AQA A-Level Maths: Pure Mathematics 2

The equation $x^3 - 3x - 1 = 0$x3−3x−1=0 has a root $\alpha$α in the interval $1.8 < x < 1.9$1.8<x<1.9.

(a) Show that $\alpha$α lies in this interval. (2 marks)

(b) Show that the equation can be rearranged into the form $x = \sqrt[3]{\,3x + 1\,}$x=33x+1​, and use the iterative formula $x_{n+1} = \sqrt[3]{\,3x_n + 1\,}$xn+1​=33xn​+1​ with $x_0 = 1.9$x0​=1.9 to find $x_1$x1​, $x_2$x2​ and $x_3$x3​, giving each value to four decimal places. (5 marks)

(c) Taking $f(x) = x^3 - 3x - 1$f(x)=x3−3x−1, apply the Newton-Raphson method once, starting from $x_0 = 1.9$x0​=1.9, to obtain a further approximation to $\alpha$α. Give your answer to four decimal places. (5 marks)

(a) Starting from the compound-angle formula $\cos(A + B) = \cos A \cos B - \sin A \sin B$cos(A+B)=cosAcosB−sinAsinB, prove that $\cos 2x = 1 - 2\sin^2 x.$cos2x=1−2sin2x. (3 marks)

(b) Hence show that $\int_0^{\pi/3} \sin^2 x \,dx = \frac{\pi}{6} - \frac{\sqrt{3}}{8}.$∫0π/3​sin2xdx=6π​−83​​. (4 marks)

(c) Using the same identity, solve the equation $\cos 2x = \sin x$cos2x=sinx for $0 \leq x < 2\pi$0≤x<2π, giving your answers in terms of $\pi$π. (3 marks)

(a) Show that $\frac{d}{dx}\Big[\ln(x^2 - 1)\Big] = \frac{2x}{x^2 - 1}.$dxd​[ln(x2−1)]=x2−12x​. (2 marks)

(b) Hence evaluate $\int_2^4 \frac{x}{x^2 - 1}\,dx,$∫24​x2−1x​dx, giving your answer in the form $\dfrac{1}{2}\ln k$21​lnk, where $k$k is an integer to be found. (6 marks)

Prove by contradiction that $\sqrt{3}$3​ is irrational.

You may assume that, for any integer $n$n, if $3$3 divides $n^2$n2 then $3$3 divides $n$n. (6 marks)

A curve is defined parametrically by $x = t^2, \qquad y = t^3 - 3t, \qquad t \in \mathbb{R}.$x=t2,y=t3−3t,t∈R.

Find an equation of the tangent to the curve at the point where $t = 2$t=2, giving your answer in the form $y = mx + c$y=mx+c. (5 marks)

(a) Find the first three terms, in ascending powers of $x$x, of the binomial expansion of $(1 + x)^{1/2}$(1+x)1/2, stating the range of values of $x$x for which the expansion is valid. (2 marks)

(b) By substituting a suitable value of $x$x, use your expansion to find an approximation for $\sqrt{1.04}$1.04​, giving your answer to four decimal places. (2 marks)