By sketching a suitable pair of graphs, show that the equation
$$\mathrm { e } ^ { 2 x } = 2 - x$$
has only one root.
Verify by calculation that this root lies between \(x = 0\) and \(x = 0.5\).
Show that, if a sequence of values given by the iterative formula
$$x _ { n + 1 } = \frac { 1 } { 2 } \ln \left( 2 - x _ { n } \right)$$
converges, then it converges to the root of the equation in part (i).
Use this iterative formula, with initial value \(x _ { 1 } = 0.25\), to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
By differentiating \(\frac { \cos x } { \sin x }\), show that if \(y = \cot x\) then \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \operatorname { cosec } ^ { 2 } x\).
By expressing \(\cot ^ { 2 } x\) in terms of \(\operatorname { cosec } ^ { 2 } x\) and using the result of part (i), show that
$$\int _ { \frac { 1 } { 4 } \pi } ^ { \frac { 1 } { 2 } \pi } \cot ^ { 2 } x \mathrm {~d} x = 1 - \frac { 1 } { 4 } \pi$$
Express \(\cos 2 x\) in terms of \(\sin ^ { 2 } x\) and hence show that \(\frac { 1 } { 1 - \cos 2 x }\) can be expressed as \(\frac { 1 } { 2 } \operatorname { cosec } ^ { 2 } x\). Hence, using the result of part (i), find
$$\int \frac { 1 } { 1 - \cos 2 x } \mathrm {~d} x$$