| Exam Board | OCR |
|---|---|
| Module | PURE |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Transformations of functions |
| Difficulty | Standard +0.3 This question involves standard inverse proportion (y = k/x), basic differentiation to find the gradient function, sketching y = k/x², recognizing that a gradient function never equals zero means no stationary points, and applying a horizontal translation. All techniques are routine for C3/C4 level with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.02r Proportional relationships: and their graphs1.02w Graph transformations: simple transformations of f(x)1.07a Derivative as gradient: of tangent to curve1.07c Sketch gradient function: for given curve |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = \dfrac{k}{x}\) | M1 | Allow any letter (except \(x, y\)) or value, for \(k\) |
| The curve passes through \(\left(1, -\frac{1}{2}\right)\) so \(k = -\frac{1}{2}\), or \(y = -\dfrac{1}{2x}\) | A1 | Allow this mark for just \(-\dfrac{1}{2x}\) oe |
| \(y = -\dfrac{1}{2x} \Rightarrow y' = \dfrac{1}{2x^2}\) | A1ft | Differentiating their \(f(x)\) correctly. Need to see their value of \(k\) substituted. Need to see \(y' =\) or \(f'(x) =\) or \(\dfrac{dy}{dx} =\) |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| [Graph: curve in 1st and 2nd quadrants only, correct shape, symmetrical, not touching axis, asymptotes clearly the axes, not finite] | B1ft | Excellent curve in 1st and 2nd quadrants only. Correct shape, symmetrical, not touching axis. Asymptote clearly the axes. Not finite. Allow slight movement away from asymptote at one end but not more. Follow through provided their curve is of the form \(y = \dfrac{k}{x^2}\) where \(k > 0\) |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| C has no stationary points as indicated by the fact that the curve for the gradient function (seen in part (a)(ii)) does not intersect (or touch) the \(x\)-axis | B1 | Curve in 5(a)(ii) must be of the form \(y = \dfrac{k}{x^2}\). Need to see idea of intersecting, touching, crossing etc \(x\)-axis only |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| [Translation method] | M1 | Their \(y = f(x)\) with \(x\) replaced by \(x \pm 2\). May be \(y = \dfrac{k}{x}\) |
| \(y = -\dfrac{1}{2(x+2)}\) | A1 | oe e.g. \(y = -\dfrac{1}{2x+4}\). Must have \(y = \ldots\) |
| [2] |
## Question 5(a)(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = \dfrac{k}{x}$ | M1 | Allow any letter (except $x, y$) or value, for $k$ |
| The curve passes through $\left(1, -\frac{1}{2}\right)$ so $k = -\frac{1}{2}$, or $y = -\dfrac{1}{2x}$ | A1 | Allow this mark for just $-\dfrac{1}{2x}$ oe |
| $y = -\dfrac{1}{2x} \Rightarrow y' = \dfrac{1}{2x^2}$ | A1ft | Differentiating their $f(x)$ correctly. Need to see their value of $k$ substituted. Need to see $y' =$ or $f'(x) =$ or $\dfrac{dy}{dx} =$ |
| **[3]** | | |
---
## Question 5(a)(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| [Graph: curve in 1st and 2nd quadrants only, correct shape, symmetrical, not touching axis, asymptotes clearly the axes, not finite] | B1ft | Excellent curve in 1st and 2nd quadrants only. Correct shape, symmetrical, not touching axis. Asymptote clearly the axes. Not finite. Allow slight movement away from asymptote at one end but not more. Follow through provided their curve is of the form $y = \dfrac{k}{x^2}$ where $k > 0$ |
| **[1]** | | |
---
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| C has no stationary points as indicated by the fact that the curve for the gradient function (seen in part (a)(ii)) does not intersect (or touch) the $x$-axis | B1 | Curve in 5(a)(ii) must be of the form $y = \dfrac{k}{x^2}$. Need to see idea of intersecting, touching, crossing etc $x$-axis **only** |
| **[1]** | | |
---
## Question 5(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| [Translation method] | M1 | Their $y = f(x)$ with $x$ replaced by $x \pm 2$. May be $y = \dfrac{k}{x}$ |
| $y = -\dfrac{1}{2(x+2)}$ | A1 | oe e.g. $y = -\dfrac{1}{2x+4}$. Must have $y = \ldots$ |
| **[2]** | | |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{a1f4ccbd-f5ed-437a-ae76-c4925ce86e25-04_700_727_260_242}
The diagram shows a curve $C$ for which $y$ is inversely proportional to $x$. The curve passes through the point $\left( 1 , - \frac { 1 } { 2 } \right)$.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Determine the equation of the gradient function for the curve $C$.
\item Sketch this gradient function on the axes in the Printed Answer Booklet.
\end{enumerate}\item The diagram indicates that the curve $C$ has no stationary points.
State what feature of your sketch in part (a)(ii) corresponds to this.
\item The curve $C$ is translated by the vector $\binom { - 2 } { 0 }$.
Find the equation of the curve after it has been translated.
\end{enumerate}
\hfill \mbox{\textit{OCR PURE Q5 [7]}}