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AQA Level 2 Certificate in Further Mathematics
Paper 1: Non-Calculator (June 2016)
💡 How to use this Interactive Exam Paper
- Try it yourself: Attempt the question on paper first.
- Show Solution: Click to reveal the step-by-step worked solution.
- Three-Layer Explanation:
- 💡 Understand: Why we are doing this.
- ✏ Method: How to do the calculation.
- 🏁 Check: What this result means.
- Non-Calculator: This is Paper 1. Arithmetic methods are shown explicitly.
Table of Contents
- Question 1 (Differentiation)
- Question 2 (Matrices)
- Question 3 (Sequences)
- Question 4 (Circle Geometry)
- Question 5 (Circle Theorems)
- Question 6 (Algebraic Identity)
- Question 7 (Quadratic Inequality)
- Question 8 (Surds & Indices)
- Question 9 (Binomial Expansion)
- Question 10 (Indices & Substitution)
- Question 11 (Coordinate Geometry)
- Question 12 (Completing the Square)
- Question 13 (Surds)
- Question 14 (Factor Theorem)
- Question 15 (Rationalising Denominators)
- Question 16 (Trigonometry)
- Question 17 (Cubic Curves)
- Question 18 (Difference of Squares)
- Question 19 (Geometry & Exact Values)
- Question 20 (Cosine Rule Proof)
Question 1 (3 marks)
Given \( y = x^2(x – 10) \)
Work out \( \frac{dy}{dx} \)
Worked Solution
Step 1: Expand the expression
💡 Why we do this: Before differentiating, it is usually easier to expand brackets so we have a simple polynomial (a sum of terms like \( ax^n \)).
\[ y = x^2(x) – x^2(10) \]
\[ y = x^3 – 10x^2 \]
Step 2: Differentiate term by term
💡 Method: Use the power rule for differentiation: if \( y = ax^n \), then \( \frac{dy}{dx} = anx^{n-1} \).
For \( x^3 \): multiply by power (3) and reduce power by 1 (to 2) \(\rightarrow 3x^2\)
For \( -10x^2 \): multiply by power (2) and reduce power by 1 (to 1) \(\rightarrow -10(2)x^1 = -20x\)
\[ \frac{dy}{dx} = 3x^2 – 20x \]
Final Answer:
\[ \frac{dy}{dx} = 3x^2 – 20x \]
Question 2 (3 marks)
\[ 4 \begin{pmatrix} 1 – 2a \\ a \end{pmatrix} = \begin{pmatrix} b \\ 12 \end{pmatrix} \]
Work out the values of \( a \) and \( b \).
Worked Solution
Step 1: Multiply the scalar into the matrix
💡 Method: Multiply every element inside the matrix by the number outside (the scalar 4).
\[ \begin{pmatrix} 4(1 – 2a) \\ 4(a) \end{pmatrix} = \begin{pmatrix} b \\ 12 \end{pmatrix} \]
\[ \begin{pmatrix} 4 – 8a \\ 4a \end{pmatrix} = \begin{pmatrix} b \\ 12 \end{pmatrix} \]
Step 2: Equate corresponding components
💡 Why we do this: Two matrices are equal only if their top elements match and their bottom elements match. This creates two simultaneous equations.
Top: \( 4 – 8a = b \) (Equation 1)
Bottom: \( 4a = 12 \) (Equation 2)
Step 3: Solve for a and b
💡 Method: Equation 2 is simpler. Solve it for \( a \), then substitute that value into Equation 1 to find \( b \).
From Equation 2:
\[ 4a = 12 \implies a = \frac{12}{4} = 3 \]
Substitute \( a = 3 \) into Equation 1:
\[ b = 4 – 8(3) \]
\[ b = 4 – 24 \]
\[ b = -20 \]
Final Answer:
\[ a = 3, \quad b = -20 \]
Question 3 (3 marks)
The \( n \)th term of a sequence is \( \frac{3n}{5n + 12} \)
(a) Work out the position of the term that has a value of \( \frac{1}{2} \)
(b) Write down the limiting value of \( \frac{3n}{5n + 12} \) as \( n \rightarrow \infty \)
Worked Solution
Part (a): Find the position (n)
💡 Why we do this: The “position” is the value of \( n \). We set the \( n \)th term expression equal to the given value and solve for \( n \).
\[ \frac{3n}{5n + 12} = \frac{1}{2} \]
Cross-multiply to remove fractions:
\[ 2(3n) = 1(5n + 12) \]
\[ 6n = 5n + 12 \]
Subtract \( 5n \) from both sides:
\[ n = 12 \]
Answer to (a): 12th term
Part (b): Find the limiting value
💡 Why we do this: As \( n \) becomes extremely large, the constant term (+12) becomes insignificant compared to the \( n \) terms. We look at the ratio of the coefficients of the highest power of \( n \).
