AQA Paper 3 2023 June — Question 5 3 marks

Exam BoardAQA
ModulePaper 3 (Paper 3)
Year2023
SessionJune
Marks3
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Mark schemeDownload PDF ↗
TopicDifferentiating Transcendental Functions
TypeFind gradient at a point - direct evaluation
DifficultyModerate -0.8 This is a straightforward differentiation and substitution question requiring only basic exponential differentiation (dy/dx = 6e^{2x}) and solving 3e^{2x} = 10 to find x, then substituting back. It's easier than average as it involves routine application of standard techniques with no problem-solving insight needed, though the substitution step adds minimal complexity beyond pure recall.
Spec1.06a Exponential function: a^x and e^x graphs and properties1.07j Differentiate exponentials: e^(kx) and a^(kx)
A curve has equation \(y = 3e^{2x}\) Find the gradient of the curve at the point where \(y = 10\) [3 marks]
Question 5:
AnswerMarks Guidance
5Obtains 2×3e 2x or 6e 2x or 2y
PI by correct answer1.1b B1
=2×3e2x
dx
dy
y =10⇒ =2×10=20
dx
Substitutesy =10or 3e 2x =10
dy
in their
dx
or
substitutes x = [0.6, 0.602] or
1 10 dy
x= ln OE in their
AnswerMarks Guidance
2  3  dx1.1a M1
Obtains 20
CAO
20 cannot come from a rounded
AnswerMarks Guidance
value for 20 seen1.1b A1
Question 5 Total3
QMarking instructions AO 
Question 5:
5 | Obtains 2×3e 2x or 6e 2x or 2y
PI by correct answer | 1.1b | B1 | dy
=2×3e2x
dx
dy
y =10⇒ =2×10=20
dx
Substitutesy =10or 3e 2x =10
dy
in their
dx
or
substitutes x = [0.6, 0.602] or
1 10 dy
x= ln OE in their
2  3  dx | 1.1a | M1
Obtains 20
CAO
20 cannot come from a rounded
value for 20 seen | 1.1b | A1
Question 5 Total | 3
Q | Marking instructions | AO | Marks | Typical solution
A curve has equation $y = 3e^{2x}$

Find the gradient of the curve at the point where $y = 10$

[3 marks]

\hfill \mbox{\textit{AQA Paper 3 2023 Q5 [3]}}
This paper (17 questions)
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Q1 1 Q2 1 Q3 1 Q4 5 Q5 3 Q6 9 Q7 14 Q8 7 Q9 12 Q10 1 Q11 1 Q12 8 Q13 4 Q14 10 Q15 11 Q16 9 Q17 6