If any of my solutions look wrong, please refer to the mark scheme. You can exit full-screen mode for the question paper and mark scheme by clicking the icon in the bottom-right corner or by pressing Esc on your keyboard.
AQA Level 2 Certificate Further Mathematics June 2018 Paper 1
📌 Mark Scheme Legend
- M: Method marks – awarded for a correct method
- A: Accuracy marks – awarded for a correct answer (after correct method)
- B: Independent marks – awarded for a correct statement/result
- ft: Follow through – marks awarded for correct work based on previous errors
- oe: Or equivalent – accept alternate forms
Table of Contents
- Question 1 (Differentiation)
- Question 2 (Coordinate Geometry)
- Question 3 (Matrices)
- Question 4 (Percentages/Algebra)
- Question 5 (Binomial Expansion)
- Question 6 (Circle Equation)
- Question 7 (Circle Geometry)
- Question 8 (Surds)
- Question 9 (Functions)
- Question 10 (Rearranging Formulae)
- Question 11 (Tangents and Curves)
- Question 12 (Coordinate Geometry)
- Question 13 (Simultaneous Equations)
- Question 14 (Indices)
- Question 15 (Quadratic Graphs)
- Question 16 (Calculus/Turning Points)
- Question 17 (Graphical Solutions)
- Question 18 (Circle Theorems Proof)
- Question 19 (Completing the Square)
- Question 20 (Trigonometry/Exact Values)
Question 1 (2 marks)
Given:
\[ y = \frac{x^6}{2} + \frac{x^4}{4} \]Work out \( \frac{dy}{dx} \).
Simplify your answer.
Worked Solution
Step 1: Understanding the Rule
💡 What are we doing?
We need to differentiate terms of the form \( ax^n \). The rule is: multiply by the power, then decrease the power by 1.
Formula: If \( y = ax^n \), then \( \frac{dy}{dx} = an x^{n-1} \).
Step 2: Differentiating Term by Term
📝 How we calculate:
We treat the coefficients \( \frac{1}{2} \) and \( \frac{1}{4} \) as constants.
For the first term \( \frac{x^6}{2} \) (or \( \frac{1}{2}x^6 \)):
\[ \frac{d}{dx}\left(\frac{1}{2}x^6\right) = \frac{1}{2} \times 6x^{6-1} = 3x^5 \]For the second term \( \frac{x^4}{4} \) (or \( \frac{1}{4}x^4 \)):
\[ \frac{d}{dx}\left(\frac{1}{4}x^4\right) = \frac{1}{4} \times 4x^{4-1} = 1x^3 = x^3 \]✅ Final Answer:
\[ \frac{dy}{dx} = 3x^5 + x^3 \]Alternative form: \( x^3(3x^2 + 1) \)
(M1) For one correct term (e.g., \( 3x^5 \) or \( x^3 \) or \( \frac{6x^5}{2} \))
(A1) For the fully correct simplified expression
Question 2 (3 marks)
\( P \) is the point \( (-12, b) \)
\( Q \) is the point \( (a, 4) \)
\( R \) is the point \( (6, -2) \)
\( Q \) is the midpoint of \( PR \).
Work out the values of \( a \) and \( b \).
Worked Solution
Step 1: Understanding the Midpoint Formula
💡 What this tells us:
The midpoint \( Q \) is exactly halfway between \( P \) and \( R \). The coordinates of a midpoint are the average of the coordinates of the endpoints.
\[ \text{Midpoint } x = \frac{x_1 + x_2}{2}, \quad \text{Midpoint } y = \frac{y_1 + y_2}{2} \]Step 2: Finding \( a \) (the x-coordinate)
📝 How we calculate:
The x-coordinate of the midpoint \( Q \) is \( a \). The x-coordinates of \( P \) and \( R \) are \( -12 \) and \( 6 \).
