Mathematics Grade 12 Paper 1 Eastern Cape Mock 2017 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 12 SEPTEMBER 2017 MATHEMATICS P1/WISKUNDE V1 MEMORANDUM MARKS/PUNTE: 150 Hierdie memorandum bestaan uit 13 bladsye./ This memorandum consists of 13 pages.

Downloaded from hlayiso.com 2 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2017) 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 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 2𝑥(𝑥 + 1) − 7(𝑥 + 1) = 0 (𝑥 + 1)(2𝑥 − 7) = 0 factors/faktore 7 𝑥-value/waarde 𝑥 = −1 or / 𝑜𝑓 𝑥 = 2 𝑥-value/waarde (3) 1.2 𝑥 2 − 5𝑥 − 1 = 0 −𝑏 ± √𝑏 2 − 4𝑎𝑐 𝑥=  substitution into correct 2𝑎 formula/substitusie in korrekte −(−5) ± √(−5)2 − 4(1)(−1) 𝑥= formule 2(1) 𝑥 = 5,19 or 𝑥 = −0,19 𝑥-values/waardes (3) 1.3 4𝑥 2 + 1 ≥ 5𝑥 4𝑥 2 − 5𝑥 + 1 ≥ 0  standard form/standaard (4𝑥 − 1)(𝑥 − 1) ≥ 0 vorm  factors/faktore + - + 1 1 1 4 𝑥 ≤ 4 1 𝑥 ≥ 1 𝑥 ≤ 4 or/ of 𝑥 ≥ 1 (4) 1.4 54𝑥+3 . 100−2𝑥+1 = 50 000 54𝑥+3 . (52 . 22 )−2𝑥+1 = 50 000 54𝑥+3 . 5−4𝑥+2 . 2−4𝑥+2 = 50 000 5−4𝑥+2 55 . 2−4𝑥 . 22 = 50 000 2−4𝑥+2 2−4𝑥 = 22 −4𝑥 = 2 −4𝑥 = 2 1  answer/antwoord 𝑥=− (4) 2 Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/SEPTEMBER 2017) MATHEMATICS P1/WISKUNDE V1 3 1.5 𝑥 = 2𝑦 …………………..(1) 𝑥 2 + 2𝑥 − 𝑦 − 𝑦 2 = 36……….(2) 𝑥 = 2𝑦 sub into (2)/vervang in (2) (2𝑦)2 + 2(2𝑦) − 𝑦 − 𝑦 2 = 36  substitution/vervanging 4𝑦 2 + 4𝑦 − 𝑦 − 𝑦 2 = 36 3𝑦 2 + 3𝑦 − 36 = 0 𝑦 2 + 𝑦 − 12 = 0  standard form/standaardvorm (𝑦 − 3)(𝑦 + 4) = 0  factors/ faktore 𝑦 = 3 or 𝑦 = −4 y − values/ 𝑦 − 𝑤𝑎𝑎𝑟𝑑𝑒𝑠 𝑥 = 6 or 𝑥 = −8 x − values/ 𝑥 − 𝑤𝑎𝑎𝑟𝑑𝑒𝑠 (5) 1.6 𝑥 2 − 𝑘𝑥 + 𝑘 − 1 = 0 ∆ = 𝑏 2 − 4𝑎𝑐  substitution/vervanging ∆ = (−𝑘)2 − 4(1)(𝑘 − 1)  simplification/vereenvoudiging ∆ = 𝑘 2 − 4𝑘 + 4 (𝑘 − 2)2 ∆ = (𝑘 − 2)2  conclusion/gevolgtrekking ∆ ≥ 0 roots are real and rational(perfect square) (4) [23] QUESTION 2/VRAAG 2 2.1.1 𝑇𝑛 = 4𝑛 − 1 𝑇𝑛 = 4𝑛 − 1 483 = 4𝑛 − 1  equating/gelykstelling 483 484 = 4𝑛  answer/antwoord 𝑛 = 121 121 terms in series/ 121terme in reeks (3) 2.1.2 121 ∑(4𝑛 − 1)  answer/antwoord 𝑛=1 (2) 2.2.1 (𝑡 − 3) − (2𝑡 − 4) = (8 − 2𝑡) − (𝑡 − 3)  