Mathematics Grade 11 P1 Eastern Cape November 2019 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 11 NOVEMBER 2019 MATHEMATICS P1/WISKUNDE V1 MARKING GUIDELINE/NASIENRIGLYN EXEMPLAR/EKSEMPLAAR MARKS/PUNTE: 150 This marking guideline consists of 13 pages./ Hierdie nasienriglyn bestaan uit 13 bladsye.

Downloaded from hlayiso.com 2 MATHEMATICS P1/WISKUNDE V1 (EC/NOVEMBER 2019) 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 is in ALLE aspekte van die nasienriglyn van toepassing. • 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 x2 − 5x − 6 = 0 ( x − 6)( x + 1) = 0  method / metode  x = 6 or / of x = −1   answers / antwoorde (3) 1.1.2 (2 x − 3)( x + 1) = 9 2 x 2 + 2 x − 3x − 3 − 9 = 0 2 x 2 − x − 12 = 0  standard form / standaardvorm −b  b 2 − 4ac x = 2a −(−1)  (−1) 2 − 4(2)(−12) =  substitution / vervanging 2(2) 1  97 = 4 = 2, 71 or / of − 2, 21   answers / antwoorde (4) 1.1.3 x − x −1 = 3 x − 3 = x −1 ( x − 3) 2 = x − 1  squaring both sides kwadreer beide kante x2 − 6 x + 9 = x −1 x 2 − 7 x + 10 = 0  standard form / standaardvorm ( x − 2)( x − 5) = 0  factors / faktore x  2 or / of x=5  both x values / beide x-waardes  selection / keuse (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/NOVEMBER 2019) MATHEMATICS P1/WISKUNDE V1 3 1.1.4 x( x + 10)  −25 x 2 + 10 x + 25  0  standard form / standaardvorm ( x + 5) 2  0  factors / faktore -5  x  R but/maar x  −5  x  R  x  −5 (4) 1.2 2x − y = 1 (1) 3 x 2 − 4 xy + y 2 = 0 (2) y = 2x −1 (3)  y = 2x −1 Substitute (3) into (2) :Vervang (3) in (2) 3 x 2 − 4 x(2 x − 1) + (2 x − 1) 2 = 0  substitution / vervanging 3x − 8 x + 4 x + 4 x − 4 x + 1 = 0 2 2 2 − x2 + 1 = 0  standard form / standaardvorm 1= x 2 or / of x −1 = 0 2  1 = x2 or / of ( x − 1)( x + 1) = 0  method/factors : metode/faktore  x = 1 or / of x = −1  both x-values / beide x-waardes y = 2(1) − 1 or / of y = 2(−1) − 1 =1 or / of y = −3  both y-values / beide y-waardes (6) OR / OF 2x − y = 1 (1) 3 x 2 − 4 xy + y 2 = 0 (2)  x = 12 ( y + 1) x = ( y + 1) 1 2 ( 3) 3[ 12 ( y + 1)]2 − 4 y[ 12 ( y + 1)] + y 2 = 0  substitution / vervanging 2 3( y4 + 2y + 14 ) − 2 y 2 − 4 y + y 2 = 0 3y2 + 6 y + 3 − 4 y2 − 8 y = 0 − y2 − 2 y + 3 = 0  standard form / standaardvorm y + 2y −3 = 0 2 ( y − 1)( y + 3) = 0  factors / faktore  y = 1 or / of y = −3  both y-values / beide y-waardes x = (1 + 1) 1 2 or / of x = 12 (−3 + 1) x = 1 or / of x = −1  both x-values / beide x-waardes (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 4 MATHEMATICS P1/WISKUNDE V1 (EC/NOVEMBER 2019) 1.3 [ x − (−3 + 2 2)][ x − (−3 − 2 2)] = 0  equation / vergelyking [( x + 3) + 2 2][( x + 3) − 2 2]  difference of 2 squares ( x + 3) 2 − (2 2) 2 = 0 verskil tussen 2 kwadrate x2 + 6x + 9 − 8 = 0  expanding / uitbreiding x2 + 6x + 1 = 0  x2 + 6 x + 1 but y intercept is 4 / ( maar y − afsnit is 4)  f ( x) = 4 x 2 + 24 x + 4  4 x 2 + 24 x + 4 (5) [27] QUESTION 2/VRAAG 2 2.1 8n−3.32 −n+1.6 2 n = 23 ( ) .(2 ) .(2.3) n −3 5 − n +1 2n  (23 ) .(25 ) n −3 − n +1 .