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GCE A Level Mathematics Pure 1 (June 2019)
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
- cso: Correct solution only
- oe: Or equivalent
Table of Contents
- Question 1 (Algebra/Factor Theorem)
- Question 2 (Trigonometry/Graphs)
- Question 3 (Differentiation)
- Question 4 (Binomial Expansion)
- Question 5 (Quadratics/Functions)
- Question 6 (Trigonometric Equations)
- Question 7 (Exponential Modelling)
- Question 8 (Integration/Area)
- Question 9 (Logarithm Proof)
- Question 10 (Logic/Proof)
- Question 11 (Sequences/Series)
- Question 12 (Trig Modelling)
- Question 13 (Integration)
- Question 14 (Parametric Differentiation)
Question 1 (3 marks)
\[ f(x) = 3x^3 + 2ax^2 – 4x + 5a \] Given that \( (x + 3) \) is a factor of \( f(x) \), find the value of the constant \( a \).
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 factor is \( (x + 3) \), which we can write as \( (x – (-3)) \).
What this tells us: We need to substitute \( x = -3 \) into the expression for \( f(x) \) and set the result equal to zero.
Since \( (x + 3) \) is a factor, \( f(-3) = 0 \).
\[ f(-3) = 3(-3)^3 + 2a(-3)^2 – 4(-3) + 5a = 0 \]Step 2: Solving for \( a \)
How we calculate: We expand the powers and simplify the linear equation to find \( a \).
Calculate powers:
\[ 3(-27) + 2a(9) + 12 + 5a = 0 \] \[ -81 + 18a + 12 + 5a = 0 \]Combine like terms (constants and \( a \) terms):
\[ (18a + 5a) + (-81 + 12) = 0 \] \[ 23a – 69 = 0 \]Solve for \( a \):
\[ 23a = 69 \] \[ a = \frac{69}{23} \] \[ a = 3 \]✓ (M1, A1)
Final Answer:
\[ a = 3 \]✓ Total: 3 marks
Question 2 (5 marks)
Figure 1 shows a plot of part of the curve with equation \( y = \cos x \) where \( x \) is measured in radians.
(a) Use the diagram above to show why the equation
\[ \cos x – 2x – \frac{1}{2} = 0 \]has only one real root, giving a reason for your answer.
(2)
Given that the root of the equation is \( \alpha \), and that \( \alpha \) is small,
(b) Use the small angle approximation for \( \cos x \) to estimate the value of \( \alpha \) to 3 decimal places.
(3)
Worked Solution
Step 1: Analyzing the Equation (Part a)
Why we do this: We need to determine the number of roots by finding intersection points. We rearrange the given equation so that one side represents the curve already drawn (\( y = \cos x \)) and the other side represents a new line we can draw.
Rearrangement:
\[ \cos x = 2x + \frac{1}{2} \]What this tells us: The roots are the x-coordinates of the intersection points between the curve \( y = \cos x \) and the line \( y = 2x + 0.5 \).
We draw the line \( y = 2x + 0.5 \) on the diagram.
- When \( x = 0 \), \( y = 0.5 \).
- When \( y = 0 \), \( 2x = -0.5 \implies x = -0.25 \).
Looking at the diagram below:
The line intersects the curve \( y = \cos x \) at exactly one point.
Therefore, the equation has only one real root.
✓ (B1)
Step 2: Small Angle Approximation (Part b)
Why we do this: The question states \( \alpha \) is small, so we can use the small angle approximation for \( \cos x \).
Formula: For small \( x \) (in radians), \( \cos x \approx 1 – \frac{x^2}{2} \).
Substitute the approximation into the original equation:
\[ \left(1 – \frac{x^2}{2}\right) – 2x – \frac{1}{2} = 0 \]Simplify:
\[ \frac{1}{2} – 2x – \frac{x^2}{2} = 0 \]Multiply by 2 to remove fractions:
\[ 1 – 4x – x^2 = 0 \] \[ x^2 + 4x – 1 = 0 \]✓ (M1)
Step 3: Solving the Quadratic
How we calculate: Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \) where \( a=1, b=4, c=-1 \).
We have two possible values:
- \( x_1 = -2 + \sqrt{5} \approx -2 + 2.236 = 0.236 \)
- \( x_2 = -2 – \sqrt{5} \approx -4.236 \)
Since \( \alpha \) is small (and from the diagram, the intersection is at positive \( x \)), we choose the value close to 0.
\[ x \approx 0.236 \]✓ (dM1, A1)
Final Answer:
\[ \alpha \approx 0.236 \]✓ Total: 5 marks
Question 3 (5 marks)
\[ y = \frac{5x^2 + 10x}{(x + 1)^2} \quad x \neq -1 \]
(a) Show that \( \frac{dy}{dx} = \frac{A}{(x + 1)^n} \) where \( A \) and \( n \) are constants to be found.
(4)
(b) Hence deduce the range of values for \( x \) for which \( \frac{dy}{dx} < 0 \).
(1)
Worked Solution
Step 1: Differentiation (Part a)
Why we do this: The function is a fraction, so we use the Quotient Rule: \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v u’ – u v’}{v^2} \).
Identify terms:
- \( u = 5x^2 + 10x \)
- \( v = (x + 1)^2 \)
Differentiate \( u \) and \( v \):
\[ u’ = 10x + 10 = 10(x + 1) \] \[ v’ = 2(x + 1)^1 \cdot (1) = 2(x + 1) \quad \text{(Chain Rule)} \]Apply Quotient Rule:
\[ \frac{dy}{dx} = \frac{(x + 1)^2 [10(x + 1)] – (5x^2 + 10x)[2(x + 1)]}{((x + 1)^2)^2} \]✓ (M1, A1)
Step 2: Simplifying the Expression
How we calculate: We can factor out \( (x + 1) \) from the numerator to simplify before expanding completely. The denominator becomes \( (x + 1)^4 \).