Divide every term by \( n \):
\[ \frac{3}{5 + \frac{12}{n}} \]
As \( n \rightarrow \infty \), the term \( \frac{12}{n} \rightarrow 0 \).
\[ \text{Limit} = \frac{3}{5 + 0} = \frac{3}{5} \]
Alternatively, \( \frac{3}{5} = 0.6 \).
Answer to (b): \( \frac{3}{5} \) or 0.6
Final Answer:
(a) \( n = 12 \)
(b) \( \frac{3}{5} \)
Question 4 (2 marks)
The equation of a circle is \( (x + 5)^2 + (y – 8)^2 = 10 \)
(a) What are the coordinates of the centre of the circle?
(b) Write down the radius of the circle.
Worked Solution
Understanding the Circle Equation
💡 Why we do this: The standard equation of a circle is \( (x – a)^2 + (y – b)^2 = r^2 \), where \( (a, b) \) is the centre and \( r \) is the radius.
Part (a): Find the centre
💡 Method: Compare \( (x + 5)^2 \) with \( (x – a)^2 \) and \( (y – 8)^2 \) with \( (y – b)^2 \).
\( x – a = x + 5 \implies -a = 5 \implies a = -5 \)
\( y – b = y – 8 \implies -b = -8 \implies b = 8 \)
Centre is \( (-5, 8) \)
Part (b): Find the radius
💡 Method: The number on the right side of the equation is \( r^2 \). We need \( r \).
\[ r^2 = 10 \]
\[ r = \sqrt{10} \]
Final Answer:
(a) \( (-5, 8) \)
(b) \( \sqrt{10} \)
Question 5 (3 marks)
\( A, B, C \) and \( D \) are points on a circle, centre \( O \).
Work out the value of \( x \).
Worked Solution
Step 1: Identify the Circle Theorems
💡 Why we do this: We have an angle at the centre (\( \angle AOC \)) and an angle at the circumference (\( \angle ADC \)) subtended by the same arc \( ABC \).
The angle at the centre is twice the angle at the circumference.
The angle subtended by the major arc \( AC \) at the centre is the reflex angle \( AOC \).
The angle subtended by the major arc \( AC \) at the circumference is \( \angle ADC \) (which is \( 3x \)).
Therefore: \( \text{Reflex } \angle AOC = 2 \times \angle ADC \)
\[ \text{Reflex } \angle AOC = 2(3x) = 6x \]
Step 2: Form an equation using angles at a point
💡 Method: Angles around the centre point \( O \) must add up to \( 360^\circ \).
We have the obtuse angle \( AOC = 2x + 48 \).
We found the reflex angle \( AOC = 6x \).
Together they complete the circle:
\[ (2x + 48) + 6x = 360 \]
Step 3: Solve for x
💡 Method: Combine like terms and solve the linear equation.
\[ 8x + 48 = 360 \]
Subtract 48 from both sides:
\[ 8x = 312 \]
Divide by 8:
39 ┌───── 8 │ 312 │-24 │ 72 │ -72 │ 0
\[ x = 39 \]
Final Answer:
\[ x = 39 \]
Question 6 (4 marks)
\( mx + 4 – 2(x + p) \equiv 6(x + 1) \)
where \( m \) and \( p \) are integers.
Work out the values of \( m \) and \( p \).
Worked Solution
Step 1: Expand both sides
💡 Why we do this: To compare the coefficients of \( x \) and the constant terms, we must first write both sides in the form \( Ax + B \).
