MADHAVA MATHEMATICS COMPETITION, December 2015 Solutions and Scheme of Marking

Transcription

1 MADHAVA MATHEMATICS COMPETITION, December 05 Solutions and Scheme of Marking NB: Part I carries 0 marks, Part II carries 30 marks and Part III carries 50 marks Part I NB Each question in Part I carries marks Let A(t) denote the area bounded by the curve y = e x, the X axis and the straight lines x = t, x = t, then lim A(t) is t A) B) C) / D) e As f(x) = e x is an even function, A(t) = as t OR A(t) = t 0 0 e x t 0 0 dydx = t e x dx = (e t ) as t e x dx = (e 0 e t ) How many triples of real numbers (x, y, z) are common solutions of the equations x + y =, xy z =? A) 0 B) C) D) infinitely many Solution: (B) xy = +z so that 4xy 4 Hence (x y) = (x+y) 4xy = 4 4xy 4 4 = 0 So x = y Thus the only solution is x =, y =, z = 0 3 For non-negative integers x, y the function f(x, y) satisfies the relations f(x, 0) = x and f(x, y + ) = f(f(x, y), y) Then which of the following is the largest? A) f(0, 5) B) f(, 3) C) f(3, ) D) f(4, ) Solution: (D) f(x, ) = f(f(x, 0), 0) = f(x, 0) = x Inductively f(x, y) = x for all integers y 0 4 Suppose p, q, r, s are,, 3, 4 in some order Let x = p + q + r + s We choose p, q, r, s so that x is as large as possible, then s is A) B) C) 3 D) 4 Solution: (C) For x to be the largest, p, q, r, s should be min{,, 3, 4}, max{,, 3, 4}, min{, 3}, max{, 3} respectively So s = 3 { 3x + x if x < 0 5 Let f(x) = x 3 + x Then f (0) is if x 0 A) 0 B) C) 3 D) None of these Solution: (D) f (x) = 3 + x, for x < 0 and f +(x) = 3x + x, for x 0 So f (0) does not exist because f (0) = 3 0 = f +(0) 6 There are 8 teams in pro-kabaddi league Each team plays against every other exactly once Suppose every game results in a win, that is, there is no draw Let w, w,, w 8 be number of wins and l, l,, l 8 be number of loses by teams T, T,, T 8, then A) w + + w 8 = 49 + (l + + l 8 ) B) w + + w 8 = l + + l 8 C) w + + w 8 = 49 (l + + l 8 ) D) None of these Solution: (B) Note that w i + l i = 7 for all i and w i l i = (w i l i ) = 0 Then wi li = (wi + l i )(w i + l i ) = 7 (w i l i ) = 0

2 7 The remainder when m+n is divided by is 8, and the remainder when m n is divided by is 6 If m > n, then the remainder when mn divided by 6 is A) B) C) 3 D) 4 Note that m + n 8(mod ) and m n 6(mod ) Adding these congruences, we get, m (mod ) This implies m (mod 6) Similarly by subtracting, we get, n ( mod 6) Thus mn (mod 6) n n + n + n 8 Let A = Select any entry and call it x Delete (n )n + (n )n + n row and column containing x to get an (n ) (n ) matrix Then select any entry from the remaining entries and call it x Delete row and column containing x to get (n ) (n ) matrix Perform n such steps Then x + x + + x n is n(n + ) A) n B) C) n(n + ) D) None of these Solution: (C) For n = 4, A = Note that S = x + x + x 3 + x 4 = a + (b + 4) + (c + 8) + (d + ) where a, b, c, d is a permutation of,, 3, 4 So S = a + b + c + d + 4 = = 34 Thus S is the same for all stated choices of x,, x n Hence taking x i s as the main diagonal elements, S = + (n + ) + (n + 3) + + [n(n ) + n] = [n + n + + n(n )] + [ n] n(n )n n(n + ) = + = n(n + ) 9 The maximum of the areas of the rectangles inscribed in the region bounded by the curve y = 3 x and X axis is A) 4 B) C) 3 D) By symmetry, let the base of the rectangle be segment with ends x, x and height y Then area A(x) = xy = x(3 x ) = 6x x 3 and A (x) = 6 6x, A (x) = x So A (x) = 0 at x = ie x = ; and A () = < 0 So A(x) is maximum at x = with maximum value A() = 4 0 How many factors of are perfect squares? A) 4 B) 0 C) 30 D) 36 Factors that are perfect squares will be of the form d = a 3 b 5 c where a = 0, or 4, b = 0,, 4 or 6, and c = 0 or Thus there are 3 4 possible divisors that are perfect squares Part II NB Each question in Part II carries 6 marks How many 5 digit palindromes are there in each of which the product of the nonzero digits is 36 and the sum of the digits is equal to 5? (A string of digits is called a palindrome if it reads the same forwards and backwards For example 04340, 6446) Solution: The first 7 digits completely determine the number Since the sum of digits is 5, the 8 th digit is odd and is a factor of 36 Note that the product of all non-zero digits ( except the digit in the 8 th place ) is a square using the definition of palindrome Hence

