ࡱ> kmjQ7 6cbjbjUU 97|7|>Fl%%%8>%&t{~&*"***999bdddddd$ :9+7\999r**i 5rrr9,**br9brryrrhZb*& P#N%fځbK0{zSYoSbr VARIANTE EXAMEN SUBIECTE EXAMEN METODE DE REZOLVARE A PROBLEMELOR PUNCTE DE VEDERE PRIVIND UNELE METODE DE DEMONSTRATIE SI REZOLVARE A PROBLEMELR SIEXERCITIILOR PROPUSE LA EXAMENUL DE BACALAUREAT SI ADMITERE LA FACULTATE Este bine stiut ca orice situatie care pleca de la o ipoteza(date care se dau) si trebuie sa ajunga la o concluzie(date care se cer), parcurge un algoritm de rezolvare prin inlantuirea de propozitii bazat pe un rationament logic. Logica ne ajuta sa rezolvam o serie de probleme care nu se pot solutiona numai pe baza gandirii spontane, prin propozitii care exprima judecati legate intre ele. De exemplu, sa urmarim inlantuirea propozitiilor: ,,Daca tu te cateri pe Everest, eu sunt martian ,,Tu te cateri pe Everest Asa dar eu sunt martian. Exemplul precedent arata un rationament, care este o inlantuire de judecati(propozitii), in care plecand de la anumite cunostinte care se dau(numite premize) se ajunge la alte cunostinte care se cer(numite concluzie). Orice rationament este corect daca si numai daca concluzia deriva din premize si nu numaidecat din ipoteza. Un rationament corect nu trebuie confundat cu adevarul concluziei. In exemplul dat, rationamentul este corect dar concluzia este falsa, decurgand din premizele false date in ipoteza. Rationamentele corecte se construiesc in orice teorie in care este valabil principiul bivalentei , pe baza operatiilor logice, bazandu-se pe tautologii. RATIONAMENT PRIN MODUS PONENS La baza acestui rationament sta implicatia logica. Rationamentul era cunoscut din andtichitate, la Diogene avand forma: ,,Daca A este atunci este si B ,,or, prima este ,,deci si prima Rationamentul preceent poate fi prezentat schematic , astfel: Observati ca cu siguranta majoritatea teroremelor studiate sunt de aceasta forma. Este important de retinut ca din orice teorema, se poate formula in mod logic din ea noi propozitii, ca: propozitia reciproca(B ! A) si propozitia contrara (non A ! non B). Noile propozitii, reciproca si contrara, devin teoreme numai daca sunt demonstrate ca fiind adevarate. Demonstratia matematica este metoda specifica de justificare a teoremelor si consta in a arata ca daca ceea ce afirma ipoteza are loc, atunci concluzia rezulta din ea in mod logic. In orice demonstratie ne putem baza numai pe axiome sau/si teoreme demonstrate anterior. Nu este admis sa fie utilizate propozitii/ proprietati care inca nu au fost demonstrate, acestea din urma putandu-se baza la randul lor pe chiar pe teorema de demonstrat. Exemplul 1. Teorema: Orice functie derivabila intr-un punct este continua in acel punct. Consideram propozitiile: A:Orice functie derivabila intr-un punct; B:Este continua in acel punct. Teorema prezentata este un rationament de tipul modus ponens, demonstratia gasindu-se in orice manual de analiza matematica. Propozita reciproca: B ! A : Orice functie continua este derivabila este o propozitie falsa. Demonstram afirmatia printr-un contraexemplu:d")"U" Functia f: R ! R , f(x) = |x| , este continua in origine, dar nu este derivabila in acest punct.  Exemplul 2. Teorema:Orice poligon convex poate fi circumscris unui cerc, daca bisectoarele unghiului poligonului sunt concurente inacelsi punct. Consideram propozitiile: A: Bisectoarele unghiurilor unui poligon convex sunt concurente in acelasi punct; B: Poligonul convex se poate circumscrie unui cerc. Rationamentul modus ponens poate fi pus in evidenta sub forma: (A si A ! B) ! B, demonstratia bazandu-se pe properietatea punctelor ce apartin bisectoarei si definitia cercului. Propozitiile reciproce B ! A si nonA ! nonB sunt deasemeni adevarate. Exemplul 3. Daca I este un interval deschis, xo(I si f,g: I ! R f d"g, sunt functii derivabile in xo astfel incat f(xo) = g(xo), atunci