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Pearson Edexcel GCE Mathematics Pure 2 (June 2024)
ℹ️ Guide to Solutions
- 💡 Concept: Explains the mathematical reasoning (“The Why”)
- ✏️ Working: Shows the step-by-step calculation (“The How”)
- 🔍 Check: Verifies the result makes sense (“The What”)
- ✅ Marks: Indicates where method (M) and accuracy (A) marks are earned
Table of Contents
- Question 1: Differentiation
- Question 2: Arithmetic Sequences
- Question 3: Transformations
- Question 4: Recurrence Relations
- Question 5: Small Angle Approximations
- Question 6: Iterative Formula
- Question 7: Vectors
- Question 8: Trigonometry
- Question 9: Projectile Motion
- Question 10: Parametric Equations
- Question 11: Integration
- Question 12: Differential Equations
- Question 13: Exponentials & Logs
- Question 14: Circles
- Question 15: Differentiation (Normals)
Question 1 (5 marks)
Given the curve with equation: \[ y = x^3 – 7x^2 + 5x – 10 \]
(a) Find in simplest form
(i) \( \frac{\mathrm{d}y}{\mathrm{d}x} \)
(ii) \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} \)
(3)
(b) Hence find the exact value of \( x \) when \( \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 0 \)
(2)
Worked Solution
Step 1: Part (a)(i) – First Derivative
💡 Strategy: We need to differentiate the function term by term using the power rule: if \( y = ax^n \), then \( \frac{\mathrm{d}y}{\mathrm{d}x} = anx^{n-1} \). Constant terms differentiate to 0.
Given \( y = x^3 – 7x^2 + 5x – 10 \)
Differentiating term by term:
- \( x^3 \to 3x^2 \)
- \( -7x^2 \to -7(2x) = -14x \)
- \( 5x \to 5 \)
- \( -10 \to 0 \)
✓ (M1 A1) Correct differentiation
Step 2: Part (a)(ii) – Second Derivative
💡 Strategy: We differentiate our answer from part (i) again with respect to \( x \).
From (i): \( \frac{\mathrm{d}y}{\mathrm{d}x} = 3x^2 – 14x + 5 \)
Differentiating again:
\[ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 3(2x) – 14(1) + 0 \] \[ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 6x – 14 \]✓ (A1ft) Correct second derivative (follow through)
Step 3: Part (b) – Solving for x
💡 Strategy: The question asks us to find \( x \) when the second derivative is zero. We set our expression from (ii) equal to 0 and solve the linear equation.
Add 14 to both sides:
\[ 6x = 14 \]Divide by 6:
\[ x = \frac{14}{6} \]Simplify the fraction:
\[ x = \frac{7}{3} \]✓ (M1) Setting second derivative to 0
✓ (A1) Correct exact value
🔍 Check: Substitute \( x = 7/3 \) back into \( 6x-14 \): \( 6(7/3) – 14 = 2(7) – 14 = 14 – 14 = 0 \). Correct.
Final Answer:
(a)(i) \( 3x^2 – 14x + 5 \)
(a)(ii) \( 6x – 14 \)
(b) \( x = \frac{7}{3} \)
Question 2 (5 marks)
Jamie takes out an interest-free loan of £8100.
Jamie makes a payment every month to pay back the loan.
Jamie repays £400 in month 1, £390 in month 2, £380 in month 3, and so on, so that the amounts repaid each month form an arithmetic sequence.
(a) Show that Jamie repays £290 in month 12.
(1)
After Jamie’s \( N \)th payment, the loan is completely paid back.
(b) Show that \( N^2 – 81N + 1620 = 0 \)
(2)
(c) Hence find the value of \( N \).
(2)
Worked Solution
Step 1: Identifying Sequence Parameters
💡 Strategy: The payments form an arithmetic sequence. We need to identify the first term (\( a \)) and the common difference (\( d \)).
Month 1: £400 \( \rightarrow a = 400 \)
Month 2: £390
Month 3: £380
The difference is \( 390 – 400 = -10 \). So, \( d = -10 \).
Step 2: Part (a) – The 12th Term
💡 Strategy: Use the formula for the \( n \)th term of an arithmetic sequence: \( u_n = a + (n-1)d \). We need to find \( u_{12} \).
✓ (B1) Correct calculation shown
Step 3: Part (b) – Sum of the Sequence
💡 Strategy: The loan is fully paid when the sum of payments equals the loan amount (£8100). Use the arithmetic sum formula: \( S_n = \frac{n}{2}(2a + (n-1)d) \).
Set sum \( S_N = 8100 \):
\[ 8100 = \frac{N}{2}(2(400) + (N-1)(-10)) \]Multiply by 2 to clear the fraction:
\[ 16200 = N(800 – 10N + 10) \] \[ 16200 = N(810 – 10N) \]Expand the bracket:
\[ 16200 = 810N – 10N^2 \]Rearrange to form a quadratic equation (move everything to the left to make \( N^2 \) positive):
\[ 10N^2 – 810N + 16200 = 0 \]Divide strictly by 10:
\[ N^2 – 81N + 1620 = 0 \](as required)
✓ (M1) Setting up sum equation
✓ (A1) Correct derivation
Step 4: Part (c) – Solving for N
💡 Strategy: Solve the quadratic equation. Since this is a calculator paper, we can use the quadratic formula or factorisation. Given the numbers, let’s look for factors of 1620 that sum to 81.
Factors of 1620: 36 and 45. Check: \( 36 + 45 = 81 \) and \( 36 \times 45 = 1620 \).
\[ (N – 36)(N – 45) = 0 \]So \( N = 36 \) or \( N = 45 \).
Logic check: Would the payment become negative? Let’s check the payment at month 45:
\[ u_{45} = 400 + 44(-10) = 400 – 440 = -40 \]A negative payment implies the bank pays Jamie, which doesn’t make sense for paying off a loan. The loan is paid off at the earlier date.
Therefore, \( N = 36 \).
