TECHNICAL MATHEMATICS PAPER 2
GRADE 12
NATIONAL SENIOR CERTIFICATE
MEMORANDUM
NOVEMBER 2019
MARKING CODES | |
A | Accuracy |
AO | Answer only |
CA | Consistent accuracy |
M | Method |
R | Rounding |
NPR | No penalty for rounding |
NPU | No penalty for unit |
S | Simplification |
F | Correct formula |
SF | Substitution in correct formula |
MARKS: 150
These marking guidelines consist of 24 pages.
NOTE:
QUESTION 1
1.1 | OED= 76° | angle A | |
1.2 | AF = √(- 2 - (-8))2 + (8 - (-4))2 | SF A | L2 |
1.3 | m = tanθ | SF A | L2 |
1.4 | MAF( -2 + (-8); 8 +(-4) | SF A | L1 |
1.5 | mAF = y2 - y1 = 2 | SF A |
QUESTION 2
2.1.1 | x2 + y2 = ( - 4 )2 + ( -3)2 | SF A | L1 |
2.1.2(a) | B( 4 ; 3) | coordinates of B A | |
2.1.2(b) | mPQ = - 4/3 | gradient of PQ | |
2.1.3 | y - 3 = - 4/3 (x - 4) OR y = mx + c | SF CA from OR | L2 |
2.2.1 | x2 + 8y2 - 32 = 0 | dividing both sides and transposing A | |
2.2.2 | CA from Q 2.2.1 |
QUESTION 3
3.1.1 | sin 3a = sin 3(32°) | value of sin 3a A | L1 | |
3.1.2 | sec 2 θ - 1 = sec 2 20° - 1 OR = 0, 21 |
OR | L1 | |
3.2.1 |
| r A | L2 | |
3.2.2 | changing to degrees A OR OR | |||
3.3.1 | 2 cosθ + sinq = 0 OR | transposing A OR (2) | ||
3.3.2 | tanθ = - 2 | ref A |
QUESTION 4
4.1.1 | cot22β - cosec22β = -1 | A (1) |
4.1.2 | tan2A×cosec2A - cos2π OR OR |
OR
OR (5) A |
4.2.1 | sec60° = 2 | 2 A (1) |
4.2.2 | cosec(180° + θ )×sin(360° - θ ) - [sin(180° + θ )]sec60° = cos2 θ | - cosecθ A |
[12] |
QUESTION 5
5.1 | 360° | ü period A (1) | ||
5.2 | a = - 2 | value of a | ||
5.3 | T(158,5° ; - 0,7) | x coordinate | ||
5.4.1 | 135°< x <180° | end points A | ||
5.4.2 | x = 45° or/of x = 135° | x = 45° | ||
[9] |
QUESTION 6
6.1 | cos 50° = 150 OR AB = 150 tan 50° OR OR 233,36 m | trig ratio A OR OR OR |
6.2 | Area of ΔACD = ½ x AC x CD sin β OR | F A OR NPR (5) |
6.3 | AD2 = AC2 + DC2 - 2AC × DCcos ACD | F A |
[12] |
QUESTION 7
7.1 | Perpendicular to the chord | ST A (1) |
7.2.1(a) | C1 = 26, 6ο [∠s in thesame segment (arc or chord)] OR | ST RE OR | ||
7.2.1(b) | A = 180ο - 90ο - 26,6ο [ sum/som∠s Δ] OR OR | ST CA from Q 7.2.1(a) OR ST CA from Q 7.2.1(a)
| ||
7.2.1(c) | B1 = 90ο - 26, 6° [ Ðin semi circle] OR OR | ST OR OR | ||
7.2.2(a) | AE = 4 units | length of AE A (1) | ||
7.2.2(b) | ED = 5 - x units | length of ED A (1) | ||
7.2.3 | 42 + (x + 5)2 = (4√5)2 One mark penalty if a candidate include x=-13 as the answer OR OR OR OR OR | M A OR OR OR OR Theorem of Pyth. A
| ||
[12] |
QUESTION 8
8.1 | 90°/ Right angle | 90°/ Right angle A (1) |
8.2.1(a) | CBO = 90ο - 30ο [tan ^ radius (diameter)] = 60ο | ST A RE A (2) |
8.2.1(b) | D2 = 30ο [tan - chord] | ST A RE A (2) |
8.2.1(c) | O1 = 60ο [∠at centre = 2∠at circumferen] OR | ST CA from Q 8.2.1 (a) OR |
8.2.1(d) | O1 = E = 60ο corresp∠s CO || FE | ST A |
8.2.2 | C1 = E = 60ο ext ∠ of cyclicquad OR | ST CA OR |
[14] |
QUESTION 9
9.1 | Parallel to the third side | ST A (1) |
9.2.1 | OGC = 90° [tan ^ radius] | ST |
9.2.2 (a) | BC : DF = 10 : 6 = 5 : 3 OR | ü proportionality A OR/OF
ü proportionality/eweredigh A ü simplified ratio/ vereenv.verhouding A AO: Full marks/Volpunte (2) |
9.2.2 (b) | EG = 2 OR | proportionality A OR |
9.2.2 (c) | OE = 6 - 2,4 = 3,6 units OR OR | length of OE CA from Q 9.2.2 (b) OR OR |
9.3 | In ΔDOE and ΔBOG OR | ST A OR (3) |
QUESTION 10
10.1.1 | BC = 20 cm – 1,5 cm | length of A (1) |
10.1.2 | 4h2 - 4dh + x2 = 0 OR OR | F A OR OR A A CA CA CA (5) |
10.1.3 | w = 64π rad/min OR»1263,13 rad/min w =2πn | conversion
|
10.1.4 | v = πDn OR OR OR OR OR OR | F A |
10.2
10.2.1 | AOB= 4 x 20°= 80° = 4/9 π OR OR | M A OR OR |
10.2.2 | s = rθ OR | F A OR |
10.2.3 | Area of a sector = rs/2 OR OR | F A OR OR (3) |
[22] |
QUESTION 11
11.1.1 | b = √2 | value of b A | |
11.1.2 | Airr= 2/3 x 19 ,125 m2 = 12, 75 m2 OR | value of Airr A OR (5) |
11.2
11.2.1 | V = l xb xh | value of h A(1) |
11.2.2 | r =½ x 3,5m= 1, 75 m | value of r A(1) |
11.2.3 | SARec.Prism = 2´(bx h) + 2 x (l x h) | SF CA from Q11.2.1 (6) |
[14] |
TOTAL:150