AQA Level 2 Certificate Further Maths Paper 2 (June 2024)

What do we know?

The line passes through a fixed point \( (-1, -1) \) and has a gradient (slope) of \( \frac{3}{2} \).

A gradient of \( \frac{3}{2} \) means for every 2 units to the right, we go 3 units up.

Using the ‘Step’ Method:

Start at \( (-1, -1) \).

• Move right 2 (to \( x=1 \)) and up 3 (to \( y=2 \)). Plot \( (1, 2) \).
• Move right 2 again (to \( x=3 \)) and up 3 (to \( y=5 \)). Plot \( (3, 5) \).
• Move backwards: Left 2 (to \( x=-3 \)) and down 3 (to \( y=-4 \)). Plot \( (-3, -4) \).

Connect the points:

Draw a straight line through all the plotted points, extending from \( x = -3 \) to \( x = 3 \).

Goal: Isolate \( n^2 \) on one side.

Subtract 2 from both sides.

Divide by 5.

Think: Which integers, when squared, are less than 7.2?

Don’t forget negative integers!

What is rate of change?

In calculus, “rate of change of \( y \) with respect to \( x \)” means the derivative, \( \frac{dy}{dx} \).

Differentiate \( y = x^4 – 3kx^2 \) term by term.

Use the rule: \( y = ax^n \rightarrow \frac{dy}{dx} = anx^{n-1} \).

We are told that when \( x = 2 \), the rate (\( \frac{dy}{dx} \)) is 23.

Substitute these values into our derivative.

Rearrange to find \( k \).

Sequence B: \( 14, 22, 30, 38 \).

Calculate the difference between terms.

Difference is \( +8 \). So the rule starts with \( 8n \).

For \( n=1 \), \( 8(1) = 8 \). We need 14, so we add 6.

We are given the relationship:

\( C_n = A_n + B_n \)

Substitute what we know:

Rearrange to make \( A_n \) the subject.

Substitute \( n = 20 \) into the formula for A.

Recall Rule: Multiply Row by Column.

Top-Left Element of result (-5) comes from: (Top Row of 1st) × (Left Col of 2nd).

Bottom-Left Element of result (0) comes from: (Bottom Row of 1st) × (Left Col of 2nd).

Concept: If a matrix multiplied by \( \mathbf{M} \) results in itself, \( \mathbf{M} \) must be the Identity Matrix \( \mathbf{I} \).

The \( 2 \times 2 \) identity matrix has 1s on the diagonal and 0s elsewhere.

The domain is the set of input values (\(x\)): \( -1 \) to \( 4 \).

The range is the set of output values (\(f(x)\)).

Since \( f(x) = 3x^2 + 2 \) is a quadratic (U-shaped curve), we need to check the endpoints and the turning point (minimum).

1. Minimum Point: \( x^2 \) is smallest when \( x=0 \). Since \( 0 \) is inside our domain (\( -1 \leqslant 0 \leqslant 4 \)), the minimum value occurs at \( x=0 \).

2. Endpoints: Calculate \( f(x) \) at \( x = -1 \) and \( x = 4 \).

The curve crosses the \( y \)-axis when \( x = 0 \).

Substitute \( x = 0 \) into the equation.

The equation is in completed square form: \( y = A – (x – B)^2 \).

The term \( (x-3)^2 \) is always positive or zero.

To get the maximum value of \( y \), we need to subtract the smallest possible amount.

The smallest value of \( (x-3)^2 \) is 0, which happens when \( x = 3 \).

Before differentiating, rewrite fractions with \( x \) in the denominator using negative indices.

\( \frac{1}{x^n} = x^{-n} \)

Multiply by the power, reduce power by 1.

Differentiate again.

Note: The derivative of \( x \) (which is \( 1x^1 \)) is just 1.

A 4-digit integer looks like: [1st] [2nd] [3rd] [4th]

Let’s count the options for each position.

Multiply the number of options for each digit together.

The point \( (1, b) \) is a maximum point (a turning point).

At any turning point (max or min), the gradient is zero.

A zero gradient means the tangent is horizontal (parallel to the x-axis).

\( y = 0 \) corresponds to where the curve crosses or touches the x-axis.

Looking at the graph:

• The curve cuts the x-axis once to the left of the y-axis.
• The curve touches the x-axis at \( x = a \) (the minimum point).

This gives 2 distinct values, not 3. (Note: \( x=a \) is a repeated root, but the statement asks for “different values”).

“Increasing” means the graph is going up as you move right.

Look at the section between \( x=0 \) (y-axis) and \( x=1 \) (the peak).

The curve is clearly rising up to the maximum point.

