p ∈ N ∩ G, atunci p ∈ H, prin urmare p + a ∈ H ¸si (p + a)
′ = a +
1
p
∈ H. Deducem
c˘a a + m ¸si a +
1
n
sunt ˆın H, deci (a + m) ∗
a +
1
n
= a +
m
n
∈ H, adic˘a q ∈ H ¸si
astfel Q ∩ G ⊂ H.
XII.122. Fie n ∈ N, n ≥ 2 ¸si polinomul f = Xn − 2nXn−1 + (2n
2 − 4)Xn−2 +
a3Xn−3 +. . .+an ∈ C[X]. Demonstrat¸i c˘a f are toate r˘ad˘acinile reale dac˘a ¸si numai
dac˘a n = 2.
Florin St˘anescu, G˘ae¸sti
Solut¸ie. Dac˘a n = 2, atunci f = X2−4X+4 are r˘ad˘acinile x1 = x2 = 2. Reciproc,
fie x1, x2, . . . , xn r˘ad˘acinile lui f, presupuse reale. Cum (x1 + x2 + . . . + xn)
2 ≤
n(x
2
1 + x
2
2 + . . . + x
2
n), rezult˘a c˘a 4n
2 ≤ 8n, deci n ≤ 2 ¸si atunci n = 2.
XII.123. Calculat¸i I =
Z arccos
√
65
9
0
tg x
√
sin xdx.
Vasile Chiriac, Bac˘au
Solut¸ie. Cu schimbarea de variabil˘a sin x = s, obt¸inem c˘a I =
Z 4
9
0
s
√
s
1 − s
2
ds.
Apoi, substitut¸ia √
s = t conduce la
I =
Z 2
3
0
2t
4
1 − t
4
dt =
−2t + arctg t −
1
2
ln 1 − t
1 + t
2
3
0
= −
4
3
+ arctg
2
3
+ ln √
5.
XII.124. Fie f : R → R o funct¸ie continu˘a cu proprietatea c˘a (f ◦ f)(x) = sin x,
∀x ∈ R. Demonstrat¸i c˘a Z 1
0
f(x)dx ≤ 1.
Dumitru Cr˘aciun, F˘alticeni
Solut¸ie. Din (f ◦f)◦f = f ◦(f ◦f), deducem c˘a sin f(x) = f(sin x), ∀x ∈ R; atunci
f(sin x) ≤ 1, ∀x ∈ R, prin urmare f(sin x) · cos x ≤ cos x, ∀x ∈
h
0,
π
2
i
. Integrˆand pe
h
0,
π
2
i
, rezult˘a c˘a Z π
2
0
f(sin x) · (sin x)
′
dx ≤ sin x
π
2
0
, adic˘a Z 1
0
f(x)dx ≤ 1.
