Mathematics Grade 11 P2 National Exemplar November 2014 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE/GRAAD 11 MATHEMATICS P2/WISKUNDE V2 NOVEMBER 2014 MEMORANDUM MARKS/PUNTE: 150 This memorandum consists of 16 pages. Hierdie memorandum bestaan uit 16 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 2 DBE/November 2014 CAPS/KABV – Memorandum NOTE: • If a candidate answers a question TWICE, only mark the FIRST attempt. • If a candidate has crossed out an attempt of a question and not redone the question, mark the crossed out version. • Consistent accuracy applies in ALL aspects of the marking memorandum. • Assuming answers/values in order to solve a problem is NOT acceptable. LET WEL: • Indien 'n kandidaat 'n vraag twee keer beantwoord, merk slegs die eerste poging. • Indien 'n kandidaat 'n antwoord doodgetrek het, maar nie oorgedoen het nie, merk die doodgetrekte antwoord. • Volgehoue akkuraatheid geld in ALLE aspekte van die memorandum. • Aanname van antwoorde/waardes om 'n probleem op te los, is ONaanvaarbaar. QUESTION/VRAAG 1 1.1.1 IQR (A) = 30 – 20  30 – 20 = 10  10 (2) 1.1.2 Data of Supermarket A is skewed to the left/Data van Supermark  A is skeef na links. comment/kommentaar OR Negatively skewed/negatief skeef (1) 1.1.3 Range/Omvang (B) = 35 – 6  35 – 6 = 29  29 (2) 1.1.4 Supermarket A  correct choice/ • Supermarket A received 25 or more deliveries on more than 7 regte keuse days whilst Supermarket B received 25 or more deliveries on  reason/rede less than 7 days/ Supermark A het op meer as 7 dae 25 of meer (2) aflewerings ontvang terwyl Supermark B op minder as 7 dae soveel aflewerings ontvang het. 1.2 x = 24,5 2 x + 293 2 x + 293 = 24,5  14 14 2 x + 293 = 343  2x + 293 = 343 2 x = 50 x = 25  25 (3) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 3 DBE/November 2014 CAPS/KABV – Memorandum QUESTION/VRAAG 2 2.1 28 days/dae  answ/antw (1) 2.2 12 days  No. of days/ 12 getal dae ∴ × 100 28 = 42,86%  percentage/ persentasie Accept/Aanvaar 12,5 days which is/dae, wat gelyk is aan 44,64% OR Accept/Aanvaar 13 days which is/dae, wat gelyk is aan 46,43% (2) 2.3 Temperature, T, in Frequency degrees Celsius 19 ≤ T < 21 2  2 and/en 6 21 ≤ T < 23 6  9 and/en 5 23 ≤ T < 25 9 25 ≤ T < 27 5 27 ≤ T < 29 4  4 and/en 2 29 ≤ T < 31 2 (3) 2.4 FREQUENCY POLYGON 10  anchored at / 9 geanker by (18 ; 0) 8 and/en 7 (32 ; 0)  points at 6 Frequency midpoints/ 5 punte by middelpunte 4  straight lines 3 joining pts/ reguitlyne 2 verbind punte 1  all points plotted/ 0 alle punte 17 19 21 23 25 27 29 31 33 35 geplot Average daily temperature (in degrees Celsius) (4) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 4 DBE/November 2014 CAPS/KABV – Memorandum QUESTION/VRAAG 3 3.1 12 + 9  substitution into mAC = gradient formula/ −9+3 