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A Level Summer 2022 Edexcel Pure Mathematics 2
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📖 Table of Contents
- Question 1 (Modulus Graphs & Equations)
- Question 2 (Exponentials & Logarithms)
- Question 3 (Periodic Sequences & Summation)
- Question 4 (Differentiation from First Principles)
- Question 5 (Trapezium Rule & Integration)
- Question 6 (Trigonometry & Newton-Raphson)
- Question 7 (Binomial Expansion)
- Question 8 (Area under Curve & Integration)
- Question 9 (Trigonometric Modelling)
- Question 10 (Functions: Inverse & Range)
- Question 11 (Algebraic Proof)
- Question 12 (Quotient Rule Differentiation)
- Question 13 (Vectors)
- Question 14 (Differential Equations & Partial Fractions)
- Question 15 (Geometric Series & Trigonometry)
- Question 16 (Parametric Equations & Normals)
Question 1 (4 marks)
Figure 1 shows a sketch of the graph with equation \( y = |3 – 2x| \).
Solve
\[ |3 – 2x| = 7 + x \]Worked Solution
Step 1: Understanding Modulus Equations
What does the modulus mean?
The equation \( |A| = B \) implies that either \( A = B \) or \( -A = B \) (equivalently \( A = -B \)).
We need to solve two separate linear equations and then check if the solutions are valid.
Step 2: Solve the positive case
Case 1: The expression inside the modulus is positive.
\[ 3 – 2x = 7 + x \]Subtract \(x\) from both sides:
\[ 3 – 3x = 7 \]Subtract 3 from both sides:
\[ -3x = 4 \] \[ x = -\frac{4}{3} \]Step 3: Solve the negative case
Case 2: The expression inside the modulus is negative (reflected part).
\[ -(3 – 2x) = 7 + x \] \[ -3 + 2x = 7 + x \]Subtract \(x\) from both sides:
\[ -3 + x = 7 \]Add 3 to both sides:
\[ x = 10 \]Step 4: Check validity
Why check? Sometimes solving modulus equations algebraically produces “extraneous” solutions, especially when the RHS contains \(x\). The RHS (\(7+x\)) must be non-negative because a modulus value is always \(\ge 0\).
Check \( x = -\frac{4}{3} \):
\[ \text{LHS} = |3 – 2(-\frac{4}{3})| = |3 + \frac{8}{3}| = |\frac{17}{3}| = \frac{17}{3} \] \[ \text{RHS} = 7 + (-\frac{4}{3}) = \frac{21}{3} – \frac{4}{3} = \frac{17}{3} \]Valid.
Check \( x = 10 \):
\[ \text{LHS} = |3 – 2(10)| = |3 – 20| = |-17| = 17 \] \[ \text{RHS} = 7 + 10 = 17 \]Valid.
Final Answer:
\[ x = -\frac{4}{3} \quad \text{or} \quad x = 10 \]✓ (4 marks)
Question 2 (4 marks)
(a) Sketch the curve with equation \( y = 4^x \) stating any points of intersection with the coordinate axes.
(b) Solve \( 4^x = 100 \) giving your answer to 2 decimal places.
Worked Solution
Part (a): Sketching the Exponential Graph
Key features of \( y = a^x \) where \( a > 1 \):
- The curve is always positive (\( y > 0 \)).
- The \(x\)-axis is an asymptote as \( x \to -\infty \).
- It increases rapidly as \( x \) increases.
- The \(y\)-intercept occurs when \( x=0 \).
When \( x=0 \), \( y = 4^0 = 1 \). So it crosses the y-axis at \((0, 1)\).
It does not cross the x-axis.
Part (b): Solving the Equation
Take logs of both sides (you can use base 10, base \(e\), or base 4):
\[ \log(4^x) = \log(100) \]Use the power law \( \log(a^b) = b \log a \):
\[ x \log 4 = \log 100 \] \[ x = \frac{\log 100}{\log 4} \] \[ x = \frac{2}{0.602…} \] \[ x \approx 3.3219… \]Final Answer:
(a) Sketch (see diagram above) with intercept at \((0, 1)\).
(b) \( x = 3.32 \) (2 d.p.)
✓ (4 marks total)
Question 3 (4 marks)
A sequence of terms \( a_1, a_2, a_3, \dots \) is defined by
\[ a_1 = 3 \] \[ a_{n+1} = 8 – a_n \](a) (i) Show that this sequence is periodic.
(ii) State the order of this periodic sequence.
(b) Find the value of
\[ \sum_{n=1}^{85} a_n \]Worked Solution
Part (a): Investigating the Sequence
Let’s calculate the first few terms to see the pattern.
Given \( a_1 = 3 \):
\[ a_2 = 8 – a_1 = 8 – 3 = 5 \] \[ a_3 = 8 – a_2 = 8 – 5 = 3 \] \[ a_4 = 8 – a_3 = 8 – 3 = 5 \]The terms are \( 3, 5, 3, 5, \dots \)
Since the terms repeat (\( a_3 = a_1 \)), the sequence is periodic.
The pattern repeats every 2 terms, so the order is 2.
Part (b): Finding the Sum
Strategy: We need to sum the first 85 terms. Since the sequence repeats every 2 terms (3, 5), we can group them into pairs.
Sum of one pair \( (3 + 5) = 8 \).
How many full pairs in 85 terms?
\[ 85 \div 2 = 42 \text{ pairs remainder } 1 \]So the sum consists of 42 pairs plus the 85th term.
The 85th term is odd-numbered. All odd-numbered terms are 3.
\[ \text{Total Sum} = (42 \times 8) + 3 \] \[ = 336 + 3 \] \[ = 339 \]Final Answer:
(a) (i) Terms are \(3, 5, 3, \dots\) repeating. (ii) Order 2.
(b) 339
✓ (4 marks total)
Question 4 (3 marks)
Given that
\[ y = 2x^2 \]use differentiation from first principles to show that
\[ \frac{dy}{dx} = 4x \]Worked Solution
Step 1: The Definition of the Derivative
Why do this? The question explicitly asks for “first principles,” so we cannot use the standard power rule. We must use the limit definition.
