| Exam Board | OCR MEI |
|---|---|
| Module | C3 (Core Mathematics 3) |
| Marks | 23 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Function Transformations |
| Type | Stationary points after transformation |
| Difficulty | Standard +0.3 This is a structured multi-part question covering standard C3 transformations, differentiation using quotient rule, and integration. Part (i) requires routine application of transformation rules, part (ii) is standard differentiation and solving f'(x)=0, part (iii) is algebraic verification, and part (iv) involves straightforward integration with logarithms. While it has multiple parts (8 marks typical), each component uses well-practiced techniques without requiring novel insight or complex problem-solving, making it slightly easier than average. |
| Spec | 1.02w Graph transformations: simple transformations of f(x)1.07i Differentiate x^n: for rational n and sums1.07n Stationary points: find maxima, minima using derivatives1.08b 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: \((x^2+3)(-1)(x+1)^{-2} + 2x(x+1)^{-1}\) |
| correct expression | A1 | 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 |
| \(\Rightarrow (x-1)(x+3) = 0\) | Some candidates get \(x^2+2x+3\), then realise this should be \(x^2+2x-3\), correct back, but not for every occurrence. Treat sympathetically. | |
| \(\Rightarrow x = 1\) 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). Must be supported, but \(-3\) could be verified by substitution into correct derivative |
| [6] |
| 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 |
| \(\int_a^b \left(x-2+\frac{4}{x}\right)dx = \left[\frac{1}{2}x^2 - 2x + 4\ln x\right]_a^b\) | B1 | \(\left[\frac{1}{2}x^2 - 2x + 4\ln x\right]\). F must show evidence of integration of at least one term |
| \(= \left(\frac{1}{2}b^2 - 2b + 4\ln b\right) - \left(\frac{1}{2}a^2 - 2a + 4\ln a\right)\) | M1 | \(F(b) - F(a)\) condone missing brackets oe (mark final answer) |
| A1 | ||
| Area is \(\int_0^1 f(x)\,dx\) | or \(f(x) = x+1-2+4/(x+1)\) | |
| So taking \(a=1\) and \(b=2\) | M1 | |
| area \(= (2-4+4\ln 2) - (\frac{1}{2}-2+4\ln 1)\) | ||
| \(= 4\ln 2 - \frac{1}{2}\) | A1 cao | must be simplified with \(\ln 1 = 0\) |
| [5] |
# Question 2:
## Part (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]** | | |
## Part (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: $(x^2+3)(-1)(x+1)^{-2} + 2x(x+1)^{-1}$ |
| correct expression | A1 | 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 |
| $\Rightarrow (x-1)(x+3) = 0$ | | Some candidates get $x^2+2x+3$, then realise this should be $x^2+2x-3$, correct back, but not for every occurrence. Treat sympathetically. |
| $\Rightarrow x = 1$ 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). Must be supported, but $-3$ could be verified by substitution into correct derivative |
| **[6]** | | |
## Part (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]** | | |
## Part (iv)
| Answer | 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 | $\left[\frac{1}{2}x^2 - 2x + 4\ln x\right]$. F must show evidence of integration of at least one term |
| $= \left(\frac{1}{2}b^2 - 2b + 4\ln b\right) - \left(\frac{1}{2}a^2 - 2a + 4\ln a\right)$ | M1 | $F(b) - F(a)$ condone missing brackets oe (mark final answer) |
| | A1 | |
| Area is $\int_0^1 f(x)\,dx$ | | or $f(x) = x+1-2+4/(x+1)$ |
| So taking $a=1$ and $b=2$ | M1 | |
| area $= (2-4+4\ln 2) - (\frac{1}{2}-2+4\ln 1)$ | | |
| $= 4\ln 2 - \frac{1}{2}$ | A1 cao | must be simplified with $\ln 1 = 0$ |
| **[5]** | | |
---2 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]{f7049002-f97a-4c83-a7d6-eba28e3b589a-1_904_937_785_604}
\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 Q2 [23]}}