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AQA Level 2 Certificate Further Mathematics Paper 2 (June 2017)
📚 Guide to this Exam
- Time allowed: 2 hours
- Calculator: Allowed for all questions
- Total marks: 105
- Structure: Click “Show Solution” to see a step-by-step breakdown.
📋 Table of Contents
- Question 1 (Sequences)
- Question 2 (Matrices)
- Question 3 (Integers/Algebra)
- Question 4 (Indices)
- Question 5 (Coordinate Midpoint)
- Question 6 (Cone Geometry)
- Question 7 (Coordinate Area)
- Question 8 (Triangle Geometry)
- Question 9 (Inequalities)
- Question 10 (Triangle Area)
- Question 11 (Piecewise Function)
- Question 12 (Factorisation)
- Question 13 (Algebraic Fractions)
- Question 14 (Perpendicular Lines)
- Question 15 (Rearranging Formulae)
- Question 16 (Indices Expressions)
- Question 17 (Differentiation)
- Question 18 (Rectangle Geometry)
- Question 19 (Trig Functions)
- Question 20 (Circle Theorems)
- Question 21 (Tangents/Normals)
- Question 22 (Intersection)
- Question 23 (Trig Identities)
- Question 24 (Completing Square)
Question 1 (3 marks)
(a) The \(n\)th term of a sequence is \( \frac{3 – 5n}{2} \)
Work out the difference between the 20th term and the 8th term. [2 marks]
(b) The \(n\)th term of another sequence is \( \frac{3n}{1 – 2n} \)
Write down the limiting value of the sequence as \( n \to \infty \) [1 mark]
Worked Solution
Part (a): Difference between terms
Step 1: Calculate the 20th and 8th terms
Substitute \( n=20 \) and \( n=8 \) into the formula.
For \( n=20 \):
\[ \frac{3 – 5(20)}{2} = \frac{3 – 100}{2} = \frac{-97}{2} = -48.5 \]For \( n=8 \):
\[ \frac{3 – 5(8)}{2} = \frac{3 – 40}{2} = \frac{-37}{2} = -18.5 \]Step 2: Find the difference
Subtract the 8th term from the 20th term (or vice versa, difference implies magnitude or \( u_{20} – u_8 \)).
Alternatively, since it is a linear sequence with common difference \( d = -2.5 \) (coefficient of \(n\)):
\[ \text{Difference} = 12 \times d = 12 \times (-2.5) = -30 \]Part (b): Limiting value
Understanding Limiting Values
As \( n \to \infty \), the constant terms (like 1) become insignificant compared to the \( n \) terms. We look at the coefficients of the highest power of \( n \).
Divide numerator and denominator by \( n \):
\[ \frac{3n}{1 – 2n} = \frac{3}{\frac{1}{n} – 2} \]As \( n \to \infty \), \( \frac{1}{n} \to 0 \).
\[ \text{Limit} = \frac{3}{0 – 2} = -1.5 \]Final Answers:
(a) \( -30 \) (or \( 30 \))
(b) \( -1.5 \) or \( -\frac{3}{2} \)
Question 2 (5 marks)
\( \mathbf{A} = \begin{pmatrix} 4 & -1 \\ 3 & -2 \end{pmatrix} \quad \mathbf{B} = \begin{pmatrix} 5 \\ 2 \end{pmatrix} \)
(a) Work out \( \mathbf{A}^2 \) [2 marks]
(b) \( k\mathbf{B} = \begin{pmatrix} 11 – 3k \\ 11 – 6k \end{pmatrix} \) where \( k \) is a constant.
Work out the value of \( k \). [2 marks]
(c) Give a reason why it is not possible to work out \( \mathbf{BA} \) [1 mark]
Worked Solution
Part (a): Matrix Multiplication
Calculate \( \mathbf{A} \times \mathbf{A} \)
Multiply rows by columns: Row 1 \(\times\) Col 1, Row 1 \(\times\) Col 2, etc.
Part (b): Scalar Multiplication
Set up the equation
Multiply matrix \( \mathbf{B} \) by the scalar \( k \).
Equate to the given matrix:
\[ \begin{pmatrix} 5k \\ 2k \end{pmatrix} = \begin{pmatrix} 11 – 3k \\ 11 – 6k \end{pmatrix} \]We can solve either component. Top row:
\[ 5k = 11 – 3k \] \[ 8k = 11 \] \[ k = \frac{11}{8} = 1.375 \]Check with bottom row:
\[ 2(1.375) = 2.75 \] \[ 11 – 6(1.375) = 11 – 8.25 = 2.75 \]Consistent.
