Page 102 - MATINF Nr. 3

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102 PROBLEME DE MATEMATICA PENTRU CONCURSURI

2
2
2
2
2
2
Solut ,ie. Avem a 4 + a 5 + a 6 = a 1 + a 2 + a 3 − 3 s , i a + a + a = 9 − (a + a + a ). Cum
6
1
5
2
3
4
2
2
2
2
2
2
2
2
(a 4 + a 5 + a 6 ) ≤ 3 (a + a + a ), obt , inem (a 1 + a 2 + a 3 − 3) ≤ 27 − 3 (a + a + a ), adic˘a
4 5 6 1 2 3
2
2
2
2
2
2
2
2
3 (a + a + a ) + (a 1 + a 2 + a 3 − 3) ≤ 27. Dar (a 1 + a 2 + a 3 ) ≤ 3 (a + a + a ), deci
3
2
1
3
1
2
2
2
2
(a 1 + a 2 + a 3 ) + (a 1 + a 2 + a 3 − 3) ≤ 27. Notˆand a 1 + a 2 + a 3 = 3s obt , inem c˘a s − s − 1 ≤ 0,
√
5 + 1
deci s ≤ = ϕ. Utilizˆand Inegalitatea mediilor, din a 1 + a 2 + a 3 = 3s rezult˘a c˘a
2
3 1
3
3
a 1 a 2 a 3 ≤ s ≤ ϕ , iar din a 4 +a 5 +a 6 = a 1 +a 2 +a 3 −3 ≤ 3(ϕ−1) = rezult˘a c˘a a 4 a 5 a 6 ≤ .
ϕ ϕ 3
1
3
Deci a 1 a 2 a 3 a 4 a 5 a 6 ≤ ϕ · = 1. Din demonstrat , ia de mai sus deducem c˘a a 1 a 2 a 3 a 4 a 5 a 6 = 1
ϕ 3
1
dac˘a s , i numai dac˘a a 1 = a 2 = a 3 = ϕ s , i a 4 = a 5 = a 6 = .
ϕ
M 24. Rezolvat ,i ˆın mult ,imea numerelor reale ecuat ,iile:
2
2
a) (4 cos x − 3)(4 cos 3x − 3) = 2 sin x;
3
3
3
3
cos x + cos 5x + cos 9x + cos 13x 3
b) = .
cos x + cos 5x + cos 9x + cos 13x 4
Marin Chirciu, Pites , ti
Solut ,ie. a) x = π + kπ s , i x = π + kπ, k ∈ Z, nu sunt solut , ii ale ecuat , iei date. Utilizˆand
2 6
cos 3x cos 3x cos 9x
2
formula 4 cos x − 3 = , pentru cos x 6= 0, ecuat , ia devine · = 2 sin x, cu
cos x cos x cos 3x
π
π
x 6= +kπ s , i x 6= +kπ, k ∈ Z. Aceasta devine succesiv: cos 9x = sin 2x; cos 9x = cos π − 2x ;
2 6 2
(4k + 1)π (4k − 1)π

9x = ± π − 2x + 2kπ, k ∈ Z, deci x = sau x = , k ∈ Z.
2 22 14
(4k + 1)π (4k + 1)π π (4k + 1)π π
x = este solut , ie ⇔ 6= + nπ s , i 6= + nπ, n ∈ Z ⇔
22 22 2 22 6
11 11n + 5 11(2m + 1) + 5
4k + 1 6= 11 + 22n s , i 4k + 1 6= + 22n (adev.) ⇔ k 6= , n ∈ Z ⇔ k 6= ,
3 2 2
m ∈ Z ⇔ k 6= 11m + 8, m ∈ Z.
(4k − 1)π
Analog, x = este solut , ie dac˘a s , i numai dac˘a k 6= 7m + 2, m ∈ Z.
14
b) Condit , ia de existent , ˘a: cos x+cos 5x+cos 9x+cos 13x 6= 0 ⇔ 2 cos 7x(cos 6x+cos 2x) 6= 0
(2k + 1)π (2k + 1)π (2k + 1)π
⇔ 4 cos 7x cos 4x cos 2x 6= 0 ⇔ x 6= , x 6= s , i x 6= , k ∈ Z.
4 8 14
3 cos x + cos 3x 3 cos 3x + cos 15x + cos 27x + cos 39x 3
3
Cum cos x = , ecuat , ia devine + = ,
4 4 4(cos x + cos 5x + cos 9x + cos 13x) 4
adic˘a cos 3x + cos 15x + cos 27x + cos 39x = 0. Notˆand 3x = y, conform rezolv˘arii condit , iei de
(2k + 1)π (2k + 1)π (2k + 1)π
existent , ˘a obt , inem c˘a x = , x = sau x = , k ∈ Z.
12 24 42
(2k + 1)π (2k + 1)π (2n + 1)π (2k + 1)π (2n + 1)π
x = este solut , ie ⇔ 6= , 6= s , i
12 12 4 12 8
(2k + 1)π (2n + 1)π
6= , n ∈ Z ⇔ 2k + 1 6= 3(2n + 1), 2(2k + 1) 6= 3(2n + 1) (adev.) s , i
12 14
7(2k + 1) 6= 6(2n + 1) (adev.) ⇔ k 6= 3n + 1, n ∈ Z.
(2k + 1)π (2k + 1)π
Analog, x = s , i x = sunt solut , ii dac˘a s , i numai dac˘a k 6= 3n + 1, n ∈ Z.
24 42