✓ (M1) Solving quadratic
✓ (A1) Selecting correct value (36)
Final Answer:
(a) Shown (£290)
(b) Shown (\( N^2 – 81N + 1620 = 0 \))
(c) \( N = 36 \)
Question 3 (4 marks)
The point \( P(3, -2) \) lies on the curve with equation \( y = f(x) \), \( x \in \mathbb{R} \).
Find the coordinates of the point to which \( P \) is mapped when the curve with equation \( y = f(x) \) is transformed to the curve with equation:
(i) \( y = f(x-2) \)
(ii) \( y = f(2x) \)
(iii) \( y = 3f(-x) + 5 \)
Worked Solution
Step 1: Part (i) – Translation
💡 Strategy: \( f(x-a) \) represents a translation by vector \( \binom{a}{0} \) (shift right by \( a \)). Here \( a = 2 \).
Original point: \( (3, -2) \)
Transformation: \( x \to x + 2 \), \( y \) stays same.
\[ (3+2, -2) = (5, -2) \]✓ (B1)
Step 2: Part (ii) – Horizontal Stretch
💡 Strategy: \( f(ax) \) represents a horizontal stretch by scale factor \( \frac{1}{a} \). Here \( a = 2 \), so scale factor is \( \frac{1}{2} \).
Original point: \( (3, -2) \)
Transformation: Multiply \( x \) coordinate by \( \frac{1}{2} \), \( y \) stays same.
\[ (3 \times \frac{1}{2}, -2) = (1.5, -2) \]✓ (B1)
Step 3: Part (iii) – Combined Transformation
💡 Strategy: We have three transformations here:
- \( f(-x) \): Reflection in the y-axis (multiply \( x \) by -1).
- \( 3f(…) \): Vertical stretch scale factor 3 (multiply \( y \) by 3).
- \( … + 5 \): Vertical translation up by 5 (add 5 to \( y \)).
Original point: \( (3, -2) \)
1. Reflection (\( f(-x) \)): \( x \) becomes \( -3 \). Point is \( (-3, -2) \).
2. Vertical Stretch (\( \times 3 \)): \( y \) becomes \( -2 \times 3 = -6 \). Point is \( (-3, -6) \).
3. Vertical Shift (\( + 5 \)): \( y \) becomes \( -6 + 5 = -1 \). Point is \( (-3, -1) \).
✓ (B1 B1) One mark for each coordinate
Final Answer:
(i) \( (5, -2) \)
(ii) \( (1.5, -2) \) or \( (\frac{3}{2}, -2) \)
(iii) \( (-3, -1) \)
Question 4 (5 marks)
A sequence \( u_1, u_2, u_3, \dots \) is defined by
\[ u_{n+1} = ku_n – 5 \] \[ u_1 = 6 \]where \( k \) is a positive constant.
Given that \( u_3 = -1 \)
(a) show that \( 6k^2 – 5k – 4 = 0 \)
(2)
(b) Hence
(i) find the value of \( k \),
(ii) find the value of \( \sum_{r=1}^{3} u_r \)
(3)
Worked Solution
Step 1: Part (a) – Generating Terms
💡 Strategy: Use the recurrence relation \( u_{n+1} = ku_n – 5 \) to find expressions for \( u_2 \) and \( u_3 \) in terms of \( k \).
Given \( u_1 = 6 \).
Find \( u_2 \):
\[ u_2 = k(u_1) – 5 = 6k – 5 \]Find \( u_3 \):
\[ u_3 = k(u_2) – 5 = k(6k – 5) – 5 \] \[ u_3 = 6k^2 – 5k – 5 \]✓ (M1) Finding \( u_3 \) in terms of \( k \)
Step 2: Part (a) – Forming the Equation
💡 Strategy: We are given that \( u_3 = -1 \). Equate our expression to -1.
Add 1 to both sides:
\[ 6k^2 – 5k – 4 = 0 \](as required)
✓ (A1) Correct derivation
Step 3: Part (b)(i) – Finding k
💡 Strategy: Solve the quadratic equation. Since \( k \) is a positive constant, we reject any negative solution.
Factorise (look for numbers multiplying to \( 6 \times -4 = -24 \) and adding to -5, i.e., -8 and 3):
\[ 6k^2 – 8k + 3k – 4 = 0 \] \[ 2k(3k – 4) + 1(3k – 4) = 0 \] \[ (2k + 1)(3k – 4) = 0 \]Solutions: \( k = -0.5 \) or \( k = \frac{4}{3} \).
Since \( k \) is positive, \( k = \frac{4}{3} \).
✓ (B1) Correct value of \( k \)
Step 4: Part (b)(ii) – Calculating Sum
💡 Strategy: Calculate the actual numerical values of \( u_1, u_2, u_3 \) using \( k = 4/3 \) and sum them.
\( u_1 = 6 \)
\( u_2 = 6(\frac{4}{3}) – 5 = 8 – 5 = 3 \)
\( u_3 = -1 \) (given)
Sum \( \sum_{r=1}^{3} u_r = u_1 + u_2 + u_3 \):
\[ 6 + 3 + (-1) = 8 \]✓ (M1) Attempting sum
✓ (A1) Correct answer
Final Answer:
(a) Shown
(b)(i) \( k = \frac{4}{3} \)
(b)(ii) 8
Question 5 (3 marks)
Given that \( \theta \) is small and in radians, use the small angle approximations to find an approximate numerical value of
\[ \frac{\theta \tan 2\theta}{1 – \cos 3\theta} \]Worked Solution
Step 1: Recalling Small Angle Approximations
💡 Strategy: When \( \theta \) is small (in radians):
- \( \sin \theta \approx \theta \)
- \( \tan \theta \approx \theta \)
- \( \cos \theta \approx 1 – \frac{\theta^2}{2} \)
We need to adapt these for \( 2\theta \) and \( 3\theta \).