(2.3) 2n 9n (3 ) 2 n  (32 ) n 23n−9.2 −5 n+5.2 2 n.32 n 23n−9.2−5n+5.22 n.32 n =  32 n 32 n = 2 3 n −9 − 5 n + 5 + 2 n  simplification / vereenvoudiging = 2 −4 1 = 16  answer / antwoord (5) 2 2.2.1 x =4 3 OR / OF 3 2  2 x =4  x  = 43 3   3  method / metode   2 3 3 = (2 )  3 2 2 2 x x 2 = 43 x 2 = 64  x = 23  x 2 = 64 or/of 𝑥 = 23 x=8 x =8  answer / antwoord (3) 2.2.2 22 x − 4 x −1 = 12 22 x − 22 x −2 = 12  2 2 x − 2 2 x − 2 = 12 22 x − 22 x.2−2 = 12 22 x (1 − 14 ) = 12  22 x (1 − 14 ) = 12 22 x. 34 = 12 22 x = 16 or / of 4 x = 16 22 x = 24 or / of 4 x = 42  22 x = 24 or / of 4x = 42 2x = 4  x=2  x=2 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/NOVEMBER 2019) MATHEMATICS P1/WISKUNDE V1 5 2.3.1 g is undefined when: g is ongedefinieerd wanneer 3− x = 0 x = 3  answer / antwoord (2) 2.3.2 x − 2x + 2 2 g ( x) = 3− x = (( x − 2 x + 1) − 1 + 2)  completing the square 3− x voltooiing van kwadraat ( x − 1) + 1 2 = 3− x For g to be equal to 0, the numerator must be equal to 0, but the minimum value for the  numerator is 1 For g to be equal to 0, the numerator must be equal to 0  g will never be equal to zero. Vir g om gelyk aan 0 te wees moet die Vir g om gelyk aan 0 te wees moet die teller gelyk teller gelyk aan 0 wees aan 0 wees, maar die minimumwaarde vir die teller is 1.  minimum value of numerator = 1 minimumwaarde van teller = 1  g sal nooit gelyk aan nul wees nie. (3) OR/OF For g ( x) to be = 0, the numerator must be = 0  For g ( x) to be = 0, the numerator must be = 0 Vir g ( x) om = 0, moet die teller = 0 x2 − 2x + 2 = 0 Vir g ( x) om = 0, moet die teller = 0 but / maar :  = b 2 − 4ac = ( −2 ) − 4 (1)( 2 ) 2 = −8  −8 No solution / Geen oplossing  no solution / geen oplossing Numerator will never be equal to 0. (3) Teller sal nooit gelyk aan 0 wees nie. [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 6 MATHEMATICS P1/WISKUNDE V1 (EC/NOVEMBER 2019) QUESTION 3/VRAAG 3 3.1.1 10 ; 7 ; 4 ; 1 ; − 2 ; − 5  −2  −5 (2) 3.1.2 Tn = 13 − 3n  13  − 3n (2) 3.1.3 − 179 = 13 − 3n  equating − 179 to 13 – 3n 3n = 192 stel –179 gelyk aan 13 – 3n  n = 64  simplifying / vereenvoudiging  answer / antwoord (3) 3.2 T2 − T1 = T3 − T2 x + 9 − (3x − 2) = 2 x + 5 − ( x + 9)  substitution / vervanging x + 9 − 3x + 2 = 2 x + 5 − x − 9  simplifying / vereenvoudiging − 2 x + 11 = x − 4 − 3x = −15 x = 5 x=5 (3) [10] QUESTION 4/VRAAG 4 4.1 -128 ; - 84 ; - 48 ; -20 44 36 28 -8 -8 0 ; 12  0  12 (2) 4.2 2a = −8 3a + b = 44 a + b + c = −128 a = − 4 3(− 4) + b = 44 − 4 + 56 + c = −128  a = − 4 b = 56 c = −180  b = 56  c = −180 Tn = − 4n + 56 n − 180 2  Tn = − 4n2 + 56n − 180 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/NOVEMBER 2019) MATHEMATICS P1/WISKUNDE V1 7 4.3 Tn = − 4n 2 + 56n − 180 = −4(n 2 −14n + 45)  common factor / gemene faktor  completing the square = −4 ( n 2 − 14n + 49 − 49 + 45 ) voltooiing van vierkant = −4 ( n − 7) 2 − 4   (n − 7)2 − 4 = − 4(n − 7) 2 + 16  + 16  biggest value for Tn is16  identifying the biggest Tn value grootste waarde vir Tn is 16 identifiseer die grootste Tn waarde (5) OR / OF Tn = − 4n 2 + 56n − 180 b b  − max at/maks by : n = − 2a 2a −56  substitution / vervanging = 2.