Part (c): Matrix Dimensions
Check compatibility
To multiply matrices, the number of columns in the first must equal the number of rows in the second.
\( \mathbf{B} \) is a \( 2 \times 1 \) matrix (2 rows, 1 column).
\( \mathbf{A} \) is a \( 2 \times 2 \) matrix.
Columns of \( \mathbf{B} \) (1) \( \neq \) Rows of \( \mathbf{A} \) (2).
Final Answers:
(a) \( \begin{pmatrix} 13 & -2 \\ 6 & 1 \end{pmatrix} \)
(b) \( k = 1.375 \) or \( \frac{11}{8} \)
(c) The number of columns in B (1) does not equal the number of rows in A (2).
Question 3 (4 marks)
(a) \( p, q \) and \( r \) are all integers greater than 1.
\( pqr = 1365 \)
Work out one possible set of values for \( p, q \) and \( r \). [2 marks]
(b) \( a \) and \( b \) are both square numbers greater than 1.
\( ab – 11b \) is also a square number.
By factorising \( ab – 11b \), work out one possible pair of values for \( a \) and \( b \).
You must show your working. [2 marks]
Worked Solution
Part (a): Prime Factors
Find the factors of 1365
Since the product of three integers is 1365, we should find its prime factors.
Number ends in 5, so divisible by 5:
\[ 1365 \div 5 = 273 \]Sum of digits of 273 is \( 2+7+3=12 \), so divisible by 3:
\[ 273 \div 3 = 91 \]Factors of 91 are 7 and 13:
\[ 91 = 7 \times 13 \]So, \( 1365 = 3 \times 5 \times 7 \times 13 \).
We need three integers \( p, q, r \) greater than 1. We can group any two prime factors together, or take three distinct primes if we treat the product as one number, but the question implies distinct single values or combinations.
Possible sets:
- \( 3, 5, 91 \) (since \( 7 \times 13 = 91 \))
- \( 3, 7, 65 \) (since \( 5 \times 13 = 65 \))
- \( 3, 13, 35 \)
- \( 5, 7, 39 \)
- \( 15, 7, 13 \)
Any set multiplying to 1365 with integers > 1 is valid.
Part (b): Square Numbers
Factorise the expression
Factor out \( b \).
We are told:
- \( b \) is a square number (> 1).
- \( a \) is a square number (> 1).
- The result \( b(a-11) \) is a square number.
For \( b \times (a-11) \) to be a square number, since \( b \) is already a square, \( (a-11) \) must also be a square number.
So we need a square number \( a \) such that \( a – 11 \) is a square number.
Let \( a = x^2 \) and \( a – 11 = y^2 \).
\[ x^2 – 11 = y^2 \] \[ x^2 – y^2 = 11 \] \[ (x – y)(x + y) = 11 \]Since 11 is prime, the only factors are 1 and 11.
\[ x – y = 1 \] \[ x + y = 11 \]Adding the equations:
\[ 2x = 12 \implies x = 6 \] \[ a = x^2 = 36 \]Check if \( a – 11 \) is square: \( 36 – 11 = 25 \) (which is \( 5^2 \)). Correct.
\( b \) can be any square number > 1. Let’s pick \( b = 4 \).
Final Answers:
(a) \( p=3, q=5, r=91 \) (or any valid combination)
(b) \( a = 36 \), \( b = 4 \) (or any square > 1 for b)
Question 4 (2 marks)
Solve \( \frac{56}{\sqrt[3]{x}} = 4 \)
Worked Solution
Rearranging the equation
Isolate the \( x \) term
Multiply by \( \sqrt[3]{x} \) and divide by 4.
Solve for \( x \)
Cube both sides to remove the cube root.
Using calculator:
\[ x = 2744 \]Final Answer:
\( x = 2744 \)
Question 5 (3 marks)
\( M \) is the midpoint of \( PQ \).
Work out the value of \( a \).
