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A Level Pure Mathematics 1 (Summer 2022)
Mark Scheme Legend
- M1: Method mark (knowing a method and attempting to apply it)
- A1: Accuracy mark (dependent on method mark)
- B1: Unconditional accuracy mark
- ft: Follow through
- awrt: Answers which round to
Table of Contents
- Question 1 (Transformations)
- Question 2 (Factor Theorem)
- Question 3 (Circle Geometry)
- Question 4 (Integration as Limit)
- Question 5 (Modelling)
- Question 6 (Cubic Graphs)
- Question 7 (Proof)
- Question 8 (Kinematics/Logarithms)
- Question 9 (Vectors)
- Question 10 (Exponential Growth)
- Question 11 (Intersection of Curves)
- Question 12 (Integration)
- Question 13 (Arithmetic Series)
- Question 14 (Trigonometry)
- Question 15 (Differentiation/Optimization)
- Question 16 (Parametric Integration)
Question 1 (4 marks)
The point \( P(-2, -5) \) lies on the curve with equation \( y = f(x) \), \( x \in \mathbb{R} \).
Find the point to which \( P \) is mapped, when the curve with equation \( y = f(x) \) is transformed to the curve with equation
- \( y = f(x) + 2 \) (1)
- \( y = |f(x)| \) (1)
- \( y = 3f(x – 2) + 2 \) (2)
Worked Solution
Part (a): Vertical Translation
What this tells us: The transformation \( y = f(x) + a \) represents a vertical translation by the vector \( \begin{pmatrix} 0 \\ a \end{pmatrix} \).
How: We add 2 to the y-coordinate of \( P(-2, -5) \).
The x-coordinate remains unchanged.
Point: \( (-2, -3) \) (B1)
Part (b): Modulus of the Function
What this tells us: The transformation \( y = |f(x)| \) makes all negative y-values positive (reflects the part of the curve below the x-axis in the x-axis).
How: We take the absolute value of the y-coordinate.
The x-coordinate remains unchanged.
Point: \( (-2, 5) \) (B1)
Part (c): Combined Transformation
What this tells us: We have a sequence of transformations for \( y = 3f(x – 2) + 2 \):
- \( f(x – 2) \): Translation in x (inside the bracket affects x).
- \( 3f(…) \): Vertical stretch.
- \( … + 2 \): Vertical translation.
How:
- \( x \to x + 2 \) (Shift right by 2): \( -2 + 2 = 0 \)
- \( y \to 3y \): \( 3 \times (-5) = -15 \)
- \( y \to y + 2 \): \( -15 + 2 = -13 \)
x-coordinate: \( -2 + 2 = 0 \) (M1)
y-coordinate: \( 3(-5) + 2 = -15 + 2 = -13 \)
Point: \( (0, -13) \) (A1)
Final Answers:
- \( (-2, -3) \)
- \( (-2, 5) \)
- \( (0, -13) \)
Question 2 (3 marks)
\( f(x) = (x – 4)(x^2 – 3x + k) – 42 \) where \( k \) is a constant.
Given that \( (x + 2) \) is a factor of \( f(x) \), find the value of \( k \).
Worked Solution
Step 1: Understanding the Factor Theorem
Why: The Factor Theorem states that if \( (x – a) \) is a factor of a polynomial \( f(x) \), then \( f(a) = 0 \).
How: Here, the factor is \( (x + 2) \), which corresponds to \( x = -2 \). We need to set \( f(-2) = 0 \) and solve for \( k \).
Step 2: Substitution
How: Substitute \( x = -2 \) into the expression for \( f(x) \).
(M1)
Step 3: Solving for k
How: Set the expression to zero and solve the linear equation.
Add 42 to both sides:
\[ -6(10 + k) = 42 \]Divide by -6:
\[ 10 + k = -7 \](M1)
Subtract 10:
\[ k = -17 \]Final Answer:
\( k = -17 \) (A1)
Question 3 (5 marks)
A circle has equation
\[ x^2 + y^2 – 10x + 16y = 80 \](a) Find
(i) the coordinates of the centre of the circle,
(ii) the radius of the circle.
(3)
Given that \( P \) is the point on the circle that is furthest away from the origin \( O \),
(b) find the exact length \( OP \).
(2)
Worked Solution
Part (a): Completing the Square
Why: The standard equation of a circle is \( (x-a)^2 + (y-b)^2 = r^2 \), where \( (a,b) \) is the centre and \( r \) is the radius. We need to complete the square for both x and y terms.
Original equation: \( x^2 – 10x + y^2 + 16y = 80 \)
Complete the square for x: \( (x – 5)^2 – 25 \)
Complete the square for y: \( (y + 8)^2 – 64 \)
Substitute back:
\[ (x – 5)^2 – 25 + (y + 8)^2 – 64 = 80 \](M1)
\[ (x – 5)^2 + (y + 8)^2 = 80 + 25 + 64 \] \[ (x – 5)^2 + (y + 8)^2 = 169 \](i) Centre: \( (5, -8) \) (A1)
(ii) Radius: \( r = \sqrt{169} = 13 \) (A1)
Part (b): Furthest Point from Origin
Why: The furthest point on a circle from an external point (the origin) lies on the line passing through the origin and the centre of the circle. The maximum distance is the distance from the origin to the centre plus the radius.
How: Calculate distance \( OC \) where \( C \) is the centre, then add \( r \).