Divide numerator and denominator by \( (x + 1) \):
\[ \frac{dy}{dx} = \frac{10(x + 1)^2 – 2(5x^2 + 10x)}{(x + 1)^3} \]Expand the numerator:
\[ \text{Numerator} = 10(x^2 + 2x + 1) – (10x^2 + 20x) \] \[ = 10x^2 + 20x + 10 – 10x^2 – 20x \] \[ = 10 \]Substitute back:
\[ \frac{dy}{dx} = \frac{10}{(x + 1)^3} \]So \( A = 10 \) and \( n = 3 \).
✓ (M1, A1)
Step 3: Determining the Range (Part b)
Why we do this: We need to find where the derivative is negative (\( < 0 \)).
Since the numerator (10) is positive, the fraction is negative only if the denominator is negative.
\[ (x + 1)^3 < 0 \]Taking the cube root:
\[ x + 1 < 0 \] \[ x < -1 \]✓ (B1)
Final Answer:
(a) \( \frac{dy}{dx} = \frac{10}{(x + 1)^3} \)
(b) \( x < -1 \)
✓ Total: 5 marks
Question 4 (6 marks)
(a) Find the first three terms, in ascending powers of \( x \), of the binomial expansion of
\[ \frac{1}{\sqrt{4 – x}} \]giving each coefficient in its simplest form.
(4)
The expansion can be used to find an approximation to \( \sqrt{2} \).
Possible values of \( x \) that could be substituted into this expansion are:
- \( x = -14 \) because \( \frac{1}{\sqrt{4 – (-14)}} = \frac{1}{\sqrt{18}} = \frac{\sqrt{2}}{6} \)
- \( x = 2 \) because \( \frac{1}{\sqrt{4 – 2}} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \)
- \( x = -\frac{1}{2} \) because \( \frac{1}{\sqrt{4 – (-\frac{1}{2})}} = \frac{1}{\sqrt{\frac{9}{2}}} = \frac{\sqrt{2}}{3} \)
(b) Without evaluating your expansion,
(i) state, giving a reason, which of the three values of \( x \) should not be used.
(1)
(ii) state, giving a reason, which of the three values of \( x \) would lead to the most accurate approximation to \( \sqrt{2} \).
(1)
Worked Solution
Step 1: Preparing for Expansion (Part a)
Why we do this: The binomial expansion formula \( (1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots \) requires the term inside the bracket to start with a ‘1’.
Strategy: Rewrite \( \frac{1}{\sqrt{4 – x}} \) as \( (4 – x)^{-\frac{1}{2}} \) and factor out the 4.
✓ (M1)
Step 2: Applying Binomial Formula
Formula: \( (1 + z)^n \approx 1 + nz + \frac{n(n-1)}{2}z^2 \).
Here, \( z = -\frac{x}{4} \) and \( n = -\frac{1}{2} \).
Don’t forget the \( \frac{1}{2} \) factor from the start:
\[ \frac{1}{2} \left[ 1 + \frac{x}{8} + \frac{3x^2}{128} \right] = \frac{1}{2} + \frac{x}{16} + \frac{3x^2}{256} \]✓ (M1, A1, A1)
Step 3: Validity of Expansion (Part b)
Why we do this: The binomial expansion for negative or fractional powers is valid only if the term ‘z’ satisfies \( |z| < 1 \).
Here \( z = -\frac{x}{4} \), so we need \( |-\frac{x}{4}| < 1 \implies |x| < 4 \).
(i) Which value to reject:
\( x = -14 \) should not be used.
Reason: \( |-14| = 14 \), which is not less than 4. The series diverges for this value.
✓ (B1)
(ii) Which value is most accurate:
\( x = -\frac{1}{2} \) should be used.
Reason: \( |-\frac{1}{2}| = 0.5 \) is closer to 0 than \( |2| = 2 \). The closer \( x \) is to 0, the faster the series converges, giving a more accurate approximation with fewer terms.
✓ (B1)
Final Answer:
(a) \( \frac{1}{2} + \frac{x}{16} + \frac{3x^2}{256} \)
(b)(i) \( x = -14 \) (outside validity range \( |x| < 4 \))
(b)(ii) \( x = -\frac{1}{2} \) (closest to zero)
✓ Total: 6 marks
Question 5 (10 marks)
\[ f(x) = 2x^2 + 4x + 9 \quad x \in \mathbb{R} \]
(a) Write \( f(x) \) in the form \( a(x + b)^2 + c \), where \( a, b \) and \( c \) are integers to be found.
(3)
(b) Sketch the curve with equation \( y = f(x) \) showing any points of intersection with the coordinate axes and the coordinates of any turning point.
(3)
(c) (i) Describe fully the transformation that maps the curve with equation \( y = f(x) \) onto the curve with equation \( y = g(x) \) where
\[ g(x) = 2(x – 2)^2 + 4x – 3 \quad x \in \mathbb{R} \](ii) Find the range of the function
\[ h(x) = \frac{21}{2x^2 + 4x + 9} \quad x \in \mathbb{R} \](4)
Worked Solution
Step 1: Completing the Square (Part a)
Method: Factor out the coefficient of \( x^2 \) from the first two terms, then complete the square.
Complete the square for \( x^2 + 2x \):
\[ x^2 + 2x = (x + 1)^2 – 1 \]Substitute back:
\[ f(x) = 2[(x + 1)^2 – 1] + 9 \] \[ f(x) = 2(x + 1)^2 – 2 + 9 \] \[ f(x) = 2(x + 1)^2 + 7 \]So \( a=2, b=1, c=7 \).
✓ (B1, M1, A1)
Step 2: Sketching the Curve (Part b)
Key Features:
- Shape: Positive \( x^2 \) coefficient implies a U-shaped parabola.
