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AQA Level 2 Certificate Further Mathematics Paper 2 (Calculator) June 2019
📝 Legend
- (M1): Method Mark – for a correct method or step
- (A1): Accuracy Mark – for a correct calculation or value
- (B1): Independent Mark – for a correct answer independent of method
- (ft): Follow Through – credit for correct work based on previous error
📚 Table of Contents
- Question 1 (Matrices)
- Question 2 (Coordinate Geometry)
- Question 3 (Inequalities)
- Question 4 (Calculus & Coordinate Geometry)
- Question 5 (Indices & Algebra)
- Question 6 (Quadratic Curves)
- Question 7 (Calculus Graphs)
- Question 8 (Sequences)
- Question 9 (Factorisation)
- Question 10 (Functions)
- Question 11 (Normals to Curves)
- Question 12 (Volume of Hemisphere)
- Question 13 (Algebraic Fractions)
- Question 14 (Cosine Rule)
- Question 15 (Rearranging Formulae)
- Question 16 (Surds & Functions)
- Question 17 (Algebraic Proof)
- Question 18 (Solving Quadratics)
- Question 19 (Geometry & Area)
- Question 20 (Circle Theorems)
- Question 21 (3D Trigonometry)
- Question 22 (Trigonometric Identities)
- Question 23 (Circle Geometry & Tangents)
- Question 24 (Stationary Points)
- Question 25 (Factor Theorem)
- Question 26 (Domain of Functions)
- Question 27 (Points of Inflection)
Question 1 (5 marks total)
1 (a)
\( a \begin{pmatrix} 3 \\ 5 \end{pmatrix} = 4 \begin{pmatrix} 2a+3 \\ b \end{pmatrix} \)
Work out the values of \( a \) and \( b \).
[3 marks]
1 (b)
\( \begin{pmatrix} m & -1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 2 & 2 \\ -2 & -1 \end{pmatrix} = \mathbf{I} \)
where \( \mathbf{I} \) is the identity matrix.
Work out the value of \( m \).
[2 marks]
Worked Solution
Part (a): Understanding Scalar Multiplication
What are we doing?
We need to multiply the vectors by the scalars (numbers) outside them and then equate the top and bottom components to form algebraic equations.
✏ Working:
Expand the scalar multiplication:
\[ \begin{pmatrix} 3a \\ 5a \end{pmatrix} = \begin{pmatrix} 4(2a+3) \\ 4b \end{pmatrix} \]This gives us two simultaneous equations:
- \( 3a = 4(2a+3) \)
- \( 5a = 4b \)
Part (a): Solving for a
Strategy:
Equation (1) only contains \( a \), so we solve that first.
✏ Working:
\[ 3a = 8a + 12 \]Subtract \( 8a \) from both sides:
\[ -5a = 12 \] \[ a = \frac{12}{-5} = -2.4 \](M1) for equation, (A1) for correct value of \( a \)
Part (a): Solving for b
Strategy:
Substitute our value of \( a \) into Equation (2).
✏ Working:
\[ 5(-2.4) = 4b \] \[ -12 = 4b \] \[ b = \frac{-12}{4} = -3 \](A1ft) Correct value of \( b \)
Part (b): Matrix Multiplication
What does \( \mathbf{I} \) mean?
The identity matrix \( \mathbf{I} \) is \( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \).
We need to multiply the two matrices and equate the result to \( \mathbf{I} \).
✏ Working:
\[ \begin{pmatrix} m & -1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 2 & 2 \\ -2 & -1 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]Perform row-by-column multiplication for the top-left element:
\[ (m \times 2) + (-1 \times -2) = 1 \] \[ 2m + 2 = 1 \](M1) Correct matrix multiplication equation
Part (b): Solving for m
✏ Working:
\[ 2m = 1 – 2 \] \[ 2m = -1 \] \[ m = -0.5 \](A1)
Check:
We can check using the top-right element: \( (m \times 2) + (-1 \times -1) = 0 \).
\( (-0.5 \times 2) + 1 = -1 + 1 = 0 \). It works.
Final Answer:
(a) \( a = -2.4 \), \( b = -3 \)
(b) \( m = -0.5 \)
Question 2 (3 marks)
Here is a sketch of quadrilateral \( PQRS \).
\( M \) is the midpoint of \( PS \).
Use gradients to show that \( MR \) is parallel to \( PQ \).
Worked Solution
Step 1: Find the coordinates of M
Why?
We are told \( M \) is the midpoint of \( PS \). To find the gradient of \( MR \), we first need the coordinates of point \( M \).
✏ Working:
Midpoint formula: \( (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}) \)
Using \( P(4, 1) \) and \( S(6, 11) \):
\[ M = \left(\frac{4+6}{2}, \frac{1+11}{2}\right) = \left(\frac{10}{2}, \frac{12}{2}\right) = (5, 6) \](M1) Correct coordinates for M
Step 2: Calculate Gradients
Strategy:
Two lines are parallel if they have the same gradient. We need to calculate the gradient of \( MR \) and the gradient of \( PQ \).
Gradient formula: \( m = \frac{y_2 – y_1}{x_2 – x_1} \)
✏ Working:
Gradient of MR: Using \( M(5, 6) \) and \( R(14, 0) \):
\[ m_{MR} = \frac{0 – 6}{14 – 5} = \frac{-6}{9} = -\frac{2}{3} \]Gradient of PQ: Using \( P(4, 1) \) and \( Q(10, -3) \):
\[ m_{PQ} = \frac{-3 – 1}{10 – 4} = \frac{-4}{6} = -\frac{2}{3} \](M1) Method for finding gradients
Step 3: Conclusion
What this tells us:
We compare the two values. If they are identical, the lines are parallel.
✏ Working:
Gradient of \( MR = -\frac{2}{3} \)
Gradient of \( PQ = -\frac{2}{3} \)
Since the gradients are equal, \( MR \) is parallel to \( PQ \).
(A1) Conclusion with supporting evidence
Final Answer:
Both gradients are \( -\frac{2}{3} \), therefore the lines are parallel.