Worked Solution
Using Midpoint Formula
Midpoint Definition
The coordinates of the midpoint \( M \) are the average of the coordinates of \( P \) and \( Q \).
\[ M_x = \frac{x_1 + x_2}{2}, \quad M_y = \frac{y_1 + y_2}{2} \]Looking at the x-coordinates:
\[ \frac{a + 4}{2} = 3a \]Looking at the y-coordinates (as a check):
\[ \frac{6 + 10}{2} = \frac{16}{2} = 8 \]This matches the y-coordinate of M given as 8.
Solving for a
Solve the linear equation
Multiply by 2:
\[ a + 4 = 6a \]Subtract \( a \) from both sides:
\[ 4 = 5a \] \[ a = \frac{4}{5} = 0.8 \]Final Answer:
\( a = 0.8 \)
Question 6 (3 marks)
A cone has vertex \( V \).
\( C \) is the centre of the base.
The slant height, \( VA \), is 20 cm.
The angle between \( VA \) and \( VC \) is \( 38^\circ \).
Work out the radius of the base.
Worked Solution
Step 1: Identify the Right-Angled Triangle
Visualize triangle VCA
The height \( VC \), radius \( CA \), and slant height \( VA \) form a right-angled triangle with the right angle at \( C \).
- Hypotenuse (\( VA \)) = 20 cm
- Angle at \( V \) (\( \angle AVC \)) = 38°
- We need to find the radius (\( CA \)), which is the side opposite the angle 38°.
Step 2: Choose Trigonometric Ratio
SOH CAH TOA
We have the Hypotenuse and want the Opposite side. We use Sine.
\[ \sin(\theta) = \frac{\text{Opposite}}{\text{Hypotenuse}} \]Step 3: Calculate
Calculator sequence: sin 38 × 20 =
Final Answer:
\( 12.3 \text{ cm} \) (to 3 significant figures)
Question 7 (4 marks)
The equation of the line through \( B \), \( P \) and \( A \) is \( 4x + 5y = 40 \).
\( BP : PA = 2 : 3 \)
Work out the area of triangle \( OBP \).
Worked Solution
Step 1: Find coordinates of A and B
Find intercepts
\( B \) is on the y-axis (where \( x=0 \)).
\( A \) is on the x-axis (where \( y=0 \)).
For B (\( x=0 \)):
\[ 4(0) + 5y = 40 \] \[ 5y = 40 \implies y = 8 \] \[ B = (0, 8) \]For A (\( y=0 \)):
\[ 4x + 5(0) = 40 \] \[ 4x = 40 \implies x = 10 \] \[ A = (10, 0) \]Step 2: Find coordinates of P
Use the Ratio
\( BP : PA = 2 : 3 \). This means \( P \) is \( \frac{2}{2+3} = \frac{2}{5} \) of the way from \( B \) to \( A \).
We need the x-coordinate of P to use as the height of the triangle (if we treat OB as the base).
x-coordinate of P:
\( x_B = 0, \quad x_A = 10 \)
\[ x_P = 0 + \frac{2}{5}(10 – 0) \] \[ x_P = \frac{2}{5} \times 10 = 4 \]y-coordinate of P (optional check):
\[ y_P = 8 + \frac{2}{5}(0 – 8) = 8 – 3.2 = 4.8 \]Step 3: Calculate Area of OBP
Area Formula
Triangle \( OBP \) has base \( OB \) along the y-axis.
Base = 8.
The perpendicular height is the x-coordinate of \( P \).
Height = 4.
Final Answer:
\( 16 \) square units
Question 8 (4 marks)
The perimeter of a triangular flower bed, \( ABC \), is marked out using 27 metres of rope.
Work out the size of angle \( BAC \).
Worked Solution
Step 1: Calculate the length of BC
Use the Perimeter
Perimeter = \( AB + AC + BC \)
Step 2: Use the Cosine Rule
Finding an Angle
We know all three sides (SSS). We use the Cosine Rule to find angle \( A \) (which is \( \angle BAC \)).
\[ a^2 = b^2 + c^2 – 2bc \cos A \]Where \( a = 12 \) (side opposite A), \( b = 8 \), \( c = 7 \).
Step 3: Solve for Angle A
Rearrange to isolate \( \cos A \):
\[ 112 \cos A = 113 – 144 \] \[ 112 \cos A = -31 \] \[ \cos A = \frac{-31}{112} \]Calculate angle \( A \):
\[ A = \cos^{-1}\left(\frac{-31}{112}\right) \] \[ A = 106.065…^\circ \]Final Answer:
\( 106.1^\circ \) (to 1 decimal place)
Question 9 (5 marks)
\( -11 < 5x \leqslant 5 \) and \( 6x + 7 \leqslant 4x + 4 \)
Show that there is exactly one integer that \( x \) can be.
