A Level H1 Mathematics Graphs Coordinate Geometry Quiz

A-Level Maths H1 Quiz - Graphs Coordinate Geometry

Answer Key and Marking Scheme

Total Marks: 50

Section A: Scatter Diagrams and Correlation (Questions 1–5) [10 marks]

1. (a) Calculate r. [2 marks]

[ \begin{aligned} S_{xx} &= \sum x^2 - \frac{(\sum x)^2}{n} = 1248 - \frac{96^2}{8} = 1248 - 1152 = 96 \[4pt] S_{yy} &= \sum y^2 - \frac{(\sum y)^2}{n} = 49,536 - \frac{624^2}{8} = 49,536 - 48,672 = 864 \[4pt] S_{xy} &= \sum xy - \frac{(\sum x)(\sum y)}{n} = 7776 - \frac{96 \times 624}{8} = 7776 - 7488 = 288 \[4pt] r &= \frac{S_{xy}}{\sqrt{S_{xx} \times S_{yy}}} = \frac{288}{\sqrt{96 \times 864}} = \frac{288}{\sqrt{82,944}} = \frac{288}{288} = 1.000 \end{aligned} ]

Answer: r = 1.000
Marking: M1 for correct substitution into formula, A1 for correct value to 3 d.p.

(b) Interpretation. [1 mark]
Answer: There is a perfect positive linear correlation between hours spent studying and test score. As study hours increase, test score increases in a perfectly linear manner.
Marking: B1 for "perfect positive" or equivalent.

2. (a) Statistical term. [1 mark]
Answer: Outlier (or anomalous point).
Marking: B1 for "outlier".

(b) Effect on r. [1 mark]
Answer: The outlier would make the value of r closer to 0 (weaker correlation) than it would be without the outlier, because it does not follow the general linear trend.
Marking: B1 for indicating r becomes weaker/closer to 0.

3. (a) Strength and direction. [1 mark]
Answer: There is a weak (or moderate) positive linear correlation between the two variables.
Marking: B1 for "weak/moderate positive".

(b) Explain why claim may be incorrect. [1 mark]
Answer: Correlation does not imply causation. A positive correlation means the variables tend to increase together, but this does not prove that increasing one variable causes the other to increase. There may be a third factor influencing both, or the relationship may be coincidental.
Marking: B1 for "correlation does not imply causation" or equivalent reasoning.

4. (a) Meaning of gradient. [1 mark]
Answer: For every increase of $1000 in advertising expenditure, sales revenue increases by approximately$3200, on average.
Marking: B1 for correct interpretation with units.

(b) Estimate sales revenue. [1 mark]
Answer: When x = 25, y = 12.4 + 3.2(25) = 12.4 + 80 = 92.4. Sales revenue is approximately $92 400.
Marking: B1 for correct substitution and answer.

(c) Reliability of estimate at x = 60. [1 mark]
Answer: The estimate would be unreliable because x = 60 ($60 000) is likely outside the range of the original data (extrapolation). The linear relationship may not hold beyond the observed data range.
Marking: B1 for "extrapolation" or "outside data range" with explanation.

5. Validity of causation conclusion. [1 mark]
Answer: The conclusion is not valid. A high correlation (r = −0.98) indicates a strong linear relationship, but it does not prove that changes in p cause changes in q. Correlation does not imply causation.
Marking: B1 for stating not valid with "correlation ≠ causation" reasoning.

