Revista Electronica MateInfo.RO ISSN 2065-6432 MAI 2010

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                                      PROBLEME PENTRU CONCURSURI (II)

                                                      Corneliu Manescu-Avram          

 1. Sa se arate ca pentru orice  numar rational x exista cel putin un numar rational y astfel incat sa fie adevarata egalitatea

                                       6(x⁵ + y⁵)  15(x⁴ + y⁴) + 10(x³ + y³)  1 = 0.

Solutie : Verificam ca perechea (x, 1  x) , x ∈ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=HGiodaa@388A@   ℚ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWefv3ySLgznfgDOjdaryqr1ngBPrginfgDObcv39gaiuaajugGbabaaaaaaaaapeGae8NgHefaaa@41B7@ , satisface egalitatea data. Intr-adevar, fie p =   = xy. Din x + y = 1, deducem x² + y² = (x + y)²  2xy = 1  2p, x³ + y³ = (x + y)³  3xy(x + y) =   = 1  3p, x⁴ + y⁴ = (x² + y²)²  2x²y² = (1  2p)²  2p² = 1  4p + 2p², x⁵ + y⁵ =                             = (x² + y²)(x³ + y³)  x²y²(x + y) = (1  2p)(1  3p)  p² = 1  5p + 5p². Rezulta identitatea

                        6(1  5p + 5p²)  15(1  4p + 2p²) + 10(1  3p)  1 = 0.

2. Sa se gaseasca bazele sistemelor de numeratie in care

a) numarul 671 se divide cu 176 ;

b) numarul 781 se divide cu 187.

Solutie : a) Fie x baza sistemului se numeratie cautat. Avem ca 6x² + 7x + 1 = (x + 1)(6x + 1) se divide cu x² + 7x + 6 = (x + 1)(x + 6), deci 6x + 1 se divide cu x + 6. Orice divizor comun al numerelor 6x + 1 si x + 6 este divizor si al numarului 6(x + 6)  (6x + 1) = 35 = 5  7. Evident insa x > 1, deci singura solutie convenabila se obtine din x + 6 = 35, deci x = 29.

b) Asemanator, 7x² + 8x + 1 = (x + 1)(7x + 1) se divide cu x² + 8x + 7 = (x + 1)(x + 7), deci       7x + 1 se divide cu x + 7. Dar 7(x + 7)  (7x + 1) = 48 = 2⁴  3 si multimea divizorilor lui 48 este D₄₈ = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}. Rezulta x + 7 ∈ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=HGiodaa@388A@  D₄₈, cu solutiile convenabile                     x ∈ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=HGiodaa@388A@  {9, 17, 41}.

3. Sa se demonstreze ca orice numar compus n, care nu este patratul unui numar prim, are cel putin un divizor prim p  

Solutie : Daca numarul n este compus si nu este patratul unui numar prim, atunci el are cel putin doi divizori primi distincti p, q, p < q si p(p + 2)  pq  n, deci p² + 2p + 1 = (p + 1)²  n + 1, de unde p + 1    si se obtine inegalitatea din enunt.

4. Fie S_k = 1^k + 2^k + ... + n^k, n, k ∈ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=HGiodaa@388A@   ℕ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWefv3ySLgznfgDOjdaryqr1ngBPrginfgDObcv39gaiuaajugGbabaaaaaaaaapeGae8xfH4eaaa@41AF@ ^*. Sa se gaseasca toate valorile lui k pentru care este adevarata urmatoarea identitate : 2S_{2k + 1} + (k  1)S_{2k – MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaacbaqcL1paqaaaaaaaaaWdbiaa=nbiaaa@383C@  1} = (k + 1) , ∀ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=bGiIaaa@37D6@ n ∈ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=HGiodaa@388A@   ℕ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWefv3ySLgznfgDOjdaryqr1ngBPrginfgDObcv39gaiuaajugGbabaaaaaaaaapeGae8xfH4eaaa@41AF@ ^*.

Solutie : Pentru n = 1 obtinem 2 + k  1 = k + 1.