Here, \( f(x) = 2x^2 \).
Step 2: Substitute and Expand
Now substitute into the formula:
\[ \frac{dy}{dx} = \lim_{h \to 0} \frac{(2x^2 + 4xh + 2h^2) – 2x^2}{h} \]Step 3: Simplify and take the limit
Cancel the \(2x^2\) terms:
\[ \frac{dy}{dx} = \lim_{h \to 0} \frac{4xh + 2h^2}{h} \]Divide by \(h\) (since \(h \neq 0\)):
\[ \frac{dy}{dx} = \lim_{h \to 0} (4x + 2h) \]As \( h \to 0 \), the term \( 2h \to 0 \):
\[ \frac{dy}{dx} = 4x \]Final Answer:
Shown: \( \frac{dy}{dx} = 4x \)
✓ (3 marks)
Question 5 (6 marks)
The table below shows corresponding values of \(x\) and \(y\) for \( y = \log_3 2x \).
The values of \(y\) are given to 2 decimal places as appropriate.
| \(x\) | 3 | 4.5 | 6 | 7.5 | 9 |
| \(y\) | 1.63 | 2 | 2.26 | 2.46 | 2.63 |
(a) Using the trapezium rule with all the values of \(y\) in the table, find an estimate for
\[ \int_3^9 \log_3 2x \, dx \]Using your answer to part (a) and making your method clear, estimate
(b) (i) \( \int_3^9 \log_3 (2x)^{10} \, dx \)
(ii) \( \int_3^9 \log_3 18x \, dx \)
Worked Solution
Part (a): The Trapezium Rule
Formula: \( \text{Area} \approx \frac{h}{2} [ y_0 + 2(y_1 + … + y_{n-1}) + y_n ] \)
Where \( h \) is the strip width: \( h = 4.5 – 3 = 1.5 \).
The first and last \(y\)-values are used once. The middle values are multiplied by 2.
\[ \text{Area} \approx \frac{1.5}{2} [ 1.63 + 2(2 + 2.26 + 2.46) + 2.63 ] \] \[ \approx 0.75 [ 1.63 + 2(6.72) + 2.63 ] \] \[ \approx 0.75 [ 1.63 + 13.44 + 2.63 ] \] \[ \approx 0.75 [ 17.7 ] \] \[ \approx 13.275 \]Rounding to a reasonable degree of accuracy (e.g., 3 s.f. or consistent with the table’s 2 d.p.): 13.3
Part (b)(i): Log Laws
We need to relate \( \log_3 (2x)^{10} \) to the original function \( \log_3 2x \).
Using the power law: \( \log_a (b^n) = n \log_a b \).
Since 10 is a constant, we can pull it out:
\[ = 10 \int_3^9 \log_3 2x \, dx \]Using the answer from (a):
\[ = 10 \times 13.275 = 132.75 \]Estimate: 133
Part (b)(ii): Log Multiplication Laws
We need to relate \( \log_3 18x \) to \( \log_3 2x \).
Notice that \( 18x = 9 \times 2x \).
Using the multiplication law: \( \log_a (AB) = \log_a A + \log_a B \).
We know that \( \log_3 9 = 2 \) (because \( 3^2 = 9 \)).
So the integral becomes:
\[ \int_3^9 (2 + \log_3 2x) \, dx \] \[ = \int_3^9 2 \, dx + \int_3^9 \log_3 2x \, dx \]The first part is the area of a rectangle of height 2 and width \( 9-3=6 \):
\[ \int_3^9 2 \, dx = [2x]_3^9 = 18 – 6 = 12 \]Adding the answer from (a):
\[ = 12 + 13.275 = 25.275 \]Estimate: 25.3
Final Answer:
(a) 13.3 (or 13.275)
(b)(i) 133
(b)(ii) 25.3
✓ (6 marks total)
Question 6 (7 marks)
Figure 2 shows a sketch of part of the curve with equation \( y = f(x) \) where
\[ f(x) = 8\sin\left(\frac{1}{2}x\right) – 3x + 9 \quad x > 0 \]and \( x \) is measured in radians.
The point \( P \), shown in Figure 2, is a local maximum point on the curve.
(a) Find the \( x \) coordinate of \( P \), giving your answer to 3 significant figures.
The curve crosses the \( x \)-axis at \( x = \alpha \), as shown in Figure 2.
Given that, to 3 decimal places, \( f(4) = 4.274 \) and \( f(5) = -1.212 \)
(b) Explain why \( \alpha \) must lie in the interval \( [4, 5] \).
(c) Taking \( x_0 = 5 \) as a first approximation to \( \alpha \), apply the Newton-Raphson method once to \( f(x) \) to obtain a second approximation to \( \alpha \).
Show your method and give your answer to 3 significant figures.
Worked Solution
Part (a): Finding the Maximum Point
Strategy: Local maximum points occur where the gradient is zero. We need to find \( f'(x) \) and set it to 0.
Differentiate with respect to \( x \):
\[ f'(x) = 8 \times \frac{1}{2}\cos\left(\frac{1}{2}x\right) – 3 \] \[ f'(x) = 4\cos\left(\frac{1}{2}x\right) – 3 \]At the maximum \( P \), \( f'(x) = 0 \):
\[ 4\cos\left(\frac{1}{2}x\right) – 3 = 0 \] \[ \cos\left(\frac{1}{2}x\right) = \frac{3}{4} \]Solve for \( x \):
\[ \frac{1}{2}x = \arccos(0.75) \]Make sure your calculator is in radians.
\[ \frac{1}{2}x = 0.7227… \] \[ x = 1.445… \]Rounding to 3 s.f.:
\[ x = 1.45 \]Part (b): Sign Change Rule
If a continuous function changes sign between two values, there must be a root between them.
