Mathematics Grade 11 Paper 1 Eastern Cape November 2016 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 11 NOVEMBER 2016 MATHEMATICS P1/WISKUNDE V1 MEMORANDUM MARKS/PUNTE: 150 This memorandum consists of 8 pages./Hierdie memorandum bestaan uit 8 bladsye.

Downloaded from hlayiso.com 2 MATHEMATICS P1 (EC/NOVEMBER 2016) 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/VRAAG 1 1.1.1 (𝑥 + 2)2 = 1 𝑥 + 2 = ±1  ±1 𝑥 = −1 or/of 𝑥 = −3  𝑥 = −1 OR/OF  𝑥 = −3 (𝑥 + 2)2 = 1 𝑥 2 + 4𝑥 + 4 − 1 = 0 𝑥 2 + 4𝑥 + 3 = 0  standard form/standaardvorm (𝑥 + 3)(𝑥 + 1) = 0  factors/faktore 𝑥 = −3 or/of 𝑥 = −1  both x-values/beide x waardes (3) 1.1.2 2𝑥 2 − 11𝑥 − 4 = 0 −𝑏 ± √𝑏 2 − 4𝑎𝑐 𝑥= 2𝑎  sub into correct formula/vervanging 11 ± √(−11)2 − 4(2)(−4) 𝑥= in korrekte formule 2(2)  simplification/vereenvoudiging 11 ± √153  𝑥 = 5,84 𝑥= 4  𝑥 = −0,34 𝑥 = 5,84 or/of 𝑥 = −0,34 (4) 1.1.3 1 𝑥2 > ;𝑥 < 0 4 2 1 𝑥 −4>0 1 (𝑥 − 2) (𝑥 + 2) > 0 1  factors/faktore  solutions/oplossings  critical values and method/kritiese 1 1 waarde en metode − 2 2 1 1 1  𝑥 < − 2 only/alleenlik 𝑥 > 2 or/of 𝑥 < − 2 1 𝑥 < −2 (4) 1.1.4 𝑥 + 5 = √3 − 3𝑥 𝑥 2 + 10𝑥 + 25 = 3 − 3𝑥  squaring both sides/kwadrering 𝑥 2 + 13𝑥 + 22 = 0 beide kante (𝑥 + 11)(𝑥 + 2) = 0  standard form/standaard vorm 𝑥 = −2 𝑜𝑟/𝑜𝑓 𝑥 = −11  both x values/beide x waardes 𝑥 = −2  𝑥 = −2 (4) Kopiereg voorbehou Blaai om asseblief

Downloaded from hlayiso.com (EC/NOVEMBER 2016) WISKUNDE V1 3 1.2.1 𝑦 2 − 9𝑥 2 (𝑦 − 3𝑥)(𝑦 + 3𝑥)  factors/faktore (1) 1.2.2 𝑦 + 3𝑥 = 2 … … … … … … … … (1) 𝑦 2 − 9𝑥 2 = 16 … … … … … … … … … . (2) From/Vanaf (2) (𝑦 − 3𝑥)(𝑦 + 3𝑥) = 16 Sub (1) into/in (2) 2(𝑦 − 3𝑥) = 16  sub 2 for (𝑦 + 3𝑥)vervanging van 2 From/Vanaf (1) vir (𝑦 + 3𝑥) 𝑦 = 2 − 3𝑥  making y the subject/maak y die 2(2 − 3𝑥 − 3𝑥) = 16 onderwerp. 4 − 12𝑥 = 16  sub y/vervanging van y. 12𝑥 = −12  x value/x-waarde 𝑥 = −1 𝑦 = 2 − 3(−1)  y value/y-waarde 𝑦=5 OR/OF From/vanaf (1) 𝑦 = 2 − 3𝑥 Sub into (2)/vervang in (2) (2 − 3𝑥)2 − 9𝑥 2 = 16 4 − 12𝑥 + 9𝑥 2 − 9𝑥 2 = 16  𝑦 = 2 − 3𝑥 4 − 12𝑥 = 16  Substitution/vervanging 12𝑥 = −12  Simplification/vereenvoudiging 𝑥 = −1  x value/x-waarde ∴𝑦=5  y value/y-waarde (5) [21] QUESTION/VRAAG 2 2.1 2 𝑎3 ( ) 2 𝑎6 =  answer/antwoord (1) 4 2.2 2𝑥−3 − 3. 2𝑥−1 2𝑥−2 1 3 2𝑥 (8 − 2)  2𝑥 common factor/gemene faktor. 1 3 = 1  −2 8 2𝑥 (4) 1  2𝑥 . 4 11 = − or/of −5 12 or/of −5,5  answer/antwoord (4) 2 2.3 10𝑥+3 = 10𝑥 . 103  sub/vervang 10𝑥 = 1,5 = 1,5 × 1000 = 1500  answer/antwoord (2) Copyright reserved Please turn over

