AQA Further Paper 2 2024 June — Question 12 5 marks

Exam BoardAQA
ModuleFurther Paper 2 (Further Paper 2)
Year2024
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear transformations
TypeFind pre-image from given image
DifficultyChallenging +1.2 This question requires finding the matrix for reflection in a non-standard line (y = x√3), then solving MN·P = Q by finding the inverse of a 2×2 matrix product. While it involves multiple steps (deriving reflection matrix, matrix multiplication, matrix inversion, solving system), each component uses standard Further Maths techniques with no novel insight required. The 5 marks and calculator-permitted nature (3 d.p.) indicate computational rather than conceptual challenge.
Spec4.03d Linear transformations 2D: reflection, rotation, enlargement, shear4.03e Successive transformations: matrix products

The transformation \(S\) is represented by the matrix \(\mathbf{M} = \begin{bmatrix} 1 & -6 \\ 2 & 7 \end{bmatrix}\) The transformation \(T\) is a reflection in the line \(y = x\sqrt{3}\) and is represented by the matrix \(\mathbf{N}\) The point \(P(x, y)\) is transformed first by \(S\), then by \(T\) The result of these transformations is the point \(Q(3, 8)\) Find the coordinates of \(P\) Give your answers to three decimal places. [5 marks]

Question 12:
AnswerMarks
12 1 3
 – 
Obtains
 2 2 
 
3 1
 
AnswerMarks Guidance
 2 2 2.2a B1
T is a reflection in the line y = x tan
3
 
1 3
 – 
So N =  2 2 
 
3 1
 
 2 2 
x 1 −6x  x−6y 
M  =    =  
y 2 7 y 2x+7y
 1 3 
− ( x−6y )+ ( 2x+7y ) 
3  x 2 2
=NM = 
 
8  y  3 1 
 ( x−6y )+ ( 2x+7y ) 
 2 2 
  1  7 3 
 3− x+ 3+  y
 
3  2  2  
=
   
8   3  7  
   +1  x+  −3 3y 
 2  2  
1.23205x+9.06218y =3
1.86603x−1.69615y =8
x=4.083,y =−0.224
P ( 4.083,−0.224 )
(3 d.p.)
Operates first M, then
their N, on column vector
x
 
y
or
Operates their N or their
3
N-1 on  
8
or
Calculates matrix
AnswerMarks Guidance
product NM3.1a M1
Obtains a correct system
of simultaneous
equations in x and y
or
Obtains correct value of
M –1 (Accept decimal
approximation.)
or
Obtains correct value of
NM
AnswerMarks Guidance
PI1.1b A1
Solves their
simultaneous equations
in x and y from
x 3
NM =  or
y 8
x 3
MN = 
y 8
or
Operates M –1 on NQ
or
Uses inverse of NM,
where
0.08927 0.47696 
(NM)−1=
0.09821 −0.06484
to obtain coordinates of
P
Condone applying
transformations in the
AnswerMarks Guidance
wrong order.1.1a M1
Obtains AWRT 4.083
and AWRT -0.224.
Condone column vector
form if x and y seen.
Accept exact values.
27+74 3
x= ,
38
14−13 3
y =
AnswerMarks Guidance
381.1b A1
Question total5
QMarking instructions AO
Question 12:
12 |  1 3
 – 
Obtains
 2 2 
 
3 1
 
 2 2  | 2.2a | B1 | π
T is a reflection in the line y = x tan
3
 
1 3
 – 
So N =  2 2 
 
3 1
 
 2 2 
x 1 −6x  x−6y 
M  =    =  
y 2 7 y 2x+7y
 1 3 
− ( x−6y )+ ( 2x+7y ) 
3  x 2 2
=NM = 
 
8  y  3 1 
 ( x−6y )+ ( 2x+7y ) 
 2 2 
  1  7 3 
 3− x+ 3+  y
 
3  2  2  
=
   
8   3  7  
   +1  x+  −3 3y 
 2  2  
1.23205x+9.06218y =3
1.86603x−1.69615y =8
x=4.083,y =−0.224
P ( 4.083,−0.224 )
(3 d.p.)
Operates first M, then
their N, on column vector
x
 
y
or
Operates their N or their
3
N-1 on  
8
or
Calculates matrix
product NM | 3.1a | M1
Obtains a correct system
of simultaneous
equations in x and y
or
Obtains correct value of
M –1 (Accept decimal
approximation.)
or
Obtains correct value of
NM
PI | 1.1b | A1
Solves their
simultaneous equations
in x and y from
x 3
NM =  or
y 8
x 3
MN = 
y 8
or
Operates M –1 on NQ
or
Uses inverse of NM,
where
0.08927 0.47696 
(NM)−1=

0.09821 −0.06484
to obtain coordinates of
P
Condone applying
transformations in the
wrong order. | 1.1a | M1
Obtains AWRT 4.083
and AWRT -0.224.
Condone column vector
form if x and y seen.
Accept exact values.
27+74 3
x= ,
38
14−13 3
y =
38 | 1.1b | A1
Question total | 5
Q | Marking instructions | AO | Marks | Typical solution
The transformation $S$ is represented by the matrix $\mathbf{M} = \begin{bmatrix} 1 & -6 \\ 2 & 7 \end{bmatrix}$

The transformation $T$ is a reflection in the line $y = x\sqrt{3}$ and is represented by the matrix $\mathbf{N}$

The point $P(x, y)$ is transformed first by $S$, then by $T$

The result of these transformations is the point $Q(3, 8)$

Find the coordinates of $P$

Give your answers to three decimal places.
[5 marks]

\hfill \mbox{\textit{AQA Further Paper 2 2024 Q12 [5]}}