   Pentru n = 2 obtinem

                               2(1 + 2^{2k + 1}) + (k  1)(1 + 2^{2k - 1}) = (k + 1)(1 + 2^k)²,

care este echivalenta cu 2^{k – MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaacbaqcL1paqaaaaaaaaaWdbiaa=nbiaaa@383C@  2}(5  k) = k + 1, deci k ∈ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=HGiodaa@388A@  {1, 2, 3} este o conditie necesara.

   Aceasta conditie este si suficienta :

                                 S₃ =  , 2S₅ + S₃ = 3 , S₇ + S₅ = 2 , ∀ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=bGiIaaa@37D6@ n ∈ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=HGiodaa@388A@   ℕ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWefv3ySLgznfgDOjdaryqr1ngBPrginfgDObcv39gaiuaajugGbabaaaaaaaaapeGae8xfH4eaaa@41AF@ ^*.

Egalitatile de mai sus pot fi demonstrate prin inductie sau se pot folosi identitatile :

   S₁ = , S₂ = ,  S₃ = , S₅ =  ,

                                 S₇ =   .

5. Sa se arate ca polinomul f ∈ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=HGiodaa@388A@   ℝ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaqeeuuDJXwAKbsr4rNCHbacfaqcLbyaqaaaaaaaaaWdbiab=1riHcaa@3D03@ [X],  f = Xⁿ + a₁X^{n – MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaacbaqcL1paqaaaaaaaaaWdbiaa=nbiaaa@383C@  1} + ...+ a_{n – MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaacbaqcL1paqaaaaaaaaaWdbiaa=nbiaaa@383C@  3}X³ + a_n, unde n  4 si a_n ≠ 0, nu are numai radacini reale.

Solutie : Fie x₁, x₂, ..., x_n radacinile lui f, care sunt nenule, deoarece a_n ≠ 0. Polinomul g (X) =      = Xⁿf  are radacinile , , ... ,  . Dar g (X) = a_nXⁿ + a_{n – MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaacbaqcL1paqaaaaaaaaaWdbiaa=nbiaaa@383C@  3}X^{n – MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaacbaqcL1paqaaaaaaaaaWdbiaa=nbiaaa@383C@  3} + ... + 1, deci

           + ... +  = 2  = 0.

   Suma patratelor unor numere reale nenule este un numar real strict pozitiv, deci radacinile lui f nu sunt toate reale.

6. Sa se gaseasca radacinile reale ale polinomului

           f = (X²  X + 1)⁴ + (X²  X + 2)⁴  (X² + X + 2)⁴  (  X²  X  1)⁴.

Solutie : Urmatoarele egalitati sunt adevarate

         X²  X + 1    X² + X + 1 =   X² + 2,

         X²  X + 1 +    X² X  1 =    X²  2X,

         (X²  X + 1)² + (  X²  X  1)² =  X⁴  3X³+ 3X² + 2,

         X²  X + 2  X²  X  2 =  2X,

         X²  X + 2 + X² + X + 2 = 2(X² + 2),

         (X²  X + 2)² + (X² + X + 2)² = 2(X² + 1)(X² + 4).

   Rezulta descompunerea

   f = X(X² + 4)[  (3X  4)(  X⁴  3X³ + 3X² + 2)  8(X⁴ + 3X² + 2)] =                                                                 =   X(X² + 4)(X  12)[14X⁴ + (X²  2)² + 32X² +24]

   Deducem ca singurele radacini reale ale lui f sunt x₁ = 0 si x₂ = 12.

Comentariu. Din f (12) = 0 se obtine egalitatea 133⁴ + 134⁴ = 59⁴ + 158⁴.

7. Sa se rezolve in ℤ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWefv3ySLgznfgDOjdaryqr1ngBPrginfgDObcv39gaiuaajugGbabaaaaaaaaapeGae8hjHOfaaa@41CA@ ₁₀₃ ecuatia x⁴ + x² +   =  .

Solutie : Folosim descompunerea

                                        X⁴ + X² + 1 = (X² + X + 1)(X²  X + 1).