(M1) Setting up equation for x or y
(A1) \( a = -3 \)
Step 3: Finding \( b \) (from the y-coordinate)
📝 How we calculate:
The y-coordinate of the midpoint \( Q \) is \( 4 \). The y-coordinates of \( P \) and \( R \) are \( b \) and \( -2 \).
Multiply both sides by 2:
\[ 8 = b – 2 \]Add 2 to both sides:
\[ b = 10 \](A1) \( b = 10 \)
✅ Final Answer:
\[ a = -3, \quad b = 10 \]Question 3 (2 marks)
Given:
\[ \mathbf{A} = \begin{pmatrix} 2 & 4 \\ 3 & -1 \end{pmatrix} \quad \text{and} \quad \mathbf{B} = \begin{pmatrix} -2 & 6 \\ 2 & 1 \end{pmatrix} \]Work out \( \mathbf{AB} \).
Worked Solution
Step 1: Matrix Multiplication Method
💡 Why we do this:
To multiply matrices, we multiply the rows of the first matrix by the columns of the second matrix.
Top-Left element = (Top Row of A) • (Left Column of B)
Step 2: Calculating Elements
Top Left: \( (2 \times -2) + (4 \times 2) = -4 + 8 = 4 \)
Top Right: \( (2 \times 6) + (4 \times 1) = 12 + 4 = 16 \)
Bottom Left: \( (3 \times -2) + (-1 \times 2) = -6 – 2 = -8 \)
Bottom Right: \( (3 \times 6) + (-1 \times 1) = 18 – 1 = 17 \)
(B1) For 2 or 3 correct elements in correct position
✅ Final Answer:
\[ \begin{pmatrix} 4 & 16 \\ -8 & 17 \end{pmatrix} \](B2) Fully correct matrix
Question 4 (4 marks)
\( P = 4x \) and \( Q = 7x \)
\( P \) increases by 25%
\( Q \) decreases by 40%
Now, \( P \) is 28 greater than \( Q \).
Work out the value of \( x \).
Worked Solution
Step 1: Calculate New Values of P and Q
💡 Strategy:
We need to apply the percentage changes to the algebraic expressions.
New P: Increase \( 4x \) by 25%.
25% of \( 4x \) is \( \frac{1}{4} \times 4x = x \).
\[ \text{New } P = 4x + x = 5x \]New Q: Decrease \( 7x \) by 40%.
10% of \( 7x \) is \( 0.7x \), so 40% is \( 4 \times 0.7x = 2.8x \).
\[ \text{New } Q = 7x – 2.8x = 4.2x \](M1) For \( 1.25 \times 4x \) (or \( 5x \))
(M1) For \( 0.6 \times 7x \) (or \( 4.2x \))
Step 2: Form Equation
💡 What the question says:
“P is 28 greater than Q” means \( P – Q = 28 \).
(M1 dep) Forming correct equation with calculated values
Step 3: Solve for x
To divide by 0.8, we can multiply by 10 then divide by 8, or think of 0.8 as \( \frac{4}{5} \).
\[ \frac{4}{5}x = 28 \] \[ 4x = 28 \times 5 \] \[ 4x = 140 \] \[ x = 35 \]✅ Final Answer:
\[ x = 35 \](A1) Correct value
Question 5 (3 marks)
In the expansion and simplification of \( (x – 3)(x^2 + 5x + k) \), the coefficient of \( x^2 \) is equal to the coefficient of \( x \).
\( k \) is a constant.
Work out the value of \( k \).
Worked Solution
Step 1: Expand the Brackets
💡 Strategy:
Multiply every term in the first bracket by every term in the second. We are specifically looking for \( x^2 \) terms and \( x \) terms.
