2019衡水名師數(shù)學(xué)專題卷專題二函數(shù)概念及其基本性質(zhì) 第I卷（選擇題）44

的定義域A,函數(shù)yln1x的定義域?yàn)锽,則AB {已知函數(shù)fx2xx{

,若f(a)f(1)0,則實(shí)數(shù)a的值等于 13

x1,xy

11 flog1x的定義域?yàn)?,則函數(shù)yf(2)的定義域?yàn)? 421,0,1,已知函數(shù)fxx24xx0,若f2a2f(a,則實(shí)數(shù)a4xx2x A.,12,B.1,C.定義在R上的奇函數(shù)fxfx4fx,且在區(qū)間0,2上是增函數(shù),(f2

f5

ff8f5

f2ff2ff5

f8

ffxRx0?時(shí),fxx22xyfx在R上的解析式為 A.fxxxB.fxxxC.fxxxD.fxxx設(shè)偶函數(shù)fx對(duì)任xRfx3則f107.5 11

f

x32時(shí),fx4xf(x在f(1)1,則滿足1f(x2)1x的取值范圍是 2,0,若偶函數(shù)fx在區(qū)間,0上單調(diào)遞減,且f30,則不等式x1fx0的解 (,1)(1,3,13,,33,3,13,yf(x1)是定義域?yàn)镽的偶函數(shù),且fx在1,f(2x1)f(x2)的解集為

,3 1 設(shè)fxx3log2x x21,則對(duì)任意實(shí)數(shù)a,?b,若ab0,則 A.fafbB.fafbC.fafbD.fafba21x2(a1)x2a21x2(a1)x2a.已知函數(shù)fx xa,若對(duì)于定義域內(nèi)的任意x,總存在x使得fxfxx 則滿足條件的實(shí)數(shù)a的取值范圍

若函數(shù)f(x)xa的單調(diào)遞增區(qū)間是[3,),則a x24x,xf(xax2bxx0為偶函數(shù),則ab 2 x2bxc的圖象經(jīng)過(guò)A0,3,B4, 2 求bcy

f(xax2bxaba0)滿足條件:f(x1f(3xf(x2xf(xf(x在[0,t已知函數(shù)f(x對(duì)一切xy都有f(xyfy)x(x2y1成立,且f(10f(0)求f(x)的解析式;[來(lái)源:P:當(dāng)0x1f(x32xa2

Qx[22g(xf(xaxP、Q至少有一個(gè)成立,求實(shí)數(shù)af(xyf(xfyx0時(shí),有f(x0f(xf(xfxm22am1x1,1a1,1恒成立,求a的取值范圍1sin2x1sin2xsinx 1sin2xsinx求函數(shù)fx)f(xx2mx2x1?xn求mng(xx2ax2(aR在(,1]xloga(nxm2)解析：由4x20得2x2,由1x0x1ABx|2x2x|x1x|2x1,選解析f(af(10f(af(1a0時(shí),2a2a1a0時(shí),a12ay 11 11 flog1x的定義域?yàn)?,即x ,所 42 421log1x2,令12x2,解得0x1y2

f(2x的定義域?yàn)?,1f(x

x24x(x2)24,x,4xx2(x2)24,xfx的圖象可知fx在上是單調(diào)增函數(shù),f2a2fa得2a2aa2a20,解2a解析：奇函fx在區(qū)間0,2上單調(diào)遞增且fxf00,已知奇函數(shù)在關(guān)于原點(diǎn)對(duì)稱的兩個(gè)區(qū)間上具有相同的單調(diào)性,故奇函數(shù)fx在區(qū)間20上單調(diào)遞增且fxf00,從而fx在2,2上單調(diào)遞增.由奇fx中任意xfxfx,且題設(shè)fx4fx,故f8f84f4f44f0;由102,故f1f0f2,即f5

f8f解析：因?yàn)閒x為奇函數(shù)且在單調(diào)遞減,要使1fx1成立x1x21,解得1x3,所以滿足1fx21x的取值范圍為1,3解析fxx3log2x∵fxx3log2x

x21定義域?yàn)镽x22x3 22x3log2x x1f2fx是奇函數(shù)fx在?上是增函數(shù),故fx在R上為增函數(shù),而ab0abfafbfafb0B.[來(lái)源:學(xué)&科&a21x2(aa21x2(a1)x2a∴a21x2(a1)x a

a210時(shí),a1a1時(shí)不等式恒成立,當(dāng)a1a21a210時(shí),a124a2120綜上所述,a的取值范圍為

axa 解析：由題意函數(shù)f(x)無(wú)最小值,f(x) 1令x

(xt,則t0f(xy2at2ta0

(x xyt,符合題意,a0時(shí),2a0a0綜上有a的取值范圍是a0答案：-3[來(lái)源。 解析xa時(shí),f(x(xaxa為減函數(shù)xa時(shí),f(xxa為增函數(shù),結(jié)合已知有a3,a 2答案：1A0,3B4,9y3x2bx 2

b93164bc 2.由1可得,該拋物線解析式 9 3 y x2 x3, 4 3 0 8 16 y

∵

x2

x30的解為:x2,x ∴公共點(diǎn)的坐標(biāo)是20或8,答案：1.∵方fx2x有兩等根,即ax2b2x0b22?0,解得b2fx1f3xx13x ∴

1a1fxx22.fxx22x的圖象的對(duì)稱軸x1,x0,t∴當(dāng)t1

fx在[0,t上是增函數(shù)f

t22t當(dāng)t1

fx在0,1上是增函數(shù),在1,t上是減函數(shù)∴f f11,t綜上,f t22t,t答案：1.x1y1,則由已知f(0f(1)1(121f(0y0f(xf(0)x(x1),[來(lái)源:學(xué)&科&又∵f(0)2∴f(x)x2x不等f(wàn)(x32xax2x232xa,即x2x1a當(dāng)0x12 1

3x2x14又x

aA{a|a 2 g(x)x2x2axx2(1a)x2ag(x在[2,2]2

2

2B{a|a3或aP、Q至少有一個(gè)成立時(shí)aAB{a|a1a答案：1.因?yàn)橛衒(xyf(xfy,令xy0f(0f(0f(0)f(00令yx可得:f(0)f(x)f(x)0,[來(lái) f(x)f(xf(xf(x是定義在[1,1]上的奇函數(shù),由題意設(shè)1x1x2f(x2f(x1f(x2f(x1f(x2x1x0f(x0f(x2f(x1f(x是在[1,1]f(x在[1,1]f(x在[1,1]f(1)f(xm22am1x1,1a1,1恒成立,只要m22am11,即m22am0恒成立.g(1) 2mm2g(am22am2amm2m2?或m

得 2mm211sin2f(xf(x0

((1sin2xsinx1)(1sin2x(sinx(1sin2xsinx1)(1sin2x(sinxf(x) 1sin2