Standard +0.3 This is a straightforward separable variables question requiring standard manipulation (separating variables, integrating both sides, applying initial condition). The integration of cot(2θ) is routine A-level material, and simplifying the final expression is mechanical. Slightly above average difficulty only due to the trigonometric functions and algebraic manipulation required.
4 The variables \(x\) and \(\theta\) are related by the differential equation
$$\sin 2 \theta \frac { \mathrm {~d} x } { \mathrm {~d} \theta } = ( x + 1 ) \cos 2 \theta$$
where \(0 < \theta < \frac { 1 } { 2 } \pi\). When \(\theta = \frac { 1 } { 12 } \pi , x = 0\). Solve the differential equation, obtaining an expression for \(x\) in terms of \(\theta\), and simplifying your answer as far as possible.
Separate variables and attempt integration of at least one side
M1
Obtain term \(\ln(x + 1)\)
A1
Obtain term \(k \ln \sin 2\theta\), where \(k = \pm 1, \pm 2\), or \(\pm \frac{1}{2}\)
M1
Obtain correct term \(-\frac{1}{2} \ln \sin 2\theta\)
A1
Evaluate a constant, or use limits \(\theta = \frac{1}{12}\pi, x = 0\) in a solution containing terms \(a\ln(x+1)\) and \(b \ln \sin 2\theta\)
M1
Obtain solution in any form, e.g. \(\ln(x+1) = \frac{1}{2}\ln \sin 2\theta - \frac{1}{2}\ln \frac{1}{2}\) (f.t. on \(k = \pm 1, \pm 2\), or \(\pm \frac{1}{2}\))
A1√
Rearrange and obtain \(x = \sqrt{(2\sin 2\theta) - 1}\), or simple equivalent
A1
[7]
Separate variables and attempt integration of at least one side | M1 |
Obtain term $\ln(x + 1)$ | A1 |
Obtain term $k \ln \sin 2\theta$, where $k = \pm 1, \pm 2$, or $\pm \frac{1}{2}$ | M1 |
Obtain correct term $-\frac{1}{2} \ln \sin 2\theta$ | A1 |
Evaluate a constant, or use limits $\theta = \frac{1}{12}\pi, x = 0$ in a solution containing terms $a\ln(x+1)$ and $b \ln \sin 2\theta$ | M1 |
Obtain solution in any form, e.g. $\ln(x+1) = \frac{1}{2}\ln \sin 2\theta - \frac{1}{2}\ln \frac{1}{2}$ (f.t. on $k = \pm 1, \pm 2$, or $\pm \frac{1}{2}$) | A1√ |
Rearrange and obtain $x = \sqrt{(2\sin 2\theta) - 1}$, or simple equivalent | A1 | [7]
4 The variables $x$ and $\theta$ are related by the differential equation
$$\sin 2 \theta \frac { \mathrm {~d} x } { \mathrm {~d} \theta } = ( x + 1 ) \cos 2 \theta$$
where $0 < \theta < \frac { 1 } { 2 } \pi$. When $\theta = \frac { 1 } { 12 } \pi , x = 0$. Solve the differential equation, obtaining an expression for $x$ in terms of $\theta$, and simplifying your answer as far as possible.
\hfill \mbox{\textit{CAIE P3 2011 Q4 [7]}}