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Level 2 Certificate Further Mathematics Paper 2 (Calculator) June 2021
Legend
- (M1) Method mark – for a correct method or step
- (A1) Accuracy mark – for a correct value or answer
- (B1) Independent mark – correct answer regardless of method
- (ft) Follow through – correct calculation using previous incorrect value
Note: This paper allows the use of a calculator.
Table of Contents
- Question 1 (Algebraic Expansion)
- Question 2 (Percentages & Algebra)
- Question 3 (Area & Algebra)
- Question 4 (Circle Area)
- Question 5 (Coordinate Geometry)
- Question 6 (Rearranging Formulas)
- Question 7 (Indices)
- Question 8 (Functions & Graphs)
- Question 9 (Algebraic Fractions)
- Question 10 (Volume of Sphere)
- Question 11 (Expanding Brackets)
- Question 12 (Linear Graphs)
- Question 13 (Differentiation & Normals)
- Question 14 (Matrix Transformations)
- Question 15 (Circle Geometry Proof)
- Question 16 (Binomial Expansion)
- Question 17 (3D Geometry & Trigonometry)
- Question 18 (Calculus Optimization)
- Question 19 (Composite Functions)
- Question 20 (Trigonometric Identities)
- Question 21 (Algebraic Identities)
Question 1 (3 marks)
Expand and simplify \( 5(2x – 1) + 4(11 – x) \)
Give your answer in the form \( a(bx + c) \) where \( a, b \) and \( c \) are integers greater than 1.
Worked Solution
Step 1: Expand the brackets
Why: We need to multiply the terms inside the parentheses by the numbers outside to remove the brackets.
✓ (M1) Expanding at least one bracket correctly
Step 2: Collect like terms
Why: Combine the \( x \) terms and the constant terms to simplify the expression.
✓ (M1) Correct simplified expression
Step 3: Factorise the result
Why: The question asks for the form \( a(bx + c) \) with integers \( > 1 \). We look for a common factor of 6 and 39.
Both 6 and 39 are divisible by 3.
\[ 3(2x + 13) \]Check: \( a=3, b=2, c=13 \). All are integers greater than 1.
Final Answer:
\[ 3(2x + 13) \]
✓ (A1)
Question 2 (4 marks total)
Part (a) (2 marks)
\( 5m \) is decreased by 40%.
The answer is \( (m + 1) \).
Work out the value of \( m \).
Worked Solution – Part (a)
Step 1: Set up the equation
Why: Decreasing by 40% is the same as finding 60% of the value (100% – 40%).
✓ (M1) Correct method to decrease by 40%
Step 2: Solve for m
Subtract \( m \) from both sides:
\[ 2m = 1 \] \[ m = 0.5 \]Answer: \( m = 0.5 \)
✓ (A1)
Part (b) (2 marks)
Solve \( \sqrt[3]{2w – 10} = 18 \)
Worked Solution – Part (b)
Step 1: Eliminate the cube root
Why: The inverse operation of a cube root is cubing.
✓ (M1) Cubing 18 correctly
Step 2: Solve the linear equation
Answer: \( w = 2921 \)
✓ (A1)
Question 3 (3 marks)
The rectangle and triangle shown have equal areas.
Work out the value of \( \frac{d}{e} \).
Give your answer in its simplest form.
Worked Solution
Step 1: Calculate the area of the rectangle
Step 2: Calculate the area of the triangle
✓ (M1) Expressions for both areas
Step 3: Equate areas and simplify
Why: The problem states the areas are equal.
Divide both sides by \( 12de \) (assuming \( d, e \neq 0 \)):
\[ \frac{24d^2e}{12de} = \frac{36de^2}{12de} \] \[ 2d = 3e \]✓ (M1) Simplifying the equation
Step 4: Rearrange to find d/e
Divide both sides by \( e \):
\[ \frac{2d}{e} = 3 \]Divide by 2:
\[ \frac{d}{e} = \frac{3}{2} \]Answer: \( 1.5 \) or \( \frac{3}{2} \)
✓ (A1)
Question 4 (3 marks)
The equations of the two circles shown are
\( x^2 + y^2 = 100 \) and \( x^2 + y^2 = 36 \)
Work out the shaded area.
