(1)df(0)f(2)與2f(1)大小不定解析當(dāng)x>1時(shí),f (x)<0,f(x)是減函數(shù),f(1)>f(2)當(dāng)x<1時(shí),f (x)>0,f(x)是增函數(shù),f(0)<f(1)因此f(0)f(2)<2f(1)4已知對(duì)任意實(shí)數(shù)x,有f(x

7、)f(x),g(x)g(x),且當(dāng)x>0,有f (x)>0,g(x)>0,則當(dāng)x<0時(shí),有(b)af (x)>0,g(x)>0bf (x)>0,g(x)<0cf (x)<0,g(x)>0df (x)<0,g(x)<0解析由已知f(x)為奇函數(shù),g(x)為偶函數(shù)x>0時(shí),f (x)>0,g(x)>0,f(x),g(x)在(0,)上遞增x<0時(shí),f(x)遞增,g(x)遞減x<0時(shí)f (x)>0,g(x)<0.5(2016·湛江一模)若函數(shù)f(x)x(br)的導(dǎo)函數(shù)在區(qū)間(1,

8、2)上有零點(diǎn),則f(x)在下列區(qū)間上單調(diào)遞增的是(d)a(2,0)b(0,1)c(1,)d(,2)解析由題意知,f(x)1,函數(shù)f(x)x(br)的導(dǎo)函數(shù)在區(qū)間(1,2)上有零點(diǎn),當(dāng)10時(shí),bx2,又x(1,2),b(1,4),令f(x)>0,解得x<或x>,即f(x)的單調(diào)遞增區(qū)間為(,),(,),b(1,4),(,2)符合題意故選d二、填空題6(2016·山東濰坊一中高二期末)函數(shù)f(x)x2sin x在(0,)上的單調(diào)遞增區(qū)間為(,).解析由f(x)12cos x>0得cos x<,又x(0,),所以<x<,故函數(shù)f(x)的單調(diào)遞增區(qū)間

9、為(,)7已知函數(shù)f(x)在(2,)上單調(diào)遞減,則a的取值范圍是(,).解析f (x),由題意得x<2時(shí),f (x)0恒成立,2a10,a.又當(dāng)a時(shí),f(x),此時(shí),函數(shù)f(x)在(2,)上不是減函數(shù),a.綜上可知,a的取值范圍為(,)三、解答題8設(shè)函數(shù)f(x)x33ax23bx的圖象與直線12xy10相切于點(diǎn)(1,11). (1)求a、b的值；(2)討論函數(shù)f(x)的單調(diào)性解析(1)f (x)3x26ax3b.因?yàn)閒(x)的圖象與直線12xy10相切于點(diǎn)(1,11),所以f(1)11,f (1)12,即,解得a1,b3.(2)由a1,b3得f (x)3x26ax3b3(x22x3)3(

10、x1)(x3)令f (x)>0,解得x<1或x>3；又令f (x)<0,解得1<x<3.故當(dāng)x(,1)時(shí),f(x)是增函數(shù)；當(dāng)x(3,)時(shí),f(x)也是增函數(shù)；當(dāng)x(1,3)時(shí),f(x)是減函數(shù)c級(jí)能力提高1已知函數(shù)f(x)x3ax23x在區(qū)間1,)上是增函數(shù),則實(shí)數(shù)a的取值范圍是_(,0_.解析f(x)x3ax23x,f (x)3x22ax3,又因?yàn)閒(x)x3ax23x在區(qū)間1,)上是增函數(shù),f (x)3x22ax30在區(qū)間1,)上恒成立,解得a0,故答案為(,02(2016·廣東汕頭高二質(zhì)檢)函數(shù)f(x)2x3ax與g(x)bx2c的圖象都過(guò)點(diǎn)p(2,0),且在點(diǎn)p處有相同的切線.(1)求實(shí)數(shù)a、b、c的值；(2)設(shè)函數(shù)f(x)f(x)g(x),求f(x)的單調(diào)區(qū)間解析(1)函數(shù)f(x)、g(x)的圖象都過(guò)點(diǎn)p(2,0),f(2)162a0,解得a8,g(2)4bc0.又f(x)、g(x)的圖象在點(diǎn)p處有相同的切線,且f(x)6x28, g(x)2bx,f(2)g(2),4b16,b4,c16.a8,b4,c16.(2)由(1)知,f(x)2x38x,g(x)4x216,f(x