AQA Level 2 Further Maths Paper 1 (Non-Calculator) 2024

💡 Why we do this: Working with ordinary integers is often easier for division than working with standard form directly, especially when we need to find a square root later.

💡 Strategy: Place the value of \( t \) into the expression and perform the division before taking the square root.

💡 Verification: We recognize 121 as a square number.

💡 What this tells us: This is a standard identity known as the “Difference of Two Squares”. It always follows the pattern \( a^2 – b^2 = (a-b)(a+b) \).

💡 Why we do this: To see what happens as \( n \) gets very large, it helps to split the fraction into separate terms.

💡 What this tells us: As \( n \) becomes infinitely large (\( n \to \infty \)), the fraction \( \frac{4}{n} \) becomes closer and closer to 0.

💡 Core Principle: Parallel lines must have the same gradient (\( m \)). To find the gradient, we rearrange both equations into the form \( y = mx + c \).

💡 Strategy: It is much easier to differentiate separate terms of the form \( ax^n \) than a fraction. We divide each term in the numerator by \( x^4 \).

💡 How: Use the power rule: multiply by the power, reduce the power by 1.

• For \( 4x^{-1} \): \( -1 \times 4x^{-2} = -4x^{-2} \)
• For \( x^3 \): \( 3 \times x^2 = 3x^2 \)

💡 Why we do this: The centre of the circle is the midpoint of the diameter \( AB \).

💡 How: The radius is the distance from the centre (0,0) to point A (or B). We use Pythagoras’ theorem.

💡 Formula: The equation of a circle with centre (0,0) is \( x^2 + y^2 = r^2 \).

💡 What this tells us: Multiplying the matrix by the point vector gives the image vector.

💡 Strategy: Always check for a common factor first. Here, every term contains \( x \).

💡 How: We need to factorise \( 2x^2 – 9x – 5 \). We need numbers that multiply to \( 2 \times -5 = -10 \) and add to \( -9 \). These are \( -10 \) and \( 1 \).

💡 Check: Set each bracket to zero.

💡 Core Principle: To find the space diagonal \( AG \) of a cuboid, we use the 3D Pythagoras formula: \( d = \sqrt{x^2 + y^2 + z^2} \).

💡 Analysis: Point P is at \( x = 2 \). Since the function is continuous, the value of the first part \( f(x) = -\frac{1}{2}x + 21 \) at \( x=2 \) gives the y-coordinate of P.

💡 Strategy: The quadratic part \( f(x) = ax^2 + bx \) passes through P(2, 20) and Q(6, 12). We form two simultaneous equations.

💡 Strategy: Since \( (3x-7) \) is a factor, we can divide the cubic by it to find the remaining quadratic factor.

💡 Core Principle: If \( (x – 2) \) is a factor, then substituting \( x = 2 \) into the polynomial must give 0.

💡 Strategy: We have two sides and the included angle (or its cosine). This is a job for the Cosine Rule: \( a^2 = b^2 + c^2 – 2bc \cos A \).

💡 Strategy: Split the L-shape into two rectangles. We can split it vertically.

• Rectangle 1 (Left): Width 7, Height \( x \). Area = \( 7x \).
• Rectangle 2 (Right): Width 8, Height \( x – 3 \). Area = \( 8(x-3) \).

💡 Critical Check: A length cannot be zero or negative. The shortest vertical side is \( x – 3 \).

💡 Strategy: Factor out the coefficient of \( x^2 \) from the first two terms first.

💡 Analysis: The line \( y = -x \) swaps coordinates and changes signs. \( (1, 0) \to (0, -1) \) and \( (0, 1) \to (-1, 0) \).

💡 Interpretation: The matrix \( \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} \) changes the sign of both x and y. This corresponds to a 180° rotation.

💡 Strategy: We need to find the angles in the quadrilateral.
1. Since \( ABCD \) is a cyclic quadrilateral, opposite angles add to 180°.
2. Angle at centre is double the angle at the circumference.

💡 Insight: \( OB \) and \( OD \) are radii, so triangles \( OBC \) and \( ODC \) are isosceles. This means angle \( OCB = \) angle \( OBC \) and angle \( OCD = \) angle \( ODC \).

The question gives the ratio \( OBC : ODC = 5 : 3 \). Let’s use \( x \).

💡 Strategy: To subtract fractions, we need a common denominator. Let’s factorise \( 3x^2 – 2x – 8 \) first.

💡 Strategy: To remove surds from the bottom, multiply the numerator and denominator by \( \sqrt{5} + \sqrt{3} \).

💡 Why we do this: Stationary points occur where the gradient (\( \frac{dy}{dx} \)) is zero.

💡 How: We find \( \frac{d^2y}{dx^2} \). If it’s positive, it’s a minimum. If negative, it’s a maximum.

💡 Strategy: We have the graph of \( y = -x^2 + 3x + 6 \). We need to solve \( x^2 – 3.5x – 3 = 0 \).
Rearrange the target equation to look like the graph equation.

💡 Insight: The intersection of the curve \( y = -x^2 + 3x + 6 \) and the line \( y = -0.5x + 3 \) gives the solutions.

Draw line \( y = -0.5x + 3 \):

• When \( x = 0, y = 3 \)
• When \( x = 6, y = 0 \)

💡 Strategy: To solve exponential equations, get the same base on both sides. Note that \( 8 = 2^3 \).

💡 Strategy: Look for common factors to simplify the fraction.

💡 Strategy: We know the 1st term (n=1) is 2, and the 3rd term (n=3) is 16. We can create simultaneous equations.