MICHAEL NOAH™ ORIGINAL PRACTICE PAPER
GRADE 12 MATHEMATICS · PAPER 2 · MOCK 001
Important: This is an original Michael Noah practice simulation, not an official DBE examination. It follows the DBE Grade 12 Paper 2 topic allocation: Statistics & Regression 20, Analytical Geometry 40, Trigonometry 50, Euclidean Geometry 40.
Instructions
- Answer all questions.
- Show calculations and give reasons in geometry where required.
- Round final numerical answers to two decimal places unless an exact value is requested.
- Diagrams are not necessarily drawn to scale.
- Do not open the memo until you finish the paper or timed section.
QUESTION 1 · STATISTICS AND REGRESSION [20]
The following 12 values are arranged in ascending order:
4, 7, 8, 9, 10, 12, 13, 14, 15, 17, 18, 22
1.1.1
Determine the median.
[2]
1.1.2
Determine the lower quartile Q₁ and upper quartile Q₃.
[2]
1.1.3
Determine the interquartile range.
[1]
1.1.4
Use the 1.5 × IQR rule to determine whether 22 is an outlier.
[3]
A study records hours studied (x) and test mark (y). The least-squares regression line is ŷ = 6.5x + 41.8 and the correlation coefficient is r = 0.94.
1.2.1
Describe the strength and direction of the linear correlation.
[2]
1.2.2
Predict the mark for a learner who studies 6 hours.
[2]
1.2.3
Interpret the gradient 6.5 in context.
[2]
1.2.4
The learner's actual mark is 78. Calculate the residual (actual − predicted).
[3]
1.2.5
Can the value r = 0.94 alone prove that extra study time causes the exact increase in marks? Explain.
[3]
QUESTION 2 · ANALYTICAL GEOMETRY [20]
Points A(−2; 3), B(4; 7) and C(6; −1) are given.
2.1
Determine the gradient of AB.
[2]
2.2
Determine the midpoint of AB.
[2]
2.3
Determine the length AB in exact form.
[3]
2.4
Determine the equation of line AB in the form y = mx + c.
[3]
2.5
Determine the equation of the perpendicular bisector of AB.
[3]
2.6
AB is a diameter of a circle. Determine the equation of this circle.
[4]
2.7
Determine, with calculation, whether C lies on this circle.
[3]
QUESTION 3 · ANALYTICAL GEOMETRY [20]
A circle has equation x² + y² − 6x + 4y − 12 = 0. Point P(6; 2) is given.
3.1
Write the equation in centre-radius form and hence state the centre and radius.
[4]
3.2
Verify that P lies on the circle.
[2]
3.3
Determine the gradient of the radius from the centre to P.
[2]
3.4
Determine the equation of the tangent to the circle at P.
[3]
3.5
Determine the x-intercept and y-intercept of this tangent.
[4]
3.6
Determine the coordinates of the point on the circle diametrically opposite P.
[3]
3.7
Determine the length of the diameter through P.
[2]
QUESTION 4 · TRIGONOMETRY [20]
4.1
Prove the identity: (1 − cos 2x) / sin 2x = tan x, for values where both sides are defined.
[5]
4.2
Solve for 0° ≤ x ≤ 360°: 2sin²x − 3sinx + 1 = 0.
[6]
4.3
Solve for 0° ≤ x ≤ 360°: cos(x − 20°) = 1/2.
[4]
4.4
Simplify to a single trigonometric expression: [sin(180° − x) cos(360° + x)] / tan(180° + x).
[5]
QUESTION 5 · TRIGONOMETRY: GRAPHS [15]
For 0° ≤ x ≤ 360°, let f(x) = 2sinx − 1 and g(x) = cos2x.
5.1
State the amplitude and range of f.
[3]
5.2
State the period of g.
[2]
5.3
Determine the x-intercepts of f.
[3]
5.4
State the coordinates of the maximum point of f in the given interval.
[2]
5.5
Determine the x-values for which g(x) = −1.
[2]
5.6
Sketch f and g on the same set of axes, showing the key turning points/intercepts needed to identify each graph.
[3]
QUESTION 6 · TRIGONOMETRY: APPLICATIONS [15]
In triangle ABC, AB = 12 cm, AC = 9 cm and ∠A = 58°.
6.1.1
Calculate BC.
[4]
6.1.2
Calculate the area of triangle ABC.
[2]
6.1.3
Calculate ∠B.
[2]
A learner observes the top of a vertical tower from point P at an angle of elevation of 32°. The learner then walks 40 m directly towards the tower to point Q, where the angle of elevation is 47°. Let the horizontal distance from Q to the base of the tower be d metres and the tower height be h metres.
