高三數(shù)學(xué)二輪復(fù)習(xí) 第2部分 必考補(bǔ)充專題 專題限時(shí)集訓(xùn)16 專題6 突破點(diǎn)16 函數(shù)的圖象和性質(zhì) 理-人教高三數(shù)學(xué)試題
專題限時(shí)集訓(xùn)(十六)函數(shù)的圖象和性質(zhì)A組高考達(dá)標(biāo) 一、選擇題1(2016·南昌一模)定義在R上的偶函數(shù)f(x)滿足:對(duì)任意的x1,x2(,0)(x1x2),都有0.則下列結(jié)論正確的是()Af(0.32)f(20.3)f(log25)Bf(log25)f(20.3)f(0.32)Cf(log25)f(0.32)f(20.3)Df(0.32)f(log25)f(20.3)A對(duì)任意的x1,x2(,0),且x1x2,都有0,f(x)在(,0)上是減函數(shù)又f(x)是R上的偶函數(shù),f(x)在(0,)上是增函數(shù)00.3220.3log25,f(0.32)f(20.3)f(log25)故選A.2(2016·安慶一模)函數(shù)f(x)cos x(x且x0)的圖象可能為()D因?yàn)閒(x)cos(x)cos xf(x),所以函數(shù)f(x)為奇函數(shù),排除A,B.當(dāng)0x1時(shí),x0,cos x0,所以f(x)0,排除C,故選D.3已知偶函數(shù)f(x)在區(qū)間0,)上單調(diào)遞增,則滿足f(2x1)f的x的取值范圍是()A.B.C. D.A偶函數(shù)滿足f(x)f(|x|),根據(jù)這個(gè)結(jié)論,有f(2x1)ff(|2x1|)f,進(jìn)而轉(zhuǎn)化為不等式|2x1|,解這個(gè)不等式即得x的取值范圍是.4(2016·青島一模)奇函數(shù)f(x)的定義域?yàn)镽,若f(x1)為偶函數(shù),且f(1)2,則f(4)f(5)的值為()A2B1 C1D2A設(shè)g(x)f(x1),f(x1)為偶函數(shù),則g(x)g(x),即f(x1)f(x1)f(x)是奇函數(shù),f(x1)f(x1)f(x1),f(x2)f(x),f(x4)f(x22)f(x2)f(x),則f(4)f(0)0,f(5)f(1)2,f(4)f(5)022,故選A.5(2016·南通三調(diào))設(shè)函數(shù)yf(x)(xR)為偶函數(shù),且xR,滿足ff,當(dāng)x2,3時(shí),f(x)x,則當(dāng)x2,0時(shí),f(x)() 【導(dǎo)學(xué)號(hào):85952060】A|x4| B|2x|C2|x1| D3|x1|DxR,滿足ff,xR,滿足ff,即f(x)f(x2)若x0,1,則x22,3,f(x)f(x2)x2,若x1,0,則x0,1函數(shù)yf(x)(xR)為偶函數(shù),f(x)x2f(x),即f(x)x2,x1,0;若x2,1,則x20,1,則f(x)f(x2)x22x4,x2,1綜上,f(x)故選D.二、填空題6(2016·寧波聯(lián)考)已知f(x)則f(f(1)_,f(f(x)1的解集為_,4f(1)1,f(f(1)f(1).f(f(x)1,f(x)1(舍去),f(x)2,x4,x,f(f(x)1的解集為,47若函數(shù)f(x)2|xa|(aR)滿足f(1x)f(1x),且f(x)在m,)上單調(diào)遞增,則實(shí)數(shù)m的最小值等于_1f(1x)f(1x),f(x)的對(duì)稱軸為x1,a1,f(x)2|x1|,f(x)的增區(qū)間為1,)m,)1,),m1,m的最小值為1.8(2016·太原模擬)已知函數(shù)f(x)若f(x1)f(x2)f(x3)(x1,x2,x3互不相等),且x1x2x3的取值范圍為(1,8),則實(shí)數(shù)m的值為_1作出f(x)的圖象,如圖所示,可令x1x2x3,則由圖知點(diǎn)(x1,0),(x2,0)關(guān)于直線x對(duì)稱,所以x1x21.