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AQA Level 2 Certificate in Further Mathematics Paper 2 (Calculator) June 2016
Using This Resource
- Attempt First: Try the question on your own before clicking “Show Solution”.
- Three-Layer Solutions:
- 💡 Understanding: The strategy and reasoning.
- ✏ Working: The mathematical steps.
- 🔍 Check: Verifying the answer makes sense.
- Calculator: This is a Calculator paper. You may use a calculator for all questions.
Table of Contents
- Question 1 (Coordinate Geometry)
- Question 2 (Quadratic Graphs)
- Question 3 (Straight Lines)
- Question 4 (Simultaneous Equations)
- Question 5 (Inequalities Logic)
- Question 6 (Calculus)
- Question 7 (Indices)
- Question 8 (Coordinate Ratio)
- Question 9 (Geometry Proof)
- Question 10 (Algebraic Fractions)
- Question 11 (Binomial Expansion)
- Question 12 (Sequences)
- Question 13 (Quadratic Sketch)
- Question 14 (Circle Geometry)
- Question 15 (Surds Equations)
- Question 16 (Rearranging Formulae)
- Question 17 (Volume Ratio)
- Question 18 (Trig Identities)
- Question 19 (Curve Sketching)
- Question 20 (Matrix Points)
- Question 21 (Algebraic Equations)
- Question 22 (3D Geometry)
- Question 23 (Matrix Transformations)
- Question 24 (Functions)
- Question 25 (Calculus Tangents)
- Question 26 (Trig Proofs)
Question 1 (3 marks)
A triangle has vertices \( A(2, 5) \), \( B(2, 0) \) and \( C(-4, 3) \).
Work out the area of triangle \( ABC \).
You may use the grid to help you.
Worked Solution
Step 1: Understand the Goal
💡 Understanding: We need to find the area of a triangle given the coordinates of its vertices. The standard formula for the area of a triangle is \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \).
The grid helps us visualize the base and height. Since points \( A \) and \( B \) have the same \( x \)-coordinate, the line \( AB \) is vertical. This makes it a convenient choice for the “base”.
Step 2: Calculate Dimensions
✏ Working:
Base (AB): Since \( A(2, 5) \) and \( B(2, 0) \) share the same x-coordinate, the length is simply the difference in y-coordinates.
\[ \text{Length } AB = 5 – 0 = 5 \]Height (h): The height is the perpendicular distance from vertex \( C(-4, 3) \) to the line containing \( AB \) (which is the line \( x=2 \)).
Distance from \( x = -4 \) to \( x = 2 \):
\[ h = 2 – (-4) = 6 \]💡 Note: We count squares horizontally from C to the line AB to find the height.
Step 3: Calculate Area
✏ Working:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \] \[ \text{Area} = \frac{1}{2} \times 5 \times 6 \] \[ \text{Area} = \frac{1}{2} \times 30 \] \[ \text{Area} = 15 \]Final Answer:
15 square units
✓ (3 marks)
Question 2 (4 marks total)
The function \( f(x) = x^2 – 4x + 1 \) has domain \( -1 \leq x \leq 5 \).
Here is the graph of \( y = f(x) \).
(a) Write down the equation of the line of symmetry of the graph. [1 mark]
(b) Use the graph to work out the solutions of \( x^2 – 4x + 1 = 5 \). Give your answers to 1 decimal place. [2 marks]
(c) Write down the range of \( f(x) \) for domain \( -1 \leq x \leq 5 \). [2 marks]
Worked Solution
Part (a): Line of Symmetry
💡 Understanding: The line of symmetry for a quadratic graph passes vertically through its turning point (vertex). Looking at the graph, the lowest point is at \( x = 2 \).
✏ Working:
The graph is symmetrical about the vertical line passing through \( x = 2 \).
Equation: \( x = 2 \)
Answer: \( x = 2 \)
Part (b): Solving graphically
💡 Understanding: We are asked to solve \( x^2 – 4x + 1 = 5 \). Since the graph shows \( y = x^2 – 4x + 1 \), we need to find the x-values where the curve meets the line \( y = 5 \).
✏ Working:
- Draw a horizontal line across the graph at \( y = 5 \).
- Mark the points where this line crosses the curve.
- Read the corresponding x-values from the x-axis.
Looking at the graph/grid:
- On the left, the intersection is slightly to the right of \( x = -1 \). It looks like \( -0.8 \).
- On the right, the intersection is slightly to the left of \( x = 5 \). It looks like \( 4.8 \).
Answer: \( x = -0.8 \) and \( x = 4.8 \)
Part (c): Range
💡 Understanding: The “range” refers to the possible output values (y-values) of the function for the given domain. We need to find the lowest and highest y-values on the graph between \( x = -1 \) and \( x = 5 \).
✏ Working:
- Minimum y-value: The lowest point of the curve (vertex) is at \( (2, -3) \). So, min \( y = -3 \).
- Maximum y-value: The highest points are at the ends of the domain. At \( x = -1 \), \( y = 6 \). At \( x = 5 \), \( y = 6 \). So, max \( y = 6 \).
The range is all values from -3 up to 6 inclusive.
\[ -3 \leq f(x) \leq 6 \]Answer: \( -3 \leq f(x) \leq 6 \) (or \( [-3, 6] \))
Question 3 (2 marks total)
\( L \) is a straight line with equation \( ax + by = c \)
where \( a \), \( b \) and \( c \) are non-zero integers.