6.2.1
Form two trigonometric equations linking h and d.
[3]
6.2.2
Calculate d.
[2]
6.2.3
Hence calculate the height h of the tower.
[2]
QUESTION 7 · EUCLIDEAN GEOMETRY [20]
In 7.1, AB is a diameter of a circle, C is a point on the circle, and AT is a tangent at A. Given ∠BAC = 34°.
7.1.1
Determine ∠ACB. Give a reason.
[2]
7.1.2
Determine ∠ABC.
[2]
7.1.3
Determine the acute angle ∠TAC.
[2]
7.1.4
Explain how the tangent-chord theorem confirms your answer to 7.1.3.
[2]
In 7.2, PQRS is a cyclic quadrilateral and PR is a diagonal. Given ∠QPS = 74°, ∠PQR = 112° and ∠QPR = 31°.
7.2.1
Determine ∠RPS.
[2]
7.2.2
Determine ∠PSR. Give a reason.
[3]
7.2.3
Determine ∠PRS.
[2]
7.2.4
Determine ∠PQS. Give a reason.
[2]
7.2.5
Determine ∠QRS. Give a reason.
[3]
QUESTION 8 · EUCLIDEAN GEOMETRY [20]
In triangle ABC, D lies on AB and E lies on AC. DE ∥ BC. AD = 6 cm, DB = 4 cm, AE = 7.5 cm and BC = 12 cm.
8.1.1
Determine AB.
[1]
8.1.2
Prove that △ADE ∼ △ABC.
[3]
8.1.3
Determine EC.
[3]
8.1.4
Determine DE.
[3]
In 8.2, P is outside a circle. PT is tangent to the circle at T. A secant from P meets the circle at Q and then R, so P, Q and R are collinear. Given ∠TPQ = 62° and ∠PTQ = 48°.
8.2.1
Determine ∠PRT. Give a reason.
[2]
8.2.2
Prove that △PTQ ∼ △PRT.
[4]
8.2.3
Hence show that PT² = PQ · PR.
[2]
8.2.4
If PQ = 5 cm and PR = 20 cm, calculate PT.
[2]
TOTAL: 150 MARKS
MEMORANDUM · PAPER 2 · MOCK 001
Marking note: Credit valid equivalent methods and correct geometric reasons. This is practice evidence, not an official DBE memo and not a standalone mastery determination.
| Item | Expected answer / method evidence | Marks |
|---|---|---|
| 1.1.1 | Median = (12 + 13)/2 = 12.5. | 2 |
| 1.1.2 | Lower half 4,7,8,9,10,12 gives Q₁=(8+9)/2=8.5. Upper half 13,14,15,17,18,22 gives Q₃=(15+17)/2=16. | 2 |
| 1.1.3 | IQR = 16 − 8.5 = 7.5. | 1 |
| 1.1.4 | Upper fence = 16 + 1.5(7.5)=27.25; lower fence = 8.5 − 11.25=−2.75. Since 22 lies within the fences, 22 is not an outlier. | 3 |
| 1.2.1 | Strong positive linear correlation. | 2 |
| 1.2.2 | ŷ=6.5(6)+41.8=80.8. | 2 |
| 1.2.3 | Within the model, each additional hour studied is associated with an increase of about 6.5 percentage points in predicted mark. | 2 |
| 1.2.4 | Residual = 78 − 80.8 = −2.8. | 3 |
| 1.2.5 | No. Correlation/regression measures association and does not by itself establish causation; other variables and study design matter. | 3 |
| 2.1 | m=(7−3)/(4−(−2))=2/3. | 2 |
| 2.2 | Midpoint = ((−2+4)/2,(3+7)/2)=(1;5). | 2 |
| 2.3 | AB=√(6²+4²)=√52=2√13. | 3 |
| 2.4 | y−3=(2/3)(x+2), hence y=(2/3)x+13/3. | 3 |
| 2.5 | Perpendicular gradient = −3/2 through (1;5): y−5=−(3/2)(x−1). | 3 |
| 2.6 | Centre (1;5), radius²=AB²/4=52/4=13. (x−1)²+(y−5)²=13. | 4 |
| 2.7 | At C(6;−1): (6−1)²+(−1−5)²=25+36=61≠13; C is not on the circle. | 3 |
| 3.1 | (x−3)²+(y+2)²=25. Centre (3;−2); radius 5. | 4 |
| 3.2 | (6−3)²+(2+2)²=9+16=25; therefore P lies on the circle. | 2 |
