| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chain Rule |
| Type | Equation of tangent line |
| Difficulty | Standard +0.3 This is a multi-part question covering standard C3 techniques: finding gradients, differentiation (including chain rule for ln x), showing a line is tangent, finding turning points, and integration. Part (iv) provides a helpful hint for the integration. While it requires multiple steps and techniques, each individual part follows routine procedures without requiring novel insight or particularly challenging problem-solving. |
| Spec | 1.03a Straight lines: equation forms y=mx+c, ax+by+c=01.07l Derivative of ln(x): and related functions1.07n Stationary points: find maxima, minima using derivatives1.08c Integrate e^(kx), 1/x, sin(kx), cos(kx)1.08d Evaluate definite integrals: between limits1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| When \(x = 1\), \(y = 1^2 - \frac{\ln 1}{8} = 1\) | B1 | |
| Gradient of \(PR = \frac{1 + 7/8}{1} = 1\frac{7}{8}\) | M1 | |
| A1 [3] | 1.9 or better |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | 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 1st B1 |
| Same as gradient of \(PR\), so \(PR\) touches curve | E1 [3] | dep gradients exact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Turning points when \(\frac{dy}{dx} = 0 \Rightarrow 2x - \frac{1}{8x} = 0\) | M1 | setting their derivative to zero |
| \(\Rightarrow 2x = \frac{1}{8x} \Rightarrow x^2 = \frac{1}{16}\) | M1 | multiplying through by \(x\) |
| \(\Rightarrow x = \frac{1}{4}\) \((x > 0)\) | A1 | 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 |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{d}{dx}(x\ln x - x) = x \cdot \frac{1}{x} + 1\cdot\ln x - 1 = \ln x\) | M1 | product rule |
| A1 | \(\ln x\) | |
| Area \(= \int_1^2 \left(x^2 - \frac{1}{8}\ln x\right)dx\) | M1 | correct integral and limits (soi) – condone no \(dx\) |
| \(= \left[\frac{1}{3}x^3 - \frac{1}{8}(x\ln x - x)\right]_1^2\) | M1 | \(\int \ln x\, dx = x\ln x - x\) used (or derived using integration by parts) |
| A1 | \(\frac{1}{3}x^3 - \frac{1}{8}(x\ln x - x)\) – bracket required | |
| \(= \left(\frac{8}{3} - \frac{1}{4}\ln 2 + \frac{1}{4}\right) - \left(\frac{1}{3} - \frac{1}{8}\ln 1 + \frac{1}{8}\right)\) | M1 | substituting correct limits |
| \(= \frac{7}{3} + \frac{1}{8} - \frac{1}{4}\ln 2\) | ||
| \(= \frac{59}{24} - \frac{1}{4}\ln 2\) | E1 [7] | must show at least one step |
# Question 3:
## Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| When $x = 1$, $y = 1^2 - \frac{\ln 1}{8} = 1$ | B1 | |
| Gradient of $PR = \frac{1 + 7/8}{1} = 1\frac{7}{8}$ | M1 | |
| | A1 [3] | 1.9 or better |
## Part (ii):
| Answer/Working | 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 1st B1 |
| Same as gradient of $PR$, so $PR$ touches curve | E1 [3] | dep gradients exact |
## Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Turning points when $\frac{dy}{dx} = 0 \Rightarrow 2x - \frac{1}{8x} = 0$ | M1 | setting their derivative to zero |
| $\Rightarrow 2x = \frac{1}{8x} \Rightarrow x^2 = \frac{1}{16}$ | M1 | multiplying through by $x$ |
| $\Rightarrow x = \frac{1}{4}$ $(x > 0)$ | A1 | 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):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{d}{dx}(x\ln x - x) = x \cdot \frac{1}{x} + 1\cdot\ln x - 1 = \ln x$ | M1 | product rule |
| | A1 | $\ln x$ |
| Area $= \int_1^2 \left(x^2 - \frac{1}{8}\ln x\right)dx$ | M1 | correct integral and limits (soi) – condone no $dx$ |
| $= \left[\frac{1}{3}x^3 - \frac{1}{8}(x\ln x - x)\right]_1^2$ | M1 | $\int \ln x\, dx = x\ln x - x$ used (or derived using integration by parts) |
| | A1 | $\frac{1}{3}x^3 - \frac{1}{8}(x\ln x - x)$ – bracket required |
| $= \left(\frac{8}{3} - \frac{1}{4}\ln 2 + \frac{1}{4}\right) - \left(\frac{1}{3} - \frac{1}{8}\ln 1 + \frac{1}{8}\right)$ | M1 | substituting correct limits |
| $= \frac{7}{3} + \frac{1}{8} - \frac{1}{4}\ln 2$ | | |
| $= \frac{59}{24} - \frac{1}{4}\ln 2$ | E1 [7] | must show at least one step |
---3 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]{e3850377-bd1a-4e3c-8424-e3db7fd3c4db-2_1018_994_481_611}
\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 Q3 [18]}}