Question 3 (4 marks)
\( -2 < a < 0 \) and \( -1 < b < 1 \)
Tick the correct box for each statement.
Worked Solution
Analysis of each statement
Context: \( a \) is a negative number between -2 and 0. \( b \) is a small number between -1 and 1 (could be positive, negative, or zero).
1. \( a^2 < 0 \)
Square numbers are always positive (or zero). Since \( a \neq 0 \), \( a^2 \) must be positive.
Therefore \( a^2 < 0 \) is Never true.
2. \( -1 < b^3 < 1 \)
If \( -1 < b < 1 \), then cubing preserves the inequality signs.
e.g., \( (0.5)^3 = 0.125 \) (in range), \( (-0.9)^3 = -0.729 \) (in range).
This is Always true.
3. \( \frac{b}{a} < 0 \)
\( a \) is always negative.
If \( b \) is positive (e.g., 0.5), \( \frac{+}{-} = \text{negative} \) (True).
If \( b \) is negative (e.g., -0.5), \( \frac{-}{-} = \text{positive} \) (False).
This is Sometimes true.
4. \( a – b > 0 \)
Rearranging: \( a > b \).
Since \( a \) is negative (e.g. -1.5) and \( b \) can be positive (e.g. 0.5), \( -1.5 > 0.5 \) is false.
Can it be true? If \( a = -0.5 \) and \( b = -0.9 \), then \( -0.5 > -0.9 \) is true.
This is Sometimes true.
(B4) 1 mark for each correct row
Final Answer:
- \( a^2 < 0 \): Never true
- \( -1 < b^3 < 1 \): Always true
- \( \frac{b}{a} < 0 \): Sometimes true
- \( a – b > 0 \): Sometimes true
Question 4 (4 marks)
\( P \) is a point on a curve.
The curve has gradient function \( \frac{x^5 – 17}{10} \)
The tangent to the curve at \( P \) is parallel to the line \( 3x – 2y = 9 \)
Work out the \( x \)-coordinate of \( P \).
Worked Solution
Step 1: Find the gradient of the line
Why?
Parallel lines have the same gradient. To find the gradient of the curve at \( P \), we first find the gradient of the line \( 3x – 2y = 9 \).
We rearrange it into the form \( y = mx + c \).
✏ Working:
\[ 3x – 2y = 9 \] \[ 3x – 9 = 2y \] \[ y = \frac{3}{2}x – \frac{9}{2} \]The gradient \( m \) is \( \frac{3}{2} \) (or 1.5).
(M1) Attempt to find gradient of line
Step 2: Equate gradients
How?
The question states the gradient function of the curve is equal to the gradient of the line at point \( P \).
✏ Working:
\[ \frac{x^5 – 17}{10} = 1.5 \](M1) Equating gradient function to 1.5
Step 3: Solve for x
✏ Working:
Multiply by 10:
\[ x^5 – 17 = 15 \]Add 17 to both sides:
\[ x^5 = 32 \](M1) Rearranging to solve for \( x^5 \)
Take the 5th root:
\[ x = \sqrt[5]{32} \] \[ x = 2 \](A1) Correct value
Check: \( 2^5 = 32 \). Correct.
Final Answer:
\( x = 2 \)
Question 5 (5 marks total)
5 (a)
Write \( \sqrt[4]{a \times a^{-9}} \) as an integer power of \( a \).
[2 marks]
5 (b)
Simplify fully \( \frac{(4cd^2)^3}{2cd^4} \)
[3 marks]
Worked Solution
Part (a): Simplifying the term inside the root
Rule: \( a^m \times a^n = a^{m+n} \)
Note that \( a = a^1 \).
✏ Working:
\[ a^1 \times a^{-9} = a^{1 + (-9)} = a^{-8} \]So the expression becomes \( \sqrt[4]{a^{-8}} \).
Part (a): Dealing with the 4th root
Rule: \( \sqrt[n]{x} = x^{\frac{1}{n}} \)
✏ Working:
\[ (a^{-8})^{\frac{1}{4}} \]Multiply the powers:
\[ -8 \times \frac{1}{4} = -2 \] \[ a^{-2} \](B2) Correct answer (B1 for simplifying inside or converting root)
Part (b): Expanding the bracket
Strategy: Apply the power of 3 to every part of \( 4cd^2 \).
\( (xy)^n = x^n y^n \)
✏ Working:
\[ (4cd^2)^3 = 4^3 \times c^3 \times (d^2)^3 \] \[ = 64 c^3 d^6 \](B1) Correct expansion of numerator
Part (b): Simplifying the fraction
✏ Working:
\[ \frac{64 c^3 d^6}{2 c d^4} \]Divide numbers: \( 64 \div 2 = 32 \)
Divide \( c \): \( c^3 \div c^1 = c^2 \)
Divide \( d \): \( d^6 \div d^4 = d^2 \)
Combine them:
\[ 32 c^2 d^2 \](B2) Fully simplified (B1 for two correct terms)
Final Answer:
(a) \( a^{-2} \)
(b) \( 32c^2d^2 \)
Question 6 (3 marks total)
Here is a sketch of the curve \( y = (2x+3)(x-2) \)
The curve intersects the \( x \)-axis at \( A \) and \( B \).
6 (a)
Complete the coordinates of \( A \) and \( B \).
\( A ( \dots\dots , 0 ) \quad B ( \dots\dots , 0 ) \)
[2 marks]
6 (b)
Write down the range of values for \( x \) for which \( (2x+3)(x-2) < 0 \)
[1 mark]
Worked Solution
Part (a): Finding Intercepts
What are we doing?
The \( x \)-axis intercepts occur where \( y = 0 \).
✏ Working:
\[ (2x+3)(x-2) = 0 \]Either \( 2x+3 = 0 \) or \( x-2 = 0 \).
1. \( 2x = -3 \implies x = -1.5 \)
2. \( x = 2 \)
From the diagram, \( A \) is on the negative side and \( B \) is on the positive side.
