全國(guó)統(tǒng)考高考數(shù)學(xué)大一輪復(fù)習(xí)第2章函數(shù)概念與基本初等函數(shù)Ⅰ第2講函數(shù)的基本性質(zhì)1備考試題文含解析

一輪復(fù)習(xí)精品資料(高中)PAGEPAGE1第二章函數(shù)概念與基本初等函數(shù)Ⅰ第二講函數(shù)的基本性質(zhì)練好題·考點(diǎn)自測(cè)1.下列說法中正確的個(gè)數(shù)是()(1)若函數(shù)y=f(x)在〖1,+∞)上是增函數(shù),則函數(shù)的單調(diào)遞增區(qū)間是〖1,+∞).(2)對(duì)于函數(shù)f(x),x∈D,若對(duì)任意x1,x2∈D(x1≠x2),有(x1-x2)〖f(x1)-f(x2)〗>0,則函數(shù)f(x)在區(qū)間D上是增函數(shù).(3)若函數(shù)y=f(x+a)是偶函數(shù),則函數(shù)y=f(x)的圖象關(guān)于直線x=a對(duì)稱.(4)若函數(shù)y=f(x+b)是奇函數(shù),則函數(shù)y=f(x)的圖象關(guān)于點(diǎn)(b,0)中心對(duì)稱.(5)已知函數(shù)y=f(x)是定義在R上的偶函數(shù),若f(x)在(-∞,0)上是減函數(shù),則f(x)在(0,+∞)上是增函數(shù).(6)若T為函數(shù)y=f(x)的一個(gè)周期,那么nT(n∈Z)也是函數(shù)f(x)的周期.A.3 B.4 C.5 D.62.〖2019北京,3,5分〗〖文〗下列函數(shù)中,在區(qū)間(0,+∞)上單調(diào)遞增的是()A.y=x12 B.y=2C.y=log12x D.y3.〖2019全國(guó)卷Ⅱ,6,5分〗〖文〗設(shè)f(x)為奇函數(shù),且當(dāng)x≥0時(shí),f(x)=ex-1,則當(dāng)x<0時(shí),f(x)=()A.e-x-1 B.e-x+1 C.-e-x-1 D.-e-x+14.〖2020山東,8,5分〗若定義在R的奇函數(shù)f(x)在(-∞,0)上單調(diào)遞減,且f(2)=0,則滿足xf(x-1)≥0的x的取值范圍是()A.〖-1,1〗∪〖3,+∞) B.〖-3,-1〗∪〖0,1〗C.〖-1,0〗∪〖1,+∞) D.〖-1,0〗∪〖1,3〗5.〖2021大同市調(diào)研測(cè)試〗已知函數(shù)f(x)=ax3+bsinx+cln(x+x2+1)+3的最大值為5,則f(x)的最小值為(A.-5 B.1 C.2 D.36.〖2020福州3月質(zhì)檢〗已知f(x)是定義在R上的偶函數(shù),其圖象關(guān)于點(diǎn)(1,0)對(duì)稱.給出以下關(guān)于f(x)的結(jié)論:①f(x)是周期函數(shù);②f(x)滿足f(x)=f(4-x);③f(x)在(0,2)上單調(diào)遞減;④f(x)=cosπx2其中正確結(jié)論的個(gè)數(shù)是()A.4 B.3 C.2 D.17.〖2018江蘇,9,5分〗函數(shù)f(x)滿足f(x+4)=f(x)(x∈R),且在區(qū)間(-2,2〗上,f(x)=cosπx2,0<x≤2拓展變式1.(1)函數(shù)f(x)=-x2-ax-(2)〖2016天津,13,5分〗已知f(x)是定義在R上的偶函數(shù),且在區(qū)間(-∞,0)上單調(diào)遞增.若實(shí)數(shù)a滿足f(2|a-1|)>f(-2),則a的取值范圍是2.(1)高斯是德國(guó)著名的數(shù)學(xué)家,近代數(shù)學(xué)奠基者之一,享有“數(shù)學(xué)王子”的稱號(hào),用其名字命名的“高斯函數(shù)”為:設(shè)x∈R,用〖x〗表示不超過x的最大整數(shù),則y=〖x〗稱為高斯函數(shù).例如:〖-2.1〗=-3,〖3.1〗=3.已知函數(shù)f(x)=2x+31+2x+1,則函數(shù)y=〖f(x)A.