Centre \( C(5, -8) \), Origin \( O(0, 0) \)
\[ OC = \sqrt{(5-0)^2 + (-8-0)^2} \] \[ OC = \sqrt{25 + 64} = \sqrt{89} \](M1)
Max distance \( OP = OC + r \)
\[ OP = \sqrt{89} + 13 \]Final Answer:
(a)(i) \( (5, -8) \)
(a)(ii) \( 13 \)
(b) \( 13 + \sqrt{89} \) (A1ft)
Question 4 (3 marks)
(a) Express \( \displaystyle \lim_{\delta x \to 0} \sum_{x=2.1}^{6.3} \frac{2}{x} \delta x \) as an integral.
(1)
(b) Hence show that \( \displaystyle \lim_{\delta x \to 0} \sum_{x=2.1}^{6.3} \frac{2}{x} \delta x = \ln k \) where \( k \) is a constant to be found.
(2)
Worked Solution
Part (a): Definition of an Integral
What this tells us: The limit of a sum as the strip width \( \delta x \) approaches zero is the definition of the definite integral.
How: The summation limits become the integral bounds, and \( \delta x \) becomes \( dx \).
(B1)
Part (b): Evaluating the Integral
Why: We know that \( \int \frac{1}{x} dx = \ln|x| \).
(M1)
Applying limits:
\[ = 2\ln(6.3) – 2\ln(2.1) \]Factor out 2:
\[ = 2(\ln(6.3) – \ln(2.1)) \]Use log laws \( \ln a – \ln b = \ln(\frac{a}{b}) \):
\[ = 2\ln\left(\frac{6.3}{2.1}\right) \] \[ = 2\ln(3) \]Use log power law \( n\ln a = \ln(a^n) \):
\[ = \ln(3^2) = \ln 9 \]Final Answer:
\( \ln 9 \) (so \( k=9 \)) (A1)
Question 5 (6 marks)
The height, \( h \) metres, of a tree, \( t \) years after being planted, is modelled by the equation
\[ h^2 = at + b \quad 0 \le t < 25 \]where \( a \) and \( b \) are constants.
Given that
- the height of the tree was 2.60 m, exactly 2 years after being planted
- the height of the tree was 5.10 m, exactly 10 years after being planted
(a) find a complete equation for the model, giving the values of \( a \) and \( b \) to 3 significant figures.
(4)
Given that the height of the tree was 7 m, exactly 20 years after being planted
(b) evaluate the model, giving reasons for your answer.
(2)
Worked Solution
Part (a): Form Simultaneous Equations
Why: We have two pairs of values for \( (t, h) \). We can substitute these into the equation to find the two unknowns \( a \) and \( b \).
Data:
- When \( t=2 \), \( h=2.60 \)
- When \( t=10 \), \( h=5.10 \)
Equation 1 (at \( t=2 \)):
\[ 2.60^2 = a(2) + b \] \[ 6.76 = 2a + b \quad \text{(1)} \]Equation 2 (at \( t=10 \)):
\[ 5.10^2 = a(10) + b \] \[ 26.01 = 10a + b \quad \text{(2)} \](A1)
Step 2: Solve Simultaneous Equations
Subtract (1) from (2):
\[ (26.01 – 6.76) = (10a – 2a) + (b – b) \] \[ 19.25 = 8a \] \[ a = \frac{19.25}{8} = 2.40625 \](dM1)
Substitute \( a \) back into (1):
\[ b = 6.76 – 2(2.40625) \] \[ b = 6.76 – 4.8125 = 1.9475 \]Rounding to 3 significant figures:
\[ a \approx 2.41, \quad b \approx 1.95 \]Complete equation:
\[ h^2 = 2.41t + 1.95 \](A1)
Part (b): Evaluate the Model
Why: We need to check if the model predicts the height accurately at \( t=20 \). We calculate the predicted height and compare it to the actual height of 7 m.
Using \( t=20 \):
\[ h^2 = 2.40625(20) + 1.9475 \] (using unrounded values for accuracy) \[ h^2 = 48.125 + 1.9475 = 50.0725 \] \[ h = \sqrt{50.0725} \approx 7.08 \, \text{m} \](M1)
Comparison:
Predicted height \( 7.08 \, \text{m} \) is very close to the actual height \( 7 \, \text{m} \) (approx 1.1% difference).
Therefore, it is a good model.
(A1)
Final Answer:
(a) \( h^2 = 2.41t + 1.95 \)
(b) Predicted \( h \approx 7.08 \text{m} \). This is close to 7m, so the model is suitable.
Question 6 (6 marks)
Figure 1 shows a sketch of a curve \( C \) with equation \( y = f(x) \) where \( f(x) \) is a cubic expression in \( x \).
The curve:
- passes through the origin
- has a maximum turning point at \( (2, 8) \)
- has a minimum turning point at \( (6, 0) \)
(a) Write down the set of values of \( x \) for which \( f'(x) < 0 \)
(1)
The line with equation \( y = k \), where \( k \) is a constant, intersects \( C \) at only one point.
(b) Find the set of values of \( k \), giving your answer in set notation.
(2)
(c) Find the equation of \( C \). You may leave your answer in factorised form.
(3)
Worked Solution
Part (a): Decreasing Function
What this tells us: \( f'(x) < 0 \) means the gradient is negative, or the curve is going downwards.
How: Looking at the graph, the curve goes down between the maximum turning point at \( x=2 \) and the minimum turning point at \( x=6 \).
