AQA AS Paper 1 2019 June — Question 9 10 marks

Exam BoardAQA
ModuleAS Paper 1 (AS Paper 1)
Year2019
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicTangents, normals and gradients
TypeFind normal line equation at given point
DifficultyModerate -0.3 Part (a) is straightforward integration with a constant found from boundary conditions (standard AS calculus). Part (b) requires multiple routine steps: finding midpoint, gradient of AB, perpendicular gradient, equation of perpendicular bisector, then verifying it matches the normal at (2,0). All techniques are standard AS-level with no novel insight required, but the multi-step nature and coordination across topics elevates it slightly above the most routine questions.
Spec1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.03b Straight lines: parallel and perpendicular relationships1.07m Tangents and normals: gradient and equations1.08a Fundamental theorem of calculus: integration as reverse of differentiation
A curve cuts the \(x\)-axis at \((2, 0)\) and has gradient function $$\frac{dy}{dx} = \frac{24}{x^3}$$
  1. Find the equation of the curve. [4 marks]
  2. Show that the perpendicular bisector of the line joining \(A(-2, 8)\) to \(B(-6, -4)\) is the normal to the curve at \((2, 0)\) [6 marks]
Question 9:
AnswerMarks Guidance
9(a)Decides to integrate
243.1a M1
24
∫ 3
𝑑𝑑
= – + c
12
2
π‘₯π‘₯
0 = – + c
12
2
c =2 3
y = – +3
12
2
π‘₯π‘₯
3
Obtains correct integrπ‘₯π‘₯al with or
AnswerMarks Guidance
without c1.1b A1
Includes c and substitutes (2, 0) to
AnswerMarks Guidance
calculate c or uses definite integral1.1a M1
Evaluates c and states correct
equation ACF
Follow through only on sign error in
AnswerMarks Guidance
integral (c = –3)1.1b A1F
AnswerMarks Guidance
9(b)Finds midpoint of AB 1.1b
Gradient of AB =
8βˆ’(βˆ’4)
βˆ’2βˆ’(βˆ’6)= 3
Gradient of perpendicular bisector
βˆ’1
3
Equation is y =
βˆ’1 2
3 𝑑𝑑+ 3
When x = 2, y = = 0
βˆ’1 2
3 Γ—2+ 3
So passes through (2,0)
Gradient of curve at (2, 0) is = 3
24
3
2
Gradient of normal is
βˆ’1
3
So perpendicular bisector is the
normal
AnswerMarks Guidance
Finds gradient of AB (unsimplified)1.1b B1
Finds gradient of curve at (2, 0)1.1b B1
Uses perpendicular gradients
AnswerMarks Guidance
property3.1a M1
Finds the correct equation of the
perpendicular bisector using (-4, 2),
or the correct equation of the
normal using (2, 0) ACF
AnswerMarks Guidance
PI by correct β€œm & c” pair3.1a M1
Completes rigorous argument to
show that the perpendicular
bisector is the normal.
Must include showing normal and
bisector have the same gradient
and that (2, 0) lies on the bisector,
or that (-4, 2) lies on the normal, or
AnswerMarks Guidance
that the equations are identical.2.1 R1
Total10
QMarking Instructions AO 
Question 9:
--- 9(a) ---
9(a) | Decides to integrate
24 | 3.1a | M1 | y = dx
24
∫ 3
𝑑𝑑
= – + c
12
2
π‘₯π‘₯
0 = – + c
12
2
c =2 3
y = – +3
12
2
π‘₯π‘₯
3
Obtains correct integrπ‘₯π‘₯al with or
without c | 1.1b | A1
Includes c and substitutes (2, 0) to
calculate c or uses definite integral | 1.1a | M1
Evaluates c and states correct
equation ACF
Follow through only on sign error in
integral (c = –3) | 1.1b | A1F
--- 9(b) ---
9(b) | Finds midpoint of AB | 1.1b | B1 | Midpoint of AB is (–4, 2)
Gradient of AB =
8βˆ’(βˆ’4)
βˆ’2βˆ’(βˆ’6)= 3
Gradient of perpendicular bisector
βˆ’1
3
Equation is y =
βˆ’1 2
3 𝑑𝑑+ 3
When x = 2, y = = 0
βˆ’1 2
3 Γ—2+ 3
So passes through (2,0)
Gradient of curve at (2, 0) is = 3
24
3
2
Gradient of normal is
βˆ’1
3
So perpendicular bisector is the
normal
Finds gradient of AB (unsimplified) | 1.1b | B1
Finds gradient of curve at (2, 0) | 1.1b | B1
Uses perpendicular gradients
property | 3.1a | M1
Finds the correct equation of the
perpendicular bisector using (-4, 2),
or the correct equation of the
normal using (2, 0) ACF
PI by correct β€œm & c” pair | 3.1a | M1
Completes rigorous argument to
show that the perpendicular
bisector is the normal.
Must include showing normal and
bisector have the same gradient
and that (2, 0) lies on the bisector,
or that (-4, 2) lies on the normal, or
that the equations are identical. | 2.1 | R1
Total | 10
Q | Marking Instructions | AO | Marks | Typical Solution
A curve cuts the $x$-axis at $(2, 0)$ and has gradient function
$$\frac{dy}{dx} = \frac{24}{x^3}$$

\begin{enumerate}[label=(\alph*)]
\item Find the equation of the curve.
[4 marks]

\item Show that the perpendicular bisector of the line joining $A(-2, 8)$ to $B(-6, -4)$ is the normal to the curve at $(2, 0)$
[6 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA AS Paper 1 2019 Q9 [10]}}
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