\( f(4) = 4.274 \) (positive)
\( f(5) = -1.212 \) (negative)
The function \( f(x) \) is continuous. Since there is a change of sign between \( x=4 \) and \( x=5 \), there must be a root \( \alpha \) in the interval \( [4, 5] \).
Part (c): Newton-Raphson Method
Formula: \( x_{n+1} = x_n – \frac{f(x_n)}{f'(x_n)} \)
We are given \( x_0 = 5 \). We need to calculate \( f(5) \) and \( f'(5) \).
From the question: \( f(5) = -1.212 \) (using more precision is better if possible).
\[ f(5) = 8\sin(2.5) – 15 + 9 = 8\sin(2.5) – 6 = -1.2123… \]Calculate \( f'(5) \):
\[ f'(5) = 4\cos(2.5) – 3 \] \[ f'(5) = 4(-0.8011…) – 3 = -3.2044… – 3 = -6.2045… \]Apply the formula:
\[ x_1 = 5 – \frac{-1.2123…}{-6.2045…} \] \[ x_1 = 5 – (0.195…) \] \[ x_1 = 4.804… \]Rounding to 3 s.f.:
\[ x_1 = 4.80 \]Final Answer:
(a) \( x = 1.45 \)
(b) Change of sign + continuous function implies root.
(c) \( 4.80 \)
✓ (7 marks total)
Question 7 (5 marks)
(a) Find the first four terms, in ascending powers of \( x \), of the binomial expansion of
\[ \sqrt{4 – 9x} \]writing each term in simplest form.
A student uses this expansion with \( x = \frac{1}{9} \) to find an approximation for \( \sqrt{3} \).
Using the answer to part (a) and without doing any calculations,
(b) state whether this approximation will be an overestimate or an underestimate of \( \sqrt{3} \), giving a brief reason for your answer.
Worked Solution
Part (a): Binomial Expansion
Strategy: The general binomial expansion formula for \( (1+x)^n \) requires the first term inside the bracket to be 1. Here we have \( \sqrt{4-9x} = (4-9x)^{1/2} \).
We must factor out the 4 first.
Now use the expansion formula \( (1+X)^n = 1 + nX + \frac{n(n-1)}{2!}X^2 + \frac{n(n-1)(n-2)}{3!}X^3 + \dots \) with \( X = -\frac{9x}{4} \) and \( n = \frac{1}{2} \).
Term 1: \( 1 \)
Term 2: \( nX = \frac{1}{2} \left( -\frac{9x}{4} \right) = -\frac{9}{8}x \)
Term 3: \( \frac{n(n-1)}{2!}X^2 = \frac{\frac{1}{2}(-\frac{1}{2})}{2} \left( -\frac{9x}{4} \right)^2 \)
\[ = -\frac{1}{8} \left( \frac{81x^2}{16} \right) = -\frac{81}{128}x^2 \]Term 4: \( \frac{n(n-1)(n-2)}{3!}X^3 = \frac{\frac{1}{2}(-\frac{1}{2})(-\frac{3}{2})}{6} \left( -\frac{9x}{4} \right)^3 \)
\[ = \frac{\frac{3}{8}}{6} \left( -\frac{729x^3}{64} \right) = \frac{1}{16} \left( -\frac{729}{64}x^3 \right) = -\frac{729}{1024}x^3 \]Combine and multiply by the outer factor of 2:
\[ 2 \left[ 1 – \frac{9}{8}x – \frac{81}{128}x^2 – \frac{729}{1024}x^3 \right] \] \[ = 2 – \frac{9}{4}x – \frac{81}{64}x^2 – \frac{729}{512}x^3 \]Part (b): Estimation Error
We need to determine if the approximation is over or under the true value.
Look at the signs of the terms in the expansion.
The expansion is \( 2 – (\text{positive term}) – (\text{positive term}) – (\text{positive term}) \dots \)
All terms after the first are negative (since \( x = 1/9 > 0 \)).
Because we are truncating the series (stopping after 4 terms), we are ignoring all the subsequent terms, which are all negative.
Therefore, we are failing to subtract these further negative quantities.
Hence, our approximation is larger than the true value.
Final Answer:
(a) \( 2 – \frac{9}{4}x – \frac{81}{64}x^2 – \frac{729}{512}x^3 \)
(b) Overestimate, because all terms in the expansion after the first are negative.
✓ (5 marks total)
Question 8 (6 marks)
Figure 3 shows a sketch of part of a curve with equation
\[ y = \frac{(x-2)(x-4)}{4\sqrt{x}} \quad x > 0 \]The region \( R \), shown shaded in Figure 3, is bounded by the curve and the \( x \)-axis.
Find the exact area of \( R \), writing your answer in the form \( a\sqrt{2} + b \), where \( a \) and \( b \) are constants to be found.
Worked Solution
Step 1: Simplify the Function
Strategy: To integrate, we need the expression in the form of powers of \( x \). We must expand the numerator and divide by the denominator.
Divide each term by \( 4x^{1/2} \):
\[ y = \frac{x^2}{4x^{1/2}} – \frac{6x}{4x^{1/2}} + \frac{8}{4x^{1/2}} \] \[ y = \frac{1}{4}x^{3/2} – \frac{3}{2}x^{1/2} + 2x^{-1/2} \]Step 2: Set up the Integral
The region \( R \) lies between the roots \( x=2 \) and \( x=4 \). Note that in this region, \( y \) is negative (below the axis). The area is the magnitude of the integral.
