| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2006 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Find stationary points - logarithmic functions |
| Difficulty | Standard +0.3 This is a structured multi-part question covering standard C3 techniques: finding x-intercepts of logarithmic functions, stationary points via differentiation, showing perpendicular tangents, and integration by parts with a guided formula. Each part is routine with clear methods, though part (iv) requires careful application of the given integral result. Slightly easier than average due to the scaffolding and standard techniques. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07n Stationary points: find maxima, minima using derivatives1.08d Evaluate definite integrals: between limits1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (i) \(2x - x\ln x = 0 \Rightarrow x(2-\ln x)=0\) | M1 | Equating to zero |
| \((x=0)\) or \(\ln x = 2 \Rightarrow\) at A, \(x = e^2\) | A1 | |
| (ii) \(\frac{dy}{dx} = 2 - x\cdot\frac{1}{x} - \ln x \cdot 1 = 1 - \ln x\) | M1, B1, A1 | Product rule; \(\frac{d}{dx}(\ln x) = \frac{1}{x}\); \(1-\ln x\) o.e. |
| \(\frac{dy}{dx} = 0 \Rightarrow 1-\ln x = 0 \Rightarrow \ln x = 1,\ x = e\) | M1, A1cao | Equating their derivative to zero |
| When \(x=e\): \(y = 2e - e\ln e = e\), so B is \((e, e)\) | B1ft | \(y = e\) |
| (iii) At A: \(\frac{dy}{dx} = 1 - \ln e^2 = 1-2 = -1\) | M1, A1cao | Substituting \(x=e^2\) (or their value) into derivative; \(-1\) and \(1\) |
| At C: \(\frac{dy}{dx} = 1 - \ln 1 = 1\) | ||
| \(1 \times -1 = -1 \Rightarrow\) tangents are perpendicular | E1 | www |
| (iv) Let \(u = \ln x\), \(\frac{dv}{dx} = x \Rightarrow v = \frac{1}{2}x^2\) | M1 | Parts: \(u=\ln x\), \(\frac{dv}{dx}=x \Rightarrow v=\frac{1}{2}x^2\) |
| \(\int x\ln x\,dx = \frac{1}{2}x^2\ln x - \int\frac{1}{2}x^2\cdot\frac{1}{x}\,dx = \frac{1}{2}x^2\ln x - \frac{1}{2}\int x\,dx\) | A1 | |
| \(= \frac{1}{2}x^2\ln x - \frac{1}{4}x^2 + c\) * | E1 | |
| \(A = \int_1^{e^2}(2x - x\ln x)\,dx = \left[x^2 - \frac{1}{2}x^2\ln x + \frac{1}{4}x^2\right]_1^{e^2}\) | B1, B1 | Correct integral and limits; \(\left[x^2 - \frac{1}{2}x^2\ln x + \frac{1}{4}x^2\right]\) o.e. |
| \(= \left(e^4 - \frac{1}{2}e^4\ln e^2 + \frac{1}{4}e^4\right) - \left(1 - \frac{1}{2}\cdot1^2\ln 1 + \frac{1}{4}\cdot1^2\right)\) | M1 | Substituting limits correctly |
| \(= \frac{3}{4}e^4 - \frac{5}{4}\) | A1cao |
## Question 7:
| Answer/Working | Marks | Guidance |
|---|---|---|
| **(i)** $2x - x\ln x = 0 \Rightarrow x(2-\ln x)=0$ | M1 | Equating to zero |
| $(x=0)$ or $\ln x = 2 \Rightarrow$ at A, $x = e^2$ | A1 | |
| **(ii)** $\frac{dy}{dx} = 2 - x\cdot\frac{1}{x} - \ln x \cdot 1 = 1 - \ln x$ | M1, B1, A1 | Product rule; $\frac{d}{dx}(\ln x) = \frac{1}{x}$; $1-\ln x$ o.e. |
| $\frac{dy}{dx} = 0 \Rightarrow 1-\ln x = 0 \Rightarrow \ln x = 1,\ x = e$ | M1, A1cao | Equating their derivative to zero |
| When $x=e$: $y = 2e - e\ln e = e$, so B is $(e, e)$ | B1ft | $y = e$ |
| **(iii)** At A: $\frac{dy}{dx} = 1 - \ln e^2 = 1-2 = -1$ | M1, A1cao | Substituting $x=e^2$ (or their value) into derivative; $-1$ and $1$ |
| At C: $\frac{dy}{dx} = 1 - \ln 1 = 1$ | | |
| $1 \times -1 = -1 \Rightarrow$ tangents are perpendicular | E1 | www |
| **(iv)** Let $u = \ln x$, $\frac{dv}{dx} = x \Rightarrow v = \frac{1}{2}x^2$ | M1 | Parts: $u=\ln x$, $\frac{dv}{dx}=x \Rightarrow v=\frac{1}{2}x^2$ |
| $\int x\ln x\,dx = \frac{1}{2}x^2\ln x - \int\frac{1}{2}x^2\cdot\frac{1}{x}\,dx = \frac{1}{2}x^2\ln x - \frac{1}{2}\int x\,dx$ | A1 | |
| $= \frac{1}{2}x^2\ln x - \frac{1}{4}x^2 + c$ * | E1 | |
| $A = \int_1^{e^2}(2x - x\ln x)\,dx = \left[x^2 - \frac{1}{2}x^2\ln x + \frac{1}{4}x^2\right]_1^{e^2}$ | B1, B1 | Correct integral and limits; $\left[x^2 - \frac{1}{2}x^2\ln x + \frac{1}{4}x^2\right]$ o.e. |
| $= \left(e^4 - \frac{1}{2}e^4\ln e^2 + \frac{1}{4}e^4\right) - \left(1 - \frac{1}{2}\cdot1^2\ln 1 + \frac{1}{4}\cdot1^2\right)$ | M1 | Substituting limits correctly |
| $= \frac{3}{4}e^4 - \frac{5}{4}$ | A1cao | |
---7 Fig. 7 shows the curve
$$y = 2 x - x \ln x , \text { where } x > 0 .$$
The curve crosses the $x$-axis at A , and has a turning point at B . The point C on the curve has $x$-coordinate 1 . Lines CD and BE are drawn parallel to the $y$-axis.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6a4c3f3b-a298-4b13-b97e-b52f8d9d527b-5_531_1262_671_536}
\captionsetup{labelformat=empty}
\caption{Fig. 7}
\end{center}
\end{figure}
(i) Find the $x$-coordinate of A , giving your answer in terms of e .\\
(ii) Find the exact coordinates of B .\\
(iii) Show that the tangents at A and C are perpendicular to each other.\\
(iv) Using integration by parts, show that
$$\int x \ln x \mathrm {~d} x = \frac { 1 } { 2 } x ^ { 2 } \ln x - \frac { 1 } { 4 } x ^ { 2 } + c$$
Hence find the exact area of the region enclosed by the curve, the $x$-axis and the lines CD and BE .
\section*{[Question 8 is printed overleaf.]}
\hfill \mbox{\textit{OCR MEI C3 2006 Q7 [18]}}