- Turning Point: From \( 2(x+1)^2 + 7 \), the minimum is at \( (-1, 7) \).
- y-intercept: Set \( x=0 \) in original equation: \( 2(0)^2 + 4(0) + 9 = 9 \). Point \( (0, 9) \).
- x-intercepts: Minimum value is 7 (above axis), so no x-intercepts.
✓ (B1, B1, B1ft)
Step 3: Transformations (Part c)
(i) Analyze \( g(x) \):
We need to compare the vertex of \( g(x) \) to \( f(x) \). Let’s complete the square for \( g(x) \).
\[ g(x) = 2(x – 2)^2 + 4x – 3 \]Wait, this isn’t fully simplified. Let’s expand and re-compress or check the vertex directly.
\[ g(x) = 2(x^2 – 4x + 4) + 4x – 3 = 2x^2 – 8x + 8 + 4x – 3 = 2x^2 – 4x + 5 \]Complete square for \( 2x^2 – 4x + 5 \):
\[ 2(x^2 – 2x) + 5 = 2((x – 1)^2 – 1) + 5 = 2(x – 1)^2 + 3 \]Comparison:
- Vertex of \( f(x) \) is \( (-1, 7) \).
- Vertex of \( g(x) \) is \( (1, 3) \).
To move from \( (-1, 7) \) to \( (1, 3) \):
- x-coordinate: \( -1 \to 1 \) (Add 2) \( \Rightarrow \) Translation +2 in x.
- y-coordinate: \( 7 \to 3 \) (Subtract 4) \( \Rightarrow \) Translation -4 in y.
Transformation: Translation by vector \( \begin{pmatrix} 2 \\ -4 \end{pmatrix} \).
✓ (M1, A1)
(ii) Range of \( h(x) \):
\[ h(x) = \frac{21}{f(x)} = \frac{21}{2(x + 1)^2 + 7} \]Since \( f(x) \) is the denominator, the extrema of \( h(x) \) occur at extrema of \( f(x) \).
- Minimum value of \( f(x) \) is 7. This gives the Maximum value of \( h(x) = \frac{21}{7} = 3 \).
- As \( x \to \pm\infty \), \( f(x) \to \infty \), so \( h(x) \to 0 \).
- Also \( f(x) \) is always positive, so \( h(x) \) is always positive.
Range:
\[ 0 < h(x) \leq 3 \]✓ (M1, A1)
Final Answer:
(a) \( f(x) = 2(x + 1)^2 + 7 \)
(b) Sketch showing U-shape, vertex \( (-1, 7) \), y-intercept 9.
(c)(i) Translation by vector \( \begin{pmatrix} 2 \\ -4 \end{pmatrix} \)
(c)(ii) \( 0 < h(x) \leq 3 \)
✓ Total: 10 marks
Question 6 (8 marks)
(a) Solve, for \( -180^\circ \leq \theta \leq 180^\circ \), the equation
\[ 5 \sin 2\theta = 9 \tan \theta \]giving your answers, where necessary, to one decimal place.
[Solutions based entirely on graphical or numerical methods are not acceptable.]
(6)
(b) Deduce the smallest positive solution to the equation
\[ 5 \sin (2x – 50^\circ) = 9 \tan (x – 25^\circ) \](2)
Worked Solution
Step 1: Using Identities (Part a)
Why we do this: The equation involves mixed trigonometric functions (\( \sin 2\theta \) and \( \tan \theta \)). We need to express everything in terms of single angles (\( \theta \)) and basic functions (\( \sin, \cos \)) to solve it.
Identities:
- \( \sin 2\theta = 2 \sin \theta \cos \theta \)
- \( \tan \theta = \frac{\sin \theta}{\cos \theta} \)
Substitute identities into the equation:
\[ 5(2 \sin \theta \cos \theta) = 9 \left( \frac{\sin \theta}{\cos \theta} \right) \] \[ 10 \sin \theta \cos \theta = \frac{9 \sin \theta}{\cos \theta} \]✓ (M1)
Step 2: Solving the Equation
Method: Multiply by \( \cos \theta \) to clear the fraction (noting \( \cos \theta \neq 0 \)) and bring all terms to one side to factorise. Critical: Do not just divide by \( \sin \theta \), as you will lose the solutions where \( \sin \theta = 0 \).
Factor out \( \sin \theta \):
\[ \sin \theta (10 \cos^2 \theta – 9) = 0 \]This gives two possibilities:
- \( \sin \theta = 0 \)
- \( 10 \cos^2 \theta – 9 = 0 \)
✓ (A1)
Step 3: Finding Solutions
Case 1: \( \sin \theta = 0 \).
In the range \( -180^\circ \leq \theta \leq 180^\circ \):
- \( \theta = -180^\circ, 0^\circ, 180^\circ \)
Case 2: \( 10 \cos^2 \theta = 9 \implies \cos^2 \theta = 0.9 \implies \cos \theta = \pm \sqrt{0.9} \approx \pm 0.9487 \).
Find reference angle: \( \cos^{-1}(\sqrt{0.9}) \approx 18.4^\circ \).
Since cosine is positive in Q1, Q4 and negative in Q2, Q3, we have solutions in all four quadrants.
Solutions for \( \cos \theta = \pm \sqrt{0.9} \):
- \( \theta \approx 18.4^\circ \)
- \( \theta \approx -18.4^\circ \)
- \( \theta \approx 180^\circ – 18.4^\circ = 161.6^\circ \)
- \( \theta \approx -(180^\circ – 18.4^\circ) = -161.6^\circ \)
All solutions: \( -180^\circ, -161.6^\circ, -18.4^\circ, 0^\circ, 18.4^\circ, 161.6^\circ, 180^\circ \).
