AQA AS Paper 1 2020 June — Question 7 6 marks

Exam BoardAQA
ModuleAS Paper 1 (AS Paper 1)
Year2020
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFunction Transformations
TypeSingle transformation application
DifficultyStandard +0.3 This question requires applying transformations to obtain new equations (straightforward: Cā‚ becomes y=(x-3)Ā², Lā‚ becomes y=x/2), solving a quadratic to find intersection points, then using the distance formula. While it involves multiple steps and combines several topics (transformations, simultaneous equations, coordinate geometry), each individual step is standard AS-level technique with no novel insight required. The 6 marks reflect the working needed rather than conceptual difficulty.
Spec1.02q Use intersection points: of graphs to solve equations1.02w Graph transformations: simple transformations of f(x)
Curve C has equation \(y = x^2\) C is translated by vector \(\begin{pmatrix} 3 \\ 0 \end{pmatrix}\) to give curve \(C_1\) Line L has equation \(y = x\) L is stretched by scale factor 2 parallel to the \(x\)-axis to give line \(L_1\) Find the exact distance between the two intersection points of \(C_1\) and \(L_1\) [6 marks]
Question 7:
AnswerMarks
7Identifies transformed functions as
(x Ā± 3)2 and 2x or , at least one
1
AnswerMarks Guidance
correct.3.1a M1
1
L has equation y =
1
1
2š‘„š‘„
(x ā€“ 3)2 =
1
2š‘„š‘„
x2 ā€“ + 9 = 0
13
2 š‘„š‘„
x = 2 or 4
1
2
y = 1 or 2
1
4
Distance = āˆš{(4 ā€“ 2)2 + (2 ā€“ 1)2}
1 1
2 4
=
5āˆš5
2š‘„š‘„
AnswerMarks Guidance
Forms correct equation1.1b A1
Solves their quadratic equation1.1a M1
Obtains correct x values1.1b A1
Applies distance formula to their x
AnswerMarks Guidance
and y values1.1a M1
Obtains correct distance for their
intersection points (non-zero
AnswerMarks Guidance
values), any equivalent exact form.1.1b A1F
Total6 4
QMarking Instructions AO 
Question 7:
7 | Identifies transformed functions as
(x Ā± 3)2 and 2x or , at least one
1
correct. | 3.1a | M1 | C has equation y = (x ā€“ 3)2
1
L has equation y =
1
1
2š‘„š‘„
(x ā€“ 3)2 =
1
2š‘„š‘„
x2 ā€“ + 9 = 0
13
2 š‘„š‘„
x = 2 or 4
1
2
y = 1 or 2
1
4
Distance = āˆš{(4 ā€“ 2)2 + (2 ā€“ 1)2}
1 1
2 4
=
5āˆš5
2š‘„š‘„
Forms correct equation | 1.1b | A1
Solves their quadratic equation | 1.1a | M1
Obtains correct x values | 1.1b | A1
Applies distance formula to their x
and y values | 1.1a | M1
Obtains correct distance for their
intersection points (non-zero
values), any equivalent exact form. | 1.1b | A1F
Total | 6 | 4
Q | Marking Instructions | AO | Marks | Typical Solution
Curve C has equation $y = x^2$

C is translated by vector $\begin{pmatrix} 3 \\ 0 \end{pmatrix}$ to give curve $C_1$

Line L has equation $y = x$

L is stretched by scale factor 2 parallel to the $x$-axis to give line $L_1$

Find the exact distance between the two intersection points of $C_1$ and $L_1$ [6 marks]

\hfill \mbox{\textit{AQA AS Paper 1 2020 Q7 [6]}}
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Q1 1 Q2 1 Q3 4 Q4 3 Q5 4 Q6 9 Q7 6 Q8 8 Q9 5 Q10 12 Q11 1 Q12 1 Q13 3 Q14 5 Q15 7 Q16 10