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專題一 函數(shù)的圖象與性質(zhì)卷卷卷2018_函數(shù)圖象的辨識T3函數(shù)圖象的辨識T7抽象函數(shù)的奇偶性與周期性T112017利用函數(shù)的單調(diào)性、奇偶性解不等式T5_分段函數(shù)、解不等式T152016函數(shù)圖象辨識T7函數(shù)圖象的對稱性T12_縱向把握趨勢卷3年2考,涉及函數(shù)圖象的識別以及函數(shù)的單調(diào)性、奇偶性與不等式的綜合問題,試題均出現(xiàn)在選擇題上,難度適中,預(yù)計2019年會重點考查分段函數(shù)的有關(guān)性質(zhì)及應(yīng)用卷3年3考,涉及函數(shù)圖象的辨識以及抽象函數(shù)的性質(zhì),其中函數(shù)圖象的識別難度較小,而函數(shù)性質(zhì)難度偏大,均出現(xiàn)在選擇題中,預(yù)計2019年會以選擇題的形式考查分段函數(shù)、函數(shù)的性質(zhì)等卷3年2考,涉及函數(shù)圖象的辨識、分段函數(shù)與不等式的綜合問題,既有選擇題,也有填空題,難度適中,預(yù)計2019年會以選擇題的形式考查函數(shù)的單調(diào)性、奇偶性等性質(zhì)橫向把握重點1.高考對此部分內(nèi)容的命題多集中于函數(shù)的概念、函數(shù)的性質(zhì)及分段函數(shù)等方面,多以選擇題、填空題形式考查,一般出現(xiàn)在第510或第1315題的位置上,難度一般主要考查函數(shù)的定義域、分段函數(shù)求值或分段函數(shù)中參數(shù)的求解及函數(shù)圖象的判斷2.此部分內(nèi)容有時也出現(xiàn)在選擇、填空中的壓軸題的位置,多與導(dǎo)數(shù)、不等式、創(chuàng)新性問題結(jié)合命題,難度較大.函數(shù)的概念及表示題組全練1(2018長春質(zhì)檢)函數(shù)y的定義域是()A1,0)(0,1)B1,0)(0,1C(1,0)(0,1 D(1,0)(0,1)解析:選D由題意得解得1x0或0x2時,f (x)f (x4),故f (x)在x2時的周期為4,則f (2 018)f (2 018)f (2 0162)f (2)e2.3設(shè)f (x)若f (a)f (a1),則f ()A2 B4C6 D8解析:選C當(dāng)0a1時,a11,f (a),f (a1)2(a11)2a,f (a)f (a1),2a,解得a或a0(舍去)f f (4)2(41)6.當(dāng)a1時,a12,f (a)2(a1),f (a1)2(a11)2a,2(a1)2a,無解綜上,f 6.4已知函數(shù)f (x)則f (f (x)2的解集為_解析:因為當(dāng)x1時,f (x)x3x2,當(dāng)x1時,f (x)2ex12,所以f (f (x)2等價于f (x)1,即2ex11,解得x1ln 2,所以f (f (x)2的解集為(,1ln 2)答案:(,1ln 2)5(2018成都模擬)設(shè)函數(shù)f :RR滿足f (0)1,且對任意x,yR都有f (xy1)f (x)f (y)f (y)x2,則f (2 018)_.解析:令xy0,則f (1)f (0)f (0)f (0)02111022.令y0,則f (1)f (x)f (0)f (0)x2.將f (0)1,f (1)2代入,得f (x)1x,所以f (2 018)2 019.