For \( \tan 2\theta \): replace \( \theta \) with \( 2\theta \)
\[ \tan 2\theta \approx 2\theta \]For \( \cos 3\theta \): replace \( \theta \) with \( 3\theta \)
\[ \cos 3\theta \approx 1 – \frac{(3\theta)^2}{2} = 1 – \frac{9\theta^2}{2} \]✓ (B1) Correct approximations stated or used
Step 2: Substitution and Simplification
💡 Strategy: Substitute these expressions into the original fraction and simplify.
Simplify numerator:
\[ \text{Numerator} = 2\theta^2 \]Simplify denominator (watch the double negative!):
\[ \text{Denominator} = 1 – 1 + \frac{9\theta^2}{2} = \frac{9\theta^2}{2} \]Combine:
\[ \frac{2\theta^2}{\frac{9\theta^2}{2}} \]✓ (M1) Correct substitution
Step 3: Calculation
💡 Strategy: Cancel the \( \theta^2 \) terms and perform the division of fractions.
✓ (A1) Correct numerical value
Final Answer:
\( \frac{4}{9} \)
Question 6 (7 marks)
Figure 1 shows a sketch of the curves with equations \( y = f(x) \) and \( y = g(x) \) where
\[ f(x) = e^{4x^2 – 1} \quad x > 0 \] \[ g(x) = 8\ln x \quad x > 0 \](a) Find
(i) \( f'(x) \)
(ii) \( g'(x) \)
(2)
Given that \( f'(x) = g'(x) \) at \( x = \alpha \)
(b) show that \( \alpha \) satisfies the equation
\[ 4x^2 + 2\ln x – 1 = 0 \](2)
The iterative formula
\[ x_{n+1} = \sqrt{\frac{1-2\ln x_n}{4}} \]is used with \( x_1 = 0.6 \) to find an approximate value for \( \alpha \).
(c) Calculate, giving each answer to 4 decimal places,
(i) the value of \( x_2 \)
(ii) the value of \( \alpha \)
(3)
Worked Solution
Step 1: Part (a) – Differentiation
💡 Strategy: Apply the chain rule for exponential functions and standard derivative for log functions.
For \( e^{u} \), derivative is \( u’ e^u \).
For \( \ln x \), derivative is \( 1/x \).
(i) \( f(x) = e^{4x^2 – 1} \)
Let \( u = 4x^2 – 1 \), then \( u’ = 8x \).
\[ f'(x) = 8x e^{4x^2 – 1} \](ii) \( g(x) = 8\ln x \)
\[ g'(x) = \frac{8}{x} \]✓ (B1 B1) Correct derivatives
Step 2: Part (b) – Equating Derivatives
💡 Strategy: Set \( f'(x) = g'(x) \) and manipulate the algebra to eliminate the exponential term using logs, aiming for the target equation.
Divide by 8:
\[ x e^{4x^2 – 1} = \frac{1}{x} \]We want to remove the \( e \), so take natural logs (ln) of both sides. First, isolate the \( e \) term is usually helpful, but let’s look at the target equation. It has \( 4x^2 \) and \( \ln x \). Taking ln immediately works:
\[ \ln(x e^{4x^2 – 1}) = \ln(\frac{1}{x}) \]Use log laws: \( \ln(ab) = \ln a + \ln b \) and \( \ln(1/x) = -\ln x \).
\[ \ln x + \ln(e^{4x^2 – 1}) = -\ln x \] \[ \ln x + (4x^2 – 1) = -\ln x \]Rearrange to match target:
\[ 4x^2 – 1 + 2\ln x = 0 \] \[ 4x^2 + 2\ln x – 1 = 0 \]✓ (M1) Equating and taking logs
✓ (A1) Deriving exact form
Step 3: Part (c) – Iteration
💡 Strategy: Use the calculator to perform the iterations. Enter 0.6, press =. Then type the formula using ‘Ans’ instead of x. Press = for \( x_2 \), keep pressing for convergence to find \( \alpha \).
\( x_1 = 0.6 \)
(i) Substitute into formula:
\[ x_2 = \sqrt{\frac{1-2\ln(0.6)}{4}} \] \[ x_2 \approx 0.7109 \] (to 4 d.p.)(ii) Continue iterating to find \( \alpha \):
\( x_3 \approx 0.6485 \)
\( x_4 \approx 0.6830 \)
… convergence is oscillatory.
Eventually \( \alpha \approx 0.6706 \).
✓ (M1) Attempting iteration
✓ (A1) Correct \( x_2 \)
✓ (A1) Correct \( \alpha \)
Final Answer:
(a)(i) \( 8x e^{4x^2 – 1} \)
(a)(ii) \( \frac{8}{x} \)
(b) Shown
(c)(i) 0.7109
(c)(ii) 0.6706
Question 7 (5 marks)
Figure 2 shows a sketch of the straight line \( l \).
Line \( l \) passes through the points \( A \) and \( B \).
Relative to a fixed origin \( O \)
- the point \( A \) has position vector \( 2\mathbf{i} – 3\mathbf{j} + 5\mathbf{k} \)
- the point \( B \) has position vector \( 5\mathbf{i} + 6\mathbf{j} + 8\mathbf{k} \)
(a) Find \( \vec{AB} \)
(1)
Given that a point \( P \) lies on \( l \) such that
\[ |\vec{AP}| = 2|\vec{BP}| \](b) find the possible position vectors of \( P \).
(4)
Worked Solution
Step 1: Part (a) – Finding Vector AB
💡 Strategy: \( \vec{AB} = \vec{OB} – \vec{OA} \).
✓ (B1) Correct vector
Step 2: Part (b) – Understanding the Ratio
💡 Strategy: The condition \( |\vec{AP}| = 2|\vec{BP}| \) means the distance from A to P is twice the distance from B to P. Since P is on the line line AB, P divides the segment AB (internally) or extends it (externally).
Case 1 (Internal): P lies between A and B such that AP:PB = 2:1.
Case 2 (External): P lies outside AB such that AP is double BP. This puts B in the middle of A and P.
