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A Level Pure Mathematics 2 (June 2019)
Mark Scheme Legend
- M marks: Method marks are awarded for ‘knowing a method and attempting to apply it’.
- A marks: Accuracy marks can only be awarded if the relevant method (M) marks have been earned.
- B marks: Unconditional accuracy marks (independent of M marks).
- ft: Follow through.
- awrt: Answers which round to.
- cso: Correct solution only.
Table of Contents
- Question 1 (Exponentials)
- Question 2 (Trapezium Rule)
- Question 3 (Sector Area)
- Question 4 (Parametric Equations)
- Question 5 (Integration Limit)
- Question 6 (Functions)
- Question 7 (Linear Modeling)
- Question 8 (Sigma Notation)
- Question 9 (Logarithmic Graphs)
- Question 10 (Vectors)
- Question 11 (Numerical Methods)
- Question 12 (Trigonometry)
- Question 13 (Optimization)
- Question 14 (Diff Eqs)
Question 1 (3 marks)
Given
\[ 2^x \times 4^y = \frac{1}{2\sqrt{2}} \]express \( y \) as a function of \( x \).
Worked Solution
Step 1: Understanding the Goal
What are we being asked to do?
We need to rearrange the equation to isolate \( y \) on one side, giving an answer in the form \( y = f(x) \). To do this, we need to combine the terms. Since the bases (2, 4, \( \frac{1}{2\sqrt{2}} \)) are all powers of 2, the best strategy is to express everything as a power of 2.
Step 2: Expressing terms as powers of 2
Why do we do this?
If we have the same base, we can equate the indices (powers). Recall the laws of indices: \( (a^m)^n = a^{mn} \) and \( a^{-n} = \frac{1}{a^n} \).
✏ Working:
Rewrite \( 4^y \):
\[ 4^y = (2^2)^y = 2^{2y} \]Rewrite the right-hand side \( \frac{1}{2\sqrt{2}} \):
\[ 2\sqrt{2} = 2^1 \times 2^{1/2} = 2^{1 + 0.5} = 2^{1.5} \text{ or } 2^{3/2} \] \[ \frac{1}{2\sqrt{2}} = \frac{1}{2^{3/2}} = 2^{-3/2} \]✓ (B1) Correct equation in powers of 2.
Step 3: Forming the equation with indices
What is the next step?
Now substitute these back into the original equation and use the multiplication law \( a^m \times a^n = a^{m+n} \).
✏ Working:
\[ 2^x \times 2^{2y} = 2^{-3/2} \] \[ 2^{x + 2y} = 2^{-3/2} \]Step 4: Solving for y
How do we isolate y?
Since the bases are the same, the powers must be equal.
✏ Working:
\[ x + 2y = -\frac{3}{2} \]✓ (M1) Setting up linear equation in x and y.
Rearrange for \( y \):
\[ 2y = -x – \frac{3}{2} \] \[ y = -\frac{1}{2}x – \frac{3}{4} \]Final Answer:
\[ y = -\frac{1}{2}x – \frac{3}{4} \](Or equivalent, e.g., \( y = -0.5x – 0.75 \))
✓ Total: 3 marks
Question 2 (4 marks)
The speed of a small jet aircraft was measured every 5 seconds, starting from the time it turned onto a runway, until the time when it left the ground.
The results are given in the table below with the time in seconds and the speed in m s\(^{-1}\).
| Time (s) | 0 | 5 | 10 | 15 | 20 | 25 |
|---|---|---|---|---|---|---|
| Speed (m s\(^{-1}\)) | 2 | 5 | 10 | 18 | 28 | 42 |
Using all of this information,
(a) Estimate the length of runway used by the jet to take off. (3)
Given that the jet accelerated smoothly in these 25 seconds,
(b) Explain whether your answer to part (a) is an underestimate or an overestimate of the length of runway used by the jet to take off. (1)
Worked Solution
Part (a): Estimating Distance
What does the question ask?
Distance is the area under a Speed-Time graph. Since we have discrete data points, we use the Trapezium Rule to estimate this area.
The Formula:
Area \( \approx \frac{h}{2} [y_0 + y_n + 2(y_1 + y_2 + \dots + y_{n-1})] \)
Where \( h \) is the width of each interval (time step), and \( y \) values are the speeds.
✏ Working:
Identify values:
- \( h = 5 \) (difference between time values)
- First value \( y_0 = 2 \)
- Last value \( y_5 = 42 \)
- Middle values: 5, 10, 18, 28
Substitute into formula:
\[ \text{Distance} \approx \frac{5}{2} [2 + 42 + 2(5 + 10 + 18 + 28)] \]✓ (M1) Attempt at Trapezium Rule with \( h=5 \).
Calculator Steps:
- Sum of middle values: \( 5 + 10 + 18 + 28 = 61 \)
- Multiply by 2: \( 61 \times 2 = 122 \)
- Add ends: \( 122 + 2 + 42 = 166 \)
- Multiply by \( \frac{5}{2} = 2.5 \): \( 166 \times 2.5 = 415 \)
✓ (M1) Correct bracket structure.
\[ = 415 \text{ m} \]Part (b): Underestimate or Overestimate?
How do we determine this?
We need to consider the shape of the graph. The data points are (0,2), (5,5), (10,10), (15,18), (20,28), (25,42).
Let’s look at the differences (approximate gradient/acceleration):
- 5-2 = 3
- 10-5 = 5
- 18-10 = 8…
The gradient is increasing, meaning the curve is convex (concave up).