✓ (M1, A1, B1)
Step 4: Deduction (Part b)
Why we do this: Compare the new equation to the original. Original: \( 5 \sin 2\theta = 9 \tan \theta \) New: \( 5 \sin (2x – 50^\circ) = 9 \tan (x – 25^\circ) \)
Notice that \( 2x – 50^\circ = 2(x – 25^\circ) \). If we let \( \theta = x – 25^\circ \), the equation becomes identical to part (a).
Let \( \theta = x – 25^\circ \).
We need the smallest positive solution for \( x \).
From part (a), possible values for \( \theta \) are \( 0^\circ, 18.4^\circ, \dots \) (and negatives).
Calculate \( x \) for each:
- If \( \theta = -18.4^\circ \): \( x = -18.4 + 25 = 6.6^\circ \)
- If \( \theta = 0^\circ \): \( x = 0 + 25 = 25^\circ \)
- If \( \theta = -161.6^\circ \): \( x = -161.6 + 25 = -136.6^\circ \) (Negative, ignore)
The smallest positive solution for \( x \) comes from the smallest \( \theta \) that satisfies \( \theta + 25 > 0 \).
\[ x = 6.6^\circ \]✓ (M1, A1)
Final Answer:
(a) \( \theta = -180^\circ, -161.6^\circ, -18.4^\circ, 0^\circ, 18.4^\circ, 161.6^\circ, 180^\circ \)
(b) \( x = 6.6^\circ \)
✓ Total: 8 marks
Question 7 (7 marks)
In a simple model, the value, \( £V \), of a car depends on its age, \( t \), in years.
The following information is available for car \( A \):
- its value when new is £20,000
- its value after one year is £16,000
(a) Use an exponential model to form, for car \( A \), a possible equation linking \( V \) with \( t \).
(4)
The value of car \( A \) is monitored over a 10-year period. Its value after 10 years is £2,000.
(b) Evaluate the reliability of your model in light of this information.
(2)
The following information is available for car \( B \):
- it has the same value, when new, as car \( A \)
- its value depreciates more slowly than that of car \( A \)
(c) Explain how you would adapt the equation found in (a) so that it could be used to model the value of car \( B \).
(1)
Worked Solution
Step 1: Setting up the Model (Part a)
Why we do this: An exponential decay model typically takes the form \( V = Ae^{kt} \) or \( V = A r^t \).
Information:
- At \( t = 0 \), \( V = 20000 \).
- At \( t = 1 \), \( V = 16000 \).
Substitute \( t = 0 \):
\[ 20000 = Ae^0 \implies A = 20000 \]Substitute \( t = 1 \):
\[ 16000 = 20000e^{k(1)} \] \[ e^k = \frac{16000}{20000} = 0.8 \]Take natural log:
\[ k = \ln(0.8) \approx -0.223 \]So the equation is:
\[ V = 20000 e^{-0.223t} \]Alternatively, using \( V = A(0.8)^t \) is also valid.
✓ (M1, M1, A1)
Step 2: Evaluating Reliability (Part b)
Why we do this: We calculate the predicted value at \( t=10 \) using our model and compare it to the actual observed value (£2000).
Using \( V = 20000 e^{-0.223(10)} \):
\[ V = 20000 e^{-2.23} \] \[ V \approx 20000 \times 0.1075 \] \[ V \approx £2150 \]Comparison:
The predicted value (£2150) is very close to the actual value (£2000).
Conclusion: The model is reliable.
✓ (M1, A1)
Step 3: Adapting for Car B (Part c)
Reasoning: Slower depreciation means the value drops less rapidly. In the term \( e^{kt} \), the constant \( k \) (which is negative) determines the rate of decay.
For slower decay, \( k \) should be closer to 0 (i.e., less negative).
Make the constant \( k \) (approx -0.223) less negative (e.g., -0.1). Alternatively, if using \( V = A r^t \), increase the base \( r \) (make it closer to 1, e.g., 0.9 instead of 0.8).
✓ (B1)
Final Answer:
(a) \( V = 20000 e^{-0.223t} \) (or equivalent)
(b) Predicted £2150 is close to £2000, so the model is reliable.
(c) Increase the value of \( k \) (make it less negative).
✓ Total: 7 marks
Question 8 (10 marks)
Figure 2 shows a sketch of part of the curve with equation \( y = x(x + 2)(x – 4) \).
The region \( R_1 \), shown shaded in Figure 2, is bounded by the curve and the negative x-axis.
(a) Show that the exact area of \( R_1 \) is \( \frac{20}{3} \).
(4)
The region \( R_2 \), also shown shaded in Figure 2, is bounded by the curve, the positive x-axis and the line with equation \( x = b \), where \( b \) is a positive constant and \( 0 < b < 4 \).
Given that the area of \( R_1 \) is equal to the area of \( R_2 \),
(b) verify that \( b \) satisfies the equation
\[ (b + 2)^2 (3b^2 – 20b + 20) = 0 \](4)
The roots of the equation \( 3b^2 – 20b + 20 = 0 \) are 1.225 and 5.442 to 3 decimal places.
The value of \( b \) is therefore 1.225 to 3 decimal places.
(c) Explain, with the aid of a diagram, the significance of the root 5.442.
(2)
Worked Solution
Step 1: Finding Area \( R_1 \) (Part a)
Why we do this: The area is the definite integral of the curve equation between the roots. \( R_1 \) is between \( x = -2 \) and \( x = 0 \).
