Mathematics Grade 12 Paper 1 EC June 2018 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 JUNE/JUNIE 2018 MATHEMATICS P1/WISKUNDE V1 MARKING GUIDELINE/NASIENRIGLYN MARKS/PUNTE: 150 This marking guideline consists of 15 pages./ Hierdie nasienriglyn bestaan uit 15 bladsye.

Downloaded from hlayiso.com 2 MATHEMATICS P1/WISKUNDE V1 (EC/JUNE 2018) NOTE/LET OP: • If a candidate answered a question TWICE, mark the FIRST attempt ONLY. Indien ʼn kandidaat ʼn vraag TWEE keer beantwoord het, merk SLEGS die EERSTE poging. • Consistent accuracy(CA) applies in ALL aspects of the memorandum. Volgehoue akkuraatheid geld deurgaans in ALLE aspekte van die memorandum. • 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 (𝑥 − 2)(3𝑥 − 1) = 0 𝑥 − 2 = 0 or/of 3𝑥 − 1 = 0 ✓✓ 𝑥-values / waardes 1 𝑥 = 2 or/of 𝑥 = 3 (2) 1.1.2 Penalise 1 mark for 2𝑥 2 + 3𝑥 − 7 = 0 incorrect rounding off./ Penaliseer 1 punt vir b  b 2  4ac x verkeerde afronding. 2a (3)  (3) 2  4(2)(7) x ✓ substitution / vervanging 2(2) 3  65 x 4  x  1, 27 or / of x  2, 77 ✓✓ 𝑥-values / waardes (3) 1.1.3  x 2  2 x  15  0 x 2  2 x  15  0 ✓ factors / faktore ( x  5)( x  3)  0 ✓ critical values with method/ + – + kritieke waardes met metode −5 3 –5 3 ✓✓ answer (accuracy) / ∴ 𝑥 < −5 𝑜𝑟/𝑜𝑓 𝑥 > 3 antwoord (akkuraatheid) (4) Copyright reserved/Kopiereg voorbehou Please turn over/Kopiereg voorbehou

Downloaded from hlayiso.com (EC/JUNIE 2018) MATHEMATICS P1/WISKUNDE V1 3 1.1.4 x 1 3x 1  3x 1 x 1  2  3 9 3x  3  1 ✓ simplifying LHS / vereenvoudiging LK  2  32  x 1 ✓ simplifying RHS / vereenvoudiging RK 1 3x   3 6  2  32 x  2 ✓ same base / dieselfde basis 32 x  2  31 2 x  2  1 ✓ equating exponents / 2 x  1 gelykstel van eksponente 1 ✓ answer / antwoord x 2 (5) 1.2 x  3 y  1....................(1) y 2  2 xy  x 2  7......(2) ✓ x subject of the formula / x onderwerp van die formule from / vanaf (1) : x  3 y  1.....(3) (3) in (2) : y 2  2 y (3 y  1)  (3 y  1) 2  7 ✓ substitution / vervanging y2  6 y2  2 y  9 y2  6 y 1 7  0 ✓ removing brackets / verwyder hakies  2 y2  4 y  6  0 ✓ standard form / standaardvorm y2  2 y  3  0  y  3 y  1  0 y  3 or / of y  1 ✓ 𝑦-values / waardes x  8 or / of x  4 ✓ x-values / waardes (6) Copyright reserved/Kopiereg voorbehou Copyright reserved/Blaai om asseblief

Downloaded from hlayiso.com 4 MATHEMATICS P1/WISKUNDE V1 (EC/JUNE 2018) 1.3 For equal roots / Vir gelyke wortels : ✓=0 n  4mn  0 2 n ( n  4 m)  0 ✓ both n-values / beide n-waardes  n  0 or / of n  4m 0  (0) 2  4m(0) n0  x ✓ substitution / vervanging 2m x0 ✓ x0 (4m)  ( 4m) 2  4m( 4m) n  4m  x  2m x  2 ✓ x  2 (5) [25] Copyright reserved/Kopiereg voorbehou Please turn over/Kopiereg voorbehou