Worked Solution
Step 1: Solve the first inequality
Divide by 5:
\[ -\frac{11}{5} < x \leqslant 1 \] \[ -2.2 < x \leqslant 1 \]Possible integers in this range: \( -2, -1, 0, 1 \)
Step 2: Solve the second inequality
Subtract \( 4x \) from both sides:
\[ 2x + 7 \leqslant 4 \]Subtract 7 from both sides:
\[ 2x \leqslant -3 \]Divide by 2:
\[ x \leqslant -1.5 \]Step 3: Combine results
Find the overlap
We need \( x \) to satisfy both conditions:
- \( x \) must be between -2.2 and 1.
- \( x \) must be less than or equal to -1.5.
Combining the inequalities:
\[ -2.2 < x \leqslant -1.5 \]Check the integers from Step 1 (\( -2, -1, 0, 1 \)):
- \( -2 \): Is \( -2 \leqslant -1.5 \)? Yes.
- \( -1 \): Is \( -1 \leqslant -1.5 \)? No.
- \( 0, 1 \): No.
Therefore, the only integer is \( -2 \).
Final Answer:
The only integer is \( -2 \).
Question 10 (3 marks)
\( ABC \) is an isosceles triangle with \( AB = AC \).
The area of \( ABC \) is \( 57.76 \text{ cm}^2 \).
Work out the length of \( AB \).
Worked Solution
Step 1: Set up the Area Formula
Use Area = \( \frac{1}{2}ab \sin C \)
Here, the angle is at \( A \), and the sides adjacent to it are \( AB \) and \( AC \).
Since the triangle is isosceles, \( AB = AC \). Let \( x = AB = AC \).
Step 2: Solve for x
We know \( \sin(150^\circ) = 0.5 \).
\[ 57.76 = \frac{1}{2} x^2 (0.5) \] \[ 57.76 = 0.25 x^2 \]Divide by 0.25 (or multiply by 4):
\[ x^2 = \frac{57.76}{0.25} \] \[ x^2 = 231.04 \] \[ x = \sqrt{231.04} \] \[ x = 15.2 \]Final Answer:
\( 15.2 \text{ cm} \)
Question 11 (6 marks)
A function \( f(x) \) is defined as:
\( f(x) = 3 – 2x \) \(\quad\) \(-2 \leqslant x < 0\)
\( f(x) = (1 + x)(3 – x) \) \(\quad\) \(0 \leqslant x < 4\)
\( f(x) = 5x – 25 \) \(\quad\) \(4 \leqslant x \leqslant 5\)
(a) Draw the graph of \( y = f(x) \) on the axes below. [4 marks]
(b) State the range of \( f(x) \). [2 marks]
Worked Solution
Part (a): Drawing the Graph
Section 1: Linear \( -2 \leqslant x < 0 \)
\( f(x) = 3 – 2x \)
- At \( x = -2 \): \( y = 3 – 2(-2) = 3 + 4 = 7 \). Plot point \( (-2, 7) \).
- At \( x = 0 \): \( y = 3 – 2(0) = 3 \). Plot point \( (0, 3) \).
Draw a straight line between these points.
Section 2: Quadratic \( 0 \leqslant x < 4 \)
\( f(x) = (1 + x)(3 – x) \)
Calculate key points:
- At \( x = 0 \): \( (1)(3) = 3 \). Connects to previous line.
- At \( x = 1 \): \( (2)(2) = 4 \). This is the vertex (midpoint of roots -1 and 3).
- At \( x = 2 \): \( (3)(1) = 3 \).
- At \( x = 3 \): \( (4)(0) = 0 \). Intersects x-axis.
- At \( x = 4 \): \( (5)(-1) = -5 \). End of this section.
Draw a smooth curve through these points.
Section 3: Linear \( 4 \leqslant x \leqslant 5 \)
\( f(x) = 5x – 25 \)
- At \( x = 4 \): \( 5(4) – 25 = -5 \). Connects to quadratic.
- At \( x = 5 \): \( 5(5) – 25 = 0 \).
Draw a straight line between \( (4, -5) \) and \( (5, 0) \).
Part (b): Determine Range
Identify Max and Min y-values
From the graph or calculations:
- Maximum height is at the vertex of the quadratic: \( y = 4 \).
