AQA Level 2 Certificate Further Mathematics Paper 1 (Non-Calculator) 2021

💡 What are we doing? To find the straight-line distance between two coordinates, we use Pythagoras’ Theorem. The difference in \(x\) and the difference in \(y\) form the sides of a right-angled triangle.

✏ Working: Be careful with the negatives when subtracting.

💡 Calculation: Square the numbers and sum them.

💡 Strategy: Before differentiating, it is much easier to multiply out the brackets to get a polynomial in standard form.

💡 Concept: The “rate of change of \( y \) with respect to \( x \)” means \( \frac{dy}{dx} \). We use the power rule: multiply by the power, then subtract 1 from the power (\( ax^n \to anx^{n-1} \)).

💡 Strategy: To visualize a line, we need its gradient (\(m\)) and y-intercept (\(c\)). Rearrange the equation to isolate \(y\).

💡 Interpretation:

• The gradient \( m = -1.5 \). This is negative, so the line goes “downhill” (top-left to bottom-right).
• The y-intercept \( c = 2.5 \). This is positive, so it crosses the y-axis above the origin.

💡 Strategy: The domain is the set of possible input values (\(x\)). We are given the output values (range) and need to work backwards.

💡 Strategy: For a quadratic \(x^2 – 4\), we can’t just substitute the endpoints because the minimum point might be inside the domain.

Domain: \( -1 < x < 3 \)

Sketch the curve or check critical points:

• At \(x = -1\): \(y = (-1)^2 – 4 = 1 – 4 = -3\)
• At \(x = 3\): \(y = 3^2 – 4 = 9 – 4 = 5\)
• Minimum Point: \(x^2\) is smallest at \(x=0\). Since \(x=0\) is inside the domain \(-1 < x < 3\), the minimum value is \(0^2 - 4 = -4\).

💡 Strategy: Set \( y = h(x) \), then rearrange to make \(x\) the subject.

💡 Strategy: Set the nth term expression equal to 5 and solve for \(n\).

💡 Concept: As \(n\) becomes very large (infinity), the constant terms (+47 and +1) become insignificant compared to the \(n\) terms. We look at the ratio of the highest power of \(n\).

💡 Logic: We are told \( \sin 220^\circ = -k \). Note that \( 220^\circ \) is in the 3rd quadrant (between 180° and 270°), where sine is negative. This implies \( k \) itself is a positive number.

Let’s find the reference angle (acute angle):

💡 Goal: We need to solve \( \sin x = k \). Since \( k = \sin 40^\circ \), we are solving \( \sin x = \sin 40^\circ \).

Since \( k \) is positive, we are looking for answers in the 1st and 2nd quadrants (where sine is positive).

💡 Strategy: Expand the right-hand side first so we can compare terms.

💡 Strategy: Substitute the expansion back into the original inequality and group terms. Notice that \(2x^2\) is on both sides, so they will cancel out.

💡 Rule: When multiplying or dividing an inequality by a negative number, flip the sign.

💡 Strategy: Look for square numbers factors in 75 and 48.

\( 75 = 25 \times 3 \)

\( 48 = 16 \times 3 \)

💡 Strategy: Combine the terms inside the bracket first, then multiply by the outside term.

💡 Strategy: We’ll multiply \( (2x – 5) \) and \( (3x – 4) \) first. Use FOIL (First, Outside, Inside, Last).

💡 Strategy: Now multiply the quadratic result by \( (x + 2) \). Multiply every term in the quadratic by \(x\), then every term by \(2\).

💡 Strategy: Combine the \(x^2\) terms and the \(x\) terms.

💡 Strategy: Find the first differences, then the second differences. The second difference determines the \(n^2\) coefficient.

💡 Formula: The coefficient of \(n^2\) is half the second difference.

💡 Strategy: Calculate \( -n^2 \) for the first few terms, subtract it from the original sequence, and find the rule for the remainder.

💡 Strategy: Multiply the rows of the first matrix by the columns of the second matrix.

💡 Knowledge: The identity matrix \( \mathbf{I} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \). Therefore \( k\mathbf{I} = \begin{pmatrix} k & 0 \\ 0 & k \end{pmatrix} \).

💡 Strategy: Compare corresponding elements.

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

We know centre \( (a, b) = (4, -2) \). We need to calculate \( r^2 \).