(a) At which point does \( L \) intersect the \( x \)-axis? [1 mark]
Circle your answer.
\( (\frac{a}{c}, 0) \) \( (\frac{c}{a}, 0) \) \( (\frac{b}{c}, 0) \) \( (\frac{c}{b}, 0) \)
(b) What is the gradient of a line parallel to \( L \)? [1 mark]
Circle your answer.
\( -\frac{b}{a} \) \( \frac{b}{a} \) \( -\frac{a}{b} \) \( \frac{a}{b} \)
Worked Solution
Part (a): x-intercept
💡 Understanding: A line intersects the \( x \)-axis when \( y = 0 \).
✏ Working:
Substitute \( y = 0 \) into \( ax + by = c \):
\[ ax + b(0) = c \] \[ ax = c \] \[ x = \frac{c}{a} \]So the point is \( (\frac{c}{a}, 0) \).
Answer: \( (\frac{c}{a}, 0) \)
Part (b): Gradient
💡 Understanding: To find the gradient, we need to rearrange the equation into the form \( y = mx + c \), where \( m \) is the gradient. Parallel lines have the same gradient.
✏ Working:
Start with \( ax + by = c \)
Subtract \( ax \) from both sides:
\[ by = -ax + c \]Divide by \( b \):
\[ y = -\frac{a}{b}x + \frac{c}{b} \]The coefficient of \( x \) is the gradient \( m = -\frac{a}{b} \).
Answer: \( -\frac{a}{b} \)
Question 4 (4 marks)
Work out the point of intersection of the lines
\[ 2x + 3y = 11 \]and
\[ 2y = 13 – 3x \]Worked Solution
Step 1: Rearrange Equations
💡 Understanding: To solve simultaneous equations, it’s usually easier if they are in the same format, e.g., \( ax + by = c \).
✏ Working:
Equation 1: \( 2x + 3y = 11 \)
Equation 2: \( 2y = 13 – 3x \)
Rearrange Equation 2 by adding \( 3x \) to both sides:
\[ 3x + 2y = 13 \quad \text{(Eq 2)} \]Step 2: Eliminate a Variable
💡 Strategy: We can multiply the equations to make the coefficients of one variable the same. Let’s make the \( y \) coefficients 6.
✏ Working:
Multiply Eq 1 by 2: \( 4x + 6y = 22 \) (A)
Multiply Eq 2 by 3: \( 9x + 6y = 39 \) (B)
Subtract (A) from (B) to eliminate \( y \):
\[ (9x + 6y) – (4x + 6y) = 39 – 22 \] \[ 5x = 17 \] \[ x = \frac{17}{5} = 3.4 \]Step 3: Find y
✏ Working:
Substitute \( x = 3.4 \) back into an original equation (e.g., Eq 1):
\[ 2(3.4) + 3y = 11 \] \[ 6.8 + 3y = 11 \] \[ 3y = 11 – 6.8 \] \[ 3y = 4.2 \] \[ y = \frac{4.2}{3} = 1.4 \]Step 4: Check
🔍 Check: Use the other equation (Eq 2).
\[ 2(1.4) = 2.8 \] \[ 13 – 3(3.4) = 13 – 10.2 = 2.8 \]They match. The solution is correct.
Answer: \( x = 3.4, y = 1.4 \)
Coordinate form: \( (3.4, 1.4) \)
Question 5 (4 marks)
\( a \), \( b \) and \( c \) are numbers such that
\[ a < 0 \] \[ b > 1 \] \[ -1 < c < 1 \]Tick the correct box for each statement.
| Statement | Always true | Sometimes true | Never true |
|---|---|---|---|
| \( a^3 < 0 \) | ⬜ | ⬜ | ⬜ |
| \( b < 10a^2 \) | ⬜ | ⬜ | ⬜ |
| \( ab > 0 \) | ⬜ | ⬜ | ⬜ |
| \( b – c > 1 \) | ⬜ | ⬜ | ⬜ |
Worked Solution
Statement 1: \( a^3 < 0 \)
We are given \( a < 0 \) (negative).
A negative number cubed is negative (e.g., \( (-2)^3 = -8 \)).
So \( a^3 \) is always negative.
Answer: Always true
Statement 2: \( b < 10a^2 \)
Given \( a < 0 \) and \( b > 1 \).
\( a^2 \) is positive. \( 10a^2 \) is positive.
Let’s test values:
- Let \( b = 2, a = -1 \): \( 10(-1)^2 = 10 \). \( 2 < 10 \) (True).
- Let \( b = 100, a = -1 \): \( 10(-1)^2 = 10 \). \( 100 < 10 \) (False).
It depends on the values chosen.
Answer: Sometimes true
Statement 3: \( ab > 0 \)
\( a \) is negative.
\( b \) is positive.
Negative \( \times \) Positive = Negative.
So \( ab \) must be less than 0.
Answer: Never true
Statement 4: \( b – c > 1 \)
Given \( b > 1 \) and \( -1 < c < 1 \).
We want to find the minimum value of \( b – c \).
To make \( b – c \) as small as possible, we need \( b \) small (just above 1) and \( c \) large (just below 1).
Let \( b = 1.1, c = 0.9 \). \( b – c = 1.1 – 0.9 = 0.2 \). This is NOT \( > 1 \).
Let \( b = 5, c = -0.5 \). \( b – c = 5 – (-0.5) = 5.5 \). This IS \( > 1 \).
