# Mathematics Grade 11 P2 Gauteng November 2022 Memo (English) > Grade 11 Mathematics Paper 2 memorandum from the Gauteng November 2022 (English), 22 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/gr-11-mathematics-p2-english-november-2022-possible-answers/268902ee-0527-4a25-9fea-c8a0d10ea1a0 - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 11 - Language: English - Document type: Memorandum - Year: 2022 - Exam period: Gauteng November - Paper: 2 - Curriculum: CAPS - Pages: 22 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 11](https://www.hlayiso.com/library/grade-11) › [Mathematics](https://www.hlayiso.com/library/grade-11/mathematics) › [Mathematics Grade 11 P2 Gauteng November 2022 Memo (English)](https://www.hlayiso.com/library/gr-11-mathematics-p2-english-november-2022-possible-answers/268902ee-0527-4a25-9fea-c8a0d10ea1a0) ## About this document Mathematics · Grade 11 · Gauteng November · 2022 · English. Memorandum, 22 pages. Read online or download the PDF. Read it free online on the page above; download the PDF from there. ## Matching paper and memo - [Gr 11 Mathematics P2 (English) November 2022 Question Paper](https://www.hlayiso.com/library/gr-11-mathematics-p2-english-november-2022-question-paper/130a0fe0-0914-40c3-aedd-67a7e297f1cd) — QP · 2022 · Eng ## Document text Text extracted from the PDF. Layout, diagrams and tables may not survive extraction. ### Page 1 Downloaded from hlayiso.com PROVINCIAL EXAMINATION NOVEMBER 2022 GRADE 11 MARKING GUIDELINES MATHEMATICS (PAPER 2) 22 pages ### Page 2 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 INSTRUCTIONS AND INFORMATION ➢ A – ACCURACY ➢ CA – CONSISTENT ACCURACY ➢ S – STATEMENT ➢ R – REASON ➢ S/R – STATEMENT with REASON NOTES: • If a candidate answered a question TWICE, mark only the first attempt. • If a candidate crossed OUT an answer and did not redo it, mark the crossed-out answer. • Consistent accuracy applies to ALL aspects of the marking guidelines. • Assuming values/answers in order to solve a question is UNACCEPTABLE. 2 ### Page 3 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 QUESTION 1 1.1 250 250 250 250 250 ✓ Box (Q1 and Q3) Commented [BK1]: Wrong values on Box and Whisker ✓ Whiskers (Min. and ⚫ ⚫ Max.) ✓ Median 245 250 255 260 265 270 275 280 285 290 295 300 305 310 315 320 325 330 335 340 (3) 1.2 The data is distributed symmetrically, a normal distribution. ✓ Explanation OR The date is slightly skewed to the right. (1) 1.3  = 23,77   (2) 1.4 (x −  ; x +  ) ✓ x = 287 = (287 – 23,77 ; 287 + 23,77) ✓ Interval = (263,23 ; 310,77) ✓6  6 municipalities are within one standard deviation. (3) 1.5 1 S-IQR = (Q3 – Q1) ✓ Q3 – Q1 2 1 1 = (306 – 266) ✓  2 2 = 20 ✓ Answer (3) [12] 3 ### Page 4 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 QUESTION 2 2.1 p=5 ✓p q = 27 ✓q r=1 ✓r (3) 2.2 CUMULATIVE FREQUENCY GRAPH (OGIVE) 34 32 30 28 26 Cumulative