(12,3) B.(0,2〗 C.{0,1,2}(2)已知函數(shù)f(x)=sinπx22x-1+2-3.〖新課標(biāo)全國(guó)Ⅰ,5分〗設(shè)函數(shù)f(x),g(x)的定義域都為R,且f(x)是奇函數(shù),g(x)是偶函數(shù),則下列結(jié)論中正確的是()A.f(x)g(x)是偶函數(shù) B.f(x)|g(x)|是奇函數(shù)C.|f(x)|g(x)是奇函數(shù) D.|f(x)g(x)|是奇函數(shù)4.〖2021陜西模擬〗若函數(shù)f(x),g(x)分別是定義在R上的偶函數(shù)、奇函數(shù),且滿足f(x)+2g(x)=ex,則()A.f(-2)<f(-3)<g(-1) B.g(-1)<f(-3)<f(-2)C.f(-2)<g(-1)<f(-3) D.g(-1)<f(-2)<f(-3)5.〖2021貴陽(yáng)市摸底測(cè)試〗已知函數(shù)f(x)的定義域?yàn)镽.當(dāng)x<0時(shí),f(x)=x3-1;當(dāng)-1≤x≤1時(shí),f(-x)=-f(x);當(dāng)x>12時(shí),f(x+12)=f(x-12).則f(8)=A.-2 B.-1 C.0 D.26.(1)〖2021山東新高考模擬〗已知函數(shù)f(x)=ex-e-xex+e-x,實(shí)數(shù)m,n滿足不等式f(2m-n)A.m+n>1 B.m+n<1C.m-n>-1 D.m-n<-1(2)〖2020廣西師大附中4月模擬〗已知定義在R上的奇函數(shù)f(x)滿足f(1+x)=-f(x),當(dāng)x∈(-1,1)時(shí),f(x)=-x2+2x,0≤x<1,ax2+bx,-1<x答案第二章函數(shù)概念與基本初等函數(shù)Ⅰ第二講函數(shù)的基本性質(zhì)1.B對(duì)于(1),函數(shù)的單調(diào)區(qū)間和函數(shù)在區(qū)間上單調(diào)是不同的,故(1)錯(cuò)誤;對(duì)于(2),對(duì)任意x1,x2∈D(x1≠x2),(x1-x2)〖f(x1)-f(x2)〗>0?x1>x2,f(x1)>f(x2)或x1<x2,f(x1)<f(x2),所以f(x)在區(qū)間D上是增函數(shù),故(2)正確;對(duì)于(3),若函數(shù)y=f(x+a)是偶函數(shù),則f(-x+a)=f(x+a),則函數(shù)y=f(x)的圖象關(guān)于直線x=a對(duì)稱,故(3)正確;對(duì)于(4),若函數(shù)y=f(x+b)是奇函數(shù),則f2.A對(duì)于冪函數(shù)y=xα,當(dāng)α>0時(shí),y=xα在(0,+∞)上單調(diào)遞增,當(dāng)α<0時(shí),y=xα在(0,+∞)上單調(diào)遞減,所以選項(xiàng)A正確;選項(xiàng)D中的函數(shù)y=1x可轉(zhuǎn)化為y=x-1,所以函數(shù)y=1x在(0,+∞)上單調(diào)遞減,故選項(xiàng)D不符合題意;對(duì)于指數(shù)函數(shù)y=ax(a>0且a≠1),當(dāng)0<a<1時(shí),y=ax在(-∞,+∞)上單調(diào)遞減,當(dāng)a>1時(shí),y=ax在(-∞,+∞)上單調(diào)遞增,而選項(xiàng)B中的函數(shù)y=2-x可轉(zhuǎn)化為y=(12)x,因此函數(shù)y=2-x在(0,+∞)上單調(diào)遞減,故選項(xiàng)B不符合題意;對(duì)于對(duì)數(shù)函數(shù)y=logax(a>0且a≠1),當(dāng)0<a<1時(shí),y=logax在(0,+∞)上單調(diào)遞減,當(dāng)a>1時(shí),y=logax在(0,+∞)上單調(diào)遞增,因此選項(xiàng)C中的函數(shù)y=log123.D解法一依題意得,當(dāng)x<0時(shí),f(x)=-f(-x)=-(e-x-1)=-e-x+1,選D.