| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2011 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Completing the square and sketching |
| Type | Solve quartic as quadratic |
| Difficulty | Moderate -0.8 This is a straightforward question requiring basic algebraic manipulation. Part (i) involves simple rearrangement of equations (equating the curve and line, factoring out x). Part (ii) is a standard quadratic-in-disguise problem using substitution u=x², followed by routine calculation of coordinates. No novel insight or complex problem-solving required—purely procedural work below average A-level difficulty. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown1.02j Manipulate polynomials: expanding, factorising, division, factor theorem |
| Answer | Marks | Guidance |
|---|---|---|
| \(2x^4 + 3x^2 - 2 = 0\) | M1, A1 [2] | First line essential; AG Factoring needed for A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\left(\frac{1}{\sqrt{2}}, \frac{2}{\sqrt{2}}\right) \left(-\frac{1}{\sqrt{2}}, -\frac{2}{\sqrt{2}}\right)\) | M1, A1, A1, A1 [3] | Reasonable attempt at solving a quadratic in \(x^2\); For a correct pair of solutions, either 2 x's or 1 x and 1 y; SC (\(\pm 0.707, \pm 1.41\)) AWRT B1 |
**(i)** $2x^5 + 3x^2 = 2x \Rightarrow 2x^5 + 3x^2 - 2x = 0$
$[x(2x^4 + 3x - 2) = 0]$
$2x^4 + 3x^2 - 2 = 0$ | M1, A1 [2] | First line essential; AG Factoring needed for A1
**(ii)** $(x^2 + 2)(2x^2 - 1) = 0$
$x = \pm \sqrt{\frac{1}{2}}$ only
$\left(\frac{1}{\sqrt{2}}, \frac{2}{\sqrt{2}}\right) \left(-\frac{1}{\sqrt{2}}, -\frac{2}{\sqrt{2}}\right)$ | M1, A1, A1, A1 [3] | Reasonable attempt at solving a quadratic in $x^2$; For a correct pair of solutions, either 2 x's or 1 x and 1 y; SC ($\pm 0.707, \pm 1.41$) AWRT B1
---\includegraphics{figure_3}
The diagram shows the curve $y = 2x^5 + 3x^3$ and the line $y = 2x$ intersecting at points $A$, $O$ and $B$.
\begin{enumerate}[label=(\roman*)]
\item Show that the $x$-coordinates of $A$ and $B$ satisfy the equation $2x^4 + 3x^2 - 2 = 0$. [2]
\item Solve the equation $2x^4 + 3x^2 - 2 = 0$ and hence find the coordinates of $A$ and $B$, giving your answers in an exact form. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2011 Q3 [5]}}