| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Find coordinates after transformation |
| Difficulty | Standard +0.3 This is a multi-part question covering standard C3 transformations, differentiation using quotient rule, algebraic manipulation, and integration. While it has many parts (4 marks worth), each component is routine: transformations follow direct rules, the quotient rule is standard, the algebraic identity is straightforward to verify, and the integration/area calculation uses basic techniques. The question requires careful work but no novel insight or problem-solving beyond applying learned procedures. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.02y Partial fractions: decompose rational functions1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08e Area between curve and x-axis: using definite integrals |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | 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/Working | 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 |
| A1 | Correct expression | |
| \(f'(x)=0 \Rightarrow 2x(x+1)-(x^2+3)=0\) | M1 | Their derivative \(= 0\), obtaining correct quadratic equation (soi) set to zero. dep 1st M1 but withhold if denominator also set to zero |
| \(\Rightarrow x^2+2x-3=0 \Rightarrow (x-1)(x+3)=0\) | ||
| \(\Rightarrow x=1\) or \(x=-3\) | ||
| When \(x=-3\), \(y=12/(-2)=-6\), so other TP is \((-3,-6)\) | B1B1cao | Must be from correct work (but see note re quadratic) |
| [6] | PR: \((x^2+3)(-1)(x+1)^{-2}+2x(x+1)^{-1}\). If formula stated correctly, allow one substitution error. 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. Must be supported, but \(-3\) could be verified by substitution into correct derivative. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | 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 |
| \(= \frac{x^2-2x+1+3}{x}\) | A1 | Correct expression |
| \(= \frac{x^2-2x+4}{x} = x - 2 + \frac{4}{x}\) | A1 | NB AG |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\int_a^b \left(x-2+\frac{4}{x}\right)dx = \left[\frac{1}{2}x^2-2x+4\ln x\right]_a^b\) | B1 | \(\left[\frac{1}{2}x^2-2x+4\ln x\right]\) |
| \(= \left(\frac{1}{2}b^2-2b+4\ln b\right)-\left(\frac{1}{2}a^2-2a+4\ln a\right)\) | M1 | \(F(b)-F(a)\), condone missing brackets, o.e. (mark final answer). F must show evidence of integration of at least one term |
| A1 | ||
| Area is \(\int_0^1 f(x)\,dx\), so taking \(a=1\) and \(b=2\): | M1 | |
| \(\text{area} = (2-4+4\ln 2)-(\frac{1}{2}-2+4\ln 1)\) | ||
| \(= 4\ln 2 - \frac{1}{2}\) | A1cao | Must be simplified with \(\ln 1 = 0\) |
| [5] | or \(f(x)=x+1-2+4/(x+1)\); \(A=\int_0^1 f(x)\,dx = \left[\frac{1}{2}x^2-x+4\ln(1+x)\right]_0^1\) M1 \(= \frac{1}{2}-1+4\ln 2 = 4\ln 2 - \frac{1}{2}\) A1 |
# Question 3:
## Part (i)
| Answer/Working | 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]** | | |
## Part (ii)
| Answer/Working | 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 |
| | A1 | Correct expression |
| $f'(x)=0 \Rightarrow 2x(x+1)-(x^2+3)=0$ | M1 | Their derivative $= 0$, obtaining correct quadratic equation (soi) set to zero. dep 1st M1 but withhold if denominator also set to zero |
| $\Rightarrow x^2+2x-3=0 \Rightarrow (x-1)(x+3)=0$ | | |
| $\Rightarrow x=1$ or $x=-3$ | | |
| When $x=-3$, $y=12/(-2)=-6$, so other TP is $(-3,-6)$ | B1B1cao | Must be from correct work (but see note re quadratic) |
| **[6]** | | PR: $(x^2+3)(-1)(x+1)^{-2}+2x(x+1)^{-1}$. If formula stated correctly, allow one substitution error. 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. Must be supported, but $-3$ could be verified by substitution into correct derivative. |
## Part (iii)
| Answer/Working | 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 |
| $= \frac{x^2-2x+1+3}{x}$ | A1 | Correct expression |
| $= \frac{x^2-2x+4}{x} = x - 2 + \frac{4}{x}$ | A1 | **NB AG** |
| **[3]** | | |
## Part (iv)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\int_a^b \left(x-2+\frac{4}{x}\right)dx = \left[\frac{1}{2}x^2-2x+4\ln x\right]_a^b$ | B1 | $\left[\frac{1}{2}x^2-2x+4\ln x\right]$ |
| $= \left(\frac{1}{2}b^2-2b+4\ln b\right)-\left(\frac{1}{2}a^2-2a+4\ln a\right)$ | M1 | $F(b)-F(a)$, condone missing brackets, o.e. (mark final answer). F must show evidence of integration of at least one term |
| | A1 | |
| Area is $\int_0^1 f(x)\,dx$, so taking $a=1$ and $b=2$: | M1 | |
| $\text{area} = (2-4+4\ln 2)-(\frac{1}{2}-2+4\ln 1)$ | | |
| $= 4\ln 2 - \frac{1}{2}$ | A1cao | Must be simplified with $\ln 1 = 0$ |
| **[5]** | | or $f(x)=x+1-2+4/(x+1)$; $A=\int_0^1 f(x)\,dx = \left[\frac{1}{2}x^2-x+4\ln(1+x)\right]_0^1$ M1 $= \frac{1}{2}-1+4\ln 2 = 4\ln 2 - \frac{1}{2}$ A1 |
---3 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]{0b4c4935-998c-404f-8fed-9b39b849168e-3_904_937_425_589}
\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 Q3 [18]}}