Mathematics Grade 12 Paper 2 KwaZulu Natal Mock 2019 Memo

Downloaded from hlayiso.com Mathematics P2 1 Preparatory Examination September 2019 NSC NATIONAL SENIOR CERTIFICATE GRADE 12 MATHEMATICS P2 MARKING GUIDELINE PREPARATORY EXAMINATION SEPTEMBER 2019 MARKS: 150 TIME: 3 hours This memorandum consists 15 of pages. Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 2 Preparatory Examination September 2019 NSC QUESTION 1 39 42 48 54 62 68 78 78 82 91 93 1.1 x = 66,82 or 66,82 thousands A ✓calculation CA ✓ answer (2) 1.2 σ = 18,3 AA ✓✓answer (2) If formula is used CA will apply. 1.3 ( x − σ; x + σ) = (48,52 ; 85,12) CA✓ substitution  6 countries CA✓ answer (2) 1.4 35 40 45 50 55 60 65 70 75 80 85 90 95 A✓min & max values A✓quartile 1 value A✓median value & form A✓quartile 3 value (4) 1.5 The data is skewed to the left/negatively skewed CA✓answer (1) 1.6.1 The sum of the data provided for the years must A✓answer (1) increase. 1.6.2 IQR will remain the same or change. A✓answer (both same or change) (1) [13] QUESTION 2 YEAR 2002 2003 2004 2005 2006 2007 2008 2009 2010 2011 2012 Agriculture (x) 4.2 3,4 3,1 2,7 2,9 3,0 2,9 3,0 2,6 2,5 2,6 Mining (y) 19,2 19,4 19,2 18,5 17,5 17,0 16,8 15,2 14,2 12,8 12,4 2.1 a= 5,65 and b= 3,65 A✓ a – value A✓ b – value (3) y= 5,65 +3,65x CA✓ equation 2.2 y= 5,65+3,65x = 5,65+3,65(1,2) A✓ substitution of 1,2 (2) CA✓ answer =10,03% 2.3 strong positive correlation of the data CA✓ strong positive r = 0,7 A✓ value of correlation coefficient (2) [7] Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 3 Preparatory Examination September 2019 NSC QUESTION 3 𝑦 L(3; 6) K(−1; 4) R M N(4; 1) 𝜃 𝑥 O P 3.1 3.1.1 6−4 m KL = 3 − (− 1) A✓ substitution in the answer only: full marks correct formula 2 1 = or 4 2 CA✓ answer (2) 3.1.2  x N + x L ; YN + YL  midpoint of R :  ;   2 2   3 + 4 6 + 1 :  ;  A✓ substitution in the  2 2  correct formula 7 7 R  ;  answer only: full marks CA✓ x CA✓ y 2 2 (3) 3.1.3 x −1 7 x −1 7 co-ordinates of M: = CA✓ = 2 2 2 2 x −1= 7 x = 8 y+4 7 y+4 7 = answer only: full marks CA✓ = 2 2 2 2 y = 7 – 4 y = 3 CA✓ x CA ✓ y  M (8; 3) (4) 3.2 The equation of NM 1 CA✓ gradient m NM = [NM || KL] 2 y − y1 = m (x − x1 ) y −1 = (x − 4) 1 CA✓ correct subst. of 2 m and N(4 ; 1) 1 1 y = x − 2 + 1 y = x − 1 CA✓ equation 2 2 (3) Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 4 Preparatory Examination September 2019 NSC 3.3 3.3.1 1 0= x −1 A✓ y = 0 2 x=2 answer only: full marks CA✓ x  P (2; 0) (2) 3.3.2 1 tan θ = answer only: full marks CA✓ tan θ = m 2 CA✓ answer θ = 26,57o (2) 3.3.3 0−4 4 m KP = =− CA✓ mKP 2 − (− 1) 3  − 4 CA✓ inclination of tan–1   = 180 − 53,13 = 126,87  3  KP KP̂N = 126,87 − 26,57 A✓ method = 100,3 CA✓ answer (4) [20] Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 5 Preparatory Examination September 2019 NSC QUESTION 4 𝑦 C B T O D 𝑥 𝑦 = −𝑥 + 5 4.1. x 2 + 2x + y 2 − 4 y = 27 A✓ completing square x 2 + 2x + 1 + y 