CAIE FP1 2019 November — Question 11 28 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2019
SessionNovember
Marks28
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSecond order differential equations
TypeParticular solution with initial conditions
DifficultyChallenging +1.8 This is a challenging Further Maths question requiring either sophisticated differential equation manipulation with substitution (showing d²w/dx² + 2dw/dx + w = -e^(-2x) from a complex given equation) or polar curve intersection and area calculation. Both alternatives demand multiple advanced techniques and careful algebraic manipulation across 14 marks total, placing it well above average difficulty but not at the extreme end for Further Maths material.
Spec4.09c Area enclosed: by polar curve4.10c Integrating factor: first order equations

Answer only one of the following two alternatives. **EITHER** It is given that \(w = \cos y\) and $$\tan y \frac{d^2 y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + 2\tan y \frac{dy}{dx} = 1 + e^{-2x} \sec y.$$
  1. Show that $$\frac{d^2 w}{dx^2} + 2\frac{dw}{dx} + w = -e^{-2x}.$$ [4]
  2. Find the particular solution for \(y\) in terms of \(x\), given that when \(x = 0\), \(y = \frac{1}{4}\pi\) and \(\frac{dy}{dx} = \frac{1}{\sqrt{3}}\). [10]
**OR** The curves \(C_1\) and \(C_2\) have polar equations, for \(0 \leq \theta \leq \frac{1}{2}\pi\), as follows: \begin{align} C_1: r &= 2(e^\theta + e^{-\theta}),
C_2: r &= e^{2\theta} - e^{-2\theta}. \end{align} The curves intersect at the point \(P\) where \(\theta = \alpha\).
  1. Show that \(e^{2\alpha} - 2e^\alpha - 1 = 0\). Hence find the exact value of \(\alpha\) and show that the value of \(r\) at \(P\) is \(4\sqrt{2}\). [6]
  2. Sketch \(C_1\) and \(C_2\) on the same diagram. [3]
  3. Find the area of the region enclosed by \(C_1\), \(C_2\) and the initial line, giving your answer correct to 3 significant figures. [5]

Answer only one of the following two alternatives.

**EITHER**

It is given that $w = \cos y$ and
$$\tan y \frac{d^2 y}{dx^2} + \left(\frac{dy}{dx}\right)^2 + 2\tan y \frac{dy}{dx} = 1 + e^{-2x} \sec y.$$

\begin{enumerate}[label=(\roman*)]
\item Show that
$$\frac{d^2 w}{dx^2} + 2\frac{dw}{dx} + w = -e^{-2x}.$$ [4]

\item Find the particular solution for $y$ in terms of $x$, given that when $x = 0$, $y = \frac{1}{4}\pi$ and $\frac{dy}{dx} = \frac{1}{\sqrt{3}}$. [10]
\end{enumerate}

**OR**

The curves $C_1$ and $C_2$ have polar equations, for $0 \leq \theta \leq \frac{1}{2}\pi$, as follows:
\begin{align}
C_1: r &= 2(e^\theta + e^{-\theta}), \\
C_2: r &= e^{2\theta} - e^{-2\theta}.
\end{align}

The curves intersect at the point $P$ where $\theta = \alpha$.

\begin{enumerate}[label=(\roman*)]
\item Show that $e^{2\alpha} - 2e^\alpha - 1 = 0$. Hence find the exact value of $\alpha$ and show that the value of $r$ at $P$ is $4\sqrt{2}$. [6]

\item Sketch $C_1$ and $C_2$ on the same diagram. [3]

\item Find the area of the region enclosed by $C_1$, $C_2$ and the initial line, giving your answer correct to 3 significant figures. [5]
\end{enumerate}

\hfill \mbox{\textit{CAIE FP1 2019 Q11 [28]}}