AQA AS Paper 1 2021 June — Question 9 7 marks

Exam BoardAQA
ModuleAS Paper 1 (AS Paper 1)
Year2021
SessionJune
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicChain Rule
TypeSecond derivative and nature determination
DifficultyStandard +0.8 This question requires finding first and second derivatives of terms with negative/fractional powers, setting dy/dx = 0, then evaluating dΒ²y/dxΒ² at the turning point. While the algebraic manipulation is standard A-level fare, the 7-mark allocation and requirement to 'fully justify' suggests students must show the second derivative is definitively positive (using the fact that a, b, c are positive constants and the turning point condition). This goes beyond routine differentiation into careful algebraic reasoning, placing it moderately above average difficulty.
Spec1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative

A curve has equation $$y = \frac{a}{\sqrt{x}} + bx^2 + \frac{c}{x^3} \quad \text{for } x > 0$$ where \(a\), \(b\) and \(c\) are positive constants. The curve has a single turning point. Use the second derivative of \(y\) to determine the nature of this turning point. You do not need to find the coordinates of the turning point. Fully justify your answer. [7 marks]

Question 9:
AnswerMarks Guidance
9Expresses all terms as powers
of x at least 2 correct. PI1.1a M1
βˆ’0.5 βˆ’3
π‘Žπ‘Žπ‘₯π‘₯ 𝑐𝑐π‘₯π‘₯
= - + 2bx – 3
d𝑦𝑦 1 βˆ’1.5 βˆ’4
2π‘Žπ‘Žπ‘₯π‘₯ 𝑐𝑐π‘₯π‘₯
dπ‘₯π‘₯
= + 2b + 12
2
d 𝑦𝑦 3 βˆ’2.5 βˆ’5
2 4π‘Žπ‘Žπ‘₯π‘₯ 𝑐𝑐π‘₯π‘₯
Asd aπ‘₯π‘₯, b, c and x are all > 0, all terms
must be positive
so is positive
2
d 𝑦𝑦
2
so turnindgπ‘₯π‘₯ point is a minimum
Differentiates at least one of
AnswerMarks Guidance
their negative powers correctly1.1a M1
Obtains completely correct
AnswerMarks Guidance
differential1.1b A1
Differentiates again, powers and
AnswerMarks Guidance
signs correct1.1a M1
Deduces that is positive
2
AnswerMarks Guidance
d 𝑦𝑦2.2a A1F
2
dπ‘₯π‘₯
Explains that positive second
differential shows that turning
AnswerMarks Guidance
point is a minimum2.4 E1F
Shows completely correct
mathematics throughout,
including coefficients of
2
d 𝑦𝑦
must refer to a, b, c and x being
2
AnswerMarks Guidance
> 0 dπ‘₯π‘₯2.1 R1
Total7
QMarking instructions AO
Question 9:
9 | Expresses all terms as powers
of x at least 2 correct. PI | 1.1a | M1 | y = + bx2 +
βˆ’0.5 βˆ’3
π‘Žπ‘Žπ‘₯π‘₯ 𝑐𝑐π‘₯π‘₯
= - + 2bx – 3
d𝑦𝑦 1 βˆ’1.5 βˆ’4
2π‘Žπ‘Žπ‘₯π‘₯ 𝑐𝑐π‘₯π‘₯
dπ‘₯π‘₯
= + 2b + 12
2
d 𝑦𝑦 3 βˆ’2.5 βˆ’5
2 4π‘Žπ‘Žπ‘₯π‘₯ 𝑐𝑐π‘₯π‘₯
Asd aπ‘₯π‘₯, b, c and x are all > 0, all terms
must be positive
so is positive
2
d 𝑦𝑦
2
so turnindgπ‘₯π‘₯ point is a minimum
Differentiates at least one of
their negative powers correctly | 1.1a | M1
Obtains completely correct
differential | 1.1b | A1
Differentiates again, powers and
signs correct | 1.1a | M1
Deduces that is positive
2
d 𝑦𝑦 | 2.2a | A1F
2
dπ‘₯π‘₯
Explains that positive second
differential shows that turning
point is a minimum | 2.4 | E1F
Shows completely correct
mathematics throughout,
including coefficients of
2
d 𝑦𝑦
must refer to a, b, c and x being
2
> 0 dπ‘₯π‘₯ | 2.1 | R1
Total | 7
Q | Marking instructions | AO | Marks | Typical solution
A curve has equation
$$y = \frac{a}{\sqrt{x}} + bx^2 + \frac{c}{x^3} \quad \text{for } x > 0$$
where $a$, $b$ and $c$ are positive constants.

The curve has a single turning point.

Use the second derivative of $y$ to determine the nature of this turning point.

You do not need to find the coordinates of the turning point.

Fully justify your answer.
[7 marks]

\hfill \mbox{\textit{AQA AS Paper 1 2021 Q9 [7]}}