| 3.3 | Gradient CP=(2−(−2))/(6−3)=4/3. | 2 |
| 3.4 | Tangent gradient = −3/4. y−2=−(3/4)(x−6). | 3 |
| 3.5 | x-intercept (26/3;0); y-intercept (0;13/2). | 4 |
| 3.6 | Opposite point = 2(3;−2)−(6;2)=(0;−6). | 3 |
| 3.7 | Diameter = 10. | 2 |
| 4.1 | 1−cos2x = 2sin²x and sin2x=2sinx cosx. Quotient = sinx/cosx = tanx, where defined. | 5 |
| 4.2 | (2sinx−1)(sinx−1)=0. sinx=1/2 or 1. Hence x=30°, 90°, 150°. | 6 |
| 4.3 | x−20°=60° or 300° (mod 360°). In the interval: x=80° or 320°. | 4 |
| 4.4 | sin(180−x)=sinx; cos(360+x)=cosx; tan(180+x)=tanx. Thus expression = cos²x, where defined. | 5 |
| 5.1 | Amplitude 2; range −3≤y≤1. | 3 |
| 5.2 | Period of cos2x = 180°. | 2 |
| 5.3 | 2sinx−1=0 ⇒ sinx=1/2 ⇒ x=30°,150°. | 3 |
| 5.4 | Maximum when sinx=1: (90°;1). | 2 |
| 5.5 | cos2x=−1 ⇒ 2x=180°,540° ⇒ x=90°,270°. | 2 |
| 5.6 | f: key points (0,−1),(90,1),(180,−1),(270,−3),(360,−1), crossing x-axis at 30°,150°. g: (0,1),(45,0),(90,−1),(135,0),(180,1),(225,0),(270,−1),(315,0),(360,1). Award for correct shape, scale/key points and identification. | 3 |
| 6.1.1 | Cosine rule: BC²=12²+9²−2(12)(9)cos58°. BC≈10.51 cm. | 4 |
| 6.1.2 | Area = ½(12)(9)sin58° ≈ 45.79 cm². | 2 |
| 6.1.3 | Using sine rule or cosine rule: ∠B≈46.55°. | 2 |
| 6.2.1 | tan47°=h/d and tan32°=h/(d+40). | 3 |
| 6.2.2 | d tan47°=(d+40)tan32°. Hence d≈55.85 m. | 2 |
| 6.2.3 | h=d tan47° ≈ 59.90 m. | 2 |
| 7.1.1 | 90°, angle subtended by a diameter at the circumference. | 2 |
| 7.1.2 | 180°−90°−34°=56°. | 2 |
| 7.1.3 | Tangent AT ⟂ radius/diameter AB at A, so acute ∠TAC=90°−34°=56°. | 2 |
| 7.1.4 | The angle between tangent AT and chord AC equals the angle in the alternate segment subtended by chord AC, namely ∠ABC=56°. | 2 |
| 7.2.1 | ∠RPS=74°−31°=43°. | 2 |
| 7.2.2 | Opposite angles of a cyclic quadrilateral are supplementary: ∠PSR=180°−112°=68°. | 3 |
| 7.2.3 | In △PRS: 180°−43°−68°=69°. | 2 |
| 7.2.4 | ∠PQS=69°, angles subtended by chord PS in the same segment are equal. | 2 |
| 7.2.5 | ∠QRS=180°−74°=106°, opposite angles of cyclic quadrilateral PQRS are supplementary. | 3 |
| 8.1.1 | AB=6+4=10 cm. | 1 |
| 8.1.2 | DE∥BC gives corresponding angles equal: ∠ADE=∠ABC and ∠AED=∠ACB. Hence △ADE∼△ABC by AA. | 3 |
| 8.1.3 | AD/AB=AE/AC=6/10=0.6. AC=7.5/0.6=12.5, so EC=5 cm. | 3 |
| 8.1.4 | DE/BC=AD/AB=0.6, hence DE=0.6(12)=7.2 cm. | 3 |
| 8.2.1 | ∠PRT=∠QRT=48°, because the angle between tangent PT and chord TQ equals the angle subtended by chord TQ in the alternate segment. | 2 |
| 8.2.2 | ∠TPQ=∠TPR=62° (P,Q,R collinear) and ∠PTQ=∠PRT=48°. Therefore △PTQ∼△PRT by AA. | 4 |
| 8.2.3 | From similarity, PT/PR = PQ/PT. Therefore PT²=PQ·PR. | 2 |
| 8.2.4 | PT²=5(20)=100, so PT=10 cm. | 2 |
Finished marking? Classify each lost mark: statistics interpretation, coordinate algebra, trig identity/equation/graph/application, or geometry theorem/reason. Repair that class before the next fresh paper.
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