Mathematics Grade 11 P1 National Exemplar November 2020 Memo

NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 11 NOVEMBER 2020 MATHEMATICS P1/WISKUNDE V1 MARKING GUIDELINE/NASIENRIGLYN EXEMPLAR/EKSEMPLAAR MARKS/PUNTE: 150 This marking guideline consists of 15 pages./ Hierdie nasienriglyn bestaan uit 15 bladsye.

2 MATHEMATICS P1/WISKUNDE V1 (EC/NOVEMBER 2020) NOTE/LET WEL:  If a candidate answers a question TWICE, mark the FIRST attempt ONLY. Indien ʼn kandidaat ʼn vraag TWEE keer beantwoord, merk SLEGS die EERSTE poging.  Consistent accuracy applies in ALL aspects of the marking guideline. Volgehoue akkuraatheid geld deurgaans in ALLE aspekte van die nasienriglyn.  If a candidate crossed out an attempt of a question and did not redo the question, mark the crossed-out attempt. Indien ʼn kandidaat ʼn poging vir ʼn vraag deurgetrek het en nie die vraag weer beantwoord het nie, merk die poging wat deurgetrek is.  The mark for substitution is awarded for substitution into the correct formula. Die punt vir substitusie word toegeken vir substitusie in die korrekte formule. QUESTION 1/VRAAG 1 1.1.1 (3x  2)( x  5)  0 2   answers / antwoorde  x   or / of x  5 (2) 3 1.1.2 3x  5 x  1  0 2 b  b 2  4ac x  2a (5)  (5) 2  4(3)(1)  substitution / vervanging  2(3) 5  37    answers / antwoorde 6 (3)   0,18 or / of 1,85 1.1.3 x  4 x2 x2  4 x  squaring both sides/ ( x  2)  (4  x) 2 kwadreer beide kante x  2  16  8 x  x 2  standard form / standaardvorm x 2  9 x  18  0  factors / faktore ( x  6)( x  3)  0  both x-values / beide x-waardes  selection / keuse  x  3 or / of x6 (5) 1.1.4 2x2  5x  3  standard form / standaardvorm 2x2  5x  3  0  factors / faktore (2 x  1)( x  3)  0 -3 1 1 3  x    3  x  2 2 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/NOVEMBER 2020) MATHEMATICS P1/WISKUNDE V1 3 1.2.1 2 2  8  2  8 3  3 8    3        27   27   27  2 2   3  answer / antwoord 4 (2)  OR/OF 9 2 OR/OF  2 3  3 2      2 3  3  3       3   2 2   3  answer / antwoord 4  (2) 9 1.2.2  12  2 3 1   2 3  2 3 1  2 3  4.3  2 3  2 3  2  2.3  2 3  2 3  2  62  answer / antwoord 4 (3) OR/OF OR/OF  36  12  2  3  1  36  12  2 3  2  62 3 2 3 2  answer / antwoord 62 3 2 3 2 (3) 4 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

