| Exam Board | Edexcel |
|---|---|
| Module | CP2 (Core Pure 2) |
| Year | 2021 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Show stationary point exists or gradient has specific property |
| Difficulty | Standard +0.8 Part (a) requires knowing the derivative of arccos and showing it's never zero—straightforward but tests a less common derivative. Part (b) involves finding a normal line equation, intercepts, and triangle area, culminating in algebraic manipulation to match a specific form with surds and π. This is a substantial multi-step problem requiring careful algebra, but follows standard techniques without requiring novel insight. Slightly above average difficulty due to the arccos derivative and algebraic complexity. |
| Spec | 1.05i Inverse trig functions: arcsin, arccos, arctan domains and graphs1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{-\lambda}{\sqrt{1-\beta x^2}}\) where \(\lambda > 0\) and \(\beta > 0\) and \(\beta \neq 1\). Alternatively \(2\cos y = x \Rightarrow \frac{dx}{dy} = \alpha \sin y \Rightarrow \frac{dy}{dx} = \frac{1}{\alpha \sin y}\) | M1 | 1.1b — Finds the correct form for \(\frac{dy}{dx}\) |
| \(\frac{dy}{dx} = \frac{-\frac{1}{2}}{\sqrt{1-\frac{1}{4}x^2}}\) or \(\frac{dy}{dx} = \frac{-1}{2\sqrt{1-\frac{1}{4}x^2}}\) o.e., or \(\frac{dy}{dx} = -\frac{1}{2\sin y}\) | A1 | 1.1b — Correct \(\frac{dy}{dx}\) |
| States \(\frac{dy}{dx} \neq 0\) therefore \(C\) has no stationary points. Tries to solve \(\frac{dy}{dx} = 0\) and ends with contradiction e.g. \(-1=0\), therefore \(C\) has no stationary points. As \(\text{cosec}\, y > 1\) therefore \(C\) has no stationary points. | A1 | 2.4 — States or shows \(\frac{dy}{dx} \neq 0\) and draws required conclusion |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dy}{dx} = \frac{-1}{2\sqrt{1-\frac{1}{4}\times 1^2}} = \left\{-\frac{1}{\sqrt{3}}\right\}\) | M1 | 1.1b — Substitutes \(x=1\) into their \(\frac{dy}{dx}\) |
| Normal gradient \(= -\frac{1}{m}\) and \(y - \frac{\pi}{3} = m_n(x-1)\). Alternatively \(\frac{\pi}{3} = m_n(1) + c \Rightarrow c = \ldots\left\{\frac{\pi}{3}-\sqrt{3}\right\}\) and then \(y = m_n x + c\) | M1 | 1.1b — Finds normal gradient and equation of normal |
| \(y=0 \Rightarrow 0 - \frac{\pi}{3} = \sqrt{3}(x_A - 1) \Rightarrow x_A = \ldots\left\{1 - \frac{\pi}{3\sqrt{3}} \text{ or } 1 - \frac{\pi\sqrt{3}}{9}\right\}\) and \(x=0 \Rightarrow y_B - \frac{\pi}{3} = \sqrt{3}(0-1) \Rightarrow y_B = \ldots\left\{\frac{\pi}{3}-\sqrt{3}\right\}\) | M1 | 3.1a — Finds where normal cuts \(x\)-axis and \(y\)-axis |
| Area \(= \frac{1}{2} \times x_A \times -y_B = \frac{1}{2}\left(1-\frac{\pi}{3\sqrt{3}}\right)\left(\sqrt{3}-\frac{\pi}{3}\right)\) | M1 | 1.1b — Finds area of triangle \(OAB = \frac{1}{2}\times x_A \times -y_B\) |