When drawing straight lines (trapeziums) between points on a curve that bends upwards, the straight lines sit above the curve.
✏ Working:
The data suggests the jet is accelerating, and the gradient is increasing (convex profile).
Therefore, the trapezia lie above the curve.
Answer: Overestimate.
✓ (B1ft) Overestimate with reason linked to convexity/shape.
Final Answer:
(a) 415 m
(b) Overestimate, because the curve is convex (trapezia lie above the curve).
✓ Total: 4 marks
Question 3 (3 marks)
Figure 1 shows a sector \( AOB \) of a circle with centre \( O \), radius 5 cm and angle \( AOB = 40^\circ \).
The attempt of a student to find the area of the sector is shown below.
(a) Explain the error made by this student. (1)
(b) Write out a correct solution. (2)
Worked Solution
Part (a): Identifying the Error
Why is the student wrong?
The formula \( A = \frac{1}{2}r^2\theta \) is derived specifically for radians. The student has substituted an angle in degrees directly into a radian formula.
Answer: The student used the angle in degrees, but the formula \( \frac{1}{2}r^2\theta \) requires the angle \( \theta \) to be in radians.
✓ (B1) Correct explanation referencing radians.
Part (b): Correct Calculation
Method 1: Convert to Radians
To convert degrees to radians, multiply by \( \frac{\pi}{180} \).
\[ 40^\circ = 40 \times \frac{\pi}{180} = \frac{2\pi}{9} \text{ radians} \]✏ Working:
\[ \text{Area} = \frac{1}{2}r^2\theta \] \[ \text{Area} = \frac{1}{2} \times 5^2 \times \frac{2\pi}{9} \]✓ (M1) Correct application of sector formula with radians.
\[ = \frac{1}{2} \times 25 \times \frac{2\pi}{9} \] \[ = \frac{25\pi}{9} \text{ cm}^2 \]Decimal approximation: \( \approx 8.73 \text{ cm}^2 \).
✓ (A1) Correct answer.
Method 2: Use Degree Formula
Alternatively, calculate as a fraction of the full circle area \( \pi r^2 \).
\[ \text{Area} = \frac{\theta}{360} \times \pi r^2 \]Final Answer:
(a) The formula requires radians, but the student used degrees.
(b) \( \frac{25\pi}{9} \text{ cm}^2 \) (or \( 8.73 \text{ cm}^2 \))
✓ Total: 3 marks
Question 4 (6 marks)
The curve \( C_1 \) with parametric equations
\[ x = 10\cos t, \quad y = 4\sqrt{2}\sin t, \quad 0 \le t < 2\pi \]meets the circle \( C_2 \) with equation
\[ x^2 + y^2 = 66 \]at four distinct points as shown in Figure 2.
Given that one of these points, \( S \), lies in the 4th quadrant, find the Cartesian coordinates of \( S \). (6)
Worked Solution
Step 1: Understanding the Problem
What is the strategy?
We need to find the coordinates where the curve \( C_1 \) and the circle \( C_2 \) intersect. The easiest way is to substitute the parametric expressions for \( x \) and \( y \) from \( C_1 \) directly into the equation for \( C_2 \). This will give us an equation in terms of \( t \).
Step 2: Substitution
✏ Working:
Substitute \( x = 10\cos t \) and \( y = 4\sqrt{2}\sin t \) into \( x^2 + y^2 = 66 \):
\[ (10\cos t)^2 + (4\sqrt{2}\sin t)^2 = 66 \]Expand the squares:
\[ 100\cos^2 t + (16 \times 2)\sin^2 t = 66 \] \[ 100\cos^2 t + 32\sin^2 t = 66 \]✓ (M1) Substituting parametric eqs into circle eq.
Step 3: Solving for t
How do we solve for t?
We have an equation with both \( \cos^2 t \) and \( \sin^2 t \). Use the identity \( \sin^2 t + \cos^2 t = 1 \) to substitute for one of them.
Let \( \cos^2 t = 1 – \sin^2 t \).
✏ Working:
\[ 100(1 – \sin^2 t) + 32\sin^2 t = 66 \] \[ 100 – 100\sin^2 t + 32\sin^2 t = 66 \] \[ 100 – 68\sin^2 t = 66 \]Rearrange:
\[ 34 = 68\sin^2 t \] \[ \sin^2 t = \frac{34}{68} = \frac{1}{2} \]✓ (M1) Using identity to get equation in single trig term.
✓ (dM1) Solving for \( \sin t \) or \( \cos t \).
Step 4: Finding the specific point S
Which quadrant?
We are told \( S \) is in the 4th quadrant. In the 4th quadrant:
- \( x \) is positive
- \( y \) is negative
Also, \( x = 10\cos t \) and \( y = 4\sqrt{2}\sin t \). For \( y \) to be negative, \( \sin t \) must be negative.
✏ Working:
From \( \sin^2 t = \frac{1}{2} \), we have \( \sin t = \pm \frac{1}{\sqrt{2}} \).
For 4th quadrant, \( \sin t = -\frac{1}{\sqrt{2}} \).
Now find \( \cos t \):
Since \( \sin^2 t = 0.5 \), then \( \cos^2 t = 0.5 \). Since \( x \) is positive in 4th quadrant, \( \cos t = +\frac{1}{\sqrt{2}} \).
Now calculate coordinates:
\[ x = 10\cos t = 10\left( \frac{1}{\sqrt{2}} \right) = \frac{10}{\sqrt{2}} = 5\sqrt{2} \] \[ y = 4\sqrt{2}\sin t = 4\sqrt{2}\left( -\frac{1}{\sqrt{2}} \right) = -4 \]✓ (M1) Substituting back to find coordinates.