Multiply \( x \) by the second bracket:
\[ x(x^2) + x(5x) + x(k) = x^3 + 5x^2 + kx \]Multiply \( -3 \) by the second bracket:
\[ -3(x^2) – 3(5x) – 3(k) = -3x^2 – 15x – 3k \]Combine all terms:
\[ x^3 + 5x^2 – 3x^2 + kx – 15x – 3k \]Step 2: Identify Coefficients
\( x^2 \) terms: \( 5x^2 – 3x^2 = 2x^2 \)
So, Coefficient of \( x^2 \) is 2.
\( x \) terms: \( kx – 15x = (k – 15)x \)
So, Coefficient of \( x \) is \( k – 15 \).
(M1) Expansion showing at least 4 correct terms
Step 3: Solve for k
💡 What the question says:
The coefficients are equal.
Add 15 to both sides:
\[ k = 17 \](M1 dep) Setting coefficients equal: \( 5-3 = k-15 \)
✅ Final Answer:
\[ k = 17 \](A1) Correct value
Question 6 (1 mark)
A circle has centre \( (-1, 2) \) and radius 5.
Which of these is the equation of the circle?
- \( (x + 1)^2 + (y – 2)^2 = 5 \)
- \( (x – 1)^2 + (y + 2)^2 = 5 \)
- \( (x + 1)^2 + (y – 2)^2 = 25 \)
- \( (x – 1)^2 + (y + 2)^2 = 25 \)
Worked Solution
Step 1: Recall the Formula
💡 Formula:
The equation of a circle with centre \( (a, b) \) and radius \( r \) is:
\[ (x – a)^2 + (y – b)^2 = r^2 \]Step 2: Substitute Values
Centre \( (a, b) = (-1, 2) \)
Radius \( r = 5 \)
Substitute these into the formula:
\[ (x – (-1))^2 + (y – 2)^2 = 5^2 \] \[ (x + 1)^2 + (y – 2)^2 = 25 \]✅ Final Answer:
\( (x + 1)^2 + (y – 2)^2 = 25 \)
(B1) Correct selection (3rd option)
Question 7 (3 marks)
Points \( A, B \) and \( C \) lie on a circle, centre \( O \).
Angle \( AOC = x + 75^\circ \)
Angle \( ABC = 2x \)
Work out the value of \( x \).
Worked Solution
Step 1: Identify the Circle Theorem
💡 Theory:
The angle subtended by an arc at the centre is double the angle subtended at the circumference.
Reflex angle \( AOC = 2 \times \text{Angle } ABC \)
Wait, looking at the diagram, Angle \( AOC \) marked is the obtuse angle (inside the quadrilateral \( OABC \) if we see it that way? No, \( O \) is the centre). The angle \( x+75 \) is at the centre.
However, angle \( ABC \) is subtended by the major arc \( AC \). The angle at the centre corresponding to the major arc is the reflex angle \( AOC \).
Or, simply: Is angle \( AOC \) and angle \( ABC \) subtended by the same arc?
Looking at the standard arrowhead shape: The angle at the centre (\( x+75 \)) is twice the angle at the circumference (reflex \( 2x \)? No.)
Let’s check the position. \( B \) is on the major arc. \( O \) is the centre. The angle at the centre is \( x + 75 \). The angle at the circumference is \( 2x \).
Theorem: Angle at centre = \( 2 \times \) Angle at circumference.
But the diagram shows \( x+75 \) as the obtuse angle inside the triangle \( AOC \)? No, it’s the angle facing downwards.
Actually, let’s look at the geometry. If \( B \) is at the bottom, and \( A, C \) are above, then \( ABC \) faces upwards. The angle \( x+75 \) is the angle at \( O \) facing downwards. These face the same way.
Therefore: \( x + 75 = 2 \times (2x) \) is incorrect if they are subtended by the minor arc? No.
Let’s re-evaluate. Angles around a point sum to 360. If \( x+75 \) is the internal angle, the reflex is \( 360 – (x+75) \).
Case 1: The angle marked \( x+75 \) and angle \( 2x \) subtend the SAME arc.
If so: \( x + 75 = 2(2x) \).
