Wednesday, 08 September 2021 06:26

## TECHNICAL MATHEMATICS PAPER 2 GRADE 12 MEMORANDUM - 2018 SEPTEMBER PREPARATORY EXAM PAPERS AND MEMOS

TECHNICAL MATHEMATICS PAPER 2
NATIONAL SENIOR CERTIFICATE
MEMORANDUM
SEPTEMBER 2018

NOTE:

• Continuous accuracy (CA) applies in ALL aspects of the marking guideline.
• After two mistakes, do not apply CA marking.
• Assuming values/answers in order to solve a problem is unacceptable.
 Symbol Explanation M Method MA Method with accuracy A Accuracy CA Consistent accuracy S Simplification or Statement R Reason

 QUESTION 1 1.1 AB = √( x2 - x1 )2  + ( y2  - y1)2= √(0 + 2)2  + (3 - 0)2= √4 + 9≈ 3, 61 ✓MA formula & substitution✓S Simplification✓CA Answer in decimal format (3) OR AB2 = OB2 + OA2 (Pyth)=22 + 32= 4 + 9AB =√13≈ 3, 61 ✓MA Pythagoras✓S Simplification✓CA Answer in decimal format (3) 1.2 mAB = y2 - y1            x2 - x1= 3 - 0   0 + 2= 3/2Equation of the line:y - y1 = m ( x - x1 )y - 0 = 3/2 ( x + 2)y = 3/2 x + 3 ✓MA formula & substitution✓M Formula equation of line✓A Substitute pt A or B✓CA Answer in standard form (4) 1.3 tanOBA = mBAtan OBA = 3/2OBA = 56, 31°OAB= 90°- 56, 31°= 33, 69° ✓A subst. into correct formula✓CA value of OBA ✓CA value of OAB (3) 1.4 Scalene triangle OR Right angled triangle ✓A (1) 1.5 a = 2b = 3Equation of ellipsex 2 + y 2 =14      9 ✓A value of✓A value of✓CA Equation of ellipse (3) [14]

 QUESTION 2 2.1 r 2 = x2 + y2= (-8)2  + 62= 64 + 36= 100r = 10B(0 ; – 10) ✓M Calculating✓CA Value of r✓A x = 0✓CA y = – 10 (neg. value)(4) 2.2 Radius of smaller circle= 10 – 4 = 6equation of smaller circlex2 + y2 = 36 ✓M radius of smaller circle✓CA Equation of smaller circle(2) 2.3 Bigger circlemradius= y2 - y1                         x2 - x1 = 6 - 0  -8 - 0=- 3/4mtangent = 4/3equation of tangfent:y - y1 = m ( x - x1 )y - 6 = 4/3( x + 8) y = 4/3 x + 32/3 + 6        3=4/3 x + 50/3 ✓MA gradient of radius of bigger circle✓CA Gradient of tangent✓A Subst. A into equation✓CA Standard form of equation(4) [10]

 QUESTION 3 3.1 3.1.1 x2 + y2 = r 2(Pyth)52 + y2 = 6, 42y = - √40, 96 - 25=-√15, 96= -3, 99... ≈-4 units ✓A Pythagoras✓A Substitution✓CA Simplification✓CA Rounded answer (4) 3.1.2 cotθ- cos ecθ x sin2 θ= 5/-4 - 6,4/-4 x (-4/6,4) 2= -5/4 + 6,4/4 x 16/40,96= 5 ✓CA cot θ✓CA cosec θ✓CA Sin θ✓CA Simplification of sin2 θ✓CA Answer (5) 3.1.3 cosθ = 5/6.4θ = 360° - cos-1 (5/6.4)= 360° - 38, 624...°= 321, 4° ✓A Ratio✓CA Reference ∠✓A 4th Quadrant✓CA Answer, rounded (4) 3.2 sin (180° + x) tan (360° + x)cos(180° - x)= (-sin x)(tan x)(-cos x)= (-sin x) sin x (-cos x)              cos x= sin2 x ✓A sin x✓A tan x✓A -cos x✓A sin x     tan x✓CA answer (5) 3.3 tan 2 5x ✓A tan 2 5x (1) 3.4 2 tan ( x - 23°) + 5 = 0tan ( x - 23°) = -5/2Ref ∠ = tan-1 (2,5) = 68,19...°x - 23° = 180° - 68,19...° OR 360°- 68,19...°x = 111,8...° + 23° OR 291,80...° + 23°= 134,8° or 314,8° ✓A RHS = -2,5 ✓CA Ref ∠✓CA 2nd quadrant✓CA 3rd quadrant ✓A Adding 23° by✓CA Answers (6) [25]

