Level 2 Certificate Further Mathematics Paper 2 (June 2023)

💡 What are we being asked to do?

We need to find the value of \(d\) that satisfies this algebraic equation. Since we have fractions on both sides, our first goal is to eliminate them to make the equation simpler.

👉 How do we remove the denominators?

We multiply the entire left side by 2 (the right denominator) and the entire right side by 5 (the left denominator).

👉 Expand the brackets:

👉 Rearrange to find d:

Subtract \(15d\) from both sides and add 6 to both sides.

💡 Strategy:

A linear sequence has the form \(dn + c\), where \(d\) is the common difference.

💡 What condition makes a term negative?

We need the term to be less than zero: \( \text{term} < 0 \).

👉 Calculate the value of the 36th term:

💡 How do we multiply matrices?

Multiply rows of the first matrix by columns of the second.

Top row calculation: \( (3 \times 1) + (5 \times 4) = t \)

Bottom row calculation: \( (u \times 1) + (2 \times 4) = 6 \)

💡 Formula: Midpoint \(x\) = \(\frac{x_1 + x_2}{2}\)

We know \(x_1 = 1\), \(x_2 = r\), and the midpoint is 5.

💡 Formula: Gradient \(m = \frac{y_2 – y_1}{x_2 – x_1}\)

We know points are \(P(1, k)\) and \(Q(9, 6)\), and gradient is 2.

💡 Concept: “Rate of change of \(y\) with respect to \(x\)” means find the derivative \(\frac{dy}{dx}\).

Using power rule: Multiply by power, reduce power by 1.

We are told the rate of change is 6.75, so set the derivative equal to 6.75.

💡 Formula: The standard equation of a circle is \( (x – a)^2 + (y – b)^2 = r^2 \), where \((a, b)\) is the centre and \(r\) is the radius.

💡 Property of Quadratics: A quadratic curve is symmetrical. The turning point lies exactly halfway between the two roots (x-intercepts).

Roots are at \(x = -4\) and \(x = p\). The midpoint is \(x = 0.5\).

💡 What does \( > 0 \) mean?

\( ax^2 + bx + c \) represents the \(y\)-value of the curve. We need to find the \(x\) values where the curve is above the x-axis (\(y > 0\)).

💡 Formula: Area of triangle = \(\frac{1}{2} \times \text{base} \times \text{height}\)

We know Area = 25 and Height (\(AD\)) = 4.

The base \(BC\) is split into parts \(BD\) and \(DC\) in the ratio \(2:3\).

Total parts = \(2 + 3 = 5\).

We need length \(DC\) (which corresponds to the 3 parts) to use in the right-angled triangle \(ADC\).

In right-angled triangle \(ADC\):

• Opposite (\(AD\)) = 4
• Adjacent (\(DC\)) = 7.5
• We need angle \(w\) (at \(C\)).

💡 SOH CAH TOA: We have Opposite and Adjacent, so use Tan.

A cuboid has 12 edges: 4 lengths, 4 widths, and 4 heights.

Dimensions are \(3x\), \(2x\), and \((x + 2y)\).

💡 Formula: Volume = Length × Width × Height

💡 Strategy: To find a maximum using calculus, differentiate \(V\) with respect to \(x\) (\(\frac{dV}{dx}\)), set it to 0, and solve.

Rearrange \(4x – 5y = 17\) into the form \(y = mx + c\) to read the gradient \(m\).

Use formula \(m = \frac{y_2 – y_1}{x_2 – x_1}\) with points \((3, 6)\) and \((-5, 16)\).

Rules:

• Parallel if gradients are equal.
• Perpendicular if product of gradients is -1 (\(m_1 \times m_2 = -1\)).

💡 Strategy: Use FOIL (First, Outer, Inner, Last) for the first part.

\((2x^3 – 9)(3x^2 + 4)\)

First expand \((x – 4)^2\), then multiply by \(x\).

Add the two results together:

\( (6x^5 + 8x^3 – 27x^2 – 36) + (x^3 – 8x^2 + 16x) \)

💡 3D Trigonometry Rule: The angle a line makes with a plane is the angle between the line (\(VA\)) and its projection on the plane (\(AX\)).

We need to find angle \( \angle VAX \).

\(ABCD\) is a square with side 15 cm. \(AC\) is the diagonal.

\(X\) is the midpoint of \(AC\).

In triangle \(VAX\):

• \( \angle VXA = 90^\circ \) (V is directly above X)
• Hypotenuse \( VA = 28 \)
• Adjacent \( AX = 10.6066… \)

Use Cosine: \( \cos(\theta) = \frac{\text{Adj}}{\text{Hyp}} \)

💡 Rule: \( x^{-n} = \frac{1}{x^n} \)

Only the \(x\) is raised to the power -7, not the 3.

\( 3x^{-7} = 3 \times \frac{1}{x^7} = \frac{3}{x^7} \)

Step 1: Simplify inside the bracket first.

Step 2: Square the result.

💡 Rule: \( \sqrt[n]{x} = x^{\frac{1}{n}} \)

💡 Strategy: Never multiply out first! Always factorise and cancel common terms.

Use the coordinates of \(P\) and \(Q\) in the equation \( g(x) = a \times b^x \).

The range is the set of possible \(y\) values.

Since the function is increasing (exponential with base > 1), the minimum y-value occurs at the minimum x-value, and maximum at maximum.

Domain: \( -1 \leqslant x \leqslant 2 \)

💡 Strategy: Since \( (2x-3) \) is a factor, we can divide the cubic expression by \( (2x-3) \) to find the remaining quadratic factor.

We need to find \( (ax^2 + bx + c) \) such that:

Solve \( 3x^2 – 8x + 2 = 0 \).

Since we need exact values and it doesn’t factorise easily, use the Quadratic Formula.

💡 Concept: A function is increasing if its gradient (derivative) is greater than or equal to 0 for all \(x\).

First, expand \( h(x) \).

We need \( h'(x) \geqslant 0 \) for all values of \( x \).

The expression \( 9ax^2 – 2a + 5 \) is a quadratic in \( x \). Since \( a > 0 \), the coefficient of \( x^2 \) (\(9a\)) is positive, so it’s a “U-shaped” parabola.

For a U-shaped parabola to be always non-negative (\(\geqslant 0\)), its minimum point must be \(\geqslant 0\).