Left Hand Side (LHS):
\[ mx + 4 – 2(x) – 2(p) \]
\[ mx + 4 – 2x – 2p \]
Group terms:
\[ (m – 2)x + (4 – 2p) \]
Right Hand Side (RHS):
\[ 6(x) + 6(1) \]
\[ 6x + 6 \]
Step 2: Equate coefficients
💡 Method: Since it is an identity (true for all \( x \)), the coefficient of \( x \) on the LHS must equal the coefficient of \( x \) on the RHS. Similarly, the constant terms must be equal.
Compare \( x \) coefficients:
\[ m – 2 = 6 \]
\[ m = 8 \]
Compare constants:
\[ 4 – 2p = 6 \]
\[ -2p = 6 – 4 \]
\[ -2p = 2 \]
\[ p = -1 \]
Final Answer:
\[ m = 8, \quad p = -1 \]
Question 7 (3 marks)
Work out the integer values of \( x \) for which \( x^2 – 20x + 96 < 0 \)
Worked Solution
Step 1: Solve the corresponding quadratic equation
💡 Why we do this: To solve an inequality, we first find the critical values (roots) where the expression equals zero. We do this by factorising.
We need factors of 96 that add up to -20.
Let’s list factors of 96:
- \( 1 \times 96 \)
- \( 2 \times 48 \)
- \( 4 \times 24 \)
- \( 8 \times 12 \) (\( 8 + 12 = 20 \))
Since the middle term is negative (-20) and the last term is positive (+96), both signs in the brackets must be negative.
\[ (x – 8)(x – 12) = 0 \]
Critical values: \( x = 8 \) and \( x = 12 \).
Step 2: Determine the inequality range
💡 Method: The quadratic \( x^2 – 20x + 96 \) represents a parabola that opens upwards (U-shaped). We want the part where it is less than 0 (below the x-axis). This happens between the roots.
Inequality: \( 8 < x < 12 \)
Step 3: List the integer values
💡 Check: The question asks for integer values. Since the inequality is strictly “less than” (<), we do not include 8 and 12.
Integers strictly between 8 and 12 are:
9, 10, 11
Final Answer:
9, 10, 11
Question 8 (3 marks)
Solve \( (3 – \sqrt{x})^{\frac{1}{3}} = -2 \)
Worked Solution
Step 1: Remove the fractional power
💡 Method: The power of \( \frac{1}{3} \) means the cube root. To eliminate it, we cube (raise to the power of 3) both sides.
\[ \left((3 – \sqrt{x})^{\frac{1}{3}}\right)^3 = (-2)^3 \]
\[ 3 – \sqrt{x} = -8 \]
(Note: \( -2 \times -2 \times -2 = -8 \))
Step 2: Isolate the square root term
💡 Method: Rearrange the equation to make \( \sqrt{x} \) positive.
Add \( \sqrt{x} \) to both sides:
\[ 3 = -8 + \sqrt{x} \]
Add 8 to both sides:
\[ 11 = \sqrt{x} \]
Step 3: Solve for x
💡 Method: Square both sides to remove the square root.
\[ 11^2 = x \]
\[ x = 121 \]
Final Answer:
\[ x = 121 \]
Question 9 (3 marks)
Expand and simplify \( (x – 5)^3 \)
Worked Solution
Step 1: Break it down into two parts
💡 Strategy: Write \( (x – 5)^3 \) as \( (x – 5)(x – 5)^2 \). First expand the squared bracket.
\[ (x – 5)^2 = (x – 5)(x – 5) \]
\[ = x^2 – 5x – 5x + 25 \]
\[ = x^2 – 10x + 25 \]
Step 2: Multiply by the third bracket
💡 Method: Multiply \( (x – 5) \) by the result from Step 1.
\[ (x – 5)(x^2 – 10x + 25) \]
Multiply \( x \) by the trinomial:
\[ x(x^2 – 10x + 25) = x^3 – 10x^2 + 25x \]
Multiply \( -5 \) by the trinomial:
\[ -5(x^2 – 10x + 25) = -5x^2 + 50x – 125 \]
Step 3: Collect like terms
💡 Check: Combine the \( x^2 \) terms and the \( x \) terms.
