志鴻優(yōu)化人教版高考數(shù)學(xué)理科一輪總復(fù)習(xí)課時(shí)規(guī)范練7一次函數(shù)二次函數(shù)(含答案詳析)

課時(shí)規(guī)范練7一次函數(shù)、二次函數(shù)課時(shí)規(guī)范練第13頁一、選擇題1.已知某二次函數(shù)的圖象與函數(shù)y=2x2的圖象的形狀相同,張口方向相反,且其極點(diǎn)為(-1,3),則此函數(shù)的剖析式為()A.y=2(x-1)2+3B.y=2(x+1)2+3C.y=-2(x-1)2+3D.y=-2(x+1)2+3答案:D剖析:設(shè)所求函數(shù)的剖析式為y=a(x+h)22+k(a≠0),由題意可知a=-2,h=1,k=3,故y=-2(x+1)+3.2.若是函數(shù)f(x)=x2+bx+c對(duì)任意的實(shí)數(shù)x,都有f(1+x)=f(-x),那么()A.f(-2)<f(0)<f(2)B.f(0)<f(-2)<f(2)C.f(2)<f(0)<f(-2)D.f(0)<f(2)<f(-2)答案:D剖析:由f(1+x)=f(-x)可知,函數(shù)的對(duì)稱軸為x=,即-,所以b=-1,則f(x)=x2-x+c,結(jié)合函數(shù)圖象可知f(0)<f(2)<f(-2),應(yīng)選D.3.若二次函數(shù)

f(x)=ax2+bx+c滿足

f(x1)=f(x2),則

f(x1+x2)等于(

)A.-

B.-C.c

D.答案:C剖析:由已知f(x1)=f(x2)且f(x)的圖象關(guān)于x=-對(duì)稱,則x1+x2=-,故f(x1+x2)=f=a·-b·+c=c.選C.4.已知函數(shù)h(x)=4x2-kx-8在[5,20]上是單調(diào)函數(shù)A.(-∞,40]B.[160,+∞)

,則k

的取值范圍是

(

)C.(-∞,40]∪[160,+∞)D.?答案:C剖析:函數(shù)h(x)的對(duì)稱軸為x=,要使h(x)在[5,20]上是單調(diào)函數(shù),應(yīng)有≤5或≥20,即k≤40或k≥160,應(yīng)選C.5.已知函數(shù)

f(x)=ax

2+(b+c)x+1(a≠0)是偶函數(shù)

,其定義域?yàn)?/p>

[a-c,b],則點(diǎn)(a,b)的軌跡是

(

)A.線段C.點(diǎn)

B.直線的一部分D.圓錐曲線答案:B剖析:因?yàn)榕己瘮?shù)的定義域關(guān)于原點(diǎn)對(duì)稱,且f(-x)=f(x),所以故a=-2b(b>0),即點(diǎn)(a,b)的軌跡是直線的一部分.6.設(shè)f(x)=|2-x2|,若0<a<b且f(a)=f(b),則a+b的取值范圍是()A.(0,2)B.(0,)C.(0,4)D.(0,2)答案:D剖析:∵f(a)=f(b),且0<a<b,則a<<b,∴|2-a2|=|2-b2|,即a2+b2=4.由(a+b)2≤2(a2+b2),易知a+b≤2(當(dāng)且僅當(dāng)a=b時(shí)取等號(hào)),又0<a<b,故0<a+b<2.二、填空題7.函數(shù)f(x)=2x2-6x+1

在區(qū)間[-1,1]上的最小值是

,最大值是

.答案:-39剖析:f(x)=2.當(dāng)x=1時(shí),f(x)min=-3;當(dāng)x=-1時(shí),f(x)max=9.8.若二次函數(shù)f(x)=x2-kx+1滿足f(1-x)=f(1+x),則f(2)=.答案:1剖析:由已知二次函數(shù)f(x)=x2-kx+1的對(duì)稱軸為x=1,即=1,k=2,∴f(2)=22-2×2+1=1.29.設(shè)二次函數(shù)f(x)=ax+2ax+1在[-3,2]上有最大值4,則實(shí)數(shù)a的值為.答案:或-3剖析:f(x)的對(duì)稱軸為x=-1.當(dāng)a>0時(shí),f(2)=4a+4a+1=8a+1,f(-3)=3a+1.∴f(2)>f(-3),即f(x)max=f(2)=8a+1=4.∴a=.當(dāng)a<0時(shí),f(x)max=f(-1)=a-2a+1=-a+1=4,∴a=-3.綜上所述,a=或a=-3.三、解答題10.若x≥0,y≥0,且x+2y=1,求2x+3y2的最小值.解:2x+3y2=2(1-2y)+3y2=3y2-4y+2,因?yàn)閤=1-2y≥0,y≥0,所以y的取值范圍為0≤y≤.設(shè)f(y)=3y2-4y+2=3.故y=時(shí),f(y)min=f,即當(dāng)y=且x=0時(shí),2x+3y2有最小值.11.已知函數(shù)f(x)=-x2+ax+在區(qū)間[0,1]上的最大值為2,求a的值.解:f(x)=-.①當(dāng)∈[0,1],即0≤a≤2時(shí),f(x)max==2,則a=3或a=-2,不合題意.②當(dāng)>1,即a>2時(shí),f(x)max=f(1)=2?a=.③當(dāng)<0,即a<0時(shí),f(x)max=f(0)=2?a=-6.綜上,f(x)在區(qū)間[0,1]上最大值為2時(shí),a=或a=-6.12.二次函數(shù)f(x)=ax2+bx+1(a>0),設(shè)f(x)=x的兩個(gè)實(shí)根為x1,x2.(1)若是b=2且|x2-x1|=2,求a的值;(2)若是x1<2<x2<4,設(shè)函數(shù)f(x)的對(duì)稱軸為x=x0,求證:x0>-1.2答案：(1)解當(dāng)b=2時(shí),f(x)=ax+2x+1(a>0),222|x2-x1|=2?(x2-x1)=4?(x1+x2)-4x1x2=4.2代入上式可得4a+4a-1=