Page 109 - MATINF Nr. 3

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PROBLEME DE MATEMATICA PENTRU CONCURSURI 109

√ √ Ä √ ä √ Ç … s − 10 å
2
2
Prin urmare, q = 5+3t −4 1 + 3t s , i S = 3 2 − 1 + 3t 2 = 3 2 − . Ar˘at˘am
2
c˘a exist˘a un triunghi ABC ce satisface ipotezele problemei s , i are aria egal˘a cu valoarea obt , inut˘a,
√ Ä √ ä
S = 3 2 − 1 + 3t . Consider˘am polinomul f(w) = (w − x)(w − y)(w − z), w ∈ R. Avem
2
√
Ä ä
2
2
2
0
3
2
2
2
f(w) = w − 3w + 3(1 − t )w − 5 + 3t − 4 1 + 3t , deci f (w) = 3(w − 2w + 1 − t ) are
√
3
2
2
r˘ad˘acinile 1−t s , i 1+t. Avem f(1−t) = 2t −6t −4+4 1 + 3t > 0 (prin calcule, inegalitatea
√
3
3
2
2
2
este echivalent˘a cu t (1 − t) (t − 4) < 0, adev.) s , i f(1 + t) = −2t − 6t − 4 + 4 1 + 3t < 0
3
2
3
(prin calcule, inegalitatea este echivalent˘a cu t (t + 6t + 9t + 4) > 0, adev.) Rezult˘a c˘a s , irul
lui Rolle asociat ecuat , iei f(w) = 0 este (−, +, −, +), deci r˘ad˘acinile x, y, z ale ecuat , iei sunt
√
2
reale (s , i distincte). Mai mult, cum f(0) = −3t − 5 + 4 1 + 3t < 0 (prin calcule, inegalitatea
2
2
2
este echivalent˘a cu 9 (t − 1) > 0, adev.), rezult˘a c˘a x, y, z > 0. Dar x + y + z = 3, deci
x, y, z ∈ (0, 3), de unde rezult˘a c˘a a, b, c > 0. De asemenea, avem x + y < 3 < 3 + z, de unde
rezult˘a c˘a a + b > c. Analog se obt , ine c˘a a + c > b s , i b + c > a, ceea ce ˆıncheie demonstrat , ia
existent , ei triunghiului ABC cu propriet˘at , ile impuse.
Z b 1
M 38. Fie a, b ∈ R, a < b. Calculat ,i integrala I = dx.
4
(x − a) + (x − b) 4
a
Daniel Jinga, Pites , ti
b − a a + b
Solut ,ie. Notˆand c = s , i aplicˆand schimbarea de variabil˘a x = t + avem I =
2
Z c 2 Z c
1 1
dt = 2 dt, funct , ia integrat˘a fiind par˘a. Astfel I =
4
4
(t + c) + (t − c) 4 (t + c) + (t − c) 4
−c 0
1 1 1
Z c Z c Z c
dt = dt = Ä √ ä Ä √ ä dt
4
2 2
t + 6t c + c 4 2 2 2 4 2 2 2 2 2 2
0 0 (t + 3c ) − 8c 0 t + 3c − 2 2c t + 3c + 2 2c

c
1 Z Å 1 1 ã 1 Z c 1
= √ √ − √ dt = √ Ä √ ä −
2
2
2
2
2
4 2c 2 0 t + 3c − 2 2c 2 t + 3c + 2 2c 2 4 2c 2 0 t + c( 2 − 1)
2
! c
Ç å
1 1 1 t 1 t
Ä √ ä 2 dt = √ √ arctg √ − √ arctg √
2
t + c( 2 + 1) 4 2c 2 c( 2 − 1) c( 2 − 1) c( 2 + 1) c( 2 + 1) 0
√ √ √ √
( 2 + 1)arctg ( 2 + 1) − ( 2 − 1)arctg ( 2 − 1) √ 3π
= √ . Astfel ˆınlocuind arctg ( 2 + 1) = s , i
4 2c 3 √ 8
√ π ( 2 + 1)π
arctg ( 2 − 1) = , obt , inem c˘a I = .
8 2(b − a) 3
π
M 39. Determinat ,i funct ,iile continue f, g : 0, → R care verific˘a simultan relat ,iile
2
π
h i
3
2
2
3
f (x) − 3f(x)g (x) = cos x, g (x) − 3f (x)g(x) = − sin x, ∀x ∈ 0,
2
s , i pentru care aria suprafet ,ei plane cuprinse ˆıntre graficele lor este:
a) maxim˘a; b) minim˘a.
Dorin M˘arghidanu, Corabia
π
Solut ,ie. Pentru orice x ∈ 0, , ˆınmult , ind a doua relat , ie din enunt , cu −i s , i adunˆand-o la
2
3
prima, rezult˘a c˘a (f(x) + ig(x)) = cos x + i sin x, deci (f(x), g(x)) ∈ (f k (x), g k (x)) | k = 0, 2 ,
x + 2kπ x + 2kπ
unde f k (x) = cos s , i g k (x) = sin . Fiecare dintre cele trei perechi verific˘a
3 î √ √
π 3 3 ó π 1 π î 3 1 ó
egalit˘at , ile din enunt , . Cun f 0 0, = , 1 , g 0 0, = 0, , f 1 0, = − , − ,
2 2 2 2 2 2 2