4 MATHEMATICS P1/WISKUNDE V1 (EC/NOVEMBER 2020) 1.3 5y  x  2 (1) x 2  3 xy  4 y  4 (2)  x  5x  2 x  5y  2 (3) Subst. / Vervang (3) into/in (2) :  substitution / vervanging  (5 y  2)  3 y (5 y  2)  4 y  4  0 2  standard form / standaardvorm 25 y 2  20 y  4  15 y 2  6 y  4 y  4  0  method;factors / metode;faktore 10 y 2  10 y  0  both x-values / beide x-waardes 10 y ( y  1)  0 y 0 or / of y 1 x  5y  2  both y-values / beide y-waardes x  5 0  2 or / of x  5 1  2 (6)  x  2 or / of x3 OR / OF  y  15  x  2  5y  x  2 1 x 2  3 xy  4 y  4 (2)  substitution / vervanging x2 1 y   x  2  3 5 5 Subst./Vervang  3 into/in  2  ,  standard form / standaardvorm x  3 x  ( x  2)   4  ( x  2)   4 2 1 5 1 5  factors / faktore x 2  53 x( x  2)  54 ( x  2)  4  both y-values / beide y-waardes 5 x 2  3 x 2  6 x  4 x  8  20  0 2 x 2  2 x  12  0  both x-values / beide x-waardes (6) x2  x  6  0 ( x  3)( x  2)  0  x  2 or / of x3 y  15  2  2  or / of y  15 (3  2) y 0 or / of y 1 1.4.1 Perimeter/Omtrek  2l  2b 280  2(2 x)  2 y  substitution / vervanging 2 y  280  4 x  simplification / vereenvoudiging  y  140  2 x Area/Oppervlakte  lb  A  2x(140  2x)  2x  y  2 x(140  2 x)  280 x  4 x 2 (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/NOVEMBER 2020) MATHEMATICS P1/WISKUNDE V1 5 1.4.2 A  280 x  4 x 2   4( x 2  70 x)   4( x 2  70 x  1225  1225)  completing the square / vierkantsvoltooiing   4[( x  35) 2  1225]   4( x  35) 2  4900  4900  The maximum area is 4900 m 2  correct conclusion / korrekte gevolgtrekking Die maksimum oppervlakte is 4900 m 2 (3) OR/OF OR/OF b x 2a  method/metode  280   35 m 2(4)  35 m  answer / antwoord  A  280 (35)  4(35) (3)  4 900 m 2 [31] QUESTION 2/VRAAG 2 2.1 3.2 x 1  2.4 x 3.2 x.2  2.22 x  3.2x.2  2.22 x  3.2 x  22 x 3.2 x  22 x  factorisation of numerator / 2.2 x (3  2 x ) faktorisering van teller  x 2 (3  2 x )  factorisation of denominator / 2 faktorisering van noemer  answer / antwoord (4) 2 2.2.1 5 x  205 2 x5  4 2  x5  4 x   4   4 5 2 2 5 5 2 2 5  x5 2   5 x  22 2  25  32  answer / antwoord (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

6 MATHEMATICS P1/WISKUNDE V1 (EC/NOVEMBER 2020) 2.2.2 12 x.3x 1  648 12 x.3x.3  648  12x.3x  216 12 x.3x  216 12.3  216 x 36 x  216  62 x  63 6 2 x  63  equating exponents / 2x  3 gelykstel van eksponente 3  answer / antwoord x 2 (4) 2.3 3x  2 f ( x)  x  10 x  25 2  x2  10x  25  0 for undefined / f is undefined when : f is ongedefinieerd wanneer vir ongedefinieerd x  10 x  25  0 2 ( x  5) 2  0 x  5  x  5  f is defined for : x  , but x  5  x  ℝ  x  5 f is gedefinieerd vir : x  , maar x  5 (4) [15] QUESTION 3/VRAAG 3 3.1.1 9 ; 5 ; 1; …;−143 Tn  13  4n  13  −4n (2) 3.1.2 Tn  13  4n T7  13  4(7)  substitution / vervanging  15  −15 (2) 3.1.3 Tn  13  4n 143  13  4n  substitution / vervanging 156   4n  n  39  answer / antwoord (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/NOVEMBER 2020) MATHEMATICS P1/WISKUNDE V1 7 3.2 Tn  an  b 16a  b  38 1  16a  b  38 and/en 41a  b  113 41a  b  113  2 25a  75  method / metode a  3 16  3  b  38  Tn  3n  10 48  b  38  b  10 Tn  3n  10 n=3  1  3n  10 9  3n n  3 (4) [10] QUESTION 4/VRAAG 4 4.1 −12 ; −8 ; 0 ; 12 4 8 12 4 4  28  48 28 ; 48 (2) 4.2 2a  4 3a  b  4 a  b  c  12 a  2 3(2)  b  4 2  2  c  12  a2  b  2 b  2 c  12  c  12  Tn  2n 2  2n  12 Tn  2n 2  2n  12 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