x³ - 10x² + 25x
- 5x² + 50x - 125
-------------------------
x³ - 15x² + 75x - 125
Final Answer:
\[ x^3 – 15x^2 + 75x – 125 \]
Question 10 (4 marks)
Given:
\[ \sqrt[4]{x} = 2 \quad \text{and} \quad y^{-2} = 25 \]
\[ x > 0 \quad \text{and} \quad y < 0 \]
Work out the value of \( \frac{x}{y} \)
Worked Solution
Step 1: Find the value of x
💡 Method: \( \sqrt[4]{x} \) means \( x^{1/4} \). To find \( x \), raise 2 to the power of 4.
\[ x = 2^4 \]
\[ x = 2 \times 2 \times 2 \times 2 = 16 \]
Step 2: Find the value of y
💡 Method: A negative power means reciprocal: \( y^{-2} = \frac{1}{y^2} \).
\[ \frac{1}{y^2} = 25 \]
Rearrange to find \( y^2 \):
\[ y^2 = \frac{1}{25} \]
Take the square root:
\[ y = \sqrt{\frac{1}{25}} = \pm \frac{1}{5} \]
Critical Check: The question states \( y < 0 \). Therefore:
\[ y = -\frac{1}{5} \]
Step 3: Calculate x divided by y
💡 Method: Divide the integer by the fraction.
\[ \frac{x}{y} = \frac{16}{-\frac{1}{5}} \]
Dividing by a fraction is the same as multiplying by its reciprocal:
\[ 16 \times -5 \]
\[ = -80 \]
Final Answer:
\[ -80 \]
Question 11 (3 marks)
\( A(1\frac{1}{5}, 3\frac{4}{5}) \), \( B(2, 1\frac{4}{5}) \) and \( C(5, 3) \) are points on a coordinate grid.
Show that the line segments \( AB \) and \( BC \) are perpendicular.
Worked Solution
Step 1: Convert coordinates to decimals
💡 Strategy: Dealing with decimals is often easier than mixed numbers for gradient calculations.
\( A(1.2, 3.8) \)
\( B(2, 1.8) \)
\( C(5, 3) \)
Step 2: Calculate the gradient of AB
💡 Method: \( m = \frac{y_2 – y_1}{x_2 – x_1} \)
\[ m_{AB} = \frac{1.8 – 3.8}{2 – 1.2} \]
\[ m_{AB} = \frac{-2}{0.8} \]
To simplify, multiply top and bottom by 10:
\[ m_{AB} = \frac{-20}{8} = -2.5 \text{ (or } -\frac{5}{2} \text{)} \]
Step 3: Calculate the gradient of BC
\[ m_{BC} = \frac{3 – 1.8}{5 – 2} \]
\[ m_{BC} = \frac{1.2}{3} \]
\[ m_{BC} = 0.4 \text{ (or } \frac{2}{5} \text{)} \]
Step 4: Check for perpendicularity
💡 Check: Two lines are perpendicular if the product of their gradients is \( -1 \).
\[ m_{AB} \times m_{BC} = -2.5 \times 0.4 \]
\[ = -1 \]
Since the product is \( -1 \), the lines are perpendicular.
Conclusion: Gradients are \( -2.5 \) and \( 0.4 \). Product is \( -1 \), therefore \( AB \perp BC \).
Question 12 (4 marks total)
You are given that \( x^2 + 6x + 2 \equiv (x + h)^2 + k \)
(a) Work out the values of \( h \) and \( k \). [2 marks]
(b) Write down the coordinates of the minimum point on the curve \( y = x^2 + 6x + 2 \). [1 mark]
(c) Solve the equation \( x^2 + 6x + 2 = 0 \). Give your answers in the form \( a \pm \sqrt{b} \). [1 mark]
Worked Solution
Part (a): Complete the Square
💡 Method: To complete the square for \( x^2 + bx \), use \( (x + \frac{b}{2})^2 – (\frac{b}{2})^2 \).
Start with \( x^2 + 6x \):
\[ (x + 3)^2 – 3^2 = (x + 3)^2 – 9 \]
Now add the constant term (+2) from the original expression:
\[ (x + 3)^2 – 9 + 2 \]
\[ (x + 3)^2 – 7 \]
Comparing to \( (x + h)^2 + k \):
\[ h = 3, \quad k = -7 \]
Part (b): Minimum Point
💡 Why we do this: For a quadratic in the form \( y = (x + h)^2 + k \), the minimum value occurs when the squared term is zero (i.e., \( x = -h \)). The minimum y-value is \( k \).