Answer: Sometimes true
Question 6 (3 marks)
For the curve \( y = f(x) \),
\[ \frac{dy}{dx} = \frac{3}{2}x – kx^4 + k \]where \( k \) is a constant.
When \( x = -2 \) the gradient of the curve is 12.
Work out the value of \( k \).
Worked Solution
Step 1: Understand Gradient Function
💡 Understanding: The expression \( \frac{dy}{dx} \) represents the gradient of the curve. We are given the formula for the gradient and a specific condition: when \( x = -2 \), the gradient is 12.
✏ Working:
Substitute \( x = -2 \) and \( \frac{dy}{dx} = 12 \) into the given equation.
\[ 12 = \frac{3}{2}(-2) – k(-2)^4 + k \]Step 2: Simplify and Solve
✏ Working:
Calculate the terms:
- \( \frac{3}{2}(-2) = -3 \)
- \( (-2)^4 = 16 \)
Substitute these values back:
\[ 12 = -3 – 16k + k \]Simplify the \( k \) terms:
\[ 12 = -3 – 15k \]Step 3: Isolate k
✏ Working:
Add 3 to both sides:
\[ 15 = -15k \]Divide by -15:
\[ k = \frac{15}{-15} = -1 \]Answer: \( k = -1 \)
Question 7 (2 marks)
Simplify fully \( \left( \frac{2}{3} x^3 y \right)^3 \)
Worked Solution
Step 1: Apply Power to Each Part
💡 Rule: When a product in brackets is raised to a power, apply that power to every factor inside: \( (ab)^n = a^n b^n \).
✏ Working:
\[ \left( \frac{2}{3} \right)^3 \times (x^3)^3 \times (y^1)^3 \]Step 2: Calculate Powers
✏ Working:
- Number: \( \left( \frac{2}{3} \right)^3 = \frac{2^3}{3^3} = \frac{8}{27} \)
- \( x \) term: \( (x^3)^3 = x^{3 \times 3} = x^9 \)
- \( y \) term: \( (y)^3 = y^3 \)
Combine them:
\[ \frac{8}{27} x^9 y^3 \]Answer: \( \frac{8}{27} x^9 y^3 \) (or \( \frac{8 x^9 y^3}{27} \))
Question 8 (3 marks)
\( D(-6, 4) \) and \( E(-2, 9) \) are joined by a straight line.
\( P \) is a point on \( DE \).
\( DP : PE = 3 : 5 \)
Work out the coordinates of \( P \).
Worked Solution
Step 1: Understand the Ratio
💡 Understanding: The ratio \( 3:5 \) means the line is divided into \( 3 + 5 = 8 \) equal parts. Point \( P \) is located \( \frac{3}{8} \) of the way from \( D \) to \( E \).
Step 2: Calculate x-coordinate
✏ Working:
Difference in x: \( -2 – (-6) = 4 \)
Fraction of difference: \( \frac{3}{8} \times 4 = 1.5 \)
Add to start (\( D \)):
\[ x_P = -6 + 1.5 = -4.5 \]Step 3: Calculate y-coordinate
✏ Working:
Difference in y: \( 9 – 4 = 5 \)
Fraction of difference: \( \frac{3}{8} \times 5 = \frac{15}{8} = 1.875 \)
Add to start (\( D \)):
\[ y_P = 4 + 1.875 = 5.875 \]Answer: \( (-4.5, 5.875) \)
Question 9 (4 marks)
\( PQ \) is parallel to \( BC \).
Prove that \( x = 3y \).
Worked Solution
Step 1: Identify Co-interior Angle
💡 Reasoning: Since lines \( PQ \) and \( BC \) are parallel, the interior angles on the same side of the transversal (line \( PB \)) add up to \( 180^\circ \). These are also called co-interior or allied angles.
✏ Working:
Let the interior angle \( \angle PBC \) be \( \theta \).
\[ x + \theta = 180^\circ \] \[ \theta = 180 – x \]Step 2: Angles Around a Point
💡 Reasoning: The angles around the point \( B \) make a full circle, so they must sum to \( 360^\circ \).
The three angles are:
- \( \theta \) (which is \( 180 – x \))
- \( y + 100 \)
- \( 2y + 80 \)
✏ Working:
\[ (180 – x) + (y + 100) + (2y + 80) = 360 \]Step 3: Solve Equation
✏ Working:
Combine number terms:
\[ 180 + 100 + 80 = 360 \]Combine algebraic terms:
\[ -x + y + 2y = -x + 3y \]So the equation becomes:
\[ 360 – x + 3y = 360 \]Subtract 360 from both sides:
\[ -x + 3y = 0 \] \[ 3y = x \]Therefore, \( x = 3y \).
Conclusion:
Proven: \( x = 3y \)
✓ (4 marks)
Question 10 (4 marks total)
(a) Simplify \( \frac{x^2 – 7x + 10}{x^2 – 2x – 15} \) [2 marks]
(b) Factorise fully \( w^5 x^3 y^2 + w^2 x^6 y^3 \) [2 marks]
Worked Solution
Part (a): Simplify Fraction
💡 Strategy: To simplify an algebraic fraction, we must factorise both the numerator and the denominator first, then cancel any common factors.
