AQA Level 2 Certificate Further Mathematics – June 2022 (Paper 2 Calculator)

Why we do this: “Factorise fully” means we must take out the highest common factor (HCF) of both the numbers and the variables.

Look at the numbers \(12\) and \(18\). The largest number that divides both is \(6\).

Look at the variables \(w\) and \(w^2\). The term \(w\) is common to both.

How to do it: Divide each term in the original expression by the common factor \(6w\) to find what goes inside the bracket.

Why we do this: The midpoint \( M \) is the average of the coordinates of \( P \) and \( Q \). We can set up an equation for the y-coordinate since we are trying to find \( a \).

Midpoint formula for y: \( y_m = \frac{y_1 + y_2}{2} \)

How to do it: Rearrange the equation to isolate \( a \).

Strategy: It’s usually easier to multiply the matrix by the scalar number (3) first, then multiply the two matrices. Or you can multiply the matrices first then multiply by 3. Let’s multiply the first matrix by 3.

Method: Multiply rows by columns.

Top-left: (Row 1 of A) • (Col 1 of B)

Top-right: (Row 1 of A) • (Col 2 of B)

Strategy: Multiply the \(2 \times 2\) matrix by the \(2 \times 1\) matrix to get a system of equations.

Why: The question tells us this equals \( \begin{pmatrix} 78 \\ 12 \end{pmatrix} \).

How: Substitute \( a = 4 \) into the bottom row equation.

Strategy: Rearrange the equation into the form \( y = mx + c \) to find the gradient \( m \).

Method: Use the point-gradient formula \( y – y_1 = m(x – x_1) \) with point \( (2, 1) \) and gradient \( -4 \).

Why: The point \( (d, 2d) \) lies on the line, so its coordinates must satisfy the equation. Substitute \( x = d \) and \( y = 2d \).

Strategy: Divide both sides by 3 to isolate \( x^2 \).

Method: If \( x^2 < 16 \), then \( x \) must be between \( -4 \) and \( 4 \).

Check: We need integers (whole numbers) that are negative and satisfy \( -4 < x < 4 \).

Note: \( -4 \) is not included because it’s strictly less than.

Strategy: Expand \( (2n+1)^2 \) and \( (2n-1)^2 \). Be careful with the subtraction sign.

Method: Collect like terms. Note that \( 4n^2 – 4n^2 = 0 \) and \( 1 – 1 = 0 \).

Reasoning: Divide by 4 and explain why the result is even.

Analysis: The number must be between 200,000 and 400,000. This means it is a 6-digit number.

The first digit must be either 2 or 3 (since 1 is too small and 5, 8, 9 are too big).

Method: Fix the first digit and calculate permutations for the remaining 5 digits.

We have 6 distinct digits in total: {1, 2, 3, 5, 8, 9}.

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

Method: We are given the rate of change is 4.

Method: Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \).

Here \( a=3, b=-10, c=-4 \).

Analysis: A “linear sequence” has a constant difference.

Term 1 to Term 2: Add \( 4k \)

Term 2 to Term 3: Add \( 4k \)

So, Term 4 = Term 3 + \( 4k \)

Method: Use the formula \( u_n = a + (n-1)d \).

Here, \( a = 30 \) and common difference \( d = 4k \).

Method: Set the expression equal to 525.

Why: The y-intercept occurs when \( x = 0 \).

Why: As \( x \) increases (gets more positive), \( 2^x \) gets larger rapidly (exponential growth).

Why: We have the hypotenuse (\(4\)) and the adjacent side (\(a-3\)) relative to angle \(x\). We should use cosine.

Method: We are given \( a > 5 \). Let’s see what this implies for \(\cos x\).

Reasoning: We know \(\cos 60^\circ = 0.5\). As the angle \(x\) decreases from \(60^\circ\) to \(0^\circ\), the cosine value increases towards 1.

Since \(\cos x > 0.5\), the angle \(x\) must be smaller than \(60^\circ\).

Also, in a triangle, a length must be positive, and the angle must be positive (\(x > 0\)).

Note: The adjacent side \(a-3\) must be less than the hypotenuse 4, which means \(a < 7\). This gives \(\cos x < 1\), which corresponds to \(x > 0^\circ\).

Why: Before differentiating, it is much easier to split the fraction into individual terms. This avoids using the quotient rule.

Method: Find \( \frac{dy}{dx} \) by multiplying by the power and reducing the power by 1.

Method: Plug in the value to find the specific gradient.

Why: The center is the midpoint of the diameter \( AB \).

Method: The radius squared (\( r^2 \)) is the squared distance from the center \( (1, 9) \) to either point \( A \) or \( B \). Let’s use \( B(4, 13) \).

Why: In a cyclic quadrilateral, opposite angles sum to \( 180^\circ \).

Opposite angles here are \( \angle PSR \) and \( \angle PQR \).

Method: We are given expressions for both angles.

Method: Substitute the expressions into the sum equation and solve for x.

Method: Find a common denominator to add the fractions.

Method: Apply the square root to the numerator and denominator separately. Remember \( \sqrt{x^6} = x^3 \).

Method: To divide by a fraction, multiply by its reciprocal (keep-change-flip).

Method: Cancel common terms and simplify coefficients.

Method: The area of a triangle is \( \frac{1}{2}ab \sin C \). We have two sides of length 16 cm. Let the angle between them be \( \theta \).

Reasoning: The triangle is isosceles with two sides of length 16 cm. The angle we found (\( \theta \)) is the angle between these equal sides (the vertex angle).

The remaining two angles (base angles) must be equal. Let them be \( y \).

Strategy: Notice that equation (1) has \( +3b \) and equation (2) has \( -3b \). Adding them will eliminate \( b \) immediately.

Strategy: We need another equation with just \( a \) and \( c \). Multiply equation (3) by 3 to get \( 3b \), then subtract equation (1).

Method: Compare Equation 4 (\( 5a + 3c = -1 \)) and Equation 5 (\( 5a + 11c = 23 \)). The \( 5a \) terms are identical, so subtract Eq 4 from Eq 5.

Method: Substitute \( a = -2 \) and \( c = 3 \) into any original equation, e.g., Equation (3).

Strategy: We need to find the sides of triangle \( APC \) to use the Cosine Rule. We already know \( AP = 25 \).

First, find the position of \( P \) on \( HG \). \( HG = AB = 40 \).

Ratio \( 3:7 \) means \( HP = \frac{3}{10} \times 40 = 12 \) and \( PG = 28 \).

Method: Use Pythagoras in the right-angled triangle \( PGC \) (on the back face).

\( PG = 28 \), \( GC = 20 \).

Method: \( AC \) is the diagonal of the base rectangle \( ABCD \). \( AB = 40 \), \( BC = 9 \).

Method: We have all three sides of triangle \( APC \): \( a = 41 \), \( p = 34.41 \), \( c = 25 \). We want angle \( P \).

Formula: \( a^2 = b^2 + c^2 – 2bc \cos A \) (rearranged for angle P).

\( AC^2 = AP^2 + PC^2 – 2(AP)(PC) \cos P \)

Method: Use FOIL (First, Outer, Inner, Last) on \( (3x + 4)(2x – 3) \).

Method: Multiply \( (6x^2 – x – 12) \) by \( (5x – 2) \).

Why: The factor theorem states that if \( (ax – b) \) is a factor, then \( f(\frac{b}{a}) = 0 \).

For \( (2x – 1) \), we solve \( 2x – 1 = 0 \Rightarrow x = 0.5 \) (or \( \frac{1}{2} \)).

Strategy: Divide \( f(x) \) by \( (2x – 1) \) to find the quadratic factor.

Method: Factorise \( x^2 + 6x + 9 \).

Reasoning: List the roots found.

Strategy: Rearrange the equation to find \( \tan x \). Remember that taking the square root gives both positive and negative results.

Method: Use the CAST diagram or graph properties.

For \( \tan x = +1.2247 \) (Positive): Quadrants 1 and 3.

For \( \tan x = -1.2247 \) (Negative): Quadrants 2 and 4.

Strategy: Convert base 16 to base 2, and handle the fraction on the RHS.

Note: \( 16 = 2^4 \).

Method: Since the bases are the same (base 2), the powers must be equal.