Fie x o radacina in ℤ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWefv3ySLgznfgDOjdaryqr1ngBPrginfgDObcv39gaiuaajugGbabaaaaaaaaapeGae8hjHOfaaa@41CA@ ₁₀₃ a polinomului X² + X +   ∈ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=HGiodaa@388A@   ℤ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWefv3ySLgznfgDOjdaryqr1ngBPrginfgDObcv39gaiuaajugGbabaaaaaaaaapeGae8hjHOfaaa@41CA@ ₁₀₃[X]. Celelalte solutii ale ecuatiei date sunt  + x,  x,    x. Acestea sunt toate solutiile, deoarece 103 este numar prim, deci ℤ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWefv3ySLgznfgDOjdaryqr1ngBPrginfgDObcv39gaiuaajugGbabaaaaaaaaapeGae8hjHOfaaa@41CA@ ₁₀₃ este corp, astfel ca ecuatia data, fiind de gradul 4, are cel mult 4 solutii. Dar 103 = 10² + 3 si

x² + x +  = , x ∈ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=HGiodaa@388A@   ℤ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWefv3ySLgznfgDOjdaryqr1ngBPrginfgDObcv39gaiuaajugGbabaaaaaaaaapeGae8hjHOfaaa@41CA@ ₁₀₃, deci ()² +  = , de unde , asadar x =  (am folosit faptul ca inversul lui  in ℤ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWefv3ySLgznfgDOjdaryqr1ngBPrginfgDObcv39gaiuaajugGbabaaaaaaaaapeGae8hjHOfaaa@41CA@ _p este   ). Solutiile ecuatiei date sunt {}.

Comentariu.  Metoda se aplica modulo orice numar prim p astfel incat p  3 este patrat perfect. Daca p = n² + 3 (n par), atunci solutiile in ℤ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWefv3ySLgznfgDOjdaryqr1ngBPrginfgDObcv39gaiuaajugGbabaaaaaaaaapeGae8hjHOfaaa@41CA@ _p ale ecuatiei date sunt

                      .

 

8. Fie P = X(X + 1)(X + 2) ... (X + 15) si S_k = X^k + (X + 1)^k + (X + 2)^k + ... + (X + 15)^k, k ∈ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=HGiodaa@388A@   ℕ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWefv3ySLgznfgDOjdaryqr1ngBPrginfgDObcv39gaiuaajugGbabaaaaaaaaapeGae8xfH4eaaa@41AF@ ^*. Sa se arate ca P + (12S₄  1700S₂ + 416704)² este patratul unui polinom cu coeficienti intregi.

Solutie : Avem

         S₂ = 16X² + 240X + 1240,

         S₄ = 16X⁴ + 480X³ + 7440X² + 57600X + 178312.

   Consideram polinoamele

           u = (X + 1)(X + 2)(X + 4)(X + 7)(X + 8)(X + 11)(X + 13)(X + 14),

           v = X(X + 3)(X + 5)(X + 6)(X + 9)(X + 10)(X + 12)(X + 15).

Este adevarata egalitatea P = uv =    , deci este suficient sa aratam ca

                                                 = 12S₄  1700s₂ + 416704.                                                 (1)

   Cu notatia Y = X² + 15X, grupand factorii egal departati de extreme, se obtine

   u = (Y + 14)(Y + 26)(Y + 44)(Y + 56) = Y⁴ + 140Y³ + 6828Y² + 134960Y + 896896,

   v = Y(Y + 36)(Y + 50)(Y + 54) = Y⁴ + 140Y³ + 6444Y² + 97200Y,

de unde

     = 192Y² + 18880Y + 448448 = 192X²(X + 15)² + 18880X(X + 15) + 448448 =

              = 192X⁴ + 5760X³ + 62080X² + 283200X + 448448.                                                    (2)

Avem si

   12S₄  1700S₂ + 416704 = 12(16X⁴ + 480X³ + 7440X² + 57600X + 178312)                         1700(16X² + 240X + 1240) + 416704 = 192X⁴ + 5760X³ + 62080X² + 283200X + 448448. (3)

   Din (2) si (3) rezulta (1). Se observa ca

    = Y⁴ + 140Y³ + 6636Y² + 116080Y + 448448 =

            = (X² + 15X)⁴ + 140(X² + 15X)³ + 6636(X² + 15X)² + 116080(X² + 15X) + 448448

are coeficienti intregi, ceea ce incheie demonstratia.

