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A-Level Pure Math Paper 1 (Edexcel 2021)
📚 Mark Scheme Legend
- M1: Method mark (knowing a method and attempting to apply it)
- A1: Accuracy mark (dependent on M mark)
- B1: Unconditional accuracy mark
- dM1: Dependent method mark
- ft: Follow through allowed
📖 Table of Contents
- Question 1 (Factor Theorem)
- Question 2 (Quadratics)
- Question 3 (Sequences)
- Question 4 (Iterative Methods)
- Question 5 (Geometric Model)
- Question 6 (Vectors)
- Question 7 (Circles)
- Question 8 (Exponential Growth)
- Question 9 (Partial Fractions)
- Question 10 (Trigonometry)
- Question 11 (Integration)
- Question 12 (Projectiles)
- Question 13 (Parametrics)
- Question 14 (Differentiation)
- Question 15 (Proof)
Question 1 (3 marks)
Given that
\[ f(x) = ax^3 + 10x^2 – 3ax – 4 \]Given that \( (x – 1) \) is a factor of \( f(x) \), find the value of the constant \( a \).
You must make your method clear.
📝 Worked Solution
Step 1: Understanding the Factor Theorem
💡 Why we do this: The Factor Theorem states that if \( (x – k) \) is a factor of a polynomial \( f(x) \), then \( f(k) = 0 \). Here, our divisor is \( (x – 1) \), so we must calculate \( f(1) \).
Since \( (x – 1) \) is a factor, we set \( x = 1 \) and equate the function to zero:
\[ f(1) = 0 \]✓ (M1)
Step 2: Substituting and Solving
🔧 How we do it: We substitute \( x = 1 \) into the given expression for \( f(x) \) and solve the resulting linear equation for \( a \).
Substitute \( x = 1 \):
\[ a(1)^3 + 10(1)^2 – 3a(1) – 4 = 0 \] \[ a(1) + 10(1) – 3a – 4 = 0 \] \[ a + 10 – 3a – 4 = 0 \]Combine like terms:
\[ 6 – 2a = 0 \]✓ (M1)
Solve for \( a \):
\[ 2a = 6 \] \[ a = 3 \]🎯 Final Answer:
\( a = 3 \)
✓ Total: 3 marks
Question 2 (4 marks)
Given that
\[ f(x) = x^2 – 4x + 5 \quad x \in \mathbb{R} \](a) express \( f(x) \) in the form \( (x + a)^2 + b \) where \( a \) and \( b \) are integers to be found.
The curve with equation \( y = f(x) \)
- meets the \( y \)-axis at the point \( P \)
- has a minimum turning point at the point \( Q \)
(b) Write down
(i) the coordinates of \( P \)
(ii) the coordinates of \( Q \)
📝 Worked Solution
Part (a): Completing the Square
💡 Why we do this: We need to convert the quadratic into vertex form. The standard method is “completing the square”, where we take half the coefficient of \( x \).
Start with \( x^2 – 4x + 5 \).
Half of the coefficient of \( x \) (which is -4) is -2. So we write:
\[ (x – 2)^2 \]Expanding this gives \( x^2 – 4x + 4 \). We need \( +5 \), so we add 1:
\[ (x – 2)^2 – 4 + 5 \] \[ f(x) = (x – 2)^2 + 1 \]✓ (M1 A1)
Part (b): Coordinates of P and Q
💡 What this tells us:
- \( P \) is the \( y \)-intercept, found where \( x = 0 \).
- \( Q \) is the minimum point. In the form \( (x+a)^2 + b \), the minimum is at \( (-a, b) \).
(i) Point P:
Substitute \( x = 0 \) into \( f(x) \):
\[ f(0) = 0^2 – 4(0) + 5 = 5 \]So, \( P = (0, 5) \)
✓ (B1)
(ii) Point Q:
From part (a), \( f(x) = (x – 2)^2 + 1 \).
The square term is minimum when \( x – 2 = 0 \), so \( x = 2 \). The minimum value is 1.
So, \( Q = (2, 1) \)
✓ (B1)
🎯 Final Answer:
(a) \( (x – 2)^2 + 1 \)
(b)(i) \( (0, 5) \)
(b)(ii) \( (2, 1) \)
Question 3 (6 marks)
The sequence \( u_1, u_2, u_3, \dots \) is defined by
\[ u_{n+1} = k – \frac{24}{u_n} \quad u_1 = 2 \]where \( k \) is an integer.
Given that \( u_1 + 2u_2 + u_3 = 0 \)
(a) show that
\[ 3k^2 – 58k + 240 = 0 \](b) Find the value of \( k \), giving a reason for your answer.
