第一篇：2014年高考數(shù)學(xué)文科(高考真題+模擬新題)分類：M單元 推理與證明
數(shù)學(xué)
M單元 推理與證明
M1 合情推理與演繹推理
16．，[2014·福建卷] 已知集合{a，b，c}＝{0，1，2}，且下列三個關(guān)系：①a≠2；②b＝2；③c≠0有且只有一個正確，則100a＋10b＋c等于________．
16．201 [解析](i)若①正確，則②③不正確，由③不正確得c＝0，由①正確得a＝1，所以b＝2，與②不正確矛盾，故①不正確．
(ii)若②正確，則①③不正確，由①不正確得a＝2，與②正確矛盾，故②不正確．(iii)若③正確，則①②不正確，由①不正確得a＝2，由②不正確及③正確得b＝0，c＝1，故③正確．
則100a＋10b＋c＝100×2＋10×0＋1＝201.14．[2014·全國新課標(biāo)卷Ⅰ] 甲、乙、丙三位同學(xué)被問到是否去過A，B，C三個城市時，甲說：我去過的城市比乙多，但沒去過B城市．乙說：我沒去過C城市．丙說：我們?nèi)巳ミ^同一城市．
由此可判斷乙去過的城市為________．
14．A [解析] 由甲沒去過B城市，乙沒去過C城市，而三人去過同一城市，可知三人去過城市A，又由甲最多去過兩個城市，且去過的城市比乙多，故乙只去過A城市．
x14．[2014·陜西卷] 已知f(x)＝x≥0，若f1(x)＝f(x)，fn＋1(x)＝f(fn(x))，n∈N＋，則1＋x
f2014(x)的表達(dá)式為________．
xx14.[解析] 由題意，得f1(x)＝f(x)＝ 1＋2014x1＋x
x
1＋xxxf2(x)＝f3(x)＝，?，x1＋2x1＋3x11＋x
由此歸納推理可得f2014(x)＝x.1＋2014x
M2 直接證明與間接證明
21．、[2014·湖南卷] 已知函數(shù)f(x)＝xcos x－sin x＋1(x＞0)．
(1)求f(x)的單調(diào)區(qū)間；
111(2)記xi為f(x)的從小到大的第i(i∈N*)個零點，證明：對一切n∈N*，有x1x2xn
321．解：(1)f′(x)＝cos x－xsin x－cos x＝－xsin x.令f′(x)＝0，得x＝kπ(k∈N*)．
當(dāng)x∈(2kπ，(2k＋1)π)(k∈N)時，sin x>0，此時f′(x)<0;
當(dāng)x∈((2k＋1)π，(2k＋2)π)(k∈N)時，sin x<0，此時f′(x)>0.故f(x)的單調(diào)遞減區(qū)間為(2kπ，(2k＋1)π)(k∈N)，單調(diào)遞增區(qū)間為((2k＋1)π，(2k＋2)π)(k∈N)．
ππ(2)由(1)知，f(x)在區(qū)間(0，π)上單調(diào)遞減．又f?＝0，故x1＝.2?
2當(dāng)n∈N*時，因為
＋f(nπ)f[（n＋1）π]＝[(－1)nnπ＋1][(－1)n1(n＋1)π＋1]＜0，且函數(shù)f(x)的圖像是連續(xù)不斷的，所以f(x)在區(qū)間(nπ，(n＋1)π)內(nèi)至少存在一個零點．又f(x)在區(qū)間(nπ，(n＋1)π)上是單調(diào)的，故
nπ＜xn＋1＜(n＋1)π.142因此，當(dāng)n＝1時，＜； x1π3
1112當(dāng)n＝2時，＋(4＋1)＜ x1x2π3
當(dāng)n≥3時，111111?4＋1＋ 2（n－1）x1x2xnπ??
111?51＜＜（n－2）（n－1）??1×2ππ?5＋?1－1＋?11?＋?＋?11?? ??2?23??n－2n－1??
1162＝?6－n－1＜＜?π3π?
1112綜上所述，對一切n∈N*，.x1x2xn3
M3數(shù)學(xué)歸納法
sin x23．、[2014·江蘇卷] 已知函數(shù)f0(x)＝(x>0)，設(shè)fn(x)為fn－1(x)的導(dǎo)數(shù)，n∈N*.x
πππ(1)求2f1?＋f2?的值； ?2?2?2?
πππ2(2)證明：對任意的n∈N*，等式?nfn－1?＋n???＝ ?44?4??2?
sin xcos xsin x23．解：(1)由已知，得f1(x)＝f′0(x)＝?′＝－，?xxx
cos xx?sin ′＝ 于是f2(x)＝f1′(x)＝?′－?x?x－sin x2cos x2sin x＋，xxx
ππ4216所以f1??＝－f2?＝－?2??2πππ
πππ故2f1?2??＝－1.?22?2?(2)證明：由已知得，xf0(x)＝sin x，等式兩邊分別對x求導(dǎo)，得f0(x)＋xf0′(x)＝cos x，π即f0(x)＋xf1(x)＝cos x＝sin?x＋?.?2?
類似可得
2f1(x)＋xf2(x)＝－sin x＝sin(x＋π)，3π3f2(x)＋xf3(x)＝－cos x＝sin?x＋?，2??
4f3(x)＋xf4(x)＝sin x＝sin(x＋2π)．
nπ下面用數(shù)學(xué)歸納法證明等式nfn－1(x)＋xfn(x)＝sin?x＋?對所有的n∈N*都成立． 2??
(i)當(dāng)n＝1時，由上可知等式成立．
kπ(ii)假設(shè)當(dāng)n＝k時等式成立，即kfk－1(x)＋xfk(x)＝sin?x.2??
因為[kfk－1(x)＋xfk(x)]′＝kfk－1′(x)＋fk(x)＋xfk′(x)＝(k＋1)fk(x)＋xfk＋1(x)，?sin?x＋kπ??′＝cos?x＋kπ·?x＋kπ′＝sin?x＋（k＋1）π?，2??2?22?????
（k＋1）π?所以(k＋1)fk(x)＋xfk＋1(x)＝sin?x＋2??，因此當(dāng)n＝k＋1時，等式也成立．
nπ綜合(i)(ii)可知，等式nfn－1(x)＋xfn(x)＝sin?x＋對所有的n∈N*都成立． 2?