21 subst in gradiënt- = formule −6 7 7  − =− 2 2 (2) 3.2 7 m =m =− (parallel lines/parallelle lyne)  gradients equal/ RNS AC 2 7 gradiënte gelyk y=− x  equation/vgl 2 (2) 3.3 2 NB = (5 5 ) 2 2 2 ( a − 9) + (7 − 9) = 125  subst into distance 2 formula/subst in a − 18a + 81 + 4 − 125 = 0 afstandformule 2  st form/st vorm a − 18a − 40 = 0  factors/faktore (a − 20)(a + 2) = 0  answ/antw a ≠ 20 ∴ a = −2 (4) 3.4 7 7 tan α = mAC = −  tan α = − 2 2 α = 180° − 74,05° = 105,95°  α = 105,95° 3 tan β = mBC = 3 2  tan β = β = 56,31° 2  β = 56,31° ∴θ = 105,95° − 56,31° = 49,64°  θ = 49,64° (5) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 5 DBE/November 2014 CAPS/KABV – Memorandum QUESTION/VRAAG 4 4.1 Perimeter/Omtrek ABCD = 4 × AD (sides of rhombus all  subst into distance equal/sye van ruit almal gelyk) form/ subst in AD = (3 + 8) 2 + (9 − 6) 2 afstandformule  130 OR/OF = 130 OR/OF 11,40 11,40 Perimeter/Omtrek = 4 130 OR/OF 45,61  4 130 OR/OF 45,61 (3) 4.2 m BD = 3  m BD = 3 mAC × mBD = −1 (diagonals of a rhombus/hoeklyne van ruit) 1 1 ∴ mAC = −  mAC = − 3 3 y − y1 = m( x − x1 ) y = mx + c 1 1 y − 6 = − ( x + 8) 6 = − (− 8) + c  subst into correct 3 3 form/subt in 1 8 18 10 korrekte formule y =− x− + OR/OF =c 3 3 3 3 1 10 1 10 y =− x+ y =− x+  answer/antw 3 3 3 3 (4) 4.3 At/By K: 1 10 3x = − x +  equate equations/ 3 3 stel vgls gelyk 9 x = − x + 10 10 x = 10 x =1  x=1 y = 3(1) = 3  y=3 K(1 ; 3) (3) OR/OF y = 3x x + 3(3 x) = 10  subst of/subst van y = 3 x into 10 x = 10 x + 3 y = 10 x =1  x=1 y = 3(1) = 3  y=3 K(1 ; 3) (3) OR/OF x = −3 y + 10 3(−3 y + 10) − y = 0  subst of/subst van − 9 y + 30 − y = 0 x = −3 y + 10 into − 10 y = −30 3x − y = 0 y=3  y=3 x = –3(3) + 10 x =1  x=1 K(1 ; 3) (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 6 DBE/November 2014 CAPS/KABV – Memorandum 4.4 xB + 3 yB + 9 =1 =3 2 2 xB + 3 = 2 yB + 9 = 6 x B = −1 y B = −3  x value/waarde B(–1 ; –3)  y value/waarde (2) OR/OF xB = −1 yB = −3 (by translation/deur translasie)  x value/waarde B(–1 ; –3)  y value/waarde (2) 4.5 6+3 9 m AB = =−  subst into gradient − 8 +1 7 formula/subst in 9−6 3 gradiëntformule mAD = = 3 + 8 11  gradient AB mAB × mAD ≠ −1  gradient AD ABCD is not a square/ is nie 'n vierkant ( BÂD ≠ 90° )  ≠ –1  S/R OR/OF (5) C(10 ; 0)  C(10 ; 0) BD 2 = (3 − (−1)) 2 + (9 − (−3)) 2  subst into distance formula/subst in = 160 afstandformule BD = 4 10 OR/OF 12,65  4 10 OR 12,65 AC 2 = (−8 − 10) 2 + (6 − 0) 2 = 360 AC = 6 10 OR/OF 18,97  6 10 OR 18,97  S/R ABCD is not a square/is nie 'n vierkant (BD ≠ AC) (5) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 7 DBE/November 2014 CAPS/KABV – Memorandum QUESTION/VRAAG 5 5.1.1 cos 203° = − cos 23°  reduction/herlei = −p  answer/antw (2) 5.1.2 sin 293° = − sin 67°  reduction/herlei = − cos 23°  co-ratio/ko-verh = −p  answ/antw ito/v p (3) 5.2 sin(360° − x). tan(− x) cos(180° + x).