tailieunhanh - Đề thi Olympic sinh viên thế giới năm 1995

" Đề thi Olympic sinh viên thế giới năm 1995 " . Đây là một sân chơi lớn để sinh viên thế giới có dịp gặp gỡ, trao đổi, giao lưu và thể hiện khả năng học toán, làm toán của mình. Từ đó đến nay, các kỳ thi Olympic sinh viênthế giới đã liên tục được mở rộng quy mô rất lớn. Kỳ thi này là một sự kiện quan. | International Competition in Mathematics for Universtiy Students in Plovdiv Bulgaria 1995 1 PROBLEMS AND SOLUTIONS First day Problem 1. 10 points Let X be a nonsingular matrix with columns X1 X2 . Xn. Let Y be a matrix with columns X2 X3 . Xn 0. Show that the matrices A YX-1 and B X-1Y have rank n 1 and have only 0 s for eigenvalues. Solution. Let J aij be the n x n matrix where ay 1 if i j 1 and aij 0 otherwise. The rank of J is n 1 and its only eigenvalues are 00s. Moreover Y XJ and A YX-1 XJX-1 B X-1 Y J. It follows that both A and B have rank n 1 with only 00s for eigenvalues. Problem 2. 15 points Let f be a continuous function on 0 1 such that for every x 2 0 1 we have f t dt - . Show that f 2 t dt -. Jx 2 Jo 3 Solution. From the inequality 1 1 1 1 0 f x x 2 dx f 2 x dx 2 xf x dx x2dx J0 Jo Jo Jo we get r1 1 1 1 1 f2 x dx 2 xf x dx x2dx 2 xf x dx --. Jo Jo Jo Jo 3 From the hypotheses we have J J f t dtdx J . This completes the proof. 3 x dx or tf t dt 2o Problem 3. 15 points Let f be twice continuously differentiable on 0 1 such that lim f0 x 1 and lim f00 x 1. Show that X 0 X 0 . f x lim o f 0 x 0. x 2 Solution. Since f0 tends to 1 and f tends to 1 as x tends to 0 there exists an interval 0 r such that f 0 x 0 and f00 x 0 for all x 2 0 r . Hence f is decreasing and f0 is increasing on 0 r . By the mean value theorem for every 0 x x0 r we obtain f x - f xo f0 x - xo 0 for some 2 x x0 . Taking into account that f0 is increasing f0 x f0 0 we get f 0 z X _ f x - f x0 x x0 f0 x x x0 f0 x 0 Taking limits as x tends to 0 we obtain r f x f x -20 lim inf lim sup f0 x . .0. f0 x Since this happens for all x0 2 0 r we deduce that lim fh 0. f 0 x 0 lim f x . exists and f 0 x Problem 4. 15 points Let F 1 1 R be the function defined by i x2 dt F L nW Show that F is one-to-one . injective and find the range . set of values of F . Solution. From the definition we have x1 F0 x ---- x 1 . ln x Therefore F0 x 0 for x 2 1 1 . Thus F is strictly increasing and hence .