1 of 22 INSTRUCTIONS: INSTRUKSIES: • This paper consists of 22 numbered pages. Hierdie vraestel bestaan uit 22 genommerde bladsye. • Write all your answers on the answer sheet provided for Section A and on the question paper for Section B. Skryf al jou antwoorde op die gegewe antwoordblad vir Afdeling A en op hierdie vraestel vir Afdeling B. • Use the left handside of the page for your rough work. Gebruik die linkerkant van die bladsy vir jou rofwerk. • No book or any written material may be brought into the examination room. Geen boek of geskrif van enige aard mag in die toetslokaal ingebring word nie. • No part of this answer sheet may be removed from the examination room. Geen deel van hierdie antwoordstel mag uitgeskeur word nie. • No calculators are allowed. Geen sakrekenaars word toegelaat nie. • Both the paper and the answer sheet must be handed in. Beide die vraestel en die antwoordblad moet ingehandig word. 2 of 22 Rofwerk / Rough work . ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... 3 of 22 Question 1 � � �� x+1 d sin = dx x+2 � �� � x+1 1 (a) cos x+2 (x + 1)(x + 2) Vraag 1 � � �� x+1 d sin = dx x+2 � �� � x+1 1 (a) cos x+2 (x + 1)(x + 2) (b) cos (1) � � x+1 (c) cos x+2 � �� � x+1 2x + 3 (d) cos x+2 (x + 2)2 �� � � 1 x+1 (e) cos x+2 (x + 2)2 (b) cos (1) � � x+1 (c) cos x+2 � �� � x+1 2x + 3 (d) cos x+2 (x + 2)2 �� � � 1 x+1 (e) cos x+2 (x + 2)2 Question 2 Vraag 2 If f is the defined by f (x) = x cos(πx), � � function 1 � = then f 4 1 � π� (a) √ 1 − 4 2 Indien f die funksie gedefinieer � deur � 1 � f (x) = x cos(πx) is, dan is f = 4 1 � π� (a) √ 1 − 4 2 (b) 0 (b) 0 (c) 1 + π (c) 1 + π (d) 3 √ 4 2 (d) 3 √ 4 2 (e) 5 √ 4 2 (e) 5 √ 4 2 Question 3 π� 3π If y sec(x) + x tan(y) = , then y � at π, 4 4 given by: 1 1 − 2π 1 + π4 (b) √ 2π − 1 √ 4 + 2π (c) 4+π √ 1 − 2π (d) 1 + π4 1 (e) √ 2π [3] Vraag 3 � (a) [3] is 3π , dan word y � by Indien y sec(x) + x tan(y) = 4 � π� gegee deur: π, 4 (a) 1 1 − 2π 1 + π4 (b) √ 2π − 1 √ 4 + 2π (c) 4+π √ 1 − 2π (d) 1 + π4 1 (e) √ 2π 4 of 22 [4] Rofwerk / Rough work . ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... 5 of 22 Question 4 9 � � � 10 −k 2 is equal to: k k=0 (a) � 3 �10 2 (b) 3 � �10 (c) 32 � �9 (d) 32 − � 1 �10 2 Vraag 4 9 � � � 10 −k 2 is gelyk aan: k (a) � 3 �10 2 − � 1 �10 2 (e) 0 (b) 3 � �10 (c) 32 � �9 (d) 32 Question 5 x lim is equal to: x→0 tan(2x) Vraag 5 x is gelyk aan: lim x→0 tan(2x) (e) 0 (a) 1 2 (a) 1 2 (b) 1 4 (b) 1 4 (c) − 1 2 [3] k=0 (c) − 1 2 (d) −1 (d) −1 (e) 1 (e) 1 6 of 22 [3] Rofwerk / Rough work . ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... 7 of 22 Question 6 Vraag 6 If f is a function such that √ x−2−2 = f � (6), lim x→6 x−6 Indien f ’n funksie is sodat √ x−2−2 = f � (6), lim x→6 x−6 then f (x) is equal to: √ (a) x √ x−2−2 (b) x−6 √ x−2 (c) x−4 √ x−2 (d) x √ (e) x − 2 dan is f (x) gelyk aan: √ (a) x √ x−2−2 (b) x−6 √ x−2 (c) x−4 √ x−2 (d) x √ (e) x − 2 (f) None of these functions. (f ) Geeneen van die funksies nie. Question 7 Vraag 7 The absolute maximum of f defined by Die absolute maksimum van f gedefinieer deur f (x) = sin(x) + cos(x) f (x) = sin(x) + cos(x) on [0, π] is: op [0, π] is: (a) 1 (a) 1 (b) 2 √ (c) 2 √ (d) 2 2 (b) 2 √ (c) 2 √ (d) 2 2 1 (e) √ 2 1 (e) √ 2 8 of 22 [3] [4] Rofwerk / Rough work . ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... 9 of 22 Question 8 Vraag 8 If lim f (x) = ∞, then f (x) could equal to: Indien lim f (x) = ∞, dan kan f (x) gelyk wees x→2 aan x→2 (a) 1 x−2 (b) ln |x − 2| (c) | ln(x − 2)| (d) 1 x2 (a) 1 x−2 (b) ln |x − 2| (c) | ln(x − 2)| (d) 1 x2 � π� (e) tan2 πx − 2 � π� (e) tan2 πx − 2 wees. Question 9 � 3 (3x + 1)2 dx is equal to: Vraag 9 � 3 (3x + 1)2 dx is gelyk aan: (a) 95 (a) 95 (b) 104 (b) 104 (c) 65 (c) 65 (d) 39 (d) 39 (e) 99 (e) 99 1 [3] 1 10 of 22 [3] Rofwerk / Rough work . ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... 