AQA Level 2 Certificate Further Mathematics Paper 1 2017 (Non-Calculator)

What are we asked to do?

We need to draw a straight line that passes through the point \( (2, 1) \) and has a steepness (gradient) of \( \frac{3}{4} \).

What does a gradient of \( \frac{3}{4} \) mean?

Gradient = \( \frac{\text{Change in } y}{\text{Change in } x} \). So, for every 4 units right, we go 3 units up.

Why do we do this?

The “gradient of the curve” tells us we need to differentiate. The derivative \( \frac{dy}{dx} \) gives us the gradient formula.

What do we know?

When \( x = -1 \), the gradient (\( \frac{dy}{dx} \)) is \( -5 \). We substitute these values into our derivative.

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

\( y = 0^2 + 7(0) – 18 = -18 \). Point: \( (0, -18) \).

x-intercepts: Set \( y = 0 \) and solve the quadratic.

\( x^2 + 7x – 18 = 0 \)

Draw a U-shaped parabola passing through \( (-9, 0) \), \( (2, 0) \), and \( (0, -18) \).

✓ (B1) Correct shape

The line of symmetry is exactly halfway between the roots (\( x \)-intercepts).

Roots are \( -9 \) and \( 2 \).

Strategy: Since all three points lie on the same straight line, the gradient between the first two points must equal the gradient between the last two points.

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

Now calculate the gradient using \( (6, -5) \) and \( (8, t) \) and set it equal to \( -1.2 \).

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

We need the coefficient of \( x^2 \) and the coefficient of \( x \).

\( x^2 \) terms: \( -kx^2 + 4x^2 \). Coefficient is \( 4 – k \).

\( x \) terms: \( -5x – 4kx \). Coefficient is \( -5 – 4k \).

✓ (M1) Identify coefficients

The question states: “Coefficient of \( x^2 \) is twice the coefficient of \( x \).”

Both terms contain \( (x+6) \). The lowest power is 3.

So, factor out \( (x+6)^3 \).

The term \( (4x+10) \) can be factorised further by taking out 2.

For a square root to be a real number, the value inside must be non-negative.

The second difference is constant at 8. This confirms it is a quadratic sequence.

The coefficient of \( n^2 \) is half the second difference: \( \frac{8}{2} = 4 \).

So the sequence starts with \( 4n^2 \).

✓ (M1) Coefficient of \( n^2 \) found

The difference sequence is: 6, 17, 28, 39.

It goes up by 11 each time. So it is \( 11n \).

Check 1st term: \( 11(1) = 11 \). We have 6. So we need to subtract 5.

Linear part: \( 11n – 5 \).

Find critical values where \( 2x^2 – x – 10 = 0 \).

Factorise: \( (2x – 5)(x + 2) = 0 \).

Critical values: \( x = 2.5 \) and \( x = -2 \).

This is a rectangle, so lengths must be positive.

Side \( 2x – 3 > 0 \Rightarrow 2x > 3 \Rightarrow x > 1.5 \).

This rules out \( x < -2 \).

Combining \( x > 1.5 \) with \( x > 2.5 \), the valid range is \( x > 2.5 \).

Circle \( C_1 \): \( (x – 3)^2 + y^2 = 36 \)

• Centre \( L = (3, 0) \)
• Radius \( LN = \sqrt{36} = 6 \)

Circle \( C_2 \): \( (x – h)^2 + y^2 = 64 \)

• Centre \( M = (h, 0) \)
• Radius \( MN = \sqrt{64} = 8 \)

We are told \( \triangle LNM \) is a right-angled triangle at \( N \) (since lines are perpendicular).

Therefore, \( LM^2 = LN^2 + MN^2 \).

Distance \( LM \) is the distance between \( (3, 0) \) and \( (h, 0) \), which is 10.

From the diagram, \( M \) is to the right of \( L \), so \( h > 3 \).

The denominators are \( (x-3) \) and \( (x-3)(x-5) \).

To make them the same, we need to multiply the top and bottom of the first fraction by \( (x-5) \).