Range: \( 2 < x < 6 \)
Set notation: \( \{x : 2 < x < 6\} \)
(B1)
Part (b): Intersection with y = k
What this tells us: \( y = k \) is a horizontal line. We want it to cross the curve exactly once.
How: Looking at the sketch:
- If \( 0 < k < 8 \), the line cuts 3 times.
- If \( k = 0 \) or \( k = 8 \), the line cuts/touches 2 times.
- If \( k > 8 \) or \( k < 0 \), the line cuts exactly once.
Values: \( k > 8 \) or \( k < 0 \)
Set notation: \( \{k : k < 0\} \cup \{k : k > 8\} \) (A1)
Part (c): Equation of the Cubic
Strategy: We know the roots (x-intercepts). The curve touches the x-axis at \( x=6 \), meaning it’s a repeated root (tangent). It passes through the origin, so \( x=0 \) is a root.
Form: \( y = Ax(x-6)^2 \)
Equation form: \( y = Ax(x-6)^2 \)
Use the maximum point \( (2, 8) \) to find \( A \):
\[ 8 = A(2)(2-6)^2 \] \[ 8 = 2A(-4)^2 \] \[ 8 = 2A(16) \] \[ 8 = 32A \] \[ A = \frac{8}{32} = \frac{1}{4} \](M1)
Equation:
\[ y = \frac{1}{4}x(x-6)^2 \](A1)
Final Answer:
(a) \( 2 < x < 6 \)
(b) \( \{k : k < 0\} \cup \{k : k > 8\} \)
(c) \( y = \frac{1}{4}x(x-6)^2 \)
Question 7 (5 marks)
(i) Given that \( p \) and \( q \) are integers such that \( pq \) is even, use algebra to prove by contradiction that at least one of \( p \) or \( q \) is even.
(3)
(ii) Given that \( x \) and \( y \) are integers such that:
- \( x < 0 \)
- \( (x + y)^2 < 9x^2 + y^2 \)
show that \( y > 4x \)
(2)
Worked Solution
Part (i): Proof by Contradiction
Why: To prove by contradiction, we assume the opposite of the statement is true and show it leads to a mathematical impossibility.
Assumption: Assume that neither \( p \) nor \( q \) is even. i.e., both \( p \) and \( q \) are odd.
Let \( p = 2m + 1 \) and \( q = 2n + 1 \) where \( m, n \) are integers.
Then:
\[ pq = (2m + 1)(2n + 1) \] \[ pq = 4mn + 2m + 2n + 1 \] \[ pq = 2(2mn + m + n) + 1 \](M1)
This is in the form \( 2k + 1 \), which represents an odd number.
However, we are given that \( pq \) is even.
This is a contradiction.
Therefore, the assumption is false, and at least one of \( p \) or \( q \) must be even.
(A1)
Part (ii): Algebraic Inequality
Strategy: Expand the bracket and simplify the inequality.
Subtract \( y^2 \) from both sides:
\[ x^2 + 2xy < 9x^2 \]Subtract \( x^2 \) from both sides:
\[ 2xy < 8x^2 \](M1)
Divide by \( 2x \):
Crucial Step: Since \( x < 0 \), dividing by \( 2x \) reverses the inequality sign.
\[ y > \frac{8x^2}{2x} \] \[ y > 4x \](A1)
Final Answer:
Proofs completed as above.
Question 8 (8 marks)
A car stops at two sets of traffic lights. The speed of the car, \( v \) ms\(^{-1}\), is modelled by the equation
\[ v = (10 – 0.4t) \ln(t + 1) \quad 0 \le t \le T \]where \( t \) seconds is the time after the car leaves the first set of traffic lights.
(a) Find the value of \( T \)
(1)
(b) Show that the maximum speed of the car occurs when
\[ t = \frac{26}{1 + \ln(t + 1)} – 1 \](4)
Using the iteration formula \( t_{n+1} = \frac{26}{1 + \ln(t_n + 1)} – 1 \) with \( t_1 = 7 \),
(c) (i) find the value of \( t_3 \) to 3 decimal places,
(ii) find, by repeated iteration, the time taken for the car to reach maximum speed.
(3)
Worked Solution
Part (a): Finding T
Why: The car stops at \( t = T \), which means \( v = 0 \).
How: Set \( v = 0 \) and solve for \( t \).
Either \( \ln(t+1) = 0 \Rightarrow t=0 \) (Start)
Or \( 10 – 0.4t = 0 \Rightarrow 10 = 0.4t \)
\[ t = \frac{10}{0.4} = 25 \]So \( T = 25 \) (B1)
Part (b): Maximum Speed
Why: Maximum speed occurs when the derivative \( \frac{dv}{dt} = 0 \).
How: Differentiate \( v \) using the Product Rule: \( v = uv \).
Let \( u = 10 – 0.4t \) \( \Rightarrow u’ = -0.4 \)
Let \( v = \ln(t + 1) \) \( \Rightarrow v’ = \frac{1}{t + 1} \)
(M1)
Set \( \frac{dv}{dt} = 0 \):
\[ \frac{10 – 0.4t}{t + 1} = 0.4 \ln(t + 1) \]Divide by 0.4:
\[ \frac{25 – t}{t + 1} = \ln(t + 1) \]Rearrange to make \( t \) the subject:
\[ 25 – t = (t + 1) \ln(t + 1) \]Wait, the target form is \( t = \frac{26}{1 + \ln(t + 1)} – 1 \). Let’s rearrange differently.