Let’s test: \( x + 75 = 4x \Rightarrow 3x = 75 \Rightarrow x = 25 \).
If \( x=25 \), Angle at centre = 100. Angle at circumference = 50. \( 100 = 2 \times 50 \). This works perfectly.
Step 2: Solve the Equation
Subtract \( x \) from both sides:
\[ 75 = 3x \] \[ x = 25 \](M1) States relation: Angle at centre = 2 × angle at circumference
(M1 dep) Equation \( x + 75 = 4x \)
✅ Final Answer:
\[ x = 25 \](A1) Correct value
Question 8 (2 marks)
Write \( (1 + 2\sqrt{5})(4 – \sqrt{5}) \) in the form \( a + b\sqrt{5} \) where \( a \) and \( b \) are integers.
Worked Solution
Step 1: Expand the Brackets (FOIL)
First: \( 1 \times 4 = 4 \)
Outer: \( 1 \times -\sqrt{5} = -\sqrt{5} \)
Inner: \( 2\sqrt{5} \times 4 = 8\sqrt{5} \)
Last: \( 2\sqrt{5} \times -\sqrt{5} = -2 \times 5 = -10 \)
(M1) Expansion with at least 3 correct terms
Step 2: Collect Like Terms
Integers: \( 4 – 10 = -6 \)
Surds: \( -\sqrt{5} + 8\sqrt{5} = 7\sqrt{5} \)
\[ -6 + 7\sqrt{5} \]✅ Final Answer:
\[ -6 + 7\sqrt{5} \](A1) Correct form
Question 9 (4 marks)
\( f(x) = 14 – x^2 \) for all real values of \( x \).
Solve \( f(2x) = 5 \).
You must show your working.
Worked Solution
Step 1: Find Expression for \( f(2x) \)
💡 How to substitute:
Replace every \( x \) in the original function with \( (2x) \). Use brackets!
(M1) Correct substitution \( 14 – (2x)^2 \)
Step 2: Set up and Solve Equation
Set \( f(2x) = 5 \):
\[ 14 – 4x^2 = 5 \]Rearrange to find \( x^2 \):
\[ 14 – 5 = 4x^2 \] \[ 9 = 4x^2 \] \[ x^2 = \frac{9}{4} \](M1 dep) Rearranging to \( 4x^2 = 9 \)
Step 3: Square Root
⚠️ Critical Point:
Remember both positive and negative roots.
✅ Final Answer:
\[ x = 1.5 \quad \text{or} \quad x = -1.5 \](A1) 1.5
(A1) -1.5
Question 10 (3 marks)
Rearrange \( \frac{1}{xy} = 4 – \frac{3}{y} \) to make \( x \) the subject.
Worked Solution
Step 1: Clear the Fractions
💡 Strategy:
Multiply every term by \( xy \) (the common denominator structure) to remove fractions.
So the equation becomes:
\[ 1 = 4xy – 3x \](M1) Clearing fractions correctly
Step 2: Factorise x
We need \( x \) on its own. It appears in two terms on the right side.
\[ 1 = x(4y – 3) \](M1 dep) Factorising x
Step 3: Isolate x
Divide by \( (4y – 3) \):
\[ x = \frac{1}{4y – 3} \]✅ Final Answer:
\[ x = \frac{1}{4y – 3} \](A1) Correct expression
Question 11 (4 marks)
A curve has equation \( y = 2x^2 + 3x – 9 \).
At a point \( P \) on the curve, the tangent is parallel to the line \( y = 4 – 5x \).
Work out the coordinates of \( P \).
You must show your working.
Worked Solution
Step 1: Identify the Gradient
💡 Connection:
If the tangent is parallel to \( y = 4 – 5x \), it must have the same gradient.
The line equation is \( y = -5x + 4 \), so the gradient \( m = -5 \).
Target Gradient = -5
Step 2: Differentiate the Curve
💡 Why?
The gradient of the curve is given by the derivative \( \frac{dy}{dx} \).
