FOR DUTCH STUDENTS! ENGLISH VERSION NEXT PAGE

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1 FOR DUTCH STUDENTS! ENGLISH VERSION NEXT PAGE Tentamen Bewijzen en Technieken 1 7 januari 211, duur 3 uur. Voeg aan het antwoord van een opgave altijd het bewijs, de berekening of de argumentatie toe. De (grafische) rekenmachine mag niet gebruikt worden bij het tentamen. Opgave 1 Bereken alle oplossingen van de volgende ongelijkheid: Opgave 2 x 3 5x 2x Ga van elk van onderstaande beweringen na of de bewering waar is of niet. Geef een bewijs als de bewering waar is en een tegenvoorbeeld als de bewering niet waar is. a) y IN x Q : xy + x 2 = 1. b) x IN y Q : xy + x 2 = 1. Opgave 3 Toon aan, met behulp van het principe van volledige inductie, dat n (k k!) = (n + 1)! 1 voor elke n IN. Zie ommezijde voor de overige opgaven. 1 Cijferbepaling: Puntentelling tentamen: opgave 1: 5 ptn; opgave 2: 6 ptn; opgave 3: 5 ptn; opgave 4: 5 ptn; opgave 5: 4 ptn; opgave 6: 6 ptn; opgave 7: 4 ptn; opgave 8: 5 ptn. Het cijfer voor het tentamen is de som van de punten gedeeld door 4. De toets telt niet meer mee.

2 Opgave 4 Bepaal alle punten P op de cirkel met middelpunt (1,2) en straal 2 waarvoor geldt dat de raaklijn aan de cirkel in P ook door (5,) gaat. Opgave 5 Laat zien dat voor elk tweetal verzamelingen A en B geldt dat Opgave 6 A B = (A B c ) (A c B) (A B). a) Geef de definitie van de verzamelingen A en B hebben dezelfde cardinaliteit (cardinality). b) Laat zien dat als A en B dezelfde cardinaliteit hebben en B en C hebben ook dezelfde cardinaliteit, dan hebben A en C ook dezelfde cardinaliteit. c) Laat zien dat de intervallen (, ) en (, 1) dezelfde cardinaliteit hebben. Opgave 7 Bepaal de afgeleide van de functie f : IR IR, gedefinieerd door f(x) = eax2 1 + bx 4 voor iedere x IR. Hierbij zijn a en b twee positieve constanten. Opgave 8 Bereken 1 x 3 x dx.

3 FOR NON-DUTCH STUDENTS! Exam Proofs and Techniques 2 January 7, 211, duration 3 hours For every exercise you have to provide the full proof, calculation and/or arguments. The (graphical) calculator can not be used during the exam. Exercise 1 Compute all solutions of the following inequality: x 3 5x 2x Exercise 2 Check for every of the statements below whether the statement is true or not. Provide a proof if the statement is true. Otherwise, provide a counterexample. a) y IN x Q : xy + x 2 = 1. b) x IN y Q : xy + x 2 = 1. Exercise 3 Show, by using the principle of induction, that for every n IN. n (k k!) = (n + 1)! 1 Exercise 4 Determine all points P on the circle with center (1,2) and radius 2 which are such that the tangent line to the circle in P contains the point (5,). Turn this page for the other exercises. 2 Grading: Grading exam: exercise 1: 5 pts; exercise 2: 6 pts; exercise 3: 5 pts; exercise 4: 5 pts; exercise 5: 4 pts; exercise 6: 6 pts; exercise 7: 4 pts; exercise 8: 5 pts. The grade for the exam is the sum of the points scored, divided by 4. The midterm does not longer count.

4 Exercise 5 Show that for every pair of sets A and B we have A B = (A B c ) (A c B) (A B). Exercise 6 a) Provide the definition of the sets A and B have the same cardinality. b) Show that if A and B have the same cardinality and B and C have the same cardinality as well, then also A and C have the same cardinality. c) Show that the intervals (, ) and (,1) have the same cardinality. Exercise 7 Determine the derivative of the function f : IR IR, defined by f(x) = eax2 1 + bx 4 for every x IR. Here a and b are two positive constants. Exercise 8 Compute 1 x 3 x dx.