(c) Find the value of \( u_3 \)
📝 Worked Solution
Part (a): Forming the Equation
💡 Why we do this: We need to express \( u_2 \) and \( u_3 \) in terms of \( k \) using the recurrence formula, then substitute them into the given equation.
Calculate \( u_2 \):
\[ u_2 = k – \frac{24}{u_1} = k – \frac{24}{2} = k – 12 \]Calculate \( u_3 \):
\[ u_3 = k – \frac{24}{u_2} = k – \frac{24}{k – 12} \]✓ (M1)
Substitute into \( u_1 + 2u_2 + u_3 = 0 \):
\[ 2 + 2(k – 12) + \left( k – \frac{24}{k – 12} \right) = 0 \]Simplify:
\[ 2 + 2k – 24 + k – \frac{24}{k – 12} = 0 \] \[ 3k – 22 – \frac{24}{k – 12} = 0 \]Multiply through by \( (k – 12) \) to remove the fraction:
\[ (3k – 22)(k – 12) – 24 = 0 \] \[ 3k^2 – 36k – 22k + 264 – 24 = 0 \] \[ 3k^2 – 58k + 240 = 0 \]✓ (dM1 A1)
Part (b): Finding k
💡 How we do it: Solve the quadratic equation. We are told \( k \) is an integer, which will help us choose the correct root.
Solve \( 3k^2 – 58k + 240 = 0 \):
Using the quadratic formula or calculator:
\[ k = 6 \quad \text{or} \quad k = \frac{40}{3} \]✓ (M1)
Since \( k \) is an integer, we must have \( k = 6 \).
Reason: \( \frac{40}{3} \) is not an integer.
✓ (A1)
Part (c): Finding u3
Substitute \( k = 6 \) into the expression for \( u_3 \):
\[ u_3 = k – \frac{24}{k – 12} \] \[ u_3 = 6 – \frac{24}{6 – 12} \] \[ u_3 = 6 – \frac{24}{-6} \] \[ u_3 = 6 – (-4) = 10 \]✓ (B1)
🎯 Final Answer:
(a) Proof complete.
(b) \( k = 6 \) (because \( k \) is an integer)
(c) \( u_3 = 10 \)
Question 4 (9 marks)
The curve with equation \( y = f(x) \) where
\[ f(x) = x^2 + \ln(2x^2 – 4x + 5) \]has a single turning point at \( x = \alpha \).
(a) Show that \( \alpha \) is a solution of the equation
\[ 2x^3 – 4x^2 + 7x – 2 = 0 \]The iterative formula
\[ x_{n+1} = \frac{1}{7}(2 + 4x_n^2 – 2x_n^3) \]is used to find an approximate value for \( \alpha \).
Starting with \( x_1 = 0.3 \)
(b) calculate, giving each answer to 4 decimal places,
(i) the value of \( x_2 \)
(ii) the value of \( x_4 \)
Using a suitable interval and a suitable function that should be stated,
(c) show that \( \alpha \) is 0.341 to 3 decimal places.
📝 Worked Solution
Part (a): Differentiation
💡 Why we do this: A turning point occurs where the derivative \( f'(x) = 0 \). We need to differentiate \( f(x) \) using the chain rule for the logarithm term.
Differentiate \( f(x) = x^2 + \ln(2x^2 – 4x + 5) \):
\[ f'(x) = 2x + \frac{1}{2x^2 – 4x + 5} \times (4x – 4) \] \[ f'(x) = 2x + \frac{4x – 4}{2x^2 – 4x + 5} \]✓ (M1 A1)
Set \( f'(x) = 0 \) for the turning point:
\[ 2x + \frac{4x – 4}{2x^2 – 4x + 5} = 0 \]Multiply through by \( (2x^2 – 4x + 5) \):
\[ 2x(2x^2 – 4x + 5) + 4x – 4 = 0 \] \[ 4x^3 – 8x^2 + 10x + 4x – 4 = 0 \] \[ 4x^3 – 8x^2 + 14x – 4 = 0 \]Divide by 2:
\[ 2x^3 – 4x^2 + 7x – 2 = 0 \]✓ (dM1 A1)
Part (b): Iteration
Using the formula \( x_{n+1} = \frac{1}{7}(2 + 4x_n^2 – 2x_n^3) \) with \( x_1 = 0.3 \):
(i) Find \( x_2 \):
\[ x_2 = \frac{1}{7}(2 + 4(0.3)^2 – 2(0.3)^3) \] \[ x_2 = \frac{1}{7}(2 + 0.36 – 0.054) \] \[ x_2 = \frac{2.306}{7} \approx 0.329428… \]To 4 d.p.: \( x_2 = 0.3294 \)
✓ (M1 A1)
(ii) Find \( x_4 \):
Calculate \( x_3 \):
\[ x_3 = \frac{1}{7}(2 + 4(0.3294…)^2 – 2(0.3294…)^3) \approx 0.3375… \]Calculate \( x_4 \):
\[ x_4 = \frac{1}{7}(2 + 4(0.3375…)^2 – 2(0.3375…)^3) \approx 0.3398… \]To 4 d.p.: \( x_4 = 0.3398 \)
✓ (A1)
Part (c): Sign Change Method
💡 Why we do this: To show a root exists between two values, we show the function changes sign between those bounds. For 0.341 to 3 d.p., the bounds are 0.3405 and 0.3415.
