| Exam Board | Edexcel |
|---|---|
| Module | P2 (Pure Mathematics 2) |
| Year | 2022 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Tangents, normals and gradients |
| Type | Determine nature of stationary points |
| Difficulty | Easy -1.2 This is a straightforward differentiation question requiring standard application of power rule for fractional indices, solving a simple equation for stationary points, and using second derivative test. All techniques are routine P2 content with no problem-solving insight needed, making it easier than average but not trivial due to the multi-part structure and fractional powers. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.07p Points of inflection: using second derivative |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = 27x^{\frac{1}{2}} - x^{\frac{3}{2}} - 20 \Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{27}{2}x^{-\frac{1}{2}} - \frac{3}{2}x^{\frac{1}{2}}\) | M1 | Power decreased by 1 in at least one term in \(x\) |
| \(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{27}{2}x^{-\frac{1}{2}} - \frac{3}{2}x^{\frac{1}{2}}\) | A1 | Correct simplified derivative; accept \(\frac{27}{2\sqrt{x}} - \frac{3}{2}\sqrt{x}\) or decimal equivalents; no need to see \(\frac{\mathrm{d}y}{\mathrm{d}x}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{27}{2}x^{-\frac{1}{2}} - \frac{3}{2}x^{\frac{1}{2}} = 0 \Rightarrow x = \ldots\) | dM1 | Sets their derivative \(= 0\), attempts to solve for \(x\); depends on M in (a); "=0" may be implied by clear attempt to solve |
| \(x = 9\) | A1 | Correct \(x\) value; ignore \(x = 0\) or negative values but A0 if extra positive values |
| \(x = 9 \Rightarrow y = \ldots\) | M1 | Substitutes their positive \(x\) value into the curve equation to find \(y\) |
| \(y = 34\) | A1 | Correct \(y\) value; ignore reference to any point at \(x = 0\) or negative values but A0 if other coordinates with positive \(x\) given |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \pm Ax^{-\frac{3}{2}} \pm Bx^{-\frac{1}{2}}\); \(\left(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)_{x=9} = -\frac{27}{4}(9)^{-\frac{3}{2}} - \frac{3}{4}(9)^{-\frac{1}{2}}\left(= -\frac{1}{2}\right)\) | M1 | Attempts second derivative achieving form \(\pm Ax^{-\frac{3}{2}} \pm Bx^{-\frac{1}{2}}\) and substitutes their \(x\) or considers sign (for \(x > 0\)) |
| \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -\frac{27}{4}x^{-\frac{3}{2}} - \frac{3}{4}x^{-\frac{1}{2}}\); \(\frac{\mathrm{d}^2y}{\mathrm{d}x^2} < 0\) for \(x > 0\) when \(x = 9\), so maximum | A1 | Correct second derivative and conclusion with correct reason; accept evaluation to \(-\frac{1}{2}\), or correct sign deduced from substitution; accept "concave down" for maximum but not just "concave" |
## Question 2:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = 27x^{\frac{1}{2}} - x^{\frac{3}{2}} - 20 \Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{27}{2}x^{-\frac{1}{2}} - \frac{3}{2}x^{\frac{1}{2}}$ | M1 | Power decreased by 1 in at least one term in $x$ |
| $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{27}{2}x^{-\frac{1}{2}} - \frac{3}{2}x^{\frac{1}{2}}$ | A1 | Correct simplified derivative; accept $\frac{27}{2\sqrt{x}} - \frac{3}{2}\sqrt{x}$ or decimal equivalents; no need to see $\frac{\mathrm{d}y}{\mathrm{d}x}$ |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{27}{2}x^{-\frac{1}{2}} - \frac{3}{2}x^{\frac{1}{2}} = 0 \Rightarrow x = \ldots$ | dM1 | Sets their derivative $= 0$, attempts to solve for $x$; depends on M in (a); "=0" may be implied by clear attempt to solve |
| $x = 9$ | A1 | Correct $x$ value; ignore $x = 0$ or negative values but A0 if extra positive values |
| $x = 9 \Rightarrow y = \ldots$ | M1 | Substitutes their positive $x$ value into the curve equation to find $y$ |
| $y = 34$ | A1 | Correct $y$ value; ignore reference to any point at $x = 0$ or negative values but A0 if other coordinates with positive $x$ given |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \pm Ax^{-\frac{3}{2}} \pm Bx^{-\frac{1}{2}}$; $\left(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\right)_{x=9} = -\frac{27}{4}(9)^{-\frac{3}{2}} - \frac{3}{4}(9)^{-\frac{1}{2}}\left(= -\frac{1}{2}\right)$ | M1 | Attempts second derivative achieving form $\pm Ax^{-\frac{3}{2}} \pm Bx^{-\frac{1}{2}}$ and substitutes their $x$ or considers sign (for $x > 0$) |
| $\frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -\frac{27}{4}x^{-\frac{3}{2}} - \frac{3}{4}x^{-\frac{1}{2}}$; $\frac{\mathrm{d}^2y}{\mathrm{d}x^2} < 0$ for $x > 0$ when $x = 9$, so **maximum** | A1 | Correct second derivative and conclusion with correct reason; accept evaluation to $-\frac{1}{2}$, or correct sign deduced from substitution; accept "concave down" for maximum but not just "concave" |
---2. In this question you must show all stages of your working.
\section*{Solutions relying entirely on calculator technology are not acceptable.}
The curve $C$ has equation
$$y = 27 x ^ { \frac { 1 } { 2 } } - x ^ { \frac { 3 } { 2 } } - 20 \quad x > 0$$
\begin{enumerate}[label=(\alph*)]
\item Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$, giving each term in simplest form.
\item Hence find the coordinates of the stationary point of $C$.
\item Find $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$ and hence determine the nature of the stationary point of $C$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P2 2022 Q2 [8]}}