| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2009 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differentiating Transcendental Functions |
| Type | Find gradient at a point - at special curve features |
| Difficulty | Standard +0.3 This is a standard C3 multi-part question requiring routine differentiation of ln x, finding gradients, and integration by parts. While part (iv) requires recognizing that d/dx(x ln x - x) helps integrate ln x, this is a common textbook technique. All steps are straightforward applications of standard methods with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations1.07n Stationary points: find maxima, minima using derivatives1.08e Area between curve and x-axis: using definite integrals1.08i Integration by parts |
| Answer | Marks | Guidance |
|---|---|---|
| Gradient of PR \(= \frac{(1 + 7/8)/1}{1} = 1 + \frac{7}{8}\) | B1, M1, A1 | [3] 1.9 or better |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dx} = 2x - \frac{1}{8x}\) | B1 | cao |
| When \(x = 1\), \(\frac{dy}{dx} = 2 - \frac{1}{8} = 1\frac{7}{8}\) | B1dep | 1.9 or better dep 1° B1 |
| Same as gradient of PR, so PR touches curve | E1 | [3] dep gradients exact |
| Answer | Marks | Guidance |
|---|---|---|
| Turning points when \(\frac{dy}{dx} = 0\) | M1 | setting their derivative to zero |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Rightarrow x = \frac{1}{4}\) (\(x > 0\)) | M1, A1 | multiplying through by \(x\); allow verification |
| When \(x = \frac{1}{4}\), \(y = \frac{1}{16} - \frac{1}{8}\ln\frac{1}{4} = -\frac{1}{16} + \frac{1}{8}\ln 4\) | M1 | substituting for \(x\) in \(y\) |
| So TP is \(\left(\frac{1}{4}, -\frac{1}{16} + \frac{1}{8}\ln 4\right)\) | A1cao | [5] o.e. but must be exact, not \(1/4^2\). Mark final answer. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{d}{dx}(x\ln x - x) = x \cdot \frac{1}{x} + 1 \cdot \ln x - 1 = \ln x\) | M1, A1 | product rule; \(\ln x\) |
| Answer | Marks | Guidance |
|---|---|---|
| Area \(= \int_{\frac{1}{4}}^{1}\left(x^2 - \frac{1}{8}\ln x\right)dx\) | M1, M1 | [7] correct integral and limits (soi) – condone no \(dx\); \(\int \ln x dx = x\ln x - x\) used (or derived using integration by parts) |
| \(= \left[\frac{1}{3}x^3 - \frac{1}{8}(x\ln x - x)\right]\) | A1 | \(\frac{1}{3}x^3 - \frac{1}{8}(x\ln x - x)\) – bracket required |
| \(= \frac{8}{3} \cdot \frac{1}{4}\ln 2 + \frac{1}{4} - \left(\frac{1}{3} \cdot \frac{1}{8}\ln 1 + \frac{1}{8}\right)\) | M1 | substituting correct limits |
| \(= \frac{7}{8} + \frac{1}{8}\ln 2\) | M1 | |
| \(= \frac{59}{24} - \frac{1}{4}\ln 2\) * | E1 | [7] must show at least one step |
## Part (i)
When $x = 1$, $y = 1^2 - (\ln 1)/8 = 1$
Gradient of PR $= \frac{(1 + 7/8)/1}{1} = 1 + \frac{7}{8}$ | B1, M1, A1 | [3] 1.9 or better
## Part (ii)
$\frac{dy}{dx} = 2x - \frac{1}{8x}$ | B1 | cao
When $x = 1$, $\frac{dy}{dx} = 2 - \frac{1}{8} = 1\frac{7}{8}$ | B1dep | 1.9 or better dep 1° B1
Same as gradient of PR, so PR touches curve | E1 | [3] dep gradients exact
## Part (iii)
Turning points when $\frac{dy}{dx} = 0$ | M1 | setting their derivative to zero
$\Rightarrow 2x - \frac{1}{8x} = 0$
$\Rightarrow 2x = \frac{1}{8x}$
$\Rightarrow x^2 = \frac{1}{16}$
$\Rightarrow x = \frac{1}{4}$ ($x > 0$) | M1, A1 | multiplying through by $x$; allow verification
When $x = \frac{1}{4}$, $y = \frac{1}{16} - \frac{1}{8}\ln\frac{1}{4} = -\frac{1}{16} + \frac{1}{8}\ln 4$ | M1 | substituting for $x$ in $y$
So TP is $\left(\frac{1}{4}, -\frac{1}{16} + \frac{1}{8}\ln 4\right)$ | A1cao | [5] o.e. but must be exact, not $1/4^2$. Mark final answer.
## Part (iv)
$\frac{d}{dx}(x\ln x - x) = x \cdot \frac{1}{x} + 1 \cdot \ln x - 1 = \ln x$ | M1, A1 | product rule; $\ln x$
---
# Question 8 (continued)
Area $= \int_{\frac{1}{4}}^{1}\left(x^2 - \frac{1}{8}\ln x\right)dx$ | M1, M1 | [7] correct integral and limits (soi) – condone no $dx$; $\int \ln x dx = x\ln x - x$ used (or derived using integration by parts)
$= \left[\frac{1}{3}x^3 - \frac{1}{8}(x\ln x - x)\right]$ | A1 | $\frac{1}{3}x^3 - \frac{1}{8}(x\ln x - x)$ – bracket required
$= \frac{8}{3} \cdot \frac{1}{4}\ln 2 + \frac{1}{4} - \left(\frac{1}{3} \cdot \frac{1}{8}\ln 1 + \frac{1}{8}\right)$ | M1 | substituting correct limits
$= \frac{7}{8} + \frac{1}{8}\ln 2$ | M1 |
$= \frac{59}{24} - \frac{1}{4}\ln 2$ * | E1 | [7] must show at least one step
---8 Fig. 8 shows the curve $y = x ^ { 2 } - \frac { 1 } { 8 } \ln x$. P is the point on this curve with $x$-coordinate 1 , and R is the point $\left( 0 , - \frac { 7 } { 8 } \right)$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{56672660-b7dc-4e10-8039-1c041e75b598-3_1022_995_479_575}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{center}
\end{figure}
(i) Find the gradient of PR.\\
(ii) Find $\frac { \mathrm { d } y } { \mathrm {~d} x }$. Hence show that PR is a tangent to the curve.\\
(iii) Find the exact coordinates of the turning point Q .\\
(iv) Differentiate $x \ln x - x$.
Hence, or otherwise, show that the area of the region enclosed by the curve $y = x ^ { 2 } - \frac { 1 } { 8 } \ln x$, the $x$-axis and the lines $x = 1$ and $x = 2$ is $\frac { 59 } { 24 } - \frac { 1 } { 4 } \ln 2$.
\hfill \mbox{\textit{OCR MEI C3 2009 Q8 [18]}}