- Minimum height is at the junction \( x=4 \): \( y = -5 \).
Final Answers:
(a) Graph drawn as above.
(b) \( -5 \leqslant f(x) \leqslant 4 \)
Question 12 (4 marks)
(a) Factorise fully \( 75 – 3x^2 \) [2 marks]
(b) Simplify fully \( (3n + 1)^2 – (3n – 1)^2 \) [2 marks]
Worked Solution
Part (a): Factorisation
Step 1: Factor out common terms
Both 75 and 3 are divisible by 3.
Step 2: Difference of Two Squares
\( 25 – x^2 \) is of the form \( a^2 – b^2 \), which factorises to \( (a-b)(a+b) \).
Part (b): Simplification
Method 1: Difference of Two Squares
Treat the expression as \( A^2 – B^2 \) where \( A = 3n+1 \) and \( B = 3n-1 \).
Method 2: Expand and Simplify
Final Answers:
(a) \( 3(5 – x)(5 + x) \)
(b) \( 12n \)
Question 13 (3 marks)
Simplify fully \( \frac{8a}{3a + 6} \times \frac{5a + 10}{3a^2} \div \frac{4}{15a^3} \)
Worked Solution
Step 1: Factorise and Change Division to Multiplication
Factorise expressions
\( 3a + 6 = 3(a + 2) \)
\( 5a + 10 = 5(a + 2) \)
Flip the fraction
Change \( \div \frac{4}{15a^3} \) to \( \times \frac{15a^3}{4} \).
Step 2: Cancel Common Terms
Cancel \( (a+2) \) from top and bottom:
\[ \frac{8a}{3} \times \frac{5}{3a^2} \times \frac{15a^3}{4} \]Combine numerators and denominators:
\[ \frac{8a \times 5 \times 15a^3}{3 \times 3a^2 \times 4} \] \[ \frac{600a^4}{36a^2} \]Step 3: Simplify Final Fraction
Divide numbers by 12:
\[ \frac{50}{3} \]Simplify powers of \( a \):
\[ \frac{a^4}{a^2} = a^2 \]Result:
\[ \frac{50a^2}{3} \]Final Answer:
\( \frac{50a^2}{3} \)
Question 14 (5 marks)
The line \( y = ax + b \) is perpendicular to the line \( x + 4y = 74 \).
The lines intersect at the point where \( x = 2 \).
Work out the values of \( a \) and \( b \).
Worked Solution
Step 1: Find gradient of given line
Rearrange into \( y = mx + c \)
Gradient \( m_1 = -0.25 \).
Step 2: Find gradient of perpendicular line (value of a)
Negative Reciprocal
For perpendicular lines, \( m_1 \times m_2 = -1 \).
Therefore, \( a = 4 \).
Step 3: Find intersection point
Substitute \( x=2 \) into first equation
Intersection point is \( (2, 18) \).
Step 4: Find value of b
Substitute point into \( y = ax + b \)
Using \( a=4 \), \( x=2 \), \( y=18 \).
Final Answers:
\( a = 4 \)
\( b = 10 \)
Question 15 (3 marks)
Rearrange \( w = \frac{8x – y}{y} \) to make \( y \) the subject.
Worked Solution
Step 1: Remove Fraction
Multiply by y
Step 2: Group y terms
Add y to both sides
We need all terms containing \( y \) on one side.
Step 3: Factorise and Solve
Isolate y
Factor \( y \) out on the left side, then divide.
Final Answer:
\( y = \frac{8x}{w + 1} \)
Question 16 (3 marks)
(a) \( a = 3^{2b} \)
Circle the correct expression for \( \frac{1}{a} \). [1 mark]
\( 3^{2b-1} \) \( 3^{-2b} \) \( -3^{2b} \) \( \left(\frac{1}{3}\right)^{2b} \)
(b) \( y = 5^x \)
Circle the correct expression for \( 25y \). [1 mark]
\( 5^{x+2} \) \( 25^x \) \( 5^{2x} \) \( 125^x \)
(c) \( w = 2^m \)
Circle the correct expression for \( w^3 \). [1 mark]
\( 8^{3m} \) \( 6^m \) \( 2^{m+3} \) \( 2^{3m} \)
Worked Solution
Part (a): Negative Indices
Answer: \( 3^{-2b} \)
Part (b): Multiplying Powers
Write 25 as a power of 5:
\[ 25 = 5^2 \] \[ 5^2 \times 5^x = 5^{2+x} = 5^{x+2} \]Answer: \( 5^{x+2} \)
Part (c): Power of a Power
Multiply the indices:
\[ 2^{m \times 3} = 2^{3m} \]Answer: \( 2^{3m} \)
Final Answers:
(a) \( 3^{-2b} \)
(b) \( 5^{x+2} \)
(c) \( 2^{3m} \)
Question 17 (10 marks)
Here is a sketch of \( y = x^3 – 6x^2 + 7 \).