Let \( g(x) = 2x^3 – 4x^2 + 7x – 2 \).
Evaluate at lower bound \( x = 0.3405 \):
\[ g(0.3405) = 2(0.3405)^3 – 4(0.3405)^2 + 7(0.3405) – 2 \] \[ g(0.3405) \approx -0.0013 \]Evaluate at upper bound \( x = 0.3415 \):
\[ g(0.3415) = 2(0.3415)^3 – 4(0.3415)^2 + 7(0.3415) – 2 \] \[ g(0.3415) \approx 0.0037 \]✓ (M1)
Conclusion:
There is a change of sign between \( x = 0.3405 \) and \( x = 0.3415 \). Since \( g(x) \) is continuous, there is a root \( \alpha \) in this interval. Therefore, \( \alpha = 0.341 \) to 3 decimal places.
✓ (A1)
Question 5 (6 marks)
A company made a profit of £20,000 in its first year of trading, Year 1.
A model for future trading predicts that the yearly profit will increase by 8% each year, so that the yearly profits will form a geometric sequence.
According to the model,
(a) show that the profit for Year 3 will be £23,328.
(b) find the first year when the yearly profit will exceed £65,000.
(c) find the total profit for the first 20 years of trading, giving your answer to the nearest £1000.
📝 Worked Solution
Part (a): Term Calculation
The sequence is geometric with \( a = 20000 \). An 8% increase means the common ratio \( r = 1.08 \).
For Year 3, we calculate the 3rd term:
\[ u_3 = ar^2 \] \[ u_3 = 20000 \times (1.08)^2 \] \[ u_3 = 20000 \times 1.1664 = 23328 \]Profit = £23,328.
✓ (B1)
Part (b): Solving Inequality
💡 Why we do this: We need to find the smallest integer \( n \) such that the \( n \)-th term is greater than 65000. We use logarithms to solve for \( n \).
✓ (M1)
Take logarithms of both sides:
\[ (n-1)\ln(1.08) > \ln(3.25) \] \[ n-1 > \frac{\ln(3.25)}{\ln(1.08)} \] \[ n-1 > 15.305… \] \[ n > 16.305… \]✓ (M1)
The first integer year is 17.
Answer: Year 17.
✓ (A1)
Part (c): Geometric Series Sum
💡 Formula: Sum of a geometric series is \( S_n = \frac{a(r^n – 1)}{r – 1} \).
Using \( n = 20 \), \( a = 20000 \), \( r = 1.08 \):
\[ S_{20} = \frac{20000(1.08^{20} – 1)}{1.08 – 1} \] \[ S_{20} = \frac{20000(4.6609… – 1)}{0.08} \] \[ S_{20} = \frac{73219…}{0.08} \approx 915239.28 \]✓ (M1)
Rounding to the nearest £1000:
£915,000
✓ (A1)
Question 6 (5 marks)
Figure 1 shows a sketch of triangle \( ABC \).
Given that
- \( \vec{AB} = -3\mathbf{i} – 4\mathbf{j} – 5\mathbf{k} \)
- \( \vec{BC} = \mathbf{i} + \mathbf{j} + 4\mathbf{k} \)
(a) find \( \vec{AC} \)
(b) show that \( \cos \angle ABC = \frac{9}{10} \)
📝 Worked Solution
Part (a): Vector Addition
💡 Method: Following the path from A to C: \( \vec{AC} = \vec{AB} + \vec{BC} \).
✓ (M1 A1)
Part (b): Angle Between Vectors
💡 Method: To find angle \( ABC \) (at vertex B), we need the dot product of the vectors originating from B: \( \vec{BA} \) and \( \vec{BC} \). Note that \( \vec{BA} = -\vec{AB} \).
