các đê thi olympic sinh viên

... Bj,j and Cn,n are relatively prime, we can choose integers Cj,n and Bj,n such that this equation is satisfied Doing this step by step for all j = n − 1, n − 2, , 1, we finally get B and C such that ... vertices if n is not divisible by Now add triangles P1 Pk Pk+2 , Pk Pk+1 Pk+2 and P1 Pk+2 Pk+3 This way we introduce two new triangles at vertices P1 and Pk so parity is preserved The vertices ... of f on [x, y] Then f ([x, y]) = [f (b), f (a)]; hence y − x = f (a) − f (b) ≤ |a − b| ≤ y − x This implies {a, b} = {x, y}, and therefore f is a monotone function Suppose f is increasing Then...

Ngày tải lên: 17/10/2013, 23:15

... using only a translation or a rotation Does this imply that f (x) = ax + b for some real numbers a and b ? Solution No The function f (x) = ex also has this property since cex = ex+log c Problem ... \{y} = 0, ¯ ¯ since A2 = So, AX annihilates the span of all the vY with X Y This implies that vX does not ¯ y lie in this span, because AX vX = v{1,2, ,k} = Therefore, the vectors vX (with X ∈ ... (this follows by applying Rolle’s theorem to the function (n) ′ ′ ex fn (x)); the same is true for the interval (−∞, xm−1 ) Hence, in each of these m − intervals, fn+1 has exactly one zero This...

Ngày tải lên: 20/10/2013, 18:15

... IMC2008, Blagoevgrad, Bulgaria is a Cauchy sequence in H (This is the crucial observation.) Indeed, for m > n, the norm ym − yn may be computed by the above ... that xn − y, xp − y = 0, so that √ (xn − y) : n ∈ N d is indeed an orthonormal system of vectors This completes the proof in the case when T = S, which we can always take if S is countable If it ... =y n→∞ n+2 n √ √ ′ ′′ (x − y), (x − y) d d ∪ √ (xn − y) : n ∈ N d is still an orthonormal system This it true for any distinct x′ , x′′ ∈ S \ T ; it follows that the entire system √ (x − y) : x...

Ngày tải lên: 20/10/2013, 18:15

... pháp luật sinh viên? Đỗ Văn Hiệp –Lớp K52A-Hóa học.Trường Đại học khoa học Tự Nhiên-ĐHQGHN Đề thi hữu ích dành cho sinh viên Hóa học - 2009 Môn thi: Hóa học hữu I Số đvht: 04 Đối tượng dự thi: K46 ... Nhiên-ĐHQGHN Đề thi hữu ích dành cho sinh viên Hóa học - 2009 Môn thi : Điện hóa học Số đvht: 02 Đối tượng dự thi: K46 ngành Hóa Thời gian làm bài: 60 phút Câu 1 Nguyên nhân sinh điện cực? Thi t lập ... Nhiên-ĐHQGHN Đề thi hữu ích dành cho sinh viên Hóa học - 2009 Môn thi : Hóa hữu II Số đvht: 04 Đối tượng dự thi: K46 A,S ngành Hóa Thời gian làm bài: 120 phút Câu Viết phương trình phản ứng sau anđêhit...

Ngày tải lên: 24/11/2013, 14:26

... Hình học Các loại Olympic khác Đề thi Chuyên đề Đáp án Toán Luyện thi Đại học Thi lớp 10 Cao học Đại học Bồi dưỡng HSG HỘI TOÁN HỌC VIỆT NAM BỘ GIÁO DỤC VÀ ĐÀO TẠO ĐÁP ÁN OLYMPIC TOÁN SINH VIÊN LẦN ... B)C = C(A + B) tức BC = CB , đpcm HỘI TOÁN HỌC VIỆT NAM BỘ GIÁO DỤC VÀ ĐÀO TẠO ĐÁP ÁN OLYMPIC TOÁN SINH VIÊN LẦN THỨ XVIII (2010) Môn: Giải tích Câu Cho hàm số f (x) = ln(x + 1) a) Chứng minh ... cho, ta thu x2s Q(x)Q(x2 ) = (x2 + 2)s Q(x3 + 2x), ∀x ∈ R Điều mâu thuẫn với giả thi t Q(0) = Vậy nên x0 = Ta giả thi t môđun |x0 | có giá trị lớn √ √ nghiệm P (x) Khi x3 + 2x0 ( x0 )3 + x0 nghiệm...

