| 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 multi-part question covering standard C3 techniques: differentiation using quotient rule, finding stationary points, verifying with second derivative, solving curve-line intersections, and computing area between curves. Part (iv) requires geometric insight about transformations and areas, which elevates it slightly above routine, but overall this is a straightforward application of well-practiced methods with clear scaffolding. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.07e Second derivative: as rate of change of gradient1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f'(x) = \frac{x \cdot 2(x-2) - (x-2)^2}{x^2}\) | M1 | Quotient (or product) rule, condone sign errors only. e.g. \(\frac{\pm x \cdot 2(x-2) \pm (x-2)^2}{x^2}\); PR: \((x-2)^2 \cdot (-x^{-2}) + (1/x) \cdot 2(x-2)\) |
| Correct expression | A1 | Correct exp, condone missing brackets here |
| \(= \frac{2x^2 - 4x - x^2 + 4x - 4}{x^2}\) | ||
| \(= (x^2-4)/x^2 = 1 - 4/x^2\) * | A1 | Simplified correctly NB AG; with correct use of brackets |
| or \(f(x) = (x^2 - 4x + 4)/x\) | ||
| \(= x - 4 + 4/x\) | M1 | Expanding bracket and dividing each term by \(x\); must be 3 terms: \((x^2-4)/x\) is M0 |
| Correctly simplified | A1 | e.g. \(x - 4 + 2/x\) is M1A0 |
| \(\Rightarrow f'(x) = 1 - 4/x^2\) * | A1 | Not from wrong working NB AG |
| \(f''(x) = 8/x^3\) | B1 | o.e. e.g. \(8x^{-3}\) or \(8x/x^4\) |
| \(f'(x) = 0\) when \(x^2 = 4\), \(x = \pm 2\) | M1 | \(x = \pm 2\) found from \(1 - 4/x^2 = 0\); allow for \(x = -2\) unsupported |
| So at Q, \(x = -2\), \(y = -8\) | A1 | \((-2, -8)\) |
| \(f''(-2)\ [= -1] < 0\) so maximum | B1dep [7] | Dep first B1. Can omit \(-1\), but if shown must be correct. Must state \(< 0\) or negative; must use 2nd derivative test |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f(1) = (-1)^2/1 = 1\); \(f(4) = (2)^2/4 = 1\) | B1 | Verifying \(f(1) = 1\) and \(f(4) = 1\); or \((x-2)^2 = x \Rightarrow x^2 - 5x + 4 = 0 \Rightarrow (x-1)(x-4) = 0\), \(x = 1, 4\) |
| \(\int_1^4 \frac{(x-2)^2}{x}\,dx = \int_1^4 (x - 4 + 4/x)\,dx\) | M1 | Expanding bracket and dividing each term by \(x\), 3 terms: \(x - 4/x\) is M0; if \(u = x-2\): \(\int \frac{u^2}{u+2}\,du = \int(u - 2 + \frac{4}{u+2})\,du\) |
| \(= \left[x^2/2 - 4x + 4\ln x\right]_1^4\) | A1 | \(x^2/2 - 4x + 4\ln x\); \(u^2/2 - 2u + 4\ln(u+2)\) |
| \(= (8 - 16 + 4\ln 4) - (\tfrac{1}{2} - 4 + 4\ln 1)\) | ||
| \(= 4\ln 4 - 4\tfrac{1}{2}\) | A1cao | |
| Area enclosed = rectangle \(-\) curve | M1 | soi |
| \(= 3 \times 1 - (4\ln 4 - 4\tfrac{1}{2}) = 7\tfrac{1}{2} - 4\ln 4\) | A1cao | o.e. but must combine numerical terms and evaluate \(\ln 1\); mark final answer |
| or 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 \(x\); must be 3 terms in \((x-2)^2\) expansion |
| A1 | \(5 - x - 4/x\) | |
| \(= \left[5x - x^2/2 - 4\ln x\right]_1^4\) | A1 | \(5x - x^2/2 - 4\ln x\) |
| \(= 20 - 8 - 4\ln 4 - (5 - \tfrac{1}{2} - 4\ln 1)\) | ||
| \(= 7\tfrac{1}{2} - 4\ln 4\) | A1cao [6] | o.e. but must combine numerical terms and evaluate \(\ln 1\); mark final answer |
| 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 [3] | Correctly simplified – not from wrong working NB AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Area is the same as that found in part (ii) | M1 | Award M1 for \(\pm\) ans to 2(ii) (unless zero) |