Vertex is at \( (-h, k) \).
\[ (-3, -7) \]
Part (c): Solve the Equation
💡 Method: Use the completed square form to solve for \( x \).
\[ (x + 3)^2 – 7 = 0 \]
Add 7 to both sides:
\[ (x + 3)^2 = 7 \]
Take the square root:
\[ x + 3 = \pm \sqrt{7} \]
Subtract 3:
\[ x = -3 \pm \sqrt{7} \]
Final Answer:
(a) \( h = 3, k = -7 \)
(b) \( (-3, -7) \)
(c) \( -3 \pm \sqrt{7} \)
Question 13 (3 marks)
Solve \( \sqrt{125} + \sqrt{20} = \sqrt{80} + \sqrt{x} \)
Worked Solution
Step 1: Simplify the surds
💡 Method: Find the largest square number factor for each number under the root.
\( \sqrt{125} = \sqrt{25 \times 5} = 5\sqrt{5} \)
\( \sqrt{20} = \sqrt{4 \times 5} = 2\sqrt{5} \)
\( \sqrt{80} = \sqrt{16 \times 5} = 4\sqrt{5} \)
Step 2: Substitute and solve for \( \sqrt{x} \)
Substitute the simplified forms back into the equation:
\[ 5\sqrt{5} + 2\sqrt{5} = 4\sqrt{5} + \sqrt{x} \]
Combine the left side:
\[ 7\sqrt{5} = 4\sqrt{5} + \sqrt{x} \]
Subtract \( 4\sqrt{5} \) from both sides:
\[ 3\sqrt{5} = \sqrt{x} \]
Step 3: Find x
💡 Method: To find \( x \), we can square both sides or write \( 3\sqrt{5} \) as a single square root.
Method A (Squaring):
\[ x = (3\sqrt{5})^2 = 3^2 \times (\sqrt{5})^2 = 9 \times 5 = 45 \]
Method B (Inside the root):
\[ 3\sqrt{5} = \sqrt{3^2 \times 5} = \sqrt{9 \times 5} = \sqrt{45} \]
So \( x = 45 \)
Final Answer:
\[ x = 45 \]
Question 14 (5 marks total)
(a) \( (x – 3) \) is a factor of \( x^3 – 8x^2 + ax + 42 \) where \( a \) is an integer.
Show that the value of \( a \) is 1. [2 marks]
(b) Hence, factorise fully \( x^3 – 8x^2 + x + 42 \). [3 marks]
Worked Solution
Part (a): Use the Factor Theorem
💡 Why we do this: If \( (x – 3) \) is a factor, then substituting \( x = 3 \) into the polynomial must equal zero.
Let \( f(x) = x^3 – 8x^2 + ax + 42 \)
Calculate \( f(3) \):
\[ (3)^3 – 8(3)^2 + a(3) + 42 = 0 \]
\[ 27 – 8(9) + 3a + 42 = 0 \]
\[ 27 – 72 + 3a + 42 = 0 \]
Combine constants:
\[ 27 + 42 – 72 = 69 – 72 = -3 \]
\[ -3 + 3a = 0 \]
\[ 3a = 3 \implies a = 1 \]
Part (b): Factorise the cubic
💡 Method: We know \( (x – 3) \) is a factor. We need to find the quadratic factor by division (or inspection), then factorise that quadratic.
Polynomial: \( x^3 – 8x^2 + x + 42 \)
Divide by \( (x – 3) \):
x² - 5x - 14
_________________
x - 3 | x³ - 8x² + x + 42
- (x³ - 3x²)
__________
-5x² + x
- (-5x² + 15x)
__________
-14x + 42
- (-14x + 42)
__________
0
Now factorise the quadratic \( x^2 – 5x – 14 \):
We need numbers that multiply to -14 and add to -5.