f  (xo)d"g (xo). Teorema data este un rationament modus ponens, luand in consideratia propozita A de la daca pana atunci, iar propozitia B in rest. Demonstratia inferentei precedente este urmatoarea: Oricarea ar fi x(I, x > xo, are loc: f(x)  f(xo) g(x)  g(xo) x xo x  xo deci, prin aplicarea limitei pentru x! xo, x > xo, se obtine: f  (xo) = fd (xo) d" gd (xo) = g (xo) Propozitia reciproca B ! A: . Daca I este un interval deschis, xo(I si f,g: I ! R, cu f(xo) = g(xo), sunt functii derivabile in xo si f  (xo) d" g (xo), atunci f d" g este falsa. Justificarea printr-un contraexemplu. Fie 0(I si functia  x2, daca x(I )"Q 0, daca x(I \Q, si g(x) = x3. Functiile f,g sunt derivabile in xo = 0, f(0) = g(0) si f  (0) = g (0) si totusi f, g nu sunt in relatia f d" g in nici o vecinatate a punctului xo = 0. In concluzie, propozitia reciproca fiind falasa, se poate afirma ca teorema data nu are teorema reciproca. Exemplul 4. Teorema directa: O functie f:I ! R, I ( R este continua intr-un punct de acumulare xo(I, daca functia f are limita in xo egala cu valoarea imaginii f(xo). Prin alegerea propozitiilor: A: Functia f are limita in xo egala cu f(xo). B: O functie f:I ! R, I ( R continua intr-un punct de acumulare xo(I. Teorema este un rationament de tip modus ponens, (A si A ! B) ! B. Se pot formula propozitiile urmatoare: Propozitia reciproca:B ! A. daca o functie f:I ! R, I ( R este continua intr-un punct de acumulare xo(I, atunci functia f are limita in xo egala cu valoarea imaginii f(xo). Propozitia contrara directei: nonA ! nonB. Daca functia f nu are limita in xo egala cu valoarea imaginii f(xo), functia f:I ! R, I ( R nu este continua in punctul de acumulare xo(I. Propozitia contrara directei: nonB ! nonA. Daca o functie f:I ! R, I ( R nu este continua intr-un punct de acumulare xo(I, atunci functia f nu are limita in xo egala cu valoarea imaginii f(xo). Prin justificarea valorii de adevar adevarul, propozitia reciproca este adevarata devenind teorema reciproca si o data cu ea devin teoreme si propozitiile contrara directa si contrara reciproca pe baza tautologiei pe care se bazeaza rationamentul prin modus ponens:`"|  (A ! B) ! (nonB ! nonA) (B ! A) ! (nonA ! nonB) Demonstratia teoremei directe. Pornind de la premiza lim f(x) = f(xo),trebuie aratat functia f este continua in xo. aceasta inseamna ca, pentru orice  > 0, exista numar strict pozitiv  = (), astfel incat oricare ar fi xo(I, x `" xo, cu | x - xo | < , sa avem: | f(x) - f(xo), | < . Dar daca x = xo, atunci | x - xo | = 0 <  si | f(x) - f(xo), | = 0 < , astfel incat restrictia x `" xo este de prisos. Conform teoremei: Functia f:I ! R este continua in punctual xo(I, daca si numai daca pentru orice numar  > 0, exista un numar  = () > 0, astfel incat oricare ar fi x(I cu | x - xo | < , sa avem | f(x) - f(xo), | < . Rezulta ca functia f este continua in xo. Demonstratia propozitiei reciproce. Din presupunerea ca f este continua in xo, pe baza definitiei, urmeaza ca pentru orice sir xn ! xo, xn(I are loc f(xn) ! f(xo). In particular, pentru sirurile xn ! xo, (xn(I) cu xn `" xo, are loc deasemenea f(xn) ! f(xo). In concluzie f(xo) este limita functiei in xo si are loc: lim f(x) = f(xo). Observatie. Demonstratia teoremelor poate urma calea analitica sau sintetica: a). In cazul analitic se porneste de la ceea ce se cere spre ceea ce este dat. Intrebarile care se pun sunt de natura: ,, ce trebuie sa stim pentru a arata ca b). in cazul sintetic se porneste de la ceea ce se da in ipoteza sau este cunoscut a fi adevarat, spre ceea ce se cere, intrebarile fiind formulate de forma: ,, ce se poate determina stiind ca. Exemplul 5. Fiind date numerele positive x1, x2, x3,, xn, se definesc:  Media aritmetica Media geometrica Media armonica Sa se arate ca are loc dubla inegalitate: Hn < Gn < An. Demonstratie. Demonstram inegalitatea Gn