答案:2 019 系統(tǒng)方法1函數(shù)定義域的求法求函數(shù)的定義域,其實質(zhì)就是以函數(shù)解析式所含運算有意義為準(zhǔn)則,列出不等式或不等式組,然后求出解集即可2分段函數(shù)問題的4種常見類型及解題策略常見類型解題策略求函數(shù)值弄清自變量所在區(qū)間,然后代入對應(yīng)的解析式,求“層層套”的函數(shù)值,要從最內(nèi)層逐層往外計算解不等式根據(jù)分段函數(shù)中自變量取值范圍的界定,代入相應(yīng)的解析式求解,但要注意取值范圍的大前提求參數(shù)“分段處理”,采用代入法列出各區(qū)間上的方程利用函數(shù)性質(zhì)求值必須依據(jù)條件找到函數(shù)滿足的性質(zhì),利用該性質(zhì)求解函數(shù)的圖象及應(yīng)用由題知法(1)(2018全國卷)函數(shù)f (x)的圖象大致為()(2)如圖,已知l1l2,圓心在l1上、半徑為1 m 的圓O在t0時與l2相切于點A,圓O沿l1以1 m/s的速度勻速向上移動,圓被直線l2所截上方圓弧長記為x,令ycos x,則y與時間t(0t1,單位:s)的函數(shù)yf (t)的圖象大致為()(3)已知函數(shù)f (x)若存在x1,x2,當(dāng)0x1x20,排除D選項又e2,1,排除C選項故選B. (2)如圖,設(shè)MON,由弧長公式知x.在RtAOM中,|AO|1t,cos1t,ycos x2cos212(1t)21.又0t1,故選B.(3)畫出函數(shù)大致圖象如圖所示由圖象知,x1,x21,x12x21,于是x1f (x2)x12x21x1,x10的解集為,f (x)單調(diào)遞增;f (x)2,所以排除C選項故選D.2.如圖,長方形ABCD的邊AB2,BC1,O是AB的中點,點P沿著邊BC,CD與DA運動,記BOPx.將動點P到A,B兩點距離之和表示為x的函數(shù)f (x),則yf (x)的圖象大致為()解析:選B當(dāng)x時,f (x)tan x,圖象不會是直線段,從而排除A、C.當(dāng)x時,f f 1,f 2.21,f f f ,從而排除D,故選B.3已知f (x)2x1,g(x)1x2.規(guī)定:當(dāng)|f (x)|g(x)時,h(x)|f (x)|;當(dāng)|f (x)|0時,f (x)單調(diào)遞增,且f (1)0,若f (x1)0,則x的取值范圍為()A(0,1)(2,) B(,0)(2,)C(,0)(3,) D(,1)(1,)(2)(2018益陽、湘潭調(diào)研)定義在R上的函數(shù)f (x),滿足f (x5)f (x),當(dāng)x(3,0時,f (x)x1,當(dāng)x(0,2時,f (x)log2x,則f (1)f (2)f (3)f (2 018)的值等于()A403 B405C806 D809(3)已知定義在R上的奇函數(shù)f (x)滿足f (x3)f (x),且當(dāng)x時,f (x)x3,則f _.解析(1)由于函數(shù)f (x)是奇函數(shù),且當(dāng)x0時f (x)單調(diào)遞增,f (1)0,所以f (1)0,故由f (x1)0,得1x11,所以0x2,故選A.(2)定義在R上的函數(shù)f (x),滿足f (x5)f (x),即函數(shù)f (x)的周期為5.又當(dāng)x(0,2時,f (x)log2x,所以f (1)log210,f (2)log221.當(dāng)x(3,0時,f (x)x1,所以f (3)f (2)1,f (4)f (1)0,f (5)f (0)1.所以f (1)f (2)f (3)f (2 018)403f (1)f (2)f (3)f (4)f (5)f (2 016)f (2 017)f (2 018)4031f (1)f (2)f (3)403011405.(3)由f (x3)f (x)知函數(shù)f (x)的周期為3,又函數(shù)f (x)為奇函數(shù),所以f f f 3.