解法二依題意得,f(-1)=-f(1)=-(e1-1)=1-e,結(jié)合選項(xiàng)知,選D.4.D解法一由題意知f(x)在(-∞,0),(0,+∞)上單調(diào)遞減,且f(-2)=f(2)=f(0)=0.當(dāng)x>0時(shí),令f(x-1)≥0,得0≤x-1≤2,∴1≤x≤3;當(dāng)x<0時(shí),令f(x-1)≤0,得-2≤x-1≤0,∴-1≤x≤1,又x<0,∴-1≤x<0;當(dāng)x=0時(shí),顯然符合題意.綜上,原不等式的解集為〖-1,0〗∪〖1,3〗,選D.解法二當(dāng)x=3時(shí),f(3-1)=0,符合題意,排除B;當(dāng)x=4時(shí),f(4-1)=f(3)<0,此時(shí)不符合題意,排除A,C.故選D.5.B由題意可知,函數(shù)f(x)的定義域?yàn)镽.令g(x)=ax3+bsinx+cln(x+x2+1),則g(-x)=a(-x)3+bsin(-x)+cln〖(-x)+(-x)2+1〗=-ax3-bsinx+cln(-x+x2+1)(x+x2+1)x+x2+1=-ax3-bsin所以函數(shù)g(x)=ax3+bsinx+cln(x+x2+1)為奇函數(shù)又f(x)max=g(x)max+3=5,所以g(x)max=2,于是g(x)min=-2,所以f(x)min=g(x)min+3=-2+3=1,故選B.6.B因?yàn)閒(x)為偶函數(shù),所以f(-x)=f(x),又其圖象關(guān)于點(diǎn)(1,0)對(duì)稱,所以f(-x)=-f(2+x),則f(x+2)=-f(x),由此可得f(x+4)=-f(x+2)=f(x),即f(x)是以4為周期的周期函數(shù),所以①正確;f(-x)=f(x)=f(x+4),把x替換成-x可得f(x)=f(4-x),所以②正確;f(x)=cosπx2是定義在R上的偶函數(shù),且(1,0)是它的圖象的一個(gè)對(duì)稱中心,所以④正確;不妨令f(x)=-cosπx2,此時(shí)f(x)滿足題意,但f所以正確結(jié)論的個(gè)數(shù)是3.故選B.7.22因?yàn)楹瘮?shù)f(x)滿足f(x+4)=f(x)(x∈R),所以4為函數(shù)f(x)的周期.因?yàn)樵趨^(qū)間(-2,2〗上,f(x)=cosπx2,0<x≤2,|x+12|,-1.(1)-3≤a≤-2由題意,得-a2≥1,(2)(12,32)因?yàn)閒(x)是定義在R上的偶函數(shù),且在區(qū)間(-∞,0)上單調(diào)遞增,所以f(x)在區(qū)間(0,+∞)上單調(diào)遞減.又f(2|a-1|)>f(-2),且f(-2)=f(2),所以-2<2|a-1|<2,則|a-1|<12,所以2.(1)Cf(x)=2x+31+2x+1=12(1+2x+1)+521+2x+1=1〖易錯(cuò)警示〗本題的易錯(cuò)點(diǎn)是沒有理解取整的意思,求出函數(shù)的值域后就迫不及待地選擇A,從而導(dǎo)致失分.有關(guān)此類新定義問題,一定要讀懂新定義的含義,并看清題干中所舉的例子的特征,才可有效避開此類錯(cuò)誤.(2)12因?yàn)閒(x)=sinπx22x-1+2-x+1,設(shè)f1(x)=2x-1+2-x+1,所以f1(x)=2x-1+12x-1≥22x-1·12x-1=2,當(dāng)且僅當(dāng)2x-1=12x-1,即x=1時(shí)取等號(hào),即當(dāng)x=1時(shí),f1(x3.B因?yàn)閒(x)為奇函數(shù),g(x)為偶函數(shù),所以f(x)g(x)為奇函數(shù),f(x)|g(x)|為奇函數(shù),|f(x)|g(x)為偶函數(shù),|f(x)g(x)|為偶函數(shù),故選B.4.D因?yàn)楹瘮?shù)f(x),g(x)分別是定義在R上的偶函數(shù)、奇函數(shù),且滿足f(x)+2g(x)=ex①,所以f(-x)+2g(-x)=e-x,即f(x)-2g(x)=e-x②.