5 Solutions Exam Proofs and Techniques, January 7, 211 Exercise 1 The inequality is equivalent with x 3 5x 2x Define the function f : IR\{ 3, 3} IR by f(x) = x3 5x 1 for every 2x 2 6 x IR\{ 3, 3}. Step 1 (find the zeros of f). For every x IR\{ 3, 3} we have f(x) = Step 2 (make a sign scheme of f). x3 5x 2x = x3 5x 2x 2 6 = 1 x 3 5x = 2x 2 6 x 3 2x 2 5x + 6 = (x 1)(x + 2)(x 3) = x = 2 or x = 1 or x = Step 3 (read solution from sign scheme of f): [ 2, 3) [1, 3) [3, ). Exercise 2 a) False. Take y = 2 IN. If the statement were true then the equation 2x + x 2 = 1, or, equivalently, x 2 + 2x 1 = would have a solution in Q. However, the only solutions are x = 1 2 and x = If one of these solutions solutions belongs to Q, then 2 Q. Contradiction. b) True. Let x IN. Then y = 1 x2 x Q is such that xy + x 2 = 1. Exercise 3 Denote the statement n (k k!) = (n + 1)! 1 by P(n). Step 1: P(1) is the statement 1 1! = 2! 1 and this statement is definitely true. Step 2: Let n N and assume that P(n) is true, i.e. n (k k!) =

6 (n + 1)! 1. Then: n+1 (k k!) = n (k k!) + (n + 1) (n + 1)! = (n + 1)! 1 + (n + 1) (n + 1)! = (n + 2)(n + 1)! 1 = (n + 2)! 1 = ((n + 1) + 1)! 1. So P(n + 1) is true. So, for every n N we have: if P(n) is true, then P(n + 1) is true. Exercise 4 Let P = (a,b) denote such a point. The line through P and center M = (1,2) has direction vector (a 1,b 2). The tangent line (through P and (5,)) has direction vector (a 5, b). As these two direction vectors must be perpendicular we have < (a 1,b 2),(a 5,b) >=, i.e. (a 1)(a 5)+(b 2)b =. So a 2 6a + b 2 2b + 5 =. Since P is an element of the circle we have (a 1) 2 + (b 2) 2 = 4 as well. So a 2 2a + b 2 4b + 1 =. So we have to solve the following system of equations: { a 2 6a + b 2 2b + 5 = a 2 2a + b 2 4b + 1 = { a 2 6a + b 2 2b + 5 = 4a + 2b + 4 = { a 2 6a + b 2 2b + 5 = { a 2 6a + (2a 2) 2 2(2a 2) + 5 = { 5a 2 18a + 13 = { (a 1)(5a 13) = So the only solutions are (1,) and ( 13 5, 16 5 ). Exercise 5 : Let a A B. If a / A then a B and hence a A c B, so definitely a (A B c ) (A c B) (A B). If a A then either a B or a / B. In the first case we have that a A B and in the second case that a A B c. In both cases we conclude that a (A B c ) (A c B) (A B). : Let a (A B c ) (A c B) (A B). If a A B c then a A and hence a A B. If a A c B then a B and hence a A B. If a A B then a A and hence a A B.

7 Exercise 6 a) There is a bijection f : A B. b) Let f : A B and g : B C be two bijections. Then g f : A C is a bijection as well (injectivity: if g(f(a)) = g(f(a )) then f(a) = f(a ) due to injectivity of g and hence a = a due to injectivity of f; surjectivity: if c C then there is a b B such that g(b) = c due to surjectivity of g, and then there is also an a A such that f(a) = b due to surjectivity of f, so g f(a) = g(f(a)) = g(b) = c.) So A and C have the same cardinality. c) Define the function f on (, ) by f(x) = 1 x+1 for every x (, ). For every x (, ) we have that x+1 > 1 and hence that f(x) = 1 x+1 (,1). So f is indeed a function from (, ) to (,1). If x 1,x 2 (, ) are such 1 that f(x 1 ) = f(x 2 ) then x 1 +1 = 1 x 2 +1, hence x = x and therefore x 1 = x 2. So f is injective. In order to prove surjectivity of f let y (,1). Take x = 1 1 y 1. Then x (, ) and f(x) = x+1 = 1 1 = y. So f is a 1+1 y bijection from (, ) to (,1). Exercise 7 Using quotient and chain rule we get f (x) = (1 + bx4 ) e ax2 2ax e ax2 4bx 3 (1 + bx 4 ) 2 = 2xeax2 (a + abx 4 2bx 2 ) (1 + bx 4 ) 2 for every x IR. Exercise 8 Using integration by substitution (y = x 2 + 1,dy = 2xdx), we get 1 x 3 1 x dx =.5x 2 x xdx = = (y 1) dy y (.5 y.5 1 y )dy = [ 1 3 y y y] y=2 y=1 = = 1 3 (2 2).