Vectors from B:
\[ \vec{BA} = -(-3\mathbf{i} – 4\mathbf{j} – 5\mathbf{k}) = 3\mathbf{i} + 4\mathbf{j} + 5\mathbf{k} \] \[ \vec{BC} = \mathbf{i} + \mathbf{j} + 4\mathbf{k} \]Find the magnitudes:
\[ |\vec{BA}| = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9+16+25} = \sqrt{50} \] \[ |\vec{BC}| = \sqrt{1^2 + 1^2 + 4^2} = \sqrt{1+1+16} = \sqrt{18} \]✓ (M1)
Calculate Dot Product:
\[ \vec{BA} \cdot \vec{BC} = (3)(1) + (4)(1) + (5)(4) \] \[ \vec{BA} \cdot \vec{BC} = 3 + 4 + 20 = 27 \]Use formula \( \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}| \cos \theta \):
\[ 27 = \sqrt{50}\sqrt{18} \cos(ABC) \] \[ 27 = \sqrt{900} \cos(ABC) \] \[ 27 = 30 \cos(ABC) \] \[ \cos(ABC) = \frac{27}{30} = \frac{9}{10} \]✓ (M1 A1)
Question 7 (9 marks)
The circle \( C \) has equation
\[ x^2 + y^2 – 10x + 4y + 11 = 0 \](a) Find
(i) the coordinates of the centre of \( C \),
(ii) the exact radius of \( C \), giving your answer as a simplified surd.
The line \( l \) has equation \( y = 3x + k \) where \( k \) is a constant.
Given that \( l \) is a tangent to \( C \),
(b) find the possible values of \( k \), giving your answers as simplified surds.
📝 Worked Solution
Part (a): Centre and Radius
Complete the square for \( x \) and \( y \):
\[ (x – 5)^2 – 25 + (y + 2)^2 – 4 + 11 = 0 \]✓ (M1)
\[ (x – 5)^2 + (y + 2)^2 – 18 = 0 \] \[ (x – 5)^2 + (y + 2)^2 = 18 \](i) Centre: \( (5, -2) \)
✓ (A1)
(ii) Radius: \( r = \sqrt{18} = 3\sqrt{2} \)
✓ (M1 A1)
Part (b): Tangent Condition
💡 Strategy: Substitute the line equation into the circle equation to form a quadratic. For a tangent, the discriminant \( b^2 – 4ac \) must be zero.
Substitute \( y = 3x + k \) into \( x^2 + y^2 – 10x + 4y + 11 = 0 \):
\[ x^2 + (3x + k)^2 – 10x + 4(3x + k) + 11 = 0 \] \[ x^2 + (9x^2 + 6kx + k^2) – 10x + 12x + 4k + 11 = 0 \]Group terms (M1):
\[ 10x^2 + (6k + 2)x + (k^2 + 4k + 11) = 0 \]✓ (A1)
Set Discriminant to 0 (\( b^2 – 4ac = 0 \)):
\[ (6k + 2)^2 – 4(10)(k^2 + 4k + 11) = 0 \] \[ 36k^2 + 24k + 4 – 40(k^2 + 4k + 11) = 0 \] \[ 36k^2 + 24k + 4 – 40k^2 – 160k – 440 = 0 \] \[ -4k^2 – 136k – 436 = 0 \]Divide by -4:
\[ k^2 + 34k + 109 = 0 \]✓ (M1)
Solve using quadratic formula:
\[ k = \frac{-34 \pm \sqrt{34^2 – 4(1)(109)}}{2} \] \[ k = \frac{-34 \pm \sqrt{1156 – 436}}{2} \] \[ k = \frac{-34 \pm \sqrt{720}}{2} \]Simplify surd \( \sqrt{720} = \sqrt{144 \times 5} = 12\sqrt{5} \):
\[ k = \frac{-34 \pm 12\sqrt{5}}{2} \] \[ k = -17 \pm 6\sqrt{5} \]✓ (A1)
Question 8 (9 marks)
A scientist is studying the growth of two different populations of bacteria.
The number of bacteria, \( N \), in the first population is modelled by the equation
\[ N = Ae^{kt} \quad t \ge 0 \]where \( A \) and \( k \) are positive constants and \( t \) is the time in hours.
Given that
- there were 1000 bacteria in this population at the start of the study
- it took exactly 5 hours for this population to double
(a) find a complete equation for the model.
(b) Hence find the rate of increase in the number of bacteria exactly 8 hours from the start. Give your answer to 2 significant figures.
The second population \( M \) is modelled by \( M = 500e^{1.4kt} \).
(c) Given that \( T \) hours after the start, the number of bacteria in both populations was the same, find the value of \( T \).