聯(lián)立①②,解得f(x)=ex+e-x2,g(x)=ex-e-x4,所以f(-2)=5.D因?yàn)楫?dāng)x>12時(shí),f(x+12)=f(x-12),所以當(dāng)x>12時(shí),f(x)的周期為1,所以f(8)=f(7×1+1)=f(1).因?yàn)楫?dāng)-1≤x≤1時(shí),f(-x)=-f(x),所以f(1)=-f(-1)=-〖(-1)36.(1)C因?yàn)閒(-x)=e-x-exe-x+ex=-ex-e-xe-x+ex=-f(x),所以f(x)為奇函數(shù),又f(x)=ex-e-xex+e-x=1-2e2x+1,所以f(2)34當(dāng)0<x<1時(shí),-1<-x<0,f(x)=-x2+2x,f(-x)=a(-x)2+b(-x)=ax2-bx,由f(-x)=-f(x),得ax2-bx=-(-x2+2x),求得a=1,b=2.又函數(shù)f(x)滿足f(1+x)=-f(x),則f(x+2)=-f(x+1)=f(x),即函數(shù)f(x)是周期為2的周期函數(shù).所以f(logba)+f(a)+f(2021b)=f(log21)+f(1)+f(20212)=f(0)+f(1)+f(1010+12)=f(0)+〖-f(0)〗+f(12)=f第二章函數(shù)概念與基本初等函數(shù)Ⅰ第二講函數(shù)的基本性質(zhì)練好題·考點(diǎn)自測(cè)1.下列說法中正確的個(gè)數(shù)是()(1)若函數(shù)y=f(x)在〖1,+∞)上是增函數(shù),則函數(shù)的單調(diào)遞增區(qū)間是〖1,+∞).(2)對(duì)于函數(shù)f(x),x∈D,若對(duì)任意x1,x2∈D(x1≠x2),有(x1-x2)〖f(x1)-f(x2)〗>0,則函數(shù)f(x)在區(qū)間D上是增函數(shù).(3)若函數(shù)y=f(x+a)是偶函數(shù),則函數(shù)y=f(x)的圖象關(guān)于直線x=a對(duì)稱.(4)若函數(shù)y=f(x+b)是奇函數(shù),則函數(shù)y=f(x)的圖象關(guān)于點(diǎn)(b,0)中心對(duì)稱.(5)已知函數(shù)y=f(x)是定義在R上的偶函數(shù),若f(x)在(-∞,0)上是減函數(shù),則f(x)在(0,+∞)上是增函數(shù).(6)若T為函數(shù)y=f(x)的一個(gè)周期,那么nT(n∈Z)也是函數(shù)f(x)的周期.A.3 B.4 C.5 D.62.〖2019北京,3,5分〗〖文〗下列函數(shù)中,在區(qū)間(0,+∞)上單調(diào)遞增的是()A.y=x12 B.y=2C.y=log12x D.y3.〖2019全國(guó)卷Ⅱ,6,5分〗〖文〗設(shè)f(x)為奇函數(shù),且當(dāng)x≥0時(shí),f(x)=ex-1,則當(dāng)x<0時(shí),f(x)=()A.e-x-1 B.e-x+1 C.-e-x-1 D.-e-x+14.〖2020山東,8,5分〗若定義在R的奇函數(shù)f(x)在(-∞,0)上單調(diào)遞減,且f(2)=0,則滿足xf(x-1)≥0的x的取值范圍是()A.〖-1,1〗∪〖3,+∞) B.〖-3,-1〗∪〖0,1〗C.〖-1,0〗∪〖1,+∞) D.〖-1,0〗∪〖1,3〗5.〖2021大同市調(diào)研測(cè)試〗已知函數(shù)f(x)=ax3+bsinx+cln(x+x2+1)+3的最大值為5,則f(x)的最小值為(A.-5 B.1 C.2 D.36.〖2020福州3月質(zhì)檢〗已知f(x)是定義在R上的偶函數(shù),其圖象關(guān)于點(diǎn)(1,0)對(duì)稱.給出以下關(guān)于f(x)的結(jié)論:①f(x)是周期函數(shù);②f(x)滿足f(x)=f(4-x);③f(x)在(0,2)上單調(diào)遞減;④f(x)=cosπx2其中正確結(jié)論的個(gè)數(shù)是()A.4 B.3 C.2 D.17.