AQA Paper 3 2024 June — Question 6 5 marks

Exam BoardAQA
ModulePaper 3 (Paper 3)
Year2024
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIndefinite & Definite Integrals
TypeFind curve from gradient
DifficultyEasy -1.2 Part (a) is a straightforward integration of polynomial and power terms using standard rules. Part (b) adds only the trivial step of finding the constant of integration using a given point. Both parts require only direct application of basic integration techniques with no problem-solving or conceptual challenge beyond routine A-level calculus.
Spec1.08b Integrate x^n: where n != -1 and sums1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)

\begin{enumerate}[label=(\alph*)] \item Find \(\int \left(6x^2 - \frac{5}{\sqrt{x}}\right) dx\) [3 marks] \item The gradient of a curve is given by $$\frac{dy}{dx} = 6x^2 - \frac{5}{\sqrt{x}}$$ The curve passes through the point \((4, 90)\). Find the equation of the curve. [2 marks]
Question 6:

AnswerMarks
6(a)1 −1
Writes term as x 2
x
AnswerMarks Guidance
PI by x 1 orx2in answer1.1b B1
 6 x 2 − d x =  6 x 2 − 5 x − 12 d x
x
= 2 x 3 − 1 0 x 12 + c
Obtains one correctly integrated
term
AnswerMarks Guidance
May be unsimplified1.1a M1
Obtains 2 x 3 − 1 0 x 12 + c
ISW
Condone omission of +c
AnswerMarks Guidance
Must be simplified1.1b A1
Subtotal3
QMarking instructions AO

AnswerMarks
6(b)Substitutes x = 4 into their
integrated expression from part
6(a), with an arbitrary constant
and sets equal to 90
AnswerMarks Guidance
PI by –183.1a M1
c = − 1 8
 y = x 2 3 − 1 0 x 12 − 1 8
Obtains y = 2 x 3 − 1 0 x 12 − 1 8
CAO
AnswerMarks Guidance
Condone f(x) for y1.1b A1
Subtotal2
Question 6 Total5
QMarking instructions AO
Question 6:
--- 6(a) ---
6(a) | 1 −1
Writes term as x 2
x
PI by x 1 orx2in answer | 1.1b | B1 |  5  ( )
 6 x 2 − d x =  6 x 2 − 5 x − 12 d x
x
= 2 x 3 − 1 0 x 12 + c
Obtains one correctly integrated
term
May be unsimplified | 1.1a | M1
Obtains 2 x 3 − 1 0 x 12 + c
ISW
Condone omission of +c
Must be simplified | 1.1b | A1
Subtotal | 3
Q | Marking instructions | AO | Marks | Typical solution
--- 6(b) ---
6(b) | Substitutes x = 4 into their
integrated expression from part
6(a), with an arbitrary constant
and sets equal to 90
PI by –18 | 3.1a | M1 | 9 0 = 2  4 3 − 1 0  4 12 + c
c = − 1 8
 y = x 2 3 − 1 0 x 12 − 1 8
Obtains y = 2 x 3 − 1 0 x 12 − 1 8
CAO
Condone f(x) for y | 1.1b | A1
Subtotal | 2
Question 6 Total | 5
Q | Marking instructions | AO | Marks | Typical solution
\begin{enumerate}[label=(\alph*)]
\item Find $\int \left(6x^2 - \frac{5}{\sqrt{x}}\right) dx$
[3 marks]

\item The gradient of a curve is given by
$$\frac{dy}{dx} = 6x^2 - \frac{5}{\sqrt{x}}$$

The curve passes through the point $(4, 90)$.

Find the equation of the curve.
[2 marks]
</end{enumerate}

\hfill \mbox{\textit{AQA Paper 3 2024 Q6 [5]}}