Technical Mathematics Grade 12 IEB Paper 2 2019 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE EXAMINATION NOVEMBER 2019 TECHNICAL MATHEMATICS: PAPER II MARKING GUIDELINES Time: 3 hours 150 marks These marking guidelines are prepared for use by examiners and sub-examiners, all of whom are required to attend a standardisation meeting to ensure that the guidelines are consistently interpreted and applied in the marking of candidates' scripts. The IEB will not enter into any discussions or correspondence about any marking guidelines. It is acknowledged that there may be different views about some matters of emphasis or detail in the guidelines. It is also recognised that, without the benefit of attendance at a standardisation meeting, there may be different interpretations of the application of the marking guidelines. IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: TECHNICAL MATHEMATICS: PAPER II – MARKING GUIDELINES Page 2 of 13 QUESTION 1 ( xB − x A ) + ( y B − y A ) 2 2 AB = 1.1 correct formula ( 200 − 100 ) + ( 0 − 250 ) 2 2 = substitution (100 ) + ( −250 ) 2 2 = = 50 29 units simplified surd form yB − y A mAB = 1.2 xB − x A 0 − 250 = 200 − 100 substitution = –2,5 simplification  x − xO y A − y O  M= A ; 1.3  2 2   100 + 0 250 + 0  M= ;  2 2  M = (50 ; 125) x-coordinate y-coordinate 1.4 tan B = m tan B = −2,5 substitution ∴ OBA ˆ ≈ 68, 2 ° ˆ angle OBA ∴ AOB ˆ =68,2° angle AOBˆ ∴θ = 180° − 68,2 ° − 68,2= ° 43,6 ° simplification IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: TECHNICAL MATHEMATICS: PAPER II – MARKING GUIDELINES Page 3 of 13 QUESTION 2 2.1 2.1.1 mLP × mLN =−1 −1× mLN =−1 OR y – y1 = m(x – x1) ∴ mLN 1 = LN gradient y – 4 = 1 (x + 4) ∴y = x + c with M(–4;4) y = x + 8 equation ∴4 =−4 + c ∴8 = c ∴y = x + 8 equation 2.1.2 ∴y =x + 8 =− x + 2 follow up from Question 2.1.1 ∴2 x = −6  x + xN y L + y N  ∴ x= −3 m L ; = ( −4; 4 )  2 2  ∴y = x + 8 ∴y =−3 + 8 =5 −3 + xN 5+y =N 4 −4 = ∴L ( −3; 5 ) 2 2 equations coordinates xN −5 =y N 3 = ∴N ( −5; 3 ) 2.1.3 y =− x + 2 coordinates x-intercept: 0 =− x + 2 x = 2 ∴ P (2; 0) coordinates 2.1.4 x 2 + y 2 r = 2 p(2 ; 0) ∴ (2) + (0) r 2 2 =substitution 2 4=r 2 ∴ x2 + y 2 = 4 equation IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: TECHNICAL MATHEMATICS: PAPER II – MARKING GUIDELINES Page 4 of 13 2.2 x-intercepts y-intercepts shape IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: TECHNICAL MATHEMATICS: PAPER II – MARKING GUIDELINES Page 5 of 13 QUESTION 3 3.1.1 x + y = 2 2 2 3.1 r ( 2 3 ) + ( −2) =r substitution 2 2 2 16 = r 2 4 units = r simplification −2 tanθ = 3.1.2 2 3 function (Ref angle) = 30° method θ = 360° − 30 =° 330° simplification sec ( a − b ) 3.2 = sec ( 2,695 − 1,112 ) substitution = sec (1,583 ) 1 = cos (1,583 ) ≈ −81,9 simplification sin 210° tan 45° cos 315° 3.3 sin 45° cos 60° − sin 30° tan 45° cos 45° = sin 45° cos 60° OR  −1   2  2  ⋅ (1) ⋅  2      sin 45° = − sin 30° × × cos 45°  2   1 cos 45° =  2  ⋅  2  sin 45° × sin 30°   = –1 = –1 