〖2018江蘇,9,5分〗函數(shù)f(x)滿足f(x+4)=f(x)(x∈R),且在區(qū)間(-2,2〗上,f(x)=cosπx2,0<x≤2拓展變式1.(1)函數(shù)f(x)=-x2-ax-(2)〖2016天津,13,5分〗已知f(x)是定義在R上的偶函數(shù),且在區(qū)間(-∞,0)上單調(diào)遞增.若實(shí)數(shù)a滿足f(2|a-1|)>f(-2),則a的取值范圍是2.(1)高斯是德國(guó)著名的數(shù)學(xué)家,近代數(shù)學(xué)奠基者之一,享有“數(shù)學(xué)王子”的稱號(hào),用其名字命名的“高斯函數(shù)”為:設(shè)x∈R,用〖x〗表示不超過x的最大整數(shù),則y=〖x〗稱為高斯函數(shù).例如:〖-2.1〗=-3,〖3.1〗=3.已知函數(shù)f(x)=2x+31+2x+1,則函數(shù)y=〖f(x)A.(12,3) B.(0,2〗 C.{0,1,2}(2)已知函數(shù)f(x)=sinπx22x-1+2-3.〖新課標(biāo)全國(guó)Ⅰ,5分〗設(shè)函數(shù)f(x),g(x)的定義域都為R,且f(x)是奇函數(shù),g(x)是偶函數(shù),則下列結(jié)論中正確的是()A.f(x)g(x)是偶函數(shù) B.f(x)|g(x)|是奇函數(shù)C.|f(x)|g(x)是奇函數(shù) D.|f(x)g(x)|是奇函數(shù)4.〖2021陜西模擬〗若函數(shù)f(x),g(x)分別是定義在R上的偶函數(shù)、奇函數(shù),且滿足f(x)+2g(x)=ex,則()A.f(-2)<f(-3)<g(-1) B.g(-1)<f(-3)<f(-2)C.f(-2)<g(-1)<f(-3) D.g(-1)<f(-2)<f(-3)5.〖2021貴陽(yáng)市摸底測(cè)試〗已知函數(shù)f(x)的定義域?yàn)镽.當(dāng)x<0時(shí),f(x)=x3-1;當(dāng)-1≤x≤1時(shí),f(-x)=-f(x);當(dāng)x>12時(shí),f(x+12)=f(x-12).則f(8)=A.-2 B.-1 C.0 D.26.(1)〖2021山東新高考模擬〗已知函數(shù)f(x)=ex-e-xex+e-x,實(shí)數(shù)m,n滿足不等式f(2m-n)A.m+n>1 B.m+n<1C.m-n>-1 D.m-n<-1(2)〖2020廣西師大附中4月模擬〗已知定義在R上的奇函數(shù)f(x)滿足f(1+x)=-f(x),當(dāng)x∈(-1,1)時(shí),f(x)=-x2+2x,0≤x<1,ax2+bx,-1<x答案第二章函數(shù)概念與基本初等函數(shù)Ⅰ第二講函數(shù)的基本性質(zhì)1.B對(duì)于(1),函數(shù)的單調(diào)區(qū)間和函數(shù)在區(qū)間上單調(diào)是不同的,故(1)錯(cuò)誤;對(duì)于(2),對(duì)任意x1,x2∈D(x1≠x2),(x1-x2)〖f(x1)-f(x2)〗>0?x1>x2,f(x1)>f(x2)或x1<x2,f(x1)<f(x2),所以f(x)在區(qū)間D上是增函數(shù),故(2)正確;對(duì)于(3),若函數(shù)y=f(x+a)是偶函數(shù),則f(-x+a)=f(x+a),則函數(shù)y=f(x)的圖象關(guān)于直線x=a對(duì)稱,故(3)正確;對(duì)于(4),若函數(shù)y=f(x+b)是奇函數(shù),則f2.A對(duì)于冪函數(shù)y=xα,當(dāng)α>0時(shí),y=xα在(0,+∞)上單調(diào)遞增,當(dāng)α<0時(shí),y=xα在(0,+∞)上單調(diào)遞減,所以選項(xiàng)A正確;選項(xiàng)D中