setting up equation/opstel van −𝑡 + 1 = −3𝑡 + 11 vergelyking 2𝑡 = 10  simplification/vereenvoudig 𝑡=5  answer/antwoord (3) 2.2.2 … ; … ; … 6 ; 2 ; −2 ; … ; … ; …  numerical values of 𝑇10 = 6 or /of 𝑇𝑛 = −4𝑛 + 46 𝑇10 ; 𝑇11 ; 𝑇12 / numeriese waardes 𝑎 + 9𝑑 = 6 𝑇1 = −4(1) + 46 van 𝑇10 ; 𝑇11 ; 𝑇12 𝑎 + 9(−4) = 6 𝑇1 = 42  difference / verskil −4 𝑎 = 42  a-value/a-waarde (3) Copyright reserved Please turn over

Downloaded from hlayiso.com 4 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2017) 2.3 𝑎𝑟 2 + 𝑎𝑟 3 = −4 𝑎 + 𝑎𝑟 = −1  setting of equations/ opstel van vergelykings 𝑎𝑟 2 (1 + 𝑟) = −4 𝑎(1 + 𝑟) = −1  common factor/gemene faktor ∴ 𝑟2 = 4 ∴ 𝑟 = ±2 𝑟 = 2 𝑎 + 𝑎(2) = −1 ∴ 𝑎 + 2𝑎 = −1 3𝑎 = −1  value of a/waarde van a 1 𝑎=−  first three terms/eerste drie 3 terme ry 1 2 4 First three terms: − 3 ; − 3 ; − 3 Eerste drie terme: (6) [17] QUESTION 3/VRAAG 3 3.1 41 ; 43 ; 47 ; 53 ; 61 ; 71 ; 83 ; 97 ; 113 ; 131 2 4 6 8 10 12 14 16 18 2 2 2 2 2 2 2 2  2nd difference/ tweede verskil 2𝑎 = 2 𝑎+𝑏 =2 𝑎 + 𝑏 + 𝑐 = 41 𝑎 = 1 𝑎=1 𝑏 = −1 𝑐 = 41 𝑏 = −1 2 𝑐 = 41 ∴ 𝑇𝑛 = 𝑛 − 𝑛 + 41 𝑇𝑛 = 𝑛2 − 𝑛 + 41 (5) 3.2 𝑇41 = 412 − 41 + 41 𝑇41 = 1681 𝑇41 = 1681 factors / faktore Factors of 1681: 1 ; 41 and 1681 Faktore van 1681 : 1 ; 41 en 1681 conclusion/ gevolgtrekking 1681 is not a prime number/ 1681 is nie ʼn priemgetal nie (3) 3.3 Consider the unit digits only/kyk na die ene syfers alleenlik. 1 ; 3 ; 7 ; 3 ;1 ;1 ; 3 ; 7 ; 3 ;1 ;  unit digits/ene syfers groups of 5/ groepe van 5  groups of 5/ 49999998 groepe van 5 = 9999999,6  conclusion/ 5 0,6 × 5 = 3 gevolgtrekking 𝑇49999998 will end in 7/ sal met ʼn 7 eindig (3) [11] Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/SEPTEMBER 2017) MATHEMATICS P1/WISKUNDE V1 5 QUESTION 4/VRAAG 4 4.1.1 𝐴 = 𝑃(1 + 𝑖)𝑛 7,2 12𝑛  sub into formula/ vervang in 𝐴 = 500 000 (1 + ) formule 1200 𝐴 = 500 000(1.006)12𝑛 12𝑛 (2) 4.1.2 𝐴 = 500 000(1.006)12𝑛 𝐴 = 500 000(1.006)12×5 𝑛 = 60 𝐴 = 𝑅 715894.21  answer/ antwoord (2) 4.1.3 𝐴 = 𝑃(1 + 𝑖)𝑛 1000000 = 500000(1.006)12𝑛  setting up equation/opstel van log 2 vergelyking 12𝑛 = using logs/ gebruik van