Section B: Regression Lines and Estimation (Questions 6–10) [12 marks]

6. (a) Regression line of y on x. [2 marks]

[ \begin{aligned} n &= 7 \[4pt] \sum x &= 22 + 25 + 28 + 30 + 32 + 35 + 38 = 210 \[4pt] \sum y &= 45 + 52 + 60 + 68 + 75 + 82 + 95 = 477 \[4pt] \sum x^2 &= 22^2 + 25^2 + 28^2 + 30^2 + 32^2 + 35^2 + 38^2 = 484 + 625 + 784 + 900 + 1024 + 1225 + 1444 = 6486 \[4pt] \sum xy &= 22(45) + 25(52) + 28(60) + 30(68) + 32(75) + 35(82) + 38(95) \ &= 990 + 1300 + 1680 + 2040 + 2400 + 2870 + 3610 = 14,890 \[4pt] b &= \frac{n\sum xy - (\sum x)(\sum y)}{n\sum x^2 - (\sum x)^2} = \frac{7(14,890) - 210(477)}{7(6486) - 210^2} \[4pt] &= \frac{104,230 - 100,170}{45,402 - 44,100} = \frac{4060}{1302} \approx 3.118... \[4pt] a &= \bar{y} - b\bar{x} = \frac{477}{7} - 3.118... \times \frac{210}{7} = 68.142... - 3.118... \times 30 \[4pt] &= 68.142... - 93.557... = -25.414... \end{aligned} ]

Answer: y = −25.4 + 3.12x (gradient and intercept to 3 s.f.)
Marking: M1 for correct method, A1 for correct values to 3 s.f.

(b) Scatter diagram. [2 marks]
Marking: B1 for correctly labelled axes with scales, B1 for accurately plotted points.

(c) Regression line on scatter diagram. [1 mark]
Marking: B1 for straight line passing through (x̄, ȳ) = (30, 68.1) with correct gradient.

(d) Estimate at 27 °C. [1 mark]
Answer: y = −25.4 + 3.12(27) = −25.4 + 84.24 = 58.84 ≈ 59 ice creams.
Marking: B1 for correct substitution and answer (accept 58–60).

7. (a) Find ȳ. [1 mark]
Answer: The regression line of y on x passes through (x̄, ȳ).
ȳ = 5.0 − 0.8(10) = 5.0 − 8.0 = −3.0.
Marking: B1 for correct substitution and answer.

(b) Why regression lines differ. [1 mark]
Answer: The regression line of y on x minimises the sum of squared vertical deviations (errors in y), while the regression line of x on y minimises the sum of squared horizontal deviations (errors in x). They are different lines unless all points lie exactly on a straight line (r = ±1).
Marking: B1 for explanation involving minimising different deviations.

8. (a) Interpret −2.5. [1 mark]
Answer: For every additional hour of training, the time taken to complete the task decreases by 2.5 minutes, on average.
Marking: B1 for correct interpretation with units.

(b) Estimate at t = 10. [1 mark]
Answer: m = 48 − 2.5(10) = 48 − 25 = 23 minutes.
Marking: B1 for correct substitution and answer.

(c) Appropriateness at t = 30. [1 mark]
Answer: It would not be appropriate because t = 30 is likely outside the range of the original training data (extrapolation). The linear relationship may not hold for such a high number of training hours; the time cannot decrease indefinitely.
Marking: B1 for "extrapolation/outside data range" with reasoning.

9. Regression line of y on x. [2 marks]

[ \begin{aligned} b &= \frac{S_{xy}}{S_{xx}} = \frac{-875}{1250} = -0.7 \[4pt] \bar{x} &= \frac{300}{15} = 20, \quad \bar{y} = \frac{450}{15} = 30 \[4pt] a &= \bar{y} - b\bar{x} = 30 - (-0.7)(20) = 30 + 14 = 44 \end{aligned} ]

Answer: y = 44 − 0.7x
Marking: M1 for correct gradient, A1 for correct equation.

10. Find the residual. [1 mark]
Answer: For x = 8, predicted y = 2.4 + 1.6(8) = 2.4 + 12.8 = 15.2.
Residual = observed − predicted = 14 − 15.2 = −1.2.
Marking: B1 for correct residual.

Section C: Coordinate Geometry and Graphs (Questions 11–15) [13 marks]

11. (a) Find dy/dx. [1 mark]
Answer: dy/dx = 3x² − 12x + 9
Marking: B1 for correct derivative.