\( A(-1.5, 0) \) and \( B(2, 0) \)
(M1) Solving equation, (A1) Correct assignment to A and B
Part (b): Solving the Inequality
Visual Interpretation:
The inequality \( < 0 \) asks: "Where is the curve below the \( x \)-axis?”
Looking at the sketch, the curve dips below the axis between the points \( A \) and \( B \).
✏ Working:
The values are between \( -1.5 \) and \( 2 \).
\[ -1.5 < x < 2 \](B1) Correct inequality (ft from part a)
Final Answer:
(a) \( A(-1.5, 0) \), \( B(2, 0) \)
(b) \( -1.5 < x < 2 \)
Question 7 (2 marks total)
7 (a)
On the grid, sketch a graph for which the rate of change of \( y \) with respect to \( x \) is always zero.
[1 mark]
7 (b)
On the grid, sketch a graph for which the rate of change of \( y \) with respect to \( x \) is always a positive constant.
[1 mark]
Worked Solution
Part (a): Rate of change is zero
Reasoning:
“Rate of change” means gradient.
A zero gradient means the line is flat (horizontal). The value of \( y \) does not change as \( x \) changes.
Sketch: A horizontal straight line.
(B1) Horizontal straight line
Part (b): Rate of change is positive constant
Reasoning:
“Positive” means the line goes up from left to right.
“Constant” means it is a straight line (not a curve).
Sketch: A straight line with a positive slope.
(B1) Straight line with positive gradient
Question 8 (7 marks total)
8 (a)
A linear sequence has first term \( 7 + 12\sqrt{5} \)
The term-to-term rule is: add \( 9 – 2\sqrt{5} \)
One term of the sequence is an integer.
Work out the value of this integer.
[2 marks]
8 (b)
The \( n \)th term of a different sequence is \( \frac{3n^2 – 1}{n^2 + 1} \)
Work out the sum of the first three terms.
[2 marks]
8 (c)
The first four terms of a quadratic sequence are:
-3 3 13 27
Work out an expression for the \( n \)th term.
[3 marks]
Worked Solution
Part (a): Linear Sequence
Strategy: We start with \( 7 + 12\sqrt{5} \) and keep adding \( 9 – 2\sqrt{5} \). We want the term where the \( \sqrt{5} \) part becomes zero.
We start with \( 12\sqrt{5} \). Each step removes \( 2\sqrt{5} \).
✏ Working:
How many steps to remove \( 12\sqrt{5} \)?
\[ 12\sqrt{5} \div 2\sqrt{5} = 6 \text{ steps} \]So we add the term 6 times.
Start value (integer part): \( 7 \)
Added value (integer part): \( 6 \times 9 = 54 \)
\[ 7 + 54 = 61 \](M1) Identifies 6 steps or correct method, (A1) 61
Part (b): Sum of terms
Strategy: Substitute \( n=1, 2, 3 \) into the formula and add the results.
✏ Working:
\( n=1 \): \( \frac{3(1)^2 – 1}{1^2 + 1} = \frac{2}{2} = 1 \)
\( n=2 \): \( \frac{3(2)^2 – 1}{2^2 + 1} = \frac{11}{5} = 2.2 \)
\( n=3 \): \( \frac{3(3)^2 – 1}{3^2 + 1} = \frac{26}{10} = 2.6 \)
Sum: \( 1 + 2.2 + 2.6 = 5.8 \)
(B1) Correct terms, (B1) Correct sum 5.8
Part (c): Quadratic Sequence
Method: Find the first and second differences.
✏ Working:
Sequence: \( -3, \quad 3, \quad 13, \quad 27 \)
1st Diff: \( \quad +6, \quad +10, \quad +14 \)
2nd Diff: \( \quad\quad +4, \quad\quad +4 \)
The coefficient of \( n^2 \) is half the second difference: \( 4 \div 2 = 2 \).
So the term starts with \( 2n^2 \).
(M1) Implies \( 2n^2 \)
Part (c): Finding the rest
✏ Working:
Write out the \( 2n^2 \) sequence:
\( n=1, 2, 3, 4 \)
\( 2n^2 \): \( 2, 8, 18, 32 \)
Subtract this from the original sequence:
Original: \( -3, \quad 3, \quad 13, \quad 27 \)
\( -2n^2 \): \( -2, \quad -8, \quad -18, \quad -32 \)
Result: \( -5, \quad -5, \quad -5, \quad -5 \)
The remainder is a constant \( -5 \).
So the formula is \( 2n^2 – 5 \).
(M1) Subtracting \( 2n^2 \), (A1) Correct expression
Final Answer:
(a) 61
(b) 5.8
(c) \( 2n^2 – 5 \)
Question 9 (2 marks)
Factorise fully \( (p+6)^{11} – (p+6)^{10} \)
Worked Solution
Step 1: Identify the common factor
Strategy: Look for the highest common factor. Both terms contain \( (p+6) \). The lowest power is 10, so we can factor out \( (p+6)^{10} \).
✏ Working:
\[ (p+6)^{11} – (p+6)^{10} = (p+6)^{10} [ (p+6)^1 – 1 ] \](M1) Factorising out \( (p+6)^{10} \)
Step 2: Simplify the bracket
✏ Working:
\[ (p+6)^{10} [ p + 6 – 1 ] \] \[ (p+6)^{10} (p + 5) \](A1) Fully simplified
Final Answer:
\( (p+6)^{10}(p+5) \)
Question 10 (4 marks total)
10 (a)
\( f(x) = x^3 – 2 \)
The domain of \( f(x) \) is \( x \leq 3 \)
Work out the range of \( f(x) \).
[2 marks]
10 (b)
\( g(x) = 5 – x^2 \)
The domain of \( g(x) \) is \( -2 \leq x \leq 1 \)
Work out the range of \( g(x) \).
[2 marks]
Worked Solution
Part (a): Range of cubic function
Reasoning: The function \( x^3 – 2 \) is an increasing function. This means the higher the \( x \) value, the higher the \( f(x) \) value.
The maximum value of \( x \) is 3, so the maximum value of \( f(x) \) will occur at \( x=3 \).