(M1) Correct differentiation \( 4x + 3 \)
Step 3: Solve for x
Set the derivative equal to the target gradient:
\[ 4x + 3 = -5 \] \[ 4x = -8 \] \[ x = -2 \](M1 dep) Equating derivative to -5
(A1) \( x = -2 \)
Step 4: Calculate y
Substitute \( x = -2 \) back into the original curve equation:
\[ y = 2(-2)^2 + 3(-2) – 9 \] \[ y = 2(4) – 6 – 9 \] \[ y = 8 – 15 \] \[ y = -7 \](A1 ft) \( y = -7 \) (follow through if method correct)
✅ Final Answer:
\[ P(-2, -7) \]Question 12 (4 marks)
In the diagram:
- \( A \) is the point \( (15, 0) \)
- \( B \) lies on the y-axis
- Angle \( ABC = 90^\circ \)
- \( \tan \theta = \frac{5}{3} \)
Work out the equation of the line \( BC \).
Worked Solution
Step 1: Find Coordinates of B
💡 Using Trigonometry:
In triangle \( OAB \), \( \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{OB}{OA} \).
We know \( OA = 15 \) and \( \tan \theta = \frac{5}{3} \).
Since \( B \) is on the y-axis, coordinates of \( B \) are \( (0, 25) \).
(M1) Finding length OB = 25 or B(0, 25)
Step 2: Find Gradient of AB
Gradient \( = \frac{\text{Change in } y}{\text{Change in } x} \)
Points are \( B(0, 25) \) and \( A(15, 0) \).
\[ m_{AB} = \frac{0 – 25}{15 – 0} = \frac{-25}{15} = -\frac{5}{3} \](M1) Correct gradient of AB
Step 3: Find Gradient of BC
💡 Perpendicular Lines:
Since \( ABC = 90^\circ \), \( AB \) is perpendicular to \( BC \).
Gradient of BC \( = \frac{-1}{\text{Gradient of AB}} \).
(M1) Negative reciprocal for perpendicular gradient
Step 4: Equation of Line BC
💡 Equation form:
\( y = mx + c \)
We know \( m = \frac{3}{5} \) and the y-intercept \( c \) is 25 (from point B).
Alternative forms accepted: \( 5y = 3x + 125 \)
(A1) Correct equation
✅ Final Answer:
\[ y = \frac{3}{5}x + 25 \]Question 13 (6 marks)
Solve the simultaneous equations:
\[ xy = 2 \] \[ y = 3x + 5 \]Do not use trial and improvement. You must show your working.
Worked Solution
Step 1: Substitution
💡 Strategy:
Substitute the expression for \( y \) from the linear equation into the non-linear equation.
Substitute \( y = 3x + 5 \) into \( xy = 2 \):
\[ x(3x + 5) = 2 \] \[ 3x^2 + 5x = 2 \]Rearrange to form a quadratic:
\[ 3x^2 + 5x – 2 = 0 \](M1) Substitution
(M1 dep) Correct 3-term quadratic
Step 2: Solve the Quadratic
Factorise \( 3x^2 + 5x – 2 \).
We need numbers multiplying to \( 3 \times -2 = -6 \) and adding to 5. (+6 and -1)
\[ 3x^2 + 6x – x – 2 = 0 \] \[ 3x(x + 2) – 1(x + 2) = 0 \] \[ (3x – 1)(x + 2) = 0 \]So:
\[ x = \frac{1}{3} \quad \text{or} \quad x = -2 \](M1 dep) Factorising or formula
(A1) Correct x-values
Step 3: Find Corresponding y-values
When \( x = \frac{1}{3} \):
\[ y = 3(\frac{1}{3}) + 5 = 1 + 5 = 6 \]When \( x = -2 \):
\[ y = 3(-2) + 5 = -6 + 5 = -1 \](A1) Correct coordinate pairs
✅ Final Answer:
\[ x = \frac{1}{3}, y = 6 \] \[ x = -2, y = -1 \](A1) All correct
Question 14 (3 marks)
Work out the value of:
\[ \left( \frac{1}{3^2} + 3^2 \right)^2 \]Worked Solution
Step 1: Simplify Inside the Bracket
So we have \( \frac{1}{9} + 9 \).