frequency of learners 24 22 20 18 16 14 12  Ground at (20 ; 0) 10  Endpoints 8  Join with free-hand 6 (3) 4 2 0 0 10 20 30 40 50 60 70 80 90 100 MARKS 2.3 M = 46  M = 46( 2) (2) 2.4 32 – 24 ✓ CF = 24( 1) =8 ✓ 32 – CF  8 learners achieved a mark of greater than 55. (2) [10] 4 ### Page 5 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 QUESTION 3 y A(0 ; 7) B(10 ; 2) D(x ; y) O   x C(7 ; –4) 3.1 AB = (0 − 10) 2 + (7 − 2) 2 ✓ Substitute into correct formula AB = 125 ✓ AB AB = 5 5 ✓ Simplified surd (3) 3.2 7 − ( −4) ✓ Gradient substitute m AC = 0−7 11 m AC = − ✓ m AC 7 (2) 3.3 11 ✓ Angle of inclination tan  = − 7 substitute (CA 4.2) RA = 57,528 …° ✓ RA  = 180° – 57,528 …° ✓ θ (subtract RA from 180°)  = 122,47° (3) 5 ### Page 6 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 3.4 tan  = 2   = 63,434 …° ˆB AC A ˆCB = 122,47° – 63,434 …° C ˆAB A ˆCB = 59,04° B ˆCD 1 m AB = 2 ˆ B = 180° – tan–1  1  – 122,74° CA 2 C ˆAB = 30,69° A ˆCD = 30,69° [alt 's, AB ‖ CD] B ˆCD = 30,69° + 59 04° B ˆCD = 89,73° OR OR IF  is not rounded tan  = 2 ✓  = 63,434 …° ˆB ✓ AC ˆ B = 122,4711 …° – 63,434 …° AC ˆB ✓ CA ˆ B = 59,036 …° AC ˆD ✓ BC 1 m AB = 2 ˆ B = 180° – tan–1  1  – 122,74 …° CA 2 ˆ B = 30,963° CA ˆ D = 30,963° AC [alt 's, AB || CD] ˆ D = 30,963° + 59 036 …° BC ˆ D = 90° BC (4) [12] 6 ### Page 7 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 QUESTION 4 y K M(4 ; 3) x O L(4 ; 3) 4.1 y = 5x + 9 and 5y + x – 19 = 0 ✓ Sub y 5(5x + 9) +x = 19 ✓ Solve for x 26x = –26 ✓ Sub x back x = –1 NO MARK for coordinates (given) y = 5(–1) + 9 y=5  K(–1 ; 4) (3) 4.2 mKL = 5 and 1 mKM = − ✓ mKL 5  mKL  mKM = –1 ✓ mKL KL ⊥ KM ✓ Definition of perpendicular ˆ M = 90°  LK gradients (3) 7 ### Page 8 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 4.3 KL = ((−1 − (−2))2 + (4 − (−1) )2 ✓ KL length Commented [BK2]: KL not ML KL = 26 ✓ KN length KN = ( −1 − 4) 2 + ( 4 − 3) 2 ✓ Sub Area formula KN = 26 ✓ Area ΔKML Area KML = 1 2 ( 26 ) ( 26 ) Area KML = 13 units2 (4) − 2 + 4 3 + ( −1)  ✓ Midpoint of NL M  4.4 ;   2 2  M (1 ; 1) 3 − ( −1) mNL = 4 − ( −2) 2 mNL = ✓ Gradient of NL 3 2 y – y1 = – (x – x1) ✓ Perpendicular gradient 3 2 y–1=– (x – 1) ✓ Substitute Midpoint 3 2 5 y=– x+ ✓ Equation 3 2 (5) 4.5 mNL = m NP = mLP ✓ Definition of collinear 2 y−3 = ✓ Sub P(7 ; y) 3 7−4 y=5 (2) 8 ### Page 9 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 4.6 Q(3 ; –2) ✓ xQ ✓ yQ (2) 4.7 K̂ = 90° [proven in 4.1] ✓ Interior angle of 90° KL = KN = 26 [proven in 4.2] ✓ Adjacent sides equal  KLQN is a square [a parallelogram with an interior  of 90° and adjacent sides equal] (2) [21] QUESTION 5 5.1 tan (180  − x ) cos (360  + x ) ✓ – tan x 2 cos (90 + x ) − tan x.cos x ✓ cos x = − 2 sin x ✓ –2 sin x sin x − .cos x = cos x ✓ – tan x identity − 2 sin x − sin x = ✓ simplify − 2 sin x 1 = ✓ answer 2 (6) 5.2 5.2.1 sin − 2 sin cos ✓ sin (1–2 cos) LHS = 2 cos2  + cos − 1 ✓ (2cos – 1)(cos + 1) sin  (1 − 2 cos ) = ✓ Sign change/negative factor (2 cos − 1) (cos + 1) − sin  (2 cos − 1) = ✓ Simplify with conclusion (2 cos − 1) (cos + 1) − sin  = cos + 1 = RHS (4) 9 ### Page 10 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 5.2.2 Undefined when ✓ 2cos – 1 = 0 2cos – 1 = 0 ✓ QI general solution 1 ✓ QII general solution Commented [BK3]: QIV not QII cos = 2 RA = 60° QI :  = 60° + k.360° ; kZ ✓ kZ Commented [BK4]: K element of integers Z QIV :  = 300° + k.360° OR when cos  = –1  = 180° + k.360° ; kZ NOTE: Accept other equivalent general solutions. cos – 1 ✓ cos  = –1 general solution Commented [BK5]: equation  = 180° + k.360° ; kZ (5) 10 ### Page 11 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 5.3 • P(k ; m) y A(b ; 1) • 6 4 α T O • x 5.3.1 b2 + 12 = 42 ✓ Pythagoras substitution b =  15 b = – 15 (QII) ✓ b = – 15 (2) 5.3.2 (a) 1 y tan = – ✓ 15 x ✓ CA x (2) (b) sin(–  – 180°) + cos(– ) ✓ sin  = sin  + cos  ✓ cos   − 15  =   +  1  ✓ Substitute x ; y and r  4   4   1 − 15  =     4  (3) 11 ### Page 12 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 5.3.3 k m k cos( – 90) = sin( – 90) = ✓ cos( – 90) = 6 6 6 k m m sin  = – cos  = ✓ sin( – 90) = 6 6 6 k 1 m  = cos  = – ✓ Complementary reductions 6 4 6 k= 3 m  − 15  2  =    ✓ Equate ratios 6  4  3 15 m= ✓ Values of k and m 2 OR OR P lies in QI, therefore both k and m are positive ✓ P in QI A 90º rotation means (x ; y) → (y ; x) ✓ (x ; y) → (y ; x) idea r2 ✓ Ratio of radii r1 6 ✓ Apply scale to x and y 4 3 ✓ Values of k and m 2  P  .1 ; . 15  = P(k ; m) 3 3 Do NOT award full marks 2 2  for answer only. 3 3 15 k= and m = 2 2 (5) [27] 12 ### Page 13 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 QUESTION 6 y 2 1 f M(48,05° ; 0,41) x −90° −60° −30° 0° 30° 60° 90° 120° 150° 180° -1 g -2 6.1 A=2 ✓ Answer (1) 6.2 120º ✓ Answer (1) 6.3 a = –1 ✓a b=3 ✓b p = 30° ✓p (3) 6.4 6.4.1 –90°  x < 60° ✓ Interval (1) 6.4.2 x = –30° or 48,05°  x  180° ✓ Interval ✓ x = –30° (2) 6.4.3 60° < x < 150° or 150° < x  180° ✓ 60º and 180º ✓ Not incl. 150º but incl. 180º (2) 6.5 t = –30° + k. 360° ; k ✓✓ t = –30° ✓ k. 360° (3) [13] 13 ### Page 14 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 QUESTION 7 T   h 1 2 G B x A 7.1 B̂1 =  =  ✓ Answer (1) Commented [BK6]: Not = but + 7.2 In TBG ✓ Correct ratio in ΔTBG h sin( + ) = ✓ BT BT h BT = ✓ Sine rule substitution sin( +  ) In ABT ✓ BT substitution  =  ✓ simplification x BT = sin  sin  x h = sin  sin ( +  ) sin  h sin  x= sin ( +  ) sin  (5) h sin  7.3 x= ✓ Substitution sin ( +  ) sin  20 sin 37 x= sin (65 + 37) sin 65 x = 13,58 m ✓ Answer The cat travelled 13,58 m (2) [8] 14 ### Page 15 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 QUESTION 8 FIGURE A FIGURE B 18 cm r Q 28° 1 3 4 3 V= r or V = r 3 3 P 8.1 1 V= (9)3 ✓ Substitution (r = 9) 3 V = 763,41cm3 ✓ Answer (2) 8.2 V= 4 r ✓ Substitute volume into 3 correct formula 4 268.08 = r ✓ Subject of the formula 3 3  286 ,08 3 =r ✓r 4π r=4 4 tan 14° = ✓ tan 14° PQ 4 PQ = ✓ PQ tan 14 PQ = 16,04 cm (5) [7] 15 ### Page 16 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 QUESTION 9 9.1 O • D M ✓ Construction F Construction: Join OF and OD In OFM and ODM i. FM = MD [GIVEN] ii. OM is a common side iii. OF = OD [Radii]  OFM  ODM [SSS] ✓R ˆ M1 = ˆ M2 [OFM  ODM] ✓R But ˆ M1 = ˆ M 2 = 180° ['s on a straight line] ✓S Commented [BK7]: + not =  ˆ M1 = ˆ M 2 = 90°  OM ⊥ FD (4) 16 ### Page 17 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 9.2 P S O • M 2 62° 1 1 R 2 1 2 Q 9.2.1 M̂ 2 = 90° [line from centre to midpoint of chord] ✓SR (2) 9.2.2 Q̂1 + Q̂2 = 90° ['s in a semi-circle] ✓SR (2) 9.2.3 P̂ = 28° [int 's of a ] ✓ S/R Ŝ = P̂ = 28° ['s in the same segment] R (2) [10] 17 ### Page 18 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 QUESTION 10 O x E D F B 10.1 B̂1 = D̂1 = x ['s in the same segment] ✓S B̂1 = Ê1 = x ['s in the same segment] ✓S (2) 10.2 10.2.1 Ô1 = 2x [ at centre = 2   at the circumf] ✓SR (2) 10.2.2 ˆ B + B̂ Ô1 = OF [ext  of ] R 1 2x = OˆFB + x  OˆFB = x OˆFB = x (2) 10.2.3 D̂2 = 90° – x ['s in a semi circle] ✓S R (2) 10.2.4 Ô2 = 180° – 2x ['s on a strt line] ✓SR Ô2 = Ê 2 = 180° – 2x ['s in the same segment] ✓SR (4) 10.3 Ê 2 = B̂ 3 + D̂2 [ext  of ] R  B̂3 = 90° – x  B̂3 = 90° – x  B̂3 = D̂2 [both 90° – x] ✓S  EB = ED [sides opp equal s] R (4) [16] 18 ### Page 19 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 QUESTION 11 L R S • O P D T 11.1 Construction: Draw diameter SP and join PT  Construction ˆ R = 90° – RS LS ˆP [tan ⊥ rad] SR SˆTR = 90° – R ˆTP ['s in a semi circle] SR ˆ P = RT RS ˆP ['s in the same segment]  LˆSR = SˆTR SR 19 ### Page 20 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 OR L R S • O P D T Construction: Draw diameter SP and join PR  Construction ˆ R = 90° – RS LS ˆP [tan ⊥ rad]  S/R S ˆRP = 90° ['s in a semi circle] SR ˆP ˆ R = 90° – RS SP [int 's of ] LˆSR = SˆPR [both 90° – RˆSP ] S ˆ R = ST SP ˆR ['s in the same segment]  S/R ˆ R = ST  LS ˆR 20 ### Page 21 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 OR L R 1 S 1 2 2 3 1 • O D T Construction: Join OS and OR  Construction Ŝ1 + Ŝ 2 = 90° [tan ⊥ radius] SR Ŝ 2 = 90° – Ŝ1 R̂ 1 = Ŝ 2 = 90° – Ŝ1 ['s opp equal radii] ✓ S/R Ô1 = 180° – 2[90° – Ŝ1 ] = 2 Ŝ1 [int 's of ] S Ô1 = Ŝ1 [at centre = 2   at circumf] S✓R Commented [BK8]: Not S but 2T  LˆSR = SˆTR (6) 21 ### Page 22 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 11.2 D C B A x T 11.2.1 tan chord theorem ✓R (1) 11.2.2 Â4 = 180° –  3 ['s strt line] ✓ SR Ĉ1 = 180° –  3 [tan chord theorem] ✓S✓R Ĉ 2 =  3 ['s strt line] ✓ SR D̂2 =  3 [tan chord theorem] ✓ SR  Ĉ 2 = D̂2 [both equal  3 ] ✓S D̂1 = F̂ [tan chord theorem/proven in 11.2.1] ✓R Â1 = Â2 [int 's in /3rd  in ] (7) [14] TOTAL: 150 22 ## Related documents - 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