📝 Worked Solution
Part (a): Constants A and k
At \( t = 0 \), \( N = 1000 \):
\[ 1000 = Ae^0 \Rightarrow A = 1000 \]✓ (B1)
At \( t = 5 \), population doubles to 2000:
\[ 2000 = 1000e^{5k} \] \[ 2 = e^{5k} \]Take ln:
\[ \ln 2 = 5k \Rightarrow k = \frac{1}{5}\ln 2 \]✓ (M1 M1)
Equation:
\[ N = 1000e^{\frac{1}{5}(\ln 2)t} \quad \text{or} \quad N = 1000e^{0.139t} \]✓ (A1)
Part (b): Rate of Increase
💡 Method: Rate of increase is the derivative \( \frac{dN}{dt} \).
Substitute \( t = 8 \) and \( k = 0.1386… \):
\[ \frac{dN}{dt} = 1000(0.1386)e^{0.1386 \times 8} \] \[ \frac{dN}{dt} \approx 138.6 \times 3.03… \approx 420 \]To 2 s.f.: 420
✓ (M1 A1)
Part (c): Equating Populations
Divide by 500:
\[ 2e^{kT} = e^{1.4kT} \]Take ln:
\[ \ln 2 + kT = 1.4kT \] \[ \ln 2 = 0.4kT \]Substitute \( k = \frac{1}{5}\ln 2 \):
\[ \ln 2 = 0.4 \left( \frac{1}{5}\ln 2 \right) T \] \[ 1 = \frac{0.4}{5} T \] \[ 1 = 0.08 T \] \[ T = \frac{1}{0.08} = 12.5 \]✓ (M1 M1 A1)
Question 9 (11 marks)
Given that \( f(x) \) can be expressed in the form
\[ \frac{A}{5x + 2} + \frac{B}{(5x + 2)^2} + \frac{C}{1 – 2x} \](a) (i) find the value of \( B \) and the value of \( C \).
(ii) show that \( A = 0 \).
(b) (i) Use binomial expansions to show that, in ascending powers of \( x \):
\[ f(x) = p + qx + rx^2 + \dots \](ii) Find the valid range.
📝 Worked Solution
Part (a): Partial Fractions
💡 Method: Multiply by the denominator to form an identity, then substitute strategic values of \( x \).
(i) Find B and C:
Let \( x = -\frac{2}{5} \) (to eliminate A and C):
\[ 50(-\frac{2}{5})^2 + 38(-\frac{2}{5}) + 9 = B(1 – 2(-\frac{2}{5})) \] \[ 8 – 15.2 + 9 = B(1 + 0.8) \] \[ 1.8 = 1.8B \Rightarrow B = 1 \]Let \( x = \frac{1}{2} \) (to eliminate A and B):
\[ 50(0.5)^2 + 38(0.5) + 9 = C(5(0.5) + 2)^2 \] \[ 12.5 + 19 + 9 = C(4.5)^2 \] \[ 40.5 = 20.25C \Rightarrow C = 2 \]✓ (M1 A1)
(ii) Show A = 0:
Compare coefficients of \( x^2 \):
LHS: 50
RHS: \( A(5)(-2) + C(25) = -10A + 25C \)
\[ 50 = -10A + 25(2) \] \[ 50 = -10A + 50 \] \[ 0 = -10A \Rightarrow A = 0 \]✓ (M1 A1)
Part (b): Binomial Expansion
We have \( f(x) = \frac{1}{(5x+2)^2} + \frac{2}{1-2x} \).
Rewrite as: \( (5x+2)^{-2} + 2(1-2x)^{-1} \).
Expand \( (5x+2)^{-2} \):
Factor out \( 2^{-2} \): \( 2^{-2}(1 + \frac{5}{2}x)^{-2} = \frac{1}{4}(1 + \frac{5}{2}x)^{-2} \)
\[ \frac{1}{4} \left[ 1 + (-2)(\frac{5x}{2}) + \frac{(-2)(-3)}{2}(\frac{5x}{2})^2 + \dots \right] \] \[ \frac{1}{4} \left[ 1 – 5x + \frac{75}{4}x^2 \right] = \frac{1}{4} – \frac{5}{4}x + \frac{75}{16}x^2 \]Expand \( 2(1-2x)^{-1} \):
\[ 2 \left[ 1 + (-1)(-2x) + \frac{(-1)(-2)}{2}(-2x)^2 + \dots \right] \] \[ 2 \left[ 1 + 2x + 4x^2 \right] = 2 + 4x + 8x^2 \]Add them together:
\[ (\frac{1}{4} + 2) + (-\frac{5}{4} + 4)x + (\frac{75}{16} + 8)x^2 \] \[ \frac{9}{4} + \frac{11}{4}x + \frac{203}{16}x^2 \]✓ (M1 M1 A1)
(ii) Range:
From \( (1 + \frac{5}{2}x) \), valid if \( |\frac{5x}{2}| < 1 \Rightarrow |x| < \frac{2}{5} \).