This can be written as a fraction: \( \frac{1}{9} + \frac{81}{9} = \frac{82}{9} \).
Wait, is there a smarter way? Let’s check the expansion first.
Step 2: Alternative – Expansion
💡 Algebraic Expansion:
\( (a + b)^2 = a^2 + 2ab + b^2 \)
Let \( a = \frac{1}{9} \) and \( b = 9 \).
Wait, let me check the question text carefully. Is it \( \frac{1}{3^2} \) or \( 3^{\frac{1}{2}} \)?
Scanning question… “Work out the value of \( \left( \frac{1}{3^2} + 3^2 \right)^2 \)”.
Wait, the OCR says “Work out the value of \( \sqrt{3}^2 + 3^2 \)”? No.
Correction: Let’s look at the mark scheme for Q14.
Mark scheme says: “Alt 1: \( 3\sqrt{3} + \dots \) oe”.
Let’s re-read the screenshot for Page 15 (Q14).
Image shows: \( \left( 3^{\frac{1}{2}} + 3^{\frac{3}{2}} \right)^2 \). Ah!
The OCR missed the fractional powers. The image clearly shows indices.
Question is: \( \left( 3^{\frac{1}{2}} + 3^{\frac{3}{2}} \right)^2 \).
Step 3: Correct Calculation
Inside bracket:
\[ 3^{\frac{1}{2}} = \sqrt{3} \] \[ 3^{\frac{3}{2}} = 3^1 \times 3^{\frac{1}{2}} = 3\sqrt{3} \]So the expression is:
\[ ( \sqrt{3} + 3\sqrt{3} )^2 \] \[ ( 4\sqrt{3} )^2 \](M1) Simplify to \( 4\sqrt{3} \) or expanding terms
Step 4: Square the Result
(M1 dep) Squaring correctly
✅ Final Answer:
\[ 48 \](A1) Correct answer
Question 15 (4 marks)
Here is the graph of \( y = 3x – x^2 \) for values of \( x \) from -1 to 4.
By drawing a suitable linear graph on the grid, work out approximate solutions to:
\[ x^2 – 4x + 2 = 0 \]Worked Solution
Step 1: Manipulate the Equation
💡 Strategy:
We have the graph of \( y = 3x – x^2 \).
We want to solve \( x^2 – 4x + 2 = 0 \).
We need to rearrange the equation so that part of it looks like our graph equation \( 3x – x^2 \).
Start with:
\[ x^2 – 4x + 2 = 0 \]We want \( -x^2 + 3x \) on one side.
Rearrange:
\[ 2 – x = -x^2 + 3x \]Or:
\[ 2 – x = 3x – x^2 \]This means we need to find where the graph \( y = 3x – x^2 \) intersects with the line \( y = 2 – x \).
(M1) Finding the linear equation \( y = 2 – x \)
Step 2: Draw the Line
📝 Plotting \( y = 2 – x \):
When \( x = 0, y = 2 \). Point (0, 2).
When \( x = 2, y = 0 \). Point (2, 0).
Draw a straight line through these points.
(M1) Drawing correct line
Step 3: Read Intersection Points
Read the x-values where the line crosses the curve.
Intersections are approximately at:
\( x \approx 0.6 \)
\( x \approx 3.4 \)
✅ Final Answer:
\[ x \approx 0.6 \quad \text{and} \quad x \approx 3.4 \](A1) 0.5 to 0.7
(A1) 3.3 to 3.5
Question 16 (4 marks)
\( y = f(x) \) is a cubic curve with a maximum and a minimum stationary point.
\[ \frac{dy}{dx} = x^2 + 2x – 3 \]- The y-coordinate of the minimum point is \( 2\frac{1}{3} \)
- The y-coordinate of the maximum point is \( 13 \)
- \( (0, 4) \) is a point on the curve
- The tangent at \( (0, 4) \) has a negative gradient
Sketch the curve on the grid below.
Show the coordinates of the stationary points.
Worked Solution
Step 1: Find x-coordinates of Stationary Points
💡 Theory:
Stationary points occur when the gradient \( \frac{dy}{dx} = 0 \).
Factorise the quadratic:
\[ (x + 3)(x – 1) = 0 \]So \( x = -3 \) or \( x = 1 \).
(M1) Setting derivative to 0 and solving
Step 2: Identify Max and Min
💡 Reasoning:
A cubic with a positive \( x^3 \) term goes up-down-up. The first stationary point (left, lower x) is the Maximum. The second (right, higher x) is the Minimum.
Alternatively, test the second derivative or check y-values given.
Maximum: At \( x = -3 \). Given y-coordinate is 13. Point: \( (-3, 13) \).
Minimum: At \( x = 1 \). Given y-coordinate is \( 2\frac{1}{3} \). Point: \( (1, 2.33) \).
Intercept: Given point \( (0, 4) \).
Step 3: Sketch the Curve
(A1) Correct stationary points matched to x-values
(A1) Smooth curve crossing negative x-axis
Question 17 (4 marks)
(a) Use the factor theorem to show that \( (x – 2) \) is a factor of \( x^3 + 8x^2 + 5x – 50 \).
(b) Hence, factorise fully \( x^3 + 8x^2 + 5x – 50 \).
Worked Solution
Part (a): Factor Theorem
💡 Theory:
If \( (x – a) \) is a factor of \( f(x) \), then \( f(a) = 0 \).
Here, we test \( x = 2 \).
Since \( f(2) = 0 \), \( (x – 2) \) is a factor.
(B1) Evaluates f(2) to 0
Part (b): Factorise Fully
💡 Strategy:
We know \( (x – 2) \) is a factor. We can use polynomial division or inspection to find the quadratic factor.
\( (x – 2)(ax^2 + bx + c) = x^3 + 8x^2 + 5x – 50 \)
By Inspection:
First term: \( x \times ax^2 = x^3 \Rightarrow a = 1 \)
Last term: \( -2 \times c = -50 \Rightarrow c = 25 \)
So we have \( (x – 2)(x^2 + bx + 25) \).
Check coefficient of \( x \):
\( -2(bx) + 25(x) \) … No, check \( x^2 \) term:
\( x(bx) – 2(x^2) = 8x^2 \)
\( bx^2 – 2x^2 = 8x^2 \)
\( b – 2 = 8 \Rightarrow b = 10 \)
Quadratic factor: \( x^2 + 10x + 25 \)
(M1) Finding quadratic \( x^2 + 10x + 25 \)
Step 3: Factorise the Quadratic
This is a perfect square:
\[ (x + 5)(x + 5) \text{ or } (x + 5)^2 \]Final Expression:
\[ (x – 2)(x + 5)^2 \](A1) Quadratic factor
(A1) Fully factorised
✅ Final Answer:
\[ (x – 2)(x + 5)^2 \]Question 18 (5 marks)
\( D, E, F \) and \( S \) are points on a circle.
\( RST \) is a tangent.
The straight line \( EDT \) is parallel to \( FS \).
\( DS = DT \).
Prove that \( FD \) is parallel to \( RST \).
Use angle \( DTS \) as \( x \) to help you.
Worked Solution
Step 1: Use Isosceles Triangle Properties
💡 Given: \( DS = DT \)
This means triangle \( DST \) is isosceles.
Base angles are equal.