log log 1.006 12𝑛 = 115.8707581 𝑛 = 9,66 𝑦𝑒𝑎𝑟𝑠 Will exceed R1 000 000 in 10 years. conclusion/gevolgtrekking Sal R1 000 000 oorskry in 10 jaar. (3) 4.2.1 −36 15 𝑖 and/𝑒𝑛 𝑛 10 000 [1 − (1 + 1200) ] sub into 𝑃𝑣 formula/ vervang in 𝑃𝑣 = 15 𝑃𝑣 formule 1200 𝑃𝑣 = 𝑅288 472,67 𝑃𝑣 = 𝑅288 472,67 𝑑𝑒𝑝𝑜𝑠𝑖𝑡/𝑜 = 𝑅350 000 − 𝑅288 472,67 𝑑𝑒𝑝𝑜𝑠𝑖𝑡/𝑜 = 𝑅61 527,33 subtracting/𝑎𝑓𝑡𝑟𝑒𝑘𝑘𝑖𝑛𝑔  answer/antwoord (5) 4.2.2 18,5 −60 18,5 𝑖 = 1200 𝑥 [1 − (1 + 1200) ] 350 000 = 18,5 𝑛 = −60 1200  substitution/substitusie 𝑥 = 𝑅 8 983,17  answer/antwoord (4) [16] Copyright reserved Please turn over

Downloaded from hlayiso.com 6 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2017) QUESTION 5/VRAAG 5 5.1 𝐴(−3; 0)  answer/antwoord (1) 5.2 𝑓(𝑥) = 𝑥 2 + 3𝑥 𝑏 𝑥=− 3 2𝑎 𝑥 = − 2 3 𝑥=− 2 3 3 2 3 𝑓 (− ) = (− ) + 3 (− ) 2 2 2 substitution/ vervanging 9 = − 4 3 9 𝑃 (− ; − )  answer/ antwoord 2 4 (3) 5.3 𝑓(−5) = 10 and / en 𝑓(−3) = 0 calculating 𝑓(−5) and 𝑓(−3) bepaling van𝑓(−5)𝑒𝑛 𝑓(−3) 10 − 0  substitution/substitusie 𝑚= −5 − (−3) 𝑚-value/waarde 𝑚 = −5 (3) 5.4 𝑥 < −3 or / of 𝑥 > 0  answer/ antwoord (2) 5.5 3 9 𝑃 (− ;− ) 2 4 3 9 (− − 2 ; − ) 2 4 1 9  answer/ antwoord (− ; − ) 2 4 or/of 𝑓(𝑥 − 2) = (𝑥 − 2)(𝑥 − 2 + 3) 𝑓(𝑥 − 2) = 𝑥 2 − 𝑥 − 2 𝑓(𝑥 − 2) = 𝑥 2 − 𝑥 − 2 (−1) 𝑥=− 2(1) 1 1 𝑥 = − 2 𝑥=− 2 (2) Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/SEPTEMBER 2017) MATHEMATICS P1/WISKUNDE V1 7 5.6 1 𝐿𝑀 = − 𝑥 + 2 − (𝑥 2 + 3𝑥) 2 𝑔(𝑥) − 𝑓(𝑥) 1 𝐿𝑀 = − 𝑥 + 2 − 𝑥 2 − 3𝑥 2  standard form/standaardvorm 7 𝐿𝑀 = −𝑥 2 − 𝑥 + 2 2 2 7 𝐿𝑀 = − (𝑥 + 𝑥 − 2) 2 completing the square/ 7 2 81 voltooing van kwadraat 𝐿𝑀 = − [(𝑥 + ) − ] 4 16 7 2 81  answer/antwoord 𝐿𝑀 = − (𝑥 + ) + 4 16 OR/OF 1 𝑔(𝑥) − 𝑓(𝑥) 𝐿𝑀 = − 𝑥 + 2 − (𝑥 2 + 3𝑥) 2 1  standard form/standaardvorm 𝐿𝑀 = − 𝑥 + 2 − 𝑥 2 − 3𝑥 2 7 𝐿𝑀 = −𝑥 2 − 𝑥 + 2 2 7 𝑑𝐿𝑀 7  𝑥 = −4 = −2𝑥 − 𝑑𝑥 2 7 −2𝑥 − = 0 2 7 81 𝑥=−  𝑦 = 16 4 7 2 7 7 𝑦 = − (− ) − (− ) + 2 4 2 4 81 𝑦= 16 7 2 81 ∴ 𝐿𝑀 = − (𝑥 + ) + 4 16 OR/OF 𝑏 𝑔(𝑥) − 𝑓(𝑥) 𝑥=− 2𝑎  standard form/standaardvorm 7 −2 𝑥=− 2(−1) 7 𝑥=− 7 4  𝑥 = −4 7 2 7 7 𝑦 = − (− ) − (− ) + 2 4 2 4 81 𝑦= 16 81  𝑦 = 16 7 2 81 ∴ 𝐿𝑀 = − (𝑥 + ) + 4 16 (4) [15] Copyright reserved Please turn over

