函數(shù)習(xí)題課課時(shí)過關(guān)能力提升基礎(chǔ)鞏固1.已知f(x)=10,xf(-2)f(-1)B.f(3)f(-2)f(-1)C.f(-2)f(3)f(-1)D.f(-1)f(3)f(-2)解析:函數(shù)f(x)=(m-1)x2+2mx+3是R上的偶函數(shù),f(-x)=(m-1)x2-2mx+3=f(x)=(m-1)x2+2mx+3,m=0,即f(x)=-x2+3.當(dāng)xf(-2)f(-3)=f(3).即f(3)f(-2)f(-1),故選B.答案:B5.若函數(shù)y=f(x)的定義域是0,3,則函數(shù)g(x)=f(3x)x-1的定義域是()A.(0,1)B.0,1C.0,1)(1,9D.0,1)解析:由題意知03x3,x-10,解得0x0時(shí),f(x)的圖象如圖所示,則f(x)的值域是.解析:當(dāng)x0時(shí),f(x)的取值范圍是(2,3.根據(jù)奇函數(shù)的性質(zhì)可得,f(x)的值域是-3,-2)(2,3.答案:-3,-2)(2,38.已知對任意實(shí)數(shù)a,b,都有f(a+b)=f(a)+f(b),且f(2)=2,f(3)=3,則f(10)=.解析:由已知可得f(5)=f(2+3)=f(2)+f(3)=5,故f(10)=2f(5)=10.答案:109.已知函數(shù)f(x)的定義域?yàn)?-2,2),函數(shù)g(x)=f(x-1)+f(3-2x).(1)求函數(shù)g(x)的定義域;(2)若f(x)是奇函數(shù),且在定義域內(nèi)單調(diào)遞減,求不等式g(x)0的解集.解:(1)函數(shù)f(x)的定義域?yàn)?-2,2),函數(shù)g(x)=f(x-1)+f(3-2x),-2x-12,-23-2x2.12x52.故函數(shù)g(x)的定義域?yàn)?2,52.(2)f(x)是奇函數(shù)且在定義域內(nèi)單調(diào)遞減,由不等式g(x)0,得f(x-1)-f(3-2x)=f(2x-3),-2x-12,-22x-32,x-12x-3.12x2.故不等式g(x)0的解集為12,2.10.已知函數(shù)f(x)=x2-2|x|.(1)判斷并證明函數(shù)f(x)的奇偶性;(2)判斷函數(shù)f(x)在區(qū)間(-1,0)內(nèi)的單調(diào)性,并加以證明.解:(1)函數(shù)f(x)是偶函數(shù).證明如下:f(x)的定義域?yàn)镽.f(-x)=(-x)2-2|-x|=x2-2|x|=f(x),函數(shù)f(x)是偶函數(shù).(2)f(x)在區(qū)間(-1,0)內(nèi)是增函數(shù).證明如下:當(dāng)x(-1,0)時(shí),f(x)=x2+2x.設(shè)-1x1x20,則x1-x2-2,即x1+x2+20.f(x1)-f(x2)=(x12-x22)+2(x1-x2)=(x1-x2)(x1+x2+2)0,f(x1)f(x2).故f(x)在區(qū)間(-1,0)內(nèi)是增函數(shù).能力提升1.函數(shù)y=1-x+1x+1的定義域?yàn)?)A.x|x1B.x|x1C.x|x1,且x-1D.x|x1,且x-1解析:函數(shù)有意義時(shí),需滿足1-x0,1+x0,解得x1,且x-1.故函數(shù)的定義域?yàn)閤|x1,且x-1.答案:D2.已知函數(shù)f(x)=1+x21-x2,則()A.f(x)是奇函數(shù),且f1x=-f(x)B.f(x)是奇函數(shù),且f1x=f(x)C.f(x)是偶函數(shù),且f1x=-f(x)D.f(x)是偶函數(shù),且f1x=f(x)解析:f(x)的定義域?yàn)閤|xR,且x1,關(guān)于原點(diǎn)對稱.又由f(-x)=1+(-x)21-(-x)2=1+x21-x2=f(x),得f(x)為偶函數(shù).又f1x=1+1x21-1x2=x2+1x2-1=-f(x),故C選項(xiàng)正確.答案:C3.如圖所示,函數(shù)的解析式為()A.y=32|x-1|,0x2B.y=32-32|x-1|,0x2C.y=32-|x-1|,0x2D.y=1-|x-1|,0x2解析:當(dāng)0x1時(shí),函數(shù)解析式為y=32x;當(dāng)10,則函數(shù)f(x)=1x的值域是.解析:因?yàn)閒(x)=1x=x2+1+x=x+122+34(x0),所以f(x)1,即f(x)的值域?yàn)?1,+).答案:(1,+)6.已知f(x)是定義在R上的偶函數(shù),且在區(qū)間(-,0)內(nèi)是增函數(shù).若f(-3)=0,則f(x)x0時(shí),f(x)3;當(dāng)x0,解得-3x0.故-3x3.答案:(-3,0)(3,+)7.已知函數(shù)f(x)=x2+bx+c,且f(1)=0.(1)若函數(shù)f(x)是偶函數(shù),求f(x)的解析式;(2)在(1)的條件下,求函數(shù)f(x)在區(qū)間-1,3上的最大值和最小值;(3)要使函數(shù)f(x)在區(qū)間-1,3上單調(diào)遞增,求b的取值范圍.解:(1)因?yàn)楹瘮?shù)f(x)是偶函數(shù),所以b=0.又因?yàn)閒(1)=0,所以1+c=0,即c=-1,所以f(x)=x2-1.(2)結(jié)合圖象得:當(dāng)x=0時(shí),f(x)min=-1;當(dāng)x=3時(shí),f(x)max=8.(3)結(jié)合圖象可知,當(dāng)-b2-1,即b2時(shí),f(x)在區(qū)間-1,3上單調(diào)遞增.8.已知函數(shù)y=f(x)的定義域?yàn)镽,對任意a,bR都有f(a+b)=f(a)+f(b),且當(dāng)x0時(shí),f(x)0恒成立.(1)證明函數(shù)y=f(x)在R上為減函數(shù);(2)討論函數(shù)y=f(x)的奇偶性;(3)若f(2+x)+f(x)x2,則x1-x20.f(a+b)=f(a)+f(b),f(x1)-f(x2)=f(x1-x2)+x2-f(x2)=f(x1-x2)+f(x2)-f(x2)=f(x1-x2).又當(dāng)x0時(shí),f(x)0恒成立,f(x1)f(x2),函數(shù)y=f(x)在R上是減函數(shù).(2)解由f(a+b)=f(a)+f(b),得f(x-x)=f(x)+