Give your answer as an integer multiple of \( \pi \).
Worked Solution
Step 1: Identify the radii
Why: The equation of a circle is \( x^2 + y^2 = r^2 \).
Outer circle: \( r^2 = 100 \Rightarrow r = 10 \)
Inner circle: \( r^2 = 36 \Rightarrow r = 6 \)
Step 2: Calculate the area of the full annulus
Why: The area between two circles is Area(outer) – Area(inner).
✓ (M1) Calculating full area difference
Step 3: Calculate shaded fraction
Why: The diagram shows the shaded region is in one quadrant (the top-right), which is \( \frac{1}{4} \) of the total circle.
✓ (M1) Using the quarter fraction
Answer: \( 16\pi \) units²
✓ (A1)
Question 5 (3 marks)
\( SQR \) is a right-angled triangle.
\( P \) is a point on \( SQ \).
Angle \( SPR = 45^\circ \).
\( M \) is the midpoint of \( QR \).
\( k \) is a constant.
Work out the coordinates of \( M \).
Worked Solution
Step 1: Determine the coordinates of S
Why: \( SQR \) is a right-angled triangle. Looking at the diagram, \( SQ \) is vertical (same x-coordinate) and \( SR \) is horizontal (same y-coordinate).
\( Q \) is at \( (1, 1) \) and \( P \) is at \( (1, 6) \). Since \( S \) is on the line extending \( QP \) upwards, \( S \) has x-coordinate 1.
\( R \) is at \( (k, 15) \). Since \( SR \) is horizontal, \( S \) has y-coordinate 15.
So, \( S = (1, 15) \).
Step 2: Use the angle SPR to find k
Why: Consider the triangle \( PSR \). \( S \) is a right angle (intersection of vertical and horizontal lines).
Length \( PS = \text{y-coord of S} – \text{y-coord of P} = 15 – 6 = 9 \).
Length \( SR = \text{x-coord of R} – \text{x-coord of S} = k – 1 \).
In triangle \( PSR \), angle \( SPR = 45^\circ \).
\[ \tan(45^\circ) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{SR}{PS} \] \[ 1 = \frac{k – 1}{9} \] \[ 9 = k – 1 \Rightarrow k = 10 \]So \( R \) is \( (10, 15) \).
✓ (M1) Finding coordinates of R
Step 3: Calculate midpoint M
Why: \( M \) is the midpoint of \( QR \).
\( Q = (1, 1) \)
\( R = (10, 15) \)
\[ M = \left( \frac{1 + 10}{2}, \frac{1 + 15}{2} \right) \] \[ M = \left( \frac{11}{2}, \frac{16}{2} \right) \] \[ M = (5.5, 8) \]✓ (M1) Midpoint formula
Answer: \( (5.5, 8) \)
✓ (A1)
Question 6 (3 marks)
Rearrange \( y = \sqrt{\frac{x + 2w}{3}} \) to make \( w \) the subject.
Worked Solution
Step 1: Square both sides
Why: We need to remove the square root to access the term with \( w \).
✓ (M1)
Step 2: Multiply by 3
✓ (M1)
Step 3: Isolate w
Subtract \( x \) from both sides:
\[ 3y^2 – x = 2w \]Divide by 2:
\[ w = \frac{3y^2 – x}{2} \]Answer: \( w = \frac{3y^2 – x}{2} \)
✓ (A1)
Question 7 (4 marks total)
Part (a) (2 marks)
\( a \) is a value greater than 1.
Work out the value of \( m \) for which \( (a^m)^4 = (a^5)^{2m} \)
Worked Solution – Part (a)
Step 1: Simplify the indices
Why: Use the power law \( (x^a)^b = x^{ab} \).
✓ (M1)
Step 2: Equate exponents
Answer: \( m = 0 \)
✓ (A1)
Part (b) (2 marks)
\( w^3 x^2 y^5 = w^{13} x^7 \)
Write \( y \) in terms of \( w \) and \( x \).
Give your answer in its simplest form.
