Mathematics Grade 10 Paper 2 KwaZulu Natal June 2019 Memo

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Downloaded from hlayiso.com Mathematics/P2 Common Test June 2019 NSC- GRADE 10 INSTRUCTIONS AND INFORMATION Read the following instructions carefully before answering the questions. Le This question paper consists of 3 questions. 2 Answer ALL the questions. ch Clearly show ALL calculations, diagrams, graphs, etc. which you have used in determining your answers. 4. Answers only will NOT necessarily be awarded full marks. > You may use an approved scientific calculator (non-programmable and non-graphical), unless stated otherwise. 6. If necessary, round off answers correct to TWO decimal places, unless stated otherwise. c Diagrams are NOT necessarily drawn to scale. 8. Write neatly and legibly. Copyright Reserved Please turn over

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Downloaded from hlayiso.com Mathematics/P2 Common Test June 2019 NSC- GRADE 10 QUESTION 2 21 Given that a@ = 24,6° and £ =132,7° calculate the value of the following (correct to TWO decimal places): i 21.1 cosa (1) . 2 led cosec2 8 (2) 22 Various options are provided as possible answers to the following questions. Choose the correct answer and write only the letter (A — D) next to the question number, for example: 2.2.4 A 221 In the diagram below of right angle triangle EFG, EF = 17cm and FG = 15cm. E 17cm F 5 8cem 15cm G Which trigonometric equation could be used to determine the value of angle £? : 17 15 A. sin E =— B. cos E = — 15 7. 15 : 8 (e, tan F = — D. sin E =— 8 17 (1) 2.2.2 In the diagram below scalene ALMN shown, M =90°. - M o N Which of the following statements is always true? A. sin L =cosL B. sin L =cosN (1) (o cos L =cos M D. sin =cosM Copyright Reserved Please turn over

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Downloaded from hlayiso.com Mathematics/P2 Common Test June 2019 NSC- GRADE 10 QUESTION 3 3.1 Complete the following: The line joining the mid-points of two sides of a triangle is ....... to the third side and equal to ee the length of the third side. (2) 3.2 In the diagram below, APOR has E, F and G the midpoints of PQ, QR and PR respectively. QG // FH . Prove: 3.2.1 QOGHF isa parallelogram G3) 822 BG GH. (3) 3.2.3 ER // PH (3) (1) TOTAL: 50 Copyright Reserved

Downloaded from hlayiso.com Downloaded from Stanmorephysics.com 1 Common Test June 2019 Mathematics P2 NSC-Marking Guideline MATHEMATICS P2 MARKING GUIDELINE COMMON TEST JUNE 2019 NATIONAL SENIOR CERTIFICATE GRADE 12 MARKS: 150 N.B. This marking guidelines consists of 14 pages. Copyright Reserved Please Turn Over

Downloaded from hlayiso.com 2 Common Test June 2019 Mathematics P2 NSC-Marking Guideline QUESTION 1 1.1 220 A 220 𝑥= ̅ = 22 10  CA answer (2) Answer only full marks 1.2 σ = 3,95  AA answer (2) If formula is used 1CA mark for substitution and 1CA mark for answer. 1.3  A minimum & maximum value  A quartile 1 value  A median value A quartile 3 value If No Diagram No marks (4) 1.4 ( x − σ; x + σ) (22 – 3,95; 22 + 3,95) (18,05; 25, 95) CA 18,05 6 runners  CA 25,95 (answer only – full marks)  CA answer (3) [11] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Downloaded from Stanmorephysics.com 3 Common Test June 2019 Mathematics P2 NSC-Marking Guideline QUESTION 2 2.1 A = 250  (A)A B = 502  (A) B (2) 2.2 2000 + 5000 + 600 + 1280 + 1150 + 18000 + 33250 CA sum 𝑥̅ = 1300 61280 CA 61280 𝑥̅ = 1300 𝑥̅ = 47,14 CA (answer only – full marks) answer (3) 2.3 65 < 𝑥 ≤ 75 AA answer (2) 2.4 CA upper AGES OF CONSUMERS limits 1400 CA grounding CUMULATIVE FREQUENCY 1200 (5; 0) 1000 CA joining 800 points with a 600 smooth curve 400 200 0 5 15 25 35 45 55 65 75 AGE (3) 2.5 Not a normal distribution. Highest frequency is found between the ages A No (2) 55 to 75. Mean < median, therefore skewed to the left. A Reason [12] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com 4 Common Test June 2019 Mathematics P2 NSC-Marking Guideline QUESTION 3 3.1 2 2 A substitution BC = √(𝑥2 − 𝑥1 ) + (𝑦2 − 𝑦1 ) = √(−2 − 1)2 + (−2 − 4)2 = 9  36 CA answer (2) = 45 = 3 5 3.2 1 1−2 4−2 A M( ; ) 2 2 2 1 A 1 M (− ; 1) 2 (2) 3.3 2 4 6 m AB = =  2 1 3 = 2  A M AB 1 mMD = – (DM ⊥ AB) CA gradient of MD 2 y = mx + c 1 1 1 1 = − (− ) + 𝑐 CA subst. (− 2 ; 1) into 2 2 eq. 1 c = 1– 4 3 = 4 1 3 𝑦=− 𝑥+ 2 4 CA answer (4) Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Downloaded from Stanmorephysics.com 5 Common Test June 2019 Mathematics P2 NSC-Marking Guideline 3.4 E is the midpoint since ME║BC.  A S/R 1+4 4+1 A substitution E( ; ) 2 2 5 5 = E( ; ) CA answer (provided 2 2 coordinates are positive) (3) 3.5 5 −2 − 2 𝑚𝐵𝐸 = 5 −2 − 2 𝑚𝐵𝐸 = 1 CA m BE (must be positive) 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) OR y = mx + c 𝑦 − (−2) = 1(𝑥 − (−2)) −2 = 1(−2) + c CA substitution 𝑦+2=𝑥+2 0=c 𝑦=𝑥 y=x CA answer (must be positive) (3) 3.6  2 1 1  A Substitution mBC = = 24 2 CA mBC = 1 2 1 mBC  tan   2 1 CA tan   2 𝜃 = 26,57° CA answer (4) [18] QUESTION 4 4.1 2 A subst. into dist. formula 𝐴𝑄 = √(−6 − 2)2 + (−7 − (−1)) 𝐴𝑄 = √(−8)2 + (−6)2 𝐴𝑄 = √64 + 36 𝐴𝑄 = √100  AQ  10 CA answer (2) Copyright Reserved Please Turn Over

