Mathematics Grade 11 P2 National Exemplar November 2020 Memo

NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 11 NOVEMBER 2020 MATHEMATICS P2/WISKUNDE V2 MARKING GUIDELINE/NASIENRIGLYN EXEMPLAR/EKSEMPLAAR MARKS/PUNTE: 150 This marking guideline consists of 14 pages./ Hierdie nasienriglyn bestaan uit 14 bladsye.

2 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2020) QUESTION 1/VRAAG 1 1.1 0  ( 2) 2  substitution / vervanging m PS   36 3  answer / antwoord OR/OF 3y  2x  6 OR/OF 3 y  2 x  6 2 2 y  x2 y  x2 3 3 2 m PS   3  answer / antwoord (2) 1.2 40 4  gradient / gradiënt mMR   10  3 7 tan RQX 𝑚 ˆ 4 tan RQX= 7 answer / antwoord ˆ RQX=29,74 (4) 1.3 ˆ =-2 tan PQX 3 tan PQX ˆ =146, 31 PQX  PQX 146,31°   146,31  29, 74 answer / antwoord   116,57 (3) 1.4 3 𝑚 mRN  2 3 2 product / produk mRN  mPS     1 2 3  1 RN  PS (3) 1.5 NR= 10  6    4  2  2 2 substitution / vervanging NR= 52 answer / antwoord QN = 13 length QN / lengte QN 1 Area =  QN  NR 2 choosing correct sides / 1 kies korrekte sye Area =  13  52 substitution / vervanging 2 Area = 13 units 2 / eenhede 2 answer / antwoord (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/SEPTEMBER 2020)) MATHEMATICS P2/WISKUNDE V2 3 1.6 y0 4 x = 0  x 3 7 substitution / vervanging 7 y  12 equation / vergelyking 12 y-coordinate / y-koördinaat y 7 OR/OF OR/OF 4 y xc equation / vergelyking 7 Subst./vervang (3; 0) 4 0=  3  c substitution / vervanging 7 12 c value of c / waarde van c 7 12 y y-coordinate / y-koördinaat 7 (4) [22] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

4 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2020) QUESTION 2/VRAAG 2 2.1  1  4   10  0  2 2 BC = substitution / vervanging BC = 25  100 answer / antwoord BC = 125  5 5 (2) 2.2 6  0 mAB  substitution / vervanging 2  6 3 mAB  answer / antwoord 4 (2) 2.3 3 mCD  mAB  gradient / gradiënt 4 y  mx  c 3 y  xc 4 Sub C  1;10  substitution / vervanging 3 10   1  c 4 43 c 4 3 43 y  x answer / antwoord (3) 4 4 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/SEPTEMBER 2020)) MATHEMATICS P2/WISKUNDE V2 5 2.4 M is the midpoint of both BD and AC / is die middelpunt van beide BC en AC  statement / stelling Midpoint of AC and BD /Middelpunt van AC en BD  1  2 10  6  M ;   substitution / vervanging  2 2   3  answer / antwoord M  ;2  2  (3) 2.5 x  4 3 y  0  ; 2  substitution / vervanging 2 2 2 x  4  3; y  0  4 x- value/waarde y-value/waarde x  7; y  4 Answer only: Full marks/ Slegs antwoord: Volpunte (3) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

6 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2020) QUESTION 3 / VRAAG 3 3.1 y B x O (– 12; – 5) diagram / diagram Therefore/d.w.s: r = 13 value of r / waarde van r sin B  cos B 5 12   substitution / vervanging 13 13 17 answer / antwoord  13 (5) 3.2 sin 43  p 3.2.1 cos133 cos(90  43)   sin 43   sin 43 answer / antwoord  p (2) 3.2.2 tan( 43)   tan 43   tan 43 p  answer / antwoord 1  p2 (3) 3.3.1 sin(360  x )  tan( x  180) sin(90  x )  sin x   sin x   tan x  cos x cos x  tan x   tan x  tan x   tan x  1  answer / antwoord (5) 3.3.2 sin 210.cos150.tan 25 tan 205.cos 315.sin135  sin 30.  cos 30.tan 25   sin 30    cos30 tan 25.cos 45.sin 45  tan 25 1 3  cos 45 .  2 2  sin 45 1 1 . 2 2 special angles / spesiale hoeke 3  answer / antwoord 2 (7) [22] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/SEPTEMBER 2020)) MATHEMATICS P2/WISKUNDE V2 7 QUESTION 4 / VRAAG 4 4.1 sin   cos  .sin   tan  cos   (1  sin 2  ) sin  (1  cos  ) factorising / faktorisering LHS = cos   cos 2   cos2  sin  (1  cos  ) = common factor / gemene faktor cos  (1  cos  ) answer / antwoord = tan  (4) 4.2 2sin x cos x  cos 2 x  0 cos x(2sin x  cos x)  0 factors / faktore  cos x  0 cos x  0 or/of 2sin x  cos x 1  tan x  1 2 cos x  0 or/of tan x  2 x  90  360.k or/of x  270  360.k  x  90  360.k or/of x  26,57  180.k  x  270  360.k  x  26,57  180.k (6) 4.3 2. sin   1 1 sin   2 1 sin   1 4 sin   4   14, 48   14, 48 or/of   165,52   165,52 (3) 4.4  x y 3 x y   45 tan    1 and/en cos( x  y )  2  2  2  x  y  30 x y  45 and/en x  y  30 2 setting up equations/ x  y  90 ...............(1) opstel van vergelykings x  y  30 ...............(2) 2 x  120 x  60  x-value/waarde y  30 y-value/waarde (5) [18] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

