Mathematics Grade 11 P2 Gauteng November 2023 Memo (English)

Downloaded from hlayiso.com PROVINCIAL EXAMINATION NOVEMBER 2023 GRADE 11 MARKING GUIDELINES MATHEMATICS (PAPER 2) 14 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 Class interval Frequency Cumulative frequency ✓ Frequency values 300 < x  400 4 4 ✓ Cumulative 400 < x  500 11 15 frequency values 500 < x  600 22 37 600 < x  700 13 50 700 < x  800 7 57 800 < x  900 3 60 (2) 1.2 70 ✓ (300 ; 0) ✓ End points 60 ✓ Shape 50 Cumulative Frequency 40 30 20 10 0 0 100 200 300 400 500 600 700 800 900 1000 Daily Sales (3) 1.3 Q2 = R580  Answer Accept R560 to R600 (2) 1.4 On 10 days they made a loss.  Answer  50 days they made a profit. (2) 1.5 To be in the top 25% of the previous months, the tuck ✓ R650 shop would need to have sold more than R650. ✓ Conclusion Therefore, R725 would have been in the top 25%. (2) [11] 3

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 QUESTION 2 2.1 2k + k + 1 + k + 2 + k − 3 + 2k − 2 ✓ 7k – 2 = 15 5 ✓ Divide by 5 7k − 2 = 15 ✓ Manipulation 5 7k – 2 = 75 7k = 77 k = 11 (3) 2.2 22 ; 12 ; 13 ; 8 ; 20  Answer  = 5,22 (2) 2.3 Mean would decrease by t units but the standard ✓ Mean decreases by t deviation would be unaffected. units ✓ Standard deviation unaffected (2) [7] QUESTION 3 3.1 0 − ( −2) ✓ Substitution into mMQ = − 2 − ( −4) correct formula 1 mMQ = − ✓ Answer 3 (2) 3.2 1 ✓ Perpendicular gradients mMQ  − = −1 MP ⊥ MQ 3 ✓ Gradient of MP mMP = 3 ✓ Answer y = 3x + 6 (3) 3.3 xR + 4 y R + ( −2) ✓ Substitution into –2 = 0= midpoint formula 2 2 ✓ xR x R = −8 yR = 2 ✓ yR R (–8 ; 2) (3) 4

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 3.4 PQ = (0 − 4) 2 + (6 − ( −2)) 2 ✓ Substitute into distance formula PQ = 80 PQ = 4 5 ✓ Simplified surd (2) 3.5 E(–4 ; –6) ✓ x E = −4 ✓ y E = −6 (2) 3.5 The diagonals of the parallelogram are perpendicular. ✓ Diagonals perpendicular ✓ Given parallelogram (2) [14] QUESTION 4 4.1 subst. x = 0 ✓ x=0 py – x – 5p = 0 ✓ Manipulation py – (0) = 5p ✓ x=5 5p y= p y=5 C(0 ; 5) (3) 4.2 py – x – 5p = 0 ✓ Standard form py = x + 5p ✓ Equate gradients x y= +5 p 1 mAC = 2 1 1  = p 2 p=2 OR 1 A(–10 ; 0) mAC = ✓ A (–10 ; 0) 2 ✓ Substitute A p(0) – (–10) – 5p = 0 10 = 5p p=2 (2) 5

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 4.3 Equation of AC: 2y – x – 10 = 0 ✓ Substitute p = 2 1 y= x+5 2 ✓ Equation of BD Equation of BD: y = – 2x – 10 1 ✓ Substitute y into other x + 5 = – 2x – 10 2 equation x + 10 = – 4x – 20 5x = – 30 x = –6 ✓ x = –6 y=2 ✓ y=2 B(–6 ; 2) (5) 4.4 1 ✓ Substitution into correct tan  = 2 formula  = 26,57 ✓ Answer (2) 4.5 AĈD = 63,43 int 's of  ✓ AĈD  = 26,56 int 's of  ✓  = ✓ Rede  ABOD is a cyclic quadrilateral (converse 's in same segment) (3) 4.6 DB is a diameter (line subtends 90) ✓ x=0  0 + 0 5 + ( −10)  5 M CD =  ;  ✓ y=–  2 2  2  5 = M CD  0 ; −   2 (2) [17] 6

