| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Areas Between Curves |
| Type | Curve-Line Intersection Area |
| Difficulty | Standard +0.3 This is a structured multi-part question covering standard C3 techniques: differentiation using quotient rule, finding stationary points, curve-line intersection, and area between curves. Part (iv) requires geometric insight about translations, but the question explicitly guides students through each step. While it requires multiple techniques, all are routine applications with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.07d Second derivatives: d^2y/dx^2 notation1.07n Stationary points: find maxima, minima using derivatives1.07q Product and quotient rules: differentiation1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f(1) = (-1)^2/1 = 1\), \(f(4) = (2)^2/4 = 1\) | B1 | Verifying \(f(1) = 1\) and \(f(4) = 1\) |
| \(\int_1^4 \frac{(x-2)^2}{x}dx = \int_1^4 (x - 4 + 4/x)dx\) | M1 | Expanding bracket and dividing each term by \(x\); 3 terms: \(x - 4/x\) is M0 |
| \(= \left[x^2/2 - 4x + 4\ln x\right]_1^4\) | A1 | \(x^2/2 - 4x + 4\ln x\) |
| \(= (8 - 16 + 4\ln 4) - (\frac{1}{2} - 4 + 4\ln 1)\) | ||
| \(= 4\ln 4 - 4\frac{1}{2}\) | A1cao | |
| Area enclosed = rectangle \(-\) curve | M1 | soi |
| \(= 3\times1 - (4\ln 4 - 4\frac{1}{2}) = 7\frac{1}{2} - 4\ln 4\) | A1cao | o.e. but must combine numerical terms and evaluate \(\ln 1\); mark final answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Area \(= \int_1^4 \left[1 - \frac{(x-2)^2}{x}\right]dx\) | M1 | No need to have limits |
| \(= \int_1^4 (5 - x - 4/x)dx\) | M1 | Expanding bracket and dividing each term by \(x\) |
| \(5 - x - 4/x\) | A1 | Must be 3 terms in \((x-2)^2\) expansion |
| \(= \left[5x - x^2/2 - 4\ln x\right]_1^4\) | A1 | \(5x - x^2/2 - 4\ln x\) |
| \(= 20 - 8 - 4\ln 4 - (5 - \frac{1}{2} - 4\ln 1)\) | ||
| \(= 7\frac{1}{2} - 4\ln 4\) | A1cao | o.e. but must combine numerical terms and evaluate \(\ln 1\); mark final answer |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \([g(x) =]\ f(x+1) - 1\) | M1 | soi [may not be stated] |
| \(= \frac{(x+1-2)^2}{x+1} - 1\) | A1 | |
| \(= \frac{x^2 - 2x + 1 - x - 1}{x+1} = \frac{x^2 - 3x}{x+1}\) | A1 | Correctly simplified – not from wrong working. NB AG |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Area is the same as that found in part (ii) | M1 | Award M1 for \(\pm\) answer to 8(ii) (unless zero) |
| \(4\ln 4 - 7\frac{1}{2}\) | A1cao | Need not justify the change of sign |
| [2] |
# Question 1:
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(1) = (-1)^2/1 = 1$, $f(4) = (2)^2/4 = 1$ | B1 | Verifying $f(1) = 1$ and $f(4) = 1$ |
| $\int_1^4 \frac{(x-2)^2}{x}dx = \int_1^4 (x - 4 + 4/x)dx$ | M1 | Expanding bracket and dividing each term by $x$; 3 terms: $x - 4/x$ is M0 |
| $= \left[x^2/2 - 4x + 4\ln x\right]_1^4$ | A1 | $x^2/2 - 4x + 4\ln x$ |
| $= (8 - 16 + 4\ln 4) - (\frac{1}{2} - 4 + 4\ln 1)$ | | |
| $= 4\ln 4 - 4\frac{1}{2}$ | A1cao | |
| Area enclosed = rectangle $-$ curve | M1 | soi |
| $= 3\times1 - (4\ln 4 - 4\frac{1}{2}) = 7\frac{1}{2} - 4\ln 4$ | A1cao | o.e. but must combine numerical terms and evaluate $\ln 1$; mark final answer |
**OR**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Area $= \int_1^4 \left[1 - \frac{(x-2)^2}{x}\right]dx$ | M1 | No need to have limits |
| $= \int_1^4 (5 - x - 4/x)dx$ | M1 | Expanding bracket and dividing each term by $x$ |
| $5 - x - 4/x$ | A1 | Must be 3 terms in $(x-2)^2$ expansion |
| $= \left[5x - x^2/2 - 4\ln x\right]_1^4$ | A1 | $5x - x^2/2 - 4\ln x$ |
| $= 20 - 8 - 4\ln 4 - (5 - \frac{1}{2} - 4\ln 1)$ | | |
| $= 7\frac{1}{2} - 4\ln 4$ | A1cao | o.e. but must combine numerical terms and evaluate $\ln 1$; mark final answer |
| | **[6]** | |
---
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $[g(x) =]\ f(x+1) - 1$ | M1 | soi [may not be stated] |
| $= \frac{(x+1-2)^2}{x+1} - 1$ | A1 | |
| $= \frac{x^2 - 2x + 1 - x - 1}{x+1} = \frac{x^2 - 3x}{x+1}$ | A1 | Correctly simplified – not from wrong working. **NB AG** |
| | **[3]** | |
---
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Area is the same as that found in part (ii) | M1 | Award M1 for $\pm$ answer to 8(ii) (unless zero) |
| $4\ln 4 - 7\frac{1}{2}$ | A1cao | Need not justify the change of sign |
| | **[2]** | |
---1 Fig. 8 shows the line $y = 1$ and the curve $y = \mathrm { f } ( x )$, where $\mathrm { f } ( x ) = \frac { ( x - 2 ) ^ { 2 } } { x }$. The curve touches the $x$-axis at $\mathrm { P } ( 2,0 )$ and has another turning point at the point Q .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d1206ce8-7716-4205-b98e-664e7ead8a25-1_961_1473_445_320}
\captionsetup{labelformat=empty}
\caption{Fig. 8}
\end{center}
\end{figure}
(i) Show that $\mathrm { f } ^ { \prime } ( x ) = 1 - \frac { 4 } { x ^ { 2 } }$, and find $\mathrm { f } ^ { \prime \prime } ( x )$.
Hence find the coordinates of Q and, using $\mathrm { f } ^ { \prime \prime } ( x )$, verify that it is a maximum point.\\
(ii) Verify that the line $y = 1$ meets the curve $y = \mathrm { f } ( x )$ at the points with $x$-coordinates 1 and 4 . Hence find the exact area of the shaded region enclosed by the line and the curve.
The curve $y = \mathrm { f } ( x )$ is now transformed by a translation with vector $\binom { - 1 } { - 1 }$. The resulting curve has equation $y = \mathrm { g } ( x )$.\\
(iii) Show that $\mathrm { g } ( x ) = \frac { x ^ { 2 } - 3 x } { x + 1 }$.\\
(iv) Without further calculation, write down the value of $\int _ { 0 } ^ { 3 } \mathrm {~g} ( x ) \mathrm { d } x$, justifying your answer.
\hfill \mbox{\textit{OCR MEI C3 Q1 [18]}}