Question

aflati nr. z-4,u-2,s<6
Care este nr ?
u-8,z-4,s>5

Answer 1

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.