Downloaded from hlayiso.com 4 MATHEMATICS P1 (EC/NOVEMBER 2016) 2.4.1 2𝑥 = 0,125 2𝑥 = 2−3  0,125 𝑎𝑠 2−3 𝑥 = −3  answer/antwoord (2) 2.4.2 9 (0,5)𝑥 . √1 + = 10 16 1 𝑥 25 25 (2) . √16 = 10  √ 16 4 1 𝑥 4  10 × 5 (2) = 10 × 5 1 𝑥  2−𝑥 (2) = 8 2−𝑥 = 23  23 ∴ 𝑥 = −3  𝑥 = −3 (5) [14] QUESTION/VRAAG 3 3.1 2𝑥(𝑥 + 1) + 𝑚 = 𝑥 2𝑥 2 + 2𝑥 + 𝑚 − 𝑥 = 0 2𝑥 2 + 𝑥 + 𝑚 = 0  standard form/standaardvorm ∆= 𝑏 2 − 4𝑎𝑐  sub into ∆/vervang in ∆ = 1 − 4(2)(𝑚) = 1 − 8𝑚  ∆= 1 − 8𝑚 For non-real roots/vir nie-reële wortels ∆< 0  ∆< 0 1 − 8𝑚 < 0 1 8𝑚 > 1  𝑚>8 (5) 1 𝑚>8 3.2 √𝑥+2  5 − 𝑥2 = 0 𝑓(𝑥 ) =  𝑥+2<0 5−𝑥 2 Undefined/ongedefinieerd  𝑥 < −2 5 − 𝑥 2 = 0 or/of 𝑥 + 2 < 0  𝑥 = √5 𝑥 = ±√5 or/of 𝑥 < −2  or/of (5) [10] QUESTION/VRAAG 4 4.1.1 22 ; 27  22  27 (2) 4.1.2 𝑇𝑛 = 5𝑛 + 2  5𝑛  2 (2) 5 4.1.3 5𝑛 + 2 = 12  5𝑛 + 2 = 125 5𝑛 = 125 − 2  125 − 2 𝑛 = 49766  𝑛 = 49766 (3) 𝑇49766 = 125 4.1.4 General Term is multiples of 5 plus 2 therefor the numbers in the sequence will always end  multiples of 5 plus 2/veelvoude van in a 2 or a 7 as multiples of 5 end with 0 or a 5 plus 2 5. Die algemene term is veelvoude van 5 plus  ending in 2 or 7/eindig in 2 of 7 2 en daarom sal die patroon altyd eindig in ʼn 2 of ʼn 5 omdat die veelvoude van 5 altyd eindig op 0 of 5 Kopiereg voorbehou Blaai om asseblief

Downloaded from hlayiso.com (EC/NOVEMBER 2016) WISKUNDE V1 5 4.2.1 39  39 (1) 4.2.2 2𝑎 = 2 𝑎=1  𝑎=1 3𝑎 + 𝑏 = 6  3𝑎 + 𝑏 = 6 𝑏=3  𝑏=3 𝑎+𝑏+𝑐 =3  𝑎+𝑏+𝑐 =3 𝑐 = −1  𝑐 = −1 (5) 𝑇𝑛 = 𝑛2 + 3𝑛 − 1 4.2.3 𝑛2 + 3𝑛 − 1 = 269  𝑛2 + 3𝑛 − 1 = 269 𝑛2 + 3𝑛 − 270 = 0 (𝑛 − 15)(𝑛 + 18) = 0  factors/faktore 𝑛 = 15 𝑇16 = 162 + 3(16) − 1  𝑛 = 15 𝑇16 = 303  𝑇16 = 303 (4) 4.3 ….. ; 6 ; 2 ; x ; -18  Method/metode -4 (𝑥 − 2) (−18 − 𝑥) (𝑥 − 2 + 4) [−18 − 𝑥 − (𝑥 − 2)] 4.3.1 𝑥 + 2 = −2𝑥 − 16  setting up of equation/opstel van 3𝑥 = −18 vergelyking 𝑥 = −6  𝑥 = −6 𝑇4 = −6  second diff/tweede verskil (4) 𝑑2 = −4 4.3.2 𝑑1 = 0  𝑑1 = 0 𝑇1 = 6  𝑇1 = 6 (3) [26] QUESTION/VRAAG 5 5.1 𝑥=8  𝑥=8 𝑦=4  𝑦=4 (2) 5.2 Domain 𝑥 ∈ 𝑅 ;𝑥 ≠ 8  domain/gebied Range 𝑦 ∈ 𝑅 ;𝑦 ≠ 4  range/terrein (2) 5.3 4  shape/vorm 3  asymptotes/asimptote 6 8  x-intercept/x-afsnit  y-intercept/y-afsnit (4) 5.4.1 8 ≥ −4 𝑥−8 8 +4≥0 𝑥−8 ∴ 𝑓(𝑥) ≥ 0  𝑥≤6  𝑥>8 𝑥 ≤ 6 or/of 𝑥 > 8  or/of (3) Copyright reserved Please turn over

Downloaded from hlayiso.com 6 MATHEMATICS P1 (EC/NOVEMBER 2016) 5.4.2 𝑓(𝑥) ≤ 3  𝑥≥0 0≤𝑥<8  𝑥<8  notation/notasie (3) 5.5 𝑦 =𝑥−8+4  𝑦 =𝑥−8+4 𝑦 =𝑥−4  𝑥−4 (2) 5.6 𝑔(𝑥) = 𝑓(𝑥 − 2) − 2  𝑥 − 10 8 𝑔(𝑥) = 𝑥−10 + 2  +2 (2) [18] QUESTION/VRAAG 6 6.1 𝑦 = 𝑎(𝑥 − 𝑥1 )(𝑥 − 𝑥2 ) 𝑦 = 𝑎(𝑥 + 5)(𝑥 + 1) 2 = 𝑎(0 + 5)(0 + 1)  (𝑥 + 5)(𝑥 + 1) 𝑎=5 2  sub of/vervang van (𝟎; 𝟐) 2  a-value/a-waarde 𝑦 = 5 (𝑥 2 + 6𝑥 + 5)  equation/vergelyking (4) 2 2 12 𝑦 = 5𝑥 + 5 𝑥 + 2 6.2 𝑦 = 𝑘. 𝑚 𝑥 2 = 𝑘. 𝑚0 𝑘=2  𝑘=2 𝑦 = 2. 𝑚 𝑥  Sub of/vervang van (1;6) 6 = 2. 𝑚  𝑚=3 𝑚=3 ∴ 𝑔(𝑥) = 2. 3𝑥 (3) 6.3 𝑦=0  answer/antwoord (1) 6.4.1 x-value of turning point/x-waarde van draaipunt. 𝑥 = −3  𝑥 = −3 decreasing/dalend 𝑥 < −3  𝑥 < −3 OR/OF (−∞ ; −3) (2) 6.4.2 0≤𝑥≤1  𝑥≥0  𝑥 ≤ −1 (2) 6.4.3 𝑥≤0  answer/antwoord (2) 6.4.4 −5 < 𝑥 < −1  𝑥 > −5  𝑥 < −1 (2) 6.5 (−5 ; 0) and 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡. (0 ; 2) 𝑦2 − 𝑦1  formula/formule 𝑚= 𝑥2 − 𝑥1  substitution/vervanging 0−2  answer/antwoord 𝑚= −5 − 0 2 𝑚= 5 (3) [19] QUESTION/VRAAG 7 7.1 𝑦 = 𝑏𝑥  𝑦 = 𝑏 𝑥−2 𝑦 = 𝑏 𝑥−2 + 4  +4 8 = 𝑏 4−2 + 4  sub of/vervang van (4;8) 4 = 𝑏2  𝑏=2 (4) 𝑏=2  7.2 𝑦 = 2𝑥−2 + 4 equation/vergelyking (1) [5] Kopiereg voorbehou Blaai om asseblief

