| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2021 |
| Session | October |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Stationary points and optimisation |
| Type | Find x-coordinates with given gradient |
| Difficulty | Standard +0.3 This is a standard A-level differentiation and integration question with routine steps: find gradient at a point, write tangent equation, find intersection, and calculate area between curve and line. All techniques are straightforward applications of core calculus methods with no novel problem-solving required. Slightly easier than average due to the guided structure and standard procedures. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07s Parametric and implicit differentiation1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(y = x^3 - 10x^2 + 27x - 23 \Rightarrow \dfrac{dy}{dx} = 3x^2 - 20x + 27\) | B1 | Correct derivative |
| \(\left(\dfrac{dy}{dx}\right)_{x=5} = 3\times5^2 - 20\times5 + 27\ (=2)\) | M1 | Substitutes \(x=5\) into derivative |
| \(y + 13 = 2(x-5)\) | M1 | Fully correct straight line method using \((5,-13)\) and their \(\dfrac{dy}{dx}\) at \(x=5\) |
| \(y = 2x - 23\) | A1 | cao. Must see full equation in required form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Both \(C\) and \(l\) pass through \((0,-23)\) and so \(C\) meets \(l\) again on the \(y\)-axis | B1 | Makes a suitable deduction. Alternative: equating \(l\) and \(C\): \(x^3-10x^2+25x=0 \Rightarrow x(x^2-10x+25)=0 \Rightarrow x=0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\pm\int\left(x^3 - 10x^2 + 27x - 23 - (2x-23)\right)dx\) | M1 | Attempt to integrate \(x^n \to x^{n+1}\) for \(\pm``C-l"\) |
| \(= \pm\left(\dfrac{x^4}{4} - \dfrac{10}{3}x^3 + \dfrac{25}{2}x^2\right)\) | A1ft | Correct integration, may be unsimplified (follow through from (a)) |
| \(\left[\dfrac{x^4}{4} - \dfrac{10}{3}x^3 + \dfrac{25}{2}x^2\right]_0^5 = \left(\dfrac{625}{4} - \dfrac{1250}{3} + \dfrac{625}{2}\right)(-0)\) | dM1 | Fully correct strategy for area; use of 5 as limit; condone omission of "\(-0\)". Depends on first M1 |
| \(= \dfrac{625}{12}\) | A1 | Correct exact value |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\pm\int\left(x^3-10x^2+27x-23\right)dx = \pm\left(\dfrac{x^4}{4}-\dfrac{10}{3}x^3+\dfrac{27}{2}x^2-23x\right)\) | M1, A1 | Attempt to integrate \(x^n\to x^{n+1}\); correct integration |
| \(\left[\dfrac{x^4}{4}-\dfrac{10}{3}x^3+\dfrac{27}{2}x^2-23x\right]_0^5 + \dfrac{1}{2}\times5(23+13) = -\dfrac{455}{12}+90\) | dM1 | Correct strategy: area of trapezium minus area under curve. Depends on first M1 |
| \(= \dfrac{625}{12}\) | A1 | Correct exact value |
## Question 7:
### Part (a):
| Answer | Mark | Guidance |
|--------|------|----------|
| $y = x^3 - 10x^2 + 27x - 23 \Rightarrow \dfrac{dy}{dx} = 3x^2 - 20x + 27$ | B1 | Correct derivative |
| $\left(\dfrac{dy}{dx}\right)_{x=5} = 3\times5^2 - 20\times5 + 27\ (=2)$ | M1 | Substitutes $x=5$ into derivative |
| $y + 13 = 2(x-5)$ | M1 | Fully correct straight line method using $(5,-13)$ and their $\dfrac{dy}{dx}$ at $x=5$ |
| $y = 2x - 23$ | A1 | cao. Must see full equation in required form |
### Part (b):
| Answer | Mark | Guidance |
|--------|------|----------|
| Both $C$ and $l$ pass through $(0,-23)$ and so $C$ meets $l$ again on the $y$-axis | B1 | Makes a suitable deduction. Alternative: equating $l$ and $C$: $x^3-10x^2+25x=0 \Rightarrow x(x^2-10x+25)=0 \Rightarrow x=0$ |
### Part (c):
| Answer | Mark | Guidance |
|--------|------|----------|
| $\pm\int\left(x^3 - 10x^2 + 27x - 23 - (2x-23)\right)dx$ | M1 | Attempt to integrate $x^n \to x^{n+1}$ for $\pm``C-l"$ |
| $= \pm\left(\dfrac{x^4}{4} - \dfrac{10}{3}x^3 + \dfrac{25}{2}x^2\right)$ | A1ft | Correct integration, may be unsimplified (follow through from (a)) |
| $\left[\dfrac{x^4}{4} - \dfrac{10}{3}x^3 + \dfrac{25}{2}x^2\right]_0^5 = \left(\dfrac{625}{4} - \dfrac{1250}{3} + \dfrac{625}{2}\right)(-0)$ | dM1 | Fully correct strategy for area; use of 5 as limit; condone omission of "$-0$". **Depends on first M1** |
| $= \dfrac{625}{12}$ | A1 | Correct exact value |
**Alternative (c):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $\pm\int\left(x^3-10x^2+27x-23\right)dx = \pm\left(\dfrac{x^4}{4}-\dfrac{10}{3}x^3+\dfrac{27}{2}x^2-23x\right)$ | M1, A1 | Attempt to integrate $x^n\to x^{n+1}$; correct integration |
| $\left[\dfrac{x^4}{4}-\dfrac{10}{3}x^3+\dfrac{27}{2}x^2-23x\right]_0^5 + \dfrac{1}{2}\times5(23+13) = -\dfrac{455}{12}+90$ | dM1 | Correct strategy: area of trapezium minus area under curve. **Depends on first M1** |
| $= \dfrac{625}{12}$ | A1 | Correct exact value |
---\begin{enumerate}
\item In this question you should show all stages of your working.
\end{enumerate}
Solutions relying entirely on calculator technology are not acceptable.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6c32000f-574f-473c-bd04-9cfe2c1bd715-16_805_1041_388_511}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a sketch of part of the curve $C$ with equation
$$y = x ^ { 3 } - 10 x ^ { 2 } + 27 x - 23$$
The point $P ( 5 , - 13 )$ lies on $C$\\
The line $l$ is the tangent to $C$ at $P$\\
(a) Use differentiation to find the equation of $l$, giving your answer in the form $y = m x + c$ where $m$ and $c$ are integers to be found.\\
(b) Hence verify that $l$ meets $C$ again on the $y$-axis.
The finite region $R$, shown shaded in Figure 2, is bounded by the curve $C$ and the line $l$.\\
(c) Use algebraic integration to find the exact area of $R$.
\hfill \mbox{\textit{Edexcel Paper 2 2021 Q7 [9]}}