By sketching a suitable pair of graphs, deduce the number of roots of the equation $x=\tan 2x$ for $0<x<2\pi .$

The equation $\left(x-0.5\right)e^{3x}=1$ has a root $\alpha .$

Show, by sketching the graph of $y=e^{3x}$ and one other suitable graph, that $\alpha$ is the only root of this equation.

The equation $x^4-1-x=0$ has a root, $\alpha ,$ between $x=1$ and $x=2.$

Show that $\alpha$ also satisfies the equation $x=\sqrt[4]{1+x}.$

The equation $x^4-1-x=0$ has a root, $\alpha ,$ between $x=1$ and $x=2.$

Show that $\alpha$ also satisfies the equation $x=\sqrt[4]{1+x}.$

Write down an iterative formula based on the above equation 

Use your iterative formula, with a starting value of $x_1=1.5,$ to find $\alpha$ correct to $2$ decimal places. Give the result of each iteration to $4$ decimal places.

The equation $cosecx=x^2$ has a root, $\alpha ,$ between $1$ and $2 .$ The equation can be rearranged either as $x={\sin }^{-1}\left(\frac{1}{x^2}\right)$ or $x=\sqrt{\frac{1}{\sin x}}.$

Write down two possible iterative formulae, one based on each given rearrangement. 

The terms of a sequence, defined by the iterative formula $x_{n+1}=\ln \left(x_n^2+4\right),$ converge to the value $\alpha .$ The first term of the sequence is $2.$

The value $\alpha$ is a root of an equation of the form $x^2=f\left(x\right).$ Find this equation.

The curve with equation $y=\frac{{\mathrm{e}}^{2x}+x}{x^3}$ has a stationary point with $x$ -coordinate lying between $1$ and $2$

.Show that the $x$-coordinate of this stationary point satisfies the equation $x=\frac{3}{2}+\frac{x}{{\mathrm{e}}^{2x}}$.

The parametric equations of a curve are $x=t^2+6,y=t^4-t^3-5t$. The curve has a stationary point for a value of $t$ that lies between $1$ and $2 .$

Show that the value of $t$ at this stationary point satisfies the equation
$t=\sqrt[3]{\frac{3t^2+5}{4}}$