Level 2 Certificate Further Mathematics Paper 1 (Non-Calculator) June 2023

Step 1: Understanding the Question

We are told that the output of the function \( f(x) \) is \(-5\). We need to find the input value \( x \) that produces this output.

Step 2: Solve the Equation

We need to isolate \( x \). First, subtract 1 from both sides.

Step 1: Understanding Composite Functions

\( fg(3) \) means we apply \( g \) first, then apply \( f \) to the result. We work from the inside out.

1. Find \( g(3) \)

2. Put that answer into \( f(x) \)

Strategy: To factorise fully, we need to find the Highest Common Factor (HCF) of the numbers and the variables.

• Numbers: Look at 6 and 21. Both are divisible by 3.
• Variables: Look at \( x^2y \) and \( xy \). Both terms contain at least one \( x \) and one \( y \).

Strategy: Divide each term by \( 3xy \) to find what goes inside the bracket.

Check: Expand the bracket to verify.

\( 3xy \times 2x = 6x^2y \) and \( 3xy \times 7 = 21xy \). Correct.

Understanding Unit Vectors: To find a transformation matrix, consider where the unit vectors \( \mathbf{i} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \) and \( \mathbf{j} = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \) land.

Reflection in \( y = -x \):

• \( (1, 0) \) reflects to \( (0, -1) \). This is the first column.
• \( (0, 1) \) reflects to \( (-1, 0) \). This is the second column.

Method: Multiply rows by columns. Remember “Across then Down”.

\( \mathbf{I} \) is the identity matrix \( \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \).

Formula: Gradient \( m = \frac{\text{change in } y}{\text{change in } x} = \frac{y_2 – y_1}{x_2 – x_1} \)

Formula: The distance \( d \) between two points is given by Pythagoras’ theorem:

Goal: We need the form \( a\sqrt{b} \). We look for a square number factor in 40.

Square numbers: 1, 4, 9, 16, 25…

40 is divisible by 4.

We need to calculate \( Y_n – X_n \). We can either expand the brackets first or factorise.

Method 1: Expansion (Standard)

Expand both double brackets and then subtract.

Subtract \( X_n \) from \( Y_n \). Be very careful with the negative sign for the terms in \( X_n \).

We need to show this is always a multiple of 3.

Why: To find the equation of a tangent (a straight line), we first need its gradient (\( m \)). The gradient of the tangent is equal to the gradient of the curve at that point, which we find by differentiating \( y \) with respect to \( x \).

How: Substitute the \( x \)-coordinate of the given point \( (1, -2) \) into the gradient function.

Method: Use the equation of a straight line formula \( y – y_1 = m(x – x_1) \) or \( y = mx + c \). We have \( m = 14 \) and the point \( (1, -2) \).

Cone: The base is a circle with radius \( r \). Area of base = \( \pi r^2 \). Height = \( x \).

Prism: The cross section is a triangle with base \( y \) and height \( y \). Area of triangle = \( \frac{1}{2} \times \text{base} \times \text{height} \). The length of the prism is \( x \).

We are told: “Volume of cone = 4 × Volume of prism”. Set up the equation.

We need to isolate \( r \). Notice that \( x \) appears on both sides and represents a length (so \( x \neq 0 \)), which means we can cancel it out.

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

Line: We are given \( 2y = x + 5 \).

We need to solve these simultaneously. It is easier to rearrange the line equation to make \( x \) the subject to avoid fractions.

\( x = 2y – 5 \)

Now substitute this expression for \( x \) into the circle equation.

Simplify the equation and solve for \( y \).

Substitute the \( y \) values back into the linear equation \( x = 2y – 5 \) to find the corresponding \( x \) values.

Multiply both sides by the denominator \( (y^2 – 2) \).

We want \( y \) as the subject, so gather all terms containing \( y^2 \) on one side and everything else on the other.

Factor out \( y^2 \) from the left-hand side.

To remove the surd from the denominator, multiply top and bottom by the conjugate of the denominator: \( 3 + \sqrt{5} \).

Expand the double brackets for both the numerator and the denominator.

Recall: \( (a-b)(a+b) = a^2 – b^2 \).

Divide the numerator by the denominator.

Why: To find the second derivative \( \frac{d^2y}{dx^2} \), we must first find the first derivative \( \frac{dy}{dx} \).

Rule: Multiply by the power and reduce the power by 1. \( \frac{d}{dx}(ax^n) = anx^{n-1} \).

Why: Differentiate \( \frac{dy}{dx} \) to get \( \frac{d^2y}{dx^2} \).

Set the second derivative equal to 55 and solve for the positive value of \( x \).

Reasoning: Think of the sine graph. It starts at 0, goes up to 1 at \( 90^\circ \), back to 0 at \( 180^\circ \), down to -1 at \( 270^\circ \), and back to 0 at \( 360^\circ \).

Step 1: Isolate tan y

Divide both sides by \( \sqrt{3} \).

Knowledge: We should know our exact trig values. \( \tan 30^\circ = \frac{1}{\sqrt{3}} \).

Since \( \tan y \) is positive, we look in the 1st and 3rd quadrants (ASTC rule).

• 1st Quadrant: \( \theta \)
• 3rd Quadrant: \( 180^\circ + \theta \)

Strategy: The denominators are \( x \), \( 3x \), and \( 1 \). The Lowest Common Multiple (LCM) is \( 3x \).

Convert each term to have \( 3x \) as the denominator.

Combine the numerators over the single denominator. Be careful with the subtraction.

Strategy: Multiply the entire inequality by \( x \). Since we are told \( x \) is positive, the inequality sign does not flip.

Method: Find two numbers that multiply to \( 3 \times 8 = 24 \) and add to \(-10\). These are \(-6\) and \(-4\).

Visualise: The quadratic \( y = 3x^2 – 10x + 8 \) is a “U-shaped” parabola. We want where it is \( \leqslant 0 \) (below the x-axis).

This occurs between the roots.

Recall: \( (a-b)^2 = a^2 – 2ab + b^2 \)

Here \( a = x^{\frac{1}{2}} \) (which is \( \sqrt{x} \)) and \( b = x^{\frac{3}{2}} \) (which is \( x\sqrt{x} \)).

Rules of indices: \( (x^m)^n = x^{mn} \) and \( x^m \times x^n = x^{m+n} \).

Equate the expanded LHS to the RHS (\( x^2 + x \)) and simplify.

Factorise the cubic expression.

Method: Use the Binomial Theorem to find the \( x^2 \) term for \( (1 + 12x)^4 \).

The general term is \( \binom{n}{r} (1)^{n-r} (12x)^r \). For \( x^2 \), we need \( r = 2 \).

Method: Find the \( x^2 \) term for \( (a + 4x)^3 \).

The expansion is \( a^3 + 3a^2(4x) + 3a(4x)^2 + (4x)^3 \).

The \( x^2 \) term comes from \( 3a(4x)^2 \).

Equate the two coefficients and solve.

Fact 1: The point \( (-2, 11) \) lies on the curve. This means when \( x = -2 \), \( y = 11 \).

Fact 2: At a stationary point, the gradient \( \frac{dy}{dx} \) is zero.

Substitute Equation 2 into Equation 1.

Strategy: We have 3 variables. Let’s express everything in terms of one variable, say \( x \), or eliminate systematically.

• (1) \( 2x + y = 13 \implies y = 13 – 2x \)
• (2) \( x + 3z = 2 \implies 3z = 2 – x \implies z = \frac{2-x}{3} \) (Fractions are messy, let’s try another route).

Better Strategy: From (2), \( x = 2 – 3z \). Substitute this into (1) to get an equation in \( y \) and \( z \).

Now substitute \( y = 6z + 9 \) into equation (3): \( z – 2y = -7 \).

Substitute \( z = -1 \) back into our expressions for \( x \) and \( y \).

Method: Expand \( c(x+d)^2 + 3 \) to compare coefficients with the left hand side.

Match the terms with the same power of \( x \).

Compare the constant terms at the end.

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

Angle \( POR = 120^\circ \).

Therefore, angle \( PQR = 120^\circ \div 2 = 60^\circ \).

Method: Use the Cosine Rule on triangle \( PQR \). We know sides 4 and 5, and the angle between them is \( 60^\circ \).

\( a^2 = b^2 + c^2 – 2bc \cos A \)

Triangle \( POR \) is an isosceles triangle with two sides equal to radius \( r \) and the angle \( 120^\circ \).

We can use the Cosine Rule again, or split the triangle into two right-angled triangles.

Splitting Method: Drop a perpendicular from O to PR. This bisects the angle \( 120^\circ \) into \( 60^\circ \) and bisects PR into \( \frac{\sqrt{21}}{2} \).

Using Sine: \( \sin(60^\circ) = \frac{\text{Opp}}{\text{Hyp}} = \frac{\sqrt{21}/2}{r} \).