Step 3: Calculating Position Vectors
Case 1 (Internal):
\( \vec{AP} = \frac{2}{3}\vec{AB} \)
\[ \vec{OP} = \vec{OA} + \frac{2}{3}\vec{AB} \] \[ \vec{OP} = \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix} + \frac{2}{3}\begin{pmatrix} 3 \\ 9 \\ 3 \end{pmatrix} \] \[ \vec{OP} = \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix} + \begin{pmatrix} 2 \\ 6 \\ 2 \end{pmatrix} = \begin{pmatrix} 4 \\ 3 \\ 7 \end{pmatrix} \]So \( P_1 = 4\mathbf{i} + 3\mathbf{j} + 7\mathbf{k} \)
Case 2 (External):
\( \vec{AP} = 2\vec{AB} \) (Going from A past B)
\[ \vec{OP} = \vec{OA} + 2\vec{AB} \] \[ \vec{OP} = \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix} + 2\begin{pmatrix} 3 \\ 9 \\ 3 \end{pmatrix} \] \[ \vec{OP} = \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix} + \begin{pmatrix} 6 \\ 18 \\ 6 \end{pmatrix} = \begin{pmatrix} 8 \\ 15 \\ 11 \end{pmatrix} \]So \( P_2 = 8\mathbf{i} + 15\mathbf{j} + 11\mathbf{k} \)
✓ (M1) Strategy for finding P
✓ (A1) First position
✓ (M1) Second strategy
✓ (A1) Second position
Final Answer:
(a) \( 3\mathbf{i} + 9\mathbf{j} + 3\mathbf{k} \)
(b) \( 4\mathbf{i} + 3\mathbf{j} + 7\mathbf{k} \) and \( 8\mathbf{i} + 15\mathbf{j} + 11\mathbf{k} \)
Question 8 (7 marks)
In this question you must show all stages of your working.
Solutions relying entirely on calculator technology are not acceptable.
(a) Prove that
\[ \frac{1}{\mathrm{cosec}\,\theta – 1} + \frac{1}{\mathrm{cosec}\,\theta + 1} \equiv 2 \tan \theta \sec \theta \](3)
(b) Hence solve, for \( 0 < x < 90^\circ \), the equation
\[ \frac{1}{\mathrm{cosec}\,2x – 1} + \frac{1}{\mathrm{cosec}\,2x + 1} = \cot 2x \sec 2x \]Give each answer, in degrees, to one decimal place.
(4)
Worked Solution
Step 1: Part (a) – Combining Fractions
Combine LHS over a common denominator:
\[ \frac{(\mathrm{cosec}\,\theta + 1) + (\mathrm{cosec}\,\theta – 1)}{(\mathrm{cosec}\,\theta – 1)(\mathrm{cosec}\,\theta + 1)} \]Simplify numerator: \( 2\mathrm{cosec}\,\theta \)
Simplify denominator (difference of squares): \( \mathrm{cosec}^2\,\theta – 1 \)
✓ (B1) Combined single fraction
Step 2: Part (a) – Identities
Use identity \( 1 + \cot^2\theta = \mathrm{cosec}^2\theta \), so \( \mathrm{cosec}^2\theta – 1 = \cot^2\theta \).
\[ \frac{2\mathrm{cosec}\,\theta}{\cot^2\theta} \]Convert to sin and cos:
\[ = 2 \left( \frac{1}{\sin\theta} \right) \div \left( \frac{\cos^2\theta}{\sin^2\theta} \right) \] \[ = \frac{2}{\sin\theta} \times \frac{\sin^2\theta}{\cos^2\theta} \] \[ = \frac{2\sin\theta}{\cos^2\theta} \] \[ = 2 \left( \frac{\sin\theta}{\cos\theta} \right) \left( \frac{1}{\cos\theta} \right) \] \[ = 2 \tan\theta \sec\theta \]QED.
✓ (M1) Pythagorean identity
✓ (A1) Completion of proof
Step 3: Part (b) – Using the Identity
💡 Strategy: Replace the LHS with the result from (a), but with angle \( 2x \). So LHS becomes \( 2\tan 2x \sec 2x \).
Rearrange to solve (bring everything to one side or factorise):
\[ 2\tan 2x \sec 2x – \cot 2x \sec 2x = 0 \] \[ \sec 2x (2\tan 2x – \cot 2x) = 0 \]Since \( \sec 2x = \frac{1}{\cos 2x} \) cannot be 0, we must have:
\[ 2\tan 2x – \cot 2x = 0 \] \[ 2\tan 2x = \frac{1}{\tan 2x} \] \[ \tan^2 2x = \frac{1}{2} \] \[ \tan 2x = \pm \sqrt{0.5} \]✓ (B1) Use identity
✓ (M1) Simplify to trig equation
Step 4: Solving for x
Range: \( 0 < x < 90 \), so \( 0 < 2x < 180 \).
1) \( \tan 2x = \sqrt{0.5} \)
\[ 2x = \tan^{-1}(\sqrt{0.5}) \approx 35.26^\circ \] \[ x \approx 17.6^\circ \]2) \( \tan 2x = -\sqrt{0.5} \)
\[ 2x = 180 – 35.26 = 144.74^\circ \] \[ x \approx 72.4^\circ \]✓ (M1) Solving for x
✓ (A1) Correct values
Final Answer:
(a) Proof shown.
(b) \( 17.6^\circ, 72.4^\circ \)
Question 9 (7 marks)
Figure 3 shows the path of a small ball.
The ball travels in a vertical plane above horizontal ground.
The ball is thrown from point A and caught at point B.
With respect to a fixed origin O, A is \( (0, 2) \) and B is \( (20, 0.8) \).
The ball reaches its maximum height when \( x = 9 \).
A quadratic function, linking \( H \) with \( x \), is used to model the path.
(a) Find \( H \) in terms of \( x \).
(4)
(b) Give one limitation of the model.
(1)
Chandra is standing directly under the path at \( x = 16 \)m.
Chandra can catch the ball if it is less than 2.5m above the ground.
(c) Use the model to determine if Chandra can catch the ball.