3.4 LHS: tan x ⋅ sin x RHS: sec x − cos x OR LHS: tan x ⋅ sin x = sin x .sin x = 1 − cos x = tan x ( cos x tan x ) cos x cos x = cos x ⋅ tan2 x 1 − cos x ( ) 2 2 sin x = cos x sec 2 x − 1 = = cos x cos x  1  sin2 x = cos x  − 1 =  cos x 2  cos x 1 ∴ LHS = RHS = − cos x cos x = sec x − cos x ∴ LHS = RHS IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: TECHNICAL MATHEMATICS: PAPER II – MARKING GUIDELINES Page 6 of 13 OR LHS : tan x ⋅ sin x OR RHS : sec x − cos x sin x sin x 1 = ⋅ = − cos x cos x 1 cos x sin2 x 1 − cos2 x = = cos x cos x 1 − cos2 x sin2 x = = cos x cos x 1 cos2 x sin x = − = sin x × cos x cos x cos x = sec x − cos x = sin x ⋅ tan x ∴ LHS = RHS ∴ LHS = RHS = cosec 2 x 2,114 for 2 x ∈ [0°;180°] 3.5 3.5.1 1 = 2,114 sin 2x 1 = sin 2x 2,114 0,473 … = sin 2x Ref angle ≈ 28,2316° =2 x 28,23° or 2= x 180° − 28,23° correct quadrants =x 14,12° = or x 75,88° both answers sin ( 360° − x ) ⋅ cos (180° − x ) ⋅ tan (180° + x ) 3.5.2 5 cos2 x ⋅ sin π 6 = ( − sin x )( − cos x )( tan x ) OR ( − sin x )( − cos x )( tan x )  1  5  ( ) cos 2 x    2 ( ) cos2 x  sin π   6   sin x  sin x ⋅ tan x ( − sin x )( − cos x )   = =  cos x   1 ( ) 1 cos x ⋅ cos2 x   2 2 = 2 tan x 2 = 2 tan x2 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: TECHNICAL MATHEMATICS: PAPER II – MARKING GUIDELINES Page 7 of 13 QUESTION 4 4.1 (360° ; 2) (0 ; 2) g f (120° ; 1) (0°;2) y-intercept f (0°;–½) y-intercept g (120°;1) turning g (180°;–2) turning f (300° ; –1) (300°;–1) turning g (360°;2) turning f (180° ; –2) 4.2 2 4.3 360° 4.4 x ∈ ( 0°;70° )∪ ( 250°;360°) give leeway with 70° (±5°) and 250° (±5°) OR 0° < x < 70° ∪ 250° < x < 360° IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: TECHNICAL MATHEMATICS: PAPER II – MARKING GUIDELINES Page 8 of 13 QUESTION 5 AC AC 2 5.1 sin 58° = ratio OR = 2 sin 58° sin 90° ∴ AC ≈ 1,7 m answer ∴ AC ≈ 1,7 m 1 = 5.2 Area of Δ ABC (1,7 m )( 2,3 m ) sin 108° formula substitution 2 ≈ 1,9 m2 5.3 BC = 2,3 m OR BC2 = 2,32 + 1,72 – 2 (2,3)(1,7) cos 108° sin 108° sin 42° BC ≈ 3,3 m BC ≈ 3,3 m (1,1) + ( 2,3 ) 2 2 5.4 = DB 2 ∴ DB ≈ 2,5 m BC2 = DC2 + DB2 − 2DC ⋅ DB ⋅ cosDˆ ( 3,3 ) =( 2 ) + ( 2,5 ) − 2 ( 2 )( 2,5 ) ⋅ cosDˆ 2 2 2 ∴ Dˆ ≈ 93,7° QUESTION 6 6.1 6.1.1 P̂2 P̂1 + Pˆ2 = 180° (Angles on a straight line) Pˆ= 60° 1 (Angle at centre = 2 × angle on circumference of circle) 60° + P = ˆ 180° 2 = ˆ 120° P 2 6.1.2 Rˆ1 = Tˆ (Angles at equal sides) and Pˆ = Rˆ + Tˆ = 180° 2 1 (int. angles of triangle) ∴120° + 2R = ˆ 180° 1 2R= 60° ˆ 1 Rˆ= 1 30° IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: TECHNICAL MATHEMATICS: PAPER II – MARKING GUIDELINES Page 9 of 13 6.2 6.2.1 Dˆ= 2 50° (Tan-chord theorem) 6.2.2 B̂1 Aˆ= 1 50° (Tan-chord theorem) Aˆ + 110° + B 1 = ˆ 180° (Interior angles of triangle) 1 50° + 110° + B̂ = 1 180° Bˆ= 20° 1 6.2.3 D̂1 B̂= 1 Cˆ= 3 20° (Angles in same segment) and Cˆ=2 20° (Given) ∴ Dˆ = 20° (Angles in same segment) 1 6.3 = 49° (Tan-chord theorem) 6.3.1 a ˆ = 78° (Tan-chord theorem) 6.3.2 PTR 32° + Tˆ= 1 78° T= ˆ 