✏ Working:
Calculate max value:
\[ f(3) = 3^3 – 2 = 27 – 2 = 25 \]Since \( x \) can be anything less than 3, \( f(x) \) can be anything less than 25.
\[ f(x) \leq 25 \](B2) Correct range
Part (b): Range of quadratic function
Reasoning: The function \( g(x) = 5 – x^2 \) is an upside-down parabola (n-shape).
We need to check the boundaries (\( x=-2, x=1 \)) AND the turning point.
The turning point (maximum) occurs at \( x=0 \), which is inside our domain \( -2 \leq x \leq 1 \).
✏ Working:
Check points:
1. Boundary \( x = -2 \): \( 5 – (-2)^2 = 5 – 4 = 1 \)
2. Boundary \( x = 1 \): \( 5 – (1)^2 = 5 – 1 = 4 \)
3. Turning point \( x = 0 \): \( 5 – 0^2 = 5 \)
The lowest value is 1. The highest value is 5.
\[ 1 \leq g(x) \leq 5 \](B2) Correct range (B1 for one correct limit)
Final Answer:
(a) \( f(x) \leq 25 \)
(b) \( 1 \leq g(x) \leq 5 \)
Question 11 (1 mark)
Here is a sketch of a quadratic curve which has a maximum point at \( (-2, 5) \)
What is the equation of the normal to the curve at the maximum point?
Circle your answer.
Worked Solution
Step 1: Understanding Tangents and Normals at Turning Points
Concept:
- At a maximum (or minimum) point, the tangent is a horizontal line.
- The normal is perpendicular to the tangent, so it must be a vertical line.
Step 2: Finding the Equation
Analysis:
The maximum point is at \( (-2, 5) \).
A vertical line passing through \( x = -2 \) has the equation \( x = -2 \).
(Note: The horizontal tangent would be \( y = 5 \)).
Answer: \( x = -2 \)
(B1) Correct selection
Final Answer: \( x = -2 \)
Question 12 (3 marks)
The diagram shows a solid hemisphere.
The diameter is \( 12a \) cm.
The volume is \( 486\pi \) cm\(^3\).
Work out the value of \( a \).
Worked Solution
Step 1: Formula and Radius
Formulas:
Volume of a sphere \( = \frac{4}{3}\pi r^3 \)
Volume of a hemisphere \( = \frac{1}{2} \times \frac{4}{3}\pi r^3 = \frac{2}{3}\pi r^3 \)
Radius: The diameter is \( 12a \), so the radius \( r = \frac{12a}{2} = 6a \).
Step 2: Set up the equation
✏ Working:
\[ \frac{2}{3}\pi r^3 = 486\pi \]Substitute \( r = 6a \):
\[ \frac{2}{3}\pi (6a)^3 = 486\pi \](M1) Setting up equation with correct volume formula
Step 3: Solve for a
✏ Working:
Cancel \( \pi \) from both sides:
\[ \frac{2}{3}(216a^3) = 486 \]Simplify the left side (\( 216 \div 3 = 72 \), \( 72 \times 2 = 144 \)):
\[ 144a^3 = 486 \] \[ a^3 = \frac{486}{144} \]Simplify fraction: Divide by 18 (or use calculator):
\[ a^3 = \frac{27}{8} \] \[ a = \sqrt[3]{\frac{27}{8}} \] \[ a = \frac{3}{2} = 1.5 \](M1) Rearranging for \( a^3 \), (A1) Correct value
Final Answer: \( a = 1.5 \)
Question 13 (4 marks)
Simplify fully \( \dfrac{x-x^3}{2x+2x^2} \)
You must show your working.
Worked Solution
Step 1: Factorise the Numerator
Strategy: First, take out common factor \( x \). Then look for “Difference of Two Squares”.
✏ Working:
\[ x – x^3 = x(1 – x^2) \]Using \( a^2 – b^2 = (a-b)(a+b) \):
\[ = x(1 – x)(1 + x) \](M1) Factorising numerator
Step 2: Factorise the Denominator
✏ Working:
Common factor is \( 2x \):
\[ 2x + 2x^2 = 2x(1 + x) \](M1) Factorising denominator
Step 3: Simplify
Strategy: Cancel common terms from top and bottom.
✏ Working:
\[ \frac{x(1-x)(1+x)}{2x(1+x)} \]Cancel \( x \) and \( (1+x) \):
\[ \frac{\cancel{x}(1-x)\cancel{(1+x)}}{2\cancel{x}\cancel{(1+x)}} \] \[ = \frac{1-x}{2} \](M1) Cancelling common factors, (A1) Final answer
Final Answer: \( \frac{1-x}{2} \)
Question 14 (3 marks)
Here is a triangle.
Use the cosine rule to work out the ratio \( b^2 : a^2 \)
Worked Solution
Step 1: Apply the Cosine Rule
Formula: \( c^2 = a^2 + b^2 – 2ab \cos C \)
Here, the side opposite the angle \( 120^\circ \) is \( b \).
The other two sides are \( a \) and \( 3a \).
✏ Working:
\[ b^2 = a^2 + (3a)^2 – 2(a)(3a) \cos(120^\circ) \](M1) Correct substitution into Cosine Rule
Step 2: Simplify the Expression
Recall: \( \cos(120^\circ) = -0.5 \)
✏ Working:
\[ b^2 = a^2 + 9a^2 – 6a^2(-0.5) \] \[ b^2 = 10a^2 – (-3a^2) \] \[ b^2 = 10a^2 + 3a^2 \] \[ b^2 = 13a^2 \](A1) Correct simplification to \( b^2 = 13a^2 \)
Step 3: Find the Ratio
✏ Working:
We need \( b^2 : a^2 \).
Substitute \( b^2 = 13a^2 \):
\[ 13a^2 : a^2 \]Divide by \( a^2 \):
\[ 13 : 1 \](A1) Correct ratio
Final Answer: \( 13 : 1 \)
Question 15 (4 marks)
Rearrange \( m = \dfrac{2p+1}{p} + \dfrac{p+5}{3p} \) to make \( p \) the subject.
