Edexcel A-Level Further Mathematics Further Pure options overview quiz quiz

$\frac{2t}{1 + t^2}$, which is in fact $\sin\theta$

$\frac{1 + t^2}{2t}$, the reciprocal of $\sin\theta$

$\frac{2t}{1 - t^2}$, which is in fact $\tan\theta$

$\frac{1 - t^2}{1 + t^2}$

$\theta = n\pi + (-1)^n\alpha$

$\theta = 2n\pi \pm \alpha$, the solution for $\cos\theta = \cos\alpha$

$\theta = n\pi + \alpha$, the solution for $\tan\theta = \tan\alpha$

$\theta = 2n\pi + \alpha$, taking only the principal value each period

$(at, 2at^2)$

$(at^2, 2at)$

$(2at, at^2)$

$(a\cos t, a\sin t)$

$(c\cos t, c\sin t)$

$(ct^2, 2ct)$

$\left(ct, \frac{c}{t}\right)$

$\left(\frac{c}{t}, ct^2\right)$

$x_{n+1} = x_n - f(x_n)f'(x_n)$

$x_{n+1} = x_n + \frac{f(x_n)}{f'(x_n)}$

$x_{n+1} = \frac{f(x_n)}{f'(x_n)}$

$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$

An even number of strips

An odd number of strips, so that the ordinates pair up neatly across the interval

Exactly three strips, fitting a single cubic curve across the whole interval

Any number of strips at all, since the rule places no restriction on the count

$1 + x + x^2$

$1 + x + \frac{x^2}{2!}$

$x + \frac{x^2}{2!} + \frac{x^3}{3!}$

$1 + \frac{x}{2} + \frac{x^2}{4}$

$x - \frac{x^2}{2!} + \frac{x^3}{3!}$

$1 - \frac{x^2}{2!} + \frac{x^4}{4!}$

$x - \frac{x^3}{3!} + \frac{x^5}{5!}$

$x + \frac{x^3}{3!} + \frac{x^5}{5!}$

$e = 1$

$e > 1$

$e = 0$ always

$0 < e < 1$

The $x$-intercept of the chord joining the endpoints

The midpoint of the interval, taken as the best estimate of the root each time

The $x$-intercept of the tangent drawn to the curve at the midpoint of the interval

The average of the two function values $f(a)$ and $f(b)$ at the endpoints

all real values of $x$ with no restriction at all

$-1 < x \le 1$

only the strictly positive values $x > 0$

the values outside the unit interval, $|x| > 1$