Downloaded from hlayiso.com (EC/JUNIE 2018) MATHEMATICS P1/WISKUNDE V1 5 QUESTION 2/VRAAG 2 2.1.1 15 ; 10 ; 7 ; 𝑥 ; 7 −5 −3 (𝑥 − 7) (7 − 𝑥) - 1st differences/1ste verskille 2 (𝑥 − 4) (−2𝑥 + 14) - 2nd differences/2de verskille ✓ 2nd differences / 2de verskille x4 2 or / of  2 x  14  x  4 ✓ equating / gelyk stel x6 or / of  3x  18 ✓ answer / antwoord x6 (3) 2.1.2 2𝑎 = 2 3𝑎 + 𝑏 = −5 𝑎+𝑏+𝑐 = 1 ✓𝑎=1 𝑎=1 3(1) + 𝑏 = −5 1 − 8 + 𝑐 = 15 ✓ 𝑏 = −8 𝑏 = −8 𝑐 = 22 ✓ 𝑐 = 22 ∴ 𝑇𝑛 = 𝑛2 − 8𝑛 + 22 ✓ answer / antwoord (4) 2.1.3 T50   50   8  50   22 ✓ substitution / vervanging 2  2122 ✓ answer / antwoord (2) 2.2.1 a  14d  74 ✓ setting up 2 equations / opstel van 2 vergelykings a  6d  34  8d  40 ✓ method / metode d 5 ✓ answer / antwoord (3) 2.2.2 a  6(5)  34 ✓ value of a / waarde van a a4 ✓ substitution into correct 40 formula / S 40   2  4   (40  1)(5)  2  vervanging in korrekte formule  4060 ✓ answer / antwoord (3) 2.2.3 40   5n  1 n 1 ✓✓ answer / antwoord (accuracy as one unit) (2) Copyright reserved/Kopiereg voorbehou Copyright reserved/Blaai om asseblief

Downloaded from hlayiso.com 6 MATHEMATICS P1/WISKUNDE V1 (EC/JUNE 2018) 2.3.1 3k Tk  15 3k ✓ factors of 15 / faktore van 15  5 3 1   3k 1 ✓ answer / antwoord 5 (2) 2.3.2 a  r n  1 Sn  r 1 ✓ a and/en r 1 24  1 5  3n  1 ✓ substitution into correct formula / 5 3 1 vervanging in korrekte formule 48   3n  1 2 1 5 5 ✓ exponential equation / 242  3n  1 eksponensiële vergelyking 243  3n 35  3n ✓ answer / antwoord n  5 (4) 2.3.3 NO / NEE. ✓ NO / NEE r  3 1 ✓ reason / rede r not in theinterval  1  r  1, r  0 (2) (nie in die interval) 2.4 1 1 1 P  9 3  9 9  9 27  ... to infinity / tot oneindigend 1  1  1 ...toinfinity / tot oneindigend  9 3 9 27 ✓ adding exponents / optel van eksponente a S  1 r ✓ sum to infinity / Som tot oneindiging 1  3 1  13 ✓ answer / antwoord 1  2 P  9 2 1 1  9 ✓ P  92 / 9 3 (4) [29] Copyright reserved/Kopiereg voorbehou Please turn over/Kopiereg voorbehou