πππππnπ令x＝nfn－1?＋fn?＝sin?＋(n∈N*)，42??4?4?4??4
πππ所以?nfn－1?＋fn???＝?44?4???(n∈N*)．
M4單元綜合5．[2014·湖南長郡中學(xué)月考] 記Sk＝1k＋2k＋3k＋?＋nk，當(dāng)k＝1，2，3，?時，觀察
111111111111下列等式：S1＝n2＋n，S2＝n3＋2＋n，S34＋3＋2，S4＝n5n4＋3－n，2232642452330
115S56＋5＋n4＋An2，?由此可以推測A＝____________． 6212
11155．－ [解析] 根據(jù)所給等式可知，各等式右邊的各項系數(shù)之和為1，所以＋126212
1A＝1，解得A＝－12
6．[2014·日照一中月考] 二維空間中圓的一維測度(周長)l＝2πr，二維測度(面積)S＝π
4r2，觀察發(fā)現(xiàn)S′＝l；三維空間中球的二維測度(表面積)S＝4πr2，三維測度(體積)V＝πr3，3
觀察發(fā)現(xiàn)V′＝S.已知四維空間中“超球”的三維測度V＝8πr3，猜想其四維測度W＝________.6．2πr4 [解析] 因為W′＝8πr3，所以W＝2πr4.7．[2014·甘肅天水一中期末] 觀察下列等式：
(1＋1)＝2×1；
(2＋1)(2＋2)＝22×1×3；
(3＋1)(3＋2)(3＋3)＝23×1×3×5.照此規(guī)律，第n個等式為________________________________________________________________________．
7．(n＋1)(n＋2)(n＋3)?(n＋n)＝2n×1×3×5×?×(2n－1)
[解析] 觀察等式規(guī)律可知第n個等式為(n＋1)(n＋2)(n＋3)?(n＋n)＝2n×1×3×5×?×(2n－1)．
8．[2014·南昌調(diào)研] 已知整數(shù)對的序列為(1，1)，(1，2)，(2，1)，(1，3)，(2，2)，(3，1)，(1，4)，(2，3)，(3，2)，(4，1)，(1，5)，(2，4)，?，則第57個數(shù)對是________．
8．(2，10)[解析] 由題意，發(fā)現(xiàn)所給序數(shù)列有如下規(guī)律：
(1，1)的和為2，共1個；
(1，2)，(2，1)的和為3，共2個；
(1，3)，(2，2)，(3，1)的和為4，共3個；
(1，4)，(2，3)，(3，2)，(4，1)的和為5，共4個；
(1，5)，(2，4)，(3，3)，(4，2)，(5，1)的和為6，共5個．
由此可知，當(dāng)數(shù)對中兩個數(shù)字之和為n時，有n－1個數(shù)對．易知第57個數(shù)對中兩數(shù)之和為12，且是兩數(shù)之和為12的數(shù)對中的第2個數(shù)對，故為(2，10)．
9．[2014·福州模擬] 已知點A(x1，ax1)，B(x2，ax2)是函數(shù)y＝ax(a>1)的圖像上任意不同的兩點，依據(jù)圖像可知，線段AB總是位于A，B兩點之間函數(shù)圖像的上方，因此有結(jié)論ax1＋ax2x1＋x2>a成立．運用類比的思想方法可知，若點A(x1，sin x1)，B(x2，sin x2)是函數(shù)y22
＝sin x(x∈(0，π))的圖像上任意不同的兩點，則類似地有________________成立．
9.sin x1＋sin x2x1＋x2 sin x1＋sin x2x1＋x2總是位于A，B兩點之間函數(shù)圖像的下方，所以有 推理與證明 M1 合情推理與演繹推理 15．B13，J3，M1[2013·福建卷] 當(dāng)x∈R，|x|<1時，有如下表達(dá)式： 12n 1＋x＋x＋?＋x.1－x 11121n1 1兩邊同時積分得：∫01dx0xdx＋∫0xdx＋?＋∫0xdx0，222221－x從而得到如下等式： 1?n＋111?121?131?1×＋?＋?＋?＋??＋?＝ln 2.22?2?3?2?n＋1?2?請根據(jù)以上材料所蘊含的數(shù)學(xué)思想方法，計算： 1111212131n?1?n＋1CCn×＋n×＋?＋n×??＝__________． 22232n＋1?2? 0n 1??3?n＋1?n0122nn 15.[解析](1＋x)＝Cn＋Cnx＋Cnx＋?＋Cnx，??－1??n＋1??2?? 11121n1012nn 兩邊同時積分得Cn∫01dx＋Cn∫0xdx＋Cn∫xdx＋?＋Cn∫0xdx＝∫(1＋x)dx，***n1n＋113n＋10 得Cn×Cn×＋n×Cn＝－1.22232n＋12n＋1 214．M1[2013·湖北卷] 古希臘畢達(dá)哥拉斯學(xué)派的數(shù)學(xué)家研究過各種多邊形數(shù)，如三角形n（n＋1）121 數(shù)1，3，6，10，?，第n個三角形數(shù)為＝n，記第n個k邊形數(shù)為N(n，k)(k≥3)，222以下列出了部分k邊形數(shù)中第n個數(shù)的表達(dá)式： 121 三角形數(shù) N(n，3)＝＋n，22正方形數(shù) N(n，4)＝n，321 五邊形數(shù) N(n，5)＝－n，22 六邊形數(shù) N(n，6)＝2n－n，?? 可以推測N(n，k)的表達(dá)式，由此計算N(10，24)＝________.11k－22 14．1 000 [解析] 觀察得k每增加1，n項系數(shù)增加n項系數(shù)減少，N(n，k)＝ 222n2 n＋(4－k)N(10，24)＝1 000.2??0，0 ?ln x，x≥1.? ＋ 2-1- ①若a>0，b>0，則ln(a)＝blna； ＋＋＋ ②若a>0，b>0，則ln(ab)＝lna＋lnb； ＋?a＋＋ ③若a>0，b>0，則ln?≥lna－lnb； ?b? ＋b＋ ④若a>0，b>0，則ln(a＋b)≤lna＋lnb＋ln 2.其中的真命題有________．(寫出所有真命題的編號) b＋bb 16．①③④ [解析] ①中，當(dāng)a≥1時，∵b>0，∴a≥1，ln(a)＝ln a＝bln a＝bln＋b＋b＋ a；當(dāng)0 ＋＋＋ ②中，當(dāng)0 aa＋＋ ≤1，即a≤b時，左邊＝0，右邊＝lna－lnb≤0bba 時，左邊＝lnln a－ln b>0，若a>b>1時，右邊＝ln a－ln b，左邊≥右邊成立；若0 1ba 時，右邊＝0, 左邊≥右邊成立；若a>1>b>0，左邊＝ln＝ln a－ln b>ln a，右邊＝ln a，b左邊≥右邊成立，∴③正確； ④中，若0 ＋ ＋ ＋＋＋ (a＋b)＝0，右邊＝ln＋a＋ln＋b＋ln 2＝ln 2>0，左邊≤ a＋b，2 (a＋b)－ln 2＝ln(a＋b)－ln 2＝a＋ba＋ba＋ba＋b又∵≤a或≤b，a，b至少有1個大于1，∴l(xiāng)nln a或lnln b，即 2222有l(wèi)n ＋ (a＋b)－ln 2＝ln(a＋b)－ln 2＝ln a＋b＋＋ ≤lna＋lnb，∴④正確． 2 14．M1[2013·陜西卷] 觀察下列等式： 2 1＝1 22 1－2＝－3 222 1－2＋3＝6 2222 1－2＋3－4＝－10 ?? 照此規(guī)律，第n個等式可為________． 14．1－2＋3－4＋?＋(－1) n＋12 n＝(－1) n＋1 n（n＋1） [解析] 結(jié)合已知所給幾項的特2 點，可知式子左邊共n項，且正負(fù)交錯，奇數(shù)項為正，偶數(shù)項為負(fù)，右邊的絕對值為左邊底 2222n＋12n 數(shù)的和，系數(shù)和最后一項正負(fù)保持一致，故表達(dá)式為1－2＋3－4＋?＋(－1)n＝(－1) ＋1 n（n＋1） M2 直接證明與間接證明 20．M2，D2，D3，D5[2013·北京卷] 已知{an}是由非負(fù)整數(shù)組成的無窮數(shù)列，該數(shù)列前n項的最大值記為An，第n項之后各項an＋1，an＋2，?的最小值記為Bn，dn＝An－Bn.-2- (1)若{an}為2，1，4，3，2，1，4，3，?，是一個周期為4的數(shù)列(即對任意n∈N，an ＋4＝an)，寫出d1，d2，d3，d4的值； (2)設(shè)d是非負(fù)整數(shù)，證明：dn＝－d(n＝1，2，3，?)的充分必要條件為{an}是公差為d的等差數(shù)列； (3)證明：若a1＝2，dn＝1(n＝1，2，3，?)，則{an}的項只能是1或者2，且有無窮多項為1.20．解：(1)d1＝d2＝1，d3＝d4＝3.(2)(充分性)因為{an}是公差為d的等差數(shù)列，且d≥0，所以a1≤a2≤?≤an≤?.因此An＝an，Bn＝an＋1，dn＝an－an＋1＝－d(n＝1，2，3，?)． (必要性)因為dn＝－d≤0(n＝1，2，3，?)