From \( (1 – 2x) \), valid if \( |-2x| < 1 \Rightarrow |x| < \frac{1}{2} \).
Strict intersection: \( |x| < \frac{2}{5} \).
✓ (B1)
Question 10 (8 marks)
(a) Given that \( 1 + \cos 2\theta + \sin 2\theta \neq 0 \), prove that
\[ \frac{1 – \cos 2\theta + \sin 2\theta}{1 + \cos 2\theta + \sin 2\theta} \equiv \tan \theta \](b) Hence solve, for \( 0 < x < 180^\circ \)
\[ \frac{1 – \cos 4x + \sin 4x}{1 + \cos 4x + \sin 4x} = 3\sin 2x \]giving your answers to one decimal place where appropriate.
📝 Worked Solution
Part (a): Trigonometric Identity
💡 Strategy: Use double angle formulae to convert \( 2\theta \) terms into single \( \theta \) terms.
Key formulae:
\( \cos 2\theta = 1 – 2\sin^2\theta \) (to cancel the 1 in numerator)
\( \cos 2\theta = 2\cos^2\theta – 1 \) (to cancel the 1 in denominator)
\( \sin 2\theta = 2\sin\theta\cos\theta \)
Numerator:
\[ 1 – (1 – 2\sin^2\theta) + 2\sin\theta\cos\theta \] \[ 2\sin^2\theta + 2\sin\theta\cos\theta \] \[ 2\sin\theta(\sin\theta + \cos\theta) \]Denominator:
\[ 1 + (2\cos^2\theta – 1) + 2\sin\theta\cos\theta \] \[ 2\cos^2\theta + 2\sin\theta\cos\theta \] \[ 2\cos\theta(\cos\theta + \sin\theta) \]Fraction:
\[ \frac{2\sin\theta(\sin\theta + \cos\theta)}{2\cos\theta(\cos\theta + \sin\theta)} \]Cancel common terms:
\[ \frac{\sin\theta}{\cos\theta} \equiv \tan\theta \]✓ (M1 A1 dM1 A1)
Part (b): Solving the Equation
Using part (a) with \( \theta = 2x \), the LHS becomes \( \tan 2x \).
\[ \tan 2x = 3\sin 2x \] \[ \frac{\sin 2x}{\cos 2x} = 3\sin 2x \]Rearrange (do not just divide by sin):
\[ \sin 2x – 3\sin 2x \cos 2x = 0 \] \[ \sin 2x (1 – 3\cos 2x) = 0 \]✓ (M1)
Case 1: \( \sin 2x = 0 \)
\[ 2x = 180^\circ \Rightarrow x = 90^\circ \](0 and 360 are outside range)
Case 2: \( \cos 2x = \frac{1}{3} \)
\[ 2x = \cos^{-1}(1/3) \approx 70.52^\circ \]Also \( 360 – 70.52 = 289.47^\circ \)
Divide by 2:
\[ x \approx 35.3^\circ, 144.7^\circ \]Final Answers:
\[ x = 35.3^\circ, 90^\circ, 144.7^\circ \]✓ (A1 A1)
Question 11 (8 marks)
Figure 2 shows a sketch of part of the curve with equation \( y = (\ln x)^2, \quad x > 0 \).
The finite region \( R \) is bounded by the curve, the line \( x = 2 \), the \( x \)-axis and the line \( x = 4 \).
The table below shows corresponding values of \( x \) and \( y \):
| x | 2 | 2.5 | 3 | 3.5 | 4 |
| y | 0.4805 | 0.8396 | 1.2069 | 1.5694 | 1.9218 |
(a) Use the trapezium rule with all values in the table to obtain an estimate for the area of \( R \) to 3 significant figures.
(b) Use algebraic integration to find the exact area of \( R \), giving your answer in the form \( y = a(\ln 2)^2 + b\ln 2 + c \).
📝 Worked Solution
Part (a): Trapezium Rule
💡 Formula: \( A \approx \frac{h}{2} [y_0 + y_n + 2(y_1 + \dots + y_{n-1})] \)
Here \( h = 0.5 \).
Round to 3 significant figures:
2.41
✓ (B1 M1 A1)
Part (b): Exact Integration
💡 Strategy: Integrate \( \int (\ln x)^2 dx \) using Integration by Parts. Let \( u = (\ln x)^2 \) and \( \frac{dv}{dx} = 1 \).
