| Exam Board | OCR MEI |
|---|---|
| Module | Further Pure with Technology (Further Pure with Technology) |
| Year | 2024 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Curve Sketching |
| Type | Parameter values from curve properties |
| Difficulty | Standard +0.8 This is a multi-part Further Maths question requiring curve sketching with technology, algebraic manipulation, limits, and optimization. While individual parts are accessible (polynomial division, finding asymptotes), the question demands sustained reasoning across multiple techniques, parameter analysis, and geometric interpretation. The optimization in part (b)(ii) requires finding roots in terms of a parameter and differentiating the resulting area expression, which is non-trivial but within reach of Further Maths students. |
| Spec | 1.02k Simplify rational expressions: factorising, cancelling, algebraic division1.02n Sketch curves: simple equations including polynomials1.02o Sketch reciprocal curves: y=a/x and y=a/x^21.02w Graph transformations: simple transformations of f(x) |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (a) | (ii) |
| 𝑎 = −0.1 have turning points. | B1 | |
| [1] | 1.1 | Allow stationary point rather than turning point. |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (a) | (iii) |
| [1] | 1.1 | |
| 1 | (a) | (iv) |
| [1] | 1.1 | Must be an equation. |
| 1 | (a) | (v) |
| Answer | Marks |
|---|---|
| So 𝐴 = 0. | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | Any valid method including CAS. |
| Answer | Marks | Guidance |
|---|---|---|
| = 0. | B1 | |
| [2] | 1.1 | |
| 1 | (a) | (vii) |
| 𝑥 approaches infinity]. | B1 | |
| [1] | 1.2 | Allow the curve tends towards the line 𝑦 = 𝑥 as 𝑥 approaches |
| Answer | Marks | Guidance |
|---|---|---|
| 1 | (b) | (ii) |
| Answer | Marks |
|---|---|
| 2 | B1FT |
| Answer | Marks |
|---|---|
| [2] | 3.1a |
| 1.1 | Use of CAS is expected to differentiate or maximise the area. |
Question 1:
1 | (a) | (ii) | The curves for the cases 𝑎 = −0.5 and
𝑎 = −0.1 have turning points. | B1
[1] | 1.1 | Allow stationary point rather than turning point.
Allow the curves for the cases 𝑎 = −0.5 and 𝑎 = −0.1 approach
the oblique asymptote 𝑦 = 𝑥 from above, whereas the curve for
the case 𝑎 = 0.5 approaches from below.
Allow the curves for the cases 𝑎 = −0.5 and 𝑎 = −0.1 are in the
positive quadrant (or equivalent) to the right of the vertical
asymptote, whereas the curve for the case 𝑎 = 0.5 does not.
1 | (a) | (iii) | 𝑎 = 1 | B1
[1] | 1.1
1 | (a) | (iv) | 𝑥 = 1 | B1
[1] | 1.1 | Must be an equation.
1 | (a) | (v) | 𝑎2 −𝑎
𝑦 = 𝑥 +𝐴+
𝑥 −1
(𝑥 +𝐴)(𝑥−1)+ 𝑎2 −𝑎
=
𝑥 −1
Equating coefficients gives
𝐴−1 = −1.
So 𝐴 = 0. | M1
A1
[2] | 1.1
1.1 | Any valid method including CAS.
Alternative working may lead to 𝐴+𝑎2 −𝑎 = 𝑎2 −𝑎. or 𝑥2 +
(𝐴−1)𝑥 −𝐴 = 𝑥2 −𝑥 and comparing coefficients.
𝑥2 −𝑥 +𝑎2 −𝑎
lim ( −𝑥)
𝑥→∞ 𝑥 −1
𝑎2 −𝑎
= lim (𝑥 + −𝑥)
𝑥→∞ 𝑥 −1
𝑎2 −𝑎
= lim ( )
𝑥→∞ 𝑥 −1
= 0. | B1
[2] | 1.1
1 | (a) | (vii) | 𝑦 = 𝑥 is an oblique asymptote [as
𝑥 approaches infinity]. | B1
[1] | 1.2 | Allow the curve tends towards the line 𝑦 = 𝑥 as 𝑥 approaches
infinity.
Condone not considering behaviour as 𝑥 approaches minus
infinity explicitly.
[4]
1 | (b) | (ii) | Differentiate
1−2𝑎
.
√1+4𝑎−4𝑎2
So
1
𝑎 =
2
√2
Maximum area .
2 | B1FT
B1
[2] | 3.1a
1.1 | Use of CAS is expected to differentiate or maximise the area.
May be implied by correct answer.
Alternative.
2
1 √1 1
×√1+4𝑎−4𝑎2 = −(𝑎− ) .
2 2 2
2
1
Since (𝑎− ) ⩾ 0 for all 𝑎.
2
2
√1 1 1
−(𝑎− ) ⩽ √ for all 𝑎.
2 2 2
1
Maximum occurs when 𝑎 =
2
Or equivalent.
Allow follow through from 1(b)(i).1 A family of curves is given by the equation
$$y = \frac { x ^ { 2 } - x + a ^ { 2 } - a } { x - 1 }$$
where the parameter $a$ is a real number.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item On the axes in the Printed Answer Booklet, sketch the curve in each of these cases.
\begin{itemize}
\item $a = - 0.5$
\item $a = - 0.1$
\item $a = 0.5$
\item State one feature of the curve for the cases $a = - 0.5$ and $a = - 0.1$ that is not a feature of the curve in the case $a = 0.5$.
\item By using a slider for $a$, or otherwise, write down the non-zero value of $a$ for which the points on the curve (\textit{) all lie on a straight line.
\item Write down the equation of the vertical asymptote of the curve (}).
\end{itemize}
The equation of the curve (*) can be written in the form $y = x + A + \frac { a ^ { 2 } - a } { x - 1 }$, where $A$ is a constant.
\item Show that $A = 0$.
\item Hence, or otherwise, find the value of
$$\lim _ { x \rightarrow \infty } \left( \frac { x ^ { 2 } - x + a ^ { 2 } - a } { x - 1 } - x \right) .$$
\item Explain the significance of the result in part (a)(vi) in terms of a feature of the curve (*).
\item In this part of the question the value of the parameter $a$ satisfies $0 < a < 1$. For values of $a$ in this range the curve intersects the $x$-axis at points X and Y . The point Z has coordinates $( 0 , - 1 )$. These three points form a triangle XYZ.
\begin{enumerate}[label=(\roman*)]
\item Determine, in terms of $a$, the area of the triangle XYZ.
\item Find the maximum area of the triangle XYZ.
\end{enumerate}
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Pure with Technology 2024 Q1 [17]}}