(2)
Worked Solution
Step 1: Part (a) – Defining the Quadratic
💡 Strategy: Since the maximum is at \( x = 9 \), the vertex form of the quadratic is most useful: \( H = a(x-9)^2 + k \).
Using \( H = a(x-9)^2 + k \).
Use point A \( (0, 2) \):
\[ 2 = a(0-9)^2 + k \] \[ 2 = 81a + k \quad \text{(1)} \]Use point B \( (20, 0.8) \):
\[ 0.8 = a(20-9)^2 + k \] \[ 0.8 = 121a + k \quad \text{(2)} \]✓ (M1) Model form
✓ (dM1) Setting up simultaneous equations
Step 2: Solving for Constants
Subtract (1) from (2):
\[ (0.8 – 2) = (121a – 81a) \] \[ -1.2 = 40a \] \[ a = \frac{-1.2}{40} = -0.03 \]Substitute back into (1) to find \( k \):
\[ k = 2 – 81(-0.03) \] \[ k = 2 – (-2.43) = 4.43 \]So, \( H = -0.03(x-9)^2 + 4.43 \).
(Alternative forms like \( ax^2+bx+c \) are acceptable if equivalent)
✓ (ddM1) Solving
✓ (A1) Correct equation
Step 3: Part (b) – Limitation
💡 Strategy: Think about real-world physics ignored by a simple quadratic model.
Valid limitations:
- Air resistance is not considered.
- Wind effects.
- The ball is not a particle (has size).
✓ (B1) Valid reason
Step 4: Part (c) – Prediction
Find \( H \) when \( x = 16 \).
\[ H = -0.03(16-9)^2 + 4.43 \] \[ H = -0.03(7)^2 + 4.43 \] \[ H = -0.03(49) + 4.43 \] \[ H = -1.47 + 4.43 \] \[ H = 2.96 \text{ m} \]Conclusion: \( 2.96 > 2.5 \), so Chandra cannot catch the ball.
✓ (M1) Calculation
✓ (A1) Correct conclusion
Final Answer:
(a) \( H = -0.03(x-9)^2 + 4.43 \) (or equivalent)
(b) E.g., Air resistance ignored.
(c) No, ball is too high (2.96m).
Question 10 (6 marks)
Figure 4 shows a sketch of the curve \( C \) with parametric equations
\[ x = (t+3)^2, \quad y = 1 – t^3, \quad -2 \le t \le 1 \]The point \( P \) with coordinates \( (4, 2) \) lies on \( C \).
(a) Using parametric differentiation, show that the tangent to \( C \) at \( P \) has equation
\[ 3x + 4y = 20 \](5)
(b) Find, according to the model, the greatest height of the slide above water level.
(1)
Worked Solution
Step 1: Part (a) – Finding t
💡 Strategy: Find the value of parameter \( t \) at the point \( P(4,2) \).
Use \( y = 2 \):
\[ 2 = 1 – t^3 \] \[ t^3 = -1 \] \[ t = -1 \]Check with \( x \): \( x = (-1+3)^2 = 2^2 = 4 \). Consistent.
✓ (B1) Correct t value
Step 2: Differentiation
\( x = (t+3)^2 \implies \frac{\mathrm{d}x}{\mathrm{d}t} = 2(t+3) \)
\( y = 1 – t^3 \implies \frac{\mathrm{d}y}{\mathrm{d}t} = -3t^2 \)
Use chain rule for gradient:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{\mathrm{d}y/\mathrm{d}t}{\mathrm{d}x/\mathrm{d}t} = \frac{-3t^2}{2(t+3)} \]✓ (M1 A1) Chain rule use
Step 3: Gradient at P
Substitute \( t = -1 \):
\[ m = \frac{-3(-1)^2}{2(-1+3)} = \frac{-3}{4} \]✓ (M1) Evaluating gradient
Step 4: Equation of Tangent
Use \( y – y_1 = m(x – x_1) \) with \( (4, 2) \) and \( m = -3/4 \).
\[ y – 2 = -\frac{3}{4}(x – 4) \]Multiply by 4:
\[ 4(y – 2) = -3(x – 4) \] \[ 4y – 8 = -3x + 12 \] \[ 3x + 4y = 20 \](as required)
✓ (A1*) Correct proof
Step 5: Part (b) – Max Height
💡 Strategy: Height is given by \( y \). Look at range of \( t \) (\( -2 \le t \le 1 \)) to maximize \( y = 1 – t^3 \).
\( y = 1 – t^3 \). To maximize \( y \), we need the most negative \( t^3 \).
Minimum \( t = -2 \).
\[ y_{max} = 1 – (-2)^3 = 1 – (-8) = 9 \]Max height = 9m.
✓ (B1) Correct height
Final Answer:
(a) Proof shown.
(b) 9m
Question 11 (5 marks)
Figure 5 shows a sketch of part of the curve \( C \) with equation
\[ y = 8x^2 e^{-3x} \quad x \ge 0 \]The finite region \( R \), shown shaded in Figure 5, is bounded by the curve \( C \), the line with equation \( x = 1 \) and the \( x \)-axis.
Find the exact area of \( R \), giving your answer in the form \( A + Be^{-3} \) where \( A \) and \( B \) are rational numbers to be found.
Worked Solution
Step 1: Setting up the Integral
💡 Strategy: The area under the curve is given by the definite integral \( \int_0^1 y \, \mathrm{d}x \).
Step 2: Integration by Parts (First Application)
💡 Strategy: We have a product of algebraic (\( x^2 \)) and exponential terms. Use Integration by Parts formula: \( \int u \frac{\mathrm{d}v}{\mathrm{d}x} dx = uv – \int v \frac{\mathrm{d}u}{\mathrm{d}x} dx \).
Let \( u = 8x^2 \) (to reduce power) and \( \frac{\mathrm{d}v}{\mathrm{d}x} = e^{-3x} \).
