| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2022 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric curves and Cartesian conversion |
| Type | Convert to Cartesian (sin/cos identities) |
| Difficulty | Standard +0.8 This P4 parametric question requires multiple non-trivial steps: using the double angle formula cos(2t) = 1 - 2sin²(t), substituting x to eliminate the parameter, finding range values from the parameter bounds, and performing partial fraction decomposition. While each technique is standard for Further Maths, the combination and algebraic manipulation required elevates this above average difficulty. |
| Spec | 1.02y Partial fractions: decompose rational functions1.03g Parametric equations: of curves and conversion to cartesian |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(y = \frac{6}{7+1-2\sin^2 t}\) | M1 | Applies \(\cos 2t = \pm1 \pm 2\sin^2 t\) to get \(y\) in terms of \(\sin t\). May be implied if \(\cos 2t\) obtained in terms of \(x\) with no explicit sight of \(\sin t\) |
| \(\sin t = \frac{x-3}{2} \Rightarrow y = \frac{6}{8-2\left(\frac{x-3}{2}\right)^2}\) | M1, A1 | Rearranges equation for \(x\) and substitutes into equation for \(y\). A1: Correct equation in terms of \(x\) and \(y\) only |
| \(\Rightarrow y = \frac{12}{16-(x-3)^2} = \frac{12}{(4-x+3)(4+x-3)} = \frac{12}{(7-x)(1+x)}\)* | M1, A1* | Simplifies and factorises denominator. May have \(\frac{1}{2}\) left in denominator. A1*: Correct result with no errors seen |
| \(t = -\frac{\pi}{2} \Rightarrow x=1,\; t=\frac{\pi}{2} \Rightarrow x=5\) so \(p=1\) and \(q=5\) | B1 | Correct values for \(p\) and \(q\) or correct inequality. Can score anywhere in response but do not allow \(1 \leq t \leq 5\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{12}{(7-x)(1+x)} = \frac{A}{7-x} + \frac{B}{1+x} \Rightarrow 12 = A(x+1) + B(7-x) \Rightarrow A=\ldots, B=\ldots\) | M1 | Correct overall method to apply partial fractions to the given fraction or equivalent correct fraction and find at least one constant. If no method shown, one correct value can imply the method. Score M0 for \(\frac{12}{(7-x)(1+x)} = \frac{A}{7-x}+\frac{B}{1+x} \Rightarrow 12=A(7-x)+B(x+1)\) |
| So \((y=)\;\frac{3}{2(x+1)} - \frac{3}{2(x-7)}\) | A1, A1 | A1: At least one correct fraction in required form e.g. \(\frac{3}{2(1+x)}\), \(-\frac{3}{2(x-7)}\). Accept \(\frac{3/2}{1+x}\), \(-\frac{6/4}{x-7}\) but not e.g. \(\frac{3}{2x+2}\), \(\frac{-3}{2x-14}\) unless correct form seen beforehand. A1: Correct expression for \(y\). No need for "\(y=\)". Note: some candidates may write \(\frac{12}{(7-x)(1+x)} = \frac{-12}{(x-7)(1+x)}\) — marks can be applied as in scheme if expression is correct |
## Question 3(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $y = \frac{6}{7+1-2\sin^2 t}$ | M1 | Applies $\cos 2t = \pm1 \pm 2\sin^2 t$ to get $y$ in terms of $\sin t$. May be implied if $\cos 2t$ obtained in terms of $x$ with no explicit sight of $\sin t$ |
| $\sin t = \frac{x-3}{2} \Rightarrow y = \frac{6}{8-2\left(\frac{x-3}{2}\right)^2}$ | M1, A1 | Rearranges equation for $x$ and substitutes into equation for $y$. A1: Correct equation in terms of $x$ and $y$ only |
| $\Rightarrow y = \frac{12}{16-(x-3)^2} = \frac{12}{(4-x+3)(4+x-3)} = \frac{12}{(7-x)(1+x)}$* | M1, A1* | Simplifies and factorises denominator. May have $\frac{1}{2}$ left in denominator. A1*: Correct result with no errors seen |
| $t = -\frac{\pi}{2} \Rightarrow x=1,\; t=\frac{\pi}{2} \Rightarrow x=5$ so $p=1$ and $q=5$ | B1 | Correct values for $p$ and $q$ or correct inequality. **Can score anywhere in response but do not allow** $1 \leq t \leq 5$ |
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## Question 3(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{12}{(7-x)(1+x)} = \frac{A}{7-x} + \frac{B}{1+x} \Rightarrow 12 = A(x+1) + B(7-x) \Rightarrow A=\ldots, B=\ldots$ | M1 | Correct overall method to apply partial fractions to **the given fraction or equivalent correct fraction** and find at least one constant. If no method shown, one correct value can imply the method. Score M0 for $\frac{12}{(7-x)(1+x)} = \frac{A}{7-x}+\frac{B}{1+x} \Rightarrow 12=A(7-x)+B(x+1)$ |
| So $(y=)\;\frac{3}{2(x+1)} - \frac{3}{2(x-7)}$ | A1, A1 | A1: At least one correct fraction in required form e.g. $\frac{3}{2(1+x)}$, $-\frac{3}{2(x-7)}$. Accept $\frac{3/2}{1+x}$, $-\frac{6/4}{x-7}$ but not e.g. $\frac{3}{2x+2}$, $\frac{-3}{2x-14}$ unless correct form seen beforehand. A1: Correct expression for $y$. No need for "$y=$". Note: some candidates may write $\frac{12}{(7-x)(1+x)} = \frac{-12}{(x-7)(1+x)}$ — marks can be applied as in scheme if expression is correct |3. The curve $C$ has parametric equations
$$x = 3 + 2 \sin t \quad y = \frac { 6 } { 7 + \cos 2 t } \quad - \frac { \pi } { 2 } \leqslant t \leqslant \frac { \pi } { 2 }$$
\begin{enumerate}[label=(\alph*)]
\item Show that $C$ has Cartesian equation
$$y = \frac { 12 } { ( 7 - x ) ( 1 + x ) } \quad p \leqslant x \leqslant q$$
where $p$ and $q$ are constants to be found.
\item Hence, find a Cartesian equation for $C$ in the form
$$y = \frac { a } { x + b } + \frac { c } { x + d } \quad p \leqslant x \leqslant q$$
where $a , b , c$ and $d$ are constants.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P4 2022 Q3 [9]}}