答案(1)A(2)B(3)類題通法函數(shù)性質(zhì)的應(yīng)用技巧奇偶性具有奇偶性的函數(shù)在關(guān)于原點對稱的區(qū)間上其圖象、函數(shù)值、解析式和單調(diào)性聯(lián)系密切,研究問題時可轉(zhuǎn)化到只研究部分(一半)區(qū)間上尤其注意偶函數(shù)f (x)的性質(zhì):f (|x|)f (x)單調(diào)性可以比較大小,求函數(shù)最值,解不等式,證明方程根的唯一性周期性利用周期性可以轉(zhuǎn)化函數(shù)的解析式、圖象和性質(zhì),把不在已知區(qū)間上的問題,轉(zhuǎn)化到已知區(qū)間上求解對稱性利用其軸對稱或中心對稱可將研究的問題,轉(zhuǎn)化到另一對稱區(qū)間上研究 應(yīng)用通關(guān)1(2018貴陽模擬)已知函數(shù)f (x),則下列結(jié)論正確的是()A函數(shù)f (x)的圖象關(guān)于點(1,2)中心對稱B函數(shù)f (x)在(,1)上是增函數(shù)C函數(shù)f (x)的圖象上至少存在兩點A,B,使得直線ABx軸D函數(shù)f (x)的圖象關(guān)于直線x1對稱解析:選A因為y2,所以該函數(shù)圖象可以由y的圖象向右平移1個單位長度,再向上平移2個單位長度得到,所以函數(shù)f (x)的圖象關(guān)于點(1,2)中心對稱,A正確,D錯誤易知函數(shù)f (x)在(,1)上單調(diào)遞減,故B錯誤易知函數(shù)f (x)的圖象是由y的圖象平移得到的,所以不存在兩點A,B使得直線ABx軸,C錯誤故選A.2(2019屆高三惠州調(diào)研)已知函數(shù)yf (x)的定義域為R,且滿足下列三個條件:對任意的x1,x24,8,當(dāng)x10恒成立;f (x4)f (x);yf (x4)是偶函數(shù)若af (6),bf (11),cf (2 017),則a,b,c的大小關(guān)系正確的是()Aabc BbacCacb Dcba解析:選B由知函數(shù)f (x)在區(qū)間4,8上為單調(diào)遞增函數(shù);由知f (x8)f (x4)f (x),即函數(shù)f (x)的周期為8,所以cf (2 017)f (25281)f (1),bf (11)f (3);由可知函數(shù)f (x)的圖象關(guān)于直線x4對稱,所以bf (3)f (5),cf (1)f (7)因為函數(shù)f (x)在區(qū)間4,8上為單調(diào)遞增函數(shù),所以f (5)f (6)f (7),即ba0)若f (xa)f (xb)(ab),則T|ab|;若f (2ax)f (x)且f (2bx)f (x)(ab),則T2|ba|.增分集訓(xùn)1定義在R上的函數(shù)yf (x)為減函數(shù),且函數(shù)yf (x1)的圖象關(guān)于點(1,0)對稱若f (x22x)f (2bb2)0,且0x2,則xb的取值范圍是()A2,0 B2,2C0,2 D0,4解析:選B設(shè)P(x,y)為函數(shù)yf (x1)的圖象上的任意一點,P關(guān)于點(1,0)對稱的點為(2x,y),f (2x1)f (x1),即f (1x)f (x1)不等式f (x22x)f (2bb2)0可化為f (x22x)f (2bb2)f (112bb2)f (b22b)函數(shù)yf (x)為定義在R上的減函數(shù),x22xb22b,即(x1)2(b1)2.0x2,或畫出可行域如圖中陰影部分所示設(shè)xbz,則bxz,由圖可知,當(dāng)直線bxz經(jīng)過點(0,2)時,z取得最小值2;當(dāng)直線bxz經(jīng)過點(2,0)時,z取得最大值2.綜上可得,xb的取值范圍是2,22(2018沈陽模擬)設(shè)f (x)是定義在R上的偶函數(shù),F(xiàn)(x)(x2)3f (x2)17,G(x),若F(x)的圖象與G(x)的圖象的交點分別為(x1,y1),(x2,y2),(xm,ym),則(xiyi)_.解析:f (x)是定義在R上的偶函數(shù),g(x)x3f (x)是定義在R上的奇函數(shù),其圖象關(guān)于原點中心對稱,函數(shù)F(x)(x2)3f (x2)17g(x2)17的圖象關(guān)于點(2,17)中心對稱又函數(shù)G(x)17的圖象也關(guān)于點(2,17)中心對稱,F(xiàn)(x)和G(x)的圖象的交點也關(guān)于點(2,17)中心對稱,x1x2xm(2)22m,y1y2ym(17)217m,(xiyi)(x1x2xm)(y1y2ym)19m.