| Exam Board | OCR MEI |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find intersection points |
| Difficulty | Standard +0.8 This is a multi-part question requiring curve sketching, differentiation for the normal, and solving a simultaneous equation where a linear normal intersects a reciprocal curve. Part (c) requires algebraic manipulation of a cubic equation, which is more demanding than standard AS exercises. The combination of techniques and the non-routine nature of finding the second intersection point elevates this above average difficulty. |
| Spec | 1.02o Sketch reciprocal curves: y=a/x and y=a/x^21.02q Use intersection points: of graphs to solve equations1.02w Graph transformations: simple transformations of f(x)1.07l Derivative of ln(x): and related functions1.07m Tangents and normals: gradient and equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Correct shape for \(y = \frac{1}{x}\), translated vertically upward | B1 | |
| Crosses \(x\)-axis at \(x = -\frac{1}{a}\) | B1 | |
| Asymptotes \(x = 0\) and \(y = a\) | B2 | B1 for one asymptote correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{dy}{dx} = -\frac{1}{x^2}\) | M1 | |
| When \(x = 2\), gradient \(= -\frac{1}{4}\) | M1 | |
| Gradient of normal \(= 4\) FT their gradient | M1 | Or gradient of given line is \(4\) and check \(4 \times -\frac{1}{4} = -1\) |
| \(y - \frac{5}{2} = 4(x-2)\) FT their gradient | M1 | Or \(y = \frac{5}{2}\) at the point on the curve where \(x = 2\) |
| \(2y = 8x - 11\) oe | A1 | At least one correct interim step or clear check that \(\left(2, \frac{5}{2}\right)\) is on given line. Any simplified form. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2\left(\frac{1}{x}+2\right)=8x-11\) | M1 | 3.1a — Substitution |
| \(8x^2-15x-2=0\) | M1 | 1.1 — Forming quadratic, condone one error |
| Other point is \(x=-\frac{1}{8}\), \(y=-6\) | A1 | 1.1 — BC \(x=2\) not needed in this case |
| [3] |
# Question 10:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Correct shape for $y = \frac{1}{x}$, translated vertically upward | B1 | |
| Crosses $x$-axis at $x = -\frac{1}{a}$ | B1 | |
| Asymptotes $x = 0$ and $y = a$ | B2 | B1 for one asymptote correct |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{dy}{dx} = -\frac{1}{x^2}$ | M1 | |
| When $x = 2$, gradient $= -\frac{1}{4}$ | M1 | |
| Gradient of normal $= 4$ FT their gradient | M1 | Or gradient of given line is $4$ and check $4 \times -\frac{1}{4} = -1$ |
| $y - \frac{5}{2} = 4(x-2)$ FT their gradient | M1 | Or $y = \frac{5}{2}$ at the point on the curve where $x = 2$ |
| $2y = 8x - 11$ oe | A1 | At least one correct interim step or clear check that $\left(2, \frac{5}{2}\right)$ is on given line. Any simplified form. |
## Question 10(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2\left(\frac{1}{x}+2\right)=8x-11$ | M1 | 3.1a — Substitution |
| $8x^2-15x-2=0$ | M1 | 1.1 — Forming quadratic, condone one error |
| Other point is $x=-\frac{1}{8}$, $y=-6$ | A1 | 1.1 — **BC** $x=2$ not needed in this case |
| **[3]** | | |
---10
\begin{enumerate}[label=(\alph*)]
\item Sketch the graph of $y = \frac { 1 } { x } + a$, where $a$ is a positive constant.
\begin{itemize}
\item State the equations of the horizontal and vertical asymptotes.
\item Give the coordinates of any points where the graph crosses the axes.
\item Find the equation of the normal to the curve $y = \frac { 1 } { x } + 2$ at the point where $x = 2$.
\item Find the coordinates of the point where this normal meets the curve again.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI AS Paper 1 Q10 [12]}}