Mathematics Grade 11 Paper 1 Eastern Cape November 2013 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 11 NOVEMBER 2013 MATHEMATICS P1/WISKUNDE V1 MEMORANDUM MARKS/PUNTE: 150 This memorandum consists of 9 pages. Hierdie memorandum bestaan uit 9 bladsye.

Downloaded from hlayiso.com 2 MATHEMATICS P1 (NOVEMBER 2013) QUESTION/VRAAG 1 1.1  standard form standaard vorm  factorisation ( )( ) faktorisering or/of  values of x (3) waardes van x 1.2 1.2.1  formula √ formule (1) 1.2.2 ( )  standard form standaard vorm √( ) ( )( )  sub into formula vervanging in ( ) formule √  values of x waardes van x √ or/of (4) 1.3 1.3.1  simplification ( ) =( ) = or/of 27x-3 vereenvoudiging  answer (2) antwoord 1.3.2 ( √ )( √ )  simplification = 4(9 – 5) vereenvoudiging = 16  answer (2) antwoord [12] QUESTION/VRAAG 2 2.1 2.1.1  cross-multiplication kruisver- menigvuldig  simplification ( ) ( ) ( ) vereenvoudiging  answer (3) antwoord 2.1.2 If roots are real, then/ As wortels reëel is, dan  statement: Δ≥0 is: 0  substitute in ( ) ( )( ) vervang in ( )  simplification vereenvoudiging  standard form standaard vorm ( )( )  factorisation faktorisering (5) 2.2 √ √ √ √  simplification √ √ vereenvoudiging  answer (2) antwoord

Downloaded from hlayiso.com (NOVEMBER 2013) WISKUNDE V1 3 2.3 =  express in exponential form = druk uit in = 30.20 eksponent-vorm =1  simplification vereenvoudiging (3)  answer/ antwoord 2.4 ( )( )  simplification vereenvoudiging  factorisation ( )( ) -3 2 faktorisering 2  or/of (4)  2.5  y the subject y die onderwerp  substituting Substitute into/Stel in = vervanging  simplification of ( ) ( ) exponents  vereenvoudiging van eksponente  value of x waarde van x and/en ( )  value of y and/en (6) waarde van y [23] QUESTION/VRAAG 3 3.1.1 x=3  answer 3.1 (2) antwoord 3.1.2 When/Wanneer ≥ ,  answer (2) antwoord 3.2 3.2.1 √   and/en  0  (5)  3.2.2 √  square both sides kwadreer beide ( ) kante  simplification vereenvoudiging  standard form standaard vorm  exclusion of -22 ( )( ) uitlating van -22 not valid/n.v.t. or/of  solution Since/Aangesien oplossing Solution/Oplossing (5) [14]

Downloaded from hlayiso.com 4 MATHEMATICS P1 (NOVEMBER 2013) QUESTION/VRAAG 4 4.1 ( )  formula formule ( )  substituting vervanging  answer (3) antwoord 4.2 4.2.1 per month/per maand/0,0125  answer/antwoord (1) 4.2.2 ( ) -1  correct formula korrekte formule (1 + 0,0125)12 -1  substituting vervanging  simplification vereenvoudiging (4)  answer/antwoord 4.2.3 ( ) ( )  substituting into ( ) correct formula vervanging in korrekte formule  values of i and n waardes van i en n (3)  answer/antwoord 4.3 ( ) ( )  formula formule  sub/verv ( ) ( ) ( )  sub/verv ( )  answer/antwoord Or/Of T0_____________T4__________T7 A= ( ) ( ) = R1 112 303,584 + R560 000 = R1 672 303,584 = R2 836 028,60 (5) [16] QUESTION/VRAAG 5 5.1 R4 000 (1)  answer/antwoord 5.2 A = P(1 – in)  value of i ( ) waarde van i 27  depreciation interest the rate of straight-line depreciation is/die rate/waardevermin- reguitlynwaardeverminderingskoers is 8,6% (2) dering rentekoers 5.3 A = P(1-i)n  value of i ( ) waarde van i  depreciation interest the rate of reducing-balance depreciation is/die rate/waardevermin- balansvermideringkoers 14,4% (2) dering [5]

Downloaded from hlayiso.com (NOVEMBER 2013) WISKUNDE V1 5 QUESTION/VRAAG 6 6.1 Figure number 1 2 3 4 6 11  Number of shaded tiles 4 16 36 64 144 484 1 mark for each column No of white tiles 1 9 25 49 121 441 completed Total number of tiles 5 25 61 113 265 925 (5) 6.2 5 25 61 113  set up of equation opstel van verg. 20 36 52  simplification vereenvoudiging  formula 16 16 formule 2a = 16 3a + b = 20 a + b + c = 5 a=8 b = 20 – 24 c = 5 – 4  second constant b = -4 c=1 difference Tn = 8n2 – 4n + 1 tweede konstante verskil Or/Of  values of a, b and c waardes van a, b en ( ) ( ) c  formula (3) formule [8] QUESTION/VRAAG 7 7.1 29 (1)  answer/antwoord 7.2      (1)   (2) Solving (1) and (2) simultaneously. Los (1) en (2) gelyktydig op. (1) x 2 (2) (7) Or/Of