答案:19m重難增分(二)新定義下的函數(shù)問題 典例細解我們將具有性質(zhì)f f (x)的函數(shù),稱為滿足“倒負”變換的函數(shù)給出下列函數(shù):f (x)ln;f (x);f (x)其中滿足“倒負”變換的函數(shù)是()ABC D解析對于,因為f lnlnf (x),所以不滿足“倒負”變換;對于,因為f f (x),所以滿足“倒負”變換;對于,因為f 即f 所以f f (x),故滿足“倒負”變換綜上可知,選C.答案C啟思維本題是在現(xiàn)有函數(shù)的圖象與性質(zhì)的基礎(chǔ)上定義的一種新的函數(shù)性質(zhì),考查在新情境下,靈活運用有關(guān)函數(shù)知識求解“新定義”類數(shù)學(xué)問題的能力求解本題的關(guān)鍵是先準(zhǔn)確寫出f 的表達式,并加以整理,再具體考慮f 與f (x)是否相等設(shè)函數(shù)f (x)的定義域為D,若f (x)滿足條件:存在a,bD(ab),使f (x)在a,b上的值域也是a,b,則稱函數(shù)f (x)為“優(yōu)美函數(shù)”若函數(shù)f (x)log2(4xt)為“優(yōu)美函數(shù)”,則t的取值范圍是()A. B.C. D.解析f (x)log2(4xt)為增函數(shù),且存在a,bD(a0),則方程m2mt0有兩個不等的實根,且兩根都大于0,所以解得0t0,f (x)ex1x2是增函數(shù)又f (1)0,函數(shù)f (x)的零點為x1,1,|1|1,02,函數(shù)g(x)x2axa3在區(qū)間0,2上有零點由g(x)0,得a(0x2),即a(x1)2(0x2),設(shè)x1t(1t3),則at2(1t3),令h(t)t2(1t3),易知h(t)在區(qū)間1,2)上是減函數(shù),在區(qū)間(2,3上是增函數(shù),2h(t)3,即2a3,故選D.3對任意實數(shù)a,b定義運算“”:ab設(shè)f (x)(x21)(4x),若函數(shù)yf (x)k的圖象與x軸恰有三個不同的交點,則實數(shù)k的取值范圍是()A(2,1) B0,1C2,0) D2,1)解析:選D當(dāng)x214x1,即x2或x3時,f (x)4x;當(dāng)x214x1,即2x3時,f (x)x21.作出f (x)的圖象如圖所示,由圖象可知,要使kf (x)有三個根,需滿足1k2,即2k0時,y2|x|x|x,函數(shù)yx在區(qū)間(0,)上是減函數(shù)故選D.2(2018貴陽模擬)若函數(shù)f (x)是定義在R上的奇函數(shù),當(dāng)x0時,f (x)log2(x2)1,則f (6)()A2 B4C2 D4解析:選C根據(jù)題意得f (6)f (6)1log2(62)1log282.故選C.3(2018長春質(zhì)檢)已知函數(shù)f (x)則函數(shù)f (x)的值域為()A1,) B(1,)C. DR解析:選B法一:當(dāng)x1時,f (x)x22(1,);當(dāng)x1時,f (x)2x1,綜上可知,函數(shù)f (x)的值域為(1,)故選B.法二:作出分段函數(shù)f (x)的圖象(圖略)可知,該函數(shù)的值域為(1,),故選B.4(2018陜西質(zhì)檢)設(shè)xR,定義符號函數(shù)sgn x則函數(shù)f (x)|x|sgn x的圖象大致是()解析:選C由符號函數(shù)解析式和絕對值運算,可得f (x)x,選C.5(2018濮陽二模)若f (x)是奇函數(shù),則f (g(2)的值為()A. BC1 D1解析:選Cf (x)是奇函數(shù),x0的解集為()A.(2,) B(2,)C.