From \( \frac{10 – 0.4t}{t + 1} – 0.4 \ln(t + 1) = 0 \):
\[ \frac{10 – 0.4t}{t + 1} = 0.4 \ln(t + 1) \] \[ 10 – 0.4t = 0.4(t + 1) \ln(t + 1) \] \[ 25 – t = (t + 1) \ln(t + 1) \] \[ 25 – t = t \ln(t + 1) + \ln(t + 1) \] \[ 25 – \ln(t + 1) = t + t \ln(t + 1) \] \[ 25 – \ln(t + 1) = t(1 + \ln(t + 1)) \]This path is getting messy. Let’s look at the target form and work backward or check algebra.
Target: \( t + 1 = \frac{26}{1 + \ln(t+1)} \)
\[ (t + 1)(1 + \ln(t+1)) = 26 \] \[ t + 1 + (t+1)\ln(t+1) = 26 \] \[ t + 1 + (t+1)\ln(t+1) = 26 \]Let’s check our derivative equation again: \( 25 – t = (t+1)\ln(t+1) \)
Add \( t+1 \) to both sides:
\[ 25 – t + (t+1) = (t+1)\ln(t+1) + (t+1) \] \[ 26 = (t+1)[\ln(t+1) + 1] \] \[ t+1 = \frac{26}{1 + \ln(t+1)} \] \[ t = \frac{26}{1 + \ln(t+1)} – 1 \](A1*)
Part (c): Iteration
(i) \( t_1 = 7 \)
\[ t_2 = \frac{26}{1 + \ln(8)} – 1 = \frac{26}{3.0794} – 1 = 7.443 \] \[ t_3 = \frac{26}{1 + \ln(8.443)} – 1 = 7.298 \](M1 A1)
(ii) Repeated iteration:
\( t_4 = 7.344 \)
\( t_5 = 7.329 \)
\( t_6 = 7.334 \)
Values converge to \( 7.33 \) s (3 s.f.)
(A1)
Final Answer:
(a) \( T = 25 \)
(c)(i) \( 7.298 \)
(c)(ii) \( 7.33 \) s
Question 9 (6 marks)
Figure 3 shows a sketch of a parallelogram \( PQRS \).
Given that:
- \( \vec{PQ} = 2\mathbf{i} + 3\mathbf{j} – 4\mathbf{k} \)
- \( \vec{QR} = 5\mathbf{i} – 2\mathbf{k} \)
(a) Show that parallelogram \( PQRS \) is a rhombus.
(2)
(b) Find the exact area of the rhombus \( PQRS \).
(4)
Worked Solution
Part (a): Rhombus Property
What this tells us: A parallelogram is a rhombus if adjacent sides have equal lengths.
How: Calculate the magnitude of \( \vec{PQ} \) and \( \vec{QR} \).
Since \( |\vec{PQ}| = |\vec{QR}| \), the adjacent sides are equal in length.
Therefore, \( PQRS \) is a rhombus.
(A1)
Part (b): Area of Rhombus
Strategy 1: Vector Product
Area = \( |\vec{PQ} \times \vec{QR}| \).
Strategy 2: Diagonals
Area = \( \frac{1}{2} |\vec{PR}| \times |\vec{QS}| \).
Let’s use the Diagonals method as it deals with vector addition.
Find diagonals:
\[ \vec{PR} = \vec{PQ} + \vec{QR} = (2\mathbf{i} + 3\mathbf{j} – 4\mathbf{k}) + (5\mathbf{i} – 2\mathbf{k}) \] \[ \vec{PR} = 7\mathbf{i} + 3\mathbf{j} – 6\mathbf{k} \] \[ \vec{QS} = \vec{QP} + \vec{PS} = -\vec{PQ} + \vec{QR} \] (since \( \vec{PS} = \vec{QR} \) in parallelogram) \[ \vec{QS} = -(2\mathbf{i} + 3\mathbf{j} – 4\mathbf{k}) + (5\mathbf{i} – 2\mathbf{k}) \] \[ \vec{QS} = 3\mathbf{i} – 3\mathbf{j} + 2\mathbf{k} \](M1)
Calculate magnitudes of diagonals:
\[ |\vec{PR}| = \sqrt{7^2 + 3^2 + (-6)^2} = \sqrt{49 + 9 + 36} = \sqrt{94} \] \[ |\vec{QS}| = \sqrt{3^2 + (-3)^2 + 2^2} = \sqrt{9 + 9 + 4} = \sqrt{22} \]Area = \( \frac{1}{2} d_1 d_2 \)
\[ \text{Area} = \frac{1}{2} \sqrt{94} \sqrt{22} = \frac{1}{2} \sqrt{2068} \] \[ \sqrt{2068} = \sqrt{4 \times 517} = 2\sqrt{517} \] \[ \text{Area} = \frac{1}{2} (2\sqrt{517}) = \sqrt{517} \](A1)
Final Answer:
Area = \( \sqrt{517} \) (approx 22.7)
Question 10 (8 marks)
The number of bees, measured in thousands, \( N_b \), is modelled by the equation
\[ N_b = 45 + 220e^{0.05t} \](a) Find the number of bees at the start of the study.
(1)
(b) Show that, exactly 10 years after the start of the study, the number of bees was increasing at a rate of approximately 18 thousand per year.
(3)
The number of wasps, measured in thousands, \( N_w \), is modelled by the equation
\[ N_w = 10 + 800e^{-0.05t} \]When \( t = T \), there are an equal number of bees and wasps.
(c) Find the value of \( T \) to 2 decimal places.