8 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2020) QUESTION 5 / VRAAG 5 5.1.1 0  y 1 or  0;1 0 1 (2) 5.1.2 Period = 720 answer (1) 5.2  shape / vorm  x-intercept / x-afsnit  y-intercept / y-afsnit  turning points / draaipunte (4) 5.3 180  x  150 or/of 30  x  180  180  x  150  30  x  180 (3) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/SEPTEMBER 2020)) MATHEMATICS P2/WISKUNDE V2 9 QUESTION 6 /VRAAG 6 6.1 PQ sin 60  PS 8 sin 60  8 PS  sin 60  PS 8 PS = 8 sin 60  PS = sin 60 16 3 PS = 16 3 3  PS = 3 (3) 6.2 PQ In PQS: tan 60  QS 8  QS = tan 60 8 8 3 QS =  m 8 3 tan 60 3  QS = m 3 8 3 QR = m 8 3 3  QR = m 3 In RQS: RS2  QR 2  QS2  2.QR.QS.cos100 formula / 2 2 2 2 8 3  8 3  8 3  8 3  formule =  3    3   2.  3  .  3  cos100         substitution / = 50,0756...... vervanging RS = 7,08 m simplification/ vereenvoudiging answer / antwoord (7) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

10 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2020) QUESTION 7 / VRAAG 7 B D C O A 7.1 ˆ =x Let/Laat A ˆ = x (angles opp = sides)/(hoeke teenoor  sye) BAD S and/en R ˆ ADO = 2 x (angle at the centre)/(middelpuntshoek ) S and/en R Similarly, if you let / Net so, as jy: Cˆ = y; ˆ = 2y then/dan: CDO S ˆ = 2 x  2 y = 2( x  y )  ADC ˆ = 2 ABC S and conclusion en gevolgtrekking (5) 7.2.1 B̂3  10 (angles opp = sides; DB = DF) S R (hoeke teenoor  sye; DB = DF) D̂ 2  20 (exterior angle of a BDF) S and/en R (buitehoek van BDF) (3) 7.2.2 ˆ = 90 (angles in a semi-circle) ABD S R (hoek in halwe sirkel ) Â = 70 (angles of a triangle) S R (hoeke van ' n driehoek ) (4) 7.2.3 Ô2  140 (angle at the centre) / (middelpuntshoek ) S R (2) 7.2.4 Ĉ1  110 (opposite angles of a c.q.)/(teenoorst. hoeke van k .v) S R OR / OF OR / OF ˆ O O ˆ O ˆ  220 (angles around a point/(omwenteling ) 1 4 3 S and/en R Ĉ1  110 (angle at the centre)/(middelpuntshoek ) S and/en R (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/SEPTEMBER 2020)) MATHEMATICS P2/WISKUNDE V2 11 7.2.5 Ê = 70 (angles in the same segment)/(hoeke in dieselfde segment ) S R OR/OF OR/OF Ê = 70 (opposite angles of a c.q.)/(teenoorst. hoeke van k .v) S R (2) 7.2.6 Ĉ2  70 (ext.  of a c.q.)/ (buitehoek van k .v) S R OR/OF OR/OF Ĉ2  110  180 (s on a straight line)/(hoeke op ' n reguitlyn) S R Ĉ2  70 (2) 7.2.7 Oˆ O ˆ  140 (vertically opp. s)/(regoorstaande e) S R 4 2 (2) [22] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

