科目：进阶数学 Further Mathematics (9231)
年份：2009 冬季
试卷：Paper 1
总页数：7
章节：9231 Further Mathematics
1. Given that
$y = x^2 \sin x,$
(i) show that the mean value of $\frac{dy}{dx}$ with respect to $x$ over the interval $0 \leqslant x \leqslant \frac{1}{2}\pi$ is $\frac{1}{2}\pi.$ [2]
(ii) find the mean value of $\frac{d^2y}{dx^2}$ with respect to $x$ over the interval $0 \leqslant x \leqslant \frac{1}{2}\pi.$ [2]
2. Relative to an origin $O$, the points $A, B, C$ have position vectors
$i, j + k, i + j + \theta k,$
respectively. The shortest distance between the lines $AB$ and $OC$ is $\frac{1}{\sqrt{2}}$. Find the value of $\theta.$ [6]
3. The curve $C$ has equation
$y = \frac{x^2 - 5x + 4}{x + 1}.$
(i) Obtain the coordinates of the points of intersection of $C$ with the axes. [2]
(ii) Obtain the equation of each of the asymptotes of $C.$ [3]
(iii6 Show that
$$\frac{d}{dx} \left[ x^{n-1} \sqrt{(4-x^2)} \right] = \frac{4(n-1)x^{n-2}}{\sqrt{(4-x^2)}} - \frac{nx^n}{\sqrt{(4-x^2)}}.$$
[3]
Let
$$I_n = \int_0^1 \frac{x^n}{\sqrt{(4-x^2)}} dx,$$
where $n \geqslant 0$. Prove that
$$n I_n = 4(n-1) I_{n-2} - \sqrt{3},$$
for $n \geqslant 2$.
[2]
Given that $I_0 = \frac{1}{6}\pi$, find $I_4$, leaving your answer in an exact form.
[4]
7 Use de Moivre’s theorem to express $\sin^6\theta$ in the form
$$a + b \cos 2\theta + c \cos 4\theta + d \cos 6\theta,$$
where $a, b, c, d$ are constants to be found.
[5]
Hence evaluate
$$\int_0^{4\pi} \sin^6 2x dx,$$
leaving your answer in terms of $\pi$.
[4]
8 (a) The curve $C_1$ has equation $y = -\ln(\cos x)$. Show that the length of the arc of9 Show that if $y$ depends on $x$ and $x = e^u$ then
$$x^2 \frac{d^2y}{dx^2} = \frac{d^2y}{du^2} - \frac{dy}{du}.$$
[4]
Given that $y$ satisfies the differential equation
$$x^2 \frac{d^2y}{dx^2} + 5x \frac{dy}{dx} + 3y = 30x^2,$$
use the substitution $x = e^u$ to show that
$$\frac{d^2y}{du^2} + 4 \frac{dy}{du} + 3y = 30e^{2u}.$$
[2]
Hence find the general solution for $y$ in terms of $x$.
10 The curve $C$ has polar equation
$$r = a \sin 3\theta,$$
where $0 \leqslant \theta \leqslant \frac{1}{3}\pi$.
(i) Show that the area of the region enclosed by $C$ is $\frac{1}{12}\pi a^2$.
[3]
(ii) Show that, at the point of $C$ at maximum distance from the initial line,
$$\tan 3\theta + 3 \tan \theta = 0.$$
[3]
(iii) Use the formula
$$\tan 3\11 Answer only one of the following two alternatives.
EITHER
Prove by induction that
$$\sum_{n=1}^{N} n^3 = \frac{1}{4} N^2 (N+1)^2.$$ [5]
Use this result, together with the formula for $\sum_{n=1}^{N} n^2$, to show that
$$\sum_{n=1}^{N} (20n^3 + 36n^2) = N(N+1)(N+3)(5N+2).$$ [3]
Let
$$S_N = \sum_{n=1}^{N} (20n^3 + 36n^2 + \mu n).$$
Find the value of the constant $\mu$ such that $S_N$ is of the form $N^2(N+1)(aN+b)$, where the constants $a$ and $b$ are to be determined. [3]
Show that, for this value of $\mu$,
$$5 + \frac{22}{N} < N^{-4} S_N < 5 + \frac{23}{N},$$
for all $N \geq 18$. [3]
OR
One of the eigenvalues of the matrix
$$A = \begin{pmatrix}
1 & -4 & 6 \\
2 & -4 & 2 \\
-3 & 4 & a
\end{pmatrix}$$
is $-2$. Find the value of $a$. [3]
Another eigenvaBLANK PAGE
BLANK PAGE
BLANK PAGE Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. University of Cambridge International Examinations is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge.