AQA AS Paper 1 Specimen — Question 9 5 marks

Exam BoardAQA
ModuleAS Paper 1 (AS Paper 1)
SessionSpecimen
Marks5
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Mark schemeDownload PDF ↗
TopicDifferentiation from First Principles
TypeExpand f(a+h) algebraically
DifficultyModerate -0.3 Part (a) is straightforward algebraic expansion and simplification. Part (b) is a standard first principles differentiation question at a specific point, requiring the limit definition but with routine algebraic manipulation. While it's a multi-step problem worth 5 marks total, both parts follow textbook procedures without requiring problem-solving insight, making it slightly easier than average for AS-level.
Spec1.07a Derivative as gradient: of tangent to curve1.07g Differentiation from first principles: for small positive integer powers of x
  1. Given that \(f(x) = x^2 - 4x + 2\), find \(f(3 + h)\) Express your answer in the form \(h^2 + bh + c\), where \(b\) and \(c \in \mathbb{Z}\). [2 marks]
  2. The curve with equation \(y = x^2 - 4x + 2\) passes through the point \(P(3, -1)\) and the point \(Q\) where \(x = 3 + h\). Using differentiation from first principles, find the gradient of the tangent to the curve at the point \(P\). [3 marks]
Question 9:
AnswerMarks
9(a)Substitutes 3 + h to obtain a
correct unsimplified expression
AnswerMarks Guidance
for f(3 + h)AO1.1a M1
or
=96hh2124h2
=h22h1
Expresses simplified answer
AnswerMarks Guidance
correctly in given formatAO1.1b A1
(b)Identifies and uses
f(xh)f(x)
to obtain an
h
expression for the gradient of
chord
Mark can be awarded for
AnswerMarks Guidance
unsimplified expression.AO1.1a M1
Gradient of chord =
h
h2 2h11
h
h2
As h0, h22
Gradient of tangent = 2
Obtains a correct and full
AnswerMarks Guidance
simplificationAO1.1b A1
Deduces that, as h approaches
f(3h)f(3)
0 the limit of
h
is 2
(Must not simply say h = 0 but
accept words rather than limit
notation)
FT ‘their’ gradient provided M1
AnswerMarks Guidance
has been awardedAO2.2a R1
Total5
QMarking Instructions AO 
Question 9:
--- 9(a) ---
9(a) | Substitutes 3 + h to obtain a
correct unsimplified expression
for f(3 + h) | AO1.1a | M1 | (3h)2 4(3h)2
or
=96hh2124h2
=h22h1
Expresses simplified answer
correctly in given format | AO1.1b | A1
(b) | Identifies and uses
f(xh)f(x)
to obtain an
h
expression for the gradient of
chord
Mark can be awarded for
unsimplified expression. | AO1.1a | M1 | f(3h)f(3)
Gradient of chord =
h
h2 2h11

h
h2
As h0, h22
Gradient of tangent = 2
Obtains a correct and full
simplification | AO1.1b | A1
Deduces that, as h approaches
f(3h)f(3)
0 the limit of
h
is 2
(Must not simply say h = 0 but
accept words rather than limit
notation)
FT ‘their’ gradient provided M1
has been awarded | AO2.2a | R1
Total | 5
Q | Marking Instructions | AO | Marks | Typical Solution
\begin{enumerate}[label=(\alph*)]
\item Given that $f(x) = x^2 - 4x + 2$, find $f(3 + h)$

Express your answer in the form $h^2 + bh + c$, where $b$ and $c \in \mathbb{Z}$.
[2 marks]

\item The curve with equation $y = x^2 - 4x + 2$ passes through the point $P(3, -1)$ and the point $Q$ where $x = 3 + h$.

Using differentiation from first principles, find the gradient of the tangent to the curve at the point $P$.
[3 marks]
\end{enumerate}

\hfill \mbox{\textit{AQA AS Paper 1  Q9 [5]}}
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