(sin 2 A + cos 2 A) (− sin x)(− tan x)  − sin x = (− cos x)(1)  − tan x  sin x   − cos x (− sin x) −   1  cos x  =  sin x  − cos x  −  2  cos x  sin x =− cos 2 x = − tan 2 x  − tan 2 x (6) 5.3.1 2 cos x + (1 + sin x) 2 LHS =  numerator/teller (1 + sin x). cos x  denominator/ noemer cos 2 x + 1 + 2 sin x + sin 2 x  multiplication/ = (1 + sin x). cos x vermenigvuldiging 1 + 1 + 2 sin x  identity/identiteit = (1 + sin x). cos x  fact/faktor 2(1 + sin x) numerator/teller = (1 + sin x). cos x 2 = cos x (5) = RHS 5.3.2 Undefined if/ongedefinieerd as: sin x = –1 or cos x = 0  90° ∴ x = 90° ; 270°  270° (2) 5.4 sin 2 x = 4 cos 2 x tan 2 x = 4  tan 2 x = 4 2 x = 75,96° + k .180°  75,96° x = 37,98° + k .90° ; k ∈ Z  37,98°  k .90°  k ∈Z (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 8 DBE/November 2014 CAPS/KABV – Memorandum 5.5.1 x2 + y2 = r 2 x 2 + ( 3) 2 = 2 2  subst x2 = 1 x = ±1 x =1 (since P lies in the 1st quadrant/aangesien P in  x=1 die 1ste kwadrant lê) (2) 5.5.2 3 sin PÔT =  correct ratio/ 2 korrekte verh PÔT = 60°  60° PÔT + α = 90° α = 90° − 60° = 30°  answer/antw (3) 5.5.3 b sin( −30°) =  correct ratio/ 20 korrekte verh b = 20 sin(−30°)  b = 20 sin(−30°) b = −10  b = –10 a cos(−30°) = 20  correct ratio/ a = 20 cos(−30°) korrekte verh a = 10 3 OR/OF 17,32  a = 10 3 OR17,32 Q(10 3 ; − 10) OR/OF Q(17,32 ; –10) (5) OR/OF OQ2 = 400  subst into distance a² + b² = 400 formula/subst in PQ² = 2² + 20² afstandformule PQ² = 404 (a − 1) 2 + (b − 3 ) 2 = 404  subst into distance formula/subst in a 2 − 2a + 1 + b 2 − 2 3b + 3 = 404 afstandformule 400 − 2a + 4 − 2 3b = 404 2a = −2 3b a = − 3b  a = − 3b (− 3b) 2 + b 2 = 400 4b 2 = 400 b 2 = 100  b = −10 b = −10 (b < 0) a = − 3 (−10) a = 10 3  a = 10 3 OR17,32 ∴ Q(10 3 ; − 10) (5) [33] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 9 DBE/November 2014 CAPS/KABV – Memorandum QUESTION/VRAAG 6 6.1 a=1  a=1 b=2  b=2 p = 45°  p = 45° (3) 6.2 x ∈ (−90° ; 0°)  extreme values/ uiterste waardes  correct notation/ korrekte notasie OR/OF (2)  extreme values/ − 90° < x < 0° uiterste waardes  correct notation/ korrekte notasie OR/OF (2)  extreme values/ between − 90° and 0° /tussen –90° en 0° uiterste waardes  correct notation/ korrekte notasie (2) 6.3 f (2 x) = cos 2(2 x) = cos 4 x  cos 4x ∴ period/periode = 90°  90° (2) 6.4 h( x) = 3 cos 2 x − 1 Minimum value/waarde = –4   –4 (2) 6.5 move 45° to the left and then reflect about the x-axis/  45° left/links skuif 45° na links en reflekteer dan om die x-as  reflection x-axis/ refleksie om x-as OR/OF The graph of g must be moved 135° to the right/  135° right/regs Die grafiek van g moet 135° na regs beweeg. (2) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 10 