(a) Use differentiation to work out the coordinates of the stationary point that is a minimum. You must show your working. [4 marks]
(b) The three roots of \( x^3 – 6x^2 + 7 = 0 \) are the \( x \)-coordinates of the points where the graph intersects the \( x \)-axis.
Show that \( x = -1 \) is one root of \( x^3 – 6x^2 + 7 = 0 \). [1 mark]
(c) Hence, work out the other two roots of \( x^3 – 6x^2 + 7 = 0 \).
Give your answers to 2 decimal places.
You must show your working. [5 marks]
Worked Solution
Part (a): Find Minimum Point
Step 1: Differentiate
Find \( \frac{dy}{dx} \) to locate stationary points (where gradient is zero).
Step 2: Solve for Stationary Points
Set \( \frac{dy}{dx} = 0 \).
So \( x = 0 \) or \( x = 4 \).
Step 3: Identify the Minimum
From the sketch, the maximum is on the y-axis (x=0) and the minimum is to the right (x=4).
Alternatively, find the second derivative:
\[ \frac{d^2y}{dx^2} = 6x – 12 \]At \( x = 4 \): \( 6(4) – 12 = 12 \) (Positive \( \to \) Minimum).
Substitute \( x = 4 \) back into original equation for \( y \):
\[ y = (4)^3 – 6(4)^2 + 7 \] \[ y = 64 – 6(16) + 7 \] \[ y = 64 – 96 + 7 = -25 \]Coordinates: \( (4, -25) \)
Part (b): Verify Root
Substitute \( x = -1 \) into \( x^3 – 6x^2 + 7 \):
\[ (-1)^3 – 6(-1)^2 + 7 \] \[ = -1 – 6(1) + 7 \] \[ = -1 – 6 + 7 = 0 \]Since the result is 0, \( x = -1 \) is a root.
Part (c): Find Other Roots
Factorise the Cubic
Since \( x = -1 \) is a root, \( (x + 1) \) is a factor.
We can perform polynomial division or algebraic inspection to find the remaining quadratic factor.
By comparing coefficients:
- \( x \cdot ax^2 = x^3 \implies a = 1 \)
- \( 1 \cdot c = 7 \implies c = 7 \)
- \( x \cdot bx + 1 \cdot x^2 = -6x^2 \) (looking at \( x^2 \) terms)
- \( b + 1 = -6 \implies b = -7 \)
So the quadratic factor is \( x^2 – 7x + 7 \).
Solve the Quadratic
Solve \( x^2 – 7x + 7 = 0 \) using the quadratic formula.
\[ x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} \]Calculating values:
\[ x_1 = \frac{7 + 4.582…}{2} = \frac{11.582…}{2} = 5.79 \] \[ x_2 = \frac{7 – 4.582…}{2} = \frac{2.417…}{2} = 1.21 \]Final Answers:
(a) \( (4, -25) \)
(c) \( x = 1.21 \) and \( x = 5.79 \)
Question 18 (4 marks)
The diagram shows a rectangle with a diagonal drawn.
The given expressions for the measurements are in centimetres.
Work out an expression for the area of the rectangle, in cm\(^2\).
Give your answer in its simplest form, in terms of \( y \).
Worked Solution
Step 1: Relate x and y using Pythagoras
The rectangle sides (\( x \) and \( 2x \)) and the diagonal (\( 4y \)) form a right-angled triangle.
\[ a^2 + b^2 = c^2 \]Step 2: Find Expression for Area
Area of a rectangle = width \(\times\) height.
Step 3: Substitute to get Area in terms of y
From Step 1, we found \( 5x^2 = 16y^2 \).