3 the 8 th digit cannot be 3 because the product in that case is So the 8 th digit is either or 9 [] Case : If the 8 th digit is then the digits in first seven places can either be a permutation of,,,3,0,0,0 or,6,0,0,0,0,0 because these are the only possibilities with sum 7 and product 6 7! Number of permutations of,,,3,0,0,0 is!3! Number of permutations of,6,0,0,0,0,0 is 7! 5! [] Case : If the 8 th digit is 9 then the digits in first seven places will be a permutation of,,0,0,0,0,0 because this is the only possibility with sum 3 and product Number of permutations of,,0,0,0,0,0 is 7! 5! Thus the number of required 5 digit palindromes with product of nonzero digits 36 and 7! sum of digits 5 is!3! +7! 5! +7! 5! [] Let H be a finite set of distinct positive integers none of which has a prime factor greater than 3 Show that the sum of the reciprocals of the elements of H is smaller than 3 Find two different such sets with sum of the reciprocals equal to 5 Solution: The given condition implies that every n H, n is of the form n = α 3 β, α, β 0 Since H is finite, k N such that α k, β k for each n H This implies n + i + 3 j + i 3 j n H i= j= i= j= ( ) = + i + 3 j + i 3 j i= j= i= j= = ( ) ( k ) k [3] = ( k+ / )( 3 k+ /3 ) < ( / )( /3 ) = (3 ) = 3 [] Let H = {,, 3, 4, 6, 8,, 4} Then = 5 [] n n H Let H = {,, 3, 4, 6, 8,, 36, 7} Then = 5 [] n n H Any other correct choices for H, also carries one mark each 3 Let A be an n n matrix with real entries such that each row sum is equal to one Find the sum of all entries of A 05 Solution: Let A be an n n matrix with real entries such that each row sum is equal to one This implies A = By repeated use of this, we get A 05 = [] So each row sum of A 05 is equal to one Hence the sum of all entries of A 05 is n [] 4 Let f : R R be a differentiable function such that f(0) = 0, f (x) > f(x) for all x R [3] 3