 QUESTION 4 4.1 In ∆AHC:AH =cosec24,5° HCAH= 100      sin24,5°≈241 m ✓A Ratio✓A Substitution✓CA Rounded answer (3) 4.2 In ∆BHC:tan B = HC            BC= 100   261= 0, 383141...B = tan-1(0, 383141...)≈ 21° ✓A Ratio✓A Substitution✓CA Rounded answer (3) 4.3 AB2 = AC 2 + BC 2 - 2AC x BC cos ACB= 2192 + 2612 - 2´ 219´ 261´cos105°= 145669, 6756AB = 381, 6669...≈ 382 m ✓A cos rule✓CA substitution✓CA simplification✓CA answer (4) 4.4 Area DABC = 1 ab sin C2= 1 ´ 261´ 219 ´sin105° 2= 3000847, 5» 3 000 848 m2 ✓A Area rule✓CA Substitution✓CA value of Area (3) [13]

 QUESTION 5 5.1 Period = 360° = 180°                2 ✓A Answer (1) 5.2 A(45°; 0) & C(225°; 0) ✓A A(45°; 0)✓A C(225°; 0) (2) 5.3 90° < x < 180° ✓A End points✓A Notation (2) 5.4 45° < x < 165° and/en 225° < x < 285° ✓A✓ A first✓A✓ A second (4) 5.5 y ∈ [-3;3] or - 3 ≤ y ≤  3 ✓A critical values✓A notation (2) [11]

 QUESTION 6 6.1 SupplementaryOR Add up to 180° completed (1) 6.2 6.2.1 E = 90°   (∠ in semi-circle) ✓S ✓R (2) 6.2.2 BDE = 45°   (co-int∠s ;BE//GD) ✓S ✓R (2) 6.2.3 BE = ED = 8 cm (sides opp. = angles) ✓S ✓R (2) 6.2.4 BGD = 90°   (∠ in semi-circle)BGDE is a rectangle (All Ðs = 90°)BGDE is a square (diagonals bisect at 45°)GD = 8 cm (all sides=) ✓SR✓SR✓SR✓S (4) 6.3 6.3.1 S1 = 41° (alt ∠s ; TS // WQ) ✓S ✓R (2) 6.3.2 V = 139° (opp ∠s of cyclic quad) ✓S ✓R (2) 6.3.3 S2 = 64°      (tan-chord) ✓S ✓R (2) [17]

 QUESTION 7 7.1 Divides the other two sides proportionally ✓ completed statement (1) 7.2 7.2.1 Let QC = x.QC = QM  (prop th; MC // PN)CN     MP     x   = 3 2,86   2x = 3 x 2,86          2= 4, 3 cm ✓ S✓R Proportionality✓S Set up proportion✓S Simplification✓S Answer (5) 7.2.2 Let NR = y.QN = QM (prop th/ewer. st; MN // PR)NR    MP7,15 = 3   y       2y = 2 x 7,15            3= 4,8 cm ✓R Proportionality MN // PR✓S Set up proportion✓S Simplification✓S Answer (4) 7.3 7.3.1 Bˆ  is common BDE = 90° = ADBDE /// DBAC (AAA) ✓S✓S✓R (3) 7.3.2 BE =    502 +1002    (Pyth)≈112 cmLet AE = x cmBD = DE    (/// Δs)BA    AC  100    = 50 x +112     80x +112 = 80 x100                   50x = 48 cm ✓S BE = 112✓S Proportionality✓S setup proportion✓S value of AE (4) 7.3.3 Area DBDE = ½ x DE x DBArea DBAC    ½ x AC x AB= 50´10080´160= 0, 39 ✓S formulae ✓S substitution✓S value of ratio (3) 7.3.4 Area AEDC = Area ABC – Area DBE= 6400 – 2500= 3900 cm2 ✓MA✓CA value of area (2) [22]

 QUESTION 8 8.1 8.1.1 reflex CAE = 2/3 x 360°= 240° ✓A Multiply by2/3 x 360° (1) 8.1.2 obtuse CAE = 360° - 240°= 120°CAB = 60° ✓S✓S (2) 8.1.3 d = s = rθ= 50 x 240° x  π                       180= 200π      3≈ 209 cm ✓A Formula✓A Multiply 180°✓A Substitution✓CA Answer✓CA Rounding (5) 8.1.4 ACP = 90° (tan ⊥ radius)G = 90°(corrsp ∠s; BG || CP)GB = sin 60° 80GB = √3 x 80          2= 40√3 ≈69 cmCP = GB = 69 cm ✓A G  = 90°  Ratio✓A Ratio✓CA Simplification ✓CA Answer✓A CP = GB (5) OR GA = AC – BP (opp sides of rectangle)= 50 – 10 = 40 cmGB » 69 cm (Pyth) CP = GB = 69 cm OR✓A G = 90°  Ratio✓A GA = 40 cm✓CA Pythagoras✓CA Answer✓A CP = GB (5) 8.1.5 Length of belt = CH + HF + FP + CP= 209 +69 + 21 + 69= 368 cm ✓A HF = CP✓CA Answer (2) 8.2 d = 19x = 134h2 - 4dh + x2= 04h2 - 4 (19) h +132 = 04h2 - 76h +169 = 0h = -b ± √b2 - 4ac               2a= 76 ± √(-76)2 - 4 (4)(169)                    2(4)= 76 ± √3072         83 cm and 16 cm ✓A Formula✓A Substitusie✓M Standard form✓A Quadratic formula✓CA Subsitutisie✓CA Answers✓A Rounding (7) [22]

 QUESTION 9 9.1 9.1.1 w = 2πn= 2π(35)= 70p » 219, 9 rad/s ✓A Formula✓A Substitution ✓CA Answer / antwoord (3) 9.1.2 D = 40 cm = 0, 4 mv = πDn= π(0, 4)(35)= 14π= 43, 98 m/s ✓A Convert to m✓A Formula✓CA Substitution✓CA Answer (4) 9.2 Vrectangluar = l x b x h= 5 x 7 x 12= 420 cm3Vcylinder = πr 2h420 = πr 2 (60)r 2 = 420  = 2,228...        60πr = 1,492...diameter = 2r≈ 2,99 cm ✓A Subst. into formula ✓CA Answer✓A Formula✓CA Substitution✓ CA Value of diameter (5) [12]

 QUESTION 10 Ar = a(o1 + on + o2 + o3 + o4 + ... + on-1)               2= 5(8 + 3 + 10 + 9 + 9 +4)         2= 187.5cm2 ORAT = a(m1 + m2 + m3 + ... + mn)= 5(8 + 10 + 10 + 9 + 9 + 9 + 9 + 4 + 4 + 3)           2           2            2          2          2= 187.5cm2 🗸formula🗸value of a🗸substitution🗸AreaOR🗸formula 🗸value of a🗸substitution🗸Area (4) TOTAL 150
Last modified on Wednesday, 08 September 2021 09:42