We need to factorise the quadratic \( x^2 – 5x + 6 \).

We look for numbers that multiply to 6 and add to -5: These are -2 and -3.

We can find the matrix columns by seeing where the unit vectors \( \mathbf{i} = \begin{pmatrix}1\\0\end{pmatrix} \) and \( \mathbf{j} = \begin{pmatrix}0\\1\end{pmatrix} \) land.

• \( (1, 0) \) rotated \( 90^\circ \) clockwise goes to \( (0, -1) \).
• \( (0, 1) \) rotated \( 90^\circ \) clockwise goes to \( (1, 0) \).

\( \mathbf{M}^2 \) means applying the transformation \( \mathbf{M} \) twice.

One application: \( 90^\circ \) clockwise.

Two applications: \( 90^\circ + 90^\circ = 180^\circ \) clockwise (or anti-clockwise, it ends up in the same place).

For \( 180^\circ \) rotation:

• \( (1, 0) \to (-1, 0) \)
• \( (0, 1) \to (0, -1) \)

Alternatively, square the matrix \( \mathbf{M} \):

A negative power means the reciprocal. So \( x^{-\frac{1}{4}} = \frac{1}{x^{\frac{1}{4}}} \).

If \( \frac{1}{x^{\frac{1}{4}}} = \frac{1}{5} \), then:

The power \( \frac{1}{4} \) is the fourth root. To remove it, we raise both sides to the power of 4.

We use the large right-angled triangle \( ABD \).

We know \( AD = 2\sqrt{3} \) (hypotenuse) and \( \angle ADB = 30^\circ \).

We need \( AB \) (opposite).

Now look at the smaller right-angled triangle \( ABC \).

We know \( AB = \sqrt{3} \) and \( \angle ACB = 45^\circ \).

Since it’s \( 45^\circ \), it is an isosceles right-angled triangle, so \( BC = AB \).

Back to the large triangle \( ABD \). We need the adjacent side \( BD \).

We need to place three specific points:

• P (Minimum): In the 3rd quadrant (bottom-left). \( x \) is negative, \( y \) is negative.
• Q (Inflection): On the positive y-axis \( (0, 3) \). The gradient is zero here (flat), but the curve continues upwards.
• R (Maximum): In the 1st quadrant (top-right). \( x \) is positive, \( y > 3 \).

The curve must follow this path:

1. Start from top-left, come down to turn at P (Minimum).
2. Go up, crossing the negative x-axis.
3. Flatten out at Q (Inflection) but keep going up.
4. Continue up to turn at R (Maximum).
5. Go down, crossing the positive x-axis.

This ensures there are exactly 3 stationary points and 3 x-axis crossings (roots).

✓ (B4) All features correct

The Sine Rule relates sides and opposite angles: \( \frac{a}{\sin A} = \frac{b}{\sin B} \).

Here, side \( y \) is opposite \( 120^\circ \) and side \( 6 \) is opposite \( x^\circ \).

We know \( \sin 120^\circ = \sin(180^\circ – 60^\circ) = \sin 60^\circ = \frac{\sqrt{3}}{2} \).

We are given \( \sin x^\circ = \frac{1}{\sqrt{12}} \).

Simplify \( \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3} \), so \( \sin x^\circ = \frac{1}{2\sqrt{3}} \).

We need two numbers that multiply to \( 2 \times 5 = 10 \) and add to \( 7 \).

The numbers are 2 and 5.

This equation is the same form as part (a), but with \( \sin \theta \) instead of \( x \).

Using the result from (a):

Case 1: \( 2\sin \theta + 5 = 0 \Rightarrow \sin \theta = -2.5 \).

Impossible, as \( \sin \theta \) must be between -1 and 1.

Case 2: \( \sin \theta + 1 = 0 \Rightarrow \sin \theta = -1 \).

First, look at \( \sqrt{300} \). We can simplify this.

Multiply numerator and denominator by the conjugate of the denominator: \( 4\sqrt{3} + 5 \).