(b) x-coordinates of stationary points. [2 marks]
Answer: Set dy/dx = 0: 3x² − 12x + 9 = 0 → x² − 4x + 3 = 0 → (x − 1)(x − 3) = 0.
x = 1 or x = 3.
Marking: M1 for setting dy/dx = 0 and solving, A1 for both x-values.

(c) Nature of stationary points. [2 marks]
Answer: d²y/dx² = 6x − 12.
At x = 1: d²y/dx² = 6(1) − 12 = −6 < 0 → maximum.
At x = 3: d²y/dx² = 6(3) − 12 = 6 > 0 → minimum.
Marking: M1 for second derivative or nature table, A1 for both correct conclusions.

12. (a) Find dy/dx. [1 mark]
Answer: y = 4x⁻¹ + x → dy/dx = −4x⁻² + 1 = 1 − 4/x²
Marking: B1 for correct derivative.

(b) Coordinates of stationary point. [2 marks]
Answer: Set dy/dx = 0: 1 − 4/x² = 0 → 4/x² = 1 → x² = 4 → x = 2 (since x > 0).
When x = 2: y = 4/2 + 2 = 2 + 2 = 4.
Stationary point: (2, 4).
Marking: M1 for solving dy/dx = 0, A1 for correct coordinates.

(c) Maximum or minimum? [1 mark]
Answer: d²y/dx² = 8/x³. At x = 2: d²y/dx² = 8/8 = 1 > 0 → minimum.
Marking: B1 for correct conclusion with justification.

13. (a) Points of intersection. [3 marks]
Answer: Equate: x² − 2x + 3 = 2x − 1 → x² − 4x + 4 = 0 → (x − 2)² = 0 → x = 2 (repeated root).
When x = 2: y = 2(2) − 1 = 3.
The line is a tangent to the curve at the point (2, 3).
Marking: M1 for equating, M1 for solving quadratic, A1 for correct coordinates and interpretation.

(b) Sketch. [2 marks]
Marking: B1 for correct shape of parabola (minimum at vertex), B1 for line and tangent point (2, 3) clearly shown.

14. (a) x-coordinate of stationary point. [2 marks]
Answer: dy/dx = 2e²ˣ − 4.
Set dy/dx = 0: 2e²ˣ − 4 = 0 → e²ˣ = 2 → 2x = ln 2 → x = (ln 2)/2.
Marking: M1 for differentiating and setting to zero, A1 for exact x-coordinate.

(b) Show it is a minimum. [1 mark]
Answer: d²y/dx² = 4e²ˣ > 0 for all x, so the stationary point is a minimum.
Marking: B1 for second derivative positive.

15. Set up definite integral. [2 marks]
Answer: First find intersection with y-axis: curve at x = 0 gives y = 5; line at x = 0 gives y = −3.
Find intersection of curve and line: x² − 4x + 5 = 2x − 3 → x² − 6x + 8 = 0 → (x − 2)(x − 4) = 0 → x = 2 or x = 4.
The region R is bounded by the y-axis (x = 0) and the first intersection x = 2.

[ \text{Area} = \int_0^2 \left[(x^2 - 4x + 5) - (2x - 3)\right] dx = \int_0^2 (x^2 - 6x + 8), dx ]

Marking: M1 for finding limits and integrand, A1 for correct integral expression.

Section D: Applications and Problem Solving (Questions 16–20) [15 marks]

16. (a) Show A = 200x − 2x². [2 marks]
Answer: Let width = x metres, length parallel to river = y metres.
Fencing used: 2x + y = 200 → y = 200 − 2x.
Area A = x × y = x(200 − 2x) = 200x − 2x².
Marking: M1 for expressing y in terms of x, A1 for deriving area expression.

(b) Value of x for maximum area. [2 marks]
Answer: dA/dx = 200 − 4x = 0 → x = 50.
d²A/dx² = −4 < 0 → maximum.
Marking: M1 for differentiating and solving, A1 for x = 50 with verification.