．所以An＝Bn＋dn≤Bn.又因為an≤An，an＋1≥Bn，所以an≤an＋1.于是，An＝an，Bn＝an＋1.因此an＋1－an＝Bn－An＝－dn＝d，即{an}是公差為d的等差數(shù)列． (3)因為a1＝2，d1＝1，所以A1＝a1＝2，B1＝A1－d1＝1.故對任意n≥1，an≥B1＝1.假設(shè){an}(n≥2)中存在大于2的項． 設(shè)m為滿足am>2的最小正整數(shù)，則m≥2，并且對任意1≤k 所以對于任意n≥1，有an≤2，即非負(fù)整數(shù)列{an}的各項只能為1或2.因為對任意n≥1，an≤2＝a1，所以An＝2.故Bn＝An－dn＝2－1＝1.因此對于任意正整數(shù)n，存在m滿足m>n，且am＝1，即數(shù)列{an}有無窮多項為1.M3 數(shù)學(xué)歸納法 M4 單元綜合1111 1．[2013·黃山質(zhì)檢] 已知n為正偶數(shù)，用數(shù)學(xué)歸納法證明1－＋－＋?＋234n＋1 1112(＋?＋)時，若已假設(shè)n＝k(k≥2為偶數(shù))時命題為真，則還需要用歸納假設(shè)n＋2n＋42n 再證n＝()時等式成立() A．k＋1B．k＋2 C．2k＋2D．2(k＋2) 1．B [解析] 根據(jù)數(shù)學(xué)歸納法的步驟可知，則n＝k(k≥2為偶數(shù))下一個偶數(shù)為k＋2，故答案為B.2．[2013·石景山期末] 在整數(shù)集Z中，被5除所得余數(shù)為k的所有整數(shù)組成一個“類”，記為[k]，即[k]＝{5n＋k|n∈Z}，k＝0，1，2，3，4.給出如下四個結(jié)論： * ①2 013∈[3]；②－2∈[2]；③Z＝[0]∪[1]∪[2]∪[3]∪[4]；④整數(shù)a，b屬于同一“類”的充要條件是a－b∈[0]． 其中，正確結(jié)論的個數(shù)為() A．1B．2C．3D． 42．C [解析] 因為2 013＝402×5＋3，所以2 013∈[3]，①正確．－2＝－1×5＋3，－2∈[3]，所以②不正確．因為整數(shù)集中的數(shù)被5除的余數(shù)可以且只可以分成五類，所以③正確．整數(shù)a，b屬于同一“類”，則整數(shù)a，b被5除的余數(shù)相同，從而a－b被5除的余數(shù)為0，反之也成立，故整數(shù)a，b屬于同一“類”的充要條件是a－b∈[0]，故④正確．所以正確的結(jié)論個數(shù)為3，選C.223344 3．[2013·汕頭期末] 已知2＋＝3＋3 4＋＝，33881515aa 6＋＝(a，t均為正實數(shù))，類比以上等式，可推測a，t的值，則a－t＝________． tt 3．－29 [解析] 類比等式可推測a＝6，t＝35，則a－t＝－29.x 4．[2013·福州期末] 已知點A(x1，ax1)，B(x2，ax2)是函數(shù)y＝a(a>1)的圖像上任意不同兩點，依據(jù)圖像可知，線段AB總是位于A、B兩點之間函數(shù)圖像的上方，因此有結(jié)論 x1＋x2 ax1＋ax2 >a2成立．運用類比思想方法可知，若點A(x1，sin x1)，B(x2，sin x2)是函數(shù)y2 ＝sin x(x∈(0，π))的圖像上的不同兩點，則類似地有________成立． sin x1＋sin x2x1＋x24.[解析] 函數(shù)y＝sin x在x∈(0，π)的圖像上任意不同兩 sin x1＋sin x2 點A，B，依據(jù)圖像可知，線段AB總是位于A，B兩點之間函數(shù)圖像的下方，所以 x1＋x2 [規(guī)律解讀] 類比推理中的結(jié)論要注意問題在變化之后的不同，要“求同存異”才能夠正確解決問題． 5．[2013·云南師大附中月考] 我們把平面內(nèi)與直線垂直的非零向量稱為直線的法向量，在平面直角坐標(biāo)系中，利用求動點軌跡方程的方法，可以求出過點A(－3，4)，且法向量為n＝(1，－2)的直線(點法式)方程為1×(x＋3)＋(－2)×(y－4)＝0，化簡得x－2y＋11＝0.類比以上方法，在空間直角坐標(biāo)系中，經(jīng)過點A(1，2，3)，且法向量為n＝(－1，－2，1)的平面(點法式)方程為________． 5．x＋2y－z－2＝0 [解析] 設(shè)B(x，y，z)為平面內(nèi)的任一點，類比得平面的方程為(－1)×(x－1)＋(－2)×(y－2)＋1×(z－3)＝0，即x＋2y－z－2＝0.* 6．[2013·黃山質(zhì)檢] 已知數(shù)列{an}滿足a1＝1，an＝logn(n＋1)(n≥2，n∈N)．定義： * 使乘積a1·a2·?·ak為正整數(shù)的k(k∈N)叫作“簡易數(shù)”．則在[1，2 012]內(nèi)所有“簡易數(shù)”的和為________． lg（n＋1） 6．2 036 [解析] ∵an＝logn(n＋1)＝，lg n lg 3lg 4lg（k＋1）lg（k＋1） ∴a1·a2·?·ak·＝＝log2(k＋1)，則“簡 lg 2lg 3lg klg 2 nn 易數(shù)”k使log2(k＋1)為整數(shù)，即滿足2＝k＋1，所以k＝2－1，則在[1，2 012]內(nèi)所有“簡 2（1－2）1210 易數(shù)”的和為2－1＋2－1＋?＋2－1＝－10＝1 023×2－10＝2 036.1－2 若 數(shù)學(xué) M單元 推理與證明 M1 合情推理與演繹推理 16．，[2014·福建卷] 已知集合{a，b，c}＝{0，1，2}，且下列三個關(guān)系：①a≠2；②b＝2；③c≠0有且只有一個正確，則100a＋10b＋c等于________． 16．201 [解析](i)若①正確，則②③不正確，由③不正確得c＝0，由①正確得a＝1，所以b＝2，與②不正確矛盾，故①不正確． (ii)若②正確，則①③不正確，由①不正確得a＝2，與②正確矛盾，故②不正確．(iii)若③正確，則①②不正確，由①不正確得a＝2，由②不正確及③正確得b＝0，c＝1，故③正確． 則100a＋10b＋c＝100×2＋10×0＋1＝201.14．[2014·全國新課標(biāo)卷Ⅰ] 甲、乙、丙三位同學(xué)被問到是否去過A，B，C三個城市時，甲說：我去過的城市比乙多，但沒去過B城市．乙說：我沒去過C城市．丙說：我們?nèi)巳ミ^同一城市． 由此可判斷乙去過的城市為________． 14．A [解析] 由甲沒去過B城市，乙沒去過C城市，而三人去過同一城市，可知三人去過城市A，又由甲最多去過兩個城市，且去過的城市比乙多，故乙只去過A城市． x14．[2014·陜西卷] 已知f(x)＝x≥0，若f1(x)＝f(x)，fn＋1(x)＝f(fn(x))，n∈N＋，則1＋x f2014(x)的表達(dá)式為________． xx14.[解析] 由題意，得f1(x)＝f(x)＝ 1＋2014x1＋x x 1＋xxxf2(x)＝f3(x)＝，…，x1＋2x1＋3x11＋x 由此歸納推理可得f2014(x)＝x.1＋2014x M2 直接證明與間接證明 21．、[2014·湖南卷] 已知函數(shù)f(x)＝xcos x－sin x＋1(x＞0)． (1)求f(x)的單調(diào)區(qū)間； 111(2)記xi為f(x)的從小到大的第i(i∈N*)個零點，證明：對一切n∈N*，有x1x2xn 321．解：(1)f′(x)＝cos x－xsin x－cos x＝－xsin x.令f′(x)＝0，得x＝kπ(k∈N*)． 當(dāng)x∈(2kπ，(2k＋1)π)(k∈N)時，sin x>0，此時f′(x)<0; 當(dāng)x∈((2k＋1)π，(2k＋2)π)(k∈N)時，sin x<0，此時f′(x)>0.故f(x)的單調(diào)遞減區(qū)間為(2kπ，(2k＋1)π)(k∈N)，單調(diào)遞增區(qū)間為((2k＋1)π，(2k＋2)π)(k∈N)． ππ(2)由(1)知，f(x)在區(qū)間(0，π)上單調(diào)遞減．又f?＝0，故x1＝.2? 2當(dāng)n∈N*時，因為 ＋f(nπ)f[（n＋1）π]＝[(－1)nnπ＋1][(－1)n1(n＋1)π＋1]＜0，且函數(shù)f(x)的圖像是連續(xù)不斷的，所以f(x)在區(qū)間(nπ，(n＋1)π)內(nèi)至少存在一個零點．又f(x)在區(qū)間(nπ，(n＋1)π)上是單調(diào)的，故 nπ＜xn＋1＜(n＋1)π.142因此，當(dāng)n＝1時，＜； x1π3 1112當(dāng)n＝2時，＋(4＋1)＜ x1x2π3 當(dāng)n≥3時，111111?4＋1＋ 2（n－1）x1x2xnπ?? 111?51＜＜（n－2）（n－1）??1×2ππ?5＋?1－1＋?11?