First, expand the equation:
\[ y = x(x + 2)(x – 4) = x(x^2 – 2x – 8) = x^3 – 2x^2 – 8x \]Integrate:
\[ \int (x^3 – 2x^2 – 8x) \, dx = \frac{x^4}{4} – \frac{2x^3}{3} – 4x^2 \]Calculate definite integral from -2 to 0:
\[ \left[ \frac{x^4}{4} – \frac{2x^3}{3} – 4x^2 \right]_{-2}^{0} \]Upper limit (0): \( 0 \)
Lower limit (-2):
\[ \frac{(-2)^4}{4} – \frac{2(-2)^3}{3} – 4(-2)^2 \] \[ = \frac{16}{4} – \frac{2(-8)}{3} – 4(4) \] \[ = 4 + \frac{16}{3} – 16 \] \[ = \frac{16}{3} – 12 = \frac{16}{3} – \frac{36}{3} = -\frac{20}{3} \]Area = Upper – Lower = \( 0 – (-\frac{20}{3}) = \frac{20}{3} \).
✓ (M1, A1)
Step 2: Setting up Equation for \( R_2 \) (Part b)
Why we do this: \( R_2 \) is the area below the axis from 0 to \( b \). Since it is below the axis, the integral will be negative. The Area is the magnitude of the integral.
Area \( R_2 = -\int_0^b y \, dx \).
We are given Area \( R_1 = \) Area \( R_2 \). So Area \( R_2 = \frac{20}{3} \).
Therefore, \( \int_0^b (x^3 – 2x^2 – 8x) \, dx = -\frac{20}{3} \).
Substitute limits 0 and \( b \):
\[ \left[ \frac{b^4}{4} – \frac{2b^3}{3} – 4b^2 \right] – [0] = -\frac{20}{3} \]Multiply by 12 to clear fractions:
\[ 3b^4 – 8b^3 – 48b^2 = -80 \] \[ 3b^4 – 8b^3 – 48b^2 + 80 = 0 \]We need to verify that \( (b + 2)^2 (3b^2 – 20b + 20) = 0 \) matches this.
Expand \( (b + 2)^2 \): \( b^2 + 4b + 4 \).
Multiply \( (b^2 + 4b + 4)(3b^2 – 20b + 20) \):
- \( b^2(3b^2 – 20b + 20) = 3b^4 – 20b^3 + 20b^2 \)
- \( 4b(3b^2 – 20b + 20) = 12b^3 – 80b^2 + 80b \)
- \( 4(3b^2 – 20b + 20) = 12b^2 – 80b + 80 \)
Sum them up:
\[ 3b^4 + (-20+12)b^3 + (20-80+12)b^2 + (80-80)b + 80 \] \[ 3b^4 – 8b^3 – 48b^2 + 80 = 0 \]This matches our integral equation. Verified.
✓ (M1, A1)
Step 3: Significance of 5.442 (Part c)
What this tells us: The integral function \( I(x) = \frac{x^4}{4} – \frac{2x^3}{3} – 4x^2 \) represents the net signed area from 0. We set this equal to \( -\frac{20}{3} \).
The roots of the equation correspond to values of \( x \) where the net area from 0 is \( -\frac{20}{3} \).
At \( x = 4 \), the curve crosses back above the axis.
The value 5.442 is a point further along the curve (past 4) where the positive area accumulated between 4 and 5.442 cancels out the “extra” negative area accumulated between \( b=1.225 \) and 4.
More simply: The net area from 0 to 5.442 is also \( -\frac{20}{3} \). This means the area above the axis between 4 and 5.442 equals the area below the axis between 1.225 and 4.
Diagram:
The net signed area between \( x=0 \) and \( x=5.442 \) is equal to \( -\frac{20}{3} \) (same as area \( R_1 \)).
✓ (B1)
Final Answer:
(a) Area is \( \frac{20}{3} \).
(b) Verified.
(c) The net area from 0 to 5.442 is also \( -\frac{20}{3} \) (meaning the positive area cancelled out the excess negative area).
✓ Total: 10 marks
Question 9 (5 marks)
Given that \( a > b > 0 \) and that \( a \) and \( b \) satisfy the equation
\[ \log a – \log b = \log(a – b) \](a) Show that
\[ a = \frac{b^2}{b – 1} \](3)
(b) Write down the full restriction on the value of \( b \), explaining the reason for this restriction.
(2)
Worked Solution
Step 1: Using Log Laws (Part a)
Why we do this: We use the subtraction law \( \log x – \log y = \log(\frac{x}{y}) \) to combine the terms on the left.
Remove logarithms (antilog both sides):
\[ \frac{a}{b} = a – b \]Step 2: Rearranging for \( a \)
How we calculate: Multiply by \( b \) and collect \( a \) terms on one side.
✓ (M1, A1)
Step 3: Finding Restrictions (Part b)
Why we do this: We analyze the formula \( a = \frac{b^2}{b-1} \) and the original conditions.
1. Denominator cannot be zero: \( b – 1 \neq 0 \implies b \neq 1 \).
2. We are given \( a > 0 \). Since \( b^2 \) is positive, \( b – 1 \) must be positive for \( a \) to be positive. So \( b > 1 \).
3. We are given \( a > b \). Let’s check if this imposes more:
\[ \frac{b^2}{b-1} > b \implies \frac{b^2}{b-1} – b > 0 \implies \frac{b^2 – b(b-1)}{b-1} > 0 \implies \frac{b}{b-1} > 0 \]Since \( b>0 \), this holds if \( b > 1 \).
Restriction: \( b > 1 \).
Reason: We need \( a > 0 \), and since \( a = \frac{b^2}{b-1} \), the denominator must be positive (as numerator \( b^2 \) is positive). Also, \( b \neq 1 \) to avoid division by zero.
✓ (B1, B1)
Final Answer:
(a) \( a = \frac{b^2}{b – 1} \)
(b) \( b > 1 \) (because \( a \) must be positive)
✓ Total: 5 marks
Question 10 (6 marks)
(i) Prove that for all \( n \in \mathbb{N} \), \( n^2 + 2 \) is not divisible by 4.