11 of 22 Question 10 � e2 ln(x) dx is equal to: x e Vraag 10 � e2 ln(x) dx is gelyk aan: x e (a) 3 2 (a) 3 2 (b) 1 2 (b) 1 2 (c) 5 2 (c) 5 2 (d) 1 (d) 1 (e) 2 (e) 2 Question 11 Vraag 11 Use Newton’s Method to find the second approximation x2 of a solution of x5 + x3 + 3x = 0 starting from x1 = 1. Gebruik Newton se Metode om die tweede approksimasie (benadering) x2 van ’n oplossing van x5 + x3 + 3x = 0 met beginpunt x1 = 1 te bepaal. Write a+b on the answer sheet where x2 = ab in lowest terms, with a and b positive integers (that is, with a and b positive integers with largest common factor 1). Skryf a+b op die antwoordblad waar x2 = ab in eenvoudigste vorm, met a en b positiewe heelgetalle (dit wil sê, met a en b positiewe heelgetalle met die grootste gemene deler 1.) 12 of 22 [3] [3] Rofwerk / Rough work . ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... 13 of 22 Section B Afdeling B Answer the questions in this section in full, showing all your working, on the paper. Antwoord die vrae in hierdie afdeling volledig, wys al jou bewerkings op die papier. Question 12 � x2 Let f (x) = sin(t2 ) dt. Vraag 12 Gestel f (x) = 4 � x2 sin(t2 ) dt. 4 (a) Find f (2). (a) Bepaal f (2). [1] (b) Find f � (x). (b) Bepaal f � (x). [3] (c) Explain what result(s) you used and why it was (they were) applicable: (c) Verduidelik watter resultaat (resultate) jy gebruik het en hoekom dit van toepassing was. [2] 14 of 22 Rofwerk / Rough work . ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... 15 of 22 Question 13 Vraag 13 Let f be a function defined by Gestel f is ’n funksie gedefinieer deur 2 5 2 f (x) = x 3 − x 3 . Dit word gegee dat 2 − 5x 3x 5 f (x) = x 3 − x 3 . Is it given that f � (x) = [10] 1 3 and f �� (x) = − 10x + 2 9x 4 3 . Complete the missing information in the table and sketch the graph of y = f (x). f � (x) = 2 − 5x 3x 1 3 en f �� (x) = − 10x + 2 4 9x 3 . Voltooi die uitgelate inligting in die tabel en skets die grafiek van y = f (x). Domain / Definisieversameling x-intercept(s) / x-snypunte y-intercept / y-snypunt Vertical asymptote / Vertikale asimptote none / geen Horizontal asymptote /Horisontale asimptote none / geen Increasing on / Toenemend op Decreasing on / Afnemend op Local maxima / Lokale maxima Local minima / Lokale minima Concave up on / Konkaaf op op Concave down on / Konkaaf af op Inflection points / Infleksiepunte 16 of 22 Rofwerk / Rough work . ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... 17 of 22 Question 14 Vraag 14 Consider the following theorem: Theorem 1 Let a and b be real numbers such that a < b and let f and g be functions. If Beskou die volgende stelling: Stelling 1 Laat a en b reële getalle wees sodaning dat a < b, en laat f en g funksies wees. As (i) f and g are continuous on [a, b], (ii) f and g are differentiable on (a, b), and (i) f en g kontinue op [a, b] is, (ii) f en g differensieerbaar op (a, b) is, en (iii) g(a) �= g(b), (iii) g(a) �= g(b) is, then dan is g � (c)(f (b) − f (a) = f � (c)(g(b) − g(a)) g � (c)(f (b) − f (a) = f � (c)(g(b) − g(a)) for some c in (a, b). vir sommige c in (a, b). (a) Find r such that, if h is defined by h(x) = f (x) − rg(x), then h(a) = h(b). (a) Bepaal r sodanig dat, as h gedefineer deur h(x) = f (x) − rg(x) is, dan is h(a) = h(b). [1] (b) Use (a) and Rolle’s theorem to prove Theorem 1. (b) Gebruik (a) en Rolle’s stelling om Stelling 1 te bewys. [3] 18 of 22 Rofwerk / Rough work . ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... 19 of 22 [1] (c) Use Theorem 1 where f (x) = sin(x) and 8 cos(c) = 2 for g(x) = 12 x2 to prove that c π � π� some c ∈ 0, . 2 (c) Gebruik Stelling 1 waar f (x) = sin(x) en 8 cos(c) = 2 g(x) = 12 x2 om te bewys dat c π � π� vir sommige c ∈ 0, . 2 20 of 22 Rofwerk / Rough work . ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... ....................................................................................................... 21 of 22 Question 15 Vraag 15 Prove that that if f is a function such that f �� exists and is negative for all x ∈ R, then f can’t lie entirely above the x-axis. Bewys dat indien f ’n funksie is sodat f �� bestaan en negatief is vir alle x ∈ R, dan kan f nie heeltemal bo die x-as lê nie. [4] Total : 60 22 of 22