Downloaded from hlayiso.com 8 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2017) QUESTION 6/VRAAG 6 6.1 𝑦 shape / vorm ℎ ℎ−1 y − intercept/ 𝑦 − 𝑎𝑓𝑠𝑛𝑖𝑡  point on graph/punt op 1• 𝑥 grafiek • 1 (3) 6.2 𝑞(𝑥) = 2𝑥  answer/antwoord (1) 6.3 ℎ−1 𝑥 = 2−𝑦 log 𝑥 −𝑦 =  interchange 𝑥 and 𝑦 log 2 log 𝑥 𝑦 = − log 2 / 𝑦 = − log 2 𝑥 /𝑦 = log 1 𝑥 ruil 𝑥 en 𝑦 2  equation/vergelyking (2) 6.4 𝑦 ≥ 0 ;𝑦 ∈ 𝑅 (1) 6.5 See 7.2.1/ sien 7.2.1 shape and x-intercept/vorm en x-afsnit (2) 6.6 log 1 𝑥 = −3 2 1 −3 ( ) =𝑥 𝑥 = 8 2 𝑥=8 0 < 𝑥 ≤ 8 ∴ 0<𝑥≤8 (2) [11] Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/SEPTEMBER 2017) MATHEMATICS P1/WISKUNDE V1 9 QUESTION 7/VRAAG 7 7.1 𝑑=5 𝑑 = 5 𝑝=2 𝑝 = 2 (2) 7.2 5−𝑥 𝑦= 𝑥−2 −(𝑥 − 2) + 3 5−𝑥 𝑦 = 𝑥−2 𝑦= (𝑥 − 2) 3 −(𝑥−2)+3 𝑦= −1 𝑦 = 𝑥−2 (𝑥−2) (2) 7.3 𝐴(5; 0) 𝑦 =𝑥−3 𝑥 =𝑦+3 𝑥 = 3 𝐴′ (0 + 3; 5 − 3) 𝑦 = 2 𝐴′ (3; 2) (2) [6] Copyright reserved Please turn over

Downloaded from hlayiso.com 10 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2017) QUESTION 8/VRAAG 8 8.1 𝑓(𝑥) = −2𝑥 2 + 𝑝 𝑓(𝑥 + ℎ) − 𝑓(𝑥) 𝑓 ′ (𝑥) = lim  formula/formule ℎ→0 ℎ −2(𝑥 + ℎ)2 + 𝑝 − (−2𝑥 2 + 𝑝)  substitution of/substitusie van = lim (𝑥 + ℎ) ℎ→0 ℎ −2(𝑥 2 + 2𝑥ℎ + ℎ2 ) + 𝑝 + 2𝑥 2 − 𝑝) = lim ℎ→0 ℎ −2𝑥 2 − 4𝑥ℎ − 2ℎ2 + 𝑝 + 2𝑥 2 − 𝑝 simplification to/ = lim ℎ→0 ℎ vereenvoudiging na (−4𝑥ℎ − 2ℎ2 ) −4𝑥ℎ − 2ℎ2 = lim ℎ→0 ℎ ℎ(−4𝑥 − 2ℎ)  common factor/gemene = lim ℎ→0 ℎ faktor = lim (−4𝑥 − 2ℎ) ℎ→0 Answer ONLY: 0 marks/ = −4𝑥 SLEGS antwoord: 0 punte  answer/antwoord Penalise 1 mark for incorrect use of formula. Must show 𝑓 ′ (𝑥). Penaliseer 1 punt vir verkeerde gebruik van formule. Moet 𝑓′(𝑥) toon. (5) 8.2 3 1 𝐷𝑥 [4 √𝑥 + + 2] 3𝑥 1 1 1 1 𝐷𝑥 [4𝑥 3 + 3 𝑥 −1 + 2] Penalise 1 mark for  4𝑥 3  3 𝑥 −1 incorrect notation. / 4 2 1 = 𝑥 −3 − 𝑥 −2 Penaliseer 1 punt vir 4 2 1 3 3 verkeerde notasie. 