| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Year | 2012 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Function Transformations |
| Type | Forward transformation (single point, multiple transformations) |
| Difficulty | Standard +0.3 This is a multi-part question covering standard C3 transformation techniques (part i), routine differentiation using quotient rule (part ii), algebraic manipulation to verify an identity (part iii), and integration with substitution (part iv). While it has multiple parts worth several marks, each component uses well-practiced techniques without requiring novel insight. The transformation questions are straightforward applications of rules, and the final area calculation, though requiring careful setup, follows a standard approach. This is slightly easier than average due to the guided structure and routine nature of each step. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.07j Differentiate exponentials: e^(kx) and a^(kx)1.08b Integrate x^n: where n != -1 and sums1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((A)\) \((0,6)\) and \((1,4)\) | B1B1 | Condone P and Q incorrectly labelled (or unlabelled) |
| \((B)\) \((-1,5)\) and \((0,4)\) | B1B1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f'(x)=\frac{(x+1)\cdot 2x-(x^2+3)\cdot 1}{(x+1)^2}\) | M1 | Quotient or product rule consistent with their derivatives, condone missing brackets; PR: \((x^2+3)(-1)(x+1)^{-2}+2x(x+1)^{-1}\); if formula stated correctly, allow one substitution error |
| A1 | correct expression; condone missing brackets if subsequent working implies they are intended | |
| \(f'(x)=0\Rightarrow 2x(x+1)-(x^2+3)=0\) | M1 | their derivative \(=0\) |
| \(\Rightarrow x^2+2x-3=0\) | A1dep | obtaining correct quadratic equation (soi); dep 1st M1 but withhold if denominator also set to zero |
| \(\Rightarrow (x-1)(x+3)=0\) | ||
| \(\Rightarrow x=1\) or \(x=-3\) | ||
| When \(x=-3\), \(y=12/(-2)=-6\) | Some candidates get \(x^2+2x+3\), then realise this should be \(x^2+2x-3\), and correct back, but not for every occurrence. Treat sympathetically. | |
| so other TP is \((-3,-6)\) | B1B1cao [6] | must be supported (but see note re quadratic); \(-3\) could be verified by substitution into correct derivative |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(f(x-1)=\frac{(x-1)^2+3}{x-1+1}\) | M1 | substituting \(x-1\) for both \(x\)'s in \(f\); allow 1 slip for M1 |
| \(=\frac{x^2-2x+1+3}{x-1+1}\) | A1 | |
| \(=\frac{x^2-2x+4}{x}=x-2+\frac{4}{x}\)* | A1 NB AG [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\int_a^b\!\left(x-2+\frac{4}{x}\right)dx=\left[\tfrac{1}{2}x^2-2x+4\ln x\right]_a^b\) | B1 | \(\left[\tfrac{1}{2}x^2-2x+4\ln x\right]\) |
| M1 | \(F(b)-F(a)\); condone missing brackets (oe, mark final answer); F must show evidence of integration of at least one term | |
| \(=\left(\tfrac{1}{2}b^2-2b+4\ln b\right)-\left(\tfrac{1}{2}a^2-2a+4\ln a\right)\) | A1 | |
| Area is \(\int_0^1 f(x)\,dx\); so taking \(a=1\) and \(b=2\) | M1 | or \(f(x)=x+1-2+4/(x+1)\) |
| area \(=(2-4+4\ln 2)-(\tfrac{1}{2}-2+4\ln 1)\) | \(A=\int_0^1 f(x)\,dx=\left[\tfrac{1}{2}x^2-x+4\ln(1+x)\right]_0^1\) M1 | |
| \(=4\ln 2-\tfrac{1}{2}\) | A1cao [5] | must be simplified with \(\ln 1=0\); \(=\tfrac{1}{2}-1+4\ln 2=4\ln 2-\tfrac{1}{2}\) A1 |
## Question 9(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(A)$ $(0,6)$ and $(1,4)$ | B1B1 | Condone P and Q incorrectly labelled (or unlabelled) |
| $(B)$ $(-1,5)$ and $(0,4)$ | B1B1 **[4]** | |