💡 Strategy: The tangent is perpendicular to the radius.

1. Find the gradient of the radius \( CA \).
2. Find the negative reciprocal to get the tangent gradient.
3. Use \( y – y_1 = m(x – x_1) \) with point \( A \).

💡 Strategy: Substitute the coordinates of \( A \) into the equation \( y = k^x \) and solve for \( k \).

💡 Strategy: Substitute \( x = -1 \) and our value of \( k \) into the equation.

💡 Strategy: Equation (3) only has \( a \) and \( c \). If we combine (1) and (2) to eliminate \( b \), we will get another equation with just \( a \) and \( c \).

💡 Strategy: Now use equations (3) and (5).

(3): \( 2a – 5c = -7 \)

(5): \( 11a + 5c = 59 \)

The \( c \) terms have opposite coefficients (\(-5\) and \(+5\)), so we can just add them.

💡 Strategy: Substitute \( a = 4 \) back into one of the equations.

💡 Check: Verify with Eq (2).

💡 Strategy: Rearrange to make \( \tan x \) the subject.

💡 Knowledge: You must know your exact trig values. Which angle has a tan of \( 1/\sqrt{3} \)?

💡 Theory: The Factor Theorem states that if \( (ax – b) \) is a factor, then \( f(\frac{b}{a}) = 0 \). Here, for \( (2x + 1) \), we substitute \( x = -\frac{1}{2} \).

💡 Strategy: We need to factorize the cubic fully. We can use polynomial division or inspection to find the quadratic factor.

Expression: \( 200x^3 + 100x^2 – 18x – 9 \)

We can also try factoring by grouping:

\( 100x^2(2x + 1) – 9(2x + 1) \)

💡 Strategy: We have the graph for \( y = x^2 – 6x + 5 \). We want to solve \( x^2 – 7x + 9 = 0 \).

Rearrange the target equation to look like the drawn graph:

We want the LHS to be \( x^2 – 6x + 5 \). Let’s adjust terms:

Substitute \( y \) for the quadratic part:

💡 Action: Plot \( y = x – 4 \).

• When \( x = 4, y = 0 \)
• When \( x = 6, y = 2 \)
• When \( x = 0, y = -4 \)

The solution is where the line crosses the curve.

💡 Formula: \( a^2 = b^2 + c^2 – 2bc \cos A \)

Here, the side opposite the angle \( 60^\circ \) is 7. The sides forming the angle are \( x \) and 3.

💡 Knowledge: \( \cos 60^\circ = 0.5 \)

💡 Reasoning: A length cannot be negative.

💡 Analysis:

1. Roots: \( y \) crosses the x-axis at \( x = 5 \). It is negative before 5 and positive after.
2. Turning Points:
   • Changes from increasing to decreasing at \( x = -1 \). This is a Maximum.
   • Changes from decreasing to increasing at \( x = 2 \). This is a Minimum.
3. Vertical Position: Since the curve is negative for all \( x < 5 \), the Maximum at \( x = -1 \) must be below the x-axis (negative y-value). This is crucial!

💡 Visual:

• Curve goes up to max at \( x=-1 \) (y is negative).
• Curve goes down to min at \( x=2 \) (y is more negative).
• Curve goes up and crosses axis at \( x=5 \).

💡 Strategy: List all digits from the date 14/09/2006 and find the unique set.

💡 Method: Use the slot method for the 4 digits. Apply the restrictions.

💡 Arithmetic:

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

Here, Base \( = 6 + 2\sqrt{7} \) and Height \( = AD \).

💡 Strategy: Divide the Area by the simplified base.

💡 Method: Multiply top and bottom by the conjugate of the denominator: \( 3 – \sqrt{7} \).

💡 Strategy: Write 8 and 4 as powers of 2 so we can use index laws.

\( 8 = 2^3 \) and \( 4 = 2^2 \)

💡 Strategy: We cannot subtract powers directly (\( 2^a – 2^b \neq 2^{a-b} \)). We must factorise out the smaller power.

💡 Circle Theorems:

• Alternate Segment Theorem: The angle between tangent \( EFG \) and chord \( FH \) is equal to the angle in the alternate segment (\( \angle FJH \) or \( \angle FKH \)). Wait, let’s use the provided Mark Scheme equations as the reliable mathematical truth for this specific problem instance.