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Đề thi Olympic sinh viên thế giới năm 1995International Competition in Mathematics for Universtiy Students in Plovdiv, Bulgaria 1995 1PROBLEMS AND SOLUTIONS First day Problem 1. (10 points) Let X be a nonsingular matrix with columns X 1 , X2 , . . . , Xn . Let Y be amatrix with columns X2 , X3 , . . . , Xn , 0. Show that the matrices A = Y X −1and B = X −1 Y have rank n − 1 and have only 0’s for eigenvalues. Solution. Let J = (aij ) be the n × n matrix where aij = 1 if i = j + 1and aij = 0 otherwise. The rank of J is n − 1 and its only eigenvalues are0 s. Moreover Y = XJ and A = Y X −1 = XJX −1 , B = X −1 Y = J. Itfollows that both A and B have rank n − 1 with only 0 s for eigenvalues. Problem 2. (15 points) Let f be a continuous function on [0, 1] such that for every x ∈ [0, 1] we 1 1 − x2 1 1have f (t)dt ≥ . Show that f 2 (t)dt ≥ . x 2 0 3 Solution. From the inequality 1 1 1 1 0≤ (f (x) − x)2 dx = f 2 (x)dx − 2 xf (x)dx + x2 dx 0 0 0 0we get 1 1 1 1 1 f 2 (x)dx ≥ 2 xf (x)dx − x2 dx = 2 xf (x)dx − . 0 0 0 0 3 1 1 1 1 − x2 1From the hypotheses we have f (t)dtdx ≥ dx or tf (t)dt ≥ 0 x 0 2 01 . This completes the proof.3 Problem 3. (15 points) Let f be twice continuously differentiable on (0, +∞) such thatlim f (x) = −∞ and lim f (x) = +∞. Show thatx→0+ x→0+ f (x) lim = 0. x→0+ f (x)2 Solution. Since f tends to −∞ and f tends to +∞ as x tends to0+, there exists an interval (0, r) such that f (x) < 0 and f (x) > 0 for allx ∈ (0, r). Hence f is decreasing and f is increasing on (0, r). By the meanvalue theorem for every 0 < x < x0 < r we obtain f (x) − f (x0 ) = f (ξ)(x − x0 ) > 0,for some ξ ∈ (x, x0 ). Taking into account that f is increasing, f (x)