d" An dupa o idee a lui Liouville, prin combinarea inductiei complete cu studiul functiilor. Etapa de verificare. Pentru n=2, inegalitatea este evident adevarata. In adevar, justificarea se poate face dupa metoda demonstratieibsintetice. Pornind de la inegalitatea, evident adevarata, (x1  x2)2 e" 0, pentru orice x1 si x2 numere pozitive si inlocuind-o succesiv cu inegalitati echivalente: (x1  x2)2 e" 0, x12  2x1x2 + x22 e" 0, x12  2x1x2 + x22 + 4x1x2 e" 4x1x2, x12 + 2 x1x2 + x22 e" 4x1x2, (x1 + x2)2 e" 4x1x2, x1  x2 e" 2 EMBED Equation.3 , de unde G1 d" A2. Etapa de demonstratie. Presupunem ca pentru n inegalitatea Gn d" An, are loc.( egalitate pentru x1 = x2 = x3 = & = xn). Pentru a demonstra adevrul inegalitatii pentru n+1 numere pozitive, sa consideram functia: Va fi suficient ca aceasta functie de xn+1 este totdeauna pozitiva sau nula (in caz de egalitate). Derivata acestei functii  este o functie cerescatroare de xn+1 care se anuleaza pentru Daca x1 = x2 = x3 = = xn, atunci xn+1 are valoarea egala cu x1. Functia f descreste pana la aceasta valoare si apoi creste, deci ea are un minim egal cu valoare care, pe baza ipotezei este pozitiva sau nula. Avem deci f e" 0. Pentru egalitate, minmul trebuie sa fie nul, ceea ce se intampla cand x1 = x2 = =x3 = & = xn, dar atunci xn+1 trebuie sa fie egal cu x1. q.e.d. Pentru a dovedi inegalitatea Hn d" Gn, se considera numerele pozitive z1-1 = z2-1 = z3-1 = & = xn-1, pentru care inegalitatea precedenta se scrie:  EMBED Equation.3  egalitatea avand loc cand z1 = z2 = =zn. Rezulta de aici ca:  EMBED Equation.3  Observatii importante 1). Inegalitatile sunt relatii de ordine tranzitive, de regula definite p R, sau submultimi ale sale. De remaecat ca in multimea numerelor complexe C nu sunt definite relatii de ordine si ca atare inegaliyatile nu au sens. 2). Pentru a demonstra9justifica0 o inegalitate se poate folosi atat metoda analitica cat si cea sintetica. Totusi este de preferat a fi fololosita demonstratia sintetica, pornind de la intrebarea ,,ce se poate determina stiid ca . 3). Principiul inductiei complete este o metoda folosita la demonstrarea inegalitatilor depinzand de n, numar natural. 4). La verificarea inegalitatilor se pot folosi numai proprietatile stabilite prin teoreme asupra acestor relatii: Se pot aduna(scadea) din ambii membri ai unei inegalitati, temeni(expresii) egali; Se pot inmulti(imparti) ambii membri ai unei inegalitati cu termeni(expresii) pozitivi. La impartire, termenii trebuie sa fie diferiti de zero(impartirea la 0 nu are sens). Impartirea inegalitatilor cu expresii care contin nedterminate sunt permise numai daca in prealabil s-a demonstrat ca pastreaza semn constant; Ridicarea la putere a unei inegalitati este permisa numai daca ambii membri sunt pozitivi. Exercitii propuse la examene bazate pe exemplul 5 Exercitiul 1. Sa se arate ca daca x, y, z sunt numere strict pozitive, atunci:  EMBED Equation.3  Solutie. Aplicand inegalitatea mediilor pentru cazul n+2, si pentru numerele de forma: EMBED Equation.3  vom obine succesiv:  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3  de hune prin adunare membru cu membru, obtinem:  EMBED Equation.3  Impartind prin 2 inegalitatea precedenta se obtine cerinta. Exercitiul 2. Sa se demonstreze ca oricare ar fi numerele reale EMBED Equation.3  are loc inegalitatea:  EMBED Equation.3  Solutie.Inegalitatea ceruta devine aproape evidenta, aplicand inegalitatea mediilor, astfel:  EMBED Equation.3  pentru numerele x, y, z, si  EMBED Equation.3  pentru numerele  EMBED Equation.3  Exercitiul 3. sa se arateca oricare ar fi numerele reale pozitive x, y, si z are loc inegalitatea:  EMBED Equation.3  Solutie. Se aplica inegalitatea mediilor pentru cazul n=3 si obtinem:  EMBED Equation.3  pentru numerele x, y si 1;  EMBED Equation.3  pentru numerele y, z si 1;  EMBED Equation.3  