的函數(shù)y=1x可轉(zhuǎn)化為y=x-1,所以函數(shù)y=1x在(0,+∞)上單調(diào)遞減,故選項(xiàng)D不符合題意;對(duì)于指數(shù)函數(shù)y=ax(a>0且a≠1),當(dāng)0<a<1時(shí),y=ax在(-∞,+∞)上單調(diào)遞減,當(dāng)a>1時(shí),y=ax在(-∞,+∞)上單調(diào)遞增,而選項(xiàng)B中的函數(shù)y=2-x可轉(zhuǎn)化為y=(12)x,因此函數(shù)y=2-x在(0,+∞)上單調(diào)遞減,故選項(xiàng)B不符合題意;對(duì)于對(duì)數(shù)函數(shù)y=logax(a>0且a≠1),當(dāng)0<a<1時(shí),y=logax在(0,+∞)上單調(diào)遞減,當(dāng)a>1時(shí),y=logax在(0,+∞)上單調(diào)遞增,因此選項(xiàng)C中的函數(shù)y=log123.D解法一依題意得,當(dāng)x<0時(shí),f(x)=-f(-x)=-(e-x-1)=-e-x+1,選D.解法二依題意得,f(-1)=-f(1)=-(e1-1)=1-e,結(jié)合選項(xiàng)知,選D.4.D解法一由題意知f(x)在(-∞,0),(0,+∞)上單調(diào)遞減,且f(-2)=f(2)=f(0)=0.當(dāng)x>0時(shí),令f(x-1)≥0,得0≤x-1≤2,∴1≤x≤3;當(dāng)x<0時(shí),令f(x-1)≤0,得-2≤x-1≤0,∴-1≤x≤1,又x<0,∴-1≤x<0;當(dāng)x=0時(shí),顯然符合題意.綜上,原不等式的解集為〖-1,0〗∪〖1,3〗,選D.解法二當(dāng)x=3時(shí),f(3-1)=0,符合題意,排除B;當(dāng)x=4時(shí),f(4-1)=f(3)<0,此時(shí)不符合題意,排除A,C.故選D.5.B由題意可知,函數(shù)f(x)的定義域?yàn)镽.令g(x)=ax3+bsinx+cln(x+x2+1),則g(-x)=a(-x)3+bsin(-x)+cln〖(-x)+(-x)2+1〗=-ax3-bsinx+cln(-x+x2+1)(x+x2+1)x+x2+1=-ax3-bsin所以函數(shù)g(x)=ax3+bsinx+cln(x+x2+1)為奇函數(shù)又f(x)max=g(x)max+3=5,所以g(x)max=2,于是g(x)min=-2,所以f(x)min=g(x)min+3=-2+3=1,故選B.6.B因?yàn)閒(x)為偶函數(shù),所以f(-x)=f(x),又其圖象關(guān)于點(diǎn)(1,0)對(duì)稱,所以f(-x)=-f(2+x),則f(x+2)=-f(x),由此可得f(x+4)=-f(x+2)=f(x),即f(x)是以4為周期的周期函數(shù),所以①正確;f(-x)=f(x)=f(x+4),把x替換成-x可得f(x)=f(4-x),所以②正確;f(x)=cosπx2是定義在R上的偶函數(shù),且(1,0)是它的圖象的一個(gè)對(duì)稱中心,所以④正確;不妨令f(x)=-cosπx2,此時(shí)f(x)滿足題意,但f所以正確結(jié)論的個(gè)數(shù)是3.故選B.7.22因?yàn)楹瘮?shù)f(x)滿足f(x+4)=f(x)(x∈R),所以4為函數(shù)f(x)的周期.因?yàn)樵趨^(qū)間(-2,2〗上,f(x)=cosπx2,0<x≤2,|x+12|,-1.(1)-3≤a≤-2由題意,得-a2≥1,(2)(12,32)因?yàn)閒(x)是定義在R上的偶函數(shù),且在區(qū)間(-∞,0)上單調(diào)遞增,所以f(x)在區(qū)間(0,+∞)上單調(diào)遞減.又f(2|a-1|)>f(-2),且f(-2)=f(2),所以-2<2|a-1|<2,則|a-1|<12,所以2.(1)Cf(x)=2x+31+2x+1=12(1+2x+1)+52