Downloaded from hlayiso.com 6 Common Test June 2019 Mathematics P2 NSC-Marking Guideline 4.2 ( x  a) 2  ( y  b) 2  r 2 A (𝑥 − 2)2 + (𝑦 + 1)2 (𝑥 − 2)2 + (𝑦 + 1)2 = 100 CA 100 (2) 4.3.1  7 1 3 3 m AQ   A  6 2 4 4 4  mQP   𝑟𝑎𝑑. ⊥ 𝑡𝑎𝑛 CA − 3 4 3 (2) 4.3.2 7 1 3 3 m AR    A 10  2 4 4 4  m PR  𝑟𝑎𝑑. ⊥ 𝑡𝑎𝑛 4 3 CA 3 (2) 4.4.1 4 mQP   3 y  y1  m ( x  x1 ) 4 y+7 =  ( x  6) CA substitution 3 4 CA answer y x  15 3 (2) 4.4.2 4 mPR  3 y  y1  m( x  x1 ) 4 y7 ( x  10) CA substitution 3 4 x 40 y  7 3 3 4 61 CA answer = x 3 3 (2) Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Downloaded from Stanmorephysics.com 7 Common Test June 2019 Mathematics P2 NSC-Marking Guideline 4.5.1 4 61 4 CA Equating 𝑥− = − 𝑥 − 15 3 3 3 8 16 𝑥= 3 3 x =2 CA x value 53 y=− 3 CA y value 53 CA both co-ordinates P(2; = − 3 ) OR The x – co-ordinate of P is 2 (ARPQ is a kite) A x = 2 4 Subst x  2 in y   x  15 3 4 y (2)  15 3 CA substitution 8 = − − 15 3  8  45 = 3 CA y value  53 = 3 CA both co-ordinates 53 (4) 𝑃 (2; − 3 ) 4.5.2 In  SPR   P̂   (ext  of SPR) A S/R A P̂      P̂     (2) Copyright Reserved Please Turn Over