＋…＋?11?? ??2?23??n－2n－1?? 1162＝?6－n－1＜＜?π3π? 1112綜上所述，對一切n∈N*，.x1x2xn3 M3數(shù)學(xué)歸納法 sin x23．、[2014·江蘇卷] 已知函數(shù)f0(x)＝(x>0)，設(shè)fn(x)為fn－1(x)的導(dǎo)數(shù)，n∈N*.x πππ(1)求2f1?＋f2?的值； ?2?2?2? πππ2(2)證明：對任意的n∈N*，等式?nfn－1?＋n???＝ ?44?4??2? sin xcos xsin x23．解：(1)由已知，得f1(x)＝f′0(x)＝?′＝－，?xxx cos xx?sin ′＝ 于是f2(x)＝f1′(x)＝?′－?x?x－sin x2cos x2sin x＋，xxx ππ4216所以f1??＝－f2?＝－?2??2πππ πππ故2f1?2??＝－1.?22?2?(2)證明：由已知得，xf0(x)＝sin x，等式兩邊分別對x求導(dǎo)，得f0(x)＋xf0′(x)＝cos x，π即f0(x)＋xf1(x)＝cos x＝sin?x＋?.?2? 類似可得 2f1(x)＋xf2(x)＝－sin x＝sin(x＋π)，3π3f2(x)＋xf3(x)＝－cos x＝sin?x＋?，2?? 4f3(x)＋xf4(x)＝sin x＝sin(x＋2π)． nπ下面用數(shù)學(xué)歸納法證明等式nfn－1(x)＋xfn(x)＝sin?x＋?對所有的n∈N*都成立． 2?? (i)當(dāng)n＝1時，由上可知等式成立． kπ(ii)假設(shè)當(dāng)n＝k時等式成立，即kfk－1(x)＋xfk(x)＝sin?x.2?? 因為[kfk－1(x)＋xfk(x)]′＝kfk－1′(x)＋fk(x)＋xfk′(x)＝(k＋1)fk(x)＋xfk＋1(x)，?sin?x＋kπ??′＝cos?x＋kπ·?x＋kπ′＝sin?x＋（k＋1）π?，2??2?22????? （k＋1）π?所以(k＋1)fk(x)＋xfk＋1(x)＝sin?x＋2??，因此當(dāng)n＝k＋1時，等式也成立． nπ綜合(i)(ii)可知，等式nfn－1(x)＋xfn(x)＝sin?x＋對所有的n∈N*都成立． 2? πππππnπ令x＝nfn－1?＋fn?＝sin?＋(n∈N*)，42??4?4?4??4 πππ所以?nfn－1?＋fn???＝?44?4??? M4單元綜合(n∈N*)． M[2014·福建卷] 閱讀下面短文，根據(jù)以下提示：1)漢語提示，2)首字母提示，3)語境提示，在每個空格內(nèi)填入一個適當(dāng)?shù)挠⒄Z單詞，所填單詞要求意義準(zhǔn)確，拼寫正確。 Many of us were raised with the saying “Waste not, want not.” None of us, 76.，can completely avoid waste in our lives.Any kind of waste is thoughtless.Whether we waste our potential talents, our own time, our limited natural 77.________(資源)，our money, or other people's time, each of us can become more aware and careful.The smallest good habits can ma It's a good feeling to know in our hearts we are doing ourin a world that is in serious trouble.By focusing on 80.________(節(jié)省)oil, water, paper, food, and clothing, we are playing a part 81.________ cutting down on waste.We must keep reminding 82.________(自己)that it is easier to get into something in the history of our evolution.It's time for us to 85.________no to waste so that our grandchildren's children will be able to develop well.We can't solve all the problems of waste, but we can encourage mindfulness.Waste not! 76．however考查副詞。我們中的許多人在成長的過程中都知道了“不浪費，就不會匱乏”。“然而”，在生活中我們沒有一個人能夠完全避免浪費。 77．resources考查名詞。resource是可數(shù)名詞，需要使用復(fù)數(shù)形式。 78．difference考查固定短語。make a(big)difference是固定短語，意為“有(很大的)作用，有(很大的)影響”。 79．best考查固定短語。do one's best是固定短語，意為“竭盡全力，全力以赴”。 80．saving 考查固定短語和非謂語動詞。focus on意為“集中于，專注于”，其中on是介詞，后接v.-ing形式。 81．in 考查固定短語。play a part/role in?是固定短語，意為“在??中發(fā)揮作用、扮演角色”。 82.ourselves 考查反身代詞。我們必須一直提醒我們自己。此處“自己”指代的是前面的主語we，故用ourselves。 83．than 考查連詞。根據(jù)前面的easier可知，此處需用連詞than。 84．done 考查固定短語和非謂語動詞。A do damage to B是固定短語，意為“A對B造成傷害”。此處do 與damage是動賓關(guān)系，故使用過去分詞。 85．say考查固定短語和非謂語動詞。現(xiàn)在到了與浪費說再見的時候了。在句型“It's time for sb to do sth.”中，需要使用不定式，故填say。 (一)[2014·福建龍巖質(zhì)檢] As children, loving our parents is an important part of life.It is our parents 1.________ create us, raise us，make us who we are and keep a roof over our heads in all kinds of weather.Here are some ways to love our parents.Firstly, tell them we love them every day.A gentle “I love you” will 2.w________a cold heart.Parents brought us into this world.3.________ them，we might still wander at an unknown corner of an unknown world.Then, show 4.________(尊敬)to them and don't get angry easily because anger helps 5.________ us nor our parents.Instead, keep 6.c________and sometimes share our feelings with them.Besides，obey their requests, 7.________ will make our attitudes better.What's more, understand that parents should be 8.f________ when they make mistakes.We should also keep company with them as much as 9.p________.Learn from them by listening to their stories as parents are the 10.________(資源)of our growth and even our teachers in one way or another.1．________ 2.________ 3.________ 4.________ 5.________ 6.________ 7.________ 8.________ 9.________ 10.