Ngày tải lên: 15/12/2013, 01:16

... and at least one is negative Hence we have at least two non-zero elements in every column of A−1 This proves part a) For part b) all b ij are zero except b1,1 = 2, bn,n = (−1)n , bi,i+1 = bi+1,i ... (bk/n) − n b bk/n b f (x)dx b(k−1)/n g(x)dx + O(ω(f, b/n) g 1) b f (x)dx 0 g(x)dx + O(ω(f, b/n) g ) This proves a) For b) we set b = π, f (x) = sin x, g(x) = (1 + 3cos x)−1 From a) and π π sin xdx ... c ∈ [a, y) such that f (c) = and f (x) > for x ∈ (c, y] For x ∈ (c, y] we have |f (x)| ≤ λf (x) This implies that the function g(x) = ln f (x) − λx is f (x) −λ ≤ Thus ln f (x)−λx ≥ not increasing...

Ngày tải lên: 21/01/2014, 21:20

... the inequality 0≤ (f (x) − x)2 dx = f (x)dx − 1 xf (x)dx + x2 dx we get f (x)dx ≥ 1 xf (x)dx − 1 This completes the proof 0 From the hypotheses we have x x2 dx = f (t)dtdx ≥ xf (x)dx − − x2 dx ... obtain −x0 ≤ lim inf x→0+ f (x) f (x) ≤ lim sup ≤ f (x) f (x) x→0+ f (x) exists and x→0+ f (x) Since this happens for all x0 ∈ (0, r) we deduce that lim f (x) = x→0+ f (x) lim Problem (15 points) Let ... ≥ δ, |x|p + |y|p = 2} Since Dδ is compact it is enough to show that f is continuous on D δ For this we show that the denominator of f is different from zero Assume x+y p the contrary Then |x +...

Ngày tải lên: 21/01/2014, 21:20

... From (i ) the first and the third integral tend to − m h n as n → ∞, hence f (0) so does the second Also n A (f (x))n dx ≤ n(f (A))n −→ (f (A) < 1) We get L = − n→∞ in this case If M = f (1) we ... consider k ≥ For any m we have (2) cosh θ = cosh ((m + 1)θ − mθ) = = cosh (m + 1)θ.cosh mθ − sinh (m + 1)θ .sinh mθ √ = cosh (m + 1)θ.cosh mθ − cosh (m + 1)θ − cosh mθ − Set cosh kθ = a, cosh (k + ... (0, 4), and T is its reflexion about the x-axis, then C(E) = > K(E) Remarks: All distances used in this problem are Euclidian Diameter of a set E is diam(E) = sup{dist(x, y) : x, y ∈ E} Contraction...

Ngày tải lên: 21/01/2014, 21:20

... contradiction it suffices to show that this sequence is strictly decreasing Now, pk − qk − (pk−1 − qk−1 ) = nk pk−1 − (nk + 1)qk−1 − pk−1 + qk−1 = (nk − 1)pk−1 − nk qk−1 and this is negative because pk−1 ... suppose α is the rational number Our aim is to show q that for some m, nm θm−1 = nm − Suppose this is not the case, so that for every m, (3) θm−1 < nm nm − For each k we write pk qk as a fraction ... real number and det ω(BA − AB) = ω n det(BA − AB) and det(BA − AB) = 0, then ω n is a real number This is possible only when n is divisible by Problem Let α be a real number, < α < a) Show that...

Ngày tải lên: 21/01/2014, 21:20

... then there n 2π are axes making k angle) If A is infinite then we can think that A = Z n and f (m) = m + for every m ∈ Z In this case we define g as a symmetry relative to , g2 as a symmetry relative ... itself It is enough to prove the theorem for every such set Let A = T (x) If A is finite, then we can think that A is the set of all vertices of a regular 2π Such rotation can be obtained as a n polygon...

Ngày tải lên: 21/01/2014, 21:20

... fi (x) − fi (x)fi (ai ) = 0, thus f (x − f1 (x)a1 − · · · − fk (x)ak ) = By the linearity of f this implies f (x) = f1 (x)f (a1 ) + · · · + fk (x)f (ak ), which gives f (x) as a linear combination ... 1−x 1−c  for x ∈ [0, c], for x ∈ [c, 1] f (f ( f (p))) = p n and n is the smallest number with this property Prove that for every n ≥ the set of n-periodic points is non-empty and finite Solution ... 1) ∈ [k + 1, m] If f (l) > l − for some l > m + 1, then l − and f (l) belong to f −1 (f (l)) and this contradicts the facts above Hence f (i) = i − for i > m + Thus we show that every function...

Ngày tải lên: 26/01/2014, 16:20

... i=1 (r) From this recursion we can compute the probabilities for small values of n and can conjecture that p n = (r) 1 + 5·6n if n ≡ r (mod )5 and pn = − 5·6n otherwise From (1), this conjecture ... fixed the opposite inequality holds ∀m This contradiction shows that g is a constant, i.e f (x) = Cx, C > Conversely, it is easy to check that the functions of this type verify the conditions of ... contains at most one marked point, delete it This decreases n + k by and the number of the marked points by at most 1, so the condition remains true Repeat this step until each row and column contains...