又1x1x2x38,所以2x39.由f(x1)f(x2)f(x3)(x1,x2,x3互不相等),結(jié)合圖象可知點(diǎn)A的坐標(biāo)為(9,3),代入函數(shù)解析式,得3log2(9m),解得m1.三、解答題9已知函數(shù)g(x)ax22ax1b(a0)在區(qū)間2,3上有最大值4和最小值1,設(shè)f(x).(1)求a,b的值;(2)若不等式f(2x)k·2x0在x1,1上有解,求實(shí)數(shù)k的取值范圍解(1)g(x)a(x1)21ba,因?yàn)閍0,所以g(x)在區(qū)間2,3上是增函數(shù),3分故解得6分(2)由已知可得f(x)x2,所以f(2x)k·2x0可化為2x2k·2x,即122·k,8分令t,則kt22t1,x1,1,則t,10分記h(t)t22t1,因?yàn)閠,故h(t)max1,所以k的取值范圍是(,1.12分10已知函數(shù)f(x)a.(1)求f(0);(2)探究f(x)的單調(diào)性,并證明你的結(jié)論;(3)若f(x)為奇函數(shù),求滿足f(ax)f(2)的x的范圍解(1)f(0)aa1.2分(2)(x)的定義域?yàn)镽,任取x1,x2R且x1x2,則f(x1)f(x2)aa.4分y2x在R上單調(diào)遞增且x1x2,02x12x2,2x12x20,2x110,2x210,f(x1)f(x2)0,即f(x1)f(x2),f(x)在R上單調(diào)遞增.8分(3)f(x)是奇函數(shù),f(x)f(x),即aa,解得a1.(或用f(0)0去解)10分f(ax)f(2),即為f(x)f(2),又因?yàn)閒(x)在R上單調(diào)遞增,所以x2.12分B組名校沖刺一、選擇題1(2016·莆田二模)已知定義在R上的奇函數(shù)滿足f(x4)f(x),且在區(qū)間0,2上是增函數(shù),則()Af(25)f(11)f(80)Bf(80)f(11)f(25)Cf(11)f(80)f(25)Df(25)f(80)f(11)Df(x4)f(x),f(x8)f(x4),f(x8)f(x),f(x)的周期為8,f(25)f(1),f(80)f(0),f(11)f(3)f(14)f(1)f(1)又奇函數(shù)f(x)在區(qū)間0,2上是增函數(shù),f(x)在區(qū)間2,2上是增函數(shù),f(25)f(80)f(11),故選D.2(2016·濟(jì)南模擬)函數(shù)f(x)ln的圖象大致是()A易知f(x)的定義域關(guān)于原點(diǎn)對(duì)稱,因?yàn)閒(x)lnlnf(x),所以函數(shù)是偶函數(shù),排除B和D;當(dāng)x時(shí),0xsin xxsin x,01,ln0,排除C,故選A.3(2016·開封模擬)設(shè)函數(shù)f(x)若f4,則b() 【導(dǎo)學(xué)號(hào):85952061】A1B. C.D.Df3×bb,當(dāng)b1,即b時(shí),f2b,即2b422,得到b2,即b;當(dāng)b1,即b時(shí),f3bb4b,即4b4,得到b,舍去綜上,b,故選D.4(2016·成都模擬)如果函數(shù)f(x)(m2)x2(n8)x1(m0,n0)在區(qū)間上單調(diào)遞減,那么mn的最大值為()A16B18 C25D.B當(dāng)m2時(shí),f(x)(n8)x1在區(qū)間上單調(diào)遞減,則n80n8,于是mn16,則mn無最大值當(dāng)m0,2)時(shí),f(x)的圖象開口向下且過點(diǎn)(0,1),要使f(x)在區(qū)間上單調(diào)遞減,需,即2nm18,又n0,則mnmm29m.