Downloaded from hlayiso.com (EC/JUNIE 2018) MATHEMATICS P1/WISKUNDE V1 7 QUESTION 3/VRAAG 3 3.1.1 P  2; 4  ✓ coordinates of P / koördinate van P Q(0;1) ✓ coordinates of Q / koördinate van Q (2) 3.1.2 y  a ( x  2) 2  4 ✓ substitution / vervanging 1  a (0  2) 2  4 3  4a ✓ answer / antwoord 3 a   4 y  bx ✓ substitution / vervanging 4b 2 ✓ answer / antwoord b  2 (4) 3.1.3 x2 / x2 ✓✓ answer / antwoord (2) 3.1.4 h( x)  2 f ( x ) is a maximum when f ( x) is a maximum  is ' n maksimum wanneer f ( x) ' n maksimum is  max value of f ( x)  4 / maks. waarde van f ( x)  4 ✓ max. value of f(x) / maks. waarde van f(x) max of h( x)  24  16 / maks van h( x)  24  16 ✓ answer / antwoord (2) 3.2.1 𝑦 ≥ 1, 𝑦 ∈ 𝑅 OR/OF 𝑦 ∈ (1; ∞) ✓ answer / antwoord (1) 3.2.2 p( x)  x 2  1 / r ( x)  x 2  2 x p ( x  1)  2  ( x  1) 2  1  2 ✓ calculation / berekening  x  2x 11 2 2  x2  2x ✓ 1 unit to the left / 1 eenheid na links Shift 1unit to theleft and 2 units down ✓ 2 units down / 2 eenhede af Skuif 1eenheid na links en 2 eenhede af OR/OF Turning Point of p(x) / Draaipunt van p(x) = (0 ; 1) Turning Point of r(x) / Draaipunt van r(x) = (-1 ; -1) Shift 1 unit to the left and 2 units down / Skuif 1 eenheid na links en 2 eenhede af (3) [14] Copyright reserved/Kopiereg voorbehou Copyright reserved/Blaai om asseblief

Downloaded from hlayiso.com 8 MATHEMATICS P1/WISKUNDE V1 (EC/JUNE 2018) QUESTION 4/VRAAG 4 4.1 3 y 5 0 1 ✓ y-intercept / y-afsnit (1) 2 4.2 3 5  0 x 1 3  5 x 1  5 x  5  3 ✓ simplification / 5 x  2 vereenvoudiging 2 x ✓ answer / antwoord 5 (2) 4.3 y ✓ asymptotes / asimptote 5 ✓ x- and y-intercepts x- en y-afsnitte 2 ✓shape / vorm -1 - 52 O x (3) 4.4 ✓ substitution / vervanging ✓ simplification / vereenvoudiging ✓ reflection / refleksie (3) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Kopiereg voorbehou

Downloaded from hlayiso.com (EC/JUNIE 2018) MATHEMATICS P1/WISKUNDE V1 9 QUESTION 5/VRAAG 5 5.1 f ( x)  log3 x ✓ interchanging x and y / x  log3 y omruil van x en y f 1 ( x)  3x ✓ answer / antwoord (2) 5.2 f -1 is a reflection of f in the line y = x / ✓✓ answer / antwoord f -1 is ’n refleksie van f in die lyn y = x (2) 5.3 y  log 3 x 2  log 3 k ✓ substitution / vervanging 2 k  3 1 ✓ answer / antwoord  9 (2) 5.4 1 0 x ✓✓ answer / antwoord 9 OR / OF log 3 x  2 and / en x  0 1 x and / en x  0 9 1 0 x (2) 9 5.5 x 1 ✓✓ answer / antwoord (2) [10] Copyright reserved/Kopiereg voorbehou Copyright reserved/Blaai om asseblief

Downloaded from hlayiso.com 10 MATHEMATICS P1/WISKUNDE V1 (EC/JUNE 2018) QUESTION 6/VRAAG 6 6.1 𝑖𝑛𝑜𝑚 𝑛 1 + 𝑖𝑒𝑓𝑓 = (1 + ) ✓ formula / formule 𝑛 23 12 𝑖𝑒𝑓𝑓 = (1 + ) −1 ✓ substitution / vervanging 1200 = 0,2558637702 effective rate / effektiewe koers =25,59% p.a ✓ answer / antwoord (3) 6.2 A P (1  i ) n ✓ i and/en n 15768,39   4,38  40 ✓ substitution / vervanging 1    400  ✓ answer / antwoord (3)  R10 200, 00 6.3.1  A log   n P ✓ A and/en P log 1  i  ✓ substitution / vervanging  50 710, 00  log   n  200 000, 00  ✓ making n subject of the formula /  24  maak n die onderwerp van die formule log 1    100  ✓ answer / antwoord (4) n  5 years 6.3.2 A  P 1  i  n 5  18  ✓ substitution / vervanging  200 000 1    100  ✓ answer / antwoord  R 457 551,55 (2) 6.3.3 Amount/Bedrag = R457 551,55 – R50 710.00 = R406 841,55 ✓ answer / antwoord (1) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Kopiereg voorbehou