Step 3: Integrate
Integrate term by term using \( \int x^n dx = \frac{x^{n+1}}{n+1} \):
\[ \int \frac{1}{4}x^{3/2} \, dx = \frac{1}{4} \cdot \frac{2}{5} x^{5/2} = \frac{1}{10}x^{5/2} \] \[ \int -\frac{3}{2}x^{1/2} \, dx = -\frac{3}{2} \cdot \frac{2}{3} x^{3/2} = -x^{3/2} \] \[ \int 2x^{-1/2} \, dx = 2 \cdot \frac{2}{1} x^{1/2} = 4x^{1/2} \]So,
\[ I = \left[ \frac{1}{10}x^{5/2} – x^{3/2} + 4x^{1/2} \right]_2^4 \]Step 4: Evaluate Limits
Upper limit (4):
\[ 4^{1/2} = 2 \] \[ \frac{1}{10}(2)^5 – (2)^3 + 4(2) = \frac{32}{10} – 8 + 8 = 3.2 \]Lower limit (2):
\[ 2^{1/2} = \sqrt{2} \] \[ \frac{1}{10}(\sqrt{2})^5 – (\sqrt{2})^3 + 4(\sqrt{2}) \] \[ (\sqrt{2})^5 = 4\sqrt{2}, \quad (\sqrt{2})^3 = 2\sqrt{2} \] \[ \frac{4\sqrt{2}}{10} – 2\sqrt{2} + 4\sqrt{2} \] \[ = 0.4\sqrt{2} + 2\sqrt{2} = 2.4\sqrt{2} = \frac{12}{5}\sqrt{2} \]Subtract:
\[ \text{Integral} = 3.2 – 2.4\sqrt{2} = \frac{16}{5} – \frac{12}{5}\sqrt{2} \]Since the area is positive and the curve is below the axis, the raw integral might be negative (Wait, let’s check values).
\( \sqrt{2} \approx 1.41 \), so \( 2.4 \times 1.41 \approx 3.38 \). \( 3.2 – 3.38 \) is negative. Correct.
Area = Absolute value = \( \frac{12}{5}\sqrt{2} – \frac{16}{5} \).
Final Answer:
\[ \text{Area} = \frac{12}{5}\sqrt{2} – \frac{16}{5} \](Also acceptable: \( 2.4\sqrt{2} – 3.2 \))
✓ (6 marks)
Question 9 (5 marks)
Figure 4 shows a sketch of a Ferris wheel.
The height above the ground, \( H \) m, of a passenger on the Ferris wheel, \( t \) seconds after the wheel starts turning, is modelled by the equation
\[ H = |A\sin(bt + \alpha)^\circ| \]where \( A \), \( b \), and \( \alpha \) are constants.
Figure 5 shows a sketch of the graph of \( H \) against \( t \), for one revolution of the wheel.
Given that
- the maximum height of the passenger above the ground is 50 m
- the passenger is 1 m above the ground when the wheel starts turning
- the wheel takes 720 seconds to complete one revolution
(a) Find a complete equation for the model, giving the exact value of \( A \), the exact value of \( b \) and the value of \( \alpha \) to 3 significant figures.
(b) Explain why an equation of the form \( H = |A\sin(bt + \alpha)^\circ| + d \) where \( d \) is a positive constant, would be a more appropriate model.
Worked Solution
Part (a): Finding Constants
Finding A: The maximum value of \( |\sin(\dots)| \) is 1. Therefore the maximum height \( H_{\max} = |A| \times 1 = |A| \).
We are given max height = 50m.
(Usually we take \( A=50 \)).
Finding b: The function \( |\sin(x)| \) has a period of 180° (half the period of \( \sin(x) \)). Wait, careful. The question says “Figure 5 shows… for one revolution”. The graph shows a single “hump”.
If the wheel makes one revolution in 720s, the argument of the sine function must change by 360° (or \( 2\pi \)) in 720s.
However, the model uses a modulus. A full sine wave goes up and down. A modulus sine wave goes up, then up again.
Let’s look at the shape. It starts low (1m), goes to 50m, and comes back down. That looks like half a sine wave period, or a full “modulus” period? No, one revolution implies the passenger goes top to bottom. In a circle, you go from bottom -> top -> bottom.
In standard sine, that is -1 to 1 to -1. In modulus, that is 1 to 0 to 1? No.
Usually height is modeled as \( H = M + R\sin(bt) \). Here we have modulus.
Let’s assume the argument \( (bt + \alpha) \) spans 180 degrees for the single hump shown? Or is the “one revolution” graph shown just part of the journey?
Re-reading: “Figure 5 shows… for one revolution”. The graph shows one single arch.
Usually, \( \sin \) goes 0 -> 1 -> 0 -> -1 -> 0. The modulus goes 0 -> 1 -> 0 -> 1 -> 0. That would be two arches per 360 degrees.
If Figure 5 is ONE revolution, and it’s ONE arch, then the sine function argument only spans 180 degrees in 720 seconds?
Wait, if the wheel rotates 360 degrees, the passenger goes up and down once. Height goes Min -> Max -> Min.
If the model is \( |50 \sin(\dots)| \), and it only produces one arch, then the argument must span 180 degrees. \( b(720) = 180 \). So \( b = 0.25 \).
Let’s check. If \( b=0.25 \), then in 720s, angle changes by 180. \( \sin \) goes from e.g. 0 to 1 to 0. Height goes 0 to 50 to 0.
Period of revolution = 720s.
Degrees turned = 360°? No, the equation model doesn’t map directly to physical rotation angle. We fit the graph.
The graph shows one complete “bump” for one revolution. This corresponds to the sine function completing half a cycle (e.g. 0 to 180).
\[ b \times 720 = 180 \] \[ b = \frac{180}{720} = \frac{1}{4} = 0.25 \]Finding \( \alpha \): Use the initial condition \( t=0, H=1 \).
Figure 5 shows the graph increasing from \( t=0 \). If \( \alpha \) is small positive, \( \sin \) increases. If \( \alpha \) is e.g. 179°, sine decreases.
\[ \alpha = \arcsin(1/50) \] \[ \alpha \approx 1.1457… \]Rounding to 3 s.f.: \( \alpha = 1.15 \)
Part (b): Validity of Model
Consider the minimum height. In the current model \( H = | \dots | \), the minimum value is 0.
The original model allows the passenger to touch the ground (H=0). In reality, a Ferris wheel capsule is suspended above the ground to allow loading/unloading.
Adding \( d \) lifts the minimum height to \( d \) meters, preventing the passenger from hitting the ground.
Final Answer:
(a) \( H = |50 \sin(0.25t + 1.15)^\circ| \)
(b) To ensure the minimum height is greater than 0 (so the capsule doesn’t hit the ground).
✓ (5 marks total)
Question 10 (8 marks)
The function \( f \) is defined by
\[ f(x) = \frac{8x + 5}{2x + 3} \quad x > -\frac{3}{2} \](a) Find \( f^{-1}\left(\frac{3}{2}\right) \).
(b) Show that
\[ f(x) = A + \frac{B}{2x + 3} \]where \( A \) and \( B \) are constants to be found.
The function \( g \) is defined by
\[ g(x) = 16 – x^2 \quad 0 \le x \le 4 \](c) State the range of \( g^{-1} \).
(d) Find the range of \( fg^{-1} \).
Worked Solution
Part (a): Inverse Function Value
Shortcut: Instead of finding the algebraic inverse \( f^{-1}(x) \) first, we can set \( f(x) = \frac{3}{2} \) and solve for \( x \).
Cross multiply:
\[ 2(8x + 5) = 3(2x + 3) \] \[ 16x + 10 = 6x + 9 \] \[ 10x = -1 \] \[ x = -0.1 \]Part (b): Algebraic Division
We need to divide \( 8x+5 \) by \( 2x+3 \).
\( \frac{8x+5}{2x+3} = \frac{4(2x+3) – 12 + 5}{2x+3} \)
\[ = \frac{4(2x+3) – 7}{2x+3} \] \[ = 4 – \frac{7}{2x+3} \]So \( A = 4 \) and \( B = -7 \).
Part (c): Range of Inverse
The range of the inverse function \( g^{-1} \) is equal to the domain of the original function \( g \).
Domain of \( g \) is \( 0 \le x \le 4 \).
Range of \( g^{-1} \) is \( 0 \le y \le 4 \).
Part (d): Range of Composite Function
We need the range of \( f(g^{-1}(x)) \).
1. Find the domain of \( f \) for this composite: This is the range of \( g^{-1} \), which is \( [0, 4] \).
2. Find the range of \( f(x) \) for the domain \( 0 \le x \le 4 \).
Using the form \( f(x) = 4 – \frac{7}{2x+3} \).
This is an increasing function (as \( x \) increases, \( 2x+3 \) increases, fraction decreases, subtracting a smaller amount means total increases).
Evaluate at endpoints:
At \( x=0 \):
\[ f(0) = \frac{5}{3} \]At \( x=4 \):
\[ f(4) = \frac{8(4)+5}{2(4)+3} = \frac{37}{11} \]So the range is \( \frac{5}{3} \le y \le \frac{37}{11} \).
Final Answer:
(a) \( -0.1 \)
(b) \( A=4, B=-7 \)
(c) \( 0 \le y \le 4 \)
(d) \( \frac{5}{3} \le y \le \frac{37}{11} \)
✓ (8 marks total)
Question 11 (4 marks)
Prove, using algebra, that
\[ n(n^2 + 5) \]is even for all \( n \in \mathbb{N} \).
Worked Solution
Step 1: Consider cases for n
Strategy: Any integer \( n \) is either even or odd. We can prove the statement by showing it holds true for both cases.
Step 2: Case 1 – n is even
Let \( n = 2k \) where \( k \) is an integer.
\[ n(n^2 + 5) = 2k((2k)^2 + 5) \] \[ = 2k(4k^2 + 5) \] \[ = 2[k(4k^2 + 5)] \]This is of the form \( 2 \times \text{integer} \), so it is even.
Step 3: Case 2 – n is odd
Let \( n = 2k + 1 \) where \( k \) is an integer.
\[ n(n^2 + 5) = (2k+1)((2k+1)^2 + 5) \] \[ = (2k+1)(4k^2 + 4k + 1 + 5) \] \[ = (2k+1)(4k^2 + 4k + 6) \]Factor out 2 from the second bracket:
\[ = (2k+1) \times 2(2k^2 + 2k + 3) \] \[ = 2[(2k+1)(2k^2 + 2k + 3)] \]This is of the form \( 2 \times \text{integer} \), so it is even.
Final Answer:
Since the expression is even when \( n \) is even and when \( n \) is odd, it is even for all \( n \in \mathbb{N} \).
✓ (4 marks)
Question 12 (6 marks)
The function \( f \) is defined by
\[ f(x) = \frac{e^{3x}}{4x^2 + k} \]where \( k \) is a positive constant.
(a) Show that
\[ f'(x) = (12x^2 – 8x + 3k) g(x) \]where \( g(x) \) is a function to be found.
Given that the curve with equation \( y = f(x) \) has at least one stationary point,
(b) find the range of possible values of \( k \).
Worked Solution
Part (a): Quotient Rule
Formula: If \( y = \frac{u}{v} \), then \( \frac{dy}{dx} = \frac{v u’ – u v’}{v^2} \).
\( u = e^{3x} \implies u’ = 3e^{3x} \)
\( v = 4x^2 + k \implies v’ = 8x \)
Factor out \( e^{3x} \) from the numerator:
\[ f'(x) = \frac{e^{3x}[3(4x^2 + k) – 8x]}{(4x^2 + k)^2} \] \[ f'(x) = \frac{e^{3x}(12x^2 + 3k – 8x)}{(4x^2 + k)^2} \] \[ f'(x) = (12x^2 – 8x + 3k) \times \frac{e^{3x}}{(4x^2 + k)^2} \]Comparing with the required form:
\[ g(x) = \frac{e^{3x}}{(4x^2 + k)^2} \]Part (b): Stationary Points
Stationary points occur when \( f'(x) = 0 \).
Since \( g(x) \) (an exponential divided by a squared term) is never zero, we must have the quadratic factor equal to zero.
For “at least one” stationary point, this quadratic equation must have real roots.