✏ Working:
Numerator: \( x^2 – 7x + 10 \)
Find two numbers that multiply to 10 and add to -7: (-2 and -5).
\[ (x – 2)(x – 5) \]Denominator: \( x^2 – 2x – 15 \)
Find two numbers that multiply to -15 and add to -2: (-5 and +3).
\[ (x – 5)(x + 3) \]Simplify:
\[ \frac{(x – 2)\cancel{(x – 5)}}{\cancel{(x – 5)}(x + 3)} = \frac{x – 2}{x + 3} \]Answer: \( \frac{x – 2}{x + 3} \)
Part (b): Factorise Fully
💡 Strategy: Look for the highest common factor (HCF) of the two terms. Compare the powers of each variable.
✏ Working:
Term 1: \( w^5 x^3 y^2 \)
Term 2: \( w^2 x^6 y^3 \)
Common Factors:
- \( w \): Lowest power is \( w^2 \)
- \( x \): Lowest power is \( x^3 \)
- \( y \): Lowest power is \( y^2 \)
HCF = \( w^2 x^3 y^2 \)
Divide each term by HCF:
- \( \frac{w^5 x^3 y^2}{w^2 x^3 y^2} = w^3 \)
- \( \frac{w^2 x^6 y^3}{w^2 x^3 y^2} = x^3 y \)
Combine:
\[ w^2 x^3 y^2 (w^3 + x^3 y) \]Answer: \( w^2 x^3 y^2 (w^3 + x^3 y) \)
Question 11 (3 marks)
The \( x^2 \) term in the expansion of \( (3x + 4)(x^2 + px + 5) \) is \( -23x^2 \)
Work out the value of \( p \).
Worked Solution
Step 1: Identify Relevant Terms
💡 Strategy: We don’t need to expand the entire expression. We only care about the terms that produce \( x^2 \).
We multiply each term in the first bracket by each term in the second.
\( (3x + 4)(x^2 + px + 5) \)
✏ Working:
Which combinations give \( x^2 \)?
- \( 3x \times px = 3px^2 \)
- \( 4 \times x^2 = 4x^2 \)
Note: \( 3x \times x^2 = 3x^3 \) (too high), \( 3x \times 5 = 15x \) (too low), etc.
Step 2: Form Equation
✏ Working:
The total \( x^2 \) term is the sum of these parts:
\[ 3px^2 + 4x^2 \]We are told this equals \( -23x^2 \).
\[ 3px^2 + 4x^2 = -23x^2 \]Divide by \( x^2 \) (equate coefficients):
\[ 3p + 4 = -23 \]Step 3: Solve for p
✏ Working:
Subtract 4 from both sides:
\[ 3p = -27 \]Divide by 3:
\[ p = -9 \]Answer: \( p = -9 \)
Question 12 (5 marks total)
Here are the first four terms of linear sequences \( X \) and \( Y \) and quadratic sequence \( Z \).
Sequence X: 7 9 11 13 …
Sequence Y: 2 5 8 11 …
Sequence Z: 14 45 88 143 …
(a) Work out the nth term of sequence \( X \). [2 marks]
(b) The nth term of sequence \( Y \) is \( 3n – 1 \).
Using your answer to part (a), or otherwise, work out the nth term of sequence \( Z \).
Give your answer in the form \( an^2 + bn + c \) where \( a \), \( b \) and \( c \) are integers. [3 marks]
Worked Solution
Part (a): Sequence X
💡 Strategy: Find the common difference between terms to get the coefficient of \( n \), then adjust to find the constant.
✏ Working:
Sequence X: 7, 9, 11, 13…
Difference: \( +2 \). So it starts with \( 2n \).
Test \( 2n \): 2, 4, 6, 8…
Compare to sequence: \( 7 – 2 = 5 \). We need to add 5.
nth term = \( 2n + 5 \)
Answer (a): \( 2n + 5 \)
Part (b): Sequence Z
💡 Analysis: Let’s check if Sequence Z is the product of Sequence X and Sequence Y. Why? Because quadratic sequences often arise from multiplying two linear sequences.
✏ Working:
Let’s check term 1:
- \( X_1 = 7 \)
- \( Y_1 = 2 \)
- \( X_1 \times Y_1 = 7 \times 2 = 14 \). (Matches \( Z_1 \))
Let’s check term 2:
- \( X_2 = 9 \)
- \( Y_2 = 5 \)
- \( X_2 \times Y_2 = 9 \times 5 = 45 \). (Matches \( Z_2 \))
It works! So Sequence Z is simply the product of the nth terms of X and Y.
Part (b): Expand the Product
✏ Working:
\[ Z_n = (2n + 5)(3n – 1) \]Expand using FOIL:
- First: \( 2n \times 3n = 6n^2 \)
- Outer: \( 2n \times -1 = -2n \)
- Inner: \( 5 \times 3n = 15n \)
- Last: \( 5 \times -1 = -5 \)
Combine terms:
\[ 6n^2 – 2n + 15n – 5 \] \[ 6n^2 + 13n – 5 \]Answer (b): \( 6n^2 + 13n – 5 \)
Question 13 (4 marks)
Here is a sketch of \( y = a + bx – 2x^2 \) where \( a \) and \( b \) are constants.
The graph intersects the \( x \)-axis at \( (-1, 0) \) and \( (\frac{7}{2}, 0) \) and the \( y \)-axis at point \( P \).
Work out the coordinates of point \( P \).
Worked Solution
Step 1: Use Roots to Form Equation
💡 Strategy: If a quadratic graph crosses the x-axis at \( p \) and \( q \), its equation includes the factors \( (x – p) \) and \( (x – q) \).
Here, roots are \( x = -1 \) and \( x = \frac{7}{2} = 3.5 \).