Downloaded from hlayiso.com (EC/JUNIE 2018) MATHEMATICS P1/WISKUNDE V1 11 QUESTION 7/VRAAG 7 Penalise 1 mark for incorrect notation in the question / 7.1 𝑓(𝑥) = 1 − 3𝑥 2 Penaliseer 1 punt vir verkeerde notasie in die vraag 𝑓(𝑥 + ℎ) = 1 − 3(𝑥 + ℎ)2 = 1 − 3(𝑥 2 + 2𝑥ℎ + ℎ2 ) = 1 − 3𝑥 2 − 6𝑥ℎ − 3ℎ2 ✓ 1 − 3𝑥 2 − 6𝑥ℎ − 3ℎ2 𝑓(𝑥 + ℎ) − 𝑓(𝑥) 𝑓′(𝑥) = lim ℎ→0 ℎ 1 − 3𝑥 2 − 6𝑥ℎ − 3ℎ2 − (1 − 3𝑥 2 ) = lim ✓ substitution / vervanging ℎ→0 ℎ −6𝑥ℎ − 3ℎ2 = lim ℎ→0 ℎ ℎ(−6𝑥 − 3ℎ) ✓ common factor / gemene faktor = lim ℎ→0 ℎ = lim (−6𝑥 − 3ℎ) ℎ→0 ✓ answer / antwoord Answer ONLY: 0 marks / = −6𝑥 SLEGS antwoord: 0 punte (4) 7.2 2  1  y x   x 1 y  x2 x ✓ y = x + 2 + x -1 y  x  2  x 1 dy ✓1 ✓−𝑥 −2  1  x 2 (3) dx 7.3 y  3x 2  2 x  1 ✓ y  6 x  2 y  6 x  2 6x  2  4 ✓ 6x  2  4 6x  6 x 1 ✓ x-coordinate / x-koördinaat y2 1; 2  ✓ answer / antwoord (4) [11] Copyright reserved/Kopiereg voorbehou Copyright reserved/Blaai om asseblief

Downloaded from hlayiso.com 12 MATHEMATICS P1/WISKUNDE V1 (EC/JUNE 2018) QUESTION 8/VRAAG 8 8.1.1 f ( x)  a( x  2)( x  23 )( x  3) ✓ substitution of x-coordinates / vervanging van x-koördinate 16  a (2  2)(2  23 )(2  3) 16 ✓ substitution of point / 16   a vervanging van punt 3 a3 ✓ value of a / waarde van a f ( x)  3( x  2)  x  23  ( x  3)  3  x  23   x 2  x  6  ✓ substitution / vervanging  3  x 3  x 2  6 x  23 x 2  23 x  4  ✓ removing brackets / verwyder hakies  3 x3  5 x 2  16 x  12 (5) 8.1.2 𝑓(𝑥) = 3𝑥 3 − 5𝑥 2 − 16𝑥 + 12 𝑓 ′ (𝑥) = 9𝑥 2 − 10𝑥 − 16 = 0 ✓ 𝑓 ′ (𝑥) = 0 (9𝑥 + 8)(𝑥 − 2) = 0 9𝑥 + 8 = 0 𝑜𝑟/𝑜𝑓 𝑥 − 2 = 0 ✓ factors / faktore 8 𝑥=− 𝑜𝑟 /𝑜𝑓 𝑥 = 2 ✓ x-values / waardes 9 4900 𝑦= (20,16) 243 8 ✓ coordinates of P / koördinate 𝐵 (− ; 20,16) van P 9 (4) 8.1.3 P (- 89 ; 20.16) y ✓ x-intercepts / x- afsnitte 12 ✓ y-intercept / y-afsnit ✓ turning pts / draaipunte ✓ shape / vorm -2 2 3 3 x Q (2 ; -16) (4) Copyright reserved/Kopiereg voorbehou Please turn over/Kopiereg voorbehou