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 QUESTION 5 5.1 5.1.1 2 ✓ Quadrant II sin  = 3 y ✓ x=– 5 2 2 2 x + (2) = 3 ✓ Substitute x=– 5 3cos2  – 1 x cos   5 2 ✓ Answer =3   –1  3  5 = –1 3 2 = 3 (4) 5.1.2 tan (– –180) ✓ –tan  = –tan  ✓ Substitution  2  ✓ Answer = –  − 5 2 = 5 (3) 5.2 5.2.1 t cos 15 = 4 t ✓ cos 15 = 4 4 cos 15 = t t ✓ t 2 − 16 sin 15 ✓ t t − 16 2 15° = 4 t (3) 5.2.2 sin 75 ✓ Complement = cos 15 ✓ Answer 4 = t (2) 7

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 5.2.3 1 – tan2 15  t 2 − 16  2  t 2 − 16  2 ✓    4  =1–      4    t 2 − 16 t 2 − 16 ✓ =1– 16 16 ✓ Correct 16 − t 2 + 16 = multiplication by 16 32 − t 2 LCD = 16 (3) 5.3 cos(90 − α ) sin( − − 540 ) ✓ tan 45 ✓ sin .– sin  tan 225 + sin . sin(180 + ) ✓ sin .sin  sin α.sin α ✓ 1 – sin2  = tan 45 + (sin .)( − sin α ) 1 ✓ sin2 α tan 2 α = 1 − sin2 α sin 2 α = cos 2 α = tan2  (5) 5.4 5.4.1 1 – cos = 2 sin2 ✓ 2 (1 – cos2) 1 – cos = 2 (1 – cos2) 1 – cos = 2 – 2 cos2 ✓ standard form 2 cos2 – cos  – 1 = 0 (2 cos + 1)(cos – 1) = 0 (2) 5.4.2 2cos + 1 = 0 or cos – 1 = 0 1 ✓ cos = – 1 2 cos = – or cos = 1 2 ✓ cos = 1 RA = 60 or  = 0 + K.360 ; K  Z ✓ Both solutions for QII :  = 120 + K.360 ; K  Z 1 cos = – QIII :  = 240 + K.360 ; K  Z 2 ✓  = 0 + K.360 ✓ +KZ (5) [27] 8

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 QUESTION 6 6.1 180 ✓ Answer (1) 6.2 1 1 ✓ Endpoints – y 2 2 ✓ Notation (2) 6.3.1 45  x  90 ✓ x  45 ✓ x < 90 (2) 6.3.2 –90< x < – 45 or 90< x < 135 ✓ –90< x < – 45 ✓ 90< x < 135 (2) 6.3.3 2 tan x + cos 2x = 2 ✓ 2 tan x – 2 = –cos 2x 2 tan x – 2 = –cos 2x 1 ✓ tan x – 1 = – cos 2x 1 2 tan x – 1 = – cos 2x 2 ✓ Value of x  f (x) = g(x) where x = 45 (3) 6.4 sin x + cos x ✓ tan x + 1 h(x) = cos x ✓ Up by 2 units sin x h(x) = +1 cos x h(x) = tan x + 1 h(x) = f (x) + 2  Vertical translation up by 2 units (2)  1  ✓ –2 cos 2x 6.5 p(x) = 4  − cos 2 x   2  ✓ Max of 2 p(x) = – 2 cos 2x  Maximum value of 2 (2) [14] 9

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 QUESTION 7 1 1 7.1 tan p = ✓ tan p = 3 3 p = 18,435 ✓ Value of p to 3 decimal places (2) 7.2 In BCD ✓ Correct substitution CD 121 into sine rule = ✓ Manipulation  18,4  sin135 sin  ✓ Length of CD  2  121 x sin 9,2 CD = sin135 CD = 27,36 m (3) [5] QUESTION 8 8.1 V = x2 h ✓ Substitute into volume 490 = x2h formula 490 ✓ h in terms of x h= 2 x (2) 8.2 A = x2 + 4xh ✓ Base of x2  490   490  A = x2 + 4x  2  ✓ 4x  2   x   x  1960 3 A = x2 + cm x (2) 8.3 490 ✓ Substitute h = 10 10 = x2 ✓ x2 = 49 x2 = 49 ✓ x=7 x=7 (x > 0) ✓ Surface area 1 960 A = (7)2 + 7 2 A = 329 cm (4) [8] 10

