Question 5 - A-Level Maths

AQA Further AS Paper 1 2018 June — Question 5 3 marks

Exam Board	AQA
Module	Further AS Paper 1 (Further AS Paper 1)
Year	2018
Session	June
Marks	3
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Linear transformations
Type	Describe 3D transformation from matrix
Difficulty	Standard +0.3 This is a straightforward matrix transformation question requiring recognition of a 3D rotation matrix combined with a scaling. Students need to identify the rotation angle (150° about z-axis) and scale factor (1) from the standard form, which is routine for Further Maths students who have learned the topic, though slightly above average difficulty due to the 3D context and the specific angle involved.
Spec	4.03d Linear transformations 2D: reflection, rotation, enlargement, shear

Describe fully the transformation given by the matrix \(\begin{pmatrix} -\frac{1}{2} & -\frac{\sqrt{3}}{2} & 0 \\ \frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \\ 0 & 0 & 1 \end{pmatrix}\) [3 marks]

Show mark scheme Show mark scheme source

Question 5:

Answer	Marks	Guidance
5	Identifies correct values of sine and cosine.	1.1a

1 √3

cos𝜃𝜃 = −2 sin𝜃𝜃 = 2

°

𝜃𝜃 = 120

Rotation about the -axis through anti-clockwise.

𝑧𝑧 120°

Selects correct angle.

Accept or or

Answer	Marks	Guidance
2𝜋𝜋 4𝜋𝜋	1.1b	A1

Deduces3 the t − ra 2 n 4 s 0 f ° orma − tio3n giving a full description.

FT their angle.

Accept or or

2𝜋𝜋 4𝜋𝜋

Condone miss−in2g4 0d°egre−e sign.

3 3

Condone missing ‘anticlockwise’.

Answer	Marks	Guidance
NMS scores 3/3	2.2a	A1F
Total	3
Q	Marking instructions	AO
6a	Identifies the sign (or just the negative) as the error. May be seen in any of the last five lines.

May be indicated within Matthew’s solution or described in words.

±

Ignore other ‘errors’ identified.

Condone identifying any of the last five lines as containing the error.

Answer	Marks	Guidance
PI	2.3	B1

± �𝑦𝑦 + 1 = 𝑒𝑒 − 𝑦𝑦

Gives a correct reason, referring to either (or ), or the operand of a logarithm, being

positive. 𝑥𝑥 𝑦𝑦

𝑒𝑒 𝑒𝑒

Answer	Marks	Guidance
Do not award if more than one error identified.	2.4	E1

so there is a contradic2tion.

𝑦𝑦−�𝑦𝑦 +1 < 0

𝑥𝑥

Answer	Marks
6b	States the correct solution of the equation.

Accept 10.0 [1787493] or

1 3 −3

ISW

Answer	Marks	Guidance
2(𝑒𝑒 −𝑒𝑒 )	1.1b	B1

arsinh 𝑥𝑥 = 3

Answer	Marks	Guidance
Total	3	𝑥𝑥 = sinh3
Q	Marking instructions	AO

Question 5:
5 | Identifies correct values of sine and cosine. | 1.1a | M1 | and
1 √3
cos𝜃𝜃 = −2 sin𝜃𝜃 = 2
°
𝜃𝜃 = 120
Rotation about the -axis through anti-clockwise.
𝑧𝑧 120°
Selects correct angle.
Accept or or
2𝜋𝜋 4𝜋𝜋 | 1.1b | A1
Deduces3 the t − ra 2 n 4 s 0 f ° orma − tio3n giving a full description.
FT their angle.
Accept or or
2𝜋𝜋 4𝜋𝜋
Condone miss−in2g4 0d°egre−e sign.
3 3
Condone missing ‘anticlockwise’.
NMS scores 3/3 | 2.2a | A1F
Total | 3
Q | Marking instructions | AO | Mark | Typical solution
6a | Identifies the sign (or just the negative) as the error. May be seen in any of the last five lines.
May be indicated within Matthew’s solution or described in words.
±
Ignore other ‘errors’ identified.
Condone identifying any of the last five lines as containing the error.
PI | 2.3 | B1 | 2 𝑥𝑥
± �𝑦𝑦 + 1 = 𝑒𝑒 − 𝑦𝑦
Gives a correct reason, referring to either (or ), or the operand of a logarithm, being
positive. 𝑥𝑥 𝑦𝑦
𝑒𝑒 𝑒𝑒
Do not award if more than one error identified. | 2.4 | E1 | It is an error because and
so there is a contradic2tion.
𝑦𝑦−�𝑦𝑦 +1 < 0
𝑥𝑥
6b | States the correct solution of the equation.
Accept 10.0 [1787493] or
1 3 −3
ISW
2(𝑒𝑒 −𝑒𝑒 ) | 1.1b | B1 | 𝑒𝑒 > 0
arsinh 𝑥𝑥 = 3
Total | 3 | 𝑥𝑥 = sinh3
Q | Marking instructions | AO | Mark | Typical solution

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Describe fully the transformation given by the matrix $\begin{pmatrix} -\frac{1}{2} & -\frac{\sqrt{3}}{2} & 0 \\ \frac{\sqrt{3}}{2} & -\frac{1}{2} & 0 \\ 0 & 0 & 1 \end{pmatrix}$
[3 marks]

\hfill \mbox{\textit{AQA Further AS Paper 1 2018 Q5 [3]}}

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