Condition for real roots: discriminant \( b^2 – 4ac \ge 0 \).
\[ a = 12, \quad b = -8, \quad c = 3k \] \[ (-8)^2 – 4(12)(3k) \ge 0 \] \[ 64 – 144k \ge 0 \] \[ 64 \ge 144k \] \[ k \le \frac{64}{144} \] \[ k \le \frac{4}{9} \]We are also given that \( k \) is a positive constant (\( k > 0 \)).
So, \( 0 < k \le \frac{4}{9} \).
Final Answer:
(a) \( g(x) = \frac{e^{3x}}{(4x^2 + k)^2} \)
(b) \( 0 < k \le \frac{4}{9} \)
✓ (6 marks total)
Question 13 (6 marks)
Relative to a fixed origin \( O \)
- the point \( A \) has position vector \( 4\mathbf{i} – 3\mathbf{j} + 5\mathbf{k} \)
- the point \( B \) has position vector \( 4\mathbf{j} + 6\mathbf{k} \)
- the point \( C \) has position vector \( -16\mathbf{i} + p\mathbf{j} + 10\mathbf{k} \)
where \( p \) is a constant.
Given that \( A \), \( B \) and \( C \) lie on a straight line,
(a) find the value of \( p \).
The line segment \( OB \) is extended to a point \( D \) so that \( \vec{CD} \) is parallel to \( \vec{OA} \).
(b) Find \( |\vec{OD}| \), writing your answer as a fully simplified surd.
Worked Solution
Part (a): Collinear Points
If \( A, B, C \) are collinear, the vector \( \vec{AB} \) must be parallel to \( \vec{BC} \) (or \( \vec{AC} \)). i.e., \( \vec{BC} = k \vec{AB} \).
\[ \vec{AB} = \vec{OB} – \vec{OA} = \begin{pmatrix} 0-4 \\ 4-(-3) \\ 6-5 \end{pmatrix} = \begin{pmatrix} -4 \\ 7 \\ 1 \end{pmatrix} \]
\[ \vec{AC} = \vec{OC} – \vec{OA} = \begin{pmatrix} -16-4 \\ p-(-3) \\ 10-5 \end{pmatrix} = \begin{pmatrix} -20 \\ p+3 \\ 5 \end{pmatrix} \]
Compare components: The x-component goes from -4 to -20 (multiply by 5). The z-component goes from 1 to 5 (multiply by 5). Consistent.
So the y-component must also be multiplied by 5.
\[ p + 3 = 5 \times 7 \] \[ p + 3 = 35 \] \[ p = 32 \]Part (b): Vector Geometry
1. \( D \) lies on \( OB \) extended. So \( \vec{OD} = \lambda \vec{OB} \).
2. \( \vec{CD} \) is parallel to \( \vec{OA} \). So \( \vec{CD} = \mu \vec{OA} \).
Let \( \vec{OD} = \lambda \begin{pmatrix} 0 \\ 4 \\ 6 \end{pmatrix} = \begin{pmatrix} 0 \\ 4\lambda \\ 6\lambda \end{pmatrix} \).
We know \( \vec{CD} = \vec{OD} – \vec{OC} \).
\[ \vec{CD} = \begin{pmatrix} 0 \\ 4\lambda \\ 6\lambda \end{pmatrix} – \begin{pmatrix} -16 \\ 32 \\ 10 \end{pmatrix} = \begin{pmatrix} 16 \\ 4\lambda – 32 \\ 6\lambda – 10 \end{pmatrix} \]This is parallel to \( \vec{OA} = \begin{pmatrix} 4 \\ -3 \\ 5 \end{pmatrix} \).
Looking at x-components: \( 16 = 4 \times 4 \). So the scale factor is 4.
Compare z-components:
\[ 6\lambda – 10 = 4 \times 5 \] \[ 6\lambda – 10 = 20 \] \[ 6\lambda = 30 \implies \lambda = 5 \](Check y-components: \( 4(5) – 32 = -12 \). \( 4 \times (-3) = -12 \). Consistent.)
So \( \vec{OD} = 5 \vec{OB} = \begin{pmatrix} 0 \\ 20 \\ 30 \end{pmatrix} \).
\[ |\vec{OD}| = \sqrt{0^2 + 20^2 + 30^2} = \sqrt{400 + 900} = \sqrt{1300} \] \[ = \sqrt{100 \times 13} = 10\sqrt{13} \]Final Answer:
(a) \( p = 32 \)
(b) \( 10\sqrt{13} \)
✓ (6 marks total)
Question 14 (10 marks)
(a) Express \( \frac{3}{(2x-1)(x+1)} \) in partial fractions.
When chemical \( A \) and chemical \( B \) are mixed, oxygen is produced.
The total volume of oxygen produced, \( V \, \text{m}^3 \), \( t \) hours after the chemicals were mixed, is modelled by the differential equation
\[ \frac{dV}{dt} = \frac{3V}{(2t-1)(t+1)} \quad V \ge 0, \ t \ge k \]where \( k \) is a constant.
Given that exactly 2 hours after the chemicals were mixed, a total volume of \( 3 \, \text{m}^3 \) of oxygen had been produced,
(b) solve the differential equation to show that
\[ V = \frac{3(2t-1)}{t+1} \]The scientist noticed that
- there was a time delay between the chemicals being mixed and oxygen being produced
- there was a limit to the total volume of oxygen produced
Deduce from the model
(c) (i) the time delay giving your answer in minutes,
(ii) the limit giving your answer in \( \text{m}^3 \)
Worked Solution
Part (a): Partial Fractions
Let \( x = -1 \):
\[ 3 = A(0) + B(-2-1) \implies 3 = -3B \implies B = -1 \]Let \( x = \frac{1}{2} \):
\[ 3 = A(\frac{3}{2}) + B(0) \implies 3 = \frac{3}{2}A \implies A = 2 \]So the expression is \( \frac{2}{2x-1} – \frac{1}{x+1} \).