(4)
Worked Solution
Part (a): Initial Value
At \( t = 0 \):
\[ N_b = 45 + 220e^0 = 45 + 220 = 265 \]265 thousand bees. (B1)
Part (b): Rate of Change
Why: “Rate of increasing” means the derivative \( \frac{dN_b}{dt} \).
(M1)
At \( t = 10 \):
\[ \frac{dN_b}{dt} = 11 e^{0.05(10)} = 11 e^{0.5} \] \[ 11 e^{0.5} \approx 18.136 \]This is approximately 18 thousand per year. (A1*)
Part (c): Intersection
How: Set \( N_b = N_w \) and solve for \( t \).
Rearrange:
\[ 35 + 220e^{0.05t} – 800e^{-0.05t} = 0 \]Multiply by \( e^{0.05t} \) to form a quadratic in \( e^{0.05t} \):
\[ 35e^{0.05t} + 220(e^{0.05t})^2 – 800 = 0 \] \[ 220(e^{0.05t})^2 + 35e^{0.05t} – 800 = 0 \](A1)
Let \( x = e^{0.05t} \). Solve \( 220x^2 + 35x – 800 = 0 \).
Divide by 5:
\[ 44x^2 + 7x – 160 = 0 \]Using quadratic formula:
\[ x = \frac{-7 \pm \sqrt{49 – 4(44)(-160)}}{88} \] \[ x = \frac{-7 \pm \sqrt{49 + 28160}}{88} = \frac{-7 \pm \sqrt{28209}}{88} \] \[ x \approx \frac{-7 + 167.95}{88} \approx 1.829 \] (Positive root required)So \( e^{0.05t} = 1.829 \)
\[ 0.05t = \ln(1.829) \] \[ t = \frac{\ln(1.829)}{0.05} \approx 12.08 \](A1)
Final Answer:
(a) 265 thousand
(c) \( T = 12.08 \)
Question 11 (7 marks)
Figure 4 shows a sketch of part of the curve \( C_1 \) with equation
\[ y = 2x^3 + 10 \quad x > 0 \]and part of the curve \( C_2 \) with equation
\[ y = 42x – 15x^2 – 7 \quad x > 0 \](a) Verify that the curves intersect at \( x = \frac{1}{2} \)
(2)
The curves intersect again at the point \( P \)
(b) Using algebra and showing all stages of working, find the exact \( x \) coordinate of \( P \)
(5)
Worked Solution
Part (a): Verification
How: Substitute \( x = \frac{1}{2} \) into both equations and check if the \( y \)-values are the same.
For \( C_1 \):
\[ y = 2\left(\frac{1}{2}\right)^3 + 10 = 2\left(\frac{1}{8}\right) + 10 = \frac{1}{4} + 10 = 10.25 \]For \( C_2 \):
\[ y = 42\left(\frac{1}{2}\right) – 15\left(\frac{1}{2}\right)^2 – 7 \] \[ y = 21 – 15\left(\frac{1}{4}\right) – 7 \] \[ y = 14 – 3.75 = 10.25 \]Since the \( y \)-values are equal (\( 10.25 \)), the curves intersect at \( x = \frac{1}{2} \). (A1)
Part (b): Finding Intersection Point P
Strategy: Set the equations equal to each other to find the intersection points. We know \( (2x – 1) \) is a factor because \( x = 1/2 \) is a root.
Rearrange to form a cubic equation equal to zero:
\[ 2x^3 + 15x^2 – 42x + 17 = 0 \](M1)
Since \( x = \frac{1}{2} \) is a root, \( (2x – 1) \) is a factor. Perform algebraic division or equating coefficients:
\[ 2x^3 + 15x^2 – 42x + 17 = (2x – 1)(Ax^2 + Bx + C) \]- Compare \( x^3 \): \( 2Ax^3 = 2x^3 \Rightarrow A = 1 \)
- Compare constants: \( -C = 17 \Rightarrow C = -17 \)
- Compare \( x^2 \): \( -x^2 + 2Bx^2 = 15x^2 \Rightarrow 2B – 1 = 15 \Rightarrow 2B = 16 \Rightarrow B = 8 \)
So, the quadratic factor is \( x^2 + 8x – 17 \). (A1)
Step 3: Solve the Quadratic
How: Use the quadratic formula to solve \( x^2 + 8x – 17 = 0 \).
(M1)
Simplify \( \sqrt{132} = \sqrt{4 \times 33} = 2\sqrt{33} \):
\[ x = \frac{-8 \pm 2\sqrt{33}}{2} \] \[ x = -4 \pm \sqrt{33} \]Since the question states \( x > 0 \) (diagram and domain), and \( -4 – \sqrt{33} < 0 \), we must check the positive root.
Wait, diagram shows \( P \) is to the right of the first intersection \( x=0.5 \). Let’s check values.
\( \sqrt{33} \approx 5.7 \). \( -4 + 5.7 = 1.7 \). This is positive.
The other root \( x = 1/2 \) is already known. The root \( x = -4-\sqrt{33} \) is negative.
So \( x = -4 + \sqrt{33} \).
(A1)
Final Answer:
\( x = -4 + \sqrt{33} \)
Question 12 (5 marks)
In this question you must show all stages of your working.
Solutions relying on calculator technology are not acceptable.
Show that
\[ \int_{1}^{e^2} x^3 \ln x \, dx = a e^8 + b \]where \( a \) and \( b \) are rational constants to be found.
Worked Solution
Step 1: Integration by Parts
Why: The integral involves a product of algebraic (\( x^3 \)) and logarithmic (\( \ln x \)) functions. This requires Integration by Parts.