Factors of -14: (-7, 2)
\[ (x – 7)(x + 2) \]
Final Answer:
\[ (x – 3)(x – 7)(x + 2) \]
Question 15 (3 marks)
Rationalise the denominator and simplify fully:
\[ \frac{6}{\sqrt{7} + 2} \]
Worked Solution
Step 1: Multiply by the conjugate
💡 Method: To remove the surd from the denominator, multiply the top and bottom by \( \sqrt{7} – 2 \) (the difference of two squares).
\[ \frac{6}{\sqrt{7} + 2} \times \frac{\sqrt{7} – 2}{\sqrt{7} – 2} \]
Step 2: Expand and simplify
Numerator:
\[ 6(\sqrt{7} – 2) = 6\sqrt{7} – 12 \]
Denominator:
\[ (\sqrt{7} + 2)(\sqrt{7} – 2) = (\sqrt{7})^2 – 2^2 \]
\[ = 7 – 4 = 3 \]
Step 3: Divide numerator by denominator
\[ \frac{6\sqrt{7} – 12}{3} \]
Divide each term by 3:
\[ \frac{6\sqrt{7}}{3} – \frac{12}{3} \]
\[ = 2\sqrt{7} – 4 \]
Final Answer:
\[ 2\sqrt{7} – 4 \quad \text{or} \quad 2(\sqrt{7} – 2) \]
Question 16 (4 marks)
Angle \( \theta \) is obtuse and \( \sin \theta = \frac{\sqrt{11}}{6} \)
Work out the value of \( \cos \theta \)
Worked Solution
Step 1: Use the Pythagorean Identity
💡 Why we do this: The fundamental identity linking sine and cosine is \( \sin^2 \theta + \cos^2 \theta = 1 \).
\[ \left( \frac{\sqrt{11}}{6} \right)^2 + \cos^2 \theta = 1 \]
\[ \frac{11}{36} + \cos^2 \theta = 1 \]
Step 2: Rearrange for \( \cos^2 \theta \)
\[ \cos^2 \theta = 1 – \frac{11}{36} \]
\[ \cos^2 \theta = \frac{36}{36} – \frac{11}{36} \]
\[ \cos^2 \theta = \frac{25}{36} \]
Step 3: Find \( \cos \theta \) considering the quadrant
💡 Critical Step: \( \cos \theta = \pm \sqrt{\frac{25}{36}} = \pm \frac{5}{6} \). We must choose the correct sign.
The question states \( \theta \) is obtuse (between \( 90^\circ \) and \( 180^\circ \)). In this quadrant, cosine is negative.
\[ \cos \theta = -\frac{5}{6} \]
Final Answer:
\[ -\frac{5}{6} \]
Question 17 (5 marks)
The diagram shows a sketch of the cubic curve \( y = \frac{1}{3}x^3 – x^2 – 3x + k \)
where \( k \) is a constant.
The \( x \)-axis is a tangent to the curve at its minimum point.
Work out the value of \( k \).
Worked Solution
Step 1: Differentiate to find turning points
💡 Why we do this: The minimum point is a turning point where the gradient (\( \frac{dy}{dx} \)) is zero.
\[ y = \frac{1}{3}x^3 – x^2 – 3x + k \]
\[ \frac{dy}{dx} = x^2 – 2x – 3 \]
Step 2: Find the x-coordinate of the minimum
💡 Method: Set \( \frac{dy}{dx} = 0 \) and solve for \( x \).
\[ x^2 – 2x – 3 = 0 \]
Factorise:
\[ (x – 3)(x + 1) = 0 \]
Turning points are at \( x = 3 \) and \( x = -1 \).
Identify Minimum: For a positive cubic (\( +x^3 \)), the shape is “up-down-up”. The first turning point (left, \( x=-1 \)) is the maximum, and the second (right, \( x=3 \)) is the minimum.
So, the minimum is at \( x = 3 \).
Step 3: Use the tangent condition to find k
💡 What this tells us: The question says the x-axis is a tangent at the minimum point. This means the minimum point sits on the x-axis, so \( y = 0 \) when \( x = 3 \).
