Mathematics Grade 11 P2 Gauteng November 2022 Memo (English)

Downloaded from hlayiso.com PROVINCIAL EXAMINATION NOVEMBER 2022 GRADE 11 MARKING GUIDELINES MATHEMATICS (PAPER 2) 22 pages

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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