We can rearrange this to find \( x^2 \):
\[ x^2 = \frac{16}{5}y^2 \]Now substitute this into the Area formula:
\[ \text{Area} = 2 \left( \frac{16}{5}y^2 \right) \] \[ \text{Area} = \frac{32}{5}y^2 \]Or in decimal form:
\[ \text{Area} = 6.4y^2 \]Final Answer:
\( \frac{32}{5}y^2 \) or \( 6.4y^2 \) cm\(^2\)
Question 19 (4 marks)
Here is a sketch of \( y = \sin x \) for \( 0^\circ \leqslant x \leqslant 360^\circ \).
\( \alpha \) is an acute angle measured in degrees.
\( \sin \alpha = k \) where \( k \) is a constant.
Write the answers to each of the following in terms of \( k \), without involving trigonometric functions.
(a) \( \sin(180^\circ – \alpha) \) [1 mark]
(b) \( \sin(360^\circ – \alpha) \) [1 mark]
(c) \( \cos \alpha \) [2 marks]
Worked Solution
Part (a): Symmetry Property
The sine graph is symmetrical about \( 90^\circ \) in the first 180 degrees.
The identity is \( \sin(180^\circ – \alpha) = \sin \alpha \).
Part (b): Periodicity/Quadrants
\( 360^\circ – \alpha \) is in the 4th quadrant, where sine is negative.
Alternatively, looking at the graph, moving back \( \alpha \) from 360 gives the negative of the height at \( \alpha \).
Part (c): Pythagorean Identity
Identity: \( \sin^2 \alpha + \cos^2 \alpha \equiv 1 \)
Since \( \alpha \) is acute, \( \cos \alpha \) is positive.
Final Answers:
(a) \( k \)
(b) \( -k \)
(c) \( \sqrt{1 – k^2} \)
Question 20 (5 marks)
Two circles overlap.
\( A, B \) and \( E \) lie on the circle, centre \( O \).
\( B, C, D \) and \( E \) lie on the other circle.
\( AOBC \) and \( AED \) are straight lines.
\( CD = CE \)
Angle \( BAE = x \)
(a) Give a reason why angle \( BEA = 90^\circ \). [1 mark]
(b) Prove that angle \( DCE = 2x \). [4 marks]
Worked Solution
Part (a): Circle Theorem
Since \( A, O, B \) lie on a straight line and pass through the centre \( O \), \( AB \) is a diameter.
Reason: Angle in a semicircle is a right angle.
Part (b): Proof
Step 1: Find angles in Triangle ABE
In \( \triangle ABE \), angle \( E = 90^\circ \) and \( A = x \).
Step 2: Use Cyclic Quadrilateral BCDE properties
Points \( B, C, D, E \) are on the second circle.
Find angle \( CBE \). Since \( ABC \) is a straight line:
Opposite angles in a cyclic quadrilateral sum to \( 180^\circ \).
Step 3: Use Isosceles Triangle CDE
We are given \( CD = CE \). This means \( \triangle CDE \) is isosceles.
Base angles are equal: \( \angle CED = \angle CDE \).
Sum of angles in \( \triangle CDE \) is 180:
\[ \angle DCE + \angle CDE + \angle CED = 180 \] \[ \angle DCE + (90 – x) + (90 – x) = 180 \] \[ \angle DCE + 180 – 2x = 180 \] \[ \angle DCE = 2x \]Conclusion:
Proven that \( \angle DCE = 2x \).
Question 21 (9 marks)
Here is a sketch of \( y = (x + 2)(4 – x) \)
The graph intersects the axes at \( A(-2, 0) \), \( B(4, 0) \) and \( C \).
(a) Work out the coordinates of \( C \). [1 mark]
(b) Work out the gradient function of the curve. [3 marks]
(c) The normal to the curve at \( C \) intersects the \( x \)-axis at \( D \).
Show that length \( BD = 2 \times \) length \( AB \). [5 marks]
Worked Solution
Part (a): Coordinates of C
y-intercept
\( C \) is on the y-axis, so \( x = 0 \).
Coordinates: \( (0, 8) \)
Part (b): Gradient Function
Expand then Differentiate
First, multiply out the brackets to get separate terms.
Now differentiate:
\[ \frac{dy}{dx} = -2x + 2 \]Part (c): Equation of the Normal
Step 1: Gradient at C
At \( C \), \( x = 0 \).
The normal is perpendicular, so:
\[ m_{\text{normal}} = -\frac{1}{2} = -0.5 \]Step 2: Equation of Normal Line
Passes through \( C(0, 8) \) with gradient \( -0.5 \).