Ngày tải lên: 26/01/2014, 16:20

... by e, this yields e + 2ef = 0, and similarly f + 2ef = 0, so that f = −2ef = e, hence e = f = g by symmetry Hence, finaly, 3e = e + f + g = 0, i.e e = f = g = For part (i) just omit some of this ... µ(p, c) = c∈S0 µ(p − 1, c) = n (1) c∈S1 The polynomial p has at most n − roots (n > is used here) This implies that µ(p , c) ≤ n − (2) c∈S0 ∪S1 If p(c) = or p(c) − = 0, then µ(p, c) − µ(p c) = or ... az for each x ∈ R+ , where + ax a ≥ Conversely, a direct verification shows that the functions of this form satisfy the initial equality Second solution As in the first solution we get that f is...

Ngày tải lên: 26/01/2014, 16:20

... bn+1 = n 2− It is a matter of simple manipulation to prove that 2f (x) > x for all x ∈ (0, 2), this implies that the sequence (2n an ) is strictly increasing The inequality 2g(x) < x for x ∈ ... permutation σ ∈ Sn such that k < whenever there is an edge from Vσ(k) to Vσ( ) It is easy to see that this permutation solves the problem Problem Let R be the set of real numbers Prove that there is ... (x + y + 1) ≥ f (x + y) + f (f (x + y)) ≥ ≥ f (x) + yf (f (x)) + f (f (x + y)) > f (f (x + y)) This contradiction completes the solution of the problem Problem For each positive integer n, let...

Ngày tải lên: 26/01/2014, 16:20

... are pairwise disjoint, so there are at most countably many of the third type b Let f be such a map Then for each value y of this map there is an x0 such that y = f (x) and f (x) = 0, because ... than δn centered at pn such that there are there are infinitely many k’s such n that pk ∈ / Ij , this can be done by induction Let n0 = and nm+1 be the j=0 nm smallest integer k > nm such that ... the countable union of nowhere dense sets (i.e the single-element sets {pn }) If T is closed then this contradicts the Baire Category Theorem Problem Prove or disprove the following statements:...

Ngày tải lên: 26/01/2014, 16:20

... bn is an integer Solution We prove by induction on n that a n /e and bn e are integers, we prove this for n = as well (For n = 0, the term 0 in the definition of the sequences must be replaced by ... are subsets of slice OBC without common interior point, and they not cover the whole slice OBC; this implies the statement In cases (b) and (c) where at least one of the signs is negative, projections ... ) ≤ L x1 − x2 Prove that ∀x1 , x2 ∈ Rn f (x1 ) − f (x2 ) ≤L f (x1 ) − f (x2 ), x1 − x2 (1) In this formula a, b denotes the scalar product of the vectors a and b Solution Let g(x) = f (x)−f...

Ngày tải lên: 26/01/2014, 16:20

... one of the k=1 sets Ak is infinite (in fact uncountable) This set Ak being infinite, there exists a sequence in Ak having distinct terms This sequence will contain a convergent subsequence (xn ... that this function has the required property Let g be an arbitrary R → R function We show that there exist real numbers x, y such that f (x, y) > g(x) + g(y) Consider the sequence (n + g(n))∞ This ... operation ∗ on the set S given by a ∗ b = c if and only if {a, b, c} ∈ F, where a = b We note that this operation is still not defined completely (we need to define a ∗ a), but nevertheless let us...

Ngày tải lên: 26/01/2014, 16:20

... assume coprime This solves the first part Since f has a quadruple pole at z = and no other poles, we have Q(z) = (z − 1)4 up to a constant factor which we can as well set equal to 1, and this determines ... 48 nodd n≥1 n≥1odd n≥1even This implies a − b + c = −1/3 These three equations easily yield a, b, c Moreover, the function f satisfies f (z) + f (−z) = 16f (z ): this follows because the branches ... us prove that the equation Pn (x) = a, where a > 0, has exactly two distinct real solutions To this end we use mathematical induction by n If n = the assertion follows directly Assuming that...

Ngày tải lên: 26/01/2014, 16:20

... Hence |grad f (u) − grad f (v)| = |au − bv| ≥ |u − v| and this inequality is strict since D ∩ Dr contains no more than one point But this contradicts the assumption that |grad f (u) − grad f ... (u)) is the gradient vector of f at the point u For a vector √ u = (a, b), |u| = a2 + b2 ) x2 y + This function satisfies 2 the conditions, since grad f (x, y) = (1 − x, y), grad f (0, 0) = (1, 0) ... rt) Now assume q = Let the minimal polynomial of r in Z[x] be ux2 + vx + w The other √ root of this polynomial is r = p−q 7, so v = −u(r+r) = −2up and w = urr = u(p2 −7q ) The discriminant is...

Ngày tải lên: 26/01/2014, 16:20

Ngày tải lên: 23/03/2014, 08:20