(4)
(ii) “Given \( x \in \mathbb{R} \), the value of \( |3x – 28| \) is greater than or equal to the value of \( (x – 9) \).”
State, giving a reason, if the above statement is always true, sometimes true or never true.
(2)
Worked Solution
Step 1: Proof by Cases (Part i)
Strategy: Consider natural numbers \( n \) as either even or odd.
Case 1: \( n \) is even.
Let \( n = 2k \). Then \( n^2 + 2 = (2k)^2 + 2 = 4k^2 + 2 \).
\( 4k^2 \) is divisible by 4, so \( 4k^2 + 2 \) leaves a remainder of 2 when divided by 4.
Case 2: \( n \) is odd.
Let \( n = 2k + 1 \). Then \( n^2 + 2 = (2k + 1)^2 + 2 = 4k^2 + 4k + 1 + 2 = 4k^2 + 4k + 3 \).
\( 4k^2 + 4k \) is divisible by 4, so \( n^2 + 2 \) leaves a remainder of 3 when divided by 4.
In both cases, \( n^2 + 2 \) is not a multiple of 4.
✓ (M1, A1)
Step 2: Testing the Statement (Part ii)
Statement: \( |3x – 28| \geq x – 9 \).
Let’s test this with specific values or sketch the graph.
Sketching \( y = |3x – 28| \) (V-shape, vertex at \( x = 9.33 \)) and \( y = x – 9 \) (line).
Try \( x = 9.4 \):
- LHS: \( |3(9.4) – 28| = |28.2 – 28| = 0.2 \)
- RHS: \( 9.4 – 9 = 0.4 \)
Is \( 0.2 \geq 0.4 \)? No. So it is not always true.
Try \( x = 0 \):
- LHS: \( |-28| = 28 \)
- RHS: \( -9 \)
Is \( 28 \geq -9 \)? Yes. So it is sometimes true.
Conclusion: Sometimes true.
✓ (M1, A1)
Final Answer:
(i) Proof complete (by cases even/odd).
(ii) Sometimes true (e.g., false for \( x=9.4 \), true for \( x=0 \)).
✓ Total: 6 marks
Question 11 (7 marks)
A competitor is running a 20 kilometre race.
She runs each of the first 4 kilometres at a steady pace of 6 minutes per kilometre. After the first 4 kilometres, she begins to slow down.
In order to estimate her finishing time, the time that she will take to complete each subsequent kilometre is modelled to be 5% greater than the time that she took to complete the previous kilometre.
Using the model,
(a) show that her time to run the first 6 kilometres is estimated to be 36 minutes 55 seconds,
(2)
(b) show that her estimated time, in minutes, to run the \( r \)th kilometre, for \( 5 \leq r \leq 20 \), is
\[ 6 \times 1.05^{r-4} \](1)
(c) estimate the total time, in minutes and seconds, that she will take to complete the race.
(4)
Worked Solution
Step 1: Calculating First 6km (Part a)
Why we do this: Break the race down into individual kilometres. The first 4 are constant. The 5th and 6th follow the geometric increase (increase by 5% means multiplying by 1.05).
Time for first 4 km:
\[ 4 \times 6 = 24 \text{ minutes} \]Time for 5th km (increase 6 mins by 5%):
\[ 6 \times 1.05 = 6.3 \text{ minutes} \]Time for 6th km (increase 5th km time by 5%):
\[ 6.3 \times 1.05 = 6.615 \text{ minutes} \]Total time for first 6 km:
\[ 24 + 6.3 + 6.615 = 36.915 \text{ minutes} \]Convert decimal minutes to seconds:
\[ 0.915 \times 60 = 54.9 \text{ seconds} \]Rounding to nearest second gives 36 minutes 55 seconds.
✓ (M1, A1)
Step 2: General Formula (Part b)
Pattern recognition:
- \( r=5 \): \( 6 \times 1.05^1 = 6 \times 1.05^{5-4} \)
- \( r=6 \): \( 6 \times 1.05^2 = 6 \times 1.05^{6-4} \)
For the \( r \)th kilometre (where \( r \geq 5 \)), the power of 1.05 is \( r-4 \).
✓ (B1)
Step 3: Total Race Time (Part c)
Strategy: Total time = (Time for first 4 km) + (Sum of times for 5th to 20th km).
The times from \( r=5 \) to \( r=20 \) form a Geometric Progression.
- First term \( a \) (5th km) = 6.3
- Common ratio \( r \) = 1.05
- Number of terms \( n \) = \( 20 – 5 + 1 = 16 \)
Sum of GP:
\[ S_{16} = \frac{a(r^n – 1)}{r – 1} = \frac{6.3(1.05^{16} – 1)}{1.05 – 1} \] \[ S_{16} = \frac{6.3(1.05^{16} – 1)}{0.05} \] \[ S_{16} \approx \frac{6.3(2.18287… – 1)}{0.05} \approx 149.04 \text{ minutes} \]Total time:
\[ 24 + 149.04 = 173.04 \text{ minutes} \]Convert to minutes and seconds:
\[ 0.04 \times 60 = 2.4 \text{ seconds} \]Total: 173 minutes 2 seconds (or 3 seconds depending on rounding).
✓ (M1, M1, A1, A1)
Final Answer:
(a) Shown.
(b) \( 6 \times 1.05^{r-4} \)
(c) 173 minutes 3 seconds
✓ Total: 7 marks
Question 12 (10 marks)
\[ f(x) = 10e^{-0.25x} \sin x, \quad x \geq 0 \]
(a) Show that the \( x \) coordinates of the turning points of the curve with equation \( y = f(x) \) satisfy the equation \( \tan x = 4 \).
(4)
Figure 3 shows a sketch of part of the curve with equation \( y = f(x) \).
(b) Sketch the graph of \( H \) against \( t \) where
\[ H(t) = |10e^{-0.25t} \sin t| \quad t \geq 0 \]showing the long-term behaviour of this curve.