Substitute \( x = 3, y = 0 \) into the original equation:
\[ 0 = \frac{1}{3}(3)^3 – (3)^2 – 3(3) + k \]
\[ 0 = \frac{1}{3}(27) – 9 – 9 + k \]
\[ 0 = 9 – 9 – 9 + k \]
\[ 0 = -9 + k \]
\[ k = 9 \]
Final Answer:
\[ k = 9 \]
Question 18 (2 marks)
Factorise fully \( x^4 – 81 \)
Worked Solution
Step 1: Difference of Two Squares (First Pass)
💡 Method: Identify that both terms are square numbers. \( x^4 = (x^2)^2 \) and \( 81 = 9^2 \).
\[ x^4 – 81 = (x^2)^2 – 9^2 \]
\[ = (x^2 – 9)(x^2 + 9) \]
Step 2: Difference of Two Squares (Second Pass)
💡 Check: The term \( (x^2 – 9) \) is also a difference of two squares (\( x^2 – 3^2 \)). The term \( (x^2 + 9) \) cannot be factorised further.
\[ x^2 – 9 = (x – 3)(x + 3) \]
Combine with the remaining factor:
\[ (x – 3)(x + 3)(x^2 + 9) \]
Final Answer:
\[ (x – 3)(x + 3)(x^2 + 9) \]
Question 19 (6 marks total)
\( ABCD \) is a square.
\( CDE \) is a straight line.
\( AC \) is \( 3\sqrt{2} \) cm and angle \( DEA = 60^\circ \)
(a) Show that the side of the square is 3 cm. [2 marks]
(b) Show that the perimeter of trapezium \( ABCE \) is \( 3(3 + \sqrt{3}) \) cm. [4 marks]
Worked Solution
Part (a): Find the side of the square
💡 Method: Let the side of the square be \( x \). In the right-angled triangle \( ADC \) (half the square), Pythagoras’ theorem applies.
\[ AD^2 + CD^2 = AC^2 \]
\[ x^2 + x^2 = (3\sqrt{2})^2 \]
\[ 2x^2 = 9 \times 2 \]
\[ 2x^2 = 18 \]
\[ x^2 = 9 \implies x = 3 \]
Side length is 3 cm.
Part (b): Perimeter of ABCE
💡 Strategy: Perimeter = \( AB + BC + CE + EA \).
We know \( AB = 3 \), \( BC = 3 \). We need to calculate \( CE \) and \( EA \).
\( CE = CD + DE = 3 + DE \).
1. Find DE using Triangle ADE:
\( ADE \) is a right-angled triangle (at D). Angle \( E = 60^\circ \). Side \( AD = 3 \).
\[ \tan(60^\circ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AD}{DE} \]
\[ \sqrt{3} = \frac{3}{DE} \]
\[ DE = \frac{3}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \sqrt{3} \]
So, \( CE = 3 + \sqrt{3} \).
2. Find EA (Hypotenuse):
\[ \sin(60^\circ) = \frac{AD}{EA} \]
\[ \frac{\sqrt{3}}{2} = \frac{3}{EA} \]
\[ EA = \frac{6}{\sqrt{3}} = 2\sqrt{3} \]
3. Total Perimeter:
\[ P = AB + BC + CE + EA \]
\[ P = 3 + 3 + (3 + \sqrt{3}) + 2\sqrt{3} \]
\[ P = 9 + 3\sqrt{3} \]
Factorise by 3:
\[ P = 3(3 + \sqrt{3}) \text{ cm} \]
Final Answer:
\( 3(3 + \sqrt{3}) \) cm
Question 20 (4 marks)
In triangle \( PQR \), \( \cos P = \frac{1}{3} \).
Sides are \( PQ = 3n \), \( PR = 2n \), and \( QR = w \).
Show that triangle \( PQR \) is isosceles.
Worked Solution
Step 1: Apply the Cosine Rule
💡 Why we do this: We have two sides and the included angle (or its cosine), and we want to find the third side \( w \) to check if it equals one of the others.
\[ a^2 = b^2 + c^2 – 2bc \cos A \]
Substitute our values:
\[ w^2 = (3n)^2 + (2n)^2 – 2(3n)(2n) \cos P \]
Step 2: Simplify the expression
\[ w^2 = 9n^2 + 4n^2 – 12n^2 \left(\frac{1}{3}\right) \]
\[ w^2 = 13n^2 – 4n^2 \]
\[ w^2 = 9n^2 \]
Step 3: Conclude the proof
💡 Check: Solve for \( w \) and compare with other sides.
\[ w = \sqrt{9n^2} = 3n \]
We have:
- \( PQ = 3n \)
- \( QR = w = 3n \)
Since two sides are equal (\( PQ = QR \)), the triangle is isosceles.
Conclusion: \( w = 3n \), which equals side \( PQ \). Therefore, isosceles.