| Exam Board | Edexcel |
|---|---|
| Module | P3 (Pure Mathematics 3) |
| Year | 2023 |
| Session | October |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Show derivative equals given algebraic form |
| Difficulty | Standard +0.3 Part (a) is a standard quotient rule application with logarithmic differentiation—routine for P3 students. Parts (b) and (c) require solving dy/dx=0 and analyzing conditions for three turning points, which involves some algebraic manipulation and understanding of function behavior, but follows predictable patterns for this topic. Overall slightly easier than average due to the structured 'show that' format guiding students through the derivative. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{dy}{dx} = \dfrac{(x^2+k)\times\dfrac{2x}{x^2+k} - 2x\ln(x^2+k)}{(x^2+k)^2}\) | M1 | Attempts quotient rule, expression of form \(\dfrac{(x^2+k)\times\frac{...}{x^2+k} - Qx\ln(x^2+k)}{(x^2+k)^2}\), \(P,Q>0\) |
| \(\dfrac{dy}{dx} = \dfrac{2x - 2x\ln(x^2+k)}{(x^2+k)^2} = \dfrac{2x(1-\ln(x^2+k))}{(x^2+k)^2}\) | M1A1 | M1: cancels \(x^2+k\) and takes factor of \(2x\). A1: fully correct, both M marks scored |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = 0\) | B1 | |
| \(\text{"1"} - \ln(x^2+k) = 0 \Rightarrow x^2 = e^{\text{"1"}} \pm k\) | M1 | Sets numerator \(= 0\) and proceeds to \(x^2 = ...\) with correct undoing of ln |
| \(x = \pm\sqrt{e-k}\) | A1ft | Follow through \(x = \pm\sqrt{e^B - k}\) for numerical \(B\) in their \(B - \ln(x^2+k)\) factor |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Upper limit is \(e\) or \(k < e\) | B1ft | Follow through on (b) of form \(x = (\pm)\sqrt{e^B \pm k}\). B0 if \(k=e\) or \(k \leq e\) given as answer without statement that upper limit is \(e\) |
# Question 5:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{dy}{dx} = \dfrac{(x^2+k)\times\dfrac{2x}{x^2+k} - 2x\ln(x^2+k)}{(x^2+k)^2}$ | M1 | Attempts quotient rule, expression of form $\dfrac{(x^2+k)\times\frac{...}{x^2+k} - Qx\ln(x^2+k)}{(x^2+k)^2}$, $P,Q>0$ |
| $\dfrac{dy}{dx} = \dfrac{2x - 2x\ln(x^2+k)}{(x^2+k)^2} = \dfrac{2x(1-\ln(x^2+k))}{(x^2+k)^2}$ | M1A1 | M1: cancels $x^2+k$ and takes factor of $2x$. A1: fully correct, both M marks scored |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = 0$ | B1 | |
| $\text{"1"} - \ln(x^2+k) = 0 \Rightarrow x^2 = e^{\text{"1"}} \pm k$ | M1 | Sets numerator $= 0$ and proceeds to $x^2 = ...$ with correct undoing of ln |
| $x = \pm\sqrt{e-k}$ | A1ft | Follow through $x = \pm\sqrt{e^B - k}$ for numerical $B$ in their $B - \ln(x^2+k)$ factor |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Upper limit is $e$ or $k < e$ | B1ft | Follow through on (b) of form $x = (\pm)\sqrt{e^B \pm k}$. B0 if $k=e$ or $k \leq e$ given as answer without statement that upper limit is $e$ |\begin{enumerate}
\item The curve $C$ has equation
\end{enumerate}
$$y = \frac { \ln \left( x ^ { 2 } + k \right) } { x ^ { 2 } + k } \quad x \in \mathbb { R }$$
where $k$ is a positive constant.\\
(a) Show that
$$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { A x \left( B - \ln \left( x ^ { 2 } + k \right) \right) } { \left( x ^ { 2 } + k \right) ^ { 2 } }$$
where $A$ and $B$ are constants to be found.
Given that $C$ has exactly three turning points,\\
(b) find the $x$ coordinate of each of these points. Give your answer in terms of $k$ where appropriate.\\
(c) find the upper limit to the value for $k$.
\hfill \mbox{\textit{Edexcel P3 2023 Q5 [7]}}