(M1) State \( DST = x \)
Step 2: Use Parallel Lines (Corresponding Angles)
💡 Given: \( EDT \) is parallel to \( FS \).
We have transversal line \( RST \).
Since \( EDT \) (or \( DT \)) is parallel to \( FS \):
\[ \text{Angle } RSF = \text{Angle } DTS = x \quad (\text{Corresponding Angles}) \](M1) State \( RSF = x \) (Corresponding)
Step 3: Use Alternate Segment Theorem
💡 Theory:
The angle between a tangent (\( RS \)) and a chord (\( SF \)) is equal to the angle subtended by the chord in the alternate segment (\( \text{Angle } FDS \)).
(M1) State \( FDS = x \) using Alternate Segment Theorem
Step 4: Conclusion
💡 Proof:
We have found that \( \text{Angle } FDS = x \).
We already established \( \text{Angle } DST = x \).
These are alternate angles (Z-angles) relative to lines \( FD \) and \( ST \).
Since alternate angles \( FDS \) and \( DST \) are equal (both are \( x \)):
\( FD \) is parallel to \( RST \).
(A1) Complete proof with reasons
✅ Proof Complete
Question 19 (4 marks)
Write \( 2x^2 – 16x + 13 \) in the form \( a(x + b)^2 + c \) where \( a, b \) and \( c \) are integers.
Worked Solution
Step 1: Factorise the Coefficient of x²
💡 Strategy:
Factorise the 2 out of the first two terms only.
(M1) Factorising 2
Step 2: Complete the Square Inside
Halve the coefficient of \( x \) (-8) to get -4.
\[ x^2 – 8x = (x – 4)^2 – 16 \]Substitute this back:
\[ 2[ (x – 4)^2 – 16 ] + 13 \](M1 dep) Correct structure \( (x-4)^2 \)
Step 3: Expand and Simplify
Multiply out the square bracket by 2:
\[ 2(x – 4)^2 – 32 + 13 \]Simplify the constant terms:
\[ 2(x – 4)^2 – 19 \](M1 dep) Expanding correct subtraction
✅ Final Answer:
\[ 2(x – 4)^2 – 19 \](A1) Correct expression
Question 20 (5 marks)
In triangle \( ABC \):
- \( AB = 6\sqrt{2} \) cm
- Angle \( ABC = 45^\circ \)
- Angle \( ACB = 60^\circ \)
Work out the value of \( x \) (length \( AC \)).
Give your answer in the form \( a\sqrt{b} \), where \( a \) and \( b \) are integers.
Worked Solution
Step 1: Choose the Sine Rule
💡 Why?
We have pairs of opposite angles and sides. We want side \( AC \) (opposite \( 45^\circ \)) and we have side \( AB \) (opposite \( 60^\circ \)).
(M1) Correct Sine Rule set up
Step 2: Substitute Exact Trigonometric Values
💡 Recall values:
\( \sin 45^\circ = \frac{1}{\sqrt{2}} \) (or \( \frac{\sqrt{2}}{2} \))
\( \sin 60^\circ = \frac{\sqrt{3}}{2} \)
(B1) Correct exact values seen
Step 3: Rearrange and Solve
Multiply both sides by \( \sin 45^\circ \):
\[ x = \frac{6\sqrt{2} \times \sin 45^\circ}{\sin 60^\circ} \] \[ x = \frac{6\sqrt{2} \times \frac{1}{\sqrt{2}}}{\frac{\sqrt{3}}{2}} \]Simplify numerator:
\[ 6\sqrt{2} \times \frac{1}{\sqrt{2}} = 6 \]So:
\[ x = \frac{6}{\frac{\sqrt{3}}{2}} \] \[ x = 6 \times \frac{2}{\sqrt{3}} = \frac{12}{\sqrt{3}} \](M1 dep) Isolating x correctly
Step 4: Rationalise the Denominator
✅ Final Answer:
\[ 4\sqrt{3} \text{ cm} \](A1) Correct final form