First application of Parts:
\[ u = (\ln x)^2 \Rightarrow \frac{du}{dx} = 2(\ln x) \cdot \frac{1}{x} \] \[ \frac{dv}{dx} = 1 \Rightarrow v = x \] \[ \int (\ln x)^2 dx = x(\ln x)^2 – \int x \cdot \frac{2\ln x}{x} dx \] \[ = x(\ln x)^2 – 2 \int \ln x \, dx \]Second application (Standard Integral):
We know \( \int \ln x \, dx = x\ln x – x \).
\[ \int (\ln x)^2 dx = x(\ln x)^2 – 2(x\ln x – x) \] \[ = x(\ln x)^2 – 2x\ln x + 2x \]✓ (M1 A1)
Apply Limits 2 to 4:
\[ \left[ x(\ln x)^2 – 2x\ln x + 2x \right]_2^4 \]Upper limit (4):
\[ 4(\ln 4)^2 – 8\ln 4 + 8 \]Since \( \ln 4 = \ln(2^2) = 2\ln 2 \):
\[ 4(2\ln 2)^2 – 8(2\ln 2) + 8 = 16(\ln 2)^2 – 16\ln 2 + 8 \]Lower limit (2):
\[ 2(\ln 2)^2 – 4\ln 2 + 4 \]Subtract Lower from Upper:
\[ (16(\ln 2)^2 – 16\ln 2 + 8) – (2(\ln 2)^2 – 4\ln 2 + 4) \] \[ = 14(\ln 2)^2 – 12\ln 2 + 4 \]✓ (dM1 ddM1 A1)
Question 12 (9 marks)
Figure 3 is a graph of the trajectory of a golf ball.
The ball is hit from a height of 3m. It reaches maximum height at horizontal distance 90m. It is at height 27m at horizontal distance 120m.
Model \( H \) as a quadratic function in \( x \).
(a) Find \( H \) in terms of \( x \).
(b) Find (i) the maximum height, (ii) horizontal distance travelled (to nearest metre).
(c) Give one limitation of the model (other than wind/air resistance).
📝 Worked Solution
Part (a): Finding the Equation
💡 Strategy: Since the maximum is at \( x = 90 \), we can use the vertex form of a quadratic: \( H = a(x – 90)^2 + k \), where \( k \) is the max height.
We have two points: \( (0, 3) \) and \( (120, 27) \).
1. Substitute \( (0, 3) \):
\[ 3 = a(0 – 90)^2 + k \Rightarrow 3 = 8100a + k \quad \text{(Eq 1)} \]2. Substitute \( (120, 27) \):
\[ 27 = a(120 – 90)^2 + k \Rightarrow 27 = 900a + k \quad \text{(Eq 2)} \]Subtract Eq 2 from Eq 1:
\[ 3 – 27 = (8100a + k) – (900a + k) \] \[ -24 = 7200a \] \[ a = -\frac{24}{7200} = -\frac{1}{300} \]✓ (M1)
Find \( k \) using Eq 2:
\[ 27 = 900(-\frac{1}{300}) + k \] \[ 27 = -3 + k \Rightarrow k = 30 \]Equation:
\[ H = -\frac{1}{300}(x – 90)^2 + 30 \]✓ (dM1 A1)
Part (b): Max Height and Distance
(i) Max Height: This is the value of \( k \) from part (a).
Max height = 30 m
✓ (B1)
(ii) Horizontal Distance:
The ball hits the ground when \( H = 0 \).
\[ 0 = -\frac{1}{300}(x – 90)^2 + 30 \] \[ \frac{1}{300}(x – 90)^2 = 30 \] \[ (x – 90)^2 = 9000 \] \[ x – 90 = \pm \sqrt{9000} \] \[ x = 90 \pm \sqrt{9000} \] \[ x = 90 \pm 94.868… \]Since \( x > 0 \), \( x = 184.868… \)
Nearest metre: 185 m
✓ (M1 A1)
Part (c): Limitations
Limitations include:
- The ball is modelled as a particle (no size/spin).
- The ground might not be flat.
✓ (B1)
Question 13 (3 marks)
A curve \( C \) has parametric equations
\[ x = \frac{t^2 + 5}{t^2 + 1}, \quad y = \frac{4t}{t^2 + 1} \quad t \in \mathbb{R} \]Show that all points on \( C \) satisfy
\[ (x – 3)^2 + y^2 = 4 \]📝 Worked Solution
Step 1: Substitute and Simplify
💡 Strategy: Substitute the expressions for \( x \) and \( y \) directly into the LHS of the Cartesian equation and simplify to match the RHS.