________ 【答案】 1．who/that 2.warm 3.Without 4.respect 5.neither 6．calm 7.which 8.forgiven/forgivable 9.possible 10.resources (二)[2014·福建泉州質(zhì)檢] A boy trembled in the cold winter, wrapping his arms around himself on a bus stop bench.He wasn't 1.________(穿著)warm clothes and the temperature was －10℃.What a heartbreaking scene!But the good 2.d________ of the ordinary people who witnessed the 11-year-old Johannes were both joyous 3.________ inspiring.A woman, sitting next to the boy, discovered he was 4.________ a school trip and was told to meet his teacher at the bus stop.She selflessly 5.c________ her own coat around his shoulders.Later, 6.________ woman at first gave him her scarf, then wrapped him in her large jacket.Throughout the day, more and 7.________ people offered Johannes their gloves and even the coats off their backs.8.________(事實上), it was a hidden camera experiment by Norwegian charity SOS Children's Village as part of their winter campaign to gather donations to send much-needed coats and blankets to 9.h________ Syrian children get through the winter.Synne Ronning, the information head of the organization, also noted that the child was a 10.v________ who was never in any danger during the filming.1．________ 2.________ 3.________ 4.________ 5.________ 6.________ 7.________ 8.________ 9.________ 10.________ 【答案】 1．wearing 2.deeds 3.and 4.on 5.covered 6.another 7.more 8.Actually 9.help 10.volunteer (三)[2014·福建莆田3月質(zhì)檢] How time flies!Three years have passed since I 1.________(進(jìn)入)the school.As a Senior Three student, it won't take long 2.________ I graduate.High school is regarded as the golden time in a person's life.Now, I have much to share 3.________ my schoolmates.Firstly，I'd like to show my appreciation to those 4.s________by me all the way，teachers，parents and 5.________(朋友)included.Without their help and advice，my life would be different.6．________，it's high time to say sorry to classmates whom I hurt or misunderstood.7.________(交際)and smiles act as bridges to friendship.More importantly，I've made up my mind to make every effort to study，for I believe hard work is the key to success.Just as the old saying 8.g________，“No pain, no gain.” Finally，I hope that every one of us can 9.a________ his/her dream in the near future.I'll attach great significance to our friendship formed at school, and I'd like to keep in contact with you after graduation.Meanwhile，I suggest all the younger fellows make 10.f________use of time，because time waits for no one.1．________ 2.________ 3.________ 4.________ 5.________ 6.________ 7.________ 8.________ 9.________ 10.________ 【答案】 1．entered 2.before 3.with 4.standing 5.friends 6．Secondly 7.Communication 8.goes 9.achieve 10.full (四)[2014·福建漳州3月質(zhì)檢] Tony Mott was just an ordinary artist.But then, 1.________ the age of 36, he had an idea which made him famous.It started when he wanted to earn some money for Christmas one year.His product was simple, a 2.s________ message—five words on a T-shirt.He took the T-shirts to a clothes shop and they sold 40 in a week.3.________(馬上), he decided to start his own business.He got the business plan right and 4.________ worked.In the last twelve months, he has sold 60，000 T-shirts worldwide.The phrases for the T-shirts come from the things he thinks 5.________ during the day and from conversations with friends at dinner.His 6.________(顧客)who include the rich and famous enjoy his imaginative phrases.Mott says，“I'm successful, but it hasn't 7.c________ my personal life.I still work at home on the same small desk 8.________ I produce all the designs.My friends, who I've 9.k________ for twenty years, are still my friends.In fact, they're as surprised about my 10.s________ as I am.” 1．________ 2.________ 3.________ 4.________ 5.________ 6.________ 7.________ 8.________ 9.