Downloaded from hlayiso.com (EC/JUNIE 2018) MATHEMATICS P1/WISKUNDE V1 13 8.2.1 f ( x)  ax3  bx 2  3 x  3 and / en g ( x)  f ( x)  12 x  4 f ( x)  3ax 2  2bx  3 f ( x)  6ax  2b ✓a=2 6a  12 and / en 2b  4 ✓b=2 a2 and / en b2 (2) 8.2.2 12 x  4  0 1 x ✓ x =  13 3 1 ✓ answer / antwoord x   (2) 3 8.2.3 1 f ( x)  0 for x    concave down / konkaaf afwaarts  ✓ answer / antwoord 3 1 f ( x)  0 for x    concave up / konkaaf opwaarts  ✓ answer / antwoord 3 1 x   is the x-coordinate of the point of inflection 3 where the concavity changes. 1 x   is die x-koordinaat van die infleksiepunt waar die 3 konkawiteit verander. (2) [19] Copyright reserved/Kopiereg voorbehou Copyright reserved/Blaai om asseblief

Downloaded from hlayiso.com 14 MATHEMATICS P1/WISKUNDE V1 (EC/JUNE 2018) QUESTION 9/VRAAG 9 9.1 EF  a  2 x ✓ EF = (a – 2x) DE  tan 600 ✓ tan ratio / tan verhouding BE DE  x tan 600 ✓ answer / antwoord  3x Area  l  b  3x  (a  2 x) ✓ substitution / vervanging No mark for the answer / (4)  3ax  2 3 x 2 Geen punt vir die antwoord 9.2 A( x)  3ax  2 3 x 2 A( x)  3a  4 3 x  0 ✓ derivative / afgeleide ✓ f ( x)  0  3a x 4 3 a ✓ answer / antwoord x 4 2 a a ✓ substitution / vervanging Area  3a    2 3   4 4 3 2 3 2  a  a 4 8 3 2 ✓ answer / antwoord  a (5) 8 [9] Copyright reserved/Kopiereg voorbehou Please turn over/Kopiereg voorbehou

Downloaded from hlayiso.com (EC/JUNIE 2018) MATHEMATICS P1/WISKUNDE V1 15 QUESTION 10/VRAAG 10 10.1.1 67 P(F and S) = / 0, 28 ✓ answer / antwoord (1) 236 10.1.2 P( M )  P(not S ) ✓ P(M) x P(not S) 120 100   236 236 ✓ answer / antwoord  0, 22 51 P( M  not S )  236  0, 22  P( M )  P(not S )  P ( M  not S ) ✓ answer / antwoord  events are independent / gebeurtenisse is onafhanklik ✓ conclusion / gevolgtrekking (4) 10.2.1 OUTCOMES UITKOMSTE ✓ 1st branch / 1ste tak ✓ 2nd branch / 2de tak ✓ outcomes / uitkomste (3) 10.2.2  x 2   2 x        0,375 ✓ setting up equation / opstel van  x2 x2  x2 x2 vergelyking  2x   2x  3   2   2    ( x  2)   ( x  2)  8 4x 3  ( x  2) 2 8 3x 2  12 x  12  32 x 3x 2  20 x  12  0 ✓ standard form / standaardvorm (3x  2)( x  6)  0 2 x or / of x  6 ✓ answer / antwoord 3 (3) [11] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Copyright reserved/Blaai om asseblief