科目：进阶数学 Further Mathematics (9231)
年份：2020 夏季
试卷：Paper 23
总页数：14
章节：9231 Further Mathematics
1. Find the general solution of the differential equation
$$\frac{d^2x}{dt^2} - 8 \frac{dx}{dt} - 9x = 9e^{8t}.$$
[6]2 Let $I_n = \int_0^1 (1 + 3x)^n e^{-3x} dx$, where $n$ is an integer.
(a) Show that $3I_n = 1 - 4^n e^{-3} + 3nI_{n-1}$. [3]
(b) Find the exact value of $I_2$. [3]3 The matrix $A$ is given by
$$A = \begin{pmatrix} 5 & -1 & 7 \\ 0 & 6 & 0 \\ 7 & 7 & 5 \end{pmatrix}.$$
(a) Find the eigenvalues of $A$. [4](b) Use the characteristic equation of $A$ to find $A^{-1}$. [4]The diagram shows the curve with equation $y = \ln x$ for $x \geq 1$, together with a set of $(N-1)$ rectangles of unit width. (a) By considering the sum of the areas of these rectangles, show that $$\ln N! > N \ln N - N + 1.$$ [5]
(b) Use a similar method to find, in terms of $N$, an upper bound for $\ln N!$. [3]
5 The curve $C$ has parametric equations
$$x = \frac{1}{2} t^2 - \ln t, \quad y = 2t + 1, \quad \text{for } \frac{1}{2} \leq t \leq 2.$$
(a) Find the exact length of $C$. [5](b) Find $\frac{d^2y}{dx^2}$ in terms of $t$, simplifying your answer. [4]6 (a) Starting from the definitions of tanh and sech in terms of exponentials, prove that
$$1 - \tanh^2\theta = \text{sech}^2\theta.$$
[3]
The variables $x$ and $y$ are such that $\tanh y = \cos\left(x + \frac{1}{4}\pi\right)$, for $-\frac{1}{4}\pi < x < \frac{3}{4}\pi$.
(b) By differentiating the equation $\tanh y = \cos\left(x + \frac{1}{4}\pi\right)$ with respect to $x$, show that
$$\frac{\mathrm{d}y}{\mathrm{d}x} = -\cos\csc\left(x + \frac{1}{4}\pi\right).$$
[4](c) Hence find the first three terms in the Maclaurin’s series for $\tanh^{-1}\left(\cos\left(x + \frac{1}{4}\pi\right)\right)$ in the form $\frac{1}{2}\ln a + bx + cx^2$, giving the exact values of the constants $a, b$ and $c$. [5]7 (a) Show that an appropriate integrating factor for
$$\left(x^2 + 1\right) \frac{\mathrm{d}y}{\mathrm{d}x} + y \sqrt{x^2 + 1} = x^2 - x \sqrt{x^2 + 1}$$
is $x + \sqrt{x^2 + 1}$. [4](b) Hence find the solution of the differential equation
$$\left(x^2 + 1\right)\frac{dy}{dx} + y\sqrt{x^2 + 1} = x^2 - x\sqrt{x^2 + 1}$$
for which $y = \ln 2$ when $x = 0$. Give your answer in the form $y = f(x)$. [7]8 (a) Use de Moivre’s theorem to show that $\sin^6 \theta = -\frac{1}{32} (\cos 6\theta - 6\cos 4\theta + 15\cos 2\theta - 10)$. [6]
It is given that $\cos^6 \theta = \frac{1}{32} (\cos 6\theta + 6\cos 4\theta + 15\cos 2\theta + 10)$.
(b) Find the exact value of $\int_0^{\frac{1}{3}\pi} \left( \cos^6\left(\frac{1}{4}x\right) + \sin^6\left(\frac{1}{4}x\right) \right) dx$. [4](c) Express each root of the equation $16c^6 + 16(1-c^2)^3 - 13 = 0$ in the form $\cos k\pi$, where $k$ is a rational number. [5]