Formula: \( \int u \frac{dv}{dx} dx = uv – \int v \frac{du}{dx} dx \).
Choice: Let \( u = \ln x \) (easier to differentiate) and \( \frac{dv}{dx} = x^3 \) (easy to integrate).
\( u = \ln x \Rightarrow \frac{du}{dx} = \frac{1}{x} \)
\( \frac{dv}{dx} = x^3 \Rightarrow v = \frac{x^4}{4} \)
Step 2: Apply Formula
(M1)
Integrate the second term:
\[ I = \left[ \frac{x^4}{4} \ln x \right]_{1}^{e^2} – \left[ \frac{x^4}{16} \right]_{1}^{e^2} \](A1)
Step 3: Apply Limits
Upper limit (\( e^2 \)):
\[ \left( \frac{(e^2)^4}{4} \ln(e^2) \right) – \left( \frac{(e^2)^4}{16} \right) \] \[ = \frac{e^8}{4} (2) – \frac{e^8}{16} \] (since \( \ln(e^2) = 2 \)) \[ = \frac{e^8}{2} – \frac{e^8}{16} \] \[ = \frac{8e^8}{16} – \frac{e^8}{16} = \frac{7e^8}{16} \](M1)
Lower limit (\( 1 \)):
\[ \left( \frac{1^4}{4} \ln(1) \right) – \left( \frac{1^4}{16} \right) \] \[ = 0 – \frac{1}{16} = -\frac{1}{16} \] (since \( \ln 1 = 0 \))Subtract Lower from Upper:
\[ I = \frac{7e^8}{16} – \left( -\frac{1}{16} \right) \] \[ I = \frac{7}{16} e^8 + \frac{1}{16} \](A1)
Final Answer:
\( a = \frac{7}{16}, b = \frac{1}{16} \)
Question 13 (7 marks)
(i) In an arithmetic series, the first term is \( a \) and the common difference is \( d \).
Show that
\[ S_n = \frac{n}{2} [2a + (n – 1)d] \](3)
(ii) James saves money over a number of weeks to buy a printer that costs £64.
He saves £10 in week 1, £9.20 in week 2, £8.40 in week 3 and so on, so that the weekly amounts he saves form an arithmetic sequence.
Given that James takes \( n \) weeks to save exactly £64,
(a) show that
\[ n^2 – 26n + 160 = 0 \](2)
(b) Solve the equation \( n^2 – 26n + 160 = 0 \)
(1)
(c) Hence state the number of weeks James takes to save enough money to buy the printer, giving a brief reason for your answer.
(1)
Worked Solution
Part (i): Proof of Sum Formula
Method: Write the sum forwards and backwards, then add the two equations.
Let the sum be \( S_n \).
\[ S_n = a + (a+d) + (a+2d) + \dots + (a + (n-1)d) \]Write in reverse:
\[ S_n = (a + (n-1)d) + (a + (n-2)d) + \dots + a \]Add the two equations:
\[ 2S_n = [2a + (n-1)d] + [2a + (n-1)d] + \dots + [2a + (n-1)d] \](M1)
There are \( n \) terms, each equal to \( 2a + (n-1)d \).
\[ 2S_n = n[2a + (n-1)d] \] \[ S_n = \frac{n}{2}[2a + (n-1)d] \](A1*)
Part (ii)(a): Modelling
Identify parameters: First term \( a = 10 \), common difference \( d = 9.20 – 10 = -0.80 \). Sum \( S_n = 64 \).
Using the sum formula:
\[ 64 = \frac{n}{2} [2(10) + (n-1)(-0.8)] \](M1)
\[ 128 = n [20 – 0.8n + 0.8] \] \[ 128 = n [20.8 – 0.8n] \] \[ 128 = 20.8n – 0.8n^2 \]Rearrange to form quadratic:
\[ 0.8n^2 – 20.8n + 128 = 0 \]Divide by 0.8:
\[ n^2 – 26n + 160 = 0 \](A1)
Part (ii)(b): Solving Quadratic
(B1)
Part (ii)(c): Interpretation
Why: The problem asks for the time to save “enough money”. At \( n=10 \), the total is exactly £64. At \( n=16 \), the total is also £64. Why? Because the terms become negative (he starts spending money/losing savings) or zero.
Check term 14: \( 10 + 13(-0.8) = -0.4 \). Terms become negative.
However, he stops once he has the money. So we take the first occurrence.
Final Answer:
\( n = 10 \) weeks.
Reason: He reaches the target amount at 10 weeks first. (Or: After 10 weeks he stops saving). (B1)
Question 14 (8 marks)
(a) Given that
\[ 2\sin(x – 60^{\circ}) = \cos(x – 30^{\circ}) \]show that
\[ \tan x = 3\sqrt{3} \](4)
(b) Hence or otherwise solve, for \( 0 \le \theta < 180^{\circ} \)
\[ 2\sin 2\theta = \cos(2\theta + 30^{\circ}) \]giving your answers to one decimal place.