Downloaded from hlayiso.com 8 Common Test June 2019 Mathematics P2 NSC-Marking Guideline 4.5.3 sin(    ) tan (   ) = cos(   ) sin  cos   cos  sin  = A expansion cos  cos   sin  sin  4 3 3 4 ( ) ( ) − (− ) ( ) = 5 5 5 5 A numerator 3 3 4 4 A denominator (− ) ( ) + ( ) ( ) 5 5 5 5 12 12 + = 25 25 A Simplification −9 16 + 25 25 24 25 = × 25 7 24 = 7 OR (4) 4 A answer 𝑡𝑎𝑛𝛽 = ∴ 𝛽 = 53.13° 3 4 𝑡𝑎𝑛𝛼 = − 3 ∴ 𝛼 = 126,87° A answer sin(126.87  53.13) tan (   ) = A simplification cos(126.87  53.13) = 3, 42857……. 24 ≅ 7 A Answer (4) [22] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Downloaded from Stanmorephysics.com 9 Common Test June 2019 Mathematics P2 NSC-Marking Guideline QUESTION 5 5.1 3 1 tan A = ; sin B = 4 3 Diagram 1 Diagram 2 A diagram 1 𝑟2 = (−4)2 + (−3)2 … . 𝑃𝑦𝑡ℎ 𝑥2 = 32 − 12 … 𝑃𝑦𝑡ℎ 𝑟2 = 25 𝑥2 = 8 𝑟=5 𝑥 = 2√2 5.1.1 cos 2𝐴 2 A identity = 2𝑐𝑜𝑠 𝐴 − 1 −4 2 = 2( ) −1 5 7 CA answer (3) = 25 5.1.2 sin(𝐴 + 𝐵) = sin 𝐴 cos 𝐵 + cos 𝐴 sin 𝐵 A Diagram 2 −3 2√2 −4 1 A expansion = ( )( ) + ( )( ) 5 3 5 3 4 + 6√2 CA answer (3) =− 15 5.2 sin 20o cos 320o + cos (–20o) sin 400o = sin 20o cos 40o + cos 20o sin 40o A reduction = sin (20o + 40o) CA simplify = sin 60o 3 = CA answer (3) 2 (provided special angle ratio) ANSWER ONLY = 0 Copyright Reserved Please Turn Over

Downloaded from hlayiso.com 10 Common Test June 2019 Mathematics P2 NSC-Marking Guideline 5.3  cos 2 90 o  θ  cos (θ)  sin (90 o θ) cos θ = sin 2 θ A numerator cos θ  cos2 θ A denominator A common sin 2 θ  factor cos θ (1  cos θ)  A difference 1  cos 2 θ of squares  cos θ (1  cos θ) A 1  cos θ simplification  (5) cos θ 1  1 cos θ  RHS 5.4.1 cos 2  cos2   sin 2  = cos   sin   cos   sin   A expansion = p.q  A answer (2) 5.4.2 1  sin 2  cos 2 sin 2   2 sin  cos   cos 2  A numerator = cos 2   sin 2  A denominator A factorise sin   cos   2 = A factorise (cos   sin  )(cos   sin  ) p CA answer (5) = q 5.5 6 cos 2 x  sin x  5  0 A identity 6 (1  sin 2 x )  sin x  5  0 6  6 sin 2 x  sin x  5  0 −6𝑠𝑖𝑛2 𝑥 + sin 𝑥 + 1 = 0 (3 sin 𝑥 + 1)(−2 sin 𝑥 + 1) = 0 CA factors 1 1 CA both ratios sin 𝑥 = − ; sin 𝑥 = 3 2 A 𝑘 ∈ ℤ 𝑥 = 199,47° + 𝑘. 360°; 𝑘 ∈ ℤ 𝑂𝑅 𝑥 = 30° + 𝑘. 360°; 𝑘 ∈ ℤ CA both 𝑥 = 340,53° + 𝑘. 360°; 𝑘 ∈ ℤ 𝑂𝑅 𝑥 = 150° + 𝑘. 360°; 𝑘 ∈ ℤ solutions of 1 sin 𝑥 = − 3 CA both solutions of 1 (6) sin 𝑥 = 2 [27] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Downloaded from Stanmorephysics.com 11 Common Test June 2019 Mathematics P2 NSC-Marking Guideline QUESTION 6 6.1 A x-intercept A t. pt. (600; 1) A shape (3) 6.2 60° < 𝑥 < 90° CA end values (2) CA notation 6.3 Graph of f moves 600 left. A shifts 600 A to the left (2) [7] QUESTION 7 7 DA A trig ratio tan 𝜃 = AB ∴ AD = AB tan 𝜃  AAD value Also AB 𝑘 = sin 𝛼 sin(180°−2𝛼) A substitute into sine AB sin 2𝛼 = 𝑘 sin 𝛼 rule 𝑘 sin 𝛼 A AB = 2 sin 𝛼 cos 𝛼 sin 2𝛼 = 2 sin 𝛼 cos 𝛼 𝑘 AB = cos 𝛼 A 𝑘. tan 𝜃 𝑘. sin 𝛼 ∴ AD = 𝐴𝐵 = 2 cos 𝛼 2 sin 𝛼 cos 𝛼 A making AB value - simplified [6] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com 12 Common Test June 2019 Mathematics P2 NSC-Marking Guideline QUESTION 8 8.1 Qˆ 4  Wˆ 2  x … (tan chord theorem) AS/R Wˆ  Wˆ  x … (WQ bisects P𝑊R.) 2 1 Qˆ1  Wˆ1  x … (tan – chord theorem) AS/R AS/R Tˆ  Wˆ  x … ( s in same segment) 2 2 AS/R Sˆ2  Wˆ1  x … s in same segment) AS/R (5) ̂2 = Q 8.2.1 T ̂1 = 𝑥 AS  TS // PR … (alternate s equal) AR (2) 8.2.2 Tˆ  Pˆ … (corresponding ∠𝑠 ; TS//PR) AASR 3 Tˆ3  Qˆ 3 … ( s in same segment) AS/R  P̂  Q̂ 3 (3) 8.2.3 In Δ TQS Tˆ2  x AS Ŝ 2  x AS  Tˆ2  Sˆ2  x AS (4)  Δ TQS isosceles … ( s opposite equal sides) AR 8.2.4 WQ ̂ P = WŜQ. . . (tan − chord theorem) AASR T1 = WŜQ … (ext  of cyclic quad) ̂ AS/R (3) [17] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Downloaded from Stanmorephysics.com 13 Common Test June 2019 Mathematics P2 NSC-Marking Guideline QUESTION 9 P1 = ̂ 9.1.1 ̂ Q1 given ̂ ̂ P1 = R tan-chord theorem AASR ̂ ∴ Q1 = R̂ ∴ TQ ∥ SR (corr ∠s 𝑎𝑟𝑒 𝑒𝑞𝑢𝑎𝑙) AR (3) 9.1.2 ̂ P1 = Q ̂1 given TS = TP tan from same point AS/R ̂ P1 = S1̂ equal ∠𝑠 opp equal sides AS/R ∴Q ̂ 1 = Ŝ1 ∴QPTS is a cyclic quad converse equal ∠𝑠 subtended by same AR chord (3) 9.1.3 QPTS is a cyclic quad ∴̂ P1 = ̂Q2 ∠𝑠 in same ⊙ segm AASR but ̂P1 = Q̂1 given ̂ ∴ Q1 = Q 2̂ AS ∴TQ bisect SQ ̂P (3) 9.2.1 In Δ LPK and Δ NPL KL̂P = LN ̂ P … tan chord theorem AASR ̂ 2 = 90° … … P ∠ in semi ⊙ AS/R P1 = ̂ ̂ P2 … both = 90o PK̂ L = NL̂P … remaining angle ∴ ΔLPK///ΔNPL . . . ∠∠∠ AR (4) 9.2.2 PL KL PK AASR = NL = PL… Δ LPK /// Δ NPL NP PL PK = PL ……. ΔLPK///ΔNPL A proportionality NP ∴ PL2 = NP. PK (3) 9.2.3 Δ NLK A answer (1) 9.2.4 ΔNLK///ΔNPL 𝐾𝑁 𝐿𝑁 ∴ = ⌈///∆′𝑠⌉ 𝐿𝑁 𝑁𝑃 A S/R 𝐿𝑁 2 = 𝐾𝑁. 𝑁𝑃 A  Substitution = 16 x 10 = 160 LN = √160 CA NL value 1 Radius = 2 √160 CA radius = 1 Area of Circle = 𝜋𝑟 2 160 1 2 = 𝜋(2 √160)2 = 125.66cm2 CA Substitution OR CA Answer (6) Copyright Reserved Please Turn Over