4 Prove that f(x) > 0 for all x > 0 Solution: By data, f f(x) (0) > f(0) = 0 So lim > 0 Hence δ > 0, such that x 0+ x f(x) > 0 for all x (0, δ) [] Now if there exists x 0 > 0 such that f(x 0 ) 0, then by intermediate value property, there exists x δ such that f(x ) = 0 Let c = inf A, A = {x x > 0, f(x) = 0} [] Clearly, as f is continuous and c is inf A, f(x) 0, for x [0, c] So c δ By the property of infimum, there exists a sequence {x n } of points which converges to c, and x n > 0 and f(x n ) = 0 By continuity, f(c) = lim n) = 0 [] n This implies that in (0, c), f(x) > 0 and f(0) = f(c) = 0 Hence by Rolle s theorem, f (b) = 0 for some b, 0 < b < c [] But then f(b) < f (b) = 0 which is a contradiction Hence f(x) > 0 for all x > 0 [] Second method As before, δ > 0, such that f(x) > 0 for all x (0, δ) [] Let S = {x f(t) > 0 for t (0, x)} Then δ S so that S is non-empty Let m = sup S If m =, we are done Let, if possible, m < Now f(x) > 0, x (0, m) By continuity, f(m) = lim t m f(t) 0 So for all x [0, m], f (x) > f(x) 0 so that f (x) > 0 Hence f is strictly increasing on [0, m], in particular, f(m) > f(m/) > 0 Since f(m) > 0, by continuity, there exists δ > 0 such that f(x) > 0 in [m, m + δ ) So, f(x) > 0, for x (0, m + δ ) Thus m + δ S, which is a contradiction since m = sup S Hence m = [3] 5 Give an example of a function which is continuous on [0, ], differentiable on (0, ) and not differentiable at the end points Justify Solution: f(x) = x x for x [0, ] [3] Then f (x) exists on (0, ), f (x) = x x x [] But f (0) = f () = [] Note: Any other correct example with justification will carry full marks Part III There are some marbles in a bowl A, B and C take turns removing one or two marbles from the bowl, with A going first, then B, then C, then A again and so on The player who takes the last marble from the bowl is the loser and the other two players are the winners If the game starts with N marbles in the bowl, for what values of N can B and C work together and force A to lose? [] [] Solution: We claim that B and C can force A to lose for all N except N = ; 3; 4; 7; or 8 At N =, A leaves At N = 3 or 4, A leaves At N = 7 or 8, A leaves 6 after which B and C must leave, 3 or 4 For N = 5 or 6, regardless of what A takes, B and C can work it so that when A s turn arrives there is only one marble left For N = 9 or 0, A must leave 7, 8 or 9 from which B and C can force 5 or 6 [6] For N = 4k where k >, A must leave either 4k or 4k from which B and C can force 4(k ) + or 4(k ) + For N = 4k+, A must leave either 4k or 4k from which B and C can force 4(k )+ or 4(k ) + For N = 4k+, A must leave either 4k+ or 4k from which B and C can force 4(k )+ or 4(k ) + For N = 4k + 3, A must leave either 4k + or 4k + from which B and C can force 4(k ) + 4

5 In all cases for N, A will always be faced with a new value of the form 4t + or 4t + on his next turn eventually forcing him to N = 5 or 6 and a loss [6] Let f : R R be a function such that f (0) exists Suppose α n, n N and both sequences {α n } and { } converge to zero Define D n = f() f(α n ) Prove that lim n D n = f (0) under EACH of the following conditions: (a) α n < 0 <, n N (b) 0 < α n < and M, n N, for some M > 0 (c) f (x) exists and is continuous for all x (, ) [3] Solution: Let ɛ > 0 be given Given that f (0) = lim x 0 f(x) f(0) x exists (a) Given that α n < 0 <, n N Since α n 0 and 0, we have f f(α n ) f(0) (0) = lim n α n and f f( ) f(0) (0) = lim n There exist n, n N such that f(α n ) f(0) α n f (0) < α n ɛ = α n ɛ, f( ) f(0) f (0) < ɛ = ɛ, n n Let n 0 = max{n, n } Then n n 0, we get f( ) f(α n ) ( )f (0) f( ) f(0) f (0) + f(α n ) f(0) α n f (0) < ( )ɛ Thus f() f(α n ) f (0) < ɛ, n n 0 n n and Hence lim n D n = f (0) [4] (b) Given that 0 < α n < and Since α n <, observe that M, n N, for some M > 0 α n M, n N Similar to part (a), there exist n, n N such that f(α n ) f(0) α n f (0) < α n ɛ = α n ɛ, n n and f( ) f(0) f (0) < ɛ = ɛ, n n Let n 0 = max{n, n } Then n n 0, we get f() f(α n ) f (0) = f() f(α n ) ( )f (0) = f() f(α n ) f (0) + α nf (0) = ( f() f(0) f (0) ) ( f(α n) f(0) α n f (0) ) f() f(0) f (0) + f(α n) f(0) α n f (0) α n < ( )ɛ + ( )ɛ Mɛ Hence lim n D n = f (0) [5] (c) Given that f (x) exists and is continuous for all x (, ) By Lagrange s Mean Value Theorem, for every positive integer n, there exists c n between 5