\( u = 8x^2 \implies \frac{\mathrm{d}u}{\mathrm{d}x} = 16x \)
\( \frac{\mathrm{d}v}{\mathrm{d}x} = e^{-3x} \implies v = -\frac{1}{3}e^{-3x} \)
Apply formula:
\[ I = \left[ -\frac{8}{3}x^2 e^{-3x} \right]_0^1 – \int_0^1 \left( -\frac{1}{3}e^{-3x} \right) (16x) \, \mathrm{d}x \] \[ I = \left[ -\frac{8}{3}x^2 e^{-3x} \right]_0^1 + \frac{16}{3} \int_0^1 x e^{-3x} \, \mathrm{d}x \]✓ (M1 A1) Correct first application
Step 3: Integration by Parts (Second Application)
💡 Strategy: We still have \( \int x e^{-3x} dx \). Apply parts again.
Let \( u = x \implies \frac{\mathrm{d}u}{\mathrm{d}x} = 1 \)
Let \( \frac{\mathrm{d}v}{\mathrm{d}x} = e^{-3x} \implies v = -\frac{1}{3}e^{-3x} \)
\[ \int x e^{-3x} dx = \left[ -\frac{1}{3}x e^{-3x} \right] – \int \left( -\frac{1}{3}e^{-3x} \right) (1) \, \mathrm{d}x \] \[ = -\frac{1}{3}x e^{-3x} + \frac{1}{3} \int e^{-3x} \, \mathrm{d}x \] \[ = -\frac{1}{3}x e^{-3x} – \frac{1}{9}e^{-3x} \]✓ (dM1) Second application
Step 4: Combining and Evaluating
Substitute back into the main expression:
\[ I = \left[ -\frac{8}{3}x^2 e^{-3x} + \frac{16}{3} \left( -\frac{1}{3}x e^{-3x} – \frac{1}{9}e^{-3x} \right) \right]_0^1 \] \[ I = \left[ -\frac{8}{3}x^2 e^{-3x} – \frac{16}{9}x e^{-3x} – \frac{16}{27}e^{-3x} \right]_0^1 \]Evaluate at upper limit \( x = 1 \):
\[ -\frac{8}{3}e^{-3} – \frac{16}{9}e^{-3} – \frac{16}{27}e^{-3} \] \[ = \left( -\frac{72}{27} – \frac{48}{27} – \frac{16}{27} \right) e^{-3} = -\frac{136}{27}e^{-3} \]Evaluate at lower limit \( x = 0 \):
\[ 0 – 0 – \frac{16}{27}(1) = -\frac{16}{27} \]Total Area = Upper – Lower:
\[ -\frac{136}{27}e^{-3} – \left( -\frac{16}{27} \right) \] \[ = \frac{16}{27} – \frac{136}{27}e^{-3} \]✓ (M1) Using limits
✓ (A1) Correct answer
Final Answer:
\( \frac{16}{27} – \frac{136}{27}e^{-3} \)
(So \( A = \frac{16}{27} \), \( B = -\frac{136}{27} \))
Question 12 (12 marks)
(a) Express \( \frac{1}{V(25-V)} \) in partial fractions.
(2)
The volume, \( V \) microlitres, of a plant cell \( t \) hours after the plant is watered is modelled by the differential equation
\[ \frac{\mathrm{d}V}{\mathrm{d}t} = \frac{1}{10}V(25-V) \]The plant cell has an initial volume of 20 microlitres.
(b) Find, according to the model, the time taken, in minutes, for the volume of the plant cell to reach 24 microlitres.
(5)
(c) Show that
\[ V = \frac{A}{e^{-kt} + B} \]where \( A, B \) and \( k \) are constants to be found.
(3)
The model predicts that there is an upper limit, \( L \) microlitres, on the volume of the plant cell.
(d) Find the value of \( L \), giving a reason for your answer.
(2)
Worked Solution
Step 1: Part (a) – Partial Fractions
Let \( V = 0 \): \( 1 = 25A \implies A = \frac{1}{25} \)
Let \( V = 25 \): \( 1 = 25B \implies B = \frac{1}{25} \)
\[ \frac{1}{V(25-V)} = \frac{1}{25V} + \frac{1}{25(25-V)} \]✓ (M1 A1) Correct fractions
Step 2: Part (b) – Separation of Variables
💡 Strategy: Separate \( V \) terms to one side and \( t \) terms to the other, then integrate.
Using part (a):
\[ \int \left( \frac{1}{25V} + \frac{1}{25(25-V)} \right) \mathrm{d}V = \int \frac{1}{10} \mathrm{d}t \]Multiply by 25 to simplify:
\[ \int \left( \frac{1}{V} + \frac{1}{25-V} \right) \mathrm{d}V = \int \frac{25}{10} \mathrm{d}t \] \[ \ln|V| – \ln|25-V| = 2.5t + C \] \[ \ln \left| \frac{V}{25-V} \right| = 2.5t + C \]✓ (M1 A1ft) Integration
Step 3: Finding Constant C
Initial condition: \( t = 0, V = 20 \).
\[ \ln \left| \frac{20}{25-20} \right| = 0 + C \] \[ \ln(4) = C \]Equation is: \( \ln \left( \frac{V}{25-V} \right) = 2.5t + \ln 4 \)
✓ (M1) Using boundary condition
Step 4: Solving for t
When \( V = 24 \):
\[ \ln \left( \frac{24}{25-24} \right) = 2.5t + \ln 4 \] \[ \ln 24 = 2.5t + \ln 4 \] \[ 2.5t = \ln 24 – \ln 4 = \ln \left( \frac{24}{4} \right) = \ln 6 \] \[ t = \frac{\ln 6}{2.5} \text{ hours} \]Question asks for minutes (\( \times 60 \)):
\[ \text{Time} = \frac{\ln 6}{2.5} \times 60 = 24 \ln 6 \] \[ \approx 43 \text{ minutes} \]✓ (dM1) Solving for t
✓ (A1) Correct value
Step 5: Part (c) – Rearranging
Exponentiate:
\[ \frac{V}{25-V} = e^{2.5t + \ln 4} = e^{2.5t} \cdot e^{\ln 4} = 4e^{2.5t} \] \[ V = 4e^{2.5t}(25-V) \] \[ V = 100e^{2.5t} – 4Ve^{2.5t} \] \[ V + 4Ve^{2.5t} = 100e^{2.5t} \] \[ V(1 + 4e^{2.5t}) = 100e^{2.5t} \] \[ V = \frac{100e^{2.5t}}{1 + 4e^{2.5t}} \]To match form \( \frac{A}{e^{-kt} + B} \), divide numerator and denominator by \( e^{2.5t} \):
\[ V = \frac{100}{e^{-2.5t} + 4} \]So \( A = 100, B = 4, k = 2.5 \).