9. Care este valoarea maxima a functiei f : ℝ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWefv3ySLgznfgDOjdaryqr1ngBPrginfgDObcv39gaiuaajugGbabaaaaaaaaapeGae8xhHifaaa@41BB@  →  ℝ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWefv3ySLgznfgDOjdaryqr1ngBPrginfgDObcv39gaiuaajugGbabaaaaaaaaapeGae8xhHifaaa@41BB@ , f(x) = ***TRANSLATION ERROR***sin 2x  sin x MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=zo+9caa@3A2D@ ?

Solutie : Functia este periodica cu perioada 2, deci este suficient s-o studiem pe intervalul      [0, 2]. Pe de alta parte, f (2) = f (x), deci ne putem restrange la intervalul [0, ].  Pe acest interval sin x  0, deci f (x) = sin x MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=zo+9caa@3A2D@ 2cos x  1 MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=zo+9caa@3A2D@ , asadar

                             f (x) =

de unde

                              f  ^‘(x) =

   Derivata se anuleaza pentru cos x = .

   Pe intervalul , avem cos x   ,  deci cos x₀ =  , unde x₀ este un punct de maxim.   

 Valoarea maxima corespunzatoare este     .

  Pe intervalul , avem cos x   , deci cos x₁ =  si x₁ este de asemenea un punct de maxim. Valoarea maxima corespunzatoare este    si ea este mai mare decat cea precedenta, deci aceasta este valoarea maxima absoluta.

10. Sa se aproximeze cu o eroare de 10 ^{– MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaacbaqcL1paqaaaaaaaaaWdbiaa=nbiaaa@383C@  3} numarul , n ∈ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=HGiodaa@388A@   ℕ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWefv3ySLgznfgDOjdaryqr1ngBPrginfgDObcv39gaiuaajugGbabaaaaaaaaapeGae8xfH4eaaa@41AF@ , n  94.

Solutie : Vom demonstra inegalitatile

                                <    n <    , n ∈ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=HGiodaa@388A@   ℕ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWefv3ySLgznfgDOjdaryqr1ngBPrginfgDObcv39gaiuaajugGbabaaaaaaaaapeGae8xfH4eaaa@41AF@ , n  94.

Adunam n si ridicam la puterea a patra. A doua inegalitate este evidenta, dupa reducerea termenilor asemenea.

   Pentru a demonstra prima inegalitate, aratam ca functia f : [4,  →  ℝ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWefv3ySLgznfgDOjdaryqr1ngBPrginfgDObcv39gaiuaajugGbabaaaaaaaaapeGae8xhHifaaa@41BB@  , f(x) =  este crescatoare si verificam prin calcul ca f(94) > 0,249.  Intr-adevar,

                                               f  ^‘(x) =     1

si vom arata ca ea este pozitiva pe intervalul [4, ). Dupa eliminarea radicalului si reducerea termenilor asemenea, obtinem inegalitatea 96x²  336x  175 > 0, care este adevarata pentru      x  4.

   Rezulta ca  = n + 0,249..., daca n  94.

11. Sa se arate ca pentru orice n ∈ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=HGiodaa@388A@   ℕ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWefv3ySLgznfgDOjdaryqr1ngBPrginfgDObcv39gaiuaajugGbabaaaaaaaaapeGae8xfH4eaaa@41AF@ , n  3, polinomul f = Xⁿ  nX + 1 are doua radacini pozitive x_n, y_n astfel incat 0 < x_n < 1 < y_n si sa se calculeze  si

Solutie : Din f (0) = 1 > 0, f (1) = 2  n < 0, tinand cont de faptul ca functia polinomiala asociata lui f este continua, rezulta ca exista x_n, y_n ∈ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=HGiodaa@388A@   ℝ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWefv3ySLgznfgDOjdaryqr1ngBPrginfgDObcv39gaiuaajugGbabaaaaaaaaapeGae8xhHifaaa@41BB@  astfel incat f (x_n) = f (y_n) = 0 si 0 < x_n < 1 < y_n.

