| Exam Board | Edexcel |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Product & Quotient Rules |
| Type | Find stationary points coordinates |
| Difficulty | Standard +0.3 This is a straightforward application of product and quotient rules with standard techniques. Part (i)(a) is a routine 'show that' differentiation using the product rule. Part (i)(b) requires setting the derivative to zero and solving—a standard stationary point problem. Part (ii) applies the quotient rule and shows the derivative is never zero, which is algebraically simple. All steps are textbook exercises with no novel insight required, making this slightly easier than average. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks |
|---|---|
| (a) (i) Applies \(vu' + uv'\) to \(x^2 \ln x\) | M1 |
| \(= \ln x \times \frac{1}{2}x^{-\frac{1}{2}} + x^{\frac{1}{2}} \times \frac{1}{x}\) | A1 |
| \(= \frac{\ln x}{2\sqrt{x}} + \frac{1}{\sqrt{x}}\) | A1 |
| (ii) Sets \(\frac{\ln x}{2\sqrt{x}} + \frac{1}{\sqrt{x}} = 0\) | M1 |
| Multiplies (or factorises) by \(\sqrt{x}\), with correct \(\ln\) work leading to \(x = e^{-2}\) | M1 |
| \(P = (e^{-2}, -2e^{-1})\) or equivalent | A1 |
| (b) Applies \(\frac{vu' - uv'}{v^2}\) to \(y = \frac{x - k}{x + k}\) with \(u = x - k\) and \(v = x + k\) | M1 |
| \(\frac{dy}{dx} = \frac{(x + k) \times 1 - (x - k) \times 1}{(x + k)^2}\) | M1 |
| \(\frac{dy}{dx} = \frac{2k}{(x + k)^2}\) | A1, A1 |
| As \(k > 0 \Rightarrow \frac{dy}{dx} > 0 \Rightarrow C\) has no turning points | B1 |
(a) (i) Applies $vu' + uv'$ to $x^2 \ln x$ | M1
$= \ln x \times \frac{1}{2}x^{-\frac{1}{2}} + x^{\frac{1}{2}} \times \frac{1}{x}$ | A1
$= \frac{\ln x}{2\sqrt{x}} + \frac{1}{\sqrt{x}}$ | A1
(ii) Sets $\frac{\ln x}{2\sqrt{x}} + \frac{1}{\sqrt{x}} = 0$ | M1
Multiplies (or factorises) by $\sqrt{x}$, with correct $\ln$ work leading to $x = e^{-2}$ | M1
$P = (e^{-2}, -2e^{-1})$ or equivalent | A1
(b) Applies $\frac{vu' - uv'}{v^2}$ to $y = \frac{x - k}{x + k}$ with $u = x - k$ and $v = x + k$ | M1
$\frac{dy}{dx} = \frac{(x + k) \times 1 - (x - k) \times 1}{(x + k)^2}$ | M1
$\frac{dy}{dx} = \frac{2k}{(x + k)^2}$ | A1, A1
As $k > 0 \Rightarrow \frac{dy}{dx} > 0 \Rightarrow C$ has no turning points | B1
---\begin{enumerate}
\item (i) (a) Show that $\frac { \mathrm { d } } { \mathrm { d } x } \left( x ^ { \frac { 1 } { 2 } } \ln x \right) = \frac { \ln x } { 2 \sqrt { } x } + \frac { 1 } { \sqrt { } x }$
\end{enumerate}
The curve with equation $y = x ^ { \frac { 1 } { 2 } } \ln x , x > 0$ has one turning point at the point $P$.\\
(b) Find the exact coordinates of $P$. Give your answer in its simplest form.\\
(ii) A curve $C$ has equation $y = \frac { x - k } { x + k }$, where $k$ is a positive constant. Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$, and show that $C$ has no turning points.\\
\hfill \mbox{\textit{Edexcel C3 2013 Q5 [11]}}