TECHNICAL MATHEMATICS
PAPER 1
GRADE 12
NSC EXAMS
PAST PAPERS AND MEMOS JUNE 2019
Marking Codes | |
A | Accuracy |
CA | Consistent accuracy |
M | Method |
R | Rounding |
NPR | No penalty for rounding |
NPU | No penalty for units omitted |
S | Simplification |
SF | Substitution in the correct formula |
NOTE:
QUESTION 1 | ||||
1.1 | 1.1.1 | (7x - 1)(x + 2) = 0 | ✓ x = 1/7 A | (2) |
1.1.2 | (x - 2)(3x - 1) = 1 3x2 - 7x + 1 = 0 x = 2,18 or x = 0,15 | ✓Simplification/Standard Form A | (4) | |
1.1.3 | -x2 - 4x + 5 ≥ 0 | ✓Factors/ Substitution in the quadratic formula M A | (3) |
1.2 | Total thickness of the pack = 151 ÷ 1 000 = 0,151 m | ✓Conversion A | (3) | |
1.3 | 1.3.1 | (y - 5x)(y + 5x) | ✓Factors A | (1) |
1.3.2 | 2y + 6x = a and y2 - 25x2 = 4 y = -3x + 2 substitute y into y2 - 25x2 = 4 (-3x + 2)2 - 25x2 - 4 = 0 9x2 - 12x + 4 - 25x2 - 4 = 0 -16x2 - 12x = 0 -x(16x + 12) = 0 x = 0 or 16x = -12 x = 0 or x = 12/16 = 3/4 if x = 0 then y = 2 if x = 3/4 then = 17/4 | ✓y - the subject A | (6) |
1.3.2 | OR | ✓y - the subject A | ||
1.4 | 1.4.1 | p ∈{1;2;3;4;5} | ✓✓All 5 correct values A | (2) |
1.4.2 | p = 6 | ✓6 A | (1) | |
1.5 | Real, rational and unequal | ✓Real and unequal A | (2) | |
1.6 | 86 = 26 + 24 + 22 + 21 - 1 mark of base 2 is not written | ✓Method M | (2) | |
[26] |
QUESTION 2 | ||||
2.1 | 2.1.1 | 3n.34 = 3n+4 | ✓3n+4 A | (1) |
2.1.2 | ✓Factors S ✓Simplification S CA | (2) | ||
2.1.3 | √32 - √72 + √18 | ✓Simplification S | (2) | |
2.1.4 | -log3243 + log31 | ✓0 A | (2) | |
2.2 | 2.2.1 | 42x2 - 3 - x = 4º 2x2 - x - 3 = 0 (2x - 3)(x + 1) = 0 x = 3/2 or x = −1 | ✓Power rule A | (6) |
2.2.2 | x = log 6 - log 2 or x = log 6 - log 2 | ✓Log law | (4) | |
2.3 | x + 2yi = (-2 + 6i)(4 - 7i) | ✓Expansion S | (4) | |
2.4 |
| ✓Finding r SF | (5) | |
[26] |
QUESTION 3 | |||
3.1 | x = 0 and y = 0 | ✓x = 0 A | (2) |
3.2 | y =1 | ✓ y = 1 A | (1) |
3.3 | ✓Shape g A | (3) | |
3.4 | x ≠ ∈ 0,x R | ✓ x ≠ 0,x R ∈ CA | (1) |
3.5 | x < of x > −1 or / 0 | ✓x<−1 CA | (2) |
[9] |
QUESTION 4 | ||||
4.1 | ✓Value of SF | (2) | ||
4.2 | m = -3 = 1 | ✓m of the line through origin and P A | (4) | |
4.3 | ✓✓Accurate answer A | (2) | ||
[8] |
QUESTION 5 | |||
5.1 | f(x) = ax2 + bx - 12 OR f(-2) = a(-2)2 + b(-2) - 12 = 0 | ✓Substitution SF | (4) |
5.2 | f(x) = x2 - 4x - 12 | ✓Method M | (3) |
OR | |||
OR |
5.3 | ✓x-intercepts A | (4) | |
[11] |
QUESTION 6 | |||
6.1 | ✓Formula A | (3) | |
6.2 | A = P(1 -i)n | ✓Formula A | (5) |
6.3 | A = P(1 -i)n | ✓Formula A | (6) |
OR | ✓Formula A | ||
[14] |
QUESTION 7 | ||||
NOTE : Deduct 1 mark for incorrect notation from 7.1 to 7.2.2 LET WEL: Trek 1 punt af vir inkorrekte notasie van 7.1 tot 7.2.2 | ||||
7.1 | ✓Formula A | (5) | ||
7.2 | 7.2.1 |
| ✓ 2x−2 A | (4) |
7.2.2 | ✓x A | (4) |
7.3 | 7.3.1 | h(x) = 3x2 - 7x + 2 h'(x) = 6x - 7 h'(0,5) = 6(0,5) - 7 mtangent = -4 | ✓6x - 7 A | (3) |
7.3.2 | h(x) = 3x2 - 7x + 2 | ✓−2 A | (4) | |
[20] | ||||
QUESTION 8 | ||||
8.1 | g(x) = x3 - x | ✓Substitution by 0 SF | (4) | |
8.2 | ✓Derivative A | (5) | ||
8.3 | −0,58 ≤ x ≤ 0,58 | ✓−0,58 CA | (3) | |
[12] |
QUESTION 9 | |||
9.1 | V = πr2h | ✓Formula A | (3) |
9.2 | SA = 2πrh + 2πr2 SA = 2πr × 275 + 2πr2 πr2 SA = 550 + 2πr2 r | ✓Formula A | (3) |
9.3 | ✓ 550r-1 A | (5) | |
[11] |
QUESTION 10 | ||||
10.1 | ✓Simplification A | (5) | ||
10.2 | 10.2.1 | ✓ - x4 A | (3) | |
10.2.2 | ✓A definite integral formula A | (5) | ||
[13] | ||||
TOTAL: | 150 |