   Avem

                         f  =    > 0,  f  =     <  0

 si f  ^‘(x) = n(x^{n – MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaacbaqcL1paqaaaaaaaaaWdbiaa=nbiaaa@383C@  1}  1) < 0 pe (0, 1),  deci f  este strict descrescatoare pe intervalul (0, 1), astfel ca x_n este unic,   < x_n <    ,  prin urmare  este un sir convergent si  = 1.

   Pe de alta parte,   f () = n  n + 1 < 0,  deoarece < n,  pentru orice n  3. Rezulta ca y_n >  si y_n este unic, deoarece f  ^‘ este strict pozitiva pe intervalul (1, ), astfel ca f este strict crescatoare pe acest interval. Avem si    =      =  =  = . Rezulta ca  =  .

   Pentru calculul limitelor de mai sus am folosit regula lui l^’Hospital.

12. Sa se determine a minim si b maxim astfel ca inegalitatea

                                           ln (1 + x)   

sa fie adevarata pentru orice x  0.

Solutie : Se studiaza functia f : (0, ) →  ℝ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWefv3ySLgznfgDOjdaryqr1ngBPrginfgDObcv39gaiuaajugGbabaaaaaaaaapeGae8xhHifaaa@41BB@ , f (x) =       si in mod evident a si b sunt respectiv marginea superioara si marginea inferioara a multimii valorilor functiei.

   Avem

                                       f  ^‘(x) =    +   < 0,  ∀ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=bGiIaaa@37D6@ x > 0.                                       (1)

Intr-adevar, ultima inegalitate este echivalenta cu

                                                   ln x   , ∀ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=bGiIaaa@37D6@ x  1,                                                   (2)

inegalitate care se obtine din cea precedenta prin transformari elementare, extragerea radacinii patrate si substitutia x + 1 → x². Functia g : [1,  →  ℝ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWefv3ySLgznfgDOjdaryqr1ngBPrginfgDObcv39gaiuaajugGbabaaaaaaaaapeGae8xhHifaaa@41BB@  , g (x) = ln x   ,  este derivabila pe intervalul [1,  si derivata ei este negativa, g ^‘(x) =     0, ∀ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=bGiIaaa@37D6@ x  1,

deci g este descrescatoare, g (x)  g (1) = 0, astfel ca (2) este adevarata, deci si (1) este adevarata.

   Functia f este descrescatoare pe intervalul (0, ),  deci

                                    b =    f (x)   = a.

Avem evident b = 0 si

                  a =       =  =  ,

unde cea de-a doua limita a fost calculata cu regula lui l^’Hospital.

   In concluzie,

                                                      ln(1 + x)  x, ∀ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=bGiIaaa@37D6@ x  0.

13. Sa se gaseasca limita sirului cu termenul general

                                                  a_n = n   .

Solutie : Functia f : (0, )   ℝ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaqeeuuDJXwAKbsr4rNCHbacfaqcLbyaqaaaaaaaaaWdbiab=1riHcaa@3D03@  definita prin f (x) =    are derivate de orice ordin. Daca exista  , atunci exista si , deoarece f= a_n. Aplicam de doua ori regula lui l^’Hospital :

   =   =  =   =  , asadar  =  .

Comentarii. a) Asemanator putem demonstra ca  =   , pentru orice a ∈ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=HGiodaa@388A@  (0, ), a ≠ 1.

b) Egalitatea din enunt poate fi scrisa sub forma e = , de unde rezulta a_n > 0 si  =  .

c) Mai general, daca f  este o functie reala de doua ori derivabila si f  ^‘(0) ≠ 0, atunci

                  =  .

14. Sa se rezolve in ℝ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaqeeuuDJXwAKbsr4rNCHbacfaqcLbyaqaaaaaaaaaWdbiab=1riHcaa@3D03@  ecuatia : sin  =  .