(2)
The function \( H(t) \) is used to model the height, in metres, of a ball above the ground \( t \) seconds after it has been kicked.
Using this model, find
(c) the maximum height of the ball above the ground between the first and second bounce.
(3)
(d) Explain why this model should not be used to predict the time of each bounce.
(1)
Worked Solution
Step 1: Differentiation (Part a)
Why we do this: Turning points occur where the gradient \( f'(x) = 0 \). We use the Product Rule because we have two functions of \( x \) multiplied: \( 10e^{-0.25x} \) and \( \sin x \).
Let \( u = 10e^{-0.25x} \) and \( v = \sin x \).
\[ u’ = -2.5e^{-0.25x}, \quad v’ = \cos x \]Apply product rule \( u’v + uv’ \):
\[ f'(x) = -2.5e^{-0.25x} \sin x + 10e^{-0.25x} \cos x \]Set \( f'(x) = 0 \):
\[ 10e^{-0.25x} \cos x – 2.5e^{-0.25x} \sin x = 0 \]Factor out \( 2.5e^{-0.25x} \) (which is never zero):
\[ 2.5e^{-0.25x} (4 \cos x – \sin x) = 0 \] \[ 4 \cos x – \sin x = 0 \] \[ 4 \cos x = \sin x \]Divide by \( \cos x \):
\[ 4 = \frac{\sin x}{\cos x} \] \[ \tan x = 4 \]✓ (M1, A1, M1, A1)
Step 2: Sketching H(t) (Part b)
Shape: \( H(t) \) is the modulus of the original function. The sine wave “bounces” off the x-axis (all loops positive). The exponential term causes the height of the bounces to decrease over time.
✓ (M1, A1)
Step 3: Calculating Max Height (Part c)
Context: Bounces occur when height is 0, i.e., \( \sin t = 0 \). This happens at \( t = 0, \pi, 2\pi, \dots \). The first bounce is at \( t = \pi \). The second bounce is at \( t = 2\pi \). We need the maximum height in the interval \( \pi < t < 2\pi \).
The maximum occurs at a turning point, satisfying \( \tan t = 4 \).
Find \( t \) in the correct quadrant (3rd quadrant for tangent to be positive? Wait. \( \sin t \) is negative in 3rd quadrant, but \( H(t) \) is modulus. The original \( f(t) \) has a minimum there, but \( H(t) \) reflects it to a maximum).
Basic angle for \( \tan t = 4 \): \( \tan^{-1}(4) \approx 1.326 \) rad.
In the interval \( \pi \) to \( 2\pi \), \( t = \pi + 1.326 \approx 3.14 + 1.326 = 4.468 \).
Substitute \( t = 4.468 \) into \( H(t) \):
\[ H(4.468) = |10e^{-0.25(4.468)} \sin(4.468)| \]Note: \( \sin(4.468) \approx \sin(256^\circ) \approx -0.97 \).
\[ H \approx |10(0.327)(-0.97)| \] \[ H \approx 3.18 \text{ m} \]✓ (M1, M1, A1)
Step 4: Model Limitation (Part d)
Reasoning: In the model \( H(t) = |10e^{-0.25t} \sin t| \), the bounces occur at fixed intervals of \( \pi \) (determined by \( \sin t \)).
In reality, as a ball loses energy and bounce height decreases, the bounces become more frequent (the time between bounces decreases).
The model predicts a constant time interval between bounces, whereas in reality, the time between bounces should decrease as the height decreases.
✓ (B1)
Final Answer:
(a) Shown.
(b) Sketch showing loops of decreasing height.
(c) 3.18 m
(d) The model assumes constant time between bounces, but real bounces get faster/more frequent.
✓ Total: 10 marks
Question 13 (11 marks)
The curve \( C \) with equation
\[ y = \frac{p – 3x}{(2x – q)(x + 3)} \quad x \in \mathbb{R}, x \neq -3, x \neq 2 \]where \( p \) and \( q \) are constants, passes through the point \( (3, \frac{1}{2}) \) and has two vertical asymptotes with equations \( x = 2 \) and \( x = -3 \).
(a) (i) Explain why you can deduce that \( q = 4 \).
(ii) Show that \( p = 15 \).
(3)
Figure 4 shows a sketch of part of the curve \( C \). The region \( R \), shown shaded in Figure 4, is bounded by the curve \( C \), the x-axis and the line with equation \( x = 3 \).
(b) Show that the exact value of the area of \( R \) is \( a \ln 2 + b \ln 3 \), where \( a \) and \( b \) are rational constants to be found.
(8)
Worked Solution
Step 1: Finding Constants (Part a)
(i) Finding \( q \): The question states there is a vertical asymptote at \( x = 2 \). This occurs when the denominator is zero. The denominator is \( (2x – q)(x + 3) \).
Substitute \( x = 2 \) into the factor \( (2x – q) \):
\[ 2(2) – q = 0 \implies 4 – q = 0 \implies q = 4 \]✓ (B1)
(ii) Finding \( p \): The curve passes through \( (3, \frac{1}{2}) \).
Substitute \( x = 3, y = \frac{1}{2} \) and \( q = 4 \) into the equation:
\[ \frac{1}{2} = \frac{p – 3(3)}{(2(3) – 4)(3 + 3)} \] \[ \frac{1}{2} = \frac{p – 9}{(2)(6)} \] \[ \frac{1}{2} = \frac{p – 9}{12} \] \[ 6 = p – 9 \implies p = 15 \]✓ (M1, A1)
Step 2: Setting up the Integral for Area (Part b)
Why we do this: The area \( R \) is bounded by \( x = 3 \), the curve, and the x-axis. We need the upper limit where the curve cuts the x-axis (i.e., \( y = 0 \)).