________ 10.________ 【答案】 1．a(chǎn)t 2.short 3.Immediately/Instantly 4.it 5.of 6．customers 7.changed 8.where 9.known 10.success (五)[2014·福建廈門3月質(zhì)檢] The old saying starts, “Give a man a fish and you feed him for a day?” Those words were taken to heart by Robert Egger, who used to be in the restaurant business.He knew from 1.________(經(jīng)驗)how much perfectly good food was 2.t________ away each day.So an organization 3.________ into being to collect leftovers, unserved food from restaurants in the neighbourhood.Volunteers put together more than 3，000 4.________(均衡的)meals a day and distribute them 5.________community centres and homeless shelters.But giving 6.a________ food was only step one.“I wanted to do more，” he says.As the rest of that old saying 7.g________，“Teach a man to fish and you feed him for a lifetime.” That's 8.________the organization also runs a training programme for people who are prepared 9.________ careers in the food service industry.They learn cooking methods.Many graduates find jobs and express their 10.s________ thanks to Egger's training programme.“Whether it is food, money or life，” Egger says, “we can't afford to waste any of them.” 1．________ 2.________ 3.________ 4.________ 5.________ 6.________ 7.________ 8.________ 9.________ 10.________ 【答案】 1．experience 2.thrown 3.came 4.balanced 5.to 6．a(chǎn)way 7.goes 8.why 9.for 10.sincere (六)[2014·福建福州質(zhì)檢] We moved away from my granny when I was eight years old.I 1.________ her terribly.Two years later my mother and father 2.________(離婚).I felt as if my world was falling apart.My mum must have sensed my longing, so she took my little brother and I back to visit my granny once in a 3.w________． Granny didn't live in a fancy house 4.________ have expensive things.But it was the little things she gave me that had always mattered.I always remember she saved her pennies in a glass jar.I am sure granny could have used those pennies 5.h________，but she saved them to give us when we came to visit.I don't remember how much we collected on our visits.Those 6.________(記憶)，of when I was a child，still give me 7.w________ feelings on days that I need them.A granny's love stays 8.________ a grandchild, down through the years, even when that child becomes a grandma.I often wonder, after all those years, when I am lucky 9.________ to find a penny 10.l________ on the ground somewhere, if it could possibly be granny giving me pennies from heaven.1．________ 2.________ 3.________ 4.________ 5.________ 6.________ 7.________ 8.________ 9.________ 10.________ 【答案】 1．missed 2.divorced 3.while 4.or 5.herself 6．memories 7.warm/wonderful 8.with 9.enough 10.lying/left (七)[2014·福建龍巖一檢] When you feel sad, you may think that the feeling will last forever.1.H________，feelings of sadness don't usually last very long—a few moments or maybe a day or two.But sometimes sad feelings can go 2.________ for a long time, hurt deeply, and make 3.________hard for you to enjoy the good things about your life.This 4.k________ of sadness that lasts a lot longer is called depression.People of all 5.________(年齡)can become depressed，6.i________ kids.Depression brings down a person's spirits and energy.It can affect 7.________ people think about themselves and their situations.If you think you have depression or you just have sadness that simply will not go away, sharing it 8.________someone who cares can help.There is always somebody to talk to when you are sad or depressed.You feel better when someone 9.________(知道)what you are going through.Plus，the other person can help you think of ways to make the situation better.But don't spend all your time 10.t________ about what is wrong.Be sure to share the good things, too.1．________ 2.________ 3.________ 4.________ 5.________ 6.________ 7.________ 8.________ 9.________ 10.