| Area \(= \frac{1}{54}\left(27\sqrt{3}-18\pi+\sqrt{3}\pi^2\right)\) \((p=27, q=-18, r=1)\) | A1 | 2.1 — Correct area |
## Question 5:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{-\lambda}{\sqrt{1-\beta x^2}}$ where $\lambda > 0$ and $\beta > 0$ and $\beta \neq 1$. Alternatively $2\cos y = x \Rightarrow \frac{dx}{dy} = \alpha \sin y \Rightarrow \frac{dy}{dx} = \frac{1}{\alpha \sin y}$ | M1 | 1.1b — Finds the correct form for $\frac{dy}{dx}$ |
| $\frac{dy}{dx} = \frac{-\frac{1}{2}}{\sqrt{1-\frac{1}{4}x^2}}$ or $\frac{dy}{dx} = \frac{-1}{2\sqrt{1-\frac{1}{4}x^2}}$ o.e., or $\frac{dy}{dx} = -\frac{1}{2\sin y}$ | A1 | 1.1b — Correct $\frac{dy}{dx}$ |
| States $\frac{dy}{dx} \neq 0$ therefore $C$ has no stationary points. Tries to solve $\frac{dy}{dx} = 0$ and ends with contradiction e.g. $-1=0$, therefore $C$ has no stationary points. As $\text{cosec}\, y > 1$ therefore $C$ has no stationary points. | A1 | 2.4 — States or shows $\frac{dy}{dx} \neq 0$ and draws required conclusion |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{-1}{2\sqrt{1-\frac{1}{4}\times 1^2}} = \left\{-\frac{1}{\sqrt{3}}\right\}$ | M1 | 1.1b — Substitutes $x=1$ into their $\frac{dy}{dx}$ |
| Normal gradient $= -\frac{1}{m}$ and $y - \frac{\pi}{3} = m_n(x-1)$. Alternatively $\frac{\pi}{3} = m_n(1) + c \Rightarrow c = \ldots\left\{\frac{\pi}{3}-\sqrt{3}\right\}$ and then $y = m_n x + c$ | M1 | 1.1b — Finds normal gradient and equation of normal |
| $y=0 \Rightarrow 0 - \frac{\pi}{3} = \sqrt{3}(x_A - 1) \Rightarrow x_A = \ldots\left\{1 - \frac{\pi}{3\sqrt{3}} \text{ or } 1 - \frac{\pi\sqrt{3}}{9}\right\}$ and $x=0 \Rightarrow y_B - \frac{\pi}{3} = \sqrt{3}(0-1) \Rightarrow y_B = \ldots\left\{\frac{\pi}{3}-\sqrt{3}\right\}$ | M1 | 3.1a — Finds where normal cuts $x$-axis and $y$-axis |
| Area $= \frac{1}{2} \times x_A \times -y_B = \frac{1}{2}\left(1-\frac{\pi}{3\sqrt{3}}\right)\left(\sqrt{3}-\frac{\pi}{3}\right)$ | M1 | 1.1b — Finds area of triangle $OAB = \frac{1}{2}\times x_A \times -y_B$ |
| Area $= \frac{1}{54}\left(27\sqrt{3}-18\pi+\sqrt{3}\pi^2\right)$ $(p=27, q=-18, r=1)$ | A1 | 2.1 — Correct area |
---\begin{enumerate}
\item The curve $C$ has equation
\end{enumerate}
$$y = \arccos \left( \frac { 1 } { 2 } x \right) \quad - 2 \leqslant x \leqslant 2$$
(a) Show that $C$ has no stationary points.
The normal to $C$, at the point where $x = 1$, crosses the $x$-axis at the point $A$ and crosses the $y$-axis at the point $B$.
Given that $O$ is the origin,\\
(b) show that the area of the triangle $O A B$ is $\frac { 1 } { 54 } \left( p \sqrt { 3 } + q \pi + r \sqrt { 3 } \pi ^ { 2 } \right)$ where $p$, $q$ and $r$ are integers to be determined.\\
(5)
\hfill \mbox{\textit{Edexcel CP2 2021 Q5 [8]}}