6 α n and such that D n = f() f(α n ) = f (c n ) Since α n 0 and 0, c n 0 It is given that f (x) is continuous Therefore lim f (c n ) = f (0) n Hence lim D n = f (0) [4] n Second method for (a), (b) [ Let g(x) = f(x) ] f(0) xf (0) on R Then g(0) = 0 and g g(x) f(x) f(0) (0) = lim = lim f (0) = 0 Let E n = g() g(α n ) Then x 0 x x 0 x E n = D n f (0) (a) Let α n < 0 < Then 0 < α n < Hence 0 < > so that 0 < < Hence ( ) ( ) lim E g( ) βn g(α n ) αn n = lim + lim = 0, n n n α n α n < Also, here as the sequences in brackets are both bounded and g (0) = 0 [4] (b) Let 0 < α n < and 0 < 0 < α n < lim E g( ) n = lim n n < M Hence ( βn < M Then > 0 and so ) ( ) g(α n ) αn lim = 0, as in (a) [5] n α n 3 Let f(x) = x 5 For x > 0, let P = (x, f(x )) Draw a tangent at the point P and let it meet the graph again at point P Then draw a tangent at P and so on Show that the ratio A( P np n+ P n+ ) is constant [] A( P n+ P n+ P n+3 ) Solution: Let f(x) = x 5 For x > 0, let P = (x, f(x )) Draw a tangent at the point P and let it meet the graph again at point P Recursively P n+ is defined We now try to calculate P in terms of P Tangent at P is given by y y = 5x 4 (x x ) ie y = 5x 4 x 4x5 This cuts the curve y = x5 at x = x, x x Hence x 5 5x 4 x + 4x 5 = 0 This is a homogeneous equation in x, x So put x = kx, k, then x 5 k5 5kx 5 + 4x 5 = 0 This implies k5 5k + 4 = 0 ie (k ) (k 3 + k + 3k + 4) = 0 Since k, k 3 + k + 3k + 4 = 0 Observe that this cubic equation has one real and two complex roots (g (x) = 3k + 4k + 3 0) The real root k must be negative [5] If x 3 = lx, then by similar argument, we see that l is again the above negative root k of k 3 + k + 3k + 4 = 0 Thus x 3 = kx = k x Hence by induction, x n+ = k n x for all n [] We now calculate A( P n P n+ P n+ ) A( P n P n+ P n+ ) = x n x 5 det n x n+ x 5 n+ x n+ x 5 n+ = k n x k 5n 5 x 5 det k n x k 5n x 5 k n+ x k 5n+5 x 5 6

7 = kn x k 5n 5 x 5 det k k 5 k k 0 = k 6n 6 x 6 D, where D = det k k 5 0 since k k k 0 Then A( P n+ P n+ P n+3 ) = k 6n x 6 D Hence the ratio A( P np n+ P n+ ) A( P n+ P n+ P n+3 ) = is constant [5] k6 4 Let p(x) be a polynomial with positive integer coefficients You can ask the question: What is p(n) for any positive integer n? What is the minimum number of questions to be asked to determine p(x) completely? Justify [3] Solution: The minimum number of questions needed is For this, let p(x) be a polynomial with positive integer coefficients say, p(x) = a 0 + a x + a x + + a k x k We can ask the question: what is p()? Let p() = N [3] Here N = a 0 + a + a + + a k > a i, i and N is a known number Also, what is p(n)? So p(n) = a 0 + a N + a N + + a k N k is a known number [4] Now express p(n) to base N, then i th digit gives a i, i because a i < N, i Thus p(x) is determined [6] Note that asking only one question ie asking for the value p(n) for a particular choice of n, is not sufficient to determine the polynomial p(x) Example Suppose p() = 9 and p(9) = 93 Now we express 93 to base 9 : 93 = (9) + 4, = (9) + 3, = 0(9) + So the remainders are, starting with the last,, 3, 4 So 93 = (9 )+3(9)+4(9 0 ) = (34) 9 So a =, a = 3, a 0 = 4 and p(x) = 4 + 3x + x 7