Downloaded from hlayiso.com (EC/NOVEMBER 2016) WISKUNDE V1 7 QUESTION/VRAAG 8 8.1 𝐴 = 𝑃(1 − 𝑖)𝑛  formula/formule 25 3  subst/vervanging 𝐴 = 7 200 (1 − )  answer/antwoord 100 𝐴 = 𝑅 3 037,50 (3) 8.2 𝐴 = 𝑃(1 + 𝑖)𝑛 𝑖 12  formula/formule 1 126,10 = 500 (1 + )  subst/vervanging 2 1 126,10 𝑖 12  simplification/vereenvoudiging = (1 + )  answer/antwoord 500 2 (4) 𝑖 = 0,14 Interest rate/rentekoers = 14% 8.3.1 𝑖𝑛 𝑚 1 + 𝑖𝑒 = (1 + )  formula/formule 𝑚 7,2 12  subst/vervanging 𝑖𝑒 = (1 + ) −1  answer/antwoord 1200 𝑖𝑒 = 0,07442 … 𝑟 = 7,442 % (3) 8.3.2 𝐴 = 𝑃(1 + 𝑖)𝑛  formula/formule 𝐴 = 120 000(1 + 7,442%)3  subst/vervanging 𝐴 = 148 834,46  answer/antwoord (3) 8.3.3 48 7,2 𝐴 = 120 000 (1 + ) 1200 𝐴 = 𝑅159 913,20 7,2 30 𝐴 = 20 000 (1 + )  R 159913,20 1200 𝐴 = 23 931,47 𝐹𝑖𝑛𝑎𝑙 𝑎𝑚𝑜𝑢𝑛𝑡 𝑅159913,20 − 23 931,47  R 23931,47 𝐹𝑖𝑛𝑎𝑙𝑒 𝑏𝑒𝑑𝑟𝑎𝑔 = 𝑅 135 981,73  subtracting/aftrek  answer/antwoord OR/OF 7,2 18 7,2 30 𝐴 = [120 000 (1 + ) − 20 000] (1 + ) 1200 1200 (4) 𝐴 = 𝑅 135 981,73 [17] Copyright reserved Please turn over

Downloaded from hlayiso.com 8 MATHEMATICS P1 (EC/NOVEMBER 2016) QUESTION/VRAAG 9 9.1 P PP 2 9  1st branch of tree diagram/1ste tak van boom 2 diagram 3 PG P 9 G  values on first 10 branch/waardes op 1ste 5 tak 9 B PB  2nd branch/2de tak 3  values on 2nd 2 9 P GP branch/waardes op 2de G 1 10 G GG tak 5 9  first four outcomes/eerste 9 vier uitkoms 5 3 B GB  last 4 outcomes/laaste 4 10 9 P uitkomste B G 2 4 9 BG 9 B BB (6) 9.2.1 5 4 𝑃(𝐵𝐵) = × 5 4 10 9  10 × 9 20 𝑃(𝐵𝐵) = = 1,333 ….  answer/antwoord (2) 90 9.2.2 2 3 3 2 𝑃 =( × )+( × ) 2 3 10 9 10 9  (10 × 9) 6 6 𝑃= + 3 2 90 90  (10 × 9) 12  simplify/vereenvoud, 𝑃= 90  adding/optel 𝑃 = 0,133  answer/antwoord (5) 9.3.1 Sport No sport Total  labels/opskrifte Geen sport Totaal  boys row/seuns ry Boys/Seuns 50 10 60  girls row/meisies ry Girls/Meisies 40 20 60  column totals/totale van Total/Totaal 90 30 120 kolomme  120 (5) 9.3.2 40 1  answer/antwoord (a) = (1) 120 3 10 1  answer/antwoord (b) = (1) 120 12 [20] TOTAL/TOTAAL: 150 Kopiereg voorbehou Blaai om asseblief