Downloaded from hlayiso.com 6 MATHEMATICS P1 (NOVEMBER 2013) (1)  Substitution into T1 (2) vervanging in T1 (3) Substitution into T2 (2) - (1) (4) vervanging in T2 (3) - (2) (5) Substitution into T3 (5) - (4) vervanging in T3  value of a  value of b waarde van b  value of c waarde van c Or/Of  Tn 1 5 11 19 29 4 6 8 10 2 2 2  Constant second 2a = 2 3a + b = 4 a+b+c=1 difference/Konstante a = 1 3+b=4 1+1+c=1 tweede verskil b=1 c = -1  value of a waarde van a  value of b waarde van b  value of c waarde van c (7)  Tn 7.3 or/of ( )  substitution into Tn vervanging in Tn  answer (3) antwoord [11] QUESTION/VRAAG 8 8.1 y = -1 (1)  y = -1 8.2 y-intercept: x = 0  value of y waarde van y  coordinate koördinaat (0 ; 1) (2) 8.3 x = 1: y = 2.31 – 1= 5  x-coordinate (1 ; 5) x-koördinaat  y-coordinate (2) y-koördinaat

Downloaded from hlayiso.com (NOVEMBER 2013) WISKUNDE V1 7 8.4  shape 5 (1 ; 5) vorm  y-intercept f y-afsnit 1  y-asymptote 0 x y-asimptote 1 -1 (3) 8.5 (1)  8.6 x=3  x-asymtote/ y=0 x-asimptote  y-asymtote/ (2) y-asimptote 8.7 ( ) =  g(0)  coordinate ( ) (2) koördinaat 8.8  –3 (2)  –3 8.9  asymptote y asymptote  y-intercept y-afsnit  one mark for each 3 branch een punt vir elke tak x -3 0 g -3 (4) 8.10 ( ) ( )  formula formule =  substitution vervanging = -1  answer (3) antwoord [22]

Downloaded from hlayiso.com 8 MATHEMATICS P1 (NOVEMBER 2013) QUESTION/VRAAG 9 9.1 9.1.1 ( ( )  factors ( )( ) ( ) faktore At (0 ; -12): -12 = a(-6)  simplification a = 2 a =2 vereenvoudiging  value of a waarde van a  equation (4) vergelyking 9.1.2 ( )  factorisation = ( ) faktorisering  completion of the ( ) square voltooiing van die = 2[( ) ] kwadraat 2( )  simplification (3) vereenvoudiging 9.2 9.2.1 ( ) ( )  factorisation ( )( ) faktorisering C(0 ; 12) and/en D(4 ; 0)  C-coordinate C-koördinaat  D-coordinate (3) D-koördinaat 9.2.2 m = -3 and/en c = 12  m and/en C ( ) (2)  g(x)=-3x+12 9.2.3 OB = or/of  ( ) ( ) ( )   length of AE ( ) lengte van AE ( ) ( ) AE = AB – EB = AE = ( ) (3) 9.2.4  (1) 9.2.5  (1) [17] QUESTION/VRAAG 10 10.1 f(x) = (x – 3)(x + 1) = x2 -2x – 3  x2  -2x (3)  -3 [3]

Downloaded from hlayiso.com (NOVEMBER 2013) WISKUNDE V1 9 QUESTION/VRAAG 11 11.1 P(M) x P(N) =  P(M) correct/korrek  P(N) correct/korrek  P(M and N) correct P(M and N) = = 0,62 P(M en N) korrek  correct rule used P(M) x P(N) P(M and/en N) korrekte reël M and N are not independent events gebruik M en N is nie onafhanklike gebeurtenisse (4) [4] QUESTION/VRAAG 12 12.1 12.1.1 A and/en B are independent/is onafhanklik Venn diagram (given/gegee)  P(A and/en B) = 0,2 P(A and/en B) = P(A) x P(B)  P(A and not B) = = 0,4 x 0,5 0,2 = 0,2 S P(A en nie B) = 2  P(B and not A) = A B 0,3 P(B en nie A) = 0,3  P(A or/of B) = 0,3 0,2 0,2 0,3 P(A or/of B) = 0,2 + 0,2 + 0,3 = 0,7 (4) 12.1.2 P(neither/nie A nor/of B) = 1 – (0,2 + 0,2 + 0,3)  answer correct = 0,3 (1) antwoord korrek 12.2 12.2.1  tree diagram shows Y YY different outcomes Y  Do you N YN each probability own a correct a cellphone elke moontlikheid Y NY Besit jy ʼn korrek N selfoon? NN N (7) 12.2.2 P(one will own a cellphone and the other not)  first product/eerste P(een sal ʼn selfoon besit en die ander nie) produk  second product/tweede produk  answer correct (3) antwoord korrek [15] TOTAL/TOTAAL: 150