To find the area of triangle \( ACD \), we can use the formula:

\[ \text{Area} = \frac{1}{2}ab \sin C \]

For triangle \( ACD \), we know side \( CD = 15 \) cm and angle \( ACD = 70^\circ \). If we find the length of side \( AC \), we can calculate the area.

We can find side \( AC \) using the Sine Rule in triangle \( ABC \). But wait, do we have enough angles?

Let’s check the diagram again carefully. The angle at \( A \) marked \( 66^\circ \) is actually angle \( BAC \), NOT angle \( CAD \). The 15cm is side \( CD \). The 70° is angle \( ACD \).

Ah, let’s re-examine the question diagram. The angle \( 66^\circ \) is marked at A. Looking at the arc, it’s inside triangle \( ABC \). So angle \( BAC = 66^\circ \).

In triangle \( ABC \):

• Angle \( ABC = 59^\circ \)
• Angle \( BAC = 66^\circ \)
• Side \( BC = 18 \) cm

We need side \( AC \) to help with the other triangle.

Use the Sine Rule on triangle \( ABC \):

\[ \frac{AC}{\sin B} = \frac{BC}{\sin A} \]

Now in triangle \( ACD \), we have:

• Side \( AC = 16.8906… \) cm
• Side \( CD = 15 \) cm
• Included Angle \( C = 70^\circ \)

Use the area formula: \( \text{Area} = \frac{1}{2} \times AC \times CD \times \sin(ACD) \)

To find intersections, we solve the equations simultaneously.

Substitute \( y = 4 – x \) into the circle equation.

Expand both squared brackets.

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

We need to find \( x \) using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \)

\( a=2, b=-18, c=37 \)

Calculate the two possible \( x \) values.

The question says “For A, the \( x \)-coordinate and \( y \)-coordinate are both positive“.

Ignore the third bracket for a moment. Expand \( (3x – 1)(4 – x) \).

Use FOIL (First, Outside, Inside, Last).

Now calculate \( (-3x^2 + 13x – 4)(2x + 5) \).

Multiply every term in the first bracket by every term in the second.

Add the results together and collect like terms.

Substitute \( x = -2 \) into \( y = 3^{-x} \).

We know Q has the same \( y \)-coordinate, so \( y = 9 \) for Q as well.

Set the equation for Q equal to 9.

Isolate the term with \( x \).

Divide both sides by 1152.

Recognize powers of 2.

\( 2^1=2, 2^2=4, \dots, 2^7=128 \).

We need to factorise \( 2x^2 + 9x – 18 \).

We are looking for two numbers that multiply to \( 2 \times -18 = -36 \) and add to \( 9 \).

The numbers are \( +12 \) and \( -3 \).

We need to factorise \( 12x^2 – 8x – 15 \).

Hint: In exam questions like this, one of the brackets is usually the same as the numerator!

Let’s check \( (2x – 3) \). If \( 2x – 3 \) is a factor, the other bracket must be \( (6x + 5) \) (since \( 2x \times 6x = 12x^2 \) and \( -3 \times 5 = -15 \)).

Let’s check: \( (2x – 3)(6x + 5) = 12x^2 + 10x – 18x – 15 = 12x^2 – 8x – 15 \). Correct!

Cancel the common factor \( (2x – 3) \).

Expand the bracket in the numerator first.

Recall: \( \sqrt{a} \times \sqrt{a} = a \).

So \( \sqrt{x^3} \times \sqrt{x^3} = x^3 \).

Divide each term by \( \sqrt{x} \), which is \( x^{1/2} \) or \( x^{0.5} \).

Subtract powers when dividing.

The question asks for integers \( b \) and \( c \). Wait, \( 2.5 \) is not an integer!

Let’s re-read: “where \( a, b \) and \( c \) are integers” (wait, the prompt text says “where \( a, b \) and \( c \) are integers” but the question image says “where \( a \) is … wait, actually the OCR text says “Give your answer in the form \( x^b + x^c \) where \( a, b \) and \( c \) are integers” – ah, looking at the image it’s actually \( \frac{a}{x^b} + x^c \) or something similar? Let me check the image carefully.)

Correction from Image: The image shows the form as \( x^{\frac{a}{b}} + x^c \) or similar? No, the OCR says “Give your answer in the form \( x^b + x^c \) where \( a, b \) and \( c \) are integers.” – wait, looking at the crop 15, it says “Give your answer in the form \( x^{\frac{a}{b}} + x^c \) where \( a, b \) and \( c \) are integers.” (The fraction a/b is small). Or maybe it’s just \( x^{5/2} + x^4 \)?

Let’s re-examine the mark scheme or standard interpretation. \( x^{2.5} \) is \( x^{5/2} \). The integers would be \( a=5, b=2, c=4 \). The prompt text for Q16 says “Give your answer in the form \( x^{\frac{a}{b}} + x^c \)”. Okay, that matches my derivation.

