Question 10 - A-Level Maths

AQA AS Paper 2 2020 June — Question 10 8 marks

Exam Board	AQA
Module	AS Paper 2 (AS Paper 2)
Year	2020
Session	June
Marks	8
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Stationary points and optimisation
Type	Determine constant from stationary point condition
Difficulty	Standard +0.3 This is a straightforward calculus question requiring students to use the turning point condition (dy/dx = 0) to find c, then solve a quadratic to find the second turning point, and finally solve an inequality. While it has multiple steps and requires careful algebraic manipulation, all techniques are standard AS-level procedures with no novel insight required, making it slightly easier than average.
Spec	1.07n Stationary points: find maxima, minima using derivatives 1.07o Increasing/decreasing: functions using sign of dy/dx

A curve has gradient function $$\frac{dy}{dx} = 3x^2 - 12x + c$$ The curve has a turning point at $(-1, 1)$

Find the coordinates of the other turning point of the curve. Fully justify your answer. [6 marks]
Find the set of values of $x$ for which $y$ is increasing. [2 marks]

Show mark scheme Show mark scheme source

Question 10:

Answer	Marks	Guidance
10(a)	States = 0 at a turning point OE
𝑑𝑑𝑦𝑦	2.4	E1

𝑑𝑑𝑦𝑦

𝑑𝑑𝑥𝑥

2 3𝑑𝑑 −c1 =2 𝑑𝑑–1+5 𝑐𝑐 = 0

Integrate to find y

y = x3 – 6x2 – 15x + k

1 = –1 – 6 + 15 + k

k = –7

= 0 gives x = –1 and x = 5

𝑑𝑑𝑦𝑦

𝑑𝑑𝑥𝑥

3 2

𝑑𝑑 = 5 −6y (=5 -1)0−7 15(5)−7

(5, –107)

𝑑𝑑𝑥𝑥

Substitutes x = –1 into = 0

Answer	Marks	Guidance
𝑑𝑑𝑦𝑦	3.1a	M1

𝑑𝑑𝑥𝑥

Answer	Marks	Guidance
Obtains correct value for c	1.1b	A1
Obtains x = 5 at other turning point	1.1b	A1

Integrates to find y, at least one

term correct and substitutes point

(–1 , 1) into their integrated

Answer	Marks	Guidance
expression to find ‘their’ k	3.1a	M1
Obtains correct y coordinate	1.1b	A1
Subtotal	6

Answer	Marks	Guidance
10(b)	Obtains lower inequality condone
inclusion of equality	1.1b	B1

Obtains upper inequality condone

Answer	Marks	Guidance
inclusion of equality	1.1b	B1
Subtotal	2
Question Total	8
Q	Marking Instructions	AO

Question 10:
--- 10(a) ---
10(a) | States = 0 at a turning point OE
𝑑𝑑𝑦𝑦 | 2.4 | E1 | At a turning point = 0
𝑑𝑑𝑦𝑦
𝑑𝑑𝑥𝑥
2
3𝑑𝑑 −c1 =2 𝑑𝑑–1+5 𝑐𝑐 = 0
Integrate to find y
y = x3 – 6x2 – 15x + k
1 = –1 – 6 + 15 + k
k = –7
= 0 gives x = –1 and x = 5
𝑑𝑑𝑦𝑦
𝑑𝑑𝑥𝑥
3 2
𝑑𝑑 = 5 −6y (=5 -1)0−7 15(5)−7
(5, –107)
𝑑𝑑𝑥𝑥
Substitutes x = –1 into = 0
𝑑𝑑𝑦𝑦 | 3.1a | M1
𝑑𝑑𝑥𝑥
Obtains correct value for c | 1.1b | A1
Obtains x = 5 at other turning point | 1.1b | A1
Integrates to find y, at least one
term correct and substitutes point
(–1 , 1) into their integrated
expression to find ‘their’ k | 3.1a | M1
Obtains correct y coordinate | 1.1b | A1
Subtotal | 6
--- 10(b) ---
10(b) | Obtains lower inequality condone
inclusion of equality | 1.1b | B1 | x < –1 and x > 5
Obtains upper inequality condone
inclusion of equality | 1.1b | B1
Subtotal | 2
Question Total | 8
Q | Marking Instructions | AO | Marks | Typical Solution

Show LaTeX source

A curve has gradient function
$$\frac{dy}{dx} = 3x^2 - 12x + c$$

The curve has a turning point at $(-1, 1)$

\begin{enumerate}[label=(\alph*)]
\item Find the coordinates of the other turning point of the curve.

Fully justify your answer.
[6 marks]

\item Find the set of values of $x$ for which $y$ is increasing.
[2 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA AS Paper 2 2020 Q10 [8]}}

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Q1 11 Q2 1 Q3 3 Q4 4 Q5 4 Q6 6 Q7 2 Q8 6 Q9 7 Q10 8 Q11 11 Q12 1 Q13 1 Q14 4 Q15 3 Q16 4 Q17 3 Q18 5 Q19 6