Part (b): Solving Differential Equation
Separate variables: Move \( V \) terms to LHS and \( t \) terms to RHS.
Using the partial fractions from (a) (replacing \( x \) with \( t \)):
\[ \ln V = \int \left( \frac{2}{2t-1} – \frac{1}{t+1} \right) \, dt \] \[ \ln V = \ln|2t-1| – \ln|t+1| + c \]Note: Integral of \( \frac{2}{2t-1} \) is \( \ln|2t-1| \) because derivative of \( 2t-1 \) is 2.
Use log laws:
\[ \ln V = \ln\left| \frac{2t-1}{t+1} \right| + c \]Use initial condition: \( t=2, V=3 \).
\[ \ln 3 = \ln\left| \frac{4-1}{3} \right| + c \] \[ \ln 3 = \ln(1) + c \] \[ \ln 3 = 0 + c \implies c = \ln 3 \]Substitute back:
\[ \ln V = \ln\left( \frac{2t-1}{t+1} \right) + \ln 3 \] \[ \ln V = \ln\left( \frac{3(2t-1)}{t+1} \right) \] \[ V = \frac{3(2t-1)}{t+1} \]Part (c): Interpretation
(i) Time delay implies \( V=0 \). Oxygen starts being produced when \( V \) becomes positive.
(ii) Limit implies \( t \to \infty \).
(i) Set \( V = 0 \):
\[ \frac{3(2t-1)}{t+1} = 0 \implies 2t-1 = 0 \implies t = 0.5 \text{ hours} \]In minutes: \( 0.5 \times 60 = 30 \) minutes.
(ii) As \( t \to \infty \):
\[ V = \frac{3(2t-1)}{t+1} = \frac{6t-3}{t+1} \]Divide top and bottom by \( t \):
\[ V = \frac{6 – 3/t}{1 + 1/t} \]As \( t \to \infty, \frac{1}{t} \to 0 \), so \( V \to \frac{6}{1} = 6 \).
Final Answer:
(a) \( \frac{2}{2x-1} – \frac{1}{x+1} \)
(c)(i) 30 minutes
(c)(ii) 6 \( \text{m}^3 \)
✓ (10 marks total)
Question 15 (10 marks)
Given that the first three terms of a geometric series are
\[ 12\cos\theta, \quad 5 + 2\sin\theta, \quad 6\tan\theta \](a) show that
\[ 4\sin^2\theta – 52\sin\theta + 25 = 0 \]Given that \( \theta \) is an obtuse angle measured in radians,
(b) solve the equation in part (a) to find the exact value of \( \theta \).
(c) show that the sum to infinity of the series can be expressed in the form
\[ k(1 – \sqrt{3}) \]where \( k \) is a constant to be found.
Worked Solution
Part (a): Geometric Series Property
For a geometric series \( a, b, c \), the common ratio \( r = \frac{b}{a} = \frac{c}{b} \). Therefore \( b^2 = ac \).
Expand LHS and simplify RHS (using \( \tan\theta = \frac{\sin\theta}{\cos\theta} \)):
\[ 25 + 20\sin\theta + 4\sin^2\theta = 72\cos\theta \frac{\sin\theta}{\cos\theta} \] \[ 25 + 20\sin\theta + 4\sin^2\theta = 72\sin\theta \]Rearrange to form a quadratic in \( \sin\theta \):
\[ 4\sin^2\theta – 52\sin\theta + 25 = 0 \]Part (b): Solving for Theta
Let \( s = \sin\theta \). \( 4s^2 – 52s + 25 = 0 \).
Using the quadratic formula or factorising:
\[ (2s – 25)(2s – 1) = 0 \]So \( s = \frac{25}{2} \) or \( s = \frac{1}{2} \).
Since \( -1 \le \sin\theta \le 1 \), \( \sin\theta = \frac{25}{2} \) is impossible.
So \( \sin\theta = \frac{1}{2} \).
Principal value: \( \frac{\pi}{6} \).
We are given \( \theta \) is obtuse (\( \frac{\pi}{2} < \theta < \pi \)).
\[ \theta = \pi – \frac{\pi}{6} = \frac{5\pi}{6} \]Part (c): Sum to Infinity
Formula: \( S_\infty = \frac{a}{1-r} \).
We need to find \( a \) and \( r \) using \( \theta = \frac{5\pi}{6} \).
First term \( a = 12\cos(5\pi/6) = 12(-\frac{\sqrt{3}}{2}) = -6\sqrt{3} \).
Second term \( = 5 + 2\sin(5\pi/6) = 5 + 2(0.5) = 6 \).
Ratio \( r = \frac{\text{2nd}}{\text{1st}} = \frac{6}{-6\sqrt{3}} = -\frac{1}{\sqrt{3}} \).
Calculate sum:
\[ S_\infty = \frac{-6\sqrt{3}}{1 – (-\frac{1}{\sqrt{3}})} = \frac{-6\sqrt{3}}{1 + \frac{1}{\sqrt{3}}} \]Multiply top and bottom by \( \sqrt{3} \):
\[ = \frac{-18}{\sqrt{3} + 1} \]Rationalise the denominator:
\[ = \frac{-18(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)} = \frac{-18(\sqrt{3}-1)}{3-1} \] \[ = \frac{-18(\sqrt{3}-1)}{2} = -9(\sqrt{3}-1) \] \[ = 9(1-\sqrt{3}) \]So \( k = 9 \).
Final Answer:
(b) \( \theta = \frac{5\pi}{6} \)
(c) \( 9(1-\sqrt{3}) \)
✓ (10 marks total)
Question 16 (12 marks)
Figure 6 shows a sketch of the curve \( C \) with parametric equations
\[ x = 2\tan t + 1 \quad y = 2\sec^2 t + 3 \quad -\frac{\pi}{4} \le t \le \frac{\pi}{3} \]The line \( l \) is the normal to \( C \) at the point \( P \) where \( t = \frac{\pi}{4} \).