46° 1 ∴ b = 46° (Tan-chord theorem) OR ˆ = 32° (angles in same segment) QPR ∴ C= 78° − 32° = 46° 6.3.3 c + 78= ° 180° (Angles on same line) ∴ c= 102° IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: TECHNICAL MATHEMATICS: PAPER II – MARKING GUIDELINES Page 10 of 13 QUESTION 7 7.1 7.1.1 QM KQ : QM = 3 : 1 (Given) ∴ KM : QM 4 : 1 =(Proportionality theorem) KM 4 ∴ = QM 1 20 units 4 1 ∴ =OR QM= × 20 units QM 1 4 ∴ 20 = 4QM = 5 units ∴ 5 units QM = 7.1.2 KP KQ KP = (Proportionality theorem PQ || LM) QM PL 3 KP = 1 4 units KP = 12 units 7.1.3 KB KM KP = (Proportionality theorem BQ || PM) QM BP 20 units 12 units = 5 units BP 20BP = 60 BP = 3 units KB + BP = KP KB + 3 = 12 KB = 9 units 7.2 7.2.1 In ∆KPM and ∆KBQ MKP ˆ = QKB ˆ (Common angle) KMPˆ = KQB ˆ (Corresponding angles BQ||PM) KPM = KBQ ˆ ˆ (Corresponding angles BQ||PM) ∴ ∆KPM ||| ∆KBQ (Angle, angle, angle) KQ BQ 7.2.2 = ( ∆KPM ||| ∆KBQ ) KM PM 3 BQ = 4 10 units 4BQ = 30 BQ = 7,5 units IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: TECHNICAL MATHEMATICS: PAPER II – MARKING GUIDELINES Page 11 of 13 QUESTION 8 8.1 v = π Dn 8,75 = π (50)n 7 0,0557 = n OR 40π w = 2π n  7  = 2π (0,0557 …) OR 2π  π   40  7 ≈ 0,35 rad/sec rad/sec 20 8.2 8.2.1 s = r θ  π  =s 28 cm ×  240° ×  180°  = 117,3 cm PC ˆ = 90° (radius⊥ tangent) 8.2.2 APC OR = tan 60° AB = BC = 28 cm (Given) 28 ∴ AC = AB + BC PC = 48,5 cm QC = 48,5 cm  = 28 cm + 28 cm = 56 cm RC 2 2 2 = tan 60° AC = AP + PC (Pyth) 12 RC = 20,8 cm ( 56 cm ) ( 28 cm ) + PC2 2 2 = ∴ TC = 20,8 cm ∴ PC = 48,5 cm Total = 117,3 + 50,3 + 2 (48,5) + ∴ QC = 48,5 cm 2 (20,8) ≈ 306,2 cm ˆ = 90° (radius⊥ tangent) ERC ED = DC = 12 cm (Given) ∴ EC = ED + DC = 12 cm + 12 cm = 24 cm (Pyth) EC2 = ER2 + RC2 ( 24 cm ) (12 cm ) + RC2 2 2 = ∴ RC = 20,8 cm ∴ TC = 20,8 cm Total belt length = 117,286 cm + 50,3 cm + 2 ( 48,5 cm ) + 2 ( 20,78 cm ) = 306,11 cm IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: TECHNICAL MATHEMATICS: PAPER II – MARKING GUIDELINES Page 12 of 13 QUESTION 9 9.1 3 2 3 2 3 2 2 2 3 2  ( = ( sh ) +  ) 2 9.1.1 Perpendicular height of ΔABE: 3 2 2  2    18 18 ( sh ) + 2 = 4 3 6 sh = units or 3,7 units 2 h OR = sin 60° 3 2 3 6 h= 2 1 9.1.2 Area ΔABE = × base × ⊥ height 2 1 3 6 = ×3 2 × 2 2 9 3 = or 7,794 2 Total area of octahedron 9 3 =8× 2 = 36 3 or 62,4 units3 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: TECHNICAL MATHEMATICS: PAPER II – MARKING GUIDELINES Page 13 of 13 9.2 Volume cylinder = π × r 2 × h π× ( 30 mm ) × 70 mm 2 = = 197 920 3372 mm3 or 63 000 π 2 Volume dome = πr3 3 2 = π ( 30 mm ) 3 3 = 56 548,668 mm3 or 18 000 π Volume centre removed π × r 2 × h π× (15 mm ) × 70 mm 2 = = 49 480,084 mm3 or 15 750 π Total volume = 197 920,3372 mm3 + 56 548,668 mm3 – 49 480,084 mm3 = 204 988,92 mm3 or 29 250 π 9.3 = a 6,5 m ÷= 5 1,3 m Area= a ( m1 + m2 + m3 + m4 + m5 )  0 + 0,8 0,8 + 1,3 1,3 + 1,1 1,1 + 0,5 0,5 + 0  = 1,3  + + + +  2 2 2 2 2  = 1,3 ( 0,4 + 1,05 + 1,2 + 0,8 + 0,25 ) = 4,81 m2 TOTAL: 150 marks IEB Copyright © 2019