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 QUESTION 9 9.1 Q 1 2 O 1 2 P R Join PO and extend ✓ Construction In ΔPOQ ✓ SR Ô1 = Q̂ + P̂1 ext  of  ✓ SR but Q̂ = P̂1 ' s opp equal radii ✓ S  Ô1 = 2P̂1 Similarly in  ROQ Ô 2 = 2P̂2 ✓ S  Ô1 + Ô 2 = 2P̂1 + 2P̂2 ✓ S ( Ô1 + Ô 2 = 2 P̂1 + P̂2 )  RÔQ = 2P̂ (6) 9.2 9.2.1 (a) D̂ = 55° ' s in the same segment ✓ S R (2) (b) Ô 2 = 110°  at centre = 2   at ✓ S R circumference (2) (c) Ê 2 = 110° – 90 ext  of  ✓ R ✓ S Ê 2 = 20° (2) 11

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 (d) Ê1 = 35° int ' s of  ✓ R ✓ S ✓R Ê 3 = 35° ' s in a semi-circle OR Ê 3 = F̂1 ' s opp equal radii ✓ S R ✓ SR Ê 3 = 35° int ' s of  (3) 9.2.2 OE = 5 equal radii ✓ SR HE = 4,7 line from centre perpendicular ✓ S ✓ R to chord OH2 = 52 – (4,7)2 Pythagoras' Theorem ✓ S OH = 1,71 units OR OE = 5 radii ✓ SR OH ✓✓ sin 20° sin 20 = 5 ✓ OH OH = 1,71 units OR HE = 4,7 line from centre ✓ S R perpendicular to chord ✓ tan 20° OH ✓ OH sin 20 = 4,7 OH = 1,71 units (4) [19] QUESTION 10 10.1 B̂1 = x alt. 's, AC ‖‖ TV ✓ S ✓R V̂1 = x ' s opp equal sides ✓ S ✓R T̂2 = x tan chord theorem ✓ S ✓R (6) 10.2 Â = 2 x opp ' s parm ✓ S ✓R B̂3 = 2 x ext  cyclic quad ✓ S ✓R  Â = B3 both equal 2x ✓ R AT = BT sides opp equal ' s 12

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 OR B̂3 = 2 x ext  cyclic quad ✓ S ✓R ✓ SR T̂1 = 2 x alt 's, AC ‖‖ TV  T̂1 = V̂1 + V̂2 both equal 2x ✓ S ✓R  BTVC is an isosceles trapezium (one pair parallel sides and one pair of base angles ✓ Logic equal)  CV = BT sides of isos trap en CV = AT opp sides parm  AT = BT (5) 10.3 In  ATB, ✓ R 5x = 180° int ' s of  ✓ S x = 36° (2) [13] QUESTION 11 11.1 D̂3 = Ĥ1 + Ê1 ext  of  ✓ S (1) 11.2 D̂3 = Ĥ1 + Ê1 ext  of  Ĥ1 = F̂2 tan chord theorem ✓ S ✓R ✓ SR and Ê1 = F̂1 tan chord theorem  D̂ 3 = F̂1 + F̂2 (  DEFH is a cyclic quadrilateral (ext  = opp int  ) ✓ R (with D̂3 = F̂1 + F̂2 ) (4) 11.3 Ĥ1 = F̂2 tan chord theorem ✓ SR ✓ S ✓R Ĥ1 = F̂1 ' s in the same segment  F̂1 = F̂2 (3) 13

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 11.4 Ĥ 3 + Ĥ 4 = Ê1 + Ê 2 ext  cyclic quad ✓ S ✓R Ĥ 4 = K̂ 1 tan chord theorem ✓ SR Ĥ 3 = K̂ 2 ' s in the same circle segment ✓ S ✓R  K̂1 + K̂ 2 = Ê1 + Ê 2 K̂1 = Ê1 given ✓ S  K̂ 2 = Ê 2 KF is a tangent converse tan chord theorem ✓ R (if K̂ 2 = Ê 2 proven) (7) [15] TOTAL: 150 14