So the factors are \( (x + 1) \) and \( (x – 3.5) \) or \( (2x – 7) \).
The equation is of the form \( y = k(x + 1)(2x – 7) \).
Step 2: Determine Constant k
💡 Check: We are given the equation form \( y = a + bx – 2x^2 \). The coefficient of \( x^2 \) is \( -2 \).
Let’s expand our factorised form and check the \( x^2 \) coefficient.
✏ Working:
Let’s use factors \( (x+1) \) and \( (x-3.5) \).
\[ y = k(x + 1)(x – 3.5) \] \[ y = k(x^2 – 3.5x + x – 3.5) \] \[ y = k(x^2 – 2.5x – 3.5) \]We know the \( x^2 \) term is \( -2x^2 \). So \( k = -2 \).
\[ y = -2(x^2 – 2.5x – 3.5) \] \[ y = -2x^2 + 5x + 7 \]Step 3: Find y-intercept (Point P)
💡 Understanding: Point P is where the graph crosses the y-axis, which occurs when \( x = 0 \).
✏ Working:
Substitute \( x = 0 \) into \( y = -2x^2 + 5x + 7 \):
\[ y = -2(0)^2 + 5(0) + 7 \] \[ y = 7 \]Answer: \( (0, 7) \)
Question 14 (4 marks)
\( P, Q, R \) and \( S \) are points on the circumference of a circle.
\( w : y = 7 : 5 \)
\( PQ = PS \)
Work out the size of angle \( x \).
Worked Solution
Step 1: Use Cyclic Quadrilateral Properties
💡 Property: Opposite angles in a cyclic quadrilateral sum to \( 180^\circ \).
Therefore, \( w + y = 180^\circ \).
✏ Working:
We are given ratio \( w : y = 7 : 5 \).
Total parts = \( 7 + 5 = 12 \).
Value of one part = \( \frac{180}{12} = 15^\circ \).
Calculate \( w \):
\[ w = 7 \times 15 = 105^\circ \]Step 2: Use Isosceles Triangle Properties
💡 Property: We are given \( PQ = PS \). This means triangle \( PQS \) is an isosceles triangle with equal sides \( PQ \) and \( PS \).
In an isosceles triangle, the base angles are equal. So \( \angle PQS = \angle PSQ \).
We know angle \( x \) represents \( \angle PQS \).
✏ Working:
In triangle \( PQS \), the angles sum to \( 180^\circ \).
\[ w + x + x = 180 \] \[ 105 + 2x = 180 \]Step 3: Solve for x
✏ Working:
\[ 2x = 180 – 105 \] \[ 2x = 75 \] \[ x = 37.5 \]Answer: \( 37.5^\circ \)
Question 15 (4 marks total)
(a) Solve \( \frac{2}{5} \sqrt{x} = 1 \) [2 marks]
(b) Solve \( x^3 = 5x^2 \) [2 marks]
Worked Solution
Part (a): Solve Surd Equation
✏ Working:
Multiply by 5:
\[ 2\sqrt{x} = 5 \]Divide by 2:
\[ \sqrt{x} = 2.5 \]Square both sides:
\[ x = 2.5^2 = 6.25 \](Alternatively: \( \sqrt{x} = \frac{5}{2} \Rightarrow x = \frac{25}{4} \))
Answer: \( x = 6.25 \)
Part (b): Solve Polynomial
💡 Warning: Do not just divide by \( x^2 \) unless you know \( x \neq 0 \). You might lose the solution \( x = 0 \).
✏ Working:
Rearrange to equal zero:
\[ x^3 – 5x^2 = 0 \]Factorise \( x^2 \):
\[ x^2(x – 5) = 0 \]Solutions are where each factor is zero:
\[ x^2 = 0 \Rightarrow x = 0 \] \[ x – 5 = 0 \Rightarrow x = 5 \]Answer: \( x = 0 \) and \( x = 5 \)
Question 16 (4 marks)
Rearrange \( y = \frac{8(w – x)}{x} \) to make \( x \) the subject.
Worked Solution
Step 1: Remove Fraction
✏ Working:
Multiply both sides by \( x \):
\[ xy = 8(w – x) \]Step 2: Expand Brackets
✏ Working:
\[ xy = 8w – 8x \]Step 3: Collect x Terms
💡 Strategy: Move all terms containing \( x \) to one side of the equation.
✏ Working:
Add \( 8x \) to both sides:
\[ xy + 8x = 8w \]Step 4: Factorise and Solve
💡 Strategy: Factor out \( x \) so it appears only once.
✏ Working:
\[ x(y + 8) = 8w \]Divide by \( (y + 8) \):
\[ x = \frac{8w}{y + 8} \]Answer: \( x = \frac{8w}{y + 8} \)
Question 17 (3 marks)
A cylinder has base radius \( x \) cm and height \( y \) cm.
A hemisphere has radius \( 6y \) cm.
The cylinder and hemisphere have equal volumes.
Work out the value of \( \frac{x}{y} \).
You must show your working.
Worked Solution
Step 1: Write Formulae
✏ Working:
Volume of cylinder: \( V_{cyl} = \pi r^2 h = \pi x^2 y \)
Volume of hemisphere: \( V_{hem} = \frac{1}{2} \times \frac{4}{3} \pi r^3 = \frac{2}{3} \pi r^3 \)
Substitute radius \( 6y \) into hemisphere formula:
\[ V_{hem} = \frac{2}{3} \pi (6y)^3 \]Step 2: Equate Volumes
✏ Working:
\[ \pi x^2 y = \frac{2}{3} \pi (216 y^3) \]Divide both sides by \( \pi \):
\[ x^2 y = \frac{2}{3} \times 216 y^3 \]Calculate number part: \( \frac{2}{3} \times 216 = 2 \times 72 = 144 \).
\[ x^2 y = 144 y^3 \]Step 3: Solve for x/y
✏ Working:
Divide by \( y \) (assuming \( y \neq 0 \)):
\[ x^2 = 144 y^2 \]Divide by \( y^2 \):
\[ \frac{x^2}{y^2} = 144 \] \[ \left( \frac{x}{y} \right)^2 = 144 \]Square root:
\[ \frac{x}{y} = \sqrt{144} = 12 \]Answer: 12
Question 18 (5 marks total)
Angle \( y \) is acute.
\[ \tan y = \frac{p + 1}{p – 1} \]where \( p \) is a constant greater than 1.
(a) Which of the statements below is correct? [1 mark]
Circle your answer.
\( y = 45^\circ \) \( y < 45^\circ \) \( y > 45^\circ \) \( y \) could be any acute angle
(b) Work out the expression for \( \sin y \).
Give your answer in the form \( \frac{ap + b}{\sqrt{cp^2 + d}} \) where \( a, b, c \) and \( d \) are integers. [4 marks]
You may use a diagram to help you.
Worked Solution
Part (a): Logic Check
💡 Reasoning: We know \( p > 1 \), so \( p + 1 > p – 1 \).
This means the numerator is larger than the denominator.
Therefore, \( \tan y > 1 \).
Since \( \tan 45^\circ = 1 \) and tan is increasing for acute angles, \( y \) must be greater than \( 45^\circ \).
Answer: \( y > 45^\circ \)
Part (b): Right-angled Triangle Method
💡 Strategy: Construct a right-angled triangle where \( \tan y = \frac{\text{Opposite}}{\text{Adjacent}} \).
- Opposite = \( p + 1 \)
- Adjacent = \( p – 1 \)
We need to find the Hypotenuse to get \( \sin y = \frac{\text{Opposite}}{\text{Hypotenuse}} \).
✏ Working:
Use Pythagoras: \( H^2 = O^2 + A^2 \)
\[ H^2 = (p+1)^2 + (p-1)^2 \]Expand brackets:
\[ H^2 = (p^2 + 2p + 1) + (p^2 – 2p + 1) \] \[ H^2 = 2p^2 + 2 \] \[ H = \sqrt{2p^2 + 2} \]Now write \( \sin y \):
\[ \sin y = \frac{p+1}{\sqrt{2p^2 + 2}} \]Answer: \( \frac{p+1}{\sqrt{2p^2 + 2}} \)
Question 19 (3 marks)
The continuous curve \( y = g(x) \) has exactly two stationary points.
The stationary points are:
- a point of inflection at \( P(1, 2) \)
- a minimum point at \( Q(a, b) \) where \( a > 1 \) and \( b < 0 \)
On the axes below, sketch the curve.
Label points \( P \) and \( Q \) on your sketch.
Worked Solution
Step 1: Analyze Stationary Points
💡 Understanding:
- Point P (1, 2): Point of inflection. The curve flattens out here (horizontal tangent) but continues in the same direction (downwards, based on the next point).
- Point Q (a, b): Minimum point. The curve turns here from going down to going up.
- Coordinates: P is in the 1st quadrant (positive x, positive y). Q is in the 4th quadrant (positive x > 1, negative y).
Step 2: Draw Sketch
✏ Drawing steps:
- Start from the top left (high y).
- Come down to P, flatten out momentarily, then continue going down.
- Cross the x-axis.
- Reach a minimum at Q (below the x-axis).
- Turn upwards and continue rising.
Question 20 (4 marks)
Under the transformation represented by \( \begin{pmatrix} -1 & -3 \\ 2 & 4 \end{pmatrix} \), the image of point \( P(a, 2) \) is point \( Q \).
Can point \( Q \) be the same as point \( P \)?
You must show your working.
Worked Solution
Step 1: Matrix Multiplication
💡 Method: Multiply the matrix by the position vector of P. The result is the position vector of Q.
✏ Working:
\[ \begin{pmatrix} -1 & -3 \\ 2 & 4 \end{pmatrix} \begin{pmatrix} a \\ 2 \end{pmatrix} = \begin{pmatrix} (-1)(a) + (-3)(2) \\ (2)(a) + (4)(2) \end{pmatrix} \] \[ = \begin{pmatrix} -a – 6 \\ 2a + 8 \end{pmatrix} \]So the coordinates of Q are \( (-a – 6, 2a + 8) \).
Step 2: Equate P and Q
💡 Condition: For Q to be the same as P, their x-coordinates must match AND their y-coordinates must match.
✏ Working:
x-coordinate equation:
\[ a = -a – 6 \] \[ 2a = -6 \] \[ a = -3 \]y-coordinate equation:
\[ 2 = 2a + 8 \] \[ -6 = 2a \] \[ a = -3 \]Step 3: Conclusion
✏ Conclusion: Since both equations give the same value \( a = -3 \), there is a consistent solution.
If \( a = -3 \), point \( P \) is \( (-3, 2) \) and maps to itself.
Answer: Yes, if \( a = -3 \).
Question 21 (6 marks)
Solve \( \frac{3}{x – 2} + \frac{2}{x – 1} = 5 \)
Do not use trial and improvement.
Write your solutions to 3 significant figures.
Worked Solution
Step 1: Combine Fractions
💡 Strategy: Make a common denominator of \( (x-2)(x-1) \).
✏ Working:
\[ \frac{3(x – 1) + 2(x – 2)}{(x – 2)(x – 1)} = 5 \]Simplify numerator:
\[ \frac{3x – 3 + 2x – 4}{(x – 2)(x – 1)} = 5 \] \[ \frac{5x – 7}{(x – 2)(x – 1)} = 5 \]Step 2: Form Quadratic Equation
✏ Working:
Multiply up the denominator:
\[ 5x – 7 = 5(x – 2)(x – 1) \]Expand brackets:
\[ 5x – 7 = 5(x^2 – x – 2x + 2) \] \[ 5x – 7 = 5(x^2 – 3x + 2) \] \[ 5x – 7 = 5x^2 – 15x + 10 \]Step 3: Rearrange to = 0
✏ Working:
\[ 0 = 5x^2 – 15x – 5x + 10 + 7 \] \[ 5x^2 – 20x + 17 = 0 \]Step 4: Solve using Quadratic Formula
✏ Working:
\( a = 5, b = -20, c = 17 \)
\[ x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \] \[ x = \frac{20 \pm \sqrt{(-20)^2 – 4(5)(17)}}{2(5)} \] \[ x = \frac{20 \pm \sqrt{400 – 340}}{10} \] \[ x = \frac{20 \pm \sqrt{60}}{10} \]Step 5: Calculate Values
✏ Working:
\( \sqrt{60} \approx 7.746 \)
\[ x_1 = \frac{20 + 7.746}{10} = 2.7746 \] \[ x_2 = \frac{20 – 7.746}{10} = 1.2254 \]Round to 3 s.f.
Answer: \( x = 2.77 \) and \( x = 1.23 \)
Question 22 (4 marks)
Pyramid \( VABCD \) has a horizontal rectangular base.
\( X \) is the centre of the base.
\( V \) is vertically above \( X \).
\( VB = VC = 17 \) cm \( AB = 22 \) cm \( BC = 16 \) cm
Work out the angle between the planes \( VBC \) and \( ABCD \).
Worked Solution
Step 1: Identify the Angle
💡 Understanding: The angle between two planes is the angle between two lines, one in each plane, both perpendicular to the line of intersection (\( BC \)).
Let \( M \) be the midpoint of \( BC \).
\( XM \) is in the base plane \( ABCD \), parallel to \( AB \), and perpendicular to \( BC \).
\( VM \) is the slant height of the face \( VBC \). Since \( \triangle VBC \) is isosceles (\( VB=VC \)), \( VM \) is perpendicular to \( BC \).
The required angle is \( \angle VMX \).
Step 2: Calculate Lengths
✏ Working:
1. Find \( XM \): \( M \) is midpoint of \( BC \). \( X \) is centre. \( XM \) is half the length of \( AB \).
\[ XM = \frac{1}{2} \times 22 = 11 \text{ cm} \]2. Find \( VM \): Consider triangle \( VBC \). \( M \) is midpoint of \( BC \), so \( MC = \frac{16}{2} = 8 \) cm. \( VC = 17 \) cm.
Use Pythagoras in \( \triangle VMC \):
\[ VM^2 = VC^2 – MC^2 \] \[ VM^2 = 17^2 – 8^2 \] \[ VM^2 = 289 – 64 = 225 \] \[ VM = 15 \text{ cm} \]Step 3: Calculate Angle
💡 Strategy: In right-angled triangle \( VXM \), we have hypotenuse \( VM = 15 \) and adjacent side \( XM = 11 \). We want angle at \( M \).
✏ Working:
\[ \cos(\angle VMX) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{XM}{VM} \] \[ \cos(\theta) = \frac{11}{15} \] \[ \theta = \cos^{-1}\left( \frac{11}{15} \right) \] \[ \theta \approx 42.83^\circ \]Answer: \( 42.8^\circ \) (to 1 d.p.)
Question 23 (5 marks)
Shape \( A \) maps to shape \( B \) by an enlargement, scale factor 3, centre the origin.
Shape \( B \) maps to shape \( C \) by a rotation through \( 180^\circ \), centre the origin.
Shape \( A \) can be mapped to shape \( C \) by a single transformation.
Use matrices to show that the single transformation is an enlargement, centre the origin.
State the scale factor of the enlargement.
Worked Solution
Step 1: Identify Matrices
💡 Recall:
- Enlargement SF 3: \( \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} \)
- Rotation 180°: \( \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \)
Step 2: Combine Matrices
💡 Order: Transformation B follows A. Matrix multiplication order is \( \mathbf{M}_2 \times \mathbf{M}_1 \). (Left matrix happens last).
✏ Working:
\[ \mathbf{M}_{\text{total}} = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} \]Multiply row by column:
\[ = \begin{pmatrix} (-1)(3) + 0 & 0 + (-1)(0) \\ 0 + (-1)(0) & 0 + (-1)(3) \end{pmatrix} \] \[ = \begin{pmatrix} -3 & 0 \\ 0 & -3 \end{pmatrix} \]Step 3: Interpret Result
💡 Understanding: The matrix \( \begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix} \) represents an enlargement with scale factor \( k \) about the origin.
Here, \( k = -3 \).
Conclusion: The single transformation is an enlargement, centre origin, with scale factor -3.
Question 24 (5 marks)
\( f(x) = \frac{x}{2x + 1} \) for positive values of \( x \).
Work out \( f(x + 1) – f(x) \).
Give your answer as a fraction in its simplest form.
Worked Solution
Step 1: Find f(x+1)
✏ Working:
Replace \( x \) with \( (x+1) \) in the original function:
\[ f(x+1) = \frac{(x+1)}{2(x+1) + 1} \] \[ f(x+1) = \frac{x+1}{2x + 2 + 1} \] \[ f(x+1) = \frac{x+1}{2x + 3} \]Step 2: Set up Subtraction
✏ Working:
\[ f(x+1) – f(x) = \frac{x+1}{2x + 3} – \frac{x}{2x + 1} \]Step 3: Common Denominator
✏ Working:
Common denominator is \( (2x+3)(2x+1) \).
\[ = \frac{(x+1)(2x+1) – x(2x+3)}{(2x+3)(2x+1)} \]Step 4: Expand Numerator
✏ Working:
Expand first part: \( (x+1)(2x+1) = 2x^2 + x + 2x + 1 = 2x^2 + 3x + 1 \)
Expand second part: \( x(2x+3) = 2x^2 + 3x \)
Subtract:
\[ (2x^2 + 3x + 1) – (2x^2 + 3x) \] \[ = 1 \]Answer: \( \frac{1}{(2x+3)(2x+1)} \) (or \( \frac{1}{4x^2 + 8x + 3} \))
Question 25 (6 marks)
The curve \( y = 2x^3 – 5 \) intersects the \( y \)-axis at \( C \).
The tangent to the curve at \( P(2, 11) \) intersects the \( y \)-axis at \( D \).
Work out the length \( CD \).
Worked Solution
Step 1: Find Coordinate C
💡 Strategy: \( C \) is the y-intercept of \( y = 2x^3 – 5 \). This occurs when \( x = 0 \).
✏ Working:
\[ y = 2(0)^3 – 5 = -5 \]So \( C = (0, -5) \).
Step 2: Find Gradient of Tangent
💡 Method: Differentiate the curve equation to find the gradient function \( \frac{dy}{dx} \).
✏ Working:
\[ y = 2x^3 – 5 \] \[ \frac{dy}{dx} = 6x^2 \]At point \( P(2, 11) \), substitute \( x = 2 \):
\[ m = 6(2)^2 = 6(4) = 24 \]The gradient of the tangent is 24.
Step 3: Equation of Tangent
✏ Working:
Using \( y – y_1 = m(x – x_1) \) with \( (2, 11) \) and \( m = 24 \):
\[ y – 11 = 24(x – 2) \] \[ y – 11 = 24x – 48 \] \[ y = 24x – 37 \]Step 4: Find Coordinate D
💡 Strategy: \( D \) is the y-intercept of the tangent line. Set \( x = 0 \).
✏ Working:
\[ y = 24(0) – 37 = -37 \]So \( D = (0, -37) \).
Step 5: Calculate Length CD
✏ Working:
Coordinate C is at \( y = -5 \).
Coordinate D is at \( y = -37 \).
Distance = Difference in y values (since both on y-axis).
\[ \text{Length} = -5 – (-37) = 32 \]Answer: 32 units
Question 26 (6 marks total)
(a) Prove that \( \sin^2 x – 3\cos^2 x \equiv 4\sin^2 x – 3 \) [2 marks]
(b) Hence, or otherwise, work out the values of \( x \) between \( 0^\circ \) and \( 360^\circ \) for which
\[ \sin^2 x – 3\cos^2 x = 0 \][4 marks]
Worked Solution
Part (a): Proof
💡 Strategy: Use the identity \( \sin^2 x + \cos^2 x = 1 \) to replace \( \cos^2 x \). We know \( \cos^2 x = 1 – \sin^2 x \).
✏ Working:
LHS: \( \sin^2 x – 3\cos^2 x \)
Substitute \( (1 – \sin^2 x) \) for \( \cos^2 x \):
\[ = \sin^2 x – 3(1 – \sin^2 x) \] \[ = \sin^2 x – 3 + 3\sin^2 x \] \[ = 4\sin^2 x – 3 \]This equals RHS. Proven.
Part (b): Solve Equation
💡 Strategy: Use the result from part (a) to simplify the equation.
✏ Working:
Replace LHS with \( 4\sin^2 x – 3 \):
\[ 4\sin^2 x – 3 = 0 \] \[ 4\sin^2 x = 3 \] \[ \sin^2 x = \frac{3}{4} \] \[ \sin x = \pm \sqrt{\frac{3}{4}} = \pm \frac{\sqrt{3}}{2} \]Part (b): Find Angles
✏ Working:
Reference angle for \( \sin x = \frac{\sqrt{3}}{2} \) is \( 60^\circ \).
Case 1: \( \sin x = +\frac{\sqrt{3}}{2} \) (Quadrants 1 & 2)
- \( x_1 = 60^\circ \)
- \( x_2 = 180 – 60 = 120^\circ \)
Case 2: \( \sin x = -\frac{\sqrt{3}}{2} \) (Quadrants 3 & 4)
- \( x_3 = 180 + 60 = 240^\circ \)
- \( x_4 = 360 – 60 = 300^\circ \)
Answer: \( 60^\circ, 120^\circ, 240^\circ, 300^\circ \)