8 MATHEMATICS P1/WISKUNDE V1 (EC/NOVEMBER 2020) 4.3 For first differences: / Vir eerste verskille 4;8;12; ...  Tn  4n Tn  4n  n  48 192  4n  answer / antwoord  n  48 (3) 192 lies between T48 and T49 192 lê tussen T48 en T49 OR/OF OR/OF Tn 1  Tn  192  2(n  1) 2  2(n  1)  12  (2n 2  2n  12)  192 2(n  1) 2  2(n  1)  12  (2n 2  2n  12)  192  4n = 192 2(n 2  2n  1)  2n  2  12  2n 2  2n  12  192 2n 2  4n  2  2n  2  12  2n 2  2n  12  192  4n  192  answer / antwoord (3) n  48 192 lies between T48 and T49 192 lê tussen T48 en T49 4.4 Pn  0  Tn  168  0 Tn  168  2n2  2n 12  168  standard form / standaardvorm 2n 2  2n  12  168 2n 2  2n  180  0  factorisation / faktorisering  9  n  10 n 2  n  90  0 -9 10 (n  10)(n  9)  0  answer / antwoord 9  n  10 (5) but / maar : n  0  Number of terms  9 Aantal terme  9 4.5 Tn  2n 2  2n  12  Tn  2(n 2  n  6)  2(n 2  n  6)  explanation / verduideliking  2  any n  0 is always even, so Tn will always be even 2  enige n  0 is altyd ewe, so Tn sal altyd ewe wees. (2) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/NOVEMBER 2020) MATHEMATICS P1/WISKUNDE V1 9 QUESTION 5/VRAAG 5 5.1 b At TP/By Draaipunt: x    method / metode 2a 6  2(1) 3  x-coordinate / x-koördinaat  y  (3) 2  6(3)  7  y-coordinate / y-koördinaat  16 (3) OR/OF f ( x)   x 2  6 x  7  ( x 2  6 x  7)  completing the square /  [( x 2  6 x  9)  9  7] vierkantsvoltooiing  [( x  3) 2  16]  ( x  3) 2  16  x-coordinate / x-koördinaat  y-coordinate / y-koördinaat  Turning point / Draaipunt :  3;16  (3) 5.2  x2  6 x  7  0  f ( x)  0 x2  6 x  7  0 ( x  7)( x  1)  0  x  7 or / of x  1  answers / antwoorde (2) 5.3  y-intercept / y-afsnit  x-intercepts / x-afsnitte  turning point / draaipunt  shape / vorm (4) 5.4 x3 answer / antwoord (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

10 MATHEMATICS P1/WISKUNDE V1 (EC/NOVEMBER 2020) 5.5 f ( x)   x 2  6 x  7 f (3)  (3) 2  6(3)  7  f (3)  20  20 f (1)  (1) 2  6(1)  7  f (1)  12  12 12  (20)  substituting into gradient formula / Average/Gemiddelde m  vervanging in gradiënt-formule 1  (3) 32  4 8  answer / antwoord (4) 5.6 f ( x)   x  6 x  7 2  ( x  3) 2  16  a  1 and/en q  16  h( x)   ( x  3)  4   16 2  p 1  ( x  1) 2  16 (2) [16] QUESTION 6/VRAAG 6 6.1 a f ( x)  q x p a  1  p  3  q 1 x3 (2) 6.2 a f ( x)  1 x3 a 6 1  substitution / vervanging 2  3 a  5  answer / antwoord (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/NOVEMBER 2020) MATHEMATICS P1/WISKUNDE V1 11 6.3 5 f ( x)  1 x3 5  y0 0 1 x3 5 1  x3  x  8  x3  5 x  8  A(8;0) (2) 6.4 A(8;0) and/en D(3;1) y y mAD  2 1 x2  x1 1 0  substitution / vervanging  3  (8) 1  mAD b  5 g ( x)  bx  c 1  substitution / vervanging  xc 5 1 1 0   8   c or / of 1  3  c 5 5 8  equation / vergelyking c  5 1 8 g ( x)  x  5 5 (4) 6.5 x  ( ; ) but / maar x  3  x  ( ; )  x  3 OR / OF (2) x  ℝ, x  3  x  ℝ  x  3 (2) 6.6 f ( x)  g ( x) 5 1 8 5 1 8 1  x   1  x  x3 5 5 x3 5 5 25  5( x  3)  x( x  3)  8( x  3) 25  5 x  15  x 2  3 x  8 x  24  standard form / standaardvorm x  6 x  16  0 2  x-values / x-waardes ( x  8)( x  2)  0 x  8 or x  2 5 y 1  coordinates / koördinate 23 2  B (2; 2) (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