(a) Using parametric differentiation, show that an equation for \( l \) is
\[ y = -\frac{1}{2}x + \frac{17}{2} \](b) Show that all points on \( C \) satisfy the equation
\[ y = \frac{1}{2}(x-1)^2 + 5 \]The straight line with equation
\[ y = -\frac{1}{2}x + k \quad \text{where } k \text{ is a constant} \]intersects \( C \) at two distinct points.
(c) Find the range of possible values for \( k \).
Worked Solution
Part (a): Parametric Differentiation
Strategy: Find the gradient of the tangent \( \frac{dy}{dx} \) using the chain rule \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \). Then find the gradient of the normal (negative reciprocal) and use \( y – y_1 = m(x – x_1) \).
\( x = 2\tan t + 1 \implies \frac{dx}{dt} = 2\sec^2 t \)
\( y = 2\sec^2 t + 3 \implies \frac{dy}{dt} = 4\sec t (\sec t \tan t) = 4\sec^2 t \tan t \)
\[ \frac{dy}{dx} = \frac{4\sec^2 t \tan t}{2\sec^2 t} = 2\tan t \]At \( P \), \( t = \frac{\pi}{4} \):
Gradient of tangent \( m_T = 2\tan(\frac{\pi}{4}) = 2(1) = 2 \).
Gradient of normal \( m_N = -\frac{1}{m_T} = -\frac{1}{2} \).
Coordinates of \( P \):
\[ x = 2\tan(\pi/4) + 1 = 3 \] \[ y = 2\sec^2(\pi/4) + 3 = 2(\sqrt{2})^2 + 3 = 2(2) + 3 = 7 \]Equation of normal:
\[ y – 7 = -\frac{1}{2}(x – 3) \] \[ y = -\frac{1}{2}x + \frac{3}{2} + 7 \] \[ y = -\frac{1}{2}x + \frac{17}{2} \]Part (b): Cartesian Equation
From \( x = 2\tan t + 1 \):
\[ \tan t = \frac{x-1}{2} \]From \( y = 2\sec^2 t + 3 \):
Using identity \( \sec^2 t = 1 + \tan^2 t \):
\[ y = 2(1 + \tan^2 t) + 3 \] \[ y = 2 + 2\tan^2 t + 3 = 2\tan^2 t + 5 \]Substitute for \( \tan t \):
\[ y = 2\left(\frac{x-1}{2}\right)^2 + 5 \] \[ y = 2\frac{(x-1)^2}{4} + 5 \] \[ y = \frac{1}{2}(x-1)^2 + 5 \]Part (c): Intersection Range
We need the line \( y = -\frac{1}{2}x + k \) to intersect the curve at TWO distinct points.
This means substituting the line eq into the curve eq and finding where the discriminant \( > 0 \).
Crucial: We must also consider the domain of \( t \). The curve is only defined for \( -\frac{\pi}{4} \le t \le \frac{\pi}{3} \).
Substitute line into curve:
\[ -\frac{1}{2}x + k = \frac{1}{2}(x-1)^2 + 5 \]Multiply by 2:
\[ -x + 2k = (x-1)^2 + 10 \] \[ -x + 2k = x^2 – 2x + 1 + 10 \] \[ x^2 – x + (11 – 2k) = 0 \]For two distinct roots, discriminant \( \Delta > 0 \):
\[ b^2 – 4ac > 0 \] \[ (-1)^2 – 4(1)(11 – 2k) > 0 \] \[ 1 – 44 + 8k > 0 \] \[ 8k > 43 \implies k > \frac{43}{8} \]Check endpoints: The curve segment is finite.
At \( t = -\frac{\pi}{4} \): \( x = 2(-1)+1 = -1 \), \( y = 2(2)+3 = 7 \). Point \( A(-1, 7) \).
At \( t = \frac{\pi}{3} \): \( x = 2(\sqrt{3})+1 \approx 4.46 \), \( y = 2(2)^2+3 = 11 \). Point \( B(1+2\sqrt{3}, 11) \).
Find \( k \) for these points:
At \( A(-1, 7) \): \( 7 = -\frac{1}{2}(-1) + k \implies 7 = 0.5 + k \implies k = 6.5 = \frac{13}{2} \).
At \( B(1+2\sqrt{3}, 11) \): \( 11 = -\frac{1}{2}(1+2\sqrt{3}) + k \implies k = 11 + 0.5 + \sqrt{3} \approx 13.2 \).
Wait, let’s visualize. The curve is a parabola opening upwards with vertex at \( (1, 5) \). The line has negative gradient \( -0.5 \).
The condition \( k > \frac{43}{8} \) (or \( 5.375 \)) ensures the line is above the tangent at the “bottom” (tangent to the full parabola).
The “bottom” tangent corresponds to the discriminant condition. The tangent point on the full parabola would be at \( 2x – 1 = 0 \implies x=0.5 \). Is \( x=0.5 \) in our range? Range of \( x \) is \( [-1, 1+2\sqrt{3}] \). Yes.
However, we are limited by the endpoints.
The line intersects twice if it cuts between the tangent point and the *lower* endpoint (in terms of \( k \)).
Let’s check the \( k \) values again.
Tangent \( k = 5.375 \).
Through \( (-1, 7) \): \( k = 6.5 \).
Through \( (1+2\sqrt{3}, 11) \): \( k \approx 13.2 \).
If \( k \) increases, the line moves up.
For \( k \) just above 5.375, we have 2 intersections.
As \( k \) increases, we eventually hit the endpoint \( (-1, 7) \) at \( k=6.5 \). Beyond this, one intersection falls off the left side of the curve.
So for 2 intersections: \( \frac{43}{8} < k \le \frac{13}{2} \).
Final Answer:
(a) \( y = -\frac{1}{2}x + \frac{17}{2} \)
(b) \( y = \frac{1}{2}(x-1)^2 + 5 \)
(c) \( \frac{43}{8} < k \le \frac{13}{2} \)
✓ (12 marks total)