12 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2020) QUESTION 8 / VRAAG 8 P E A O C D F 8.1 ENG construction Draw diameter FP and join PA ˆ =x Let EFD S and R ˆ = 90 (tan  radius) OFD ˆ = 90  x  OFE S and R ˆ = 90  x (angles in the same segment)  PAE S and R ˆ = 90 (angles in a semi circle) PAF ˆ =x  EAF conclusion ˆ =x ˆ =A  EFD 8.1 AFR AFR Teken middellyn FP en verbind PA  konstruksie ˆ =x Laat EFD ˆ = 90 (raaklyn  radius ) OFD  S en R ˆ = 90  x  OFE  ˆ PAE = 90  x (hoeke in dieselfde segment )  S en R  S en R ˆ = 90 (hoeke in ' n halwe sirkel ) PAF ˆ =x  EAF  gevolgtrekking ˆ =x ˆ =A  EFD (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/SEPTEMBER 2020)) MATHEMATICS P2/WISKUNDE V2 13 A E 1 2 3 B O 1 2 D 2 x 1 C F 8.2.1 A = 𝑥 corresponding angles; AB S R ∥ DC / 𝑜𝑜𝑟𝑒𝑒𝑛𝑘𝑜𝑚𝑠𝑡𝑖𝑔𝑒 ℎ𝑜𝑒𝑘𝑒; AB || DC E 𝑥 tan-chord / 𝑟𝑎𝑎𝑘𝑙𝑦𝑛-𝑘𝑜𝑜𝑟𝑑 S R D 𝑥 angles opposite = sides / ℎ𝑜𝑒𝑘𝑒 𝑡𝑒𝑒𝑛𝑜𝑜𝑟 sye S R E 𝑥 alternate angles, AB S R ∥ DC / 𝑣𝑒𝑟𝑤𝑖𝑠𝑠𝑒𝑙𝑒𝑛𝑑𝑒 ℎ𝑜𝑒𝑘𝑒; AB || DC C E 𝑥 exterior angle of a c.q. / 𝑏𝑢𝑖𝑡𝑒ℎ𝑜𝑒𝑘 𝑣𝑎𝑛 ′𝑛 𝑘. 𝑣 S R (10) 8.2.2 B = 180° 𝑥 opposite angles of a c.q. S R 𝑡𝑒𝑒𝑛𝑜𝑜𝑟𝑠𝑡. ℎ𝑜𝑒𝑘𝑒 𝑣𝑎𝑛 ′𝑛 𝑘. 𝑣 A+B=𝑥 180° 𝑥 180° ∴ AD ∥ BC co-interior angles formed = 𝑘𝑜-𝑏𝑖𝑛𝑛𝑒 ℎ𝑜𝑒𝑘𝑒 𝑔𝑒𝑣𝑜𝑟𝑚 = 180 R ∴ ABCD is a parallelogram opp. sides ∥ ABCD is ′n parallelogram 𝑡𝑒𝑒𝑛𝑜𝑜𝑟𝑠𝑡. 𝑠𝑦𝑒 || R (4) [19] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

14 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2020) QUESTION 9 / VRAAG 9 9.1.1 perpendicular to the chord / loodreg op die koord answer/antwoord (1) 9.1.2 interior opposite angle / teenoorstaande binnehoek answer/antwoord (1) 9.2 F 1 x O 1 2 3 1 2 L C E H 9.2.1 E E 90° line from centre S R 𝑙𝑦𝑛 𝑣𝑎𝑛𝑎𝑓 𝑑𝑖𝑒 𝑚𝑖𝑑𝑑𝑒𝑙𝑝𝑢𝑛𝑡 FCH 90° angles in a semi-circle S R ℎ𝑜𝑒𝑘𝑒 𝑖𝑛 '𝑛 ℎ𝑎𝑙𝑤𝑒 𝑠𝑖𝑟𝑘𝑒𝑙 ∴ FCH E ∴ FC ∥ OE corresponding angles formed are R 𝑜𝑜𝑟𝑒𝑒𝑛𝑘𝑜𝑚𝑠𝑡𝑖𝑔𝑒 ℎ𝑜𝑒𝑘𝑒 𝑤𝑎𝑡 𝑔𝑒𝑣𝑜𝑟𝑚 𝑤𝑜𝑟𝑑 𝑖𝑠 (5) 9.2.2 ˆ LFO = 90 (tan  radius) / (raaklyn  radius ) S and/en R Ê 2  90 (proven) / (reeds bewys ) S and/en R  OFLE is a c.q. (converse exterior angle of a c.q.) R (omgekeerde buitehoek van k .v stelling ) (3) 9.2.3 Ĥ  x (tan - chord) / (raaklyn - koord ) S R Ô1  2 x (angle at the centre) S (middelpuntshoek ) R (4) [14] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief