AQA Level 2 Further Mathematics Paper 2 (Specimen 2020)

What are we finding? We need the area of triangle \( OPQ \).

Strategy: To find the area of a right-angled triangle, we need the base and height lengths. \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \] The vertices are \( O(0,0) \), \( P \) (on the y-axis), and \( Q \) (intersection of the lines).

Finding P: The point \( P \) is on the y-axis and the line \( y = 6 \). This means its x-coordinate is 0.

Finding Q: The point \( Q \) is where the lines \( y = 2x \) and \( y = 6 \) intersect. We solve these equations simultaneously.

Why? The equation of a circle with centre \( (0,0) \) is \( x^2 + y^2 = r^2 \). We are given the circumference, so we need to find the radius \( r \) first.

The midpoint \( M(x_m, y_m) \) of two points \( A(x_1, y_1) \) and \( B(x_2, y_2) \) is the average of their coordinates:

Rule: When dividing an inequality by a negative number, you MUST flip the inequality sign.

Reasoning: We need numbers which, when squared, are greater than or equal to 16. These are large positive numbers (like 5, 6) or “large” negative numbers (like -5, -6) because \( (-5)^2 = 25 \).

Symmetry Property: A quadratic graph is symmetrical. The axis of symmetry passes through the maximum point (vertex). The roots A and B are equidistant from the axis of symmetry.

What does this mean? The line \( y = k \) is a horizontal line. For it to intersect the curve exactly once, it must touch the graph at the turning point (the maximum).

Step 1: Set up the equation

We multiply row by column: \( \mathbf{A} \times \mathbf{B} = \mathbf{C} \)

Step 2: Form equations

Top row multiplication:

\( (4 \times s) + (-1 \times -5) = -1 \)

Bottom row multiplication:

\( (-7 \times s) + (2 \times -5) = t \)

Step 3: Solve for s and t

Concept: \( \mathbf{I} \) is the identity matrix \( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \). If \( \mathbf{AD} = \mathbf{I} \), then \( \mathbf{D} \) is the inverse of \( \mathbf{A} \), or we can just multiply and equate.

Why? Parallel lines have the same gradient. We need to rearrange the given equation into \( y = mx + c \) form to see the gradient \( m \).

Method: We know \( m = 0.5 \) and the line passes through \( (4, 0) \). Use \( y – y_1 = m(x – x_1) \) or substitute into \( y = mx + c \).

Rule: To divide by a fraction, multiply by its reciprocal (flip the second fraction).

Strategy: Find a common denominator. The denominators are \( 2x^2 \) and \( 3x \). The lowest common multiple is \( 6x^2 \).

Theorem: The angle at the centre is twice the angle at the circumference subtended by the same arc.

Angle at centre (\( \angle BOC \)) = \( x + 62^\circ \)

Angle at circumference (\( \angle BAC \)) = \( 2x – 50^\circ \)

Reasoning: Triangle \( OBC \) is an isosceles triangle because \( OB \) and \( OC \) are both radii of the circle. Therefore, the base angles are equal (\( \angle OBC = \angle OCB = y \)).

Why? It is much easier to differentiate separate terms than a fraction. Divide each term in the numerator by \( 2x^4 \).

First Derivative (\( \frac{dy}{dx} \)): Multiply by the power and decrease power by 1.

Second Derivative (\( \frac{d^2y}{dx^2} \)): Differentiate again.

Why? The question states that the terms are equal for the value \( k \). We equate the two expressions, replacing \( n \) with \( k \).

Method: Cross-multiply to eliminate fractions.

Strategy: Factorise or use the quadratic formula. We need factors of -60 that add to -28.

Factors of 60: 1&60, 2&30, …

Values 2 and 30 work: \( -30 + 2 = -28 \).

Step 1: Find missing side length

We need the length of the left vertical side. Looking at the right side profile, it goes down \( 15x \), then in, then down \( y \). So the total height on the left is \( 15x + y \).

Check widths: Bottom \( 20x \) + Step \( 10x \) = Top \( 30x \). This matches.

Strategy: Split the shape into two rectangles. Let’s split it vertically where the step occurs.

Rectangle 1 (Left): Width \( 20x \), Height \( (15x + y) \)

Rectangle 2 (Right Top): Width \( 10x \), Height \( 15x \)

Method: To find a maximum, differentiate with respect to \( x \), set to zero, and solve.

Rule: \( gf(4) \) means calculate \( f(4) \) first, then put the result into \( g \).

Rule: \( fg(x) \) means substitute the entire function \( g(x) \) into \( x \) in \( f(x) \).

Formula: \( \frac{\sin A}{a} = \frac{\sin B}{b} \)

Pair 1: Angle \( x \) is opposite side \( 2y \).

Pair 2: Angle \( 18^\circ \) is opposite side \( y \).

Important: The calculator gives the acute angle (principal value). Since the question specifies \( x \) is obtuse (between 90° and 180°), we must use the property \( \sin(180 – x) = \sin x \).

Substitution: Substitute \( x = 0 \) and \( y = 3 \) into \( y = ab^{-x} \).

Strategy: Look for common factors first, then difference of two squares.

Strategy: The denominator already has a factor of \( 2x \). We need to factorise the quadratic \( 6x^2 – x – 35 \).

We need factors of \( 6 \times -35 = -210 \) that add to \( -1 \).

Factors are \( -15 \) and \( 14 \).

Action: Multiply every term by \( x^{\frac{1}{2}} \) (or \( \sqrt{x} \)).

Rules of indices: \( x^a \times x^b = x^{a+b} \).

We have 5 digits: {2, 4, 6, 7, 8}.

Constraint 1: The number must be odd. Look at the digits: only 7 is odd. So the last digit MUST be 7.

Constraint 2: The number must be greater than 30,000. The first digit must be 4, 6, 7, or 8.

Concept: If \( (ax + b) \) is a factor, then \( f(-\frac{b}{a}) = 0 \).

For \( (3x + 1) \), we substitute \( x = -\frac{1}{3} \).

Strategy: We know one factor. We can perform polynomial division or compare coefficients to find the quadratic factor.

\( 3x^3 – 2x^2 – 7x – 2 = (3x + 1)(Ax^2 + Bx + C) \)

Step 2: Factorise the quadratic

We need factors of -2 that add to -1. (-2 and 1)

Strategy: We need to find angle \( M \) in triangle \( VMD \). To do this, we need the lengths of all three sides: \( VD \), \( VM \), and \( DM \).

We are given \( VD = 10 \text{ cm} \).

Consider triangle \( VBC \). This is an isosceles triangle with sides \( 10, 10 \) and base \( 6 \) (\( BC = 6 \)). \( M \) is the midpoint of \( BC \), so \( VMB \) is a right-angled triangle.

Consider the rectangular base \( ABCD \). \( D \) is a corner and \( M \) is on \( BC \). Let’s look at triangle \( DCM \) in the base.

\( DC = AB = 8 \text{ cm} \).

\( CM = 3 \text{ cm} \) (half of 6).

Triangle \( DCM \) is right-angled at \( C \).

We have sides \( \sqrt{91} \), \( \sqrt{73} \), and \( 10 \). We want angle \( M \) (opposite side \( VD = 10 \)).

Formula: \( a^2 = b^2 + c^2 – 2bc \cos A \)

Here: \( VD^2 = VM^2 + DM^2 – 2(VM)(DM) \cos M \)

Strategy: Expand \( (2n + 3)^3 \) using the binomial expansion or by multiplying out brackets.

Recall \( (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \).

Here \( a = 2n \) and \( b = 3 \).

Condition: To show an expression is divisible by 9, we must factorise 9 out of the entire expression.