12 MATHEMATICS P1/WISKUNDE V1 (EC/NOVEMBER 2020) 6.7 8  x  3 or x  0  8  x  3  x  0 OR / OF OR / OF x [8;  3)  [0; ) 𝑥 ∈ [[−8; −3)   [0; ∞)]  (3) [19] QUESTION 7/VRAAG 7 7.1.1 y  4  answer / antwoord (1) 7.1.2 y  (4; )  answer / antwoord (1) OR / OF OR / OF  answer / antwoord (1) y  4 7.1.3 x 1 g ( x)     4 2 0 1  y-value / y-waarde y   4 2  substitution / vervanging  3 x 1 0  4 2   x 4  21  answer / antwoord 22  2 x (3) 2  x x  2 Intercepts / Afsnitte :(0;  3) and / en ( 2;0) 7.1.4 x  2  answer / antwoord (1) 7.2 y  asymptote above x-axis q asimptoot bo x-as  y-intercept positive y-afsnit positief  shape / vorm x h (3) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/NOVEMBER 2020) MATHEMATICS P1/WISKUNDE V1 13 QUESTION 8/VRAAG 8 8.1  i  m ieff  1  nom   1  formula / formule  m  12  0.095   substitution / vervanging  1   1  12   0,099247 ...  r  9,92 %  answer / antwoord (3) 8.2 A  P (1  i ) n R 764 050 ,60  P 1  0,08   A  R764050,60 5 P 764050 .60  substitution / vervanging 1  0.08 5  R 520 000  answer / antwoord (3) 8.3.1   0,075  48  0,11 12 A  28 000 1    R 7 300  1   48   12   4   0,075   28 0001   12  12   0,11   R 37 760 ,78  R 7 300 1    −R7 300  4  12  0,11    1  12  0,11    R 30 460 ,781    4   4   simplification / vereenvoudiging  R 42 181,59  answer / antwoord (5) 8.3.2 A  P(1  i ) n 60  0, 08  A  R42181,59 1    substitution into correct formula  12  vervanging in korrekte formule  R62844, 06 R80 000  R62844, 06  R62 844,06  R17155,94  A  P(1  i ) n  R17 155,964 60  0, 08  R17155,94  P 1    12   method / metode 17155,94 P 60  0, 08  1    12   R11515, 25  He needs to deposit R11 515 / Hy moet R11 515 deponeer  answer / antwoord (5) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

14 MATHEMATICS P1/WISKUNDE V1 (EC/NOVEMBER 2020) QUESTION 9/VRAAG 9 9.1 If A and B are independent, then: As A en B onafhanklik is, dan: P(A and/en B) = P(A)  P(B) P(A) = 1 P(not/nie A) = 1 0,45  0,55 = 0,55 11 = 20 P(A or/of B) = P(A) + P(B) – P(A and/en B) 0,685 = 0,55 + 0,3 – P(A and/en B)  substitution / vervanging ∴P(A and/en B) = 0,165 33 =  answer / antwoord 200 P(A)  P(B) = 0,55  0,3  P(A)  P(B) = 0,165 33 = 200 ∴ A and B are independent events. /  conclusion / gevolgtrekking A en B is onafhanklike gebeurtenisse. (5) 9.2.1 S N R 9 5 c (8)  a = 10 12 a (10) b (13)  b =13 c=8 4  d =14 d (14) A a = 10 ; b = 13 ; c = 8 ; d = 14 (4) 9.2.2 39 5 P  A or/of (N and/en R)   39 5    75 75 75 75 44   answer / antwoord 75 (3)  0,59 [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/NOVEMBER 2020) MATHEMATICS P1/WISKUNDE V1 15 QUESTION 10/VRAAG 10 10 Let the total number of balls be t. Laat die totale aantal balle t wees. 5 4  and/en t t 1 Green/Groen: 5 G 4 Blue/Blou: t – 5 t 1 5 G t B G B B P(GG) = P(G)  P(G)  equation / vergelyking 5 4 5   t t  1 18 20 5  standard form / standaardvorm  t (t  1) 18 5t 2  5t  360 5t 2  5t  360  0  factorisation / faktorisering t 2  t  72  0 t=9 (t  9)(t  8)  0  t  9 or t  8  There are 9 balls. Daar is 9 balle. (6) TOTAL / TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief