MICHAEL NOAH™ ORIGINAL PRACTICE PAPER
GRADE 12 MATHEMATICS · PAPER 1 · MOCK 001
Important: This is an original Michael Noah practice simulation, not an official DBE examination. It follows the DBE Grade 12 Mathematics Paper 1 topic allocation: Algebra 25, Patterns 25, Functions 35, Finance 15, Differential Calculus 35, Probability 15.
Instructions
- Answer all questions.
- Show all calculations, diagrams and reasoning needed to earn method marks.
- Round final numerical answers to two decimal places unless the question requires an exact value.
- Use a non-programmable calculator where appropriate.
- Do not open the memo until you finish the paper or timed section.
QUESTION 1 · ALGEBRA, EQUATIONS AND INEQUALITIES [25]
1.1
Solve: 2x² − 5x − 3 = 0.
[4]
1.2
Solve for real x: √(2x + 3) = x. Check any possible extraneous solution.
[5]
1.3
Solve the inequality: (x − 4)(x + 1) ≤ 0.
[4]
1.4
The equation x² + (k − 2)x + k = 0 has equal roots. Determine the possible values of k.
[5]
1.5
The graphs y = x² − 4x + 3 and y = 2x − 3 intersect. Determine the coordinates of both intersection points in exact form.
[7]
QUESTION 2 · PATTERNS, SEQUENCES AND SERIES [25]
2.1
The arithmetic sequence is 7; 12; 17; …
2.1.1 Determine Tn. [3]
2.1.2 Determine T30. [2]
2.1.3 Which term has value 247? [3]
2.1.1 Determine Tn. [3]
2.1.2 Determine T30. [2]
2.1.3 Which term has value 247? [3]
[8]
2.2
The quadratic sequence is 2; 7; 14; 23; …
2.2.1 Write the next two terms. [2]
2.2.2 Derive a formula for Tn. [6]
2.2.3 Determine T20. [1]
2.2.1 Write the next two terms. [2]
2.2.2 Derive a formula for Tn. [6]
2.2.3 Determine T20. [1]
[9]
2.3
The geometric sequence is 243; 81; 27; …
2.3.1 Determine Tn. [3]
2.3.2 Determine the sum to infinity. [3]
2.3.3 Determine the number of the first term whose value is less than 1. [2]
2.3.1 Determine Tn. [3]
2.3.2 Determine the sum to infinity. [3]
2.3.3 Determine the number of the first term whose value is less than 1. [2]
[8]
QUESTION 3 · FUNCTIONS AND GRAPHS [18]
Let f(x) = x² − 6x + 5.
3.1
Determine all x- and y-intercepts.
[4]
3.2
Determine the coordinates of the turning point.
[4]
3.3
Sketch f, showing the intercepts and turning point.
[4]
3.4
State the domain and range of f.
[3]
3.5
Use the graph or algebra to solve f(x) ≥ 0.
[3]
QUESTION 4 · FUNCTIONS AND GRAPHS [17]
Let g(x) = 2/(x − 1) + 3.
4.1
State the equations of the asymptotes.
[2]
4.2
Determine the x- and y-intercepts.
[4]
4.3
State the range of g.
[2]
4.4
Determine g⁻¹(x).
[5]
4.5
Describe the geometric relationship between the graphs of g and g⁻¹.
[2]
4.6
Determine the x-coordinates of the points where g(x) = x.
[2]
QUESTION 5 · FINANCE, GROWTH AND DECAY [15]
5.1
R25 000 is invested at a nominal rate of 9.6% p.a. compounded monthly for 4 years. Calculate the future value.
[5]
5.2
A loan of R120 000 is repaid by equal monthly payments over 5 years. Interest is 11% p.a. compounded monthly. Payments start one month after the loan is granted. Calculate the monthly payment.
[6]
5.3
Convert a nominal rate of 10.5% p.a. compounded monthly to the equivalent effective annual rate.
[4]
QUESTION 6 · DIFFERENTIAL CALCULUS [20]
Let h(x) = x³ − 6x² + 9x + 4.
6.1
Determine h′(x).
[3]
6.2
Determine the x-values of the stationary points.
[4]
6.3
Determine the coordinates of the stationary points.
[4]
6.4
Classify each stationary point as a local maximum or local minimum. Give a reason.
[4]
6.5
Determine the point of inflection.
[3]
6.6
Determine the y-intercept of h.
[2]
QUESTION 7 · DIFFERENTIAL CALCULUS / OPTIMISATION [15]
A rectangular sheet of cardboard measures 30 cm by 20 cm. Squares of side x cm are cut from each corner. The sides are folded up to form an open box.
7.1
Show that the volume is V(x) = 600x − 100x² + 4x³.
[3]
7.2
Determine V′(x).
[3]
7.3
Write the equation that must be solved for a stationary volume.
[2]
7.4
Hence determine the value of x in the physical domain that maximises the volume. Give x to two decimal places.
[4]
7.5
Calculate the maximum volume to two decimal places.
[3]
QUESTION 8 · COUNTING PRINCIPLE AND PROBABILITY [15]
A bag contains 5 red counters, 3 blue counters and 2 green counters.
8.1
One counter is selected at random. Find P(not blue).
[2]
8.2
Two counters are selected without replacement. Find P(both red).
[3]
8.3
Two counters are selected without replacement. Find the probability of selecting one red and one blue counter, in any order.
[4]
8.4
How many 6-digit codes can be made from digits 0–9 if no digit may be repeated and the first digit may not be 0?
[3]
8.5
A 4-digit PIN is made using digits 0–9, with repetition allowed. How many PINs contain exactly one digit 7?
[3]
TOTAL: 150 MARKS
MEMORANDUM · MOCK 001
Marking note: This memo gives expected mathematical evidence for this original practice paper. Valid alternative methods should receive appropriate method credit. Do not turn a single paper score into a mastery claim.
| Item | Expected answer / method evidence | Marks |
|---|---|---|
| 1.1 | (2x + 1)(x − 3)=0; x=3 or x=−1/2. | 4 |
| 1.2 | x≥0. Square: 2x+3=x² ⇒ x²−2x−3=0 ⇒ (x−3)(x+1)=0. x=3; reject x=−1 because it does not satisfy the original equation. | 5 |
| 1.3 | Critical values −1 and 4. Product ≤0 on −1 ≤ x ≤ 4. | 4 |
| 1.4 | Equal roots: Δ=0. (k−2)²−4k=0 ⇒ k²−8k+4=0 ⇒ k=4±2√3. | 5 |
| 1.5 | x²−4x+3=2x−3 ⇒ x²−6x+6=0 ⇒ x=3±√3. Then y=2x−3 ⇒ y=3±2√3. Points: (3+√3, 3+2√3) and (3−√3, 3−2√3). | 7 |
| 2.1 | Tn=7+(n−1)5=5n+2; T30=152; 5n+2=247 ⇒ n=49. | 8 |
| 2.2 | Next terms 34; 47. First differences 5,7,9; second difference 2 ⇒ a=1. Substitution gives Tn=n²+2n−1. T20=439. | 9 |
| 2.3 | r=1/3. Tn=243(1/3)^(n−1). S∞=243/(1−1/3)=364.5. T6=1, so first term below 1 is T7. | 8 |
| 3.1 | f(x)=(x−1)(x−5): x-intercepts (1,0),(5,0); y-intercept (0,5). | 4 |
| 3.2 | x=−b/(2a)=3; f(3)=−4 ⇒ (3,−4). | 4 |
| 3.3 | Upward-opening parabola through (1,0),(5,0),(0,5), turning at (3,−4), with correct symmetry. | 4 |
| 3.4 | Domain: x∈R. Range: y≥−4. | 3 |
| 3.5 | x≤1 or x≥5. | 3 |
| 4.1 | x=1 and y=3. | 2 |
| 4.2 | x-intercept: (1/3,0). y-intercept: (0,1). | 4 |
| 4.3 | y∈R, y≠3. | 2 |
| 4.4 | Swap x,y and solve: g⁻¹(x)=1+2/(x−3). | 5 |
| 4.5 | The inverse is the reflection of g in the line y=x. | 2 |
| 4.6 | x=2/(x−1)+3 ⇒ (x−3)(x−1)=2 ⇒ x²−4x+1=0 ⇒ x=2±√3. | 2 |
| 5.1 | FV=25 000(1+0.096/12)^48 ≈ R36 647.60. | 5 |
| 5.2 | i=0.11/12, n=60. X=120000i/[1−(1+i)^−60] ≈ R2 609.09 per month. | 6 |
| 5.3 | i_eff=(1+0.105/12)^12−1≈0.110203 ⇒ 11.02% p.a. effective. | 4 |
| 6.1 | h′(x)=3x²−12x+9=3(x−1)(x−3). | 3 |
| 6.2 | h′(x)=0 ⇒ x=1 or x=3. | 4 |
| 6.3 | h(1)=8; h(3)=4 ⇒ (1,8) and (3,4). | 4 |
| 6.4 | h″(x)=6x−12. h″(1)<0 ⇒ local maximum at (1,8); h″(3)>0 ⇒ local minimum at (3,4). | 4 |
| 6.5 | h″(x)=0 ⇒ x=2; h(2)=6 ⇒ (2,6). | 3 |
| 6.6 | h(0)=4 ⇒ (0,4). | 2 |
| 7.1 | V=x(30−2x)(20−2x)=600x−100x²+4x³. | 3 |
| 7.2 | V′(x)=600−200x+12x². | 3 |
| 7.3 | 12x²−200x+600=0 (equivalent simplified equation accepted). | 2 |
| 7.4 | 3x²−50x+150=0 ⇒ x=(50±10√7)/6. Physical domain 0<x<10 gives x≈3.92 cm. | 4 |
| 7.5 | V(3.9237…)≈1 056.31 cm³. | 3 |
| 8.1 | 7/10. | 2 |
| 8.2 | (5/10)(4/9)=2/9. | 3 |
| 8.3 | (5/10)(3/9)+(3/10)(5/9)=1/3. | 4 |
| 8.4 | 9×9×8×7×6×5=136 080. | 3 |
| 8.5 | Choose position of 7: 4 ways. Other three positions each have 9 choices excluding 7: 4×9³=2 916. | 3 |
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