LHS: \( (x – 3)^2 + y^2 \)
First, simplify \( x – 3 \):
\[ x – 3 = \frac{t^2 + 5}{t^2 + 1} – 3 \] \[ = \frac{t^2 + 5 – 3(t^2 + 1)}{t^2 + 1} \] \[ = \frac{t^2 + 5 – 3t^2 – 3}{t^2 + 1} \] \[ = \frac{2 – 2t^2}{t^2 + 1} \]✓ (M1)
Now square it:
\[ (x – 3)^2 = \left( \frac{2(1 – t^2)}{t^2 + 1} \right)^2 = \frac{4(1 – t^2)^2}{(t^2 + 1)^2} \]Square \( y \):
\[ y^2 = \left( \frac{4t}{t^2 + 1} \right)^2 = \frac{16t^2}{(t^2 + 1)^2} \]Add them:
\[ (x – 3)^2 + y^2 = \frac{4(1 – 2t^2 + t^4) + 16t^2}{(t^2 + 1)^2} \] \[ = \frac{4 – 8t^2 + 4t^4 + 16t^2}{(t^2 + 1)^2} \] \[ = \frac{4t^4 + 8t^2 + 4}{(t^2 + 1)^2} \]Factor out 4:
\[ = \frac{4(t^4 + 2t^2 + 1)}{(t^2 + 1)^2} \]Recognise perfect square:
\[ = \frac{4(t^2 + 1)^2}{(t^2 + 1)^2} \] \[ = 4 \]RHS = 4. Proven.
✓ (M1 A1)
Question 14 (4 marks)
Given that
\[ y = \frac{x – 4}{2 + \sqrt{x}} \quad x > 0 \]show that
\[ \frac{dy}{dx} = \frac{1}{A\sqrt{x}} \]where \( A \) is a constant to be found.
📝 Worked Solution
Step 1: Simplify the Expression
💡 Insight: Instead of using the quotient rule immediately, notice that the numerator \( x – 4 \) is a difference of two squares: \( (\sqrt{x})^2 – 2^2 \).
Substitute this into the equation for \( y \):
\[ y = \frac{(\sqrt{x} – 2)(\sqrt{x} + 2)}{2 + \sqrt{x}} \]Cancel the common term \( (\sqrt{x} + 2) \):
\[ y = \sqrt{x} – 2 \] \[ y = x^{1/2} – 2 \]✓ (M1 A1)
Step 2: Differentiate
Now differentiate \( y = x^{1/2} – 2 \):
\[ \frac{dy}{dx} = \frac{1}{2}x^{-1/2} \] \[ \frac{dy}{dx} = \frac{1}{2\sqrt{x}} \]This matches the form \( \frac{1}{A\sqrt{x}} \) with \( A = 2 \).
✓ (M1 A1)
Note: You could also use the Quotient Rule, but simplifying first is much faster.
Question 15 (6 marks)
(i) Use proof by exhaustion to show that for \( n \in \mathbb{N}, n \le 4 \)
\[ (n + 1)^3 > 3^n \](ii) Given that \( m^3 + 5 \) is odd, use proof by contradiction to show, using algebra, that \( m \) is even.
📝 Worked Solution
Part (i): Proof by Exhaustion
💡 Method: Check the statement for every integer \( n \) from 1 to 4.
n = 1:
\[ (1+1)^3 = 2^3 = 8 \] \[ 3^1 = 3 \] \[ 8 > 3 \quad \text{(True)} \]n = 2:
\[ (2+1)^3 = 3^3 = 27 \] \[ 3^2 = 9 \] \[ 27 > 9 \quad \text{(True)} \]n = 3:
\[ (3+1)^3 = 4^3 = 64 \] \[ 3^3 = 27 \] \[ 64 > 27 \quad \text{(True)} \]n = 4:
\[ (4+1)^3 = 5^3 = 125 \] \[ 3^4 = 81 \] \[ 125 > 81 \quad \text{(True)} \]The statement holds for all \( n \in \{1, 2, 3, 4\} \).
✓ (M1 A1)
Part (ii): Proof by Contradiction
💡 Strategy: To prove “if \( m^3 + 5 \) is odd then \( m \) is even” by contradiction, we assume the opposite: assume \( m \) is odd.
Assumption: Assume \( m \) is odd.
If \( m \) is odd, we can write \( m = 2k + 1 \) for some integer \( k \).
✓ (M1)
Now consider \( m^3 + 5 \):
\[ m^3 + 5 = (2k + 1)^3 + 5 \] \[ = (8k^3 + 12k^2 + 6k + 1) + 5 \] \[ = 8k^3 + 12k^2 + 6k + 6 \]✓ (M1)
Factor out 2:
\[ = 2(4k^3 + 6k^2 + 3k + 3) \]Since this expression has a factor of 2, \( m^3 + 5 \) must be even.
✓ (A1)
Conclusion:
This contradicts the given information that \( m^3 + 5 \) is odd. Therefore, the assumption that \( m \) is odd is false. Hence, \( m \) must be even.
✓ (A1)
🎉 Exam Complete!
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