Edexcel AS Paper 1 2023 June — Question 16 6 marks

Exam BoardEdexcel
ModuleAS Paper 1 (AS Paper 1)
Year2023
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicApplied differentiation
TypeFind curve equation from derivative
DifficultyStandard +0.3 This is a straightforward integration problem with three conditions to find three unknowns. Students must integrate a simple expression involving powers of x, apply the stationary point condition (f'(4)=0), and use two boundary conditions. All steps are routine AS-level techniques with no novel insight required, making it slightly easier than average.
Spec1.08a Fundamental theorem of calculus: integration as reverse of differentiation

  1. A curve has equation \(y = \mathrm { f } ( x ) , x \geqslant 0\)
Given that
  • \(\mathrm { f } ^ { \prime } ( x ) = 4 x + a \sqrt { x } + b\), where \(a\) and \(b\) are constants
  • the curve has a stationary point at \(( 4,3 )\)
  • the curve meets the \(y\)-axis at - 5
    find \(\mathrm { f } ( x )\), giving your answer in simplest form.

Question 16:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Sets \(f'(4) = 0 \Rightarrow 16 + 2a + b = 0\)M1 Key step setting up equation in \(a\) and \(b\). Condone slips.
Integrates \(f'(x) = 4x + a\sqrt{x} + b \Rightarrow \{f(x) =\} 2x^2 + \frac{2}{3}ax^{\frac{3}{2}} + bx\ \{+c\}\)M1, A1ft Award for \(x^n \to x^{n+1}\) or \(b \to bx\). Allow ft on \(b\) in terms of \(a\). May be left unsimplified but indices must be processed.
Deduces that \(c = -5\)B1 Must be the constant term in \(f(x)\). Note deducing \(b=-5\) is B0.
Full and complete method using \(f'(4)=0\) and \(f(4)=3\) to find values of \(a\) and \(b\). Note: \(a=-15\) and \(b=14\)ddM1 Dependent on both previous M marks. Must include \(f'(4)=0\) giving \(16+2a+b=0\) and \(f(4)=3\) giving \(32+\frac{16}{3}a+4b-5=3\)
\(\{f(x)=\}\ 2x^2 - 10x^{\frac{3}{2}} + 14x - 5\)A1 Exact simplified equivalent accepted, e.g. use of \(x\sqrt{x}\) in place of \(x^{\frac{3}{2}}\)
## Question 16:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Sets $f'(4) = 0 \Rightarrow 16 + 2a + b = 0$ | M1 | Key step setting up equation in $a$ and $b$. Condone slips. |
| Integrates $f'(x) = 4x + a\sqrt{x} + b \Rightarrow \{f(x) =\} 2x^2 + \frac{2}{3}ax^{\frac{3}{2}} + bx\ \{+c\}$ | M1, A1ft | Award for $x^n \to x^{n+1}$ or $b \to bx$. Allow ft on $b$ in terms of $a$. May be left unsimplified but indices must be processed. |
| Deduces that $c = -5$ | B1 | Must be the constant term in $f(x)$. Note deducing $b=-5$ is B0. |
| Full and complete method using $f'(4)=0$ and $f(4)=3$ to find values of $a$ and $b$. Note: $a=-15$ and $b=14$ | ddM1 | Dependent on **both** previous M marks. Must include $f'(4)=0$ giving $16+2a+b=0$ and $f(4)=3$ giving $32+\frac{16}{3}a+4b-5=3$ |
| $\{f(x)=\}\ 2x^2 - 10x^{\frac{3}{2}} + 14x - 5$ | A1 | Exact simplified equivalent accepted, e.g. use of $x\sqrt{x}$ in place of $x^{\frac{3}{2}}$ |

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\begin{enumerate}
  \item A curve has equation $y = \mathrm { f } ( x ) , x \geqslant 0$
\end{enumerate}

Given that

\begin{itemize}
  \item $\mathrm { f } ^ { \prime } ( x ) = 4 x + a \sqrt { x } + b$, where $a$ and $b$ are constants
  \item the curve has a stationary point at $( 4,3 )$
  \item the curve meets the $y$-axis at - 5\\
find $\mathrm { f } ( x )$, giving your answer in simplest form.
\end{itemize}

\hfill \mbox{\textit{Edexcel AS Paper 1 2023 Q16 [6]}}