課時(shí)規(guī)范練7一次函數(shù)、二次函數(shù)一、選擇題1.已知某二次函數(shù)的圖象與函數(shù)y=2x2的圖象的形狀一樣,開口方向相反,且其頂點(diǎn)為(-1,3),則此函數(shù)的解析式為()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解析:設(shè)所求函數(shù)的解析式為y=a(x+h)2+k(a0),由題意可知a=-2,h=1,k=3,故y=-2(x+1)2+3.2.如果函數(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解析: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),故選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.cD.來源:答案:C解析:由已知f(x1)=f(x2)且f(x)的圖象關(guān)于x=-對(duì)稱,來源: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),則k的取值范圍是()A.(-,40B.160,+)C.(-,40160,+)D.答案:C解析: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k40或k160,故選C.5.已知函數(shù)f(x)=ax2+(b+c)x+1(a0)是偶函數(shù),其定義域?yàn)閍-c,b,則點(diǎn)(a,b)的軌跡是()A.線段B.直線的一部分C.點(diǎn)D.圓錐曲線答案:B解析: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.若函數(shù)f(x)=x2-|x+a|為偶函數(shù),則實(shí)數(shù)a的值為()來源:A.0B.1C.2D.3答案:A解析:f(-x)=f(x),(-x)2-|-x+a|=x2-|x+a|,|-x+a|=|x+a|,(-x+a)2=(x+a)2,即4ax=0,a=0.二、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解析: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解析: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=1.9.設(shè)二次函數(shù)f(x)=ax2+2ax+1在-3,2上有最大值4,則實(shí)數(shù)a的值為. 答案:或-3解析: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=-3.10.若函數(shù)f(x)=(x+a)(bx+2a)(a,bR)是偶函數(shù),且它的值域?yàn)?-,4,則該函數(shù)的解析式f(x)=. 答案:-2x2+4解析:f(x)=(x+a)(bx+2a)=bx2+(2a+ab)x+2a2是偶函數(shù),則其圖象關(guān)于y軸對(duì)稱,2a+ab=0,b=-2或a=0(舍去).又f(x)=-2x2+2a2且值域?yàn)?-,4,2a2=4,f(x)=-2x2+4.11.當(dāng)x(1,2)時(shí),不等式x2+mx+4<0恒成立,則m的取值范圍是. 答案:m-5解析:不等式x2+mx+4<0對(duì)x(1,2)恒成立,mx<-x2-4對(duì)x(1,2)恒成立,即m<-對(duì)x(1,2)恒成立,令y=x+,則函數(shù)y=x+在x(1,2)上是減函數(shù),4<y<5,-5<-<-4,m-5.三、解答題12.若x0,y0,且x+2y=1,求2x+3y2的最小值.解:2x+3y2=2(1-2y)+3y2=3y2-4y+2,因?yàn)閤=1-2y0,y0,所以y的取值范圍為0y.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.13.二次函數(shù)f(x)滿足f(x+1)-f(x)=2x,且f(0)=1.在區(qū)間-1,1上,y=f(x)圖象恒在直線y=2x+m上方,試確定實(shí)數(shù)m的取值范圍.解:由f(0)=1,可設(shè)f(x)=ax2+bx+1(a0),故f(x+1)-f(x)=a(x+1)2+b(x+1)+1-(ax2+bx+1)=2ax+a+b,由題意得解得故f(x)=x2-x+1.由題意得x2-x+1>2x+m,即x2-3x+1>m對(duì)x-1,1恒成立,設(shè)g(x)=x2-3x+1,則問題可轉(zhuǎn)化為g(x)min>m,又g(x)在-1,1上遞減,故g(x)min=g(1)=-1,故m<-1.14.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)=-.當(dāng)0,1,即0a2時(shí),f(x)max=2,則a=3或a=-2,不合題意.當(dāng)>1,即a>2時(shí),f(x)max=f(1)=2a=.當(dāng)<0,即a<0時(shí),f(x)max=f(0)=2a=-6.綜上,f(x)在區(qū)間0,1上最大值為2時(shí),a=或a=-6.15.二次函數(shù)f(x)=ax2+bx+1(a>0),設(shè)f(x)=x的兩個(gè)實(shí)根為x1,x2.(1)如果b=2且|x2-x1|=2,求a的值;(2)如果x1<2<x2<4,設(shè)函數(shù)f(x)的對(duì)稱軸為x=x0,求證:x0>-1.來源:(1)解:當(dāng)b=2時(shí),f(x)=ax2+2x+1(a>0),方程f(x)=x為ax2+x+1=0.|x2-x1|=2(x2-x1)2=4(x1+x2)2-4x1x2=4.由韋達(dá)定理可知,x1+x2=-,x1x2=.代入上式可得4a2+4a-1=0,解得a=或a=(舍去).(2)證明:因?yàn)閍x2+(b-1)x+1=0(a>0)的兩根滿足x1<2<x2<4,設(shè)g(x)=ax2+(b-1)x+1,所以來源:即則2a-b>0.又因?yàn)楹瘮?shù)f(x)的對(duì)稱軸為x=x0,故x0=->-1.四、選做題1.(20xx浙江高考)已知a,b,cR,函數(shù)f(x)=ax2+bx+c.若f(0)=f(4)>f(1),則()A.a>0,4a+b=0B.a<0,4a+b=0C.a>0,2a+b=0D.a<0,2a+b=0答案:A解析:由f(0)=f(4)知二次函數(shù)f(x)=ax2+bx+c對(duì)稱軸為x=2,即-=2.所以4a+b=0.又f(0)>f(1)且f(0),f(1)在對(duì)稱軸同側(cè),故函數(shù)f(x)在(-,2上單調(diào)遞減,則拋物線開口方向朝上,知a>0,故選A.2.已知函數(shù)f(x)=-2x2+bx+c在x=1時(shí)有最大值1,0<m<n,并且xm,n時(shí),f(x)的取值范圍為.則的值為. 答案:解析:由題意知f(x)=-2(x-1)2+1,f(x)1,1,即m1,f(x)在m,n上單調(diào)遞減,f(m)=-2(m-1)2+1=且f(n)=-2(n-1)2+1=.3.已知函數(shù)f(x)=x2+|x-a|+1,aR.(1)試判斷f(x)的奇偶性;(2)若-a,求f(x)的最小值.解:(1)當(dāng)a=0時(shí),函數(shù)f(-x)=(-x)2+|-x|+1=f(x),此時(shí),f(x)為偶函數(shù).當(dāng)a0時(shí),f(a)=a2+1,f(-a)=a2+2|a|+1,f(a)f(-a),f(a)-f(-a),此時(shí),f(x)既不是奇函數(shù),也不是偶函數(shù).(2)當(dāng)xa時(shí),f(x)=x2-x+a+1=+a+,a,故函數(shù)f(x)在(-,a上單調(diào)遞減,從而函數(shù)f(x)在(-,a上的最小值為f(a)=a2+1.當(dāng)xa時(shí),函數(shù)f(x)=x2+x-a+1=-a+,a-,故函數(shù)f(x)在a,+)上單調(diào)遞增,從而函數(shù)f(x)在a,+)上的最小值為f(a)=a2+1.綜上得,當(dāng)-a時(shí),函數(shù)f(x)的最小值為a2+1.