Final Answer:
\[ S = (5\sqrt{2}, -4) \](Or decimal approx: \( (7.07, -4) \))
✓ Total: 6 marks
Question 5 (3 marks)
Figure 3 shows a sketch of the curve with equation \( y = \sqrt{x} \).
The point \( P(x, y) \) lies on the curve.
The rectangle, shown shaded on Figure 3, has height \( y \) and width \( \delta x \).
Calculate
\[ \lim_{\delta x \to 0} \sum_{x=4}^{9} \sqrt{x} \, \delta x \]Worked Solution
Step 1: Recognizing the Definition of Integration
What does this limit represent?
The expression \( \sum y \, \delta x \) represents the sum of the areas of rectangles under the curve. As the width of the rectangles (\( \delta x \)) tends to 0, this sum becomes the definite integral of the function between the specified limits.
So, we need to calculate:
\[ \int_{4}^{9} \sqrt{x} \, dx \]✓ (B1) Identifying the integral form.
Step 2: Evaluating the Integral
✏ Working:
First, write \( \sqrt{x} \) as a power:
\[ \sqrt{x} = x^{1/2} \]Now integrate using the power rule \( \int x^n \, dx = \frac{x^{n+1}}{n+1} \):
\[ \int_{4}^{9} x^{1/2} \, dx = \left[ \frac{x^{3/2}}{3/2} \right]_{4}^{9} \] \[ = \left[ \frac{2}{3}x^{3/2} \right]_{4}^{9} \]✓ (M1) Correct integration to \( x^{3/2} \).
Step 3: Substituting Limits
How do we calculate the value?
Substitute upper limit (9) and lower limit (4) and subtract.
Recall that \( x^{3/2} = (\sqrt{x})^3 \).
✏ Working:
\[ = \frac{2}{3}(9)^{3/2} – \frac{2}{3}(4)^{3/2} \]Calculate powers:
\[ 9^{3/2} = (\sqrt{9})^3 = 3^3 = 27 \] \[ 4^{3/2} = (\sqrt{4})^3 = 2^3 = 8 \]Substitute back:
\[ = \frac{2}{3}(27) – \frac{2}{3}(8) \] \[ = 18 – \frac{16}{3} \] \[ = \frac{54}{3} – \frac{16}{3} = \frac{38}{3} \]Final Answer:
\[ \frac{38}{3} \](or \( 12\frac{2}{3} \) or 12.67)
✓ Total: 3 marks
Question 6 (10 marks)
Figure 4 shows a sketch of the graph of \( y = g(x) \), where
\[ g(x) = \begin{cases} (x-2)^2 + 1 & x \le 2 \\ 4x – 7 & x > 2 \end{cases} \](a) Find the value of \( gg(0) \). (2)
(b) Find all values of \( x \) for which \( g(x) > 28 \). (4)
The function \( h \) is defined by \( h(x) = (x-2)^2 + 1 \), \( x \le 2 \).
(c) Explain why \( h \) has an inverse but \( g \) does not. (1)
(d) Solve the equation \( h^{-1}(x) = -\frac{1}{2} \). (3)
Worked Solution
Part (a): Composite Function
What is gg(0)?
This means \( g(g(0)) \). We first calculate \( g(0) \), and then put that result back into \( g \).
✏ Working:
Find \( g(0) \): Since \( 0 \le 2 \), use the first part of the function.
\[ g(0) = (0-2)^2 + 1 = (-2)^2 + 1 = 4 + 1 = 5 \]Now find \( g(5) \): Since \( 5 > 2 \), use the second part of the function.
\[ g(5) = 4(5) – 7 = 20 – 7 = 13 \]Answer: 13
✓ (M1) Attempt to find g(0) and subst into g. (A1) 13.
Part (b): Solving Inequality
Strategy:
We need to solve \( g(x) > 28 \) for both parts of the domain separately.
Case 1: \( x \le 2 \)
\[ (x-2)^2 + 1 > 28 \] \[ (x-2)^2 > 27 \]Square root both sides (remembering positive and negative roots):
\[ x-2 > \sqrt{27} \quad \text{or} \quad x-2 < -\sqrt{27} \] \[ x > 2 + 3\sqrt{3} \quad \text{or} \quad x < 2 - 3\sqrt{3} \]Since we must satisfy \( x \le 2 \), and \( 2+3\sqrt{3} > 2 \), we discard the positive root.
So, \( x < 2 - 3\sqrt{3} \).
Case 2: \( x > 2 \)
\[ 4x – 7 > 28 \] \[ 4x > 35 \] \[ x > \frac{35}{4} \]Check if this satisfies \( x > 2 \). Yes, \( 8.75 > 2 \).
Final Answer:
\[ x < 2 - 3\sqrt{3} \quad \text{or} \quad x > \frac{35}{4} \]✓ Total: 4 marks
Part (c): Existence of Inverse
What is the condition for an inverse?
A function has an inverse if and only if it is one-to-one.
Answer:
\( h(x) \) is defined for \( x \le 2 \), which is one side of the parabola vertex, making it one-to-one.
\( g(x) \) is many-to-one (e.g., horizontal line test cuts graph twice for values > 1), so it does not have an inverse.
✓ (B1) Explanation referencing one-to-one vs many-to-one.
Part (d): Inverse Equation
Understanding the Notation:
\( h^{-1}(x) = -\frac{1}{2} \) means that if we put \( -\frac{1}{2} \) into the function \( h \), we get \( x \).