(4)
Worked Solution
Part (a): Compound Angle Formulae
Formulas:
- \( \sin(A-B) = \sin A \cos B – \cos A \sin B \)
- \( \cos(A-B) = \cos A \cos B + \sin A \sin B \)
LHS:
\[ 2(\sin x \cos 60^{\circ} – \cos x \sin 60^{\circ}) \] \[ = 2(\frac{1}{2}\sin x – \frac{\sqrt{3}}{2}\cos x) = \sin x – \sqrt{3}\cos x \]RHS:
\[ \cos x \cos 30^{\circ} + \sin x \sin 30^{\circ} \] \[ = \frac{\sqrt{3}}{2}\cos x + \frac{1}{2}\sin x \](M1)
Equate:
\[ \sin x – \sqrt{3}\cos x = \frac{\sqrt{3}}{2}\cos x + \frac{1}{2}\sin x \]Group terms:
\[ \sin x – \frac{1}{2}\sin x = \frac{\sqrt{3}}{2}\cos x + \sqrt{3}\cos x \] \[ \frac{1}{2}\sin x = \frac{3\sqrt{3}}{2}\cos x \]Divide by \( \cos x \) and multiply by 2:
\[ \tan x = 3\sqrt{3} \](A1*)
Part (b): Solving Trigonometric Equation
Analysis: The equation is \( 2\sin 2\theta = \cos(2\theta + 30^{\circ}) \). Note the change from \( -30^{\circ} \) to \( +30^{\circ} \). We cannot directly use the result from (a) without modification.
Let’s expand from scratch or use similar logic.
(M1)
Group sine terms:
\[ 2\sin 2\theta + \frac{1}{2}\sin 2\theta = \frac{\sqrt{3}}{2}\cos 2\theta \] \[ \frac{5}{2}\sin 2\theta = \frac{\sqrt{3}}{2}\cos 2\theta \]Divide by \( \cos 2\theta \):
\[ \tan 2\theta = \frac{\sqrt{3}}{5} \] \[ \tan 2\theta \approx 0.3464 \](A1)
Step 3: Finding Angles
Range: \( 0 \le \theta < 180^{\circ} \Rightarrow 0 \le 2\theta < 360^{\circ} \).
Inverse tan:
\[ 2\theta = \tan^{-1}\left(\frac{\sqrt{3}}{5}\right) \approx 19.106^{\circ} \]Second quadrant solution (ASTC) – Tan is positive in 1st and 3rd.
\[ 2\theta = 19.1^{\circ}, \quad 19.1^{\circ} + 180^{\circ} = 199.1^{\circ} \]Divide by 2:
\[ \theta = \frac{19.1}{2} = 9.6^{\circ} \] \[ \theta = \frac{199.1}{2} = 99.6^{\circ} \]Rounding to 1 d.p.:
\( \theta = 9.6^{\circ}, 99.6^{\circ} \) (A1)
Final Answer:
\( \theta = 9.6^{\circ}, 99.6^{\circ} \)
Question 15 (10 marks)
Figure 5 shows the design for a solid toy that looks like a piece of cheese.
The toy is modelled so that
- face \( ABC \) is a sector of a circle with radius \( r \) cm and centre \( A \)
- angle \( BAC = 0.8 \) radians
- faces \( ABC \) and \( DEF \) are congruent
- edges \( AD, CF \) and \( BE \) are perpendicular to faces \( ABC \) and \( DEF \)
- edges \( AD, CF \) and \( BE \) have length \( h \) cm
Given that the volume of the toy is 240 cm\(^3\),
(a) show that the surface area of the toy, \( S \) cm\(^2\), is given by
\[ S = 0.8r^2 + \frac{1680}{r} \]making your method clear.
(4)
Using algebraic differentiation,
(b) find the value of \( r \) for which \( S \) has a stationary point.
(4)
(c) Prove, by further differentiation, that this value of \( r \) gives the minimum surface area of the toy.
(2)
Worked Solution
Part (a): Volume and Surface Area
Formulas:
- Area of Sector \( = \frac{1}{2}r^2\theta \)
- Arc Length \( = r\theta \)
- Volume of Prism \( = \text{Area} \times \text{Length} \)
Given \( \theta = 0.8 \).
Volume \( V = \text{Area of Sector} \times h \)
\[ V = \left(\frac{1}{2}r^2(0.8)\right) \times h = 0.4r^2 h \]Given \( V = 240 \):
\[ 0.4r^2 h = 240 \] \[ h = \frac{240}{0.4r^2} = \frac{600}{r^2} \](M1)
Surface Area \( S = 2 \times (\text{Sector Area}) + (\text{Two Rectangles}) + (\text{Curved Face}) \)
- 2 Sectors: \( 2 \times 0.4r^2 = 0.8r^2 \)
- 2 Rectangles (sides): \( 2 \times (r \times h) = 2rh \)
- Curved Face: \( (\text{Arc Length}) \times h = (0.8r)h = 0.8rh \)
(M1)
Substitute \( h = \frac{600}{r^2} \):
\[ S = 0.8r^2 + 2.8r\left(\frac{600}{r^2}\right) \] \[ S = 0.8r^2 + \frac{2.8 \times 600}{r} \] \[ S = 0.8r^2 + \frac{1680}{r} \](A1*)
Part (b): Stationary Point
Strategy: Find \( \frac{dS}{dr} \) and set to zero.
(M1)
Set \( \frac{dS}{dr} = 0 \):
\[ 1.6r = \frac{1680}{r^2} \] \[ 1.6r^3 = 1680 \] \[ r^3 = \frac{1680}{1.6} = 1050 \] \[ r = \sqrt[3]{1050} \approx 10.16 \]Exact value: \( \sqrt[3]{1050} \) (A1)
Part (c): Test for Minimum
Strategy: Find the second derivative \( \frac{d^2S}{dr^2} \) and evaluate at the stationary point.