Worked Solution
Step 1: Combine Fractions
Strategy: Find a common denominator to combine the right-hand side. The denominators are \( p \) and \( 3p \), so the common denominator is \( 3p \).
✏ Working:
Multiply the first fraction by 3 top and bottom:
\[ m = \frac{3(2p+1)}{3p} + \frac{p+5}{3p} \] \[ m = \frac{6p + 3 + p + 5}{3p} \] \[ m = \frac{7p + 8}{3p} \](M1) Combining into single fraction
Step 2: Isolate p terms
Strategy: Multiply by \( 3p \) to remove the fraction, then group \( p \) terms on one side.
✏ Working:
\[ 3pm = 7p + 8 \]Subtract \( 7p \) from both sides:
\[ 3pm – 7p = 8 \](M1) Grouping \( p \) terms
Step 3: Factorise and Solve
✏ Working:
Factor out \( p \):
\[ p(3m – 7) = 8 \]Divide by the bracket:
\[ p = \frac{8}{3m – 7} \](M1) Factorising, (A1) Correct final formula
Final Answer: \( p = \frac{8}{3m – 7} \)
Question 16 (3 marks)
The curve \( y = 2\sqrt{x-a} + 5 \) passes through the point \( (1, 8) \).
Work out the value of \( a \).
Worked Solution
Step 1: Substitute the coordinates
Strategy: The point \( (1, 8) \) lies on the curve, so substituting \( x=1 \) and \( y=8 \) into the equation must satisfy it.
✏ Working:
\[ 8 = 2\sqrt{1-a} + 5 \](M1) Substituting \( x \) and \( y \)
Step 2: Rearrange to isolate the square root
✏ Working:
Subtract 5 from both sides:
\[ 3 = 2\sqrt{1-a} \]Divide by 2:
\[ 1.5 = \sqrt{1-a} \](M1) Isolating term with \( a \)
Step 3: Solve for a
✏ Working:
Square both sides:
\[ 1.5^2 = 1 – a \] \[ 2.25 = 1 – a \]Rearrange for \( a \):
\[ a = 1 – 2.25 \] \[ a = -1.25 \](A1) Correct answer
Final Answer: \( a = -1.25 \)
Question 17 (5 marks)
Show that \( (x+1)(x+3)(x+4) – x(x^2+7x+11) \) can be written in the form \( (x+a)(x+b) \) where \( a \) and \( b \) are positive integers.
Worked Solution
Step 1: Expand the first part (triple brackets)
Strategy: Expand two brackets first, then multiply by the third.
✏ Working:
Expand \( (x+1)(x+3) \):
\[ x^2 + 3x + x + 3 = x^2 + 4x + 3 \]Now multiply by \( (x+4) \):
\[ (x^2 + 4x + 3)(x + 4) \] \[ = x(x^2 + 4x + 3) + 4(x^2 + 4x + 3) \] \[ = x^3 + 4x^2 + 3x + 4x^2 + 16x + 12 \]Collect terms:
\[ = x^3 + 8x^2 + 19x + 12 \](M1) Expansion of two brackets, (M1) Full expansion
Step 2: Expand the second part and subtract
✏ Working:
Expand \( x(x^2+7x+11) \):
\[ x^3 + 7x^2 + 11x \]Now perform the subtraction:
\[ (x^3 + 8x^2 + 19x + 12) – (x^3 + 7x^2 + 11x) \]Group like terms:
\[ (x^3 – x^3) + (8x^2 – 7x^2) + (19x – 11x) + 12 \] \[ = x^2 + 8x + 12 \](A1) Correct quadratic
Step 3: Factorise the result
Strategy: Find two numbers that multiply to 12 and add to 8.
✏ Working:
Factors of 12: 1&12, 2&6, 3&4.
2 + 6 = 8.
\[ x^2 + 8x + 12 = (x+2)(x+6) \]This is in the form \( (x+a)(x+b) \) with \( a=2, b=6 \) (or vice versa).
(A1) Factorisation
Final Answer: \( (x+2)(x+6) \)
Question 18 (3 marks)
Solve \( 4(x-5)^2 = k^2 \) where \( k \) is a constant.
Give your answers in their simplest form in terms of \( k \).
Worked Solution
Step 1: Isolate the squared term
✏ Working:
\[ (x-5)^2 = \frac{k^2}{4} \](M1) Division by 4
Step 2: Square root both sides
Important: Remember the \( \pm \) symbol when taking a square root to find variable values.
✏ Working:
\[ x – 5 = \pm \sqrt{\frac{k^2}{4}} \] \[ x – 5 = \pm \frac{k}{2} \](A1) Taking square root correctly
Step 3: Solve for x
✏ Working:
\[ x = 5 \pm \frac{k}{2} \]The two solutions are:
\[ x = 5 + \frac{k}{2} \quad \text{and} \quad x = 5 – \frac{k}{2} \]Alternatively written as \( x = \frac{10+k}{2} \) and \( x = \frac{10-k}{2} \).
(A1) Both answers in simplest form
Final Answer: \( 5 + \frac{k}{2} \) and \( 5 – \frac{k}{2} \)
Question 19 (6 marks total)
\( ABC \) is a right-angled triangle.
\( ACD \) is an isosceles triangle.
All dimensions are in centimetres.
19 (a)
Show that \( AC = 5x \).
[1 mark]
19 (b)
Work out an expression, in cm\(^2\), for the area of quadrilateral \( ABCD \).
Give your answer in the form \( px^2 \) where \( p \) is an integer.
[5 marks]
Worked Solution
Part (a): Pythagoras Theorem
Why? \( \triangle ABC \) is a right-angled triangle.
✏ Working:
\[ AC^2 = AB^2 + BC^2 \] \[ AC^2 = (3x)^2 + (4x)^2 \] \[ AC^2 = 9x^2 + 16x^2 = 25x^2 \] \[ AC = \sqrt{25x^2} = 5x \](B1) Correct working shown
Part (b): Area of Triangle ABC
✏ Working:
\[ \text{Area}_{ABC} = \frac{1}{2} \times \text{base} \times \text{height} \] \[ = \frac{1}{2} \times 4x \times 3x \] \[ = 6x^2 \](M1) Correct area for ABC
Part (b): Area of Triangle ACD
Strategy: \( \triangle ACD \) is isosceles with sides \( 6.5x, 6.5x \) and base \( 5x \). We need the vertical height relative to the base \( AC \).