So, \( x = h(-\frac{1}{2}) \).
✏ Working:
\[ x = h\left(-\frac{1}{2}\right) \]Substitute \( -\frac{1}{2} \) into \( h(x) = (x-2)^2 + 1 \):
\[ x = \left(-\frac{1}{2} – 2\right)^2 + 1 \] \[ x = \left(-\frac{5}{2}\right)^2 + 1 \] \[ x = \frac{25}{4} + 1 \] \[ x = \frac{29}{4} \text{ or } 7.25 \]✓ (A1) 7.25.
Question 7 (7 marks)
A small factory makes bars of soap.
On any day, the total cost to the factory, £\( y \), of making \( x \) bars of soap is modelled to be the sum of two separate elements:
- a fixed cost
- a cost that is proportional to the number of bars of soap that are made that day
(a) Write down a general equation linking \( y \) with \( x \), for this model. (1)
The bars of soap are sold for £2 each.
On a day when 800 bars of soap are made and sold, the factory makes a profit of £500.
On a day when 300 bars of soap are made and sold, the factory makes a loss of £80.
Using the above information,
(b) Show that \( y = 0.84x + 428 \). (3)
(c) With reference to the model, interpret the significance of the value 0.84 in the equation. (1)
Assuming that each bar of soap is sold on the day it is made,
(d) Find the least number of bars of soap that must be made on any given day for the factory to make a profit that day. (2)
Worked Solution
Part (a): The Equation
A “fixed cost” plus a “proportional cost” describes a straight line graph.
Answer: \( y = kx + c \) (or \( y = mx + c \)) where \( k, c \) are constants.
✓ (B1) Correct form.
Part (b): Finding Constants
Concept: Profit = Revenue – Cost
We are given information about profit/loss, but our equation is for Cost (\( y \)).
Revenue = Selling Price \(\times\) Quantity = \( 2x \).
Cost = \( y \).
Profit = \( 2x – y \).
✏ Working:
Scenario 1: 800 bars, Profit £500.
\[ 500 = 2(800) – y_1 \] \[ 500 = 1600 – y_1 \implies y_1 = 1100 \]So, when \( x = 800 \), \( y = 1100 \).
Scenario 2: 300 bars, Loss £80 (Profit = -80).
\[ -80 = 2(300) – y_2 \] \[ -80 = 600 – y_2 \implies y_2 = 680 \]So, when \( x = 300 \), \( y = 680 \).
Find gradient (k):
\[ k = \frac{1100 – 680}{800 – 300} = \frac{420}{500} = \frac{42}{50} = 0.84 \]Find intercept (c):
\[ y = 0.84x + c \] \[ 680 = 0.84(300) + c \] \[ 680 = 252 + c \] \[ c = 680 – 252 = 428 \]Equation: \( y = 0.84x + 428 \).
✓ (M1) Using profit formula to find y values. (dM1) Solving for constants. (A1) Correct proof.
Part (c): Interpretation
Answer: The cost of making each extra bar of soap (Marginal Cost).
✓ (B1) Marginal cost or cost per bar.
Part (d): Calculating Break-even
To make a profit, Revenue > Cost.
✏ Working:
\[ 2x > 0.84x + 428 \] \[ 1.16x > 428 \] \[ x > \frac{428}{1.16} \] \[ x > \frac{42800}{116} \] \[ x > 368.96… \]Since \( x \) must be an integer (bars of soap), the least number is 369.
Answer: 369 bars.
✓ (A1) 369.
Question 8 (6 marks)
(i) Find the value of
\[ \sum_{r=4}^{\infty} 20 \times \left(\frac{1}{2}\right)^r \](3)
(ii) Show that
\[ \sum_{n=1}^{48} \log_5 \left(\frac{n+2}{n+1}\right) = 2 \](3)
Worked Solution
Part (i): Geometric Series
Identifying the Series:
This is a geometric series with common ratio \( \frac{1}{2} \). Since \( |r| < 1 \), the sum to infinity exists.
Formula: \( S_\infty = \frac{a}{1-r} \).
Warning: The sum starts from \( r=4 \), not \( r=1 \) or \( r=0 \). We must find the first term \( a \) by substituting \( r=4 \).
✏ Working:
First term \( a \) (when \( r=4 \)):
\[ a = 20 \times \left(\frac{1}{2}\right)^4 = 20 \times \frac{1}{16} = \frac{20}{16} = \frac{5}{4} = 1.25 \]Common ratio \( r = \frac{1}{2} \).
\[ S_\infty = \frac{1.25}{1 – 0.5} = \frac{1.25}{0.5} = 2.5 \]Answer: 2.5
✓ (M1) Correct formula. (M1) Correct ‘a’. (A1) 2.5.
Part (ii): Method of Differences
Strategy:
Use log laws: \( \log(\frac{a}{b}) = \log a – \log b \).
Write out the first few terms to see the cancellation pattern.
✏ Working:
\[ \log_5 \left(\frac{n+2}{n+1}\right) = \log_5 (n+2) – \log_5 (n+1) \]Let’s sum from \( n=1 \) to \( 48 \):
- \( n=1: \log_5 3 – \log_5 2 \)
- \( n=2: \log_5 4 – \log_5 3 \)
- \( n=3: \log_5 5 – \log_5 4 \)
- …
- \( n=48: \log_5 50 – \log_5 49 \)
Notice the diagonal cancellation: \( \log_5 3 \) cancels with \( -\log_5 3 \), etc.
