| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Parametric integration |
| Type | Show integral then evaluate area |
| Difficulty | Standard +0.3 This is a standard C4 parametric equations question requiring routine techniques: solving for parameter value, finding tangent equation using dy/dx = (dy/dt)/(dx/dt), setting up area integral using A = ∫y dx/dt dt, and evaluating a basic trigonometric integral. All steps follow textbook methods with no novel insight required, making it slightly easier than average. |
| Spec | 1.03g Parametric equations: of curves and conversion to cartesian1.07s Parametric and implicit differentiation1.08h Integration by substitution |
| Answer | Marks |
|---|---|
| (a) \(\cos 2t = \frac{1}{2}\), \(2t = \frac{\pi}{3}\), \(t = \frac{\pi}{6}\) | M1 A1 |
| (b) \(\frac{dx}{dt} = -2\sin 2t\), \(\frac{dy}{dt} = -\cosec t \cot t\) | M1 |
| \(\frac{dy}{dx} = \frac{-\cosec t \cot t}{-2\sin 2t}\) | M1 A1 |
| \(t = \frac{\pi}{6}\), \(y = 2\), grad \(= 2\) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(y = 2x + 1\) | A1 | |
| (c) \(x = 0 \Rightarrow t = \frac{\pi}{4}\) | B1 | |
| \(\therefore\) area \(= \int_{\frac{\pi}{4}}^{\frac{\pi}{6}} \cos t \times (-2\sin 2t) dt\) | M1 | |
| \(= -\int_{\frac{\pi}{4}}^{\frac{\pi}{6}} \cos t \times 4\sin t \cos t dt = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} 4\cos t dt\) | M1 A1 | |
| (d) \(= [4\sin t]_{\frac{\pi}{6}}^{\frac{\pi}{4}} = 2\sqrt{2} - 2 = 2(\sqrt{2} - 1)\) | M2 A1 | (14 marks) |
**(a)** $\cos 2t = \frac{1}{2}$, $2t = \frac{\pi}{3}$, $t = \frac{\pi}{6}$ | M1 A1 |
**(b)** $\frac{dx}{dt} = -2\sin 2t$, $\frac{dy}{dt} = -\cosec t \cot t$ | M1 |
$\frac{dy}{dx} = \frac{-\cosec t \cot t}{-2\sin 2t}$ | M1 A1 |
$t = \frac{\pi}{6}$, $y = 2$, grad $= 2$ | M1 |
$\therefore y - 2 = 2\left(x - \frac{1}{2}\right)$
$y = 2x + 1$ | A1 |
**(c)** $x = 0 \Rightarrow t = \frac{\pi}{4}$ | B1 |
$\therefore$ area $= \int_{\frac{\pi}{4}}^{\frac{\pi}{6}} \cos t \times (-2\sin 2t) dt$ | M1 |
$= -\int_{\frac{\pi}{4}}^{\frac{\pi}{6}} \cos t \times 4\sin t \cos t dt = \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} 4\cos t dt$ | M1 A1 |
**(d)** $= [4\sin t]_{\frac{\pi}{6}}^{\frac{\pi}{4}} = 2\sqrt{2} - 2 = 2(\sqrt{2} - 1)$ | M2 A1 | (14 marks)
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**Total: 75 marks**7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e877dc80-4cfc-4c8b-9640-9b186cd7ab13-12_556_860_246_452}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows the curve with parametric equations
$$x = \cos 2 t , \quad y = \operatorname { cosec } t , \quad 0 < t < \frac { \pi } { 2 } .$$
The point $P$ on the curve has $x$-coordinate $\frac { 1 } { 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of the parameter $t$ at $P$.
\item Show that the tangent to the curve at $P$ has the equation
$$y = 2 x + 1$$
The shaded region is bounded by the curve, the coordinate axes and the line $x = \frac { 1 } { 2 }$.
\item Show that the area of the shaded region is given by
$$\int _ { \frac { \pi } { 6 } } ^ { \frac { \pi } { 4 } } k \cos t \mathrm {~d} t$$
where $k$ is a positive integer to be found.
\item Hence find the exact area of the shaded region.\\
7. continued\\
7. continued
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 Q7 [14]}}