From \( y = \frac{15 – 3x}{(2x – 4)(x + 3)} \), \( y = 0 \) when numerator is 0:
\[ 15 – 3x = 0 \implies x = 5 \]So we integrate from \( x = 3 \) to \( x = 5 \).
Integral needed:
\[ \int_{3}^{5} \frac{15 – 3x}{(2x – 4)(x + 3)} \, dx \]Step 3: Partial Fractions
Method: Decompose the integrand.
\[ \frac{15 – 3x}{(2x – 4)(x + 3)} = \frac{A}{2x – 4} + \frac{B}{x + 3} \]Let \( x = -3 \):
\[ 15 – 3(-3) = B(2(-3) – 4) \] \[ 24 = -10B \implies B = -2.4 \]Let \( x = 2 \):
\[ 15 – 3(2) = A(2 + 3) \] \[ 9 = 5A \implies A = 1.8 \]Rewrite integrand:
\[ \frac{1.8}{2x – 4} – \frac{2.4}{x + 3} \] \[ = \frac{0.9}{x – 2} – \frac{2.4}{x + 3} \]✓ (M1, A1)
Step 4: Integration and Evaluation
Integration Rule: \( \int \frac{1}{x+k} \, dx = \ln|x+k| \).
Substitute upper limit (5):
\[ (0.9 \ln 3 – 2.4 \ln 8) \]Substitute lower limit (3):
\[ (0.9 \ln 1 – 2.4 \ln 6) \]Note \( \ln 1 = 0 \).
\[ \text{Area} = (0.9 \ln 3 – 2.4 \ln 8) – (0 – 2.4 \ln 6) \] \[ = 0.9 \ln 3 – 2.4 \ln(2^3) + 2.4 \ln(2 \times 3) \] \[ = 0.9 \ln 3 – 7.2 \ln 2 + 2.4(\ln 2 + \ln 3) \] \[ = 0.9 \ln 3 – 7.2 \ln 2 + 2.4 \ln 2 + 2.4 \ln 3 \]Collect terms:
\[ \ln 3: 0.9 + 2.4 = 3.3 \] \[ \ln 2: -7.2 + 2.4 = -4.8 \] \[ \text{Area} = 3.3 \ln 3 – 4.8 \ln 2 \]Or in fractions: \( \frac{33}{10} \ln 3 – \frac{48}{10} \ln 2 \).
✓ (M1, M1, A1)
Final Answer:
(a) \( q = 4, p = 15 \)
(b) \( \text{Area} = -4.8 \ln 2 + 3.3 \ln 3 \) (or \( \frac{33}{10} \ln 3 – \frac{24}{5} \ln 2 \))
✓ Total: 11 marks
Question 14 (7 marks)
The curve \( C \), in the standard Cartesian plane, is defined by the equation
\[ x = 4 \sin 2y \quad -\frac{\pi}{4} < y < \frac{\pi}{4} \]The curve \( C \) passes through the origin \( O \).
(a) Find the value of \( \frac{dy}{dx} \) at the origin.
(2)
(b) (i) Use the small angle approximation for \( \sin 2y \) to find an equation linking \( x \) and \( y \) for points close to the origin.
(ii) Explain the relationship between the answers to (a) and (b)(i).
(2)
(c) Show that, for all points \( (x, y) \) lying on \( C \),
\[ \frac{dy}{dx} = \frac{1}{a \sqrt{b – x^2}} \]where \( a \) and \( b \) are constants to be found.
(3)
Worked Solution
Step 1: Differentiation (Part a)
Why we do this: The equation is given as \( x \) in terms of \( y \). It is easier to find \( \frac{dx}{dy} \) first and then invert it.
At the origin \( (0,0) \), \( y = 0 \):
\[ \frac{dy}{dx} = \frac{1}{8 \cos(0)} = \frac{1}{8(1)} = \frac{1}{8} \]✓ (M1, A1)
Step 2: Small Angle Approximation (Part b)
(i) For small \( \theta \), \( \sin \theta \approx \theta \). Here \( \theta = 2y \).
✓ (B1)
(ii) Relationship:
The line \( y = \frac{1}{8}x \) has a gradient of \( \frac{1}{8} \). This matches the value of \( \frac{dy}{dx} \) at the origin found in part (a). The small angle approximation gives the equation of the tangent at the origin.
✓ (B1)
Step 3: General Derivative in Terms of x (Part c)
Why we do this: We have \( \frac{dy}{dx} = \frac{1}{8 \cos 2y} \), but we need it in terms of \( x \). We use the identity \( \cos^2 \theta + \sin^2 \theta = 1 \) and the original equation \( x = 4 \sin 2y \).
From \( x = 4 \sin 2y \), we have:
\[ \sin 2y = \frac{x}{4} \]Use identity \( \cos 2y = \sqrt{1 – \sin^2 2y} \):
\[ \cos 2y = \sqrt{1 – \left(\frac{x}{4}\right)^2} = \sqrt{1 – \frac{x^2}{16}} \]Substitute into the derivative expression:
\[ \frac{dy}{dx} = \frac{1}{8 \sqrt{1 – \frac{x^2}{16}}} \] \[ = \frac{1}{8 \sqrt{\frac{16 – x^2}{16}}} \] \[ = \frac{1}{8 \frac{\sqrt{16 – x^2}}{4}} \] \[ = \frac{1}{2 \sqrt{16 – x^2}} \]So \( a = 2 \) and \( b = 16 \).
✓ (M1, A1, A1)
Final Answer:
(a) \( \frac{1}{8} \)
(b) \( y = \frac{1}{8}x \); The gradient of this line is the gradient of the curve at the origin.
(c) \( \frac{dy}{dx} = \frac{1}{2 \sqrt{16 – x^2}} \)
✓ Total: 7 marks