________ 【答案】 1．However 2.on 3.it 4.kind 5.ages 6.including 7．how 8.with 9.knows 10.talking (八)[2014·福建四校聯(lián)考] Dear future pen pal,My name is Tudor, which is a very Welsh name.Maybe you have never been to Wales, but perhaps you know where it is.It is a small country, but with a big heart, I think.My family live 1.________ a farm, where we mainly have sheep, but there are 2.a________ some cows, which I like very much.In this part of the country it is quite mountainous.They're not big, 3.h________ mountains, but pretty impressive.Well, I'm just fifteen now, and my future work will be here in these 4.________(田野).My brother is an engineer, so no farming for him, and my sister is now 5.________ and has a small child.So, here I am, an uncle 6.a________，and with three years' school still ahead.I'm very 7.________ in things about China.In fact I'm learning some Chinese from a Chinese student.He 8.________(鼓勵)me to find a pen pal in Beijing and that's 9.________ I'm writing this letter.I've put my 10.a________ on the top of the letter.Please write and tell me about yourself.Best wishes.Yours，Tudor 1．________ 2.________ 3.________ 4.________ 5.________ 6.________ 7.________ 8.________ 9.________ 10.________ 【答案】 1．on 2.also 3.high 4.fields 5.married 6.already 7．interested 8.encouraged 9.why 10.address (九)[2014·福建三校聯(lián)考] For 25 years Terry Cemm was a policeman, but for the last 17 years he has been 1.________(行走)up and down five miles of beach every day, looking for things that might be 2.u________ to someone.Terry's a beachcomber(海灘拾荒者)． Nearly everything in his cottage has come 3.________ the sea—chairs, tables, even tins of food.“I even found a box of beer just before Christmas.That was nice，” he remembers.He finds lots of bottles with messages in them, 4.m________ from children.They all get a 5.r________ if there's an address in the bottle.“Shoes? If you find one, you'll find the 6.________ the next week，” he says.But does he really 7.m________ a living? “Half a living，” he smiles.“Anyway I have my police pension.But I don't actually need money.My life is rich in 8.________(多樣性)．” Terry is happy.“You have to find a way to live a simple life.” “Some people say I'm mad，” says Terry.“9.________ there are many more who'd like to do 10.________ I do.Look at me.I've got everything I could possibly want.” 1．________ 2.________ 3.________ 4.________ 5.________ 6.________ 7.________ 8.________ 9.________ 10.________ 【答案】 1．walking 2.useful 3.from 4.mainly 5.reply 6．other 7.make 8.variety 9.But 10.what (十)[2014·福建福州模擬] A rich man was near death and was very upset.He had worked so hard for his money 1.________ he dreamt he could take it with him to heaven.So he 2.________(祈禱)his dream would come true.An angel appeared and said no.The man begged the angel to speak to God to see whether he might 3.________ the rules.The angel reappeared and said that God could permit him to take one suitcase.4.________(激動地)，the man gathered his suitcase and filled it with pure gold bars.Afterwards, he died and showed up in heaven to greet St.Peter.5.S________ the suitcase，St.Peter said，“Hold 6.________； you can't bring that here！”The man explained that he had God's 7.p________.St.Peter checked it out and said，“You are right.You are allowed 8.o________ suitcase，but I'm supposed to check its contents 9.________ letting it through.” Inspecting the things that the man found too 10.________(珍貴的)to leave behind, St.Peter exclaimed, “You brought pavement? As you can see, the street of heaven is made of gold！” 1．________ 2.________ 3.________ 4.________ 5.________ 6.________ 7.________ 8.________ 9.________ 10.________ 【答案】 1．that 2.prayed 3.break 4.Excitedly 5.Seeing/Shown 6．on 7.permission 8.one 9.before 10.precious 2012高考文科試題解析分類匯編：推理和證明 1.【2012高考全國文12】正方形ABCD的邊長為1，點E在邊AB上，點F在邊BC上，AE?BF? 13。