Total Surface Area = Curved Area + Base Area.

Base Area = \( \pi r^2 = \pi (7)^2 = 49\pi \).

Curved Area = \( \pi rl = 7\pi l \).

Solve for \( l \).

Sketch the right-angled triangle formed by height (\(h\)), radius (\(r\)), and slant height (\(l\)).

• Hypotenuse = \( l = 25 \)
• Adjacent (base) = \( r = 7 \)
• We want the angle between slant height and base (let’s call it \( \theta \)).

We have Adjacent and Hypotenuse -> Use Cosine.

We need the angle between plane \( MNB \) and plane \( HDCG \) (the back face).

Wait, plane \( HDCG \) is vertical. Plane \( MNB \) is slanted.

Actually, the question is slightly simpler. The plane \( HDCG \) is parallel to the plane \( ABFE \)? No, it’s perpendicular to the base.

Let’s look at the triangle \( NCB \) on the base.

The “line of greatest slope” or the perpendicular distance is the key.

The angle \( 36^\circ \) is the angle between the vertical plane \( MNB \) (wait, the question says \( MN \) is vertical. So \( MNB \) is a vertical plane? No, \( MN \) is a vertical line. \( M, N, B \) define a plane. Since \( MN \) is vertical, the plane \( MNB \) is perpendicular to the floor \( ABCD \).)

Ah, let’s re-read carefully: “The acute angle between the planes \( MNB \) and \( HDCG \) is \( 36^\circ \)”.

Plane \( HDCG \) is the back face. Plane \( MNB \) passes through \( N \) (on the back edge \( DC \)) and \( B \) (front right corner).

Since \( MN \) is vertical, the plane \( MNB \) is a vertical wall standing on the line \( NB \).

Plane \( HDCG \) is the back wall standing on line \( DC \).

So the angle between the two vertical planes is simply the angle between their floor lines, \( NB \) and \( DC \).

Therefore, we are looking for the angle \( BNC \).

Wait, let’s check the geometry. The angle between two vertical planes is the angle between their horizontal traces.

Trace of \( MNB \) on floor is \( NB \).

Trace of \( HDCG \) on floor is \( DC \).

So the angle is \( \angle BNC = 36^\circ \).

Multiply both sides by \( (5m – 3k^3) \).

We want \( k \) as the subject, so get all terms with \( k^3 \) on one side and everything else on the other.

Add \( 3mk^3 \) to both sides. Add \( 7 \) to both sides.

Pull \( k^3 \) out as a common factor on the right side.

Divide by the bracket, then take the cube root.

Using the Binomial Expansion for \( (A + B)^n \):

\( \binom{n}{r} A^{n-r} B^r \)

Here \( n=5 \), \( A=3 \), \( B=ax \).

For \( x^2 \), we need \( (ax)^2 \), so \( r=2 \).

For \( x^4 \), we need \( (ax)^4 \), so \( r=4 \).

We are given: \( 8 \times \text{coeff } x^2 = \text{coeff } x^4 \).

The centre of the circle is \( (3, 4) \).

The point on the circle is \( (5, 9) \).

Gradient \( m = \frac{y_2 – y_1}{x_2 – x_1} \).

The tangent is perpendicular to the radius.

Product of gradients is -1.

Use \( y – y_1 = m(x – x_1) \) with point \( (5, 9) \).

Point A (y-intercept): Set \( x = 0 \).

Point B (x-intercept): Set \( y = 0 \).

Triangle \( AOB \) is a right-angled triangle with base \( OB \) and height \( OA \).

Area = \( \frac{1}{2} \times \text{base} \times \text{height} \).

Let \( y = f(x) \). Swap \( x \) and \( y \), then solve for \( y \).

Substitute \( f(x) \) into \( g(x) \).

Replace every \( x \) in \( g(x) \) with \( \left(\frac{2-x}{3}\right) \).

Expand the brackets carefully.

Remember \( \left(\frac{A}{B}\right)^2 = \frac{A^2}{B^2} \).

Combine the results from Step 1 and Step 3.

The angle between tangent \( PB \) and chord \( AB \) is \( y \).

By the Alternate Segment Theorem, the angle in the alternate segment is equal to \( y \).

So, angle \( ACB = y \).

We are given \( AC = BC \).

This means triangle \( ABC \) is isosceles.

Therefore, base angles are equal: \( \angle CAB = \angle CBA \).

Angles on a straight line add to \( 180^\circ \).

\( PAC \) is a straight line, so \( \angle PAB + \angle CAB = 180^\circ \).

Sum of angles in triangle \( PAB \) is \( 180^\circ \).

The angles are:

• \( \angle APB = x \)
• \( \angle ABP = y \)
• \( \angle PAB = 90 + \frac{y}{2} \)