3 𝑥 −3  − 3 𝑥 −2 (4) [9] Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/SEPTEMBER 2017) MATHEMATICS P1/WISKUNDE V1 11 QUESTION 9/VRAAG 9 𝑓(𝑥) = (𝑥 − 1)2 (𝑥 + 3) 9.1 𝑓(𝑥) = 𝑥 3 + 𝑥 2 − 5𝑥 + 3  𝑓(𝑥) = 𝑥 3 + 𝑥 2 − 5𝑥 + 3 𝑓 ′ (𝑥) = 3𝑥 2 + 2𝑥 − 5  𝑓 ′ (𝑥) = 0 3𝑥 2 + 2𝑥 − 5 = 0  factors/faktore (3𝑥 + 5)(𝑥 − 1) = 0 5 𝑥 = − 3 or / of 𝑥 = 1  𝑥-values/waardes 𝑓(1) = 0  𝑦-values/waardes 5 256 𝑓 (− ) = 3 27 (5) 9.2 256 shape / vorm 27 𝑥 − intercepts/𝑥 − 𝑎𝑓𝑠𝑛𝑖𝑡𝑡𝑒 −3 𝑦 − intercept/𝑦 − 𝑎𝑓𝑠𝑛𝑖𝑡 5 1 − 3  stationary points/draaipunte (4) 9.3 𝑓 ′′ (𝑥) = 6𝑥 + 2  𝑓 ′′ (𝑥) = 6𝑥 + 2 6𝑥 + 2 = 0 1  𝑥 = −3 1 𝑥=− 3 128 20 128 20  𝑦 = 27 / 4,74 / 4 27 𝑦= / 4,74 / 4 27 27 (3) 9.4 256  answer 0<𝑘< 27 (2) 9.5 𝑓 ′ (𝑥) = 3𝑥 2 + 2𝑥 − 5 3𝑥 2 + 2𝑥 − 5 = −5 𝑓 ′ (𝑥) = −5 3𝑥 2 + 2𝑥 = 0 𝑥(3𝑥 + 2) = 0  factors/ faktore 2 𝑥 = 0 or/of 𝑥 = − 3 2 2 175 𝑥 = − 3 𝑓 (− ) = 3 27 2 175 𝑦 = −5𝑥 + 𝑐 𝑓 (− 3) = 27 175 2 = −5 (− ) + 𝑐 27 3  substitution/vervanging 85 𝑐= 27  answer/antwoord 85 𝑦 = −5𝑥 + 27 (6) [20] Copyright reserved Please turn over

Downloaded from hlayiso.com 12 MATHEMATICS P1/WISKUNDE V1 (EC/SEPTEMBER 2017) QUESTION 10/VRAAG 10 10.1 243 = 2(𝑥 × 2𝑥) + 2(2𝑥 × ℎ) + 2(𝑥 × ℎ) 243 = 4𝑥 2 + 4𝑥ℎ + 2𝑥ℎ TSA equation and sub/ TOA 243 = 4𝑥 2 + 6𝑥ℎ vergelyking met vervanging simplification/ 243 − 4𝑥 2 vereenvoudiging ℎ= (2) 6𝑥 81 2𝑥 ℎ= − 2𝑥 3 10.2 81 2𝑥 𝑉 = 2𝑥 × 𝑥 × ( − )  sub into volume formula/ 2𝑥 3 4 3 vervanging in volume formule 𝑉 = 81𝑥 − 𝑥 3 (1) 10.3 𝑑𝑉 = 81 − 4𝑥 2 𝑑𝑥 81 − 4𝑥 2 81 − 4𝑥 2 = 0 81 − 4𝑥 2 = 0 81 𝑥2 = 81 4 𝑥 2 = 4 9 𝑥 = = 4.5 2 answer/antwoord (4) [7] Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/SEPTEMBER 2017) MATHEMATICS P1/WISKUNDE V1 13 QUESTION 11/VRAAG 11 11.1 9 × 9 × 9 9 × 9 × 9 × 5 × 4 = 14580 5 × 4  14580 (3) 11.2.1 12!  12! = 119750400 2!.2! 2! .2!  119750400 (3) 11.2.2 10! 10! 1  2! 2! = = 0,015  119750400 119750400 66  answer/antwoord (3) [9] 11.3.1 1 𝐻= 2 1 𝑀= 1 3 𝑇= 2  first branch with values/ 1 eerste tak met waardes 𝑇= 2 2  top part of second branch 𝐹= 3 with values / boonste gedeelte 1 𝐻= van tweede tak met waardes 2  bottom part of second branch with values / onderste gedeelte van tweede tak met waardes (3) 11.3.2 2 𝑃(𝐹) = 2 3 𝑃(𝐹) = 3 (1) 11.3.3 1 1 𝑃(𝑀/𝐻) = × 3 2 1 𝑃(𝑀/𝐻) = 6 1 𝑃(𝑀/𝐻) = 6 (2) [15] TOTAL/TOTAAL: 150 Copyright reserved Please turn over