Let’s drop a perpendicular from \( D \) to the midpoint of \( AC \). Let’s call the height \( h \).
✏ Working:
Half of base \( AC \) is \( 2.5x \).
Using Pythagoras on the half-triangle:
\[ h^2 + (2.5x)^2 = (6.5x)^2 \] \[ h^2 + 6.25x^2 = 42.25x^2 \] \[ h^2 = 36x^2 \] \[ h = 6x \](M1) Calculation of height
Now area of \( \triangle ACD \):
\[ \text{Area}_{ACD} = \frac{1}{2} \times \text{base} \times h \] \[ = \frac{1}{2} \times 5x \times 6x \] \[ = 15x^2 \](M1) Correct area for ACD
Part (b): Total Area
✏ Working:
\[ \text{Total Area} = \text{Area}_{ABC} + \text{Area}_{ACD} \] \[ = 6x^2 + 15x^2 \] \[ = 21x^2 \](A1) Final answer
Final Answer: \( 21x^2 \) cm\(^2\)
Question 20 (5 marks total)
\( A, B, C, D \) are points on a circle.
\( D, E, F \) are points on a different circle, centre \( C \).
\( DCE \), \( ADF \) and \( BCF \) are straight lines.
Angle \( DEF = x \)
20 (a)
Prove that angle \( BAD = 2x \).
[3 marks]
20 (b)
In the case when \( AB \) is parallel to \( DE \), work out the size of angle \( x \).
[2 marks]
Worked Solution
Part (a): Step 1 – Circle 2 Properties
Look at Circle 2 (Centre C):
We are given \( \angle DEF = x \).
\( DCE \) is a straight line passing through the centre \( C \), so \( DE \) is a diameter.
Theorem: The angle in a semi-circle is \( 90^\circ \). Therefore \( \angle DFE = 90^\circ \).
Alternatively, use the “Angle at centre” theorem. The angle at centre \( C \) subtended by arc \( DF \) is \( \angle DCF \).
Actually, a simpler property: \( \angle DCF \) is the angle at the centre? No, \( D, C, E \) is a line.
Let’s use the property that \( \triangle CDE \) is isosceles? No. \( CD \) and \( CF \) are radii.
\( \triangle CDF \) is isosceles because \( CD = CF \) (radii).
\( \angle CFE = x \) (Triangle logic? No).
Let’s try Angles at Centre:
Angle at circumference \( \angle DEF = x \). This angle subtends arc \( DF \).
Therefore, angle at centre \( \angle DCF = 2x \).
(M1) Establishing \( \angle DCF = 2x \) or similar link
Part (a): Step 2 – Link to Circle 1
Look at Circle 1:
\( BCF \) is a straight line, so \( \angle BCD \) and \( \angle DCF \) are on a straight line.
\( \angle BCD = 180^\circ – 2x \).
✏ Working:
\( ABCD \) is a cyclic quadrilateral (all points on Circle 1).
Theorem: Opposite angles in a cyclic quadrilateral sum to \( 180^\circ \).
\( \angle BAD + \angle BCD = 180^\circ \)
\[ \angle BAD + (180 – 2x) = 180 \] \[ \angle BAD – 2x = 0 \] \[ \angle BAD = 2x \](A1) Proof complete with reasons
Part (b): Parallel Lines
Given: \( AB \) is parallel to \( DE \).
Since \( DCE \) is a straight line, \( AB \) is parallel to \( DC \).
This creates alternate angles (Z-angles) or allied angles (C-angles).
✏ Working:
Extend line \( AD \). \( \angle ADC \) and \( \angle DAB \) are allied angles? Not quite.
Consider the transversal \( AD \). Or consider quadrilateral \( ABCD \).
If \( AB \parallel DC \), then \( ABCD \) is an isosceles trapezium (since it’s cyclic).
This means \( \angle BAD = \angle CDA \) and \( \angle ABC = \angle BCD \).
Wait, simpler approach:
If \( AB \parallel DE \), then \( \angle BAD \) and \( \angle ADC \) sum to \( 180^\circ \) (Interior angles)? No, that’s if \( AD \) is transversal.
Let’s use the Z-angle with transversal \( AC \)? We don’t have line \( AC \).
Let’s use the transversal \( AD \). \( \angle ADE \) corresponds to?
Let’s look at \( \angle CDE \). It’s \( 180^\circ \) (straight line). No, \( DCE \) is the line.
Angle \( \angle BCD = 180 – 2x \).
Since \( AB \parallel DC \), interior angles sum to 180:
\( \angle ABC + \angle BCD = 180 \)
\( \angle ABC + 180 – 2x = 180 \implies \angle ABC = 2x \).
Also in cyclic quad, \( \angle ADC + \angle ABC = 180 \).
\( \angle ADC + 2x = 180 \implies \angle ADC = 180 – 2x \).
Now look at the straight line \( ADF \).
\( \angle CDF = 180 – \angle ADC = 180 – (180 – 2x) = 2x \).
In Circle 2, \( \triangle CDF \) is isosceles (\( CD=CF \)).
So \( \angle CFD = \angle CDF = 2x \).
Sum of angles in \( \triangle CDF \):
\( \angle DCF + 2x + 2x = 180 \).
We know \( \angle DCF = 2x \) (from Part a).
So \( 2x + 2x + 2x = 180 \).
\[ 6x = 180 \] \[ x = 30^\circ \](M1) Setting up equation, (A1) Correct answer
Final Answer: \( x = 30^\circ \)
Question 21 (4 marks)
\( ABCDEFGH \) is a cuboid.
\( BC = 15 \) cm, \( CD = 12 \) cm, \( DH = 8 \) cm.
Work out the size of the angle between the line \( CE \) and the plane \( CDHG \).
Work out the answer to 1 decimal place.
Worked Solution
Step 1: Identify the Triangle
Strategy: To find the angle between a line and a plane, we project the line onto the plane.
The line is \( CE \).
The plane is \( CDHG \) (the back face).
\( C \) is already on the plane.
We need to project \( E \) onto the plane. The point on the plane closest to \( E \) is \( H \) (since \( EH \) is perpendicular to the back face).
So the projection of \( CE \) is \( CH \).
We need to find the angle \( \angle ECH \) in the right-angled triangle \( EHC \).
Step 2: Calculate Lengths
✏ Working:
Length EH: This corresponds to the depth \( BC \).
\[ EH = 15 \text{ cm} \]Length HC: This is the diagonal of the rectangle \( CDHG \).
Sides of \( CDHG \) are \( CD = 12 \) and \( DH = 8 \).
\[ HC^2 = 12^2 + 8^2 \] \[ HC^2 = 144 + 64 = 208 \] \[ HC = \sqrt{208} \](M1) Calculating \( HC^2 \) or \( HC \)
Step 3: Calculate the Angle
✏ Working:
In \( \triangle EHC \) (right-angled at H):
- Opposite (\( EH \)) = 15
- Adjacent (\( HC \)) = \( \sqrt{208} \)
(M1) Correct trig ratio setup
\[ x = \tan^{-1}\left(\frac{15}{\sqrt{208}}\right) \] \[ x = \tan^{-1}(1.040…) \] \[ x = 46.12…^\circ \](A1) Correct calculation
Final Answer: \( 46.1^\circ \) (1 d.p.)
Question 22 (5 marks total)
22 (a)
Show that \( \dfrac{2\sin^2 x – 1 + \cos^2 x}{\sin x \cos x} \) is equivalent to \( \tan x \).
[3 marks]
22 (b)
Hence solve \( \dfrac{2\sin^2 x – 1 + \cos^2 x}{\sin x \cos x} = -1 \) for \( 0^\circ \leq x \leq 360^\circ \).
[2 marks]
Worked Solution
Part (a): Simplify the Numerator
Identity: \( \sin^2 x + \cos^2 x = 1 \).
We can rewrite the numerator using this.
✏ Working:
Numerator: \( 2\sin^2 x – 1 + \cos^2 x \)
Group terms:
\[ \sin^2 x + (\sin^2 x + \cos^2 x) – 1 \]Substitute \( 1 \):
\[ \sin^2 x + 1 – 1 \] \[ = \sin^2 x \](M1) Using identity to simplify numerator
Part (a): Simplify the Fraction
✏ Working:
\[ \frac{\sin^2 x}{\sin x \cos x} \]Cancel \( \sin x \) from top and bottom:
\[ \frac{\sin x}{\cos x} \](M1) Simplification
Identity: \( \frac{\sin x}{\cos x} \equiv \tan x \)
So, LHS \( = \tan x \).
(A1) Complete proof
Part (b): Solve the Equation
Strategy: Replace the complicated fraction with \( \tan x \) (from part a).
✏ Working:
\[ \tan x = -1 \]Reference angle: \( \tan^{-1}(1) = 45^\circ \).
Since \( \tan x \) is negative, \( x \) is in the 2nd and 4th quadrants.
2nd Quadrant: \( 180^\circ – 45^\circ = 135^\circ \)
4th Quadrant: \( 360^\circ – 45^\circ = 315^\circ \)
(M1) Identifying correct quadrants/angles, (A1) Correct answers
Final Answer: \( x = 135^\circ, 315^\circ \)
Question 23 (6 marks total)
A circle has centre \( C \) and equation \( (x-1)^2 + (y+3)^2 = 25 \).
\( P(4, -7) \) and \( Q \) are points on the circle.
The tangent at \( Q \) is parallel to the \( x \)-axis.
The tangents at \( P \) and \( Q \) intersect at point \( R \).
23 (a)
Write down the coordinates of \( C \).
[1 mark]
23 (b)
Show that the equation of the tangent at \( Q \) is \( y = 2 \).
[1 mark]
23 (c)
Work out the \( x \)-coordinate of \( R \).
[4 marks]
Worked Solution
Part (a): Centre Coordinates
Formula: For \( (x-a)^2 + (y-b)^2 = r^2 \), centre is \( (a, b) \).
Answer: \( C(1, -3) \)
(B1)
Part (b): Tangent at Q
Reasoning: The tangent at \( Q \) is parallel to the \( x \)-axis (horizontal). This means \( Q \) must be the highest or lowest point on the circle.
From the diagram, \( Q \) is the highest point.
✏ Working:
Centre \( y = -3 \). Radius \( = \sqrt{25} = 5 \).
Max \( y \) value \( = -3 + 5 = 2 \).
Therefore, the horizontal line passing through \( Q \) is \( y = 2 \).
(B1) Showing calculation
Part (c): Tangent at P
Strategy:
- Find the gradient of the radius \( CP \).
- The tangent is perpendicular to the radius, so find the negative reciprocal gradient.
- Find the equation of the tangent at \( P \).
- Intersect this with the tangent at \( Q \) (\( y=2 \)).
✏ Working:
1. Gradient CP: \( C(1, -3) \), \( P(4, -7) \)
\[ m_{CP} = \frac{-7 – (-3)}{4 – 1} = \frac{-4}{3} \](M1) Gradient of radius
2. Gradient of Tangent:
\[ m_{tangent} = \frac{-1}{-4/3} = \frac{3}{4} \](M1) Gradient of tangent
3. Equation of Tangent: using \( P(4, -7) \)
\[ y – y_1 = m(x – x_1) \] \[ y – (-7) = \frac{3}{4}(x – 4) \] \[ y + 7 = \frac{3}{4}x – 3 \] \[ y = \frac{3}{4}x – 10 \](M1) Equation of tangent
Part (c): Intersection
✏ Working:
We know the tangent at \( Q \) is \( y = 2 \).