pentru numerele x, z si 1. Adunand relatiile membru cu membru, se obtine inegalitaea din enunt. Evident, egalitatea se obtine x = y. Exercitiul 4. Pentru numerele reale x,y si z pozitive asa incat x + y + z =1,sa se arate ca:  EMBED Equation.3  Solutie. Se transforma membrul stang, efectuand calculele si grupand convenabil: E= EMBED Equation.3   EMBED Equation.3  (vezi ipoteza x + y + z =1). Aplicand inegalitatea mediilor pentru numerele pozitive x, y si z cu conditia din ipoteza, vom obtine:  EMBED Equation.3  deci  EMBED Equation.3  Aplicand inegalitatea mediilor (aritmetica si armonica) pentru aceleasi numere, se obtine: EMBED Equation.3  deci  EMBED Equation.3  Inlocuind in E ultimele doua inegalitati vom obtine: E EMBED Equation.3  1 +2"27+9 = 64. Exercitiul 5. Fie a,b si c numere reale pozitive. Sa se arate ca:  EMBED Equation.3  Solutia. Adunam in ambii membri 3 si distribuim la fiecare fractie din membrul stang pe 1:  EMBED Equation.3  se obtine dupa efectuarea calculelor: EMBED Equation.3  Notand prin x = b + x, y = a + c si z = a + b, inegalitatea precedenta devine: EMBED Equation.3  (*) Aplicand inegalitatea mediilor(pentru mentru media aritmetica si geometrica) pentru numerele x, y si z si respectiv  EMBED Equation.3  si  EMBED Equation.3 , se obtine:  EMBED Equation.3 , ceea ce trebuia aratat. Observatie. Inegalitatea (*) se poate justifica si folosind inegalitatea Cauchy-Schwartz-Buniacovski:  EMBED Equation.3   EMBED Equation.3  Alte exercitii 6). a). Sa se arate ca pentru orice a =1 +b, ( b > 0, a,b ( R+) si numarul natural n e"2, are loc inegalitatea:  EMBED Equation.3 ; b).Sa se arate, folosind rezultatul de la a) ca pentru orice numar natural n e"2, are loc:  EMBED Equation.3 . Solutii. a). Din ipoteza ca b > 0 si a = 1 + b, prin ridicare la puterea n , se obtine:  EMBED Equation.3   EMBED Equation.3 . b). Inlocuind in inegalitatea de la a),  EMBED Equation.3 , se obtine:  EMBED Equation.3 . 7). Fie numerele reale a1> a2>>an, b1>b2>>bn. Sa se arate ca daca n e"2,atunci: a1b1 + a2b2 + & + anbn > a1bn + a2bn-1 + & + anb1. Solutie. Cazul 1. Pentru n = 2k, relatia precedenta devine: a1(b1  bn ) +& +ak(bk  bn-k+1)  ak+1(bk  bn-k+1)  &  an(b1  b2) > 0, de unde (a1  an ) (b1  bn) + & +(ak  ak+1) (bk  bn-k=1) > 0. Si tinand cont ca fiecare paranteaza este pozitiva (vezi conditiile din ipoteza), rezulta inegalitatea ceruta. Cazul 2. pentru n= 2k+1,inegalitatea devine: a1b1 + + ak-1bk-1 + akbk + ak+1bk+1 + + anbn > >a1bn + + ak-1bk-1 + akbk + ak+1bk+1 + + anb1. Inegalitatea s-a redus la cazul precedent, dupa reducerea termenilor akbk. 8). Daca a si b sunt numere reale nenule , demonstrati ca:  EMBED Equation.3 . Solutie. Notand cu  EMBED Equation.3  prin ridicare la patrat, rezulta  EMBED Equation.3 , de unde inegalitatea ceruta devine: 3(x2  2)  8x + 10 e" 0. Dupa efectuarea calculelor si rezolvarea inecuatieise 3x2  8x + 4 e" 0 se obtine  EMBED Equation.3  sau x e" 2. Dar  EMBED Equation.3 , deci daca ab > 0, atunci x e" 2 care conduce la (a  b)2e" 0, evidenta. Daca ab > 0, atunci  EMBED Equation.3 , deoarece EMBED Equation.3 . 9). Sa se arate ca: EMBED Equation.3 , unde a si b sunt numere reale. Solutie. Daca ab > 0, atunci |a + b| = |a| + |b| si succesiv  EMBED Equation.3  Daca ab < 0, atunci |a + b| = max{|a| , |b|} si fie de exemplu |a| > |b|, deci |a + b| < |a|, de hune rezulta:  EMBED Equation.3 . 10). Sa se demonstreze ca pentru numerele reale pozitive a,b si pentru orice n intreg , are loc:  EMBED Equation.3 . Solutie. Pentru n = 0 se obtine egalitate. Pentru n e" 1, avem pe baza inegalitatii mediilor:  EMBED Equation.3  Fie acum n =  k, cu k . 0. Avem:  EMBED Equation.3   EMBED Equation.3   EMBED Equation.3 , deoarece se poate demonstra prin recurenta sau cu ajutorul binomului ca:  EMBED Equation.3  2. 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