Solutie : Avem evident x  0, deci x  1 si   1, de unde x  5. Vom studia functia    f : [1, 5] →  ℝ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaqeeuuDJXwAKbsr4rNCHbacfaqcLbyaqaaaaaaaaaWdbiab=1riHcaa@3D03@  definita prin f (x) = sin    . Functia are derivate de orice ordin pe intervalul (1, 5) si se anuleaza pentru x ∈ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=HGiodaa@388A@ {2, 3, 4}. Vom arata ca functia f  nu mai are alte zerouri pe [1, 5]. Se verifica simplu ca 1 si 5 nu sunt zerouri pentru functia f,  deci ramane sa cautam zerourile pe (1, 5). Daca functia f  are cel putin patru zerouri pe intervalul (1, 5), atunci, conform teoremei lui Rolle,  f  ^‘ are cel putin trei zerouri, f  ^‘’ are cel putin doua zerouri si f ^‘’’ are cel putin un zero. Dar

                         (x) =  cos       < 0,

∀ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=bGiIaaa@37D6@ x ∈ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=HGiodaa@388A@  (1, 5), contradictie. Rezulta ca ecuatia data are numai solutiile x ∈ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=HGiodaa@388A@  {2, 3, 4}.

15. Sa se determine functiile polinomiale cu proprietatea ca

                      f (b)  f (a) = (b  a)f  ^‘, ∀ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=bGiIaaa@37D6@ a, b ∈ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=HGiodaa@388A@   ℝ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaqeeuuDJXwAKbsr4rNCHbacfaqcLbyaqaaaaaaaaaWdbiab=1riHcaa@3D03@ ₊ .

Solutie : Daca f  este o functie polinomiala de grad cel mult 3,  f  = a₀x³ + a₁x² + a₂x + a₃,          cu a₁ = 0, atunci f (b)  f (a) = (b  a)[a₀(a² + ab + b²) + a₂] si f  ^‘(x) = 3a₀x² + a₂ , deci

                                   = a₀(a² + ab + b²) + a₂ ,

astfel ca egalitatea din enunt este adevarata.

   Reciproc, daca egalitatea din enunt este adevarata pentru o functie polinomiala de grad n  3,

 f = a₀xⁿ + a₁x^{n – MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaacbaqcL1paqaaaaaaaaaWdbiaa=nbiaaa@383C@  1} + .. + a_n, a₀ ≠ 0, atunci

                   = a₀x^{n – MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaacbaqcL1paqaaaaaaaaaWdbiaa=nbiaaa@383C@  1} + a₁x^{n – MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaacbaqcL1paqaaaaaaaaaWdbiaa=nbiaaa@383C@  2} + .. + a_{n – MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaacbaqcL1paqaaaaaaaaaWdbiaa=nbiaaa@383C@  1} , x > 0,

                    = na₀+ (n  1)a₁+ ... + a_{n – MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaacbaqcL1paqaaaaaaaaaWdbiaa=nbiaaa@383C@  1} , x  0

de unde, prin identificarea coeficientilor, se obtine 3^{n – MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaacbaqcL1paqaaaaaaaaaWdbiaa=nbiaaa@383C@  1} = n², astfel ca n = 3. In acest caz avem si a₁ =   a₁ , deci a₁ = 0.

   Rezulta ca o functie polinomiala asociata unui polinom cu coeficienti reali are proprietatea respectiva daca si numai daca grad f  3 si (0) = 0.

16. Sa se demonstreze ca daca f : [0, 1] →  ℝ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaqeeuuDJXwAKbsr4rNCHbacfaqcLbyaqaaaaaaaaaWdbiab=1riHcaa@3D03@  este o functie de clasa C¹, atunci

                   .

Solutie : Se integreaza prin parti :

                     = xf(x) MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=zo+9caa@3A2D@   = f(1)   .

   Rezulta

              =     =  .

   Din inegalitatea lui Cauchy obtinem

        .

   Prima integrala din dreapta este egala cu  , deci se obtine inegalitatea din enunt.

Observatie. Exista functii pentru care are loc egalitatea, de exemplu f(x) = x²  x.