Worked Solution – Part (b)
Step 1: Isolate y⁵
✓ (M1) Correct division of indices
Step 2: Find the 5th root
Answer: \( y = w^2 x \)
✓ (A1)
Question 8 (4 marks)
A function \( f \) is given by
\( f(x) = 4x \) for \( x < 0 \)
\( f(x) = x^2 – 8x \) for \( 0 \leqslant x \leqslant 8 \)
\( f(x) = 16 – 2x \) for \( x > 8 \)
A sketch of \( y = f(x) \) is shown.
Work out all the values of \( x \) for which \( f(x) = -12 \).
Worked Solution
Step 1: Check the first part of the function (\( x < 0 \))
Is \( x < 0 \)? Yes. So \( x = -3 \) is a solution.
✓ (B1) First value
Step 2: Check the middle part (\( 0 \leqslant x \leqslant 8 \))
Factorise:
\[ (x – 2)(x – 6) = 0 \] \[ x = 2 \text{ or } x = 6 \]Are these in range \( [0, 8] \)? Yes. So \( x = 2 \) and \( x = 6 \) are solutions.
✓ (B2) Correct quadratic solutions
Step 3: Check the third part (\( x > 8 \))
Is \( x > 8 \)? Yes. So \( x = 14 \) is a solution.
✓ (B1) Fourth value
Answer: \( -3, 2, 6, 14 \)
Question 9 (4 marks total)
Part (a) (1 mark)
Circle the expression that is equivalent to \( \frac{1}{a} + \frac{1}{b} \)
\( \frac{2}{a+b} \quad \frac{ab}{b+a} \quad \frac{2}{ab} \quad \frac{b+a}{ab} \)
Worked Solution – Part (a)
Find a common denominator (\( ab \)):
\[ \frac{1}{a} + \frac{1}{b} = \frac{b}{ab} + \frac{a}{ab} = \frac{b+a}{ab} \]Answer: \( \frac{b+a}{ab} \)
✓ (B1)
Part (b) (3 marks)
Simplify fully \( \frac{6c^4 – c^3}{36c^2 – 1} \)
Worked Solution – Part (b)
Step 1: Factorise the numerator
Take out common factor \( c^3 \):
\[ 6c^4 – c^3 = c^3(6c – 1) \]Step 2: Factorise the denominator
Why: This is a difference of two squares \( (A^2 – B^2) = (A-B)(A+B) \).
✓ (B1) Correct factorisation
Step 3: Simplify
Cancel the \( (6c – 1) \) terms:
\[ \frac{c^3}{6c + 1} \]Answer: \( \frac{c^3}{6c + 1} \)
✓ (B2)
Question 10 (3 marks)
The radius of a sphere, in cm, is \( \frac{3k}{2} \)
The volume of the sphere, in cm³, is \( 972\pi \)
Work out the value of \( k \).
(Volume of a sphere = \( \frac{4}{3}\pi r^3 \))
Worked Solution
Step 1: Substitute values into the formula
✓ (M1) Setting up equation
Step 2: Simplify and solve
Divide by \( \pi \):
\[ 972 = \frac{4}{3} \left( \frac{27k^3}{8} \right) \] \[ 972 = \frac{1}{3} \left( \frac{27k^3}{2} \right) \] \[ 972 = \frac{9k^3}{2} \] \[ 972 = 4.5 k^3 \]✓ (M1) Rearranging for k³
Step 3: Final calculation
Answer: \( k = 6 \)
✓ (A1)
Question 11 (3 marks)
Expand and simplify fully \( (5x + 3y^2)(4x – y^2) \)
Worked Solution
Step 1: Multiply terms (FOIL method)
✓ (M1) 3 out of 4 terms correct
Step 2: Collect like terms
Why: The terms \( -5xy^2 \) and \( 12xy^2 \) are like terms.
✓ (A1) Correct simplified expression
Answer: \( 20x^2 + 7xy^2 – 3y^4 \)
✓ (A1)
Question 12 (5 marks)
\( A \) and \( B \) are points on the line \( y = 3x + 2 \)
\( B \), \( C \) and \( D(5, 0) \) are points on the line \( L \).
\( OA : AC = 1 : 4 \)
Work out the x-coordinate of \( B \).
Worked Solution
Step 1: Find coordinates of A and C
Why: \( A \) is on the line \( y=3x+2 \) and on the y-axis (from diagram). \( C \) is on the y-axis.
For \( A \), let \( x=0 \) in \( y=3x+2 \): \( y=2 \). So \( A \) is \( (0, 2) \).
\( OA \) represents the distance from origin to \( A \), which is 2 units.
Ratio \( OA : AC = 1 : 4 \).
\[ \frac{2}{AC} = \frac{1}{4} \Rightarrow AC = 8 \]\( C \) is 8 units above \( A \). \( y_C = 2 + 8 = 10 \).
So \( C \) is \( (0, 10) \).
✓ (B1) Coordinates of C
Step 2: Find the equation of Line L
Why: Line L passes through \( C(0, 10) \) and \( D(5, 0) \).
Gradient \( m = \frac{y_2 – y_1}{x_2 – x_1} = \frac{0 – 10}{5 – 0} = \frac{-10}{5} = -2 \).
Equation: \( y = -2x + c \). Since y-intercept is 10:
\[ y = -2x + 10 \]✓ (M1) Equation of line L
Step 3: Find intersection point B
Why: \( B \) is where the two lines cross. Set their equations equal.
✓ (A1) Correct x-coordinate
Answer: \( 1.6 \)
Question 13 (4 marks)
\( P \) is the point on the curve \( y = ax^3 + 10x^2 \) where \( x = 2 \).
The gradient of the normal to the curve at \( P \) is \( -\frac{1}{4} \).
Work out the value of \( a \).
Worked Solution
Step 1: Find gradient of the tangent
Why: The normal and tangent are perpendicular. \( m_{\text{normal}} \times m_{\text{tangent}} = -1 \).
✓ (M1)
Step 2: Differentiate the curve equation
✓ (M1) Differentiation
Step 3: Equate gradients at x = 2
Substitute \( x = 2 \) into derivative:
\[ 3a(2)^2 + 20(2) = 12a + 40 \]Set equal to tangent gradient 4:
\[ 12a + 40 = 4 \] \[ 12a = -36 \] \[ a = -3 \]Answer: \( a = -3 \)
✓ (A1)
Question 14 (3 marks total)
Part (a) (1 mark)
\( \mathbf{A} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \)
Describe geometrically the single transformation represented by \( \mathbf{A} \).
Worked Solution – Part (a)
Applying to unit vectors:
\( \begin{pmatrix} 1 \\ 0 \end{pmatrix} \to \begin{pmatrix} 1 \\ 0 \end{pmatrix} \) (x stays same)
\( \begin{pmatrix} 0 \\ 1 \end{pmatrix} \to \begin{pmatrix} 0 \\ -1 \end{pmatrix} \) (y becomes negative)
This is a Reflection in the x-axis.
✓ (B1)
Part (b) (2 marks)
\( \mathbf{B} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \)
Describe geometrically the single transformation represented by \( \mathbf{B}^2 \).
Worked Solution – Part (b)
Step 1: Calculate B²
✓ (M1) Matrix multiplication
Step 2: Identify transformation
\( x \to -x \) and \( y \to -y \).
This is a Rotation of 180° about the origin (or Enlargement scale factor -1).
Answer: Rotation 180° about the origin
✓ (B1)
Question 15 (5 marks)
\( A \), \( B \) and \( C \) are points on a circle, centre \( O \).
\( ACD \) is a straight line.
Angle \( BCD = w \).
Prove that \( w = x + 90^\circ \).
Worked Solution
Step 1: Analyze triangle OAB
Why: \( OA \) and \( OB \) are radii, so triangle \( OAB \) is isosceles.
\( \angle OBA = \angle OAB = x \) (base angles of isosceles triangle).
Sum of angles in triangle \( OAB \):
\[ \angle AOB = 180 – (x + x) = 180 – 2x \]✓ (M1) Finding angle AOB
Step 2: Find angle ACB
Why: The angle at the centre is double the angle at the circumference.
✓ (M1) Circle theorem application
Step 3: Use the straight line property
Why: Angles on a straight line add to 180°.
✓ (M1) Straight line equation
Step 4: Rearrange to prove
QED
✓ (A1) Complete proof
Proven: \( w = x + 90^\circ \)
Question 16 (3 marks)
The coefficient of \( x^4 \) in the expansion of \( (a + 2x)^6 \) is 1500.
Work out the two possible values of \( a \).
Worked Solution
Step 1: Find the general term for x⁴
Why: In binomial expansion \( (A+B)^n \), the term is \( \binom{n}{r} A^{n-r} B^r \). Here \( n=6 \), we want \( x^4 \) so we look at \( (2x)^4 \).
Term involves \( \binom{6}{4} a^2 (2x)^4 \).
\[ \binom{6}{4} = 15 \] \[ (2x)^4 = 16x^4 \]Term is: \( 15 \times a^2 \times 16x^4 = 240a^2 x^4 \)
✓ (M1) Establishing coefficient expression
Step 2: Solve for a
✓ (M1) Equation for a²
Step 3: Final values
Answer: \( 2.5 \) and \( -2.5 \)
✓ (A1)
Question 17 (5 marks)
\( ABCDEFGH \) is a cube with side length 32 cm.
\( M \) and \( N \) are points on \( DH \) and \( CG \) respectively.
The angle marked at \( M \) (between \( MN \) and \( MG \)) is \( 28^\circ \).
Work out the size of the angle that the line \( BM \) makes with the plane \( ABCD \).
Worked Solution
Step 1: Calculate the height of M from the top face
Why: In triangle \( MNG \), \( N \) is at the same height as \( M \) (horizontal line). \( G \) is the corner above \( N \).
\( MN = 32 \) (width of cube).
\( \tan(28^\circ) = \frac{\text{Opp}}{\text{Adj}} = \frac{GN}{MN} \).
\[ GN = 32 \times \tan(28^\circ) \]Step 2: Calculate height MD
Why: \( M \) is on \( DH \). \( H \) is at top, \( D \) is at bottom. \( D, C, G, H \) is a rectangle. \( NC = MD \).
Height of cube \( CG = 32 \).
\( NC = CG – GN = 32 – 32 \tan(28^\circ) \).
\( MD = NC = 32(1 – \tan 28^\circ) \).
Step 3: Calculate the diagonal BD
Why: We need the base of the right-angled triangle \( MDB \) to find the angle at \( B \).
Base \( ABCD \) is a square of side 32.
\[ BD = \sqrt{32^2 + 32^2} = 32\sqrt{2} \]Step 4: Calculate the required angle
Let the angle be \( \theta \). In right-angled triangle \( MDB \):
\[ \tan \theta = \frac{MD}{BD} \] \[ \tan \theta = \frac{32(1 – \tan 28^\circ)}{32\sqrt{2}} \] \[ \tan \theta = \frac{1 – \tan 28^\circ}{\sqrt{2}} \] \[ \tan \theta \approx \frac{1 – 0.5317}{1.414} \] \[ \tan \theta \approx \frac{0.4683}{1.414} \approx 0.331 \] \[ \theta = \tan^{-1}(0.331…) \] \[ \theta \approx 18.3^\circ \]Answer: \( 18.3^\circ \)
✓ (A1) [Accept 18.3 to 18.4]
Question 18 (5 marks)
\( y = 12x + \frac{3}{x} \)
Show that \( y \) has a minimum value when \( x = 0.5 \).
Worked Solution
Step 1: Differentiate y
Why: To find stationary points (max/min), we need the derivative \( \frac{dy}{dx} \).
Rewrite \( \frac{3}{x} \) as \( 3x^{-1} \).
\[ y = 12x + 3x^{-1} \] \[ \frac{dy}{dx} = 12 – 3x^{-2} = 12 – \frac{3}{x^2} \]✓ (M1) Correct differentiation
Step 2: Show stationary point at x = 0.5
Set \( \frac{dy}{dx} = 0 \) and substitute \( x = 0.5 \):
\[ 12 – \frac{3}{(0.5)^2} = 12 – \frac{3}{0.25} \] \[ = 12 – 12 = 0 \]So \( x = 0.5 \) is a stationary point.