Step 3: Find D (x-intercept)
At \( D \), \( y = 0 \).
So \( D \) is at \( (16, 0) \).
Step 4: Compare Lengths
\( A(-2, 0) \), \( B(4, 0) \), \( D(16, 0) \).
Length \( AB = 4 – (-2) = 6 \)
Length \( BD = 16 – 4 = 12 \)
\( 2 \times \text{length } AB = 2 \times 6 = 12 \)
Therefore, \( BD = 2 \times AB \).
Conclusion:
Shown that \( BD = 12 \) and \( 2 \times AB = 12 \).
Question 22 (5 marks)
The equation of a circle is \( (x – 2)^2 + (y – 1)^2 = 16 \)
The equation of a line is \( y = 2x + 1 \)
The circle and the line intersect at two points.
Work out the coordinates of the two points.
You must show your working. Do not use trial and improvement.
Worked Solution
Step 1: Substitution
Substitute \( y = 2x + 1 \) into the circle equation.
Step 2: Expand and Solve for x
Collect like terms:
\[ 5x^2 – 4x + 4 = 16 \] \[ 5x^2 – 4x – 12 = 0 \]Factorise (or use quadratic formula):
We need numbers that multiply to \( 5 \times -12 = -60 \) and add to -4. (-10 and 6).
\[ 5x^2 – 10x + 6x – 12 = 0 \] \[ 5x(x – 2) + 6(x – 2) = 0 \] \[ (5x + 6)(x – 2) = 0 \]Solutions for \( x \):
\[ x = 2 \quad \text{or} \quad x = -\frac{6}{5} = -1.2 \]Step 3: Find y coordinates
Substitute \( x \) values back into linear equation \( y = 2x + 1 \).
When \( x = 2 \):
\[ y = 2(2) + 1 = 5 \]Point 1: \( (2, 5) \)
When \( x = -1.2 \):
\[ y = 2(-1.2) + 1 \] \[ y = -2.4 + 1 = -1.4 \]Point 2: \( (-1.2, -1.4) \)
Final Answer:
\( (2, 5) \) and \( (-1.2, -1.4) \)
Question 23 (3 marks)
In this question, \( \tan x \neq 0 \) and \( \sin x \neq 0 \).
Show that \( \frac{1}{\tan^2 x} – \frac{1}{\sin^2 x} \) is a constant.
Worked Solution
Step 1: Convert tan to sin and cos
We know \( \tan x = \frac{\sin x}{\cos x} \), so \( \frac{1}{\tan x} = \frac{\cos x}{\sin x} \).
Therefore \( \frac{1}{\tan^2 x} = \frac{\cos^2 x}{\sin^2 x} \).
Step 2: Combine Fractions
Step 3: Use Identity
We know \( \sin^2 x + \cos^2 x = 1 \).
So \( \cos^2 x – 1 = -\sin^2 x \).
Conclusion:
The expression simplifies to \( -1 \), which is a constant.
Question 24 (5 marks)
Write \( 12x^2 – 60x + 5 \) in the form \( a(bx + c)^2 + d \) where \( a, b, c \) and \( d \) are integers.
Worked Solution
Step 1: Understand the Target Form
The form \( a(bx + c)^2 + d \) implies we need to complete the square, but the coefficient inside the bracket is \( b \), not necessarily 1. Also \( a \) is outside.
Since we want integers, we should look at how \( 12x^2 \) can be formed from \( a(bx)^2 \). \( a \cdot b^2 = 12 \).
Possible integer combinations for \( ab^2 = 12 \):
- \( b=1 \implies a=12 \)
- \( b=2 \implies a=3 \) (since \( 3 \times 4 = 12 \))
Let’s try to factor out 3 first, which leaves \( 4x^2 \), a perfect square.
Step 2: Complete the Square for \( 4x^2 – 20x \)
We want \( (2x + c)^2 = 4x^2 + 4cx + c^2 \).
Matching the middle term: \( 4c = -20 \implies c = -5 \).
Therefore:
\[ 4x^2 – 20x = (2x – 5)^2 – 25 \]Step 3: Substitute Back
Check if \( a, b, c, d \) are integers.
\( a = 3, b = 2, c = -5, d = -70 \). Yes.
Final Answer:
\( 3(2x – 5)^2 – 70 \)