Page 141 - MATINF Nr. 13-14

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PROBLEME DE MATEMATICA PENTRU CONCURSURI 141

2
Cum f(−3) = 82 − 3(m + 6) < 0 s , i lim f(x) = +∞, rezult˘a c˘a f are cel put , in cˆate o
x→±∞
a
r˘ad˘acin˘ real˘a ˆın intervalele (−∞, −3) s , i (−3, +∞).
ˆ a
In concluzie, polinomul f are exact dou˘ r˘ad˘acini reale.
Solut ,ia 2 (Daniel V˘acaru, Pites , ti). Pentru x ∈ R, avem

2
3
0
2
f (x) = 4x + 24mx + 48mx + (m + 6) ,
00
2

2

2
f (x) = 12x + 48mx + 48m = 12 x + 4mx + 4m = 12 (x + 2m) + 4m(1 − m) ≥ 0.
0
0
a
Deducem c˘ funct , ia f este strict cresc˘atoare pe R, prin urmare f are cel mult o r˘ad˘cin˘ real˘a.
a
a
0
0
a
Dar f este polinom de grad impar, deci f are exact o r˘ad˘acin˘ real˘a. Deducem c˘ polinomul f
a
are cel mult dou˘a r˘ad˘acini reale. Continuˆand ca la solut , ia anterioar˘a, se obt , ine c˘a f are exact
dou˘ r˘ad˘acini reale.
a
M 222. Calculat ,i
Z x
e sin 2x − 4ctg 2x π
dx, x ∈ kπ, kπ + , k ∈ Z.
x
2 + e sin 2x 2
Mih´aly Bencze, Bras , ov
Solut ,ie. Avem
2
2
4 cos 2x cos x − sin x
x
x
e − e −
Z x Z Z
e sin 2x − 4ctg 2x sin 2x sin x cos x
2
2
2
dx = dx = dx
2 + e sin 2x 2 1
x
x
+ e x e +
sin 2x sin x cos x
1 1
x
e −
Z 2 + 2
x
= sin x cos x dx = ln |e + tg x + ctg x| + C.
x
e + tg x + ctg x
M 223. Calculat ,i
√
Z 10
I = x · [x] · {x} dx
0
a
(unde [x] s , i {x} reprezint˘ partea ˆıntreag˘a, respectiv partea fract ,ionar˘ a num˘arului real x).
a
Dorin M˘arghidanu, Corabia
Solut ,ie. Avem
√
10
Z
I = x · [x] · (x − [x]) dx
0
√
Z 1 Z 2 Z 3 Z 10
= x · 0 · (x − 0) dx + x · 1 · (x − 1) dx + x · 2 · (x − 2) dx + x · 3 · (x − 3) dx
0 1 2 3
√
x x 2x 9x 7 3 38 9
Å 3 2 ã 2 Å 3 ã 3 Å 2 ã 10 √
3
= − + − 2x 2 + x − = − + − 10 + 10 10 − 27 −
3 2 3 2 3 2 3 2
√ 1 2 3
= 10 10 − 28.