而g(m)m29m在0,2)上為增函數(shù),m0,2)時(shí),g(m)g(2)16,mn16,故m0,2)時(shí),mn無最大值當(dāng)m2時(shí),f(x)的圖象開口向上且過點(diǎn)(0,1),要使f(x)在區(qū)間上單調(diào)遞減,需2,即2mn12,而2mn2,mn18,當(dāng)且僅當(dāng)即時(shí),取“”,此時(shí)滿足m2.故(mn)max18.故選B.二、填空題5(2016·合肥二模)在平面直角坐標(biāo)系xOy中,若直線y2a與函數(shù)y|xa|1的圖象只有一個(gè)交點(diǎn),則a的值為_函數(shù)y|xa|1的圖象如圖所示,因?yàn)橹本€y2a與函數(shù)y|xa|1的圖象只有一個(gè)交點(diǎn),故2a1,解得a.6(2016·泉州二模)若函數(shù)f(x)(a0,且a1)的值域是4,),則實(shí)數(shù)a的取值范圍是_(1,2當(dāng)x2時(shí),f(x)x6,f(x)在(,2上為減函數(shù),f(x)4,)當(dāng)x2時(shí),若a(0,1),則f(x)3logax在(2,)上為減函數(shù),f(x)(,3loga2),顯然不滿足題意,a1,此時(shí)f(x)在(2,)上為增函數(shù),f(x)(3loga2,),由題意可知(3loga2,)4,),則3loga24,即loga21,1a2.三、解答題7已知奇函數(shù)f(x)的定義域?yàn)?,1,當(dāng)x1,0)時(shí),f(x)x.(1)求函數(shù)f(x)在0,1上的值域;(2)若x(0,1,yf2(x)f(x)1的最小值為2,求實(shí)數(shù)的值解(1)設(shè)x(0,1,則x1,0),所以f(x)x2x.又因?yàn)閒(x)為奇函數(shù),所以f(x)f(x),所以當(dāng)x(0,1時(shí),f(x)f(x)2x,所以f(x)(1,2又f(0)0,所以當(dāng)x0,1時(shí)函數(shù)f(x)的值域?yàn)?1,20.4分(2)由(1)知當(dāng)x(0,1時(shí),f(x)(1,2,所以f(x),令tf(x),則t1,g(t)f2(x)f(x)1t2t121.8分當(dāng),即1時(shí),g(t)g無最小值當(dāng)1即12時(shí),g(t)ming12.解得±2舍去當(dāng)1,即2時(shí),g(t)ming(1)2,解得4.綜上所述,4.12分8函數(shù)f(x)是定義在R上的偶函數(shù),且對(duì)任意實(shí)數(shù)x,都有f(x1)f(x1)成立,已知當(dāng)x1,2時(shí),f(x)logax.(1)求x1,1時(shí),函數(shù)f(x)的表達(dá)式;(2)求x2k1,2k1(kZ)時(shí),函數(shù)f(x)的表達(dá)式;(3)若函數(shù)f(x)的最大值為,在區(qū)間1,3上,解關(guān)于x的不等式f(x).解(1)因?yàn)閒(x1)f(x1),且f(x)是R上的偶函數(shù),所以f(x2)f(x),所以f(x)3分(2)當(dāng)x2k1,2k時(shí),f(x)f(x2k)loga(2x2k),同理,當(dāng)x(2k,2k1時(shí),f(x)f(x2k)loga(2x2k),所以f(x)6分(3)由于函數(shù)是以2為周期的周期函數(shù),故只需要考查區(qū)間1,1,當(dāng)a1時(shí),由函數(shù)f(x)的最大值為,知f(0)f(x)maxloga2,即a4.當(dāng)0a1時(shí),則當(dāng)x±1時(shí),函數(shù)f(x)取最大值為,即loga(21),舍去綜上所述a4.9分當(dāng)x1,1時(shí),若x1,0,則log4(2x),所以2x0;若x(0,1,則log4(2x),所以0x2,所以此時(shí)滿足不等式的解集為(2,2)因?yàn)楹瘮?shù)是以2為周期的周期函數(shù),所以在區(qū)間1,3上,f(x)的解集為(,4),綜上所得不等式的解集為(2,2)(,4).12分