TECHNICAL MATHEMATICS PAPER 2
GRADE 12
NATIONAL SENIOR CERTIFICATE EXAMINATIONS
MEMORANDUM
MAY/JUNE 2021
Marking Codes | |
A | Accuracy |
AO | Answer Only |
CA | Consistent accuracy |
I | Identity |
F | Correct Formula |
M | Method |
NPR | No penalty for rounding |
NPU | No penalty for units |
R | Rounding |
RE | Reason |
S | Simplification |
SF | Substitution in correct formula |
ST | Statement |
ST/RE | Statement with Reason |
NOTE: if the candidate used (-3 ; 1) follow the marking guidelines in the addendum
QUESTION 1
NOTE: if the candidate used (–3 ; 1) follow the marking guidelines in the addendum | ||
1.1 | a = 1 b = –1 | value of A |
1.2 | KL =√(xK - xL)2 + (yK - yL)2 = √(1-(-3))2 + (7-(-1))2 =√80 OR 4√5 OR ≈ 8,94 (2) | SF Length |
1.3 | M(xK + xL) ; (yK + yL) OR | x value y value [Penalty of one mark if not simplified] |
1.4 | mKL = yL - yK x - xL = - 1 - 7 - 3 - 1 = 2 | SF A gradient CA AO Full marks (2) |
1.5 | tanθ= m = 2 θ ≈ 63,4° Penalty for rounding | CA from Q 1.4 gradient CA value of θ(rounded)CA AO Full marks(2) |
1.6 | y = 2x + c 1= 2(-5) + c c = 11 y = 2x +11 OR y - y1 = m ( x - x1 ) y - 1 = 2 ( x - (-5)) y = 2x +10 + 1 y = 2x + 11 | ü gradien CA SF (–5; 1) A equation CA OR gradient CA SF (–5; 1) A equation CA(3) |
1.7 | y = 3/2x + 17/2 OR OR | M LHS ≠ RHS CA conclusion CA OR M mNew A conclusion CA OR M equation CA conclusion g CA(2) [15] |
QUESTION 2
2.1.1 | r2 = x2 + y2 = (-5)2 + (12)2 = 169\ x2 + y2 =169 OR x2 + y2 =132 x = ± √169 - y2 OR y = ± √169 - x2 | SF A equation CA AO Full marks (2) |
2.1.2 | t = √169 =13 | value of t CA(1) |
2.1.3 | mOB = -12/5 mtang = 5/12 y = mx + c OR y - y1 = m(x - x1 ) 12 = 5/12 (-5) + c OR y - 12 = ( x - 5/12 (x -(-5)) c = 169/12 y = 5/12x + 169/12 OR x.x1 + y.y1 = r2 x (-5) + y (12) = 169 12y = 5x +169 y = 5/12x + 169/12 | gradient A gradient CA substitution (–5; 1)A equation CA OR substitution(–5; 1) 169 CA S CA equation CA (4) |
2.2 | both x-intercepts A both y-intercepts A elliptical shape CA (3)[10] |
QUESTION 3
3.1.1 | (√13)2 = (3) 2 + (m ) 2 13 = 9 + m 2 m 2 = 4 OR m = √(13)2 - 32 m = 2 | value of m A AO Full marks (1) | |||
3.1.2 | sec2 β + tan 2 β = (√13)2 + (2/3)2 3 = 13/9 + 4/9 = 17/9 OR sec2 β + tan 2 β = 1 + tan 2 β + tan 2 β = 1 + 2 tan 2 β = 1 + (2/3)2 = 1 + 8/9 = 17/9 | CA from/ vanaf Q/V3.1.1 ratio of b A ratio of tan b CA simplification CA value of b + tan 2 b CA OR ratio of tan b CA value of sec2 β + tan 2 β CA(4) | |||