2 − 4 y + 4 = 27 + 1 + 4 LHS (x + 1)2 + ( y − 2)2 = 32 A✓ LHS  T (–1; 2) CA✓ x CA✓ y (4) OR 2 x= = −1 −2 AA✓✓ x −4 y= =2 AA✓✓ y −2 (4) 4.2 mTB  m BD = −1 [tangent ⊥ radius] mTB = 1 A✓ mTB = 1 y = mx + c 2 = 1 . (–1) + c A✓ subst of m andT c = 3 A✓ value of c  Equation of TB is y = x + 3 (3) 4.3 equating TB and BD: x + 3 = −x + 5 A✓equating 2x = 2 CA✓ x = 1 x =1 y = 1 + 3 or y = −1 + 5 CA✓ y = 4 B(1 ; 4) CA provided first (3) quadrant. Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 6 Preparatory Examination September 2019 NSC 4.4 ( x + 1) 2 + ( y − 2) 2 = r 2 CA✓ subst T in circle eq (1 + 1) 2 + ( 4 − 2) 2 = r 2 CA ✓ subst B 8 = r2 CA✓equation ( x + 1) 2 + ( y − 2) 2 = 8 (3) 4.5 Draw a horizontal line TE with E on BD: Point D(5 ; 0) and Point E: 2 = –x + 5 A✓ coordinates of D  E(3 ; 2) CA✓coordinates of E 1 Area of trap TEDO = (TE + OD) ⊥ h CA✓ length of TE 2 1 = (4 + 5)  2 CA✓ subst in area of 2 trap = 9 units 1 Area of ΔTBE = (TB  BE) 2 = 1 2 8 8( )( ) CA✓subst in correct = 4 units area of  formula Area of OTBD = 9 + 4 A✓ method = 13 units2 CA✓ answer (7) [20] Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 7 Preparatory Examination September 2019 NSC QUESTION 5 5.1.1 − 3 A answer tan  = =− 3 1 (1) 5.1.2 𝑟 2 = (1)2 + (−√3) 2 Pyth 𝑟2 = 4 A𝑟 = 2 𝑟=2 sin(− ) = − sin A reduction − 3 3 = −  or  CA answer  2  2 (3) 5.1.3 sin( − 60) = sin  cos 60 − cos sin 60 Aexpansion  3  1   1  3  =  −   −     2   2  2  CA both special ratios  2    CA value of cos  − 3 3 −2 3 − 3 = − = = 4 4 4 2 CAanswer (4) 5.2 tan(180° − 𝜃) sin(90° + 𝜃) cos 300° sin( 𝜃 − 360°) −tan 𝜃 . 𝑐𝑜𝑠𝜃 A −tan 𝜃 A 𝑐𝑜𝑠𝜃 = A cos 60° A 𝑠𝑖𝑛𝜃 cos 60° . sin 𝜃 sin − . cos Aidentity = cos 1 . sin 2 CA answer = −2 (6) 5.3.1 cos 2 x − 1 LHS = sin 2 x 1 − 2 sin 2 x − 1 A 1 − 2𝑠𝑖𝑛2 𝑥 = 2 sin x cos x A 2 sin 𝑥 cos 𝑥 sin x =− cos x A simplification = − tan x = RHS (3) 5.3.2 x = 90° ; 180° ; 270° A 90° A 180° A 270° (3) Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 8 Preparatory Examination September 2019 NSC 5.3.3 1 − tan 2 x = 4 1 tan 2 x = − 1 A tan 2 x = − 4 4 2 x = 165,96 + k .180 A165,96° x = 82,98 + k .90 ; k  Z CA 82,98° A𝑘 ∈ 𝑍 (4) [24] QUESTION 6 6.1 y 2 f g 1 0,87 x –120o –90o – 60o –30o 0o 30o 60o 90o 120o 150o 180o –1 (–120° ; –0,87) –2 A x-intercepts A shape A (–30° ; 1) & (150° ; -1) A (–120° ; 0) & (180° ; -0,87) (4) o 6.2 Period = 360 A answer (1) 6.3.1 – 30o < x < 150o CA endpoints A correct interval (2) 6.3.2 x  (0 ; 60 ) o o CA endpoints A correct interval (2) 6.4 ( y = cos x + 30 + 60 o o ) A cos ( x + 90o ) y = cos ( x + 90o ) CA answer y = − sin x (2) [11] Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 9 Preparatory Examination September 2019 NSC QUESTION 7 Q 𝑑 P θ 𝑑 S R 7.1 𝑄𝑆 2 = 𝑑2 + 𝑑2 − 2𝑑. 𝑑. cos 𝜃 A subst into 𝑄𝑆 2 = 2𝑑2 − 2𝑑 2 . cos 𝜃 cos-rule 𝑄𝑆 2 = 2𝑑2 (1 − cos 𝜃) A common 𝑄𝑆 = 𝑑√2(1 − cos 𝜃) factor (2) 7.2 𝑅̂ = 180° − 𝜃 opp. ∠𝑠 cyclic quad suppl = 𝑄𝑆̂𝑅 equal sides, equal angles 𝑆𝑄̂ 𝑅 = 2𝜃 − 180° sum ∠𝑠 ∆ A 𝑆𝑄̂ 𝑅 1 ∆𝑄𝑅𝑆 = . 