Problems

1960 Paper 4 Q301

D: 1500.0 B: 1474.1

The roots of $x^2 - sx + p = 0$ are $\alpha$ and $\beta$. By considering $$\frac{1}{1-\alpha y} + \frac{1}{1-\beta y},$$ or otherwise, show that $$\alpha^n + \beta^n = \sum_{0 \leqslant 2r \leqslant n} (-p)^r s^{n-2r} \{ 2(r,n-2r)-(r,n-2r-1) \}$$ where $(i,j)$ denotes the binomial coefficient $(i+j)!/i!j!$ if $i$ and $j$ are positive, and is defined to be zero if $j = -1$.

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1960 Paper 4 Q302

D: 1500.0 B: 1500.0

$f(x)$ is a polynomial of degree $n > 0$, and $f'(x)$ is its derivative. Every (real or complex) root $x$ of $f'(x) = 0$ also satisfies $f(x) = 0$. Prove that $f(x) = 0$ has a single root of multiplicity $n$.

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1960 Paper 4 Q303

D: 1500.0 B: 1500.0

A sequence of integers $a_n$ is defined by $$a_1 = 2,$$ $$a_{n+1} = a_n^2 - a_n + 1 \quad (n > 0).$$ Prove:

$\sum_{n=1}^{\infty} \frac{1}{a_n} = 1$;
If $m \neq n$, then $a_m$ and $a_n$ have no common factor greater than 1.

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1960 Paper 4 Q304

D: 1500.0 B: 1500.0

$A$, $B$ and $C$ are the three angles of a triangle. Show that $$\begin{vmatrix} \sin A & \sin B & \sin C \\ \cos A & \cos B & \cos C \\ \sin^3 A & \sin^3 B & \sin^3 C \end{vmatrix} = 0.$$

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1960 Paper 4 Q305

D: 1500.0 B: 1500.0

If $x_1, x_2, \ldots, x_n$ denote the complex $n$th roots of unity, evaluate $$\prod_{i< j} (x_i - x_j).$$

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1960 Paper 4 Q306

D: 1500.0 B: 1500.0

Let $f(x)$ be a real differentiable function defined for $a < x < b$ and suppose that $$f(a) = f(b) = 0.$$ Show that there is at least one number $\xi$ in $a < \xi < b$ for which $$|f'(\xi)| \geq \frac{4}{(b-a)^2} \left| \int_a^b |f(x)| dx \right|.$$

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1960 Paper 4 Q307

D: 1500.0 B: 1500.0

Evaluate the following integrals, where $z$ is any real number and $n$ is any positive integer: $$\int_{-1}^{+1} \frac{dx}{\sqrt{(1-2xz+z^2)^{}}}, \quad \int_0^{\pi} \frac{\sin nx}{\sin x} dx.$$

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1960 Paper 4 Q308

D: 1500.0 B: 1500.0

Solve completely the following differential equations:

$y' = y + e^{-x}$;
$x^3y'' + xy = 1$;
$y' = \frac{2x+3y}{3x+8y}$.

Solution:

We can see the general solution of $y'=y$ is $Ae^x$, therefore we are looking for a particular integral of the form $y = Ce^{-x}$, so $-Ce^{-x} = 2e^{-x} \Rightarrow C = -2$. Therefore the general solution is $y = Ae^x-2e^{-x}$
First consider $x^2y'' + y = 0$. Consider solutions of the form $y = x^t$, so $(t(t-1)+1)x^t = 0$, ie $t^2 - t + 1 = 0 \Rightarrow t = \frac{1 \pm \sqrt{-3}}{2}$, ie the general solution is $Ax^{\omega} + B x^{\omega^2}$. Now we seek a particular integral for $x^2 y'' + y = \frac{1}{x}$. Trying $y = \frac{C}{x}$ we find $\frac{2C}{x} + \frac{C}{x} = \frac{1}{x} \Rightarrow C = \frac{1}{3}$.Therefore the general solution is $y = Ae^x - 2e^{-x} + \frac{1}{3x}$
\begin{align*} && \frac{\d y}{\d x} &= \frac{2x+3y}{3x+8y} \\ y = vx: &&v + x\frac{\d v}{\d x} &= \frac{2+3v}{3+8v} \\ \Rightarrow && \frac{\d v}{\d x} &= \frac{2-8v^2}{x (3+8v)} \\ \Rightarrow && \int \frac{2}{x} \d x &= \int \frac{3+8v}{(1-2v)(1+2v)} \d v \\ &&&=\frac12 \int \left ( \frac{7}{1-2v} - \frac{1}{1+2v} \right) \d v \\ \Rightarrow && 8\ln |x| &= -7\ln|1-2v| - \ln|1+2v| + C \\ \Rightarrow && 8 \ln |x| &= - \ln |(1-2v)^7(1+2v)| +C \\ \Rightarrow && C &= \ln \left (x^8\left(1-2\frac{y}{x}\right)^7\left(1+2\frac{y}{x}\right) \right) \\ \Rightarrow && K &= (x-2y)^7(x+2y) \end{align*} Alternative method: Notice that $\dot{y} = 2x+3y$ and $\dot{x} = 3x+8y$, which is the equation $\begin{pmatrix} \dot{x} \\ \dot{y} \end{pmatrix} = \begin{pmatrix} 3 & 8 \\ 2 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$. This has eigenvalues $\lambda = 7,-1$ and eigenvectors $\binom{2}{1}$ and $\binom{-2}{1}$, therefore $\begin{pmatrix} x \\ y \end{pmatrix} = Ae^{7t}\begin{pmatrix} 2 \\ 1 \end{pmatrix} + Be^{-t}\begin{pmatrix} -2 \\ 1 \end{pmatrix}$. So we can write: \begin{align*} && x(t) &= 2Ae^{7t}-2Be^{-t} \\ && y(t) &= Ae^{7t}+Be^{-t} \\ \Rightarrow && 4Ae^{7t} &= x+2y \\ && 4Be^{-t} &= 2y-x \\ \Rightarrow && (2y-x)^7(x+2y) &= K \end{align*}

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1960 Paper 4 Q309

D: 1500.0 B: 1500.0

If $y_m(x)$ is defined as a function of $x$ by the equation $$y_m(x) = (-1)^m e^{x^2} \frac{d^m}{dx^m} e^{-x^2},$$ show that $y_m$ is a polynomial in $x$ and that $$y_{n+1}(x) = -\frac{d}{dx} y_n(x) + 2xy_n(x).$$ Deduce, by induction on $n$ or otherwise, that $$\frac{d^2}{dx^2} y_n(x) - 2x \frac{d}{dx} y_n(x) + 2ny_n(x) = 0.$$

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1960 Paper 4 Q310

D: 1500.0 B: 1500.0

Let $$f_m(x) = \frac{x}{2} \left[ \sin x - \frac{\sin 2x}{2} + \frac{\sin 3x}{3} - \ldots + (-1)^{m+1} \frac{\sin mx}{m} \right].$$ By considering $df_m(x)/dx$, or otherwise, show that $$(-1)^m f_m(x) > 0$$ for $0 < x < \pi/(2m+1)$. Show also that $$(-1)^m f_m\left(\frac{\pi}{m+\frac{1}{3}}\right) < 0.$$

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