---
## Question 9(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f'(x)=\frac{(x+1)\cdot 2x-(x^2+3)\cdot 1}{(x+1)^2}$ | M1 | Quotient or product rule consistent with their derivatives, condone missing brackets; PR: $(x^2+3)(-1)(x+1)^{-2}+2x(x+1)^{-1}$; if formula stated correctly, allow one substitution error |
| | A1 | correct expression; condone missing brackets if subsequent working implies they are intended |
| $f'(x)=0\Rightarrow 2x(x+1)-(x^2+3)=0$ | M1 | their derivative $=0$ |
| $\Rightarrow x^2+2x-3=0$ | A1dep | obtaining correct quadratic equation (soi); dep 1st M1 but withhold if denominator also set to zero |
| $\Rightarrow (x-1)(x+3)=0$ | | |
| $\Rightarrow x=1$ or $x=-3$ | | |
| When $x=-3$, $y=12/(-2)=-6$ | | Some candidates get $x^2+2x+3$, then realise this should be $x^2+2x-3$, and correct back, but not for every occurrence. Treat sympathetically. |
| so other TP is $(-3,-6)$ | B1B1cao **[6]** | must be supported (but see note re quadratic); $-3$ could be verified by substitution into correct derivative |
---
## Question 9(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $f(x-1)=\frac{(x-1)^2+3}{x-1+1}$ | M1 | substituting $x-1$ for both $x$'s in $f$; allow 1 slip for M1 |
| $=\frac{x^2-2x+1+3}{x-1+1}$ | A1 | |
| $=\frac{x^2-2x+4}{x}=x-2+\frac{4}{x}$* | A1 **NB AG** **[3]** | |
---
## Question 9(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int_a^b\!\left(x-2+\frac{4}{x}\right)dx=\left[\tfrac{1}{2}x^2-2x+4\ln x\right]_a^b$ | B1 | $\left[\tfrac{1}{2}x^2-2x+4\ln x\right]$ |
| | M1 | $F(b)-F(a)$; condone missing brackets (oe, mark final answer); F must show evidence of integration of at least one term |
| $=\left(\tfrac{1}{2}b^2-2b+4\ln b\right)-\left(\tfrac{1}{2}a^2-2a+4\ln a\right)$ | A1 | |
| Area is $\int_0^1 f(x)\,dx$; so taking $a=1$ and $b=2$ | M1 | or $f(x)=x+1-2+4/(x+1)$ |
| area $=(2-4+4\ln 2)-(\tfrac{1}{2}-2+4\ln 1)$ | | $A=\int_0^1 f(x)\,dx=\left[\tfrac{1}{2}x^2-x+4\ln(1+x)\right]_0^1$ M1 |
| $=4\ln 2-\tfrac{1}{2}$ | A1cao **[5]** | must be simplified with $\ln 1=0$; $=\tfrac{1}{2}-1+4\ln 2=4\ln 2-\tfrac{1}{2}$ A1 |9 Fig. 9 shows the curve $y = \mathrm { f } ( x )$, which has a $y$-intercept at $\mathrm { P } ( 0,3 )$, a minimum point at $\mathrm { Q } ( 1,2 )$, and an asymptote $x = - 1$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{7b77c646-2bc5-4166-b22e-3c1229abd722-5_906_944_333_566}
\captionsetup{labelformat=empty}
\caption{Fig. 9}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Find the coordinates of the images of the points P and Q when the curve $y = \mathrm { f } ( x )$ is transformed to\\
(A) $y = 2 \mathrm { f } ( x )$,\\
(B) $y = \mathrm { f } ( x + 1 ) + 2$.
You are now given that $\mathrm { f } ( x ) = \frac { x ^ { 2 } + 3 } { x + 1 } , x \neq - 1$.
\item Find $\mathrm { f } ^ { \prime } ( x )$, and hence find the coordinates of the other turning point on the curve $y = \mathrm { f } ( x )$.
\item Show that $\mathrm { f } ( x - 1 ) = x - 2 + \frac { 4 } { x }$.
\item Find $\int _ { a } ^ { b } \left( x - 2 + \frac { 4 } { x } \right) \mathrm { d } x$ in terms of $a$ and $b$.
Hence, by choosing suitable values for $a$ and $b$, find the exact area enclosed by the curve $y = \mathrm { f } ( x )$, the $x$-axis, the $y$-axis and the line $x = 1$.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI C3 2012 Q9 [18]}}