𝑄𝑆. 𝑄𝑅 sin 𝑆𝑄̂ 𝑅 2 1 = . 𝑑√2(1 − cos 𝜃). 𝑑√2(1 − cos 𝜃) sin(2𝜃 − 180°) Asubst into 2 1 area = 𝑑2 . 2(1 − cos 𝜃)(− sin 2𝜃) rule 2 = −𝑑2 (1 − cos 𝜃). sin 2𝜃 Asimplify & reduction (3) [5] Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 10 Preparatory Examination September 2019 NSC QUESTION 8 8.1 F B C 1 2 65 O 1 2 D 3 A 40 E 8.1.1 Dˆ 2 + D̂3 = 90O (tan ⊥ radius) AS AR  D̂2 = 90o − 40o = 50o A answer (3) 8.1.2 D̂1 = Ĉ2 = 65 (angles opposite equal sides) o AS/R AS AR D̂1 + D̂2 = FB̂C (ext angle of a cyclic quad) 65o + 50o = FB̂C  FB̂C = 115o CAanswer (4) Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 11 Preparatory Examination September 2019 NSC 8.2 R 32o S 2 O 1 T 21 1 2 15o P 1 2 M 8.2.1 Ŝ = 32o (s in the same segment) AS AR (2) 8.2.2 ˆ = 64o ( at centre = 2 ×  at circumference) AS AR (2) 2 8.2.3 Tˆ = 90o (line from centre to midpoint of chord) AS/R 1 Ô1 + T̂1 + M̂ 2 = 180O (sum of angles of ∆) AMethod Ô1 + 90∘ + 15∘ = 180° Ô1 = 75∘ Aanswer (3) [14] Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 12 Preparatory Examination September 2019 NSC QUESTION 9 P 1 2 x x 1 1 Q R 2 2 3 4 W 2 S 1 1 3 2 T V 9.1 Pˆ1 = Qˆ1 = x (tan chord theorem) AS A R Qˆ1 = S = x (corresp  ’s (PQ//VS) A S AR Sˆ = x (4) 9.2 Pˆ2 = Qˆ 2 (tan chord theorem) A S AR Qˆ 2 = Vˆ2 (alt  ’s (PQ//VS) A S AR  Pˆ2 = Vˆ2  PQTV is a cyclic quadrilateral (converse of angles in the same seg.) A R (5) 9.3 Qˆ 3 = Pˆ1 (  ’s in the same segment) A S AR  Qˆ 3 = Pˆ2  TQ is a tangent (converse: tan chord theorem) A R (3) [12] Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 13 Preparatory Examination September 2019 NSC QUESTION 10 10.1 A D P 1 Q E F B C Construct AP = DE and AQ = DF and draw PQ A construction In  APQ and DEF: AP = DE AQ = DF Â = D̂ ∆APQ  ∆DEF (SAS) A ∆APQ  ∆DEF P̂1 = Ê, But B̂ = Ê … A SAS (given) Pˆ1 = Bˆ PQ || BC (corresponding angles =) AS AR AB AC  = (proportionality theorem; PQ || BC) AS AR AP AQ AB AC  = (construction) DE DF (7) Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 14 Preparatory Examination September 2019 NSC 10.2 D E F H G 10.2.1 Ĝ (tan chord theorem) A S/R EFˆH (alt  ’s, DE||FH) A S R FEˆ H = Gˆ (  ’s opposite = sides) A S/R (4) 10.2.2(a) In Δ’s DEF and DGE: 1) D̂ is common A S 2) DEˆ F = Gˆ (proven) A S 3) 𝐷𝐹̂ 𝐸 = 𝐷𝐸̂ 𝐺 (sum of angles in triangle) A R ΔDEF ||| ΔDGE (  ) (3) 10.2.2(b) Let DÊF = x : EF̂D = 2 x (ext . angle of Δ) A S A R D̂ = 2 x (s opposite = sides) A S 2 x + 2 x + x = 180 A Equation  x = 36 A value of x o  D̂ = 2 x = 72 (5) Copyright Reserved Please turn over

Downloaded from hlayiso.com Mathematics P2 15 Preparatory Examination September 2019 NSC 10.2.3 DE DF = (ΔDEF ||| ΔDGE) A S DG DE FG DF A S = ( DE = FG ) DG FG 2 k = A substitution k +2 2 k 2 + 2k = 4 (3) 10.2.4 GH GF 2 = = (Prop. Intercept theorem DE//FH) GE GD k + 2 A S AR (2) [24] TOTAL: 150 Copyright Reserved Please turn over