Substitute \( y = 2 \) into the tangent equation:
\[ 2 = \frac{3}{4}x – 10 \] \[ 12 = \frac{3}{4}x \] \[ 48 = 3x \] \[ x = 16 \](A1) Correct x-coordinate
Final Answer: \( x = 16 \)
Question 24 (4 marks)
Show that the curve \( y = \frac{3}{5}x^5 + x^4 \) has exactly two stationary points.
Worked Solution
Step 1: Differentiate
Strategy: Stationary points occur when the gradient (\( \frac{dy}{dx} \)) is zero.
✏ Working:
\[ \frac{dy}{dx} = \frac{3}{5}(5x^4) + 4x^3 \] \[ \frac{dy}{dx} = 3x^4 + 4x^3 \](M1) Differentiation method, (A1) Correct derivative
Step 2: Solve for zero
✏ Working:
\[ 3x^4 + 4x^3 = 0 \]Factorise common term \( x^3 \):
\[ x^3(3x + 4) = 0 \](M1) Factorising
Solutions:
- \( x^3 = 0 \implies x = 0 \)
- \( 3x + 4 = 0 \implies 3x = -4 \implies x = -\frac{4}{3} \)
Step 3: Conclusion
Verification:
We found exactly two real values for \( x \) where the gradient is zero. Therefore, there are exactly two stationary points.
(A1) Correct values with statement
Final Answer: The derivative \( x^3(3x+4) = 0 \) yields exactly two solutions: \( x=0 \) and \( x=-\frac{4}{3} \).
Question 25 (3 marks)
\( f(x) = x^3 – 10x – c \) where \( c \) is a positive integer.
\( (x+c) \) is a factor of \( f(x) \).
Use the factor theorem to work out the value of \( c \).
Worked Solution
Step 1: Apply the Factor Theorem
Theorem: If \( (x+c) \) is a factor, then \( f(-c) = 0 \).
✏ Working:
Substitute \( x = -c \) into \( f(x) \):
\[ (-c)^3 – 10(-c) – c = 0 \] \[ -c^3 + 10c – c = 0 \] \[ -c^3 + 9c = 0 \](M1) Setting up equation
Step 2: Solve for c
✏ Working:
Factorise:
\[ c(9 – c^2) = 0 \] \[ c(3 – c)(3 + c) = 0 \](M1) Factorising
Possible values: \( c = 0 \), \( c = 3 \), \( c = -3 \).
Step 3: Select valid value
Constraint: The question states \( c \) is a positive integer.
Answer: \( c = 3 \)
(A1) Correct value selected
Final Answer: \( c = 3 \)
Question 26 (4 marks)
\( f(x) \) is a function with domain all values of \( x \).
\( f(x) = \sqrt{x^2 + 6x – a} \) where \( a \) is a constant.
Work out the possible values of \( a \).
Give your answer as an inequality.
Worked Solution
Step 1: Understanding Domain Constraints
Reasoning: For a square root function \( \sqrt{y} \) to be defined for all real numbers, the term inside the square root (\( y \)) must be non-negative (greater than or equal to zero) for all values of \( x \).
So, we need \( x^2 + 6x – a \geq 0 \) for all \( x \).
Step 2: Completing the Square or Discriminant
Method 1: Completing the Square
The quadratic \( x^2 + 6x – a \) is a “happy” parabola (U-shaped). Its minimum point must be on or above the x-axis.
✏ Working:
\[ x^2 + 6x – a = (x+3)^2 – 9 – a \]The minimum value of this expression is \( -9 – a \) (when \( x = -3 \)).
For the expression to be always \( \geq 0 \), the minimum value must be \( \geq 0 \).
\[ -9 – a \geq 0 \](M1) Completing square, (M1) Setting inequality
Step 3: Solve the Inequality
✏ Working:
\[ -9 \geq a \] \[ a \leq -9 \](A1) Correct inequality
Method 2: Discriminant
For a quadratic \( Ax^2 + Bx + C \) to be always positive (never cross x-axis) or touch it once, the discriminant \( b^2 – 4ac \) must be \( \leq 0 \).
\( A=1, B=6, C=-a \).
\[ 6^2 – 4(1)(-a) \leq 0 \] \[ 36 + 4a \leq 0 \] \[ 4a \leq -36 \] \[ a \leq -9 \]Final Answer: \( a \leq -9 \)
Question 27 (3 marks)
The curve \( y = f(x) \) has \( \dfrac{dy}{dx} = (x+2)^6 + (x+2)^4 \).
The curve has exactly one stationary point at \( P \) where \( x = -2 \).
Use the expression for \( \dfrac{dy}{dx} \) to show that \( P \) is a point of inflection.
Worked Solution
Step 1: Understanding Points of Inflection
Definition: A stationary point is a point of inflection if the gradient does not change sign as we pass through it. It goes from positive-zero-positive or negative-zero-negative.
We need to test the gradient slightly to the left and slightly to the right of \( x = -2 \).
Step 2: Test values
✏ Working:
1. Test a value to the left (\( x < -2 \)): Let \( x = -3 \).
\[ \frac{dy}{dx} = (-3+2)^6 + (-3+2)^4 \] \[ = (-1)^6 + (-1)^4 \] \[ = 1 + 1 = 2 \quad (\text{positive}) \](M1) Evaluating gradient for \( x < -2 \)
2. Test a value to the right (\( x > -2 \)): Let \( x = -1 \).
\[ \frac{dy}{dx} = (-1+2)^6 + (-1+2)^4 \] \[ = (1)^6 + (1)^4 \] \[ = 1 + 1 = 2 \quad (\text{positive}) \](M1) Evaluating gradient for \( x > -2 \)
Step 3: Conclusion
Analysis:
The gradient is positive before \( P \) and positive after \( P \).
Alternatively, note that \( (x+2)^6 \) and \( (x+2)^4 \) are even powers, so they are always \( \geq 0 \). The sum is strictly positive for all \( x \neq -2 \).
Conclusion:
Since the gradient is positive on both sides of the stationary point, \( P \) is a point of inflection.
(A1) Clear statement with valid reasoning
Final Answer: The gradient is positive for \( x < -2 \) and positive for \( x > -2 \). Since the sign of the gradient does not change, \( P \) is a point of inflection.