The only terms remaining are the very first negative term (\(-\log_5 2\)) and the very last positive term (\(\log_5 50\)).
\[ \text{Sum} = \log_5 50 – \log_5 2 \]Combine logs again:
\[ = \log_5 \left(\frac{50}{2}\right) = \log_5 25 \]Since \( 25 = 5^2 \):
\[ = 2 \]✓ (M1) Splitting log. (M1) Showing cancellation. (A1) Proof.
Question 9 (9 marks)
A research engineer is testing the effectiveness of the braking system of a car when it is driven in wet conditions.
The engineer measures and records the braking distance, \( d \) metres, when the brakes are applied from a speed of \( V \) km h\(^{-1}\).
Graphs of \( d \) against \( V \) and \( \log_{10} d \) against \( \log_{10} V \) were plotted.
The results are shown above together with a data point from each graph.
(a) Explain how Figure 6 would lead the engineer to believe that the braking distance should be modelled by the formula \( d = k V^n \) where \( k \) and \( n \) are constants, and show that \( k \approx 0.017 \). (3)
Using the information given in Figure 5, with \( k = 0.017 \),
(b) Find a complete equation for the model giving the value of \( n \) to 3 significant figures. (3)
Sean is driving this car at 60 km h\(^{-1}\) in wet conditions when he notices a large puddle in the road 100 m ahead. It takes him 0.8 seconds to react before applying the brakes.
(c) Use your formula to find out if Sean will be able to stop before reaching the puddle. (3)
Worked Solution
Part (a): The Linear Graph
Understanding Log-Log Plots:
Take logs of the model equation \( d = k V^n \):
\[ \log_{10} d = \log_{10}(k V^n) \] \[ \log_{10} d = \log_{10} k + n \log_{10} V \]This is in the form \( Y = mX + c \) where \( Y = \log d \), \( X = \log V \), \( m = n \), and \( c = \log k \).
Explanation: Because Figure 6 shows a straight line for \( \log d \) against \( \log V \), the relationship is of the form \( d = k V^n \).
Finding k: The y-intercept is given as -1.77.
\[ \text{Intercept } c = \log_{10} k = -1.77 \] \[ k = 10^{-1.77} \] \[ k \approx 0.01698… \approx 0.017 \]✓ (M1) Log equation. (A1) Comparison to line. (B1) Calculating k.
Part (b): Finding n
We are given a point from Figure 5: \( (V, d) = (30, 20) \).
Substitute this into \( d = 0.017 V^n \).
✏ Working:
\[ 20 = 0.017 \times 30^n \] \[ \frac{20}{0.017} = 30^n \] \[ 1176.47… = 30^n \]Take logs:
\[ \log(1176.47) = n \log(30) \] \[ n = \frac{\log(1176.47)}{\log(30)} \] \[ n \approx 2.079… \]Answer: \( d = 0.017 V^{2.08} \) (n to 3 s.f.)
✓ (M1) Substituting point. (M1) Solving for n. (A1) Equation.
Part (c): Total Stopping Distance
Two components to stopping:
1. Thinking Distance (distance travelled during reaction time).
2. Braking Distance (given by formula).
Unit Conversion: Speed is in km/h, time in seconds, distance in meters.
\[ 60 \text{ km/h} = \frac{60 \times 1000}{3600} \text{ m/s} = \frac{60000}{3600} = \frac{50}{3} \approx 16.67 \text{ m/s} \]Thinking Distance:
\[ \text{Distance} = \text{Speed} \times \text{Time} = 16.67 \times 0.8 = 13.33 \text{ m} \]Braking Distance:
Use formula with \( V = 60 \) (formula uses km/h):
\[ d = 0.017 \times 60^{2.08} \] \[ d \approx 0.017 \times 4987.6 \] \[ d \approx 84.79 \text{ m} \]Total Distance:
\[ 13.33 + 84.79 = 98.12 \text{ m} \]Conclusion:
The puddle is 100m away. Since \( 98.12 < 100 \), Sean stops before the puddle.
✓ (M1) Calculating d. (M1) Adding thinking distance. (A1) Correct conclusion.
Question 10 (6 marks)
Figure 7 shows a sketch of triangle \( OAB \).
The point \( C \) is such that \( \vec{OC} = 2\vec{OA} \).
The point \( M \) is the midpoint of \( AB \).
The straight line through \( C \) and \( M \) cuts \( OB \) at the point \( N \).
Given \( \vec{OA} = \mathbf{a} \) and \( \vec{OB} = \mathbf{b} \),
(a) Find \( \vec{CM} \) in terms of \( \mathbf{a} \) and \( \mathbf{b} \). (2)
(b) Show that \( \vec{ON} = \left(2 – \frac{3}{2}\lambda\right)\mathbf{a} + \frac{1}{2}\lambda\mathbf{b} \), where \( \lambda \) is a scalar constant. (2)
(c) Hence prove that \( ON:NB = 2:1 \). (2)
Worked Solution
Part (a): Finding Vector CM
Vector Path: \( \vec{CM} = \vec{CO} + \vec{OA} + \vec{AM} \).