(2,) D(,2)解析:選B函數(shù)f (x)exexex滿足f (x)f (x),f (x)為奇函數(shù)且是單調(diào)遞增函數(shù),關(guān)于x的不等式f (2x1)f (x1)0,即為f (2x1)f (x1),2x1x1,解得x2,故選B.12(2018陜西二模)已知函數(shù)f (x)ex2(x0)與g(x)ln(xa)2的圖象上存在關(guān)于y軸對稱的點,則a的取值范圍是()A. B(,e)C. D.解析:選B由題意知,方程f (x)g(x)0在(0,)上有解,即ex2ln(xa)20在(0,)上有解,即函數(shù)yex的圖象與yln(xa)的圖象在(0,)上有交點,函數(shù)yln(xa)的圖象是由函數(shù)yln x的圖象向左平移a個單位得到的,當(dāng)yln x向左平移且平移到過點(0,1)后開始,兩函數(shù)的圖象有交點,把點(0,1)代入yln(xa)得,1ln a,ae,a3時滿足f (x)f (x3)f (x6),函數(shù)f (x)的周期為6.f (2 009)f (33465)f (5)f (1)當(dāng)x0時f (x)log2(1x),f (1)1,f (2 009)f (1)1.答案:116已知函數(shù)f (x)e|x|,函數(shù)g(x)對任意的x1,m(m1),都有f (x2)g(x),則m的取值范圍是_解析:作出函數(shù)yh(x)e|x2|和yg(x)的圖象,如圖所示,由圖可知當(dāng)x1時,h(1)g(1),又當(dāng)x4時,h(4)e24時,由ex24e5x,得e2x74,即2x7ln 4,解得xln 2,又m1,11),若對于任意a,b,cR,都有f (a)f (b)f (c)成立,則實數(shù)m的取值范圍是_解析:因為f (x)1,所以當(dāng)m1時,函數(shù)f (x)在R上是減函數(shù),函數(shù)f (x)的值域為(1,m),所以f (a)f (b)2,f (c)f (c)對任意的a,b,cR恒成立,所以m2,所以1f (c)1,滿足題意當(dāng)m2m,f (c)1,所以2m1,所以m,所以m1.綜上可知,m2,故所求實數(shù)m的取值范圍是.答案:20已知函數(shù)f (x)若f (x)的值域為R,則實數(shù)a的取值范圍是_解析:依題意,當(dāng)x1時,f (x)1log2x單調(diào)遞增,f (x)1log2x在區(qū)間1,)上的值域是1,)因此,要使函數(shù)f (x)的值域是R,則需函數(shù)f (x)在(,1)上的值域M(,1)當(dāng)a10,即a1時,函數(shù)f (x)在(,1)上單調(diào)遞減,函數(shù)f (x)在(,1)上的值域M(a3,),顯然此時不能滿足M(,1),因此a0,即a1時,函數(shù)f (x)在(,1)上單調(diào)遞增,函數(shù)f (x)在(,1)上的值域M(,a3),由M(,1)得解得12的解集為()A(2,) B.(2,)C.(,) D(,)解析:選B因為f (x)是R上的偶函數(shù),且在(,0上是減函數(shù),所以f (x)在0,)上是增函數(shù)因為f (1)2,所以f (1)2,所以f (log2x)2f (|log2x|)f (1)|log2x|1log2x1或log2x2或0x.故選B.2(2019屆高三太原模擬)已知函數(shù)f (x)是偶函數(shù),f (x1)是奇函數(shù),且對于任意x1,x20,1,且x1x2,都有(x1x2)f (x1)f (x2)bc BbacCbca Dcab解析:選B法一:因為函數(shù)f (x)是偶函數(shù),f (x1)是奇函數(shù),所以f (x)f (x),f (x1)f (x1),所以f (x1)f (x1),所以f (x)f (x2),所以f (x)f (x4),所以af f f ,bf f ,cf f ,又對于任意x1,x20,1,且x1x2,都有(x1x2)f (x1)f (x2)0,所以f (x)在0,1上是減函數(shù),因為ac,故選B.法二:因為函數(shù)f (x)是偶函數(shù),f (x1)是奇函數(shù),且對于任意x1,x20,1,且x1x2,都有(x1x2)f (x1)f (x2)0,即f (x)在0,1上是減函數(shù),不妨取f (x)cosx,則af coscos,bf coscos,cf coscos,因為函數(shù)ycos x在0,1上是減函數(shù),且ac,故選B.3(2018全國卷)設(shè)函數(shù)f (x)則滿足f (x1)f (2x)的x的取值范圍是()A(,1 B(0,)C(1,0) D(,0)解析:選D法一:當(dāng)即x1時,f (x1)f (2x),即為2(x1)22x,即(x1)2x,解得x1.因此不等式的解集為(,1當(dāng)時,不等式組無解當(dāng)即1x0時,f (x1)f (2x),即為122x,解得x0時,f (x1)1,f (2x)1,不合題意綜上,不等式f (x1)f (2x)的解集為(,0)法二:f (x)函數(shù)f (x)的圖象如圖所示結(jié)合圖象知,要使f (x1)f (2x),則需或x0.給出下列命題:f (221)1;函數(shù)yf (x)圖象的一條對稱軸方程為x4;函數(shù)yf (x)在6,4上為減函數(shù);方程f (x)0在6,6上有4個根其中正確的命題個數(shù)為()A1 B2C3 D4解析:選D令x2,由f (x4)f (x)f (2)得f (2)0.因為函數(shù)yf (x)是R上的偶函數(shù),所以f (2)f (2)0,所以f (x4)f (x),即函數(shù)yf (x)是以4為周期的周期函數(shù),所以f (221)f (5541)f (1)因為f (3)1,所以f (3)f (1)1,從而f (221)1,正確因為函數(shù)圖象關(guān)于y軸對稱,函數(shù)的周期為4,所以函數(shù)yf (x)圖象的一條對稱軸方程為x4,正確因為當(dāng)x1,x20,2,且x1x2時,都有0,設(shè)x1x2,則f (x1)f (x2),易知函數(shù)yf (x)在0,2上是增函數(shù)根據(jù)圖象的對稱性,易知函數(shù)yf (x)在2,0上是減函數(shù),又根據(jù)周期性,易知函數(shù)yf (x)在6,4上為減函數(shù),正確因為f (2)f (2)0,由函數(shù)f (x)的單調(diào)性及周期性,可知在6,6上有且僅有f (2)f (2)f (6)f (6)0,即方程f (x)0在6,6上有4個根綜上所述,四個命題都正確故選D.5(2018長沙模擬)定義運算:xy例如:343,(2)44,則函數(shù)f (x)x2(2xx2)的最大值為_解析:由已知得f (x)x2(2xx2)畫出函數(shù)f (x)的大致圖象(圖略)可知,函數(shù)f (x)的最大值為4.答案:46(2019屆高三石家莊檢測)已知定義域為R的函數(shù)f (x)是奇函數(shù),當(dāng)x0時,f (x)|xa2|a2,且對xR,恒有f (x1)f (x),則實數(shù)a的取值范圍為_解析:定義域為R的函數(shù)f (x)是奇函數(shù),當(dāng)x0時,f (x)|xa2|a2作出函數(shù)f (x)的圖象如圖所示當(dāng)xf (sin x1m)恒成立,則實數(shù)m的取值范圍為_解析:因為f (x2)是偶函數(shù),所以函數(shù)f (x)的圖象關(guān)于x2對稱又f (x)在(,2)上為增函數(shù),則f (x)在(2,)上為減函數(shù),所以不等式f (2sin x2)f (sin x1m)恒成立等價于|2sin x22|sin x1m2|,即|2sin x|sin x1m|,兩邊同時平方,得3sin2x2(1m)sin x(1m)20,即(3sin x1m)(sin x1m)0,即或即或即或即m4,故實數(shù)m的取值范圍為(,2)(4,)答案:(,2)(4,)- 1.請仔細閱讀文檔,確保文檔完整性,對于不預(yù)覽、不比對內(nèi)容而直接下載帶來的問題本站不予受理。
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