3.2.1 | cosθ = ½ θ = 60° | value of θ A(1) | |||
3.2.2 | tan a = - 1 ref ∠ = 45° a = 180° - 45° a = 135° | ref. 2nd quadrant value AO Full marks AA CA (3) | |||
3.2.3 | cos ( a - θ) = cos(135° - 60° ) = cos 75° ≈ 0 , 26 OR √6 - √2 4 | substitution CA value of (a - θ ) CA NPR (2) AO Full marks | |||
3.3 | 2 tan x + 0 , 924 = 0 2 tan x = - 0 , 924 tan x = - 0 , 462 ref∠ ≈ 24 ,8° x ≈ 180° - 24 ,8° or x ≈ 360° - 24 ,8° x » 155, 2° or x ≈ 335, 2° | S ref∠ x ≈ 155, 2° CA x ≈ 335, 2° NPR |
QUESTION 4
4.1 | cosθ (tanθ + cotθ ) =cosθ (sinθ + cosθ) cosθ sinθ = cos θ (sin2 θ+ cos 2θ) cosθ × sinθ = cosθ( 1 ) cosθ× sinθ = 1 OR cosecθ sinθ OR cosθ (tanθ + cotθ ) = cosθ × tanθ+ cosθ ×cotθ = cosθ × sinθ + cosθ× cosθ cosθ sinθ = sinθ + cos2θ sinθ = sin2θ + cos2θ sinθ = 1 OR cosecθ sinθ OR cosθ( tanθ+ 1 ) tanθ cosθ x(tan2 θ +1) tanθ cosθ x (sec2θ) tanθ = cosθ × ( 1 . cosθ) cos2θ sinθ = 1 OR cosecθ sinθ | (5) |
4.2 | sin 2 (180° + B) ×cosec(π- B) sec( 2π - B) ×cos (180° - B) = sin 2 B×cosecB sec B×(- cos B) = sin2B × 1= 1 sinB - 1 ×cos B cosB = - sin B OR sin 2 (180° + B) × 1 sin (π - B) 1 ×cos (180° - B) cos ( 2π - B) sin 2 B× 1= 1 sin B 1 ×( - cos B) cos B sin 2 B × 1 sinB - 1 ×cos B cosB = - sin B | sin 2 B A |
QUESTION 5
5.1 | ||||||
5.2.1 | x = 90° and x = 270° | 90° 270°(2) | ||||
5.2.2 | x ∈ (90ο ; 135°] or x =180° OR 90° < x ≤ 135° or x =180° | x ∈ (90ο ; 135°] |
QUESTION 6
6.1 | PR 2 = QR 2 + PQ 2 - 2QR×PQcos Q = ( 750 ) 2 + (1200 ) 2 - 2 (750 ) (1200 ) cos 60° = 1 102 50 PR = 1 050 m | ü cosine rule SF value/PR A (3) | |
6.2 | S = 120° | size of S A(1) | |
6.3 | PS = PR sin R1 sin S PS = 1 050 sin 40,5° sin120° PS = 1 050 sin 40 , 5° sin120° PS ≈ 787, 41m | sine rule SF value of PS NPR (3) | |
6.4 | Area ΔQPR =½QR ×QPsin Q = 1 (750)(1 200)sin 60° ≈389711, 43 m 2 | area rule A SF A value of CA(3) [10] |
QUESTION 7
7.1 | are equal | answer A(1) |
7.2 | ||
7.2.1(a) | PTS = R1 = 56° (∠s in thesame segment) OSR =R1 =56°(∠s opp. = sides) TPS = PTS = 56º ∠s opp. = sides = OR= chords subtend | ST RE A ST RE A ST A(5) |
7.2.1(b) | PSR =90° (∠s in semicircle) P1 + 90°+ 56° =180° (sum of ∠s of Δ) P1 = 34° OR O1 =112° ∠ at centre = 2 x ∠at circum P1 = S1 =34° (∠s opp. = sides) | ST RE A value of P1 CA OR ST RE A value of P1 CA(3) |