(−4)  n=7 =7  substituting n = 7 into Tn T7 = − 4(7) 2 + 56(7) − 180 vervang n = 7 in Tn =16  answer / antwoord (5) 4.4 k  −16   answer / antwoord (1 mark if/punt as k  −16 ) (2) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 8 MATHEMATICS P1/WISKUNDE V1 (EC/NOVEMBER 2019) QUESTION 5/VRAAG 5 5.1 g ( x) = −a x − 3 = −a 1  substituting − 3 and 1 vervang –3 en 1 a=3  answer / antwoord (2) 5.2 x = −3 and / en y = −1  x = −3  y = −1 (2) 5.3 y  0 ; y R  answer / antwoord (1) 5.4 −3 −1 = 0  substituting y = 0 x+3 −3 vervang y = 0 =1 x+3 −3= x+3  x = −6  x = −6 −3 −1 = y 0+3  y = −2  y = −2 (3) 5.5 f:  x and y-intercepts x- en y-afsnitte  asymptotes / asimptote  shape / vorm g:  y-intercept / y-afsnit  asymptote / asimptoot  shape / vorm (6) 5.6 y = x+k − 1 = −3 + k  substituting (−3; − 1) k = 2 vervang (–3 ; –1) y = x+2  y = x+2 (2) OR / OF y = (x − (−3) ) − 1  substituting (−3;−1) = (x + 3) − 1 vervang (–3 ; –1) y = x+2  y = x+2 (2) 5.7 b =1  answer / antwoord (2) [18] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/NOVEMBER 2019) MATHEMATICS P1/WISKUNDE V1 9 QUESTION 6/VRAAG 6 6.1 g ( x) = x + 4 0= x+4  substitution / vervanging  x = −4 P(−4;0)  answer / antwoord (2) 6.2 Q (2 ; 0)  answer / antwoord (2) 6.3 Average gradient / Gemiddelde gradiënt y2 − y1 = x2 − x1 9−0  substitution / vervanging = −1 − (−4) =3  answer / antwoord (2) 6.4 f ( x) = ax + bx + c 2 y = a ( x − x1 )( x − x2 ) = a ( x + 4)( x − 2)  substituting − 4 and 2 vervanging van – 4 en 2 9 = a (−1 + 4)(−1 − 2)  substituting / vervang (−1; 9) 9 = −9a  a = −1  a = −1  y = −( x + 4)( x − 2)  substituting −1 and expanding vervang – 1 en uitbreiding = −( x 2 + 2 x − 8) = −x2 − 2x + 8  b = −2  b = −2 c =8  c =8 (6) OR / OF y = a( x + p)2 + q y = a( x + 1) 2 + 9  substituting / vervang (−1; 9) 0 = a(− 4 + 1) 2 + 9 or / of 0 = a(2 + 1) 2 + 9  substituting / vervang (−4 ; 0) −9a = 9 or/of (2 ; 0)  a = −1  a = −1  y = −( x + 1) 2 + 9  substituting −1 and expanding vervang – 1 en uitbreiding = −( x 2 + 2 x + 1) + 9 = − x2 − 2 x − 1 + 9 = − x2 − 2 x + 8  b = −2  b = −2 c =8  c =8 (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 10 MATHEMATICS P1/WISKUNDE V1 (EC/NOVEMBER 2019) 6.5 x + 4 = −x2 − 2x + 8  g ( x) = f ( x) x 2 + 3x − 4 = 0  standard form / standaardvorm ( x + 4)( x − 1) = 0  factors / faktore  x = − 4 or / of x = 1  y = 1+ 4  choosing / kies x = 1 =5  y=5  S (1; 5) (5) 6.6 Length of TV at max / Lengte van TV by maks: = yT − yV = − x 2 − 2 x + 8 − ( x + 4)  − x 2 − 2 x + 8 − ( x + 4) = − x 2 − 3x + 4 = −( x 2 + 3x − 4)  factoring out −1 / = −( x + 3x + − − 4) 2 9 9 gemene faktor – 1 4 4 = − ( x + 32 ) 2 − 254   completing the square voltooiing van vierkant = −( x + 32 ) 2 + 254  − ( x + 32 ) 2 + 254  Length TV at max / Lengte TV by maks: = 6, 25 units / eenhede  answer / antwoord (5) OR / OF Length of TV at max / Lengte van TV by maks: = yT − yV = − x 2 − 2 x + 8 − ( x + 4)  − x 2 − 2 x + 8 − ( x + 4) = − x 2 − 3x + 4 x = − 2ba  x = − 2ba = − 2(( −−3)1) = − 23  x = − 32 Length of TV at max / Lengte van TV by maks: y = −( 32 ) 2 − 3(− 32 ) + 4  substitution / vervanging = 6, 25 units / eenhede  answer / antwoord (5) 6.7 − 4  x  0 or / of x  2  − 4  x  0  x  2 OR / OF x  (−4; 0)  (2; )  x   (−4; 0) (2; ) (Accuracy / Akkuraatheid) (3) [25] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/NOVEMBER 2019) MATHEMATICS P1/WISKUNDE V1 11 QUESTION 7/VRAAG 7 7.1 A = P (1 − i ) n  formula / formule 7 210 = P (1 − 0,134)5  n=5  substitution into correct formula vervanging in korrekte formule 7 210 = 0, 4870678 P  P = R14802,87  answer / antwoord (4) 7.2  i  n ieff = 1 + nom  − 1  n  4  0,082   n = 4  substitution/vervanging = 1 +  −1  4   0,0846  reff  8,46%  answer / antwoord (3) 7.3.1 A = P (1 + i ) n  i = 0,12103 and/en  n = 12 = R 20 000 (1 + ) 0 ,103 12 12  substitution into correct formula vervanging in korrekte formule  R 22 160,09  answer / antwoord (4) 7.3.2 Let the money that he withdrew be x / Laat die geld wat hy onttrek het x wees Balance at start of 2nd year / Balans aan die begin van 2de jaar R 22160,09 + R15 000 = R37160,09  R 22 160,09 + R15 000 ( ) (  R37160,09 1 + 0,115 24 − x   1 + 0,168 8  = R30183,64  R37160,09(1 + 0,12115)24 ( 12 )  ( 4 )  )   subtracting x / trek x af (46718,49558 − x) = 30183,64  (1+ ) 0,168 8 4 (R37160,09(1+ 240,115 12 ) )− x(1+ ) 0,168 8 4 − x = 30183,64 − 46718,49558  equating to / stel gelyk aan (1+ ) 0,168 8 4 R 30 183,64 − x = −24999,9939  x = R 25000,00  answer / antwoord (6) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com 12 MATHEMATICS P1/WISKUNDE V1 (EC/NOVEMBER 2019) QUESTION 8/VRAAG 8 8.1.1 Let Red balls be R and Yellow be Y Laat Rooi balle R en Geel balle Y wees  first branch / eerste tak  probabilities of first branch waarskynlikhede van eerste tak  second branches / tweede takke  probabilities of second branches waarskynlikhede van tweede takke  outcomes / uitkomste (5) 8.1.2 P(different colours/ verskillende kleure) = P(RY) or / of P(YR)  P(RY) + P(YR) = P(RY) + P(YR) = ( 83  75 ) + ( 85  73 )  83  75  85  73 15 = or / of 0,54  answer / antwoord 28 OR / OF (4) P(different colors) = 1 − P(same colour) P (verskillende kleure) = 1 − P(dieselfde kleur ) = 1 − ( P(RR) + P(YY) )  1 − (P(RR) + P(YY)) = 1 − ( ( 83  72 ) + ( 85  74 ) )  83  72  85  74 = 1 − 13 28 15 = or / of 0,54  answer / antwoord 28 (4) 8.2.1 Let x betheprobabilitythat bothSA&NZreachfinals Laat xdiewaarskynlikheid wees dat beideSA & NZ die finaal haal  0,06  0,35 − x & 0,5 − x  x (3) 8.2.2 0,35 − x + x + 0,5 − x + 0,06 = 1  equation/vergelyking  x = 0,09  answer/antwoord (2) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com (EC/NOVEMBER 2019) MATHEMATICS P1/WISKUNDE V1 13 QUESTION 9/VRAAG 9 9.1 a = 83 b = 240  a = 83  b = 240 (2) 9.2 If M and F are independent, then As M en F onafhanklik is, dan is: P(M)  P(F) = P(M and/en F)  correct rule / korrekte reël P(M)  P(F) = 125 500  260 500  P(M) = 125 500  P(F) = 500 260 13 = or 0,13  100 13 or / of 0,13 100 65 13 P(M and/en F) = = or / of 0,13  P(M and/en F) 500 100 = 500 65 or / of 0,13  P(M)  P(F) = P(M and/en F) and conclusion /  M and F are independent events. en gevolgtrekking M en F is onafhanklike gebeurtenisse. (4) 9.3 P(M/F or/of PS/F) = P(M/F) + P(PS/F) − P(M/F and/en PS/F)  correct rule / korrekte reël 65 114 65 114 = + −0  + 500 500 500 500 179 = or / of 0,36  answer / antwoord 500 (3) [9] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief