科目：进阶数学 Further Mathematics (9231)
年份：2019 冬季
试卷：Paper 11
总页数：23
章节：9231 Further Mathematics
1 The curve $C$ has equation $y = x^a$ for $0 \leqslant x \leqslant 1$, where $a$ is a positive constant. Find, in terms of $a$, the coordinates of the centroid of the region enclosed by $C$, the line $x = 1$ and the $x$-axis. [6]
2 It is given that $y = \ln(ax + 1)$, where $a$ is a positive constant. Prove by mathematical induction that, for every positive integer $n$,
$$\frac{d^n y}{dx^n} = (-1)^{n-1} \frac{(n-1)! a^n}{(ax+1)^n}.$$
[6]3 The integral $I_n$, where $n$ is a positive integer, is defined by
$$I_n = \int_{\frac{1}{2}}^{1} x^{-n} \sin \pi x \, dx.$$
(i) Show that
$$n(n + 1)I_{n+2} = 2^{n+1}n + \pi - \pi^2 I_n.$$ [5]
(ii) Find $I_5$ in terms of $\pi$ and $I_1$. [2]4 The line $y = 2x + 1$ is an asymptote of the curve $C$ with equation
$$y = \frac{x^2 + 1}{ax + b}.$$
(i) Find the values of the constants $a$ and $b$. [3]
(ii) State the equation of the other asymptote of $C$. [1]
(iii) Sketch $C$. [Your sketch should indicate the coordinates of any points of intersection with the $y$-axis. You do not need to find the coordinates of any stationary points.] [3]5 Let $S_N = \sum_{r=1}^{N} (5r + 1)(5r + 6)$ and $T_N = \sum_{r=1}^{N} \frac{1}{(5r + 1)(5r + 6)}$.
(i) Use standard results from the List of Formulae (MF10) to show that
$$S_N = \frac{1}{3} N (25N^2 + 90N + 83).$$
[3]
(ii) Use the method of differences to express $T_N$ in terms of $N$. [4](iii) Find $\lim_{N \to \infty} \left(N^{-3} S_N T_N\right)$.
[2]6 With $O$ as the origin, the points $A, B, C$ have position vectors
$$\mathbf{i} - \mathbf{j}, \quad 2\mathbf{i} + \mathbf{j} + 7\mathbf{k}, \quad \mathbf{i} - \mathbf{j} + \mathbf{k}$$
respectively.
(i) Find the shortest distance between the lines $OC$ and $AB$. [5](ii) Find the cartesian equation of the plane containing the line $OC$ and the common perpendicular of the lines $OC$ and $AB$. [4]
7 The equation $x^3 + 2x^2 + x + 7 = 0$ has roots $\alpha, \beta, \gamma$.
(i) Use the relation $x^2 = -7y$ to show that the equation
$$49y^3 + 14y^2 - 27y + 7 = 0$$
has roots $\frac{\alpha}{\beta\gamma}, \frac{\beta}{\gamma\alpha}, \frac{\gamma}{\alpha\beta}$. [4](ii) Show that $\frac{\alpha^2}{\beta^2 \gamma^2} + \frac{\beta^2}{\gamma^2 \alpha^2} + \frac{\gamma^2}{\alpha^2 \beta^2} = \frac{58}{49}$.
(iii) Find the exact value of $\frac{\alpha^3}{\beta^3 \gamma^3} + \frac{\beta^3}{\gamma^3 \alpha^3} + \frac{\gamma^3}{\alpha^3 \beta^3}$.8 The matrix $M$ is defined by
$$M = \begin{pmatrix} 2 & m & 1 \\ 0 & m & 7 \\ 0 & 0 & 1 \end{pmatrix},$$
where $m \neq 0, 1, 2$.
(i) Find a matrix $P$ and a diagonal matrix $D$ such that $M = PDP^{-1}$. [7](ii) Find $M^7P$. [3]
9 (i) Use de Moivre’s theorem to show that
$$\sec 6\theta = \frac{\sec^6\theta}{32 - 48\sec^2\theta + 18\sec^4\theta - \sec^6\theta}.$$
[6](ii) Hence obtain the roots of the equation $$3x^6 - 36x^4 + 96x^2 - 64 = 0$$ in the form sec $q\pi$, where $q$ is rational. [5]
10 The matrix $A$ is defined by
$$A = \begin{pmatrix}
1 & 5 & 1 \\
1 & -2 & -2 \\
2 & 3 & \theta
\end{pmatrix}.$$
(i) (a) Find the rank of $A$ when $\theta \neq -1$. [3]
(b) Find the rank of $A$ when $\theta = -1$. [1]
Consider the system of equations
$$x + 5y + z = -1,$$
$$x - 2y - 2z = 0,$$
$$2x + 3y + \theta z = \theta.$$
(ii) Solve the system of equations when $\theta \neq -1$. [3](iii) Find the general solution when $\theta = -1$. [3]
(iv) Show that if $\theta = -1$ and $\phi \neq -1$ then $Ax = \begin{pmatrix} -1 \\ 0 \\ \phi \end{pmatrix}$ has no solution. [2]11 Answer only one of the following two alternatives.
EITHER
It is given that $w = \cos y$ and
$$\tan y \frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^2 + 2 \tan y \frac{dy}{dx} = 1 + e^{-2x} \sec y.$$
(i) Show that
$$\frac{d^2w}{dx^2} + 2 \frac{dw}{dx} + w = -e^{-2x}.$$ [4]
(ii) Find the particular solution for $y$ in terms of $x$, given that when $x = 0, y = \frac{1}{3}\pi$ and $\frac{dy}{dx} = \frac{1}{\sqrt{3}}$. [10]Turn over
OR
The curves $C_1$ and $C_2$ have polar equations, for $0 \leqslant \theta \leqslant \frac{1}{2}\pi$, as follows:
$$C_1: r = 2(e^{\theta} + e^{-\theta})$$
$$C_2: r = e^{2\theta} - e^{-2\theta}$$
The curves intersect at the point $P$ where $\theta = \alpha$.
(i) Show that $e^{2\alpha} - 2e^{\alpha} - 1 = 0$. Hence find the exact value of $\alpha$ and show that the value of $r$ at $P$ is $4\sqrt{2}$. [6](ii) Sketch $C_1$ and $C_2$ on the same diagram. (iii) Find the area of the region enclosed by $C_1, C_2$ and the initial line, giving your answer correct to 3 significant figures. [5]
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