Downloaded from hlayiso.com 14 Common Test June 2019 Mathematics P2 NSC-Marking Guideline From Question No. 9.2.2: PL2  NP.PK = 10 cm x 6 cm = 60 cm2 NL2 = PL2 + PN2 … Pythagoras A PL2 = 60 cm2 = 60 + 100 (Δ LPN) = 160 cm2 A Pythagoras ∴ NL = √160𝑐𝑚 1 1 1 NL  diameter  160 cm 2 2 2 CA NL value CA radius = Area of circle = r 2 1 160 2 2 1 = 𝜋 × (2 √160) 𝑐𝑚2 CA substitution = 125,66 cm2 OR 40𝜋 𝑐𝑚2 CA answer [23] QUESTION 10 10.1 AE 4 A S/R = … … prop theorem ; EB//FC EF 6 AE AC EF = CD … … prop theorem; EC//FD A S/R 4 10 = 6 CD CD = 15 𝑢𝑛𝑖𝑡𝑠 CA answer (3) 10.2 ∆FEC 3 =5 same height ∆CFA A S/R Area ∆CFA 10 2 Area ∆FAD = 25 = 5 same height A S/R Area ∆FEC Area ∆FEC Area ∆CFA = × Area ∆FAD Area ∆CFA Area ∆FAD 3 2 CA simplify = × 5 5 6 = 25 CA answer (4) [7] TOTAL MARKS: 150 Copyright Reserved Please Turn Over