| \(4\ln 4 - 7\tfrac{1}{2}\) | A1cao [2] | Need not justify the change of sign |
## Question 2(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f'(x) = \frac{x \cdot 2(x-2) - (x-2)^2}{x^2}$ | M1 | Quotient (or product) rule, condone sign errors only. e.g. $\frac{\pm x \cdot 2(x-2) \pm (x-2)^2}{x^2}$; PR: $(x-2)^2 \cdot (-x^{-2}) + (1/x) \cdot 2(x-2)$ |
| Correct expression | A1 | Correct exp, condone missing brackets here |
| $= \frac{2x^2 - 4x - x^2 + 4x - 4}{x^2}$ | | |
| $= (x^2-4)/x^2 = 1 - 4/x^2$ * | A1 | Simplified correctly **NB AG**; with correct use of brackets |
| **or** $f(x) = (x^2 - 4x + 4)/x$ | | |
| $= x - 4 + 4/x$ | M1 | Expanding bracket and dividing each term by $x$; must be 3 terms: $(x^2-4)/x$ is M0 |
| Correctly simplified | A1 | e.g. $x - 4 + 2/x$ is M1A0 |
| $\Rightarrow f'(x) = 1 - 4/x^2$ * | A1 | Not from wrong working **NB AG** |
| $f''(x) = 8/x^3$ | B1 | o.e. e.g. $8x^{-3}$ or $8x/x^4$ |
| $f'(x) = 0$ when $x^2 = 4$, $x = \pm 2$ | M1 | $x = \pm 2$ found from $1 - 4/x^2 = 0$; allow for $x = -2$ unsupported |
| So at Q, $x = -2$, $y = -8$ | A1 | $(-2, -8)$ |
| $f''(-2)\ [= -1] < 0$ so maximum | B1dep **[7]** | Dep first B1. Can omit $-1$, but if shown must be correct. Must state $< 0$ or negative; must use 2nd derivative test |
---
## Question 2(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(1) = (-1)^2/1 = 1$; $f(4) = (2)^2/4 = 1$ | B1 | Verifying $f(1) = 1$ and $f(4) = 1$; or $(x-2)^2 = x \Rightarrow x^2 - 5x + 4 = 0 \Rightarrow (x-1)(x-4) = 0$, $x = 1, 4$ |
| $\int_1^4 \frac{(x-2)^2}{x}\,dx = \int_1^4 (x - 4 + 4/x)\,dx$ | M1 | Expanding bracket and dividing each term by $x$, 3 terms: $x - 4/x$ is M0; if $u = x-2$: $\int \frac{u^2}{u+2}\,du = \int(u - 2 + \frac{4}{u+2})\,du$ |
| $= \left[x^2/2 - 4x + 4\ln x\right]_1^4$ | A1 | $x^2/2 - 4x + 4\ln x$; $u^2/2 - 2u + 4\ln(u+2)$ |
| $= (8 - 16 + 4\ln 4) - (\tfrac{1}{2} - 4 + 4\ln 1)$ | | |
| $= 4\ln 4 - 4\tfrac{1}{2}$ | A1cao | |
| Area enclosed = rectangle $-$ curve | M1 | soi |
| $= 3 \times 1 - (4\ln 4 - 4\tfrac{1}{2}) = 7\tfrac{1}{2} - 4\ln 4$ | A1cao | o.e. but must combine numerical terms and evaluate $\ln 1$; mark final answer |
| **or** 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 $x$; must be 3 terms in $(x-2)^2$ expansion |
| | A1 | $5 - x - 4/x$ |
| $= \left[5x - x^2/2 - 4\ln x\right]_1^4$ | A1 | $5x - x^2/2 - 4\ln x$ |
| $= 20 - 8 - 4\ln 4 - (5 - \tfrac{1}{2} - 4\ln 1)$ | | |
| $= 7\tfrac{1}{2} - 4\ln 4$ | A1cao **[6]** | o.e. but must combine numerical terms and evaluate $\ln 1$; mark final answer |
---
## Question 2(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 **[3]** | Correctly simplified – not from wrong working **NB AG** |
---
## Question 2(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Area is the same as that found in part (ii) | M1 | Award M1 for $\pm$ ans to 2(ii) (unless zero) |
| $4\ln 4 - 7\tfrac{1}{2}$ | A1cao **[2]** | Need not justify the change of sign |
---2 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]{6ea594c5-52ba-4467-a098-cb66004b5a38-1_959_1469_748_317}
\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 Q2 [18]}}