高中數(shù)學(xué)導(dǎo)數(shù)及其應(yīng)用C組測試題新人教A版選修1

1、(數(shù)學(xué)選修1-1)第三章 導(dǎo)數(shù)及其應(yīng)用提高訓(xùn)練C組一、選擇題 TOC o 1-5 h z .若 f (x) =sin a -cosx ,則 f (a)等于()A. sin a B. cosotC . since + cosotD. 2sin a.若函數(shù)f (x) =x2+bx+c的圖象的頂點(diǎn)在第四象限，則函數(shù) f(x)的圖象是().已知函數(shù)f (x) =x3 + ax2 x-1在(-qo/Hc)上是單調(diào)函數(shù)，則實(shí)數(shù)a的取值范圍是()a.(,-問U73,z) b , -V3J3C.(-二，-13) (.3, 二)D . (- . 3, /3).對于R上可導(dǎo)的任意函數(shù) f(x),若滿足(x-1)f

2、(x)之0,則必有()A.f(0) f (2)：二 2f (1) B.f (0)f(2) 2f (1)C. f (0) f (2) -2f (1) D.f(0)f(2) 2f (1).若曲線y=x4的一條切線l與直線x+4y8 = 0垂直，則l的方程為()A. 4x-y 3 = 0 B , x+4y-5=0 C . 4x-y + 3 = 0 D . x + 4y+3=06.函數(shù)f(x)的定義域?yàn)殚_區(qū)間(a,b),導(dǎo)函數(shù)f(x)在(a,b)內(nèi)的圖象如圖所示,則函數(shù)f(x)在開區(qū)間A. 1個(gè) B . 2個(gè)二、填空題2.右函數(shù)f (x)= x(x- c)在x= 2處有極大值，則常數(shù) c的值為;.函數(shù)

3、y = 2x+sin x的單調(diào)增區(qū)間為 。.設(shè)函數(shù) “乂)=8$(屆十中)(0平兀)，若(刈十(刈為奇函數(shù)，則=312. 一一.設(shè) f(x)=x-x -2x+5 ,當(dāng) x w -1,2時(shí)，f (x) m恒成立，則實(shí)數(shù) m 的2取值范圍為。.對正整數(shù)n ,設(shè)曲線y=xn(1_x)在x = 2處的切線與y軸交點(diǎn)的縱坐標(biāo)為 an,則數(shù)列的前n項(xiàng)和的公式是n 1一三、解答題.求函數(shù)y = (1+cos2x)3的導(dǎo)數(shù)。.求函數(shù)y = J2x+4 - Jx +3的值域。.已知函數(shù)f (x) =x3+ax2+bx+c在x =|與x = 1時(shí)都取得極值求a,b的值與函數(shù)f (x)的單調(diào)區(qū)間(2)若對xW-1,

4、2,不等式f (x) 0,b0, f (x)=2x+b,直線過第一、三、四象限 2B f(x) =3x2 +2ax-1 0 在(口,也)恒成立，A=4a212E0= -V3aV3C當(dāng)x小時(shí)，f(x) 0 ,函數(shù)f (x)在(1,也)上是增函數(shù)；當(dāng)x1時(shí)，f(x)M0, f(x)在(,1)上是減函數(shù)，故 f (x)當(dāng)x = 1時(shí)取得最小值，即有f(0) _ f ,f(2) _ f 彳導(dǎo) f(0)f(2) ,2f(1)A 與直線x+4y 8 = 0垂直的直線l為4x y+m = 0 ,即y = x4在某一點(diǎn)白導(dǎo)數(shù)為 4 ,而y = 4x3,所以y=x4在(1,1)處導(dǎo)數(shù)為4,此點(diǎn)的切線為 4x-y

5、-3 = 0A 極小值點(diǎn)應(yīng)有先減后增的特點(diǎn)，即 f (x)0二、填空題6 f(x) =3x2 4cx+c2,f(2)=c2 8c+12 = 0,c=2,或6, c = 2時(shí)取極小值34) y = 2 +cosx 0對于任何實(shí)數(shù)都成立 f (x) - -sin( 3x )(、. 3x ) - - 3sin( 3x )6f (x) f (x) = 2cos( . 3x )3要使f (x) + f (x)為奇函數(shù)，需且僅需 k + = kn +, k w Z , TOC o 1-5 h z 323T3T即：中=內(nèi)+-,k= Z。又0c中,所以k只能取0 ,從而中=二。66(7,收)xW1,2時(shí)，f(

6、x)max=72n2y與=2nJL(n+2 ),切線方程為：y +2n = 2n,(n+ 2 )(x 2),令x=0,求出切線與y軸交點(diǎn)的縱坐標(biāo)為 y0 =(n+1 )2n，所以一a-=2n,則 n 1 TOC o 1-5 h z a2 1 -2n數(shù)列一n-1的前n項(xiàng)和Sn =2-2n 11-2三、解答題1 .解：y = (1 cos2x)3 = (2cos2 x)3 = 8cos6 x55y =48cos x (cosx) =48cos x (-sinx)5=-48sin xcos x。2.解：函數(shù)的定義域?yàn)?2 x-4 2 Jx311,2x-4. 4x12當(dāng)x22時(shí)，y 0,即2,+s)是

7、函數(shù)的遞增區(qū)間，當(dāng) x = 2時(shí)，ymm = -1所以值域?yàn)?,一)。3.解：(1) f (x) =x3+ax2+bx+c, f (x) =3x2+2ax + b TOC o 1-5 h z ,212 41由 f () = a+b=0, f (1) = 3 + 2a + b = 0得 a = _,b = 23932f(x)=3x2x2=(3x+2)(x1),函數(shù) f(x)的單調(diào)區(qū)間如下表：x()231(1,F, If (x)+00+f(x)極大值J極小值 TOC o 1-5 h z 22所以函數(shù)f(x)的遞增區(qū)間是(七一一)與(1,+g)，遞減區(qū)間是(，1);33一.1 o2222(2) f (x) =x -x 2x +c, xw -1,2,當(dāng) x = -一時(shí)，f (一)=一 十c23327為極大值，而f(2) =2 c,則f(2) =2-c為最大值，要使f(x)；c2,x- -1,2恒成立，則只需要 c2 f(2) =2十c,得c2。x2 ax b4.斛：設(shè) g(