✓ (M1) Removing logs
✓ (M1) Rearranging
✓ (A1) Correct form
Step 6: Part (d) – Upper Limit
💡 Strategy: Consider what happens as \( t \to \infty \).
As \( t \to \infty \), \( e^{-2.5t} \to 0 \).
\[ V \to \frac{100}{0 + 4} = 25 \]\( L = 25 \) microlitres.
✓ (B1) Value 25
✓ (B1) Valid reason
Final Answer:
(a) \( \frac{1}{25V} + \frac{1}{25(25-V)} \)
(b) 43 minutes
(c) \( V = \frac{100}{e^{-2.5t} + 4} \)
(d) \( L = 25 \) (denominator tends to 4)
Question 13 (9 marks)
The world human population, \( P \) billions, is modelled by the equation
\[ P = ab^t \]where \( a \) and \( b \) are constants and \( t \) is the number of years after 2004.
Using the estimated population figures for the years from 2004 to 2007, a graph is plotted of \( \log_{10} P \) against \( t \).
The points lie approximately on a straight line with
- gradient 0.0054
- intercept 0.81 on the \( \log_{10} P \) axis
(a) Estimate, to 3 decimal places, the value of \( a \) and the value of \( b \).
(4)
In the context of the model,
(b) (i) interpret the value of the constant \( a \),
(ii) interpret the value of the constant \( b \).
(2)
(c) Use the model to estimate the world human population in 2030.
(2)
(d) Comment on the reliability of the answer to part (c).
(1)
Worked Solution
Step 1: Part (a) – Linearising the Model
💡 Strategy: Take logs of \( P = ab^t \) to match the straight line equation \( y = mx + c \).
Compare with \( y = c + mx \) where \( y = \log_{10} P \) and \( x = t \).
- Intercept \( c = \log_{10} a = 0.81 \)
- Gradient \( m = \log_{10} b = 0.0054 \)
✓ (M1) Correct log structure
Step 2: Calculating Constants
Find \( a \):
\[ \log_{10} a = 0.81 \implies a = 10^{0.81} \approx 6.457 \]Find \( b \):
\[ \log_{10} b = 0.0054 \implies b = 10^{0.0054} \approx 1.013 \]✓ (A1 A1) Correct values to 3dp
Step 3: Part (b) – Interpretations
(i) \( a \) is the value of \( P \) when \( t = 0 \) (year 2004).
Interpretation: The world population in 2004 was approximately 6.46 billion.
(ii) \( b \) is the multiplier for each year.
Interpretation: The population increases by approximately 1.3% each year (or factor of 1.013).
✓ (B1 B1) Correct interpretations
Step 4: Part (c) – Estimation
Year 2030 is \( 2030 – 2004 = 26 \) years after start.
\[ t = 26 \] \[ P = 6.457 \times (1.013)^{26} \] \[ P \approx 9.03 \text{ billion} \]✓ (M1) Substituting t=26
✓ (A1) Correct answer
Step 5: Part (d) – Reliability
We are predicting for 2030 using data from 2004-2007.
Comment: Unreliable because it involves significant extrapolation (2030 is far outside the data range).
✓ (B1) Extrapolation mentioned
Final Answer:
(a) \( a = 6.457 \), \( b = 1.013 \)
(b) (i) 2004 Population (ii) Annual growth factor
(c) 9 billion
(d) Unreliable (extrapolation)
Question 14 (8 marks)
The circle \( C_1 \) has equation
\[ x^2 + y^2 – 6x + 14y + 33 = 0 \](a) Find
(i) the coordinates of the centre of \( C_1 \)
(ii) the radius of \( C_1 \)
(3)
A different circle \( C_2 \)
- has centre with coordinates \( (-6, -8) \)
- has radius \( k \), where \( k \) is a constant
Given that \( C_1 \) and \( C_2 \) intersect at 2 distinct points,
(b) find the range of values of \( k \), writing your answer in set notation.
(5)
Worked Solution
Step 1: Part (a) – Centre and Radius
💡 Strategy: Complete the square for \( x \) and \( y \) terms to get the standard circle form \( (x-a)^2 + (y-b)^2 = r^2 \).
(i) Centre: \( (3, -7) \)
(ii) Radius \( r_1 = \sqrt{25} = 5 \)
✓ (B1) Centre
✓ (M1 A1) Method & Radius
Step 2: Part (b) – Distance Between Centres
💡 Strategy: Two circles intersect if the distance \( d \) between their centres satisfies \( |r_1 – r_2| < d < r_1 + r_2 \).
Centre \( C_1 \): \( P(3, -7) \), \( r_1 = 5 \).
Centre \( C_2 \): \( Q(-6, -8) \), \( r_2 = k \).
Distance \( d = PQ \):
\[ d = \sqrt{(3 – (-6))^2 + (-7 – (-8))^2} \] \[ d = \sqrt{9^2 + 1^2} = \sqrt{81 + 1} = \sqrt{82} \]✓ (M1 A1) Correct distance
Step 3: Part (b) – Inequalities for Intersection
Condition 1 (Touching externally): \( r_1 + r_2 > d \implies 5 + k > \sqrt{82} \implies k > \sqrt{82} – 5 \).
Wait, for intersection (2 points), we need \( k + 5 > \sqrt{82} \) AND \( |k-5| < \sqrt{82} \).
Let’s visualise. Circles touch externally if \( r_1 + r_2 = d \). They touch internally if \( |r_1 – r_2| = d \).
Intersection occurs strictly between these limits.
Lower Limit (External touch): \( 5 + k = \sqrt{82} \implies k = \sqrt{82} – 5 \).
Upper Limit (Internal touch): \( k – 5 = \sqrt{82} \implies k = \sqrt{82} + 5 \). (Assuming \( k > 5 \))
Or \( 5 – k = \sqrt{82} \implies k = 5 – \sqrt{82} \) (Reject as \( k>0 \)).
Range: \( \sqrt{82} – 5 < k < \sqrt{82} + 5 \).
✓ (dM1) Setting up inequalities
✓ (A1) Correct limits
Step 4: Set Notation
✓ (A1) Correct notation
Final Answer:
(a) \( (3, -7) \), Radius 5
(b) \( \{ k : \sqrt{82} – 5 < k < \sqrt{82} + 5 \} \)
Question 15 (12 marks)
The curve \( C \) has equation
\[ (x+y)^3 = 3x^2 – 3y – 2 \](a) Find an expression for \( \frac{\mathrm{d}y}{\mathrm{d}x} \) in terms of \( x \) and \( y \).
(5)
The point \( P(1, 0) \) lies on \( C \).
(b) Show that the normal to \( C \) at \( P \) has equation
\[ y = -2x + 2 \](2)
(c) Prove that the normal to \( C \) at \( P \) does not meet \( C \) again.
You should use algebra for your proof and make your reasoning clear.
(5)
Worked Solution
Step 1: Part (a) – Implicit Differentiation
💡 Strategy: Differentiate both sides with respect to \( x \). Remember to multiply by \( \frac{\mathrm{d}y}{\mathrm{d}x} \) whenever differentiating a \( y \) term (Chain Rule).
LHS: \( \frac{\mathrm{d}}{\mathrm{d}x} (x+y)^3 = 3(x+y)^2 (1 + \frac{\mathrm{d}y}{\mathrm{d}x}) \)
RHS: \( \frac{\mathrm{d}}{\mathrm{d}x} (3x^2 – 3y – 2) = 6x – 3\frac{\mathrm{d}y}{\mathrm{d}x} \)
Equate:
\[ 3(x+y)^2 (1 + \frac{\mathrm{d}y}{\mathrm{d}x}) = 6x – 3\frac{\mathrm{d}y}{\mathrm{d}x} \]Expand LHS:
\[ 3(x+y)^2 + 3(x+y)^2 \frac{\mathrm{d}y}{\mathrm{d}x} = 6x – 3\frac{\mathrm{d}y}{\mathrm{d}x} \]Group \( \frac{\mathrm{d}y}{\mathrm{d}x} \) terms:
\[ 3(x+y)^2 \frac{\mathrm{d}y}{\mathrm{d}x} + 3\frac{\mathrm{d}y}{\mathrm{d}x} = 6x – 3(x+y)^2 \] \[ \frac{\mathrm{d}y}{\mathrm{d}x} [3(x+y)^2 + 3] = 6x – 3(x+y)^2 \] \[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{6x – 3(x+y)^2}{3(x+y)^2 + 3} \]Simplify by dividing by 3:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2x – (x+y)^2}{(x+y)^2 + 1} \]✓ (M1 A1) LHS Diff
✓ (A1) RHS Diff
✓ (M1 A1) Rearranging
Step 2: Part (b) – Normal Gradient
At \( P(1, 0) \), substitute \( x=1, y=0 \) into the derivative:
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2(1) – (1+0)^2}{(1+0)^2 + 1} \] \[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2 – 1}{1 + 1} = \frac{1}{2} \]The tangent gradient is \( m_T = 1/2 \).
The normal gradient \( m_N = -1 / m_T = -2 \).
Equation of normal: \( y – y_1 = m_N(x – x_1) \)
\[ y – 0 = -2(x – 1) \] \[ y = -2x + 2 \](as required)
✓ (M1) Gradient calculation
✓ (A1) Correct proof
Step 3: Part (c) – Intersection
💡 Strategy: Substitute the line equation \( y = 2-2x \) into the curve equation \( (x+y)^3 = 3x^2 – 3y – 2 \) and solve for \( x \). We know \( x=1 \) is a solution; we check for others.
Substitute \( y = 2-2x \):
LHS: \( (x + (2-2x))^3 = (2-x)^3 \)
RHS: \( 3x^2 – 3(2-2x) – 2 = 3x^2 – 6 + 6x – 2 = 3x^2 + 6x – 8 \)
Equate:
\[ (2-x)^3 = 3x^2 + 6x – 8 \]Expand LHS: \( 8 – 12x + 6x^2 – x^3 \)
\[ 8 – 12x + 6x^2 – x^3 = 3x^2 + 6x – 8 \]Rearrange to form a cubic equation (= 0):
\[ x^3 – 3x^2 + 18x – 16 = 0 \]We know \( x=1 \) is a root (since P is on the line and curve). Use synthetic division or algebraic division by \( (x-1) \).
\[ (x-1)(x^2 – 2x + 16) = 0 \]Now check the quadratic \( x^2 – 2x + 16 \) for roots using the discriminant \( \Delta = b^2 – 4ac \).
\[ \Delta = (-2)^2 – 4(1)(16) \] \[ \Delta = 4 – 64 = -60 \]Since \( \Delta < 0 \), the quadratic has no real roots.
Therefore, \( x=1 \) is the only real solution. The normal does not meet \( C \) again.
✓ (M1) Substitution
✓ (dM1) Cubic equation
✓ (ddM1) Factoring out (x-1)
✓ (A1) Discriminant logic
✓ (A1) Conclusion
Final Answer:
(a) \( \frac{2x – (x+y)^2}{(x+y)^2 + 1} \)
(b) Shown
(c) Proof shown (discriminant < 0)