DBE/November 2014 CAPS/KABV – Memorandum QUESTION/VRAAG 7 7.1.1 AĈB = 180° − 150° = 30°  30° AB 6  subst into sine rule/ = subst in sin-reël sin 30° sin 110°  AB = 3,19 AB = 3,19m (3) 7.1.2 Area ∆ABC: 1 = .AB.BC. sin B 2 1  subst into correct = × 3,19 × 6 × sin 40° form/subst in 2 korrekte formule 2 = 6,15m  6,15 (2) 7.1.3 1 Volume of pyramid = area of base × ⊥ height  correct formula/ 3 korrekte formule 1  subst = × 6,15 × 8 3 = 16,4m 3  answer/antw (3) 7.2 radius of base/radius van basis: r r = tan 36°  = tan 36° h h r = 2 tan 36° = 1,45m  r = 2 tan 36° = 1,45 slant height/skuinshoogte: l 1  = S 1 h cos 36° = h cos 36° 2  S= = 2,47 m 2 cos 36° S= = 2,47 cos 36°  2   subst into correct SA = π(2tan 36°)2 + π(2tan 36°)   form/subst in  cos 36°  korrekte formule = 17,92 m2  answer/antw OR/OF Surface area of cone = area of base + area of curved surface buite-opp van keël = opp van basis + opp van geboë opp = πr 2 + πrS = π (1,45) 2 + π (1,45)(2,47)  subst into correct form/subst in = 17,86m 2 korrekte formule  answer/antw (6) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 11 DBE/November 2014 CAPS/KABV – Memorandum QUESTION/VRAAG 8 8.1 M 2 T 1 37° O 2 1 S P 8.1.1 M̂ + M̂ = 90° (∠ in semi circle/∠ in halfsirkel or/of diameter  S/R 1 2 subtends right∠/midlyn onderspan regte∠ / 1 or/of ∠ ) 2  S M̂ = 53° (2) 1 OR/OF Ô 2 = 74° (∠ at centre/midpt = 2 ×∠ at circum/by omtrek)  S/R Ô1 = 106° (∠s on a str line/∠e op reguitlyn)  S M̂ = 53° (∠ at centre/midpt = 2 ×∠ at circum/by omtrek) 1 (2) 8.1.2 Ô = 2 × M̂ (∠ at centre/midpt = 2 ×∠ at circum/by omtrek)  S/R 1 1  S Ô1 = 106° (2) OR/OF Ô 2 = 74° (∠ at centre/midpt = 2 ×∠ at circum/by omtrek)  S/R Ô1 = 106° (∠s on a str line/∠e op reguitlyn)  S (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 12 DBE/November 2014 CAPS/KABV – Memorandum 8.2 K O 106° 1 T 2 1 M L 8.2.1 Ô = 360° − 106° = 254° (∠s round a pt or ∠s in a rev)  S 2 (∠e om 'n pt of ∠e omw) 1 T̂ = × Ô (∠ at centre/midpt = 2 ×∠ at circum/by omtrek) 1 2 2  S = 127°  R (3) 8.2.2 KO = OL (radii equal/radiusse gelyk)  S KM = ML (Tans from common/same pt/rklyne van dies pt)  S/R ∴ KOLM is a kite (adj sides of quad are =/aangr sye v vh =)  S/R (3) 8.2.3 OK̂M = 90° (tan/rkl ⊥ radius or/of tan/rkl ⊥ diam/midlyn)  S/R OL̂M = 90° (tan/rkl ⊥ radius or/of tan/rkl ⊥ diam/midlyn) OK̂M + OL̂M = 180° OKML = cyc quad/kdvh  S (opp ∠s quad supp or converse opp ∠s of cyclic quad)/  R (tos ∠e vierh supp of omgek tos ∠e van kdvh) (3) 8.2.4 M̂ + Ô1 = 180° (opp ∠s of cyclic quad/tos ∠e van kdvh)  R  S M̂ = 74° (2) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 13 DBE/November 2014 CAPS/KABV – Memorandum QUESTION/VRAAG 9 K 12 L 2 1 20° 1 M 2 4 3 1 1 2 2 N P 9.1 NK̂M = K̂ 1 = 20° (equal chords; equal ∠s) S (gelyke koorde; gelyke ∠e) R (2) 9.2 Alternate ∠s are equal/verwiss ∠e gelyk  R (1) 9.3 NM = LM (radii)  S NM = KL (given/gegee)  S ∴ KL = LM (2) 9.4.1 MK̂L = K̂ = 20° (∠s/e opp equal sides/to gelyke sye)  S/R 2  S KL̂M = L̂ 2 = 140° (∠s sum in ∆ /∠e som in ∆)  S KN̂M = N̂ 1 = 180° − 140° = 40° (opp ∠s of cyclic quad/  R tos ∠e van kdvh) (4) 9.4.2 KM̂N = M̂ 2 = 180° − (20° + 40°) = 120° (∠s sum in ∆ /  S ∠e som in ∆) LM̂N = M̂ 1 + M̂ 2 = 20° + 120° = 140° LP̂N = P̂1 + P̂2 = 70° (∠ at centre = 2 ×∠ at circumference)  S R (∠ by midpt = 2 ×∠ by omtrek) (3) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 14 DBE/November 2014 CAPS/KABV – Memorandum QUESTION/VRAAG 10 10.1 Construction: Draw diameter AD and join DB.  construction/ Konstruksie: Trek middellyn AD en verbind DB konstruksie D C B O P A T Proof/Bewys: BÂT + DÂB = 90° (tangent/rklyn ⊥ radius)  S R DB̂C + CB̂A = 90° (∠ in semi circle/halfsirkel)  S R DÂB + AD̂B = 90° (∠s/e of/van ∆) BÂT = AD̂B BĈA = AD̂B (∠s in same segment/∠e in dies segment)  S/R BÂT = BĈA (6) OR/OF Construct diameter AD and join DC.  construction/ Konstrueer middellyn AD en verbind DC. konstruksie D C B O P A T Proof/Bewys: BÂT + DÂB = 90° (tangent/rklyn ⊥ radius)  S R DĈB + BĈA = 90° (∠ in semi circle/halfsirkel)  S R BĈA = 90° − DĈB DÂB = 90° − BÂT DĈB = DÂB (∠s in same segment/∠e in dies segment)  S/R BÂT = BĈA (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 15 DBE/November 2014 CAPS/KABV – Memorandum OR/OF Construction: Draw radii OA and OB.  construction/ Konstruksie: Trek radiusses OA en OB. konstruksie C B O P A T Proof/Bewys: OÂB + BÂT = 90° (tangent/rklyn ⊥ radius)  S R OÂB = 90° − BÂT OB̂A = 90° − BÂT (∠s opp = sides/∠e to = sye)  S AÔB = 180° − 2(90° − BÂT) (∠s/e of/van ∆) AÔB = 2BÂT  S AÔB = 2Ĉ (∠ at centre = 2 ×∠ at circumference/  S/R BÂT = BĈA ∠ by midpt = 2 ×∠ by omtrek) (6) 10.2 P 1 2 1 S 2 3 1 2 Q 2 1 3 R T Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 16 DBE/November 2014 CAPS/KABV – Memorandum 10.2.1(a) Tan chord theorem/rklyn-koordstelling  R (1) 10.2.1(b) ∠s in same segment/ ∠e in dieselfde segment  R (1) 10.2.2 R̂ = P̂ + T̂ (ext ∠ of ∆/ buite∠ v ∆)  S 1 2 P̂2 = Q̂ 2 (from/vanaf 10.2.1(b)) Q̂1 = T̂ (from/vanaf 10.2.1(a)) ∴ Q̂ + Q̂ = P̂ + T̂  S 1 2 2 ∴ Q̂ + Q̂ = R̂  S 1 2 1 ∴PQ = PR (sides opp =∠s/sye to =∠e)  S/R ∴∆PQR = isosceles triangle/gelykbenige driehoek (4) 10.2.3 R̂ 2 = Q̂1 (∠s in same segment/∠e in dies segment)  S/R T̂ = Q̂1 (from/vanaf 10.2.1(a)) R̂ 2 = T̂  S PR is a tangent to circle RST at R (converse tan chord th) PR is 'n rklyn aan sirkel RST by R (omgekeerde rkl-kdst)  R (3) OR/OF P̂1 = 180° − (Q̂1 + Q̂ 2 + R̂ 1 ) (∠s/e of/van ∆)  S R̂ 2 = Q̂1 (∠s in same segment/∠e in dies segment)  S/R Q̂1 = T̂ (from/vanaf 10.2.1(a))  R ∴ R̂ 2 = T̂ (3) TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou