| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Equation of tangent line |
| Difficulty | Standard +0.3 This is a straightforward AS-level question requiring chain rule differentiation of x^(1/2), finding a tangent equation at a given point, and computing an area using integration. All steps are routine applications of standard techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y=\frac{1}{3}x^2-2\sqrt{x}+3 \Rightarrow \frac{dy}{dx}=\frac{2}{3}x-x^{-\frac{1}{2}}\) | M1 | \(x^n \rightarrow x^{n-1}\) seen at least once; \(...x^2\rightarrow...x^1\), \(...x^{\frac{1}{2}}\rightarrow...x^{-\frac{1}{2}}\), \(3\rightarrow0\) |
| A1 | \(\frac{2}{3}x - x^{-\frac{1}{2}}\) or any unsimplified equivalent (indices must be processed); accept \(0.6\dot{6}x\) but not rounded values | |
| \(x=4 \Rightarrow y=\frac{13}{3}\) | B1 | Correct \(y\) coordinate of \(P\); may be seen embedded in equation of \(l\) |
| \(\left(\frac{dy}{dx}\right)_{x=4}=\frac{2}{3}\times4-4^{-\frac{1}{2}}\left(=\frac{13}{6}\right)\) \(\therefore y-\frac{13}{3}=\frac{13}{6}(x-4)\) | M1 | Fully correct strategy for equation of \(l\); look for \(y-"\frac{13}{3}"="\frac{13}{6}"(x-4)\) where gradient comes from differentiating and substituting \(x=4\); if \(y=mx+c\) must proceed to \(c=...\); do not allow perpendicular gradient |
| \(13x-6y-26=0\) | A1* | Obtains printed answer with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int\left(\frac{x^2}{3}-2\sqrt{x}+3\right)dx = \frac{x^3}{9}-\frac{4}{3}x^{\frac{3}{2}}+3x(+c)\) | M1 | \(x^n\rightarrow x^{n+1}\) seen at least once |
| A1 | \(\frac{x^3}{9}-\frac{4}{3}x^{\frac{3}{2}}+3x\); accept exact decimals \(\frac{1}{9}\) \((0.\dot{1})\) and \(-\frac{4}{3}\) \((-1.\dot{3})\) but not rounded values | |
| \(y=0 \Rightarrow x=2\) | B1 | Deduces correct value for \(x\) at intersection of \(l\) with \(x\)-axis; may be seen on figure |
| Area of \(R\) is \(\left[\frac{x^3}{9}-\frac{4}{3}x^{\frac{3}{2}}+3x\right]_0^4 - \frac{1}{2}\times(4-"2")\times"\frac{13}{3}"=\frac{76}{9}-\frac{13}{3}\) | M1 | Fully correct strategy for area: correct attempt at triangle area using their values (could use integration); correct attempt at area under curve using 0 and 4; two values subtracted |
| \(=\frac{37}{9}\) | A1 | \(\frac{37}{9}\) or exact equivalent e.g. \(4\frac{1}{9}\) or \(4.\dot{1}\) but not \(4.111...\) |
# Question 10:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y=\frac{1}{3}x^2-2\sqrt{x}+3 \Rightarrow \frac{dy}{dx}=\frac{2}{3}x-x^{-\frac{1}{2}}$ | M1 | $x^n \rightarrow x^{n-1}$ seen at least once; $...x^2\rightarrow...x^1$, $...x^{\frac{1}{2}}\rightarrow...x^{-\frac{1}{2}}$, $3\rightarrow0$ |
| | A1 | $\frac{2}{3}x - x^{-\frac{1}{2}}$ or any unsimplified equivalent (indices must be processed); accept $0.6\dot{6}x$ but not rounded values |
| $x=4 \Rightarrow y=\frac{13}{3}$ | B1 | Correct $y$ coordinate of $P$; may be seen embedded in equation of $l$ |
| $\left(\frac{dy}{dx}\right)_{x=4}=\frac{2}{3}\times4-4^{-\frac{1}{2}}\left(=\frac{13}{6}\right)$ $\therefore y-\frac{13}{3}=\frac{13}{6}(x-4)$ | M1 | Fully correct strategy for equation of $l$; look for $y-"\frac{13}{3}"="\frac{13}{6}"(x-4)$ where gradient comes from differentiating and substituting $x=4$; if $y=mx+c$ must proceed to $c=...$; do not allow perpendicular gradient |
| $13x-6y-26=0$ | A1* | Obtains printed answer with no errors |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int\left(\frac{x^2}{3}-2\sqrt{x}+3\right)dx = \frac{x^3}{9}-\frac{4}{3}x^{\frac{3}{2}}+3x(+c)$ | M1 | $x^n\rightarrow x^{n+1}$ seen at least once |
| | A1 | $\frac{x^3}{9}-\frac{4}{3}x^{\frac{3}{2}}+3x$; accept exact decimals $\frac{1}{9}$ $(0.\dot{1})$ and $-\frac{4}{3}$ $(-1.\dot{3})$ but not rounded values |
| $y=0 \Rightarrow x=2$ | B1 | Deduces correct value for $x$ at intersection of $l$ with $x$-axis; may be seen on figure |
| Area of $R$ is $\left[\frac{x^3}{9}-\frac{4}{3}x^{\frac{3}{2}}+3x\right]_0^4 - \frac{1}{2}\times(4-"2")\times"\frac{13}{3}"=\frac{76}{9}-\frac{13}{3}$ | M1 | Fully correct strategy for area: correct attempt at triangle area using their values (could use integration); correct attempt at area under curve using 0 and 4; two values subtracted |
| $=\frac{37}{9}$ | A1 | $\frac{37}{9}$ or exact equivalent e.g. $4\frac{1}{9}$ or $4.\dot{1}$ but not $4.111...$ |
---10.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d31369fa-9532-4a09-b67d-a3a3cbf7d586-30_639_878_246_596}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
In this question you must show all stages of your working.\\
Solutions relying on calculator technology are not acceptable.\\
Figure 2 shows a sketch of part of the curve $C$ with equation
$$y = \frac { 1 } { 3 } x ^ { 2 } - 2 \sqrt { x } + 3 \quad x \geqslant 0$$
The point $P$ lies on $C$ and has $x$ coordinate 4\\
The line $l$ is the tangent to $C$ at $P$.
\begin{enumerate}[label=(\alph*)]
\item Show that $l$ has equation
$$13 x - 6 y - 26 = 0$$
The region $R$, shown shaded in Figure 2, is bounded by the $y$-axis, the curve $C$, the line $l$ and the $x$-axis.
\item Find the exact area of $R$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel AS Paper 1 2022 Q10 [10]}}