動點P從E出發(fā)沿直線向F運動，每當(dāng)碰到正方形的邊時反彈，反彈時反 射角等于入射角，當(dāng)點P第一次碰到E時，P與正方形的邊碰撞的次數(shù)為 （A）8（B）6（C）4（D） 3【答案】B 【命題意圖】本試題主要考查了反射原理與三角形相似知識的運用。通過相似三角形，來確定反射后的點的落的位置，結(jié)合圖像分析反射的次數(shù)即可。 【解析】解：結(jié)合已知中的點E,F的位置，進(jìn)行作圖，推理可知，在反射的過程中，直線是平行的，那么利用平行關(guān)系，作圖，可以得到回到EA點時，需要碰撞8次即可。 ?2?n??...?sin2.【2012高考上海文18】若Sn?sin?sinn?N?），則在S1,S2,...,S100777 中，正數(shù)的個數(shù)是（） A、16B、72C、86D、100 【答案】C 【解析】依據(jù)正弦函數(shù)的周期性，可以找其中等于零或者小于零的項.【點評】本題主要考查正弦函數(shù)的圖象和性質(zhì)和間接法解題.解決此類問題需要找到規(guī)律，從題目出發(fā)可以看出來相鄰的14項的和為0，這就是規(guī)律，考查綜合分析問題和解決問題的能力.3.【2012高考江西文5】觀察下列事實|x|+|y|=1的不同整數(shù)解（x,y）的個數(shù)為4，|x|+|y|=2的不同整數(shù)解（x,y）的個數(shù)為8，|x|+|y|=3的不同整數(shù)解（x,y）的個數(shù)為12 ….則|x|+|y|=20的不同整數(shù)解（x，y）的個數(shù)為 A.76B.80C.86D.92 【答案】B 【解析】本題主要為數(shù)列的應(yīng)用題，觀察可得不同整數(shù)解的個數(shù)可以構(gòu)成一個首先為4，公差為4的等差數(shù)列，則所求為第20項，可計算得結(jié)果.4.【2012高考陜西文12】觀察下列不等式 1? 1?121 22??321 ?532，?5 31?1 22?132?142 …… 照此規(guī)律，第五個不等式為.．．． 【答案】1? ? ? ? ? ? 116 .【解析】觀察不等式的左邊發(fā)現(xiàn)，第n個不等式的左邊=1?1?1??? 2?n?1??1n?1 ?n?1?，右邊=5.【2012 k，所以第五個不等式為1? ? ? ?2 ? ? ? 116 ． 表示為 高考湖南文 k?1 16】對于 n?N，將n n?ak?2?ak?1?2???a1?2?a0?2,當(dāng)i?k時ai?1，當(dāng)0?i?k?1時ai為0 或1，定義bn如下：在n的上述表示中，當(dāng)a0,a1，a2，…，ak中等于1的個數(shù)為奇數(shù)時，bn=1；否則bn=0.（1）b2+b4+b6+b8=__； （2）記cm為數(shù)列{bn}中第m個為0的項與第m+1個為0的項之間的項數(shù)，則cm的最大值是___.【答案】（1）3；（2）2.010【解析】（1）觀察知1?a0?2,a0?1,b1?1；2?1?2?0?2,a1?1,a0?0,b2?1； 10210 一次類推3?1?2?1?2,b3?0；4?1?2?0?2?0?2,b4?1； 5?1?2?0?2?1?2,b5?0；6?1?2?1?2?0?2，b6?0，b7?1,b8?1，210210 b2+b4+b6+b8=３；（2）由（1）知cm的最大值為２.【點評】本題考查在新環(huán)境下的創(chuàng)新意識，考查運算能力，考查創(chuàng)造性解決問題的能力.需要在學(xué)習(xí)中培養(yǎng)自己動腦的習(xí)慣，才可順利解決此類問題.6.【2012高考湖北文17】傳說古希臘畢達(dá)哥拉斯學(xué)派的數(shù)學(xué)家經(jīng)常在沙灘上面畫點或用小石子表示數(shù)。他們研究過如圖所示的三角形數(shù)： 將三角形數(shù)1,3，6,10，…記為數(shù)列{an}，將可被5整除的三角形數(shù)按從小到大的順序組成一個新數(shù)列{bn}，可以推測： （Ⅰ）b2012是數(shù)列{an}中的第______項；（Ⅱ）b2k-1=______。（用k表示）【答案】（Ⅰ）5030；（Ⅱ） 5k?5k?1? n(n?1)2 【解析】由以上規(guī)律可知三角形數(shù)1,3,6,10,…,的一個通項公式為an?，寫出其若 干項有：1,3,6,10,15,21,28,36,45,55,66,78,91,105,110,發(fā)現(xiàn)其中能被5整除的為10,15,45,55,105,110,故b1?a4,b2?a5,b3?a9,b4?a10,b5?a14,b6?a15.從而由上述規(guī)律可猜想：b2k?a5k? b2k?1?a5k?1? (5k?1)(5k?1?1) ? 5k(5k?1) （k為正整數(shù)），5k(5k?1)，故b2012?a2?1006?a5?1006?a5030,即b2012是數(shù)列{an}中的第5030項.【點評】本題考查歸納推理，猜想的能力.歸納推理題型重在猜想，不一定要證明，但猜想需要有一定的經(jīng)驗與能力，不能憑空猜想.來年需注意類比推理以及創(chuàng)新性問題的考查.7.【2102高考北京文20】（本小題共13分）設(shè)A是如下形式的2行3列的數(shù)表，滿足性質(zhì)P：a，b，c，d，e，f∈[-1,1],且a+b+c+d+e+f=0.記ri（A）為A的第i行各數(shù)之和（i=1,2），Cj（A）為第j列各數(shù)之和（j=1，2，3）；記k（A）為|r1(A)|, |r2(A)|, |c1(A)|，|c2(A)|，|c3(A)|中的最小值。對如下數(shù)表A，求k（A）的值 設(shè)數(shù)表A形如 其中-1≤d≤0，求k（A）的最大值； （Ⅲ）對所有滿足性質(zhì)P的2行3列的數(shù)表A，求k(A)的最大值。 【考點定位】此題作為壓軸題難度較大，考查學(xué)生分析問題解決問題的能力，考查學(xué)生嚴(yán)謹(jǐn)?shù)倪壿嬎季S能力。 （1）因為r1(A)=1.2，r2(A)??1.2，c1(A)?1.1，c2(A)?0.7，c3(A)??1.8，所以 k(A)?0.7 （2）r1(A)?1?2d，r2(A)??1?2d，c1(A)?c2(A)?1?d，c3(A)??2?2d.因為?1?d?0，所以|r1(A)|=|r2(A)|?d?0，|c3(A)|?d?0.所以k(A)?1?d?1.當(dāng)d?0時，k(A)取得最大值1.（3 任意改變A的行次序或列次序，或把A中的每個數(shù)換成它的相反數(shù)，所得數(shù)表A*仍滿足性 * 質(zhì)P，并且k(A)?k(A)，因此，不妨設(shè)r1(A)?0,c1(A)?0,c2(A)?0，由k(A)的定義 知 3k，?1(A k(? A)(A? r(?)c A)?(A,k?，(A?) c 從) ?(A c而?) a (A (?)kb)?r1 ?(a?b?c?d?e?f)?(a?b?f)?a?b?f?3 因此k(A)?1，由（2）知，存在滿足性質(zhì)P的數(shù)表A，使k(A)?1，故k(A)的最大值為1。 8.【2102高考福建文20】20.（本小題滿分13分） 某同學(xué)在一次研究性學(xué)習(xí)中發(fā)現(xiàn)，以下五個式子的值都等于同一個常數(shù)。（1）sin213°+cos217°-sin13°cos17°（2）sin215°+cos215°-sin15°cos15°（3）sin218°+cos212°-sin18°cos12° （4）sin2（-18°）+cos248°-sin2（-18°）cos248°（5）sin2（-25°）+cos255°-sin2（-25°）cos255° Ⅰ 試從上述五個式子中選擇一個，求出這個常數(shù) Ⅱ 根據(jù)（Ⅰ）的計算結(jié)果，將該同學(xué)的發(fā)現(xiàn)推廣位三角恒等式，并證明你的結(jié)論。 考點：三角恒等變換。難度：中。 分析：本題考查的知識點恒等變換公式的轉(zhuǎn)換及其應(yīng)用。解答： （I）選擇（2）：sin15?cos15?sin15cos15?1? sin30? （II）三角恒等式為：sin??cos(30??)?sin?cos(30??)? sin??cos(30??)?sin?cos(30??) ?sin2???34sin?? 234 ??cos?? sin?)?sin?2 ?? sin?)第二篇：2013高考數(shù)學(xué)_(真題+模擬新題分類)_推理與證明_理
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