(c) Maximum area. [1 mark]
Answer: A = 200(50) − 2(50)² = 10 000 − 5000 = 5000 m².
Marking: B1 for correct area.

17. (a) Population in 2020. [1 mark]
Answer: When t = 0: P = 120e⁰ = 120 thousand = 120 000.
Marking: B1 for correct population.

(b) Year population exceeds 200 000. [2 marks]
Answer: 200 = 120e⁰·⁰⁴ᵗ → e⁰·⁰⁴ᵗ = 200/120 = 5/3 → 0.04t = ln(5/3) → t = ln(5/3)/0.04 ≈ 12.77.
Population exceeds 200 000 during year t = 13, i.e., 2020 + 13 = 2033.
Marking: M1 for setting up equation and using logs, A1 for correct year.

(c) Rate of increase in 2030. [2 marks]
Answer: t = 10 in 2030. dP/dt = 120 × 0.04 × e⁰·⁰⁴ᵗ = 4.8e⁰·⁴.
dP/dt = 4.8 × e⁰·⁴ ≈ 4.8 × 1.4918 ≈ 7.16 thousand per year.
Marking: M1 for differentiating and substituting t = 10, A1 for correct rate.

18. (a) Equation of tangent at x = 1. [3 marks]
Answer: y = ln(2x + 1). When x = 1: y = ln 3.
dy/dx = 2/(2x + 1). At x = 1: dy/dx = 2/3.
Tangent: y − ln 3 = (2/3)(x − 1) → y = (2/3)x − 2/3 + ln 3.
Marking: M1 for y-coordinate, M1 for gradient, A1 for correct equation.

(b) Equation of normal at x = 1. [1 mark]
Answer: Gradient of normal = −3/2.
Normal: y − ln 3 = (−3/2)(x − 1) → y = (−3/2)x + 3/2 + ln 3.
Marking: B1 for correct normal equation.

19. (a) Vertical asymptote. [1 mark]
Answer: x = 1 (denominator zero).
Marking: B1 for x = 1.

(b) Horizontal asymptote. [1 mark]
Answer: As x → ±∞, y = (2x + 1)/(x − 1) = (2 + 1/x)/(1 − 1/x) → 2/1 = 2.
Horizontal asymptote: y = 2.
Marking: B1 for y = 2.

(c) Axis intercepts. [2 marks]
Answer: x-intercept (y = 0): 0 = (2x + 1)/(x − 1) → 2x + 1 = 0 → x = −1/2. Point: (−1/2, 0).
y-intercept (x = 0): y = (0 + 1)/(0 − 1) = −1. Point: (0, −1).
Marking: B1 for each intercept.

20. (a) Units for maximum profit. [2 marks]
Answer: dP/dx = 80 − 4x = 0 → x = 20.
d²P/dx² = −4 < 0 → maximum.
x = 20 hundred = 2000 units.
Marking: M1 for differentiating and solving, A1 for correct number of units.

(b) Maximum profit. [1 mark]
Answer: P = 80(20) − 2(20)² − 200 = 1600 − 800 − 200 = 600.
Maximum profit = $600 000.
Marking: B1 for correct profit.

(c) Range of x for profit. [2 marks]
Answer: Profit when P > 0: 80x − 2x² − 200 > 0 → −2x² + 80x − 200 > 0 → x² − 40x + 100 < 0.
Roots: x = [40 ± √(1600 − 400)]/2 = [40 ± √1200]/2 = [40 ± 20√3]/2 = 20 ± 10√3.
x ≈ 20 − 17.32 = 2.68 or x ≈ 20 + 17.32 = 37.32.
Since 0 ≤ x ≤ 30, profit when 2.68 < x ≤ 30, i.e., approximately 2.68 < x ≤ 30 hundred units.
Marking: M1 for setting up inequality and finding roots, A1 for correct range within domain.

END OF ANSWER KEY