7.2.1(c) | 34°+ P2 =56° P2 = 22° S3 = P2 = 22° (∠s in same segment) OR S1 + S2 + S3 = 90° (∠ in the semi-circle) S1 + S2 = 180°- 112° (sum of ∠s of Δ) = 68° S3 = 90° -68° = 22° OR O2 + O3 =112° ∠at centre = 2 x ∠at circum. S2 = T2 = 34° [∠s opp. = sides S3 = 90° - 68° = 22° ∠ in the semi-circle | ST CA ST CA RE A OR ST CA ST A ST CA OR ST ST ST CA A CA(3) |
7.2.2 | O3 = 44° OR OR | ST OR OR |
QUESTION 8
8.1 | M = 98° ≠ 90° OR | ST M = 98° ≠ 90° A OR |
8.2.1 | P2 + 98°=180° (Opp. ∠s of cyclic quad.) P2 =82° | ST / RE A P2 =82° A(2) |
8.2.2 | P1 +82°=180° ( ∠s on straight line) OR | ST / RE A OR |
8.2.3 | L1 = 27° (tan-chord theorem) | ST A RE A(2) |
8.3.1 | K is common OR Equiangular | ST A ST A ST/RE A(3) |
8.3.2 | KL = KP (|||∠s) KN KL KL2 =KN.KP | ST A RE A(2) |
8.4 | KL2 =KN.KP (6)2 = 13.KP KP ≈ 2,77 units | subst A value of KP (2) |
8.5 | K + 27°+ 98°=180° (∠s of/van D) OR | ST/RE CA OR |
[18] |
QUESTION 9
9.1.1 | AB = AC (Prop. theorem; DE || BC) DB EC 1,8 = 3 DB 2 DB = 2/3 x 1 , 8 m DB=1,2 m | ST/RE A length of DB A (2) |
9.1.2 | AD= 1,8/3 = 0,6 m or AD =1,8 -1, 2 = 0, 6 m DF = 3/2 (0,6 m) = 0,9m | M CA length of DF CA(2) |
9.2 | CF = 1 =1 (BF = FC; F is the midpoint of BC) OR | ST A OR |
[7] |
QUESTION 10
10.1.1(a) | BC = 6,95 - 4 = 2,95m | height of segment A NPU (1) |
10.1.1(b) | h = 2, 95 m and d =13, 9 m OR | formula A OR |
10.1.2(a) | angle of sector, FOG = 20% x 2π OR OR 1, 26 rad OR | A OR OR |
10.1.2(b) | A = r 2θ OR | Formula A OR |
10.2.1 | n = 18 | M n (in rev) A value of n CA NPU NPR AO Full marks(2) |
10.2.2 | D = 2 x 10 m = 20 m OR OR | Formula A SF CA circum.velocity CA NPU NPR (3) |
10.2.3 | w = 2πn OR | Formula SF CA ang.velocity CA NPU NPR (3) |
[19] |
QUESTION 11
11.1.1 | Area = length´breadth OR | M A length A AO Full marks(2) | ||
11.1.2 | p = 15m | values of p CA(1) | ||
11.1.3 | AT = a(o1 + on+ o2 + o3 + ....+ on-1) OR OR | formula A OR OR | ||
11.2 | ||||
11.2.1 | 1 l = 1 000 cm3 1,5l = 1500 cm3 | value of A | ||
11.2.2 | TSA/TBO = 4( ½ side length of base x slant height) + (sidelength )2 =4(½ x 3 x 3.81) + (3 x 3) = 22.86 + 9 = 31.86cm2 | F A SF A value of TSA CA NPR/NPU AO Full marks | ||
11.2.3 | Volume of pyramid = 1/3 (length x breadth) x ⊥ Height = 1/3 (3 x 3) x 3,5 = 10, 5 cm3 number of small pyramids = 1500 10,5 ≈142,86 142Remaining milk= 1500 - (142 x 10, 5) OR 0,86 x 10,5 = 9cm3OR 9 ml | SF A value of V pyramid CA M CA value of CA NPU/NPR(4)[17] |
TOTAL: 150