✓ (M1) Verifying derivative is zero
Step 3: Determine nature of the point
Why: Use the second derivative \( \frac{d^2y}{dx^2} \). Positive = Minimum.
At \( x = 0.5 \):
\[ \frac{6}{(0.5)^3} = \frac{6}{0.125} = 48 \]Since \( 48 > 0 \), it is a minimum.
✓ (A1) Proof complete
Question 19 (3 marks total)
Part (a) (1 mark)
\( f(x) = (x + 2)^3 \)
\( g \) is a function such that \( gf(x) = (x + 2)^{12} \)
Work out an expression for \( g(x) \).
Worked Solution – Part (a)
We know \( f(x) = (x+2)^3 \).
We want \( g(f(x)) = ((x+2)^3)^4 = (x+2)^{12} \).
So \( g \) raises the input to the power of 4.
\[ g(x) = x^4 \]Answer: \( g(x) = x^4 \)
Part (b) (2 marks)
\( h(x) = x^2 + 5 \)
\( k \) is a function such that \( hk(x) = 4x^2 + 5 \)
Work out an expression for \( kh(x) \).
Worked Solution – Part (b)
Step 1: Find k(x)
\( h(k(x)) = (k(x))^2 + 5 \).
We are given \( hk(x) = 4x^2 + 5 \).
\[ (k(x))^2 + 5 = 4x^2 + 5 \] \[ (k(x))^2 = 4x^2 \] \[ k(x) = 2x \] (assuming positive for standard function notation)✓ (M1)
Step 2: Find kh(x)
Answer: \( 2x^2 + 10 \)
✓ (A1)
Question 20 (4 marks)
Show that \( \frac{2\sin x + \cos x}{\tan x} – \frac{1}{\sin x} \) can be written in the form \( a\cos x + b\sin x \)
where \( a \) and \( b \) are integers.
Worked Solution
Step 1: Substitute tan x
Use \( \tan x = \frac{\sin x}{\cos x} \):
\[ \frac{2\sin x + \cos x}{\frac{\sin x}{\cos x}} – \frac{1}{\sin x} \] \[ = \frac{\cos x(2\sin x + \cos x)}{\sin x} – \frac{1}{\sin x} \]✓ (M1)
Step 2: Combine fractions
✓ (M1) Common denominator
Step 3: Use Identity
Use \( \sin^2 x + \cos^2 x = 1 \Rightarrow \cos^2 x – 1 = -\sin^2 x \):
\[ \frac{2\sin x \cos x – \sin^2 x}{\sin x} \]✓ (M1) Pythagorean identity
Step 4: Simplify
Divide each term by \( \sin x \):
\[ \frac{2\sin x \cos x}{\sin x} – \frac{\sin^2 x}{\sin x} \] \[ = 2\cos x – \sin x \]Answer: \( 2\cos x – 1\sin x \) (so \( a=2, b=-1 \))
✓ (A1)
Question 21 (6 marks)
\( 3x^2 + 2bx + 8a \) can be written in the form \( 3(x + a)^2 + b + 2 \)
Work out the two possible pairs of values of \( a \) and \( b \).
Worked Solution
Step 1: Expand the completed square form
✓ (M1) Expansion
Step 2: Equate coefficients
Why: Since the expressions are identical, coefficients of \( x \) must match and constants must match.
Coefficient of \( x \): \( 2b = 6a \Rightarrow b = 3a \)
Constant term: \( 8a = 3a^2 + b + 2 \)
✓ (M1) Setting up simultaneous equations
Step 3: Solve the system
Substitute \( b = 3a \) into the second equation:
\[ 8a = 3a^2 + (3a) + 2 \] \[ 3a^2 – 5a + 2 = 0 \]✓ (M1) Quadratic in ‘a’
Factorise:
\[ (3a – 2)(a – 1) = 0 \] \[ a = \frac{2}{3} \text{ or } a = 1 \]Step 4: Find corresponding b values
If \( a = 1 \), \( b = 3(1) = 3 \).
If \( a = \frac{2}{3} \), \( b = 3(\frac{2}{3}) = 2 \).
Answers:
\( a = 1, b = 3 \)
\( a = \frac{2}{3}, b = 2 \)
✓ (A1) Both pairs correct