We know:
- \( \vec{OC} = 2\mathbf{a} \implies \vec{CO} = -2\mathbf{a} \)
- \( \vec{OA} = \mathbf{a} \)
- \( M \) is midpoint of \( AB \), so \( \vec{AM} = \frac{1}{2}\vec{AB} \)
- \( \vec{AB} = \vec{AO} + \vec{OB} = -\mathbf{a} + \mathbf{b} \)
✏ Working:
\[ \vec{AM} = \frac{1}{2}(\mathbf{b} – \mathbf{a}) \] \[ \vec{CM} = \vec{CA} + \vec{AM} = (\vec{CO} + \vec{OA}) + \vec{AM} \]Alternatively directly from C:
\[ \vec{CM} = \vec{CO} + \vec{OM} \]Find \( \vec{OM} \):
\[ \vec{OM} = \vec{OA} + \vec{AM} = \mathbf{a} + \frac{1}{2}(\mathbf{b} – \mathbf{a}) = \frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b} \]Now find \( \vec{CM} \):
\[ \vec{CM} = \vec{CO} + \vec{OM} = -2\mathbf{a} + \left(\frac{1}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\right) \] \[ \vec{CM} = -\frac{3}{2}\mathbf{a} + \frac{1}{2}\mathbf{b} \]✓ (M1) Valid vector path. (A1) Correct expression.
Part (b): Vector Equation for ON
Collinearity: Since \( C, M, N \) are collinear, \( \vec{CN} = \lambda \vec{CM} \).
We can write \( \vec{ON} = \vec{OC} + \vec{CN} \).
✏ Working:
\[ \vec{ON} = \vec{OC} + \lambda \vec{CM} \] \[ \vec{ON} = 2\mathbf{a} + \lambda \left(-\frac{3}{2}\mathbf{a} + \frac{1}{2}\mathbf{b}\right) \]Group terms:
\[ \vec{ON} = \left(2 – \frac{3}{2}\lambda\right)\mathbf{a} + \frac{1}{2}\lambda\mathbf{b} \]✓ (M1) Using line equation. (A1) Shown.
Part (c): Finding the Ratio
Key Insight: \( N \) lies on the line \( OB \).
This means \( \vec{ON} \) must be parallel to \( \mathbf{b} \).
Therefore, the component of \( \mathbf{a} \) must be zero.
✏ Working:
Coefficient of \( \mathbf{a} = 0 \):
\[ 2 – \frac{3}{2}\lambda = 0 \] \[ 2 = \frac{3}{2}\lambda \implies \lambda = \frac{4}{3} \]Now find the coefficient of \( \mathbf{b} \) (which gives \( \vec{ON} \)):
\[ \vec{ON} = \frac{1}{2}\left(\frac{4}{3}\right)\mathbf{b} = \frac{2}{3}\mathbf{b} \]Interpretation:
\( \vec{ON} = \frac{2}{3}\vec{OB} \).
This means \( N \) is \( \frac{2}{3} \) of the way along \( OB \).
So \( ON = \frac{2}{3}OB \), which implies \( NB = \frac{1}{3}OB \).
Ratio \( ON:NB = \frac{2}{3} : \frac{1}{3} = 2:1 \).
✓ (M1) Setting a-component to 0. (A1) Proof.
Question 11 (11 marks)
Figure 8 shows a sketch of the curve \( C \) with equation \( y = x^x \), \( x > 0 \).
(a) Find, by firstly taking logarithms, the x coordinate of the turning point of \( C \). (5)
The point \( P(\alpha, 2) \) lies on \( C \).
(b) Show that \( 1.5 < \alpha < 1.6 \). (2)
A possible iteration formula that could be used in an attempt to find \( \alpha \) is
\[ x_{n+1} = 2x_n^{1-x_n} \]Using this formula with \( x_1 = 1.5 \),
(c) find \( x_4 \) to 3 decimal places, (2)
(d) describe the long-term behaviour of \( x_n \). (2)
Worked Solution
Part (a): Stationary Point
Why take logarithms?
We cannot differentiate \( x^x \) using standard rules like the power rule. Logarithmic differentiation helps bring the power down.
✏ Working:
\[ y = x^x \] \[ \ln y = \ln(x^x) = x \ln x \]Differentiate implicitly with respect to \( x \):
\[ \frac{d}{dx}(\ln y) = \frac{d}{dx}(x \ln x) \] \[ \frac{1}{y} \frac{dy}{dx} = 1 \cdot \ln x + x \cdot \frac{1}{x} \] \[ \frac{1}{y} \frac{dy}{dx} = \ln x + 1 \] \[ \frac{dy}{dx} = y(1 + \ln x) = x^x(1 + \ln x) \]For a turning point, \( \frac{dy}{dx} = 0 \):
\[ x^x(1 + \ln x) = 0 \]Since \( x^x \neq 0 \) for \( x > 0 \):
\[ 1 + \ln x = 0 \] \[ \ln x = -1 \] \[ x = e^{-1} = \frac{1}{e} \]✓ (M1) Taking logs. (M1) Implicit diff. (A1) Derivative. (M1) Setting to 0. (A1) Answer.
Part (b): Sign Change Method
What are we solving?
Point \( P(\alpha, 2) \) is on \( y = x^x \). So we solve \( x^x = 2 \).
Let \( f(x) = x^x – 2 \). Look for a sign change between 1.5 and 1.6.
✏ Working:
\[ f(1.5) = 1.5^{1.5} – 2 \approx 1.837 – 2 = -0.16… \] \[ f(1.6) = 1.6^{1.6} – 2 \approx 2.121 – 2 = +0.12… \]Conclusion:
There is a sign change (negative to positive) and the function is continuous, so there is a root \( \alpha \) in the interval \( (1.5, 1.6) \).
✓ (M1) Attempting values. (A1) Conclusion.
Part (c): Iteration
✏ Working:
\( x_{n+1} = 2x_n^{1-x_n} \)
\( x_1 = 1.5 \)
\( x_2 = 2(1.5)^{1-1.5} = 2(1.5)^{-0.5} \approx 1.63299… \)
\( x_3 = 2(1.63299…)^{1-1.63299…} \approx 1.46626… \)
\( x_4 = 2(1.46626…)^{1-1.46626…} \approx 1.67313… \)
Answer: 1.673 (3 d.p.)
✓ (M1) Attempting iteration. (A1) 1.673.
Part (d): Long-term Behaviour
Look at the sequence values: 1.5, 1.63, 1.47, 1.67, 1.41, 1.73…
It goes up, down, up, down, and the gap widens.
Answer: The sequence oscillates and diverges (does not converge). It does not approach a limit.
✓ (B1) Oscillates. (B1) Diverges/periodic behavior.
Question 12 (7 marks)
(a) Prove
\[ \frac{\cos 3\theta}{\sin \theta} + \frac{\sin 3\theta}{\cos \theta} \equiv 2 \cot 2\theta \](4)
(b) Hence solve, for \( 90^\circ < \theta < 180^\circ \), the equation
\[ \frac{\cos 3\theta}{\sin \theta} + \frac{\sin 3\theta}{\cos \theta} = 4 \]giving any solutions to one decimal place. (3)
Worked Solution
Part (a): Trigonometric Proof
Strategy: Combine the fractions on the LHS into a single fraction.
✏ Working:
\[ \text{LHS} = \frac{\cos 3\theta \cos \theta + \sin 3\theta \sin \theta}{\sin \theta \cos \theta} \]Numerator: Use the addition formula \( \cos(A-B) = \cos A \cos B + \sin A \sin B \).
\[ \cos 3\theta \cos \theta + \sin 3\theta \sin \theta = \cos(3\theta – \theta) = \cos 2\theta \]Denominator: Recognise \( \sin \theta \cos \theta = \frac{1}{2} \sin 2\theta \).
\[ \text{LHS} = \frac{\cos 2\theta}{\frac{1}{2} \sin 2\theta} \] \[ = 2 \frac{\cos 2\theta}{\sin 2\theta} \] \[ = 2 \cot 2\theta \quad \text{(RHS)} \]✓ (M1) Common denominator. (A1) Numerator identity. (dM1) Denominator identity. (A1) Completion.
Part (b): Solving the Equation
✏ Working:
Using part (a), the equation becomes:
\[ 2 \cot 2\theta = 4 \] \[ \cot 2\theta = 2 \] \[ \tan 2\theta = \frac{1}{2} \]Solve for \( 2\theta \):
Base angle: \( \tan^{-1}(0.5) \approx 26.565^\circ \).
Given range for \( \theta \) is \( 90^\circ < \theta < 180^\circ \).
So range for \( 2\theta \) is \( 180^\circ < 2\theta < 360^\circ \).
In this range (3rd quadrant), tan is positive.
\[ 2\theta = 180^\circ + 26.565^\circ = 206.565^\circ \] \[ \theta = \frac{206.565}{2} = 103.28…^\circ \]Answer: \( 103.3^\circ \)
✓ (M1) tan2theta = 0.5. (dM1) Finding angle in range. (A1) 103.3.
Question 13 (10 marks)
A manufacturer produces a storage tank.
The tank is modelled in the shape of a hollow circular cylinder closed at one end with a hemispherical shell at the other end as shown in Figure 9.
The walls of the tank are assumed to have negligible thickness.
The cylinder has radius \( r \) metres and height \( h \) metres and the hemisphere has radius \( r \) metres.
The volume of the tank is 6 m\(^3\).
(a) Show that, according to the model, the surface area of the tank, in m\(^2\), is given by
\[ A = \frac{12}{r} + \frac{5}{3}\pi r^2 \](4)
The manufacturer needs to minimise the surface area of the tank.
(b) Use calculus to find the radius of the tank for which the surface area is a minimum. (4)
(c) Calculate the minimum surface area of the tank, giving your answer to the nearest integer. (2)
Worked Solution
Part (a): Deriving the Area Formula
Breakdown of Shapes:
Volume = Volume of Cylinder + Volume of Hemisphere.
Surface Area = Area of Cylinder Base + Curved Area of Cylinder + Curved Area of Hemisphere.
1. Volume Constraint:
\[ V = \pi r^2 h + \frac{1}{2} \left( \frac{4}{3} \pi r^3 \right) = 6 \] \[ \pi r^2 h + \frac{2}{3} \pi r^3 = 6 \]Rearrange to find \( h \):
\[ \pi r^2 h = 6 – \frac{2}{3} \pi r^3 \] \[ h = \frac{6}{\pi r^2} – \frac{2}{3} r \]2. Surface Area Formula:
Base (circle) + Cylinder Curve (\( 2\pi r h \)) + Hemisphere Curve (\( 2\pi r^2 \)):
\[ A = \pi r^2 + 2\pi r h + 2\pi r^2 \] \[ A = 3\pi r^2 + 2\pi r h \]Substitute \( h \):
\[ A = 3\pi r^2 + 2\pi r \left( \frac{6}{\pi r^2} – \frac{2}{3} r \right) \] \[ A = 3\pi r^2 + \frac{12}{r} – \frac{4}{3} \pi r^2 \]Combine \( r^2 \) terms:
\[ A = \frac{12}{r} + \left( 3 – \frac{4}{3} \right) \pi r^2 \] \[ A = \frac{12}{r} + \frac{5}{3} \pi r^2 \quad \text{(Shown)} \]✓ (M1) Volume eq. (A1) Expression for h. (M1) Subst into Area. (A1) Proof.
Part (b): Minimization
Strategy: Find \( \frac{dA}{dr} \) and set to 0.
✏ Working:
\[ A = 12r^{-1} + \frac{5}{3} \pi r^2 \] \[ \frac{dA}{dr} = -12r^{-2} + \frac{10}{3} \pi r \]Set \( \frac{dA}{dr} = 0 \):
\[ \frac{10}{3} \pi r = \frac{12}{r^2} \] \[ 10 \pi r^3 = 36 \] \[ r^3 = \frac{36}{10\pi} = \frac{3.6}{\pi} \] \[ r = \sqrt[3]{\frac{3.6}{\pi}} \] \[ r \approx 1.046… \text{ m} \]✓ (M1) Differentiating. (A1) Correct derivative. (M1) Setting to 0. (A1) 1.05m.
Part (c): Minimum Area
✏ Working:
Substitute \( r \approx 1.046 \) back into \( A \):
\[ A = \frac{12}{1.046} + \frac{5}{3} \pi (1.046)^2 \] \[ A \approx 11.47 + 5.73 \] \[ A \approx 17.2 \]Answer: 17 m\(^2\).
✓ (M1) Substitution. (A1) 17.
Question 14 (15 marks)
(a) Use the substitution \( u = 4 – \sqrt{h} \) to show that
\[ \int \frac{dh}{4 – \sqrt{h}} = -8\ln|4 – \sqrt{h}| – 2\sqrt{h} + k \]where \( k \) is a constant. (6)
A team of scientists is studying a species of slow growing tree.
The rate of change in height of a tree in this species is modelled by the differential equation
\[ \frac{dh}{dt} = \frac{t^{0.25}(4 – \sqrt{h})}{20} \]where \( h \) is the height in metres and \( t \) is the time, measured in years, after the tree is planted.
(b) Find, according to the model, the range in heights of trees in this species. (2)
One of these trees is one metre high when it is first planted.
According to the model,
(c) Calculate the time this tree would take to reach a height of 12 metres, giving your answer to 3 significant figures. (7)
Worked Solution
Part (a): Integration by Substitution
Setup: \( u = 4 – \sqrt{h} = 4 – h^{1/2} \).
Find \( dh \) in terms of \( du \).
✏ Working:
\[ \frac{du}{dh} = -\frac{1}{2}h^{-1/2} = -\frac{1}{2\sqrt{h}} \] \[ dh = -2\sqrt{h} \, du \]Since \( u = 4 – \sqrt{h} \), we have \( \sqrt{h} = 4 – u \).
\[ dh = -2(4-u) \, du \]Substitute into Integral:
\[ \int \frac{-2(4-u)}{u} \, du \] \[ = \int \left( \frac{-8 + 2u}{u} \right) \, du \] \[ = \int \left( -\frac{8}{u} + 2 \right) \, du \] \[ = -8 \ln|u| + 2u + c \]Back Substitute:
\[ = -8 \ln|4 – \sqrt{h}| + 2(4 – \sqrt{h}) + c \] \[ = -8 \ln|4 – \sqrt{h}| + 8 – 2\sqrt{h} + c \]Combine constants (\( 8+c = k \)):
\[ = -8 \ln|4 – \sqrt{h}| – 2\sqrt{h} + k \]✓ (M1) Finding dh. (M1) Substituted integral. (A1) Integrated form. (A1) Final result.
Part (b): Max Height
Growth stops when \( \frac{dh}{dt} = 0 \).
✏ Working:
\[ 4 – \sqrt{h} = 0 \] \[ \sqrt{h} = 4 \implies h = 16 \]Since growth rate is positive for \( \sqrt{h} < 4 \) (i.e., \( h < 16 \)), the tree grows towards 16m.
Answer: \( 0 < h \le 16 \).
✓ (M1) Setting rate to 0. (A1) Correct range.
Part (c): Solving Differential Equation
Separate Variables: Group \( h \) terms on LHS, \( t \) terms on RHS.
✏ Working:
\[ \int \frac{1}{4-\sqrt{h}} \, dh = \int \frac{t^{0.25}}{20} \, dt \]Use result from (a):
\[ -8 \ln|4 – \sqrt{h}| – 2\sqrt{h} = \frac{1}{20} \frac{t^{1.25}}{1.25} + c \] \[ -8 \ln|4 – \sqrt{h}| – 2\sqrt{h} = \frac{1}{25} t^{1.25} + c \]Initial Conditions: \( t=0, h=1 \).
\[ -8 \ln(4-1) – 2(1) = 0 + c \] \[ c = -8 \ln 3 – 2 \]Find t when h=12:
\[ -8 \ln(4-\sqrt{12}) – 2\sqrt{12} = \frac{1}{25} t^{1.25} + (-8 \ln 3 – 2) \]Left side value:
\[ 4-\sqrt{12} \approx 0.5359 \] \[ -8 \ln(0.5359) – 2(3.464) \approx 4.99 – 6.93 = -1.94 \]Right side constant \( c \approx -8(1.099) – 2 = -10.79 \).
\[ -1.94 = 0.04 t^{1.25} – 10.79 \] \[ 8.85 = 0.04 t^{1.25} \] \[ t^{1.25} = 221.25 \] \[ t = (221.25)^{1/1.25} = (221.25)^{0.8} \] \[ t \approx 75.2 \]Answer: 75.2 years.
✓ (M1) Separation. (M1) Integration. (M1) Finding c. (A1) Answer.