科目：进阶数学 Further Mathematics (9231)
年份：2021 冬季
试卷：Paper 23
总页数：14
章节：9231 Further Mathematics
1. Find the Maclaurin’s series for $e^x \tan x$ from first principles up to and including the term in $x^2$. [5]
2 The matrix $A$ is given by
$$A = \begin{pmatrix}
-1 & 2 & 12 \\
0 & 1 & 0 \\
0 & 0 & 3
\end{pmatrix}.$$
Use the characteristic equation of $A$ to show that
$$A^4 = pA^2 + qI,$$
where $p$ and $q$ are integers to be determined. [6]3 The curve $C$ has equation
$$xy^3 - 4x^3y = 3.$$
(a) Show that, at the point $(-1, 1)$ on $C$, $\frac{dy}{dx} = 11$. [3](b) Find the value of $\frac{d^2y}{dx^2}$ at the point $(-1, 1)$. [5]The diagram shows the curve with equation $y = \frac{\ln x}{x^2}$ for $x \geq 2$, together with a set of $(N-2)$ rectangles of unit width.
(a) By considering the sum of the areas of these rectangles, show that
$$\sum_{r=1}^{N} \frac{\ln r}{r^2} < \frac{2+3\ln 2}{4} - \frac{1+\ln N}{N}.$$
[7](b) Use a similar method to find, in terms of $N$, a lower bound for $\sum_{r=1}^{N} \frac{\ln r}{r^2}$. [3]5 Find the particular solution of the differential equation
$$\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + y = 4\cos x,$$
given that, when $x = 0$, $y = -4$ and $\frac{dy}{dx} = 3.$ [11]9 © UCLES 2021 9231/23/O/N/21 [Turn over]
6 (a) Use de Moivre’s theorem to show that
$$\text{cosec} 5\theta = \frac{\text{cosec}^5\theta}{5 \text{cosec}^4\theta - 20 \text{cosec}^2\theta + 16}.$$
[6](b) Hence obtain the roots of the equation
$$x^5 - 10x^4 + 40x^2 - 32 = 0$$
in the form $\text{cosec}(q\pi)$, where $q$ is rational. [4]7 (a) Show that an appropriate integrating factor for
$$\sqrt{x^2 - 1} \frac{dy}{dx} + y = x^2 - x\sqrt{x^2 - 1}$$
is $x + \sqrt{x^2 - 1}$. [4](b) Hence find the solution of the differential equation
$$\sqrt{x^2 - 1} \frac{dy}{dx} + y = x^2 - x\sqrt{x^2 - 1}$$
for which $y = 1$ when $x = \frac{5}{4}$. Give your answer in the form $y = f(x)$. [7]8 (a) Starting from the definition of cosh in terms of exponentials, prove that
$$2 \cosh^2 A = \cosh 2A + 1.$$
[3]
The curve $C$ has parametric equations
$$x = 2 \cosh 2t + 3t, \quad y = \frac{3}{2} \cosh 2t - 4t, \quad \text{for } -\frac{1}{2} \leq t \leq \frac{1}{2}.$$
The area of the surface generated when $C$ is rotated through $2\pi$ radians about the $y$-axis is denoted by $A$.
(b) (i) Show that $A = 10\pi \int_{-\frac{1}{2}}^{\frac{1}{2}} (2 \cosh 2t + 3t) \cosh 2t \, dt.$
[4](ii) Hence find $A$ in terms of $\pi$ and $e$. [7]