17. Sa se gaseasca polinoamele P ∈ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=HGiodaa@388A@   ℤ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWefv3ySLgznfgDOjdaryqr1ngBPrginfgDObcv39gaiuaajugGbabaaaaaaaaapeGae8hjHOfaaa@41CA@ [X] pentru care

                                                dx ∈ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=HGiodaa@388A@   ℚ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWefv3ySLgznfgDOjdaryqr1ngBPrginfgDObcv39gaiuaajugGbabaaaaaaaaapeGae8NgHefaaa@41B7@ .

Solutie : Fie P = a₀ + a₁X + a₂X² + ... + a_nXⁿ. Restul impartirii lui P la X² + 1 este un polinom de grad cel mult 1

                                      P = (X² + 1)Q + pX + q, Q ∈ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=HGiodaa@388A@   ℤ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWefv3ySLgznfgDOjdaryqr1ngBPrginfgDObcv39gaiuaajugGbabaaaaaaaaapeGae8hjHOfaaa@41CA@ [X],  p, q ∈ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=HGiodaa@388A@   ℤ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWefv3ySLgznfgDOjdaryqr1ngBPrginfgDObcv39gaiuaajugGbabaaaaaaaaapeGae8hjHOfaaa@41CA@ .

Avem P(i) + P(  i) = 2q si P(X) + P(  X) = (X² + 1)[Q(X) + Q(  X)] + 2q.

   Integrala definita pe intervalul [0, 1] a unei functii polinomiale cu coeficienti intregi este un numar rational. Avem si

                                            dx = arctg x  = 

care este un numar transcendent, deci irational. Rezulta ca integrala data este un numar rational daca si numai daca q = 0, prin urmare

          P(i) + P(  i) = 2(a₀  a₂ + a₄  a₆ + ...) = 0,

unde suma se extinde asupra tuturor coeficientilor de rang par ai lui f.

   Am obtinut astfel urmatorul rezultat : integrala data este un numar rational daca si numai daca suma coeficientilor cu indicele multiplu de 4 ai polinomului este egala cu suma coeficientilor cu indicele multiplu de 4 plus 2.

18. Se da functia f : ℝ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWefv3ySLgznfgDOjdaryqr1ngBPrginfgDObcv39gaiuaajugGbabaaaaaaaaapeGae8xhHifaaa@41BB@  →  ℝ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWefv3ySLgznfgDOjdaryqr1ngBPrginfgDObcv39gaiuaajugGbabaaaaaaaaapeGae8xhHifaaa@41BB@  definita prin f (x) = xⁿ + ax, unde n ∈ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=HGiodaa@388A@   ℕ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWefv3ySLgznfgDOjdaryqr1ngBPrginfgDObcv39gaiuaajugGbabaaaaaaaaapeGae8xfH4eaaa@41AF@ , n  2 si a ∈ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaccaqcLbyaqaaaaaaaaaWdbiab=HGiodaa@388A@   ℝ MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWefv3ySLgznfgDOjdaryqr1ngBPrginfgDObcv39gaiuaajugGbabaaaaaaaaapeGae8xhHifaaa@41BB@ . Sa se arate ca egalitatea

                               =    

este adevarata daca si numai daca n = 2 si a =  1.

Solutie : Avem

                               =   =   +    ,

             =   =   + 2a + a².

   Se obtine

                               = 

care este o egalitate de numere pozitive, deci este echivalenta cu

                                        (a + 1)² +    = 0.

   Aceasta suma de numere pozitive este nula daca si numai daca fiecare termen este nul. Cum    n  2, se obtine n = 2 si a =  1.

 

CATEDRA DE MATEMATICA, GRUPUL SCOLAR DE TRANSPORTURI – MathType@MTEF@5@5@+=feaagyart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLnhiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq=Jc9vqaqpepm0xbba9pwe9Q8fs0=yqaqpepae9pg0FirpepeKkFr0xfr=xfr=xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaacbaqcLbuaqaaaaaaaaaWdbiaa=nbiaaa@3780@  PLOIESTI

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