(M1)
Since \( r > 0 \) (it’s a length), \( r^3 > 0 \), so \( \frac{3360}{r^3} > 0 \).
Therefore \( \frac{d^2S}{dr^2} > 0 \) for all \( r > 0 \).
Positive second derivative indicates a minimum.
(A1)
Final Answer:
(a) \( S = 0.8r^2 + \frac{1680}{r} \)
(b) \( r = \sqrt[3]{1050} \) (or approx 10.2 cm)
Question 16 (9 marks)
Figure 6 shows a sketch of the curve \( C \) with parametric equations
\[ x = 8\sin^2 t \quad y = 2\sin 2t + 3\sin t \quad 0 \le t \le \frac{\pi}{2} \]The region \( R \), shown shaded in Figure 6, is bounded by \( C \), the \( x \)-axis and the line with equation \( x = 4 \).
(a) Show that the area of \( R \) is given by
\[ \int_{0}^{a} (8 – 8\cos 4t + 48\sin^2 t \cos t) \, dt \]where \( a \) is a constant to be found.
(5)
(b) Hence, using algebraic integration, find the exact area of \( R \).
(4)
Worked Solution
Part (a): Setting up the Integral
Formula: Area under parametric curve \( = \int y \frac{dx}{dt} dt \).
Step 1: Find limits for \( t \).
- Lower limit (at origin): \( x = 0 \Rightarrow 8\sin^2 t = 0 \Rightarrow \sin t = 0 \Rightarrow t = 0 \).
- Upper limit (at \( x=4 \)): \( 8\sin^2 t = 4 \Rightarrow \sin^2 t = \frac{1}{2} \Rightarrow \sin t = \frac{1}{\sqrt{2}} \Rightarrow t = \frac{\pi}{4} \).
So \( a = \frac{\pi}{4} \).
Step 2: Find \( \frac{dx}{dt} \).
\[ x = 8\sin^2 t = 8(\sin t)^2 \] \[ \frac{dx}{dt} = 8 \cdot 2\sin t \cdot \cos t = 16\sin t \cos t \]Or use \( \sin 2t \): \( \frac{dx}{dt} = 8\sin 2t \).
Step 3: Form the integral \( \int y \frac{dx}{dt} dt \).
\[ \text{Area} = \int_{0}^{\pi/4} (2\sin 2t + 3\sin t)(16\sin t \cos t) \, dt \](M1)
Use identity \( \sin 2t = 2\sin t \cos t \):
\[ \text{Area} = \int_{0}^{\pi/4} (2(2\sin t \cos t) + 3\sin t)(16\sin t \cos t) \, dt \] \[ = \int_{0}^{\pi/4} (4\sin t \cos t + 3\sin t)(16\sin t \cos t) \, dt \]Expand:
\[ = \int_{0}^{\pi/4} (64\sin^2 t \cos^2 t + 48\sin^2 t \cos t) \, dt \](A1)
We need to match the target form: \( 8 – 8\cos 4t + 48\sin^2 t \cos t \).
The second term \( 48\sin^2 t \cos t \) matches.
Focus on \( 64\sin^2 t \cos^2 t \). Use double angle identities.
\[ \sin t \cos t = \frac{1}{2}\sin 2t \] \[ 64(\sin t \cos t)^2 = 64\left(\frac{1}{2}\sin 2t\right)^2 = 64\left(\frac{1}{4}\sin^2 2t\right) = 16\sin^2 2t \]Use \( \sin^2 A = \frac{1 – \cos 2A}{2} \):
\[ 16\sin^2 2t = 16\left(\frac{1 – \cos 4t}{2}\right) = 8(1 – \cos 4t) = 8 – 8\cos 4t \](M1)
Combine terms:
\[ \text{Area} = \int_{0}^{\pi/4} (8 – 8\cos 4t + 48\sin^2 t \cos t) \, dt \](A1*)
With \( a = \frac{\pi}{4} \). (B1)
Part (b): Evaluating the Integral
Strategy: Integrate term by term.
- \( \int 8 dt = 8t \)
- \( \int 8\cos 4t dt = 2\sin 4t \)
- \( \int 48\sin^2 t \cos t dt \): Let \( u = \sin t \), \( du = \cos t dt \). This is reverse chain rule.
Third term integral:
\[ \int 48\sin^2 t \cos t \, dt = 48 \frac{\sin^3 t}{3} = 16\sin^3 t \]Total Integral:
\[ I = [ 8t – 2\sin 4t + 16\sin^3 t ]_{0}^{\pi/4} \](M1 A1)
Substitute Upper Limit (\( \pi/4 \)):
- \( 8(\pi/4) = 2\pi \)
- \( 2\sin(4 \times \pi/4) = 2\sin(\pi) = 0 \)
- \( 16\sin^3(\pi/4) = 16\left(\frac{1}{\sqrt{2}}\right)^3 = 16\left(\frac{1}{2\sqrt{2}}\right) = \frac{8}{\sqrt{2}} = 4\sqrt{2} \)
Value at upper limit: \( 2\pi – 0 + 4\sqrt{2} \)
Substitute Lower Limit (0):
- \( 8(0) = 0 \)
- \( 2\sin(0) = 0 \)
- \( 16\sin^3(0) = 0 \)
Value at lower limit: 0
Result:
\[ \text{Area} = 2\pi + 4\sqrt{2} \](A1)
Final Answer:
\( a = \frac{\pi}{4} \)
Area = \( 2\pi + 4\sqrt{2} \)