16 problems found
Solution:
The distinct points \(P_{1},P_{2},P_{3},Q_{1},Q_{2}\) and \(Q_{3}\) in the Argand diagram are represented by the complex numbers \(z_{1},z_{2},z_{3},w_{1},w_{2}\) and \(w_{3}\) respectively. Show that the triangles \(P_{1}P_{2}P_{3}\) and \(Q_{1}Q_{2}Q_{3}\) are similar, with \(P_{i}\) corresponding to \(Q_{i}\) (\(i=1,2,3\)) and the rotation from \(1\) to \(2\) to \(3\) being in the same sense for both triangles, if and only if \[ \frac{z_{1}-z_{2}}{z_{2}-z_{3}}=\frac{w_{1}-w_{2}}{w_{1}-w_{3}}. \] Verify that this condition may be written \[ \det\begin{pmatrix}z_{1} & z_{2} & z_{3}\\ w_{1} & w_{2} & w_{3}\\ 1 & 1 & 1 \end{pmatrix}=0. \]
The function \(\mathrm{f}\) is defined for \(x<2\) by \[ \mathrm{f}(x)=2| x^{2}-x|+|x^{2}-1|-2|x^{2}+x|. \] Find the maximum and minimum points and the points of inflection of the graph of \(\mathrm{f}\) and sketch this graph. Is \(\mathrm{f}\) continuous everywhere? Is \(\mathrm{f}\) differentiable everywhere? Find the inverse of the function \(\mathrm{f}\), i.e. expressions for \(\mathrm{f}^{-1}(x),\) defined in the various appropriate intervals.
Solution: \[ f(x) = 2|x(x-1)| + |(x-1)(x+1)|-2|x(x+1)| \] Therefore the absolute value terms will change behaviour at \(x = -1, 0, 1\). Then \begin{align*} f(x) &= \begin{cases} 2(x^2-x)+(x^2-1)-2(x^2+x) & x \leq -1 \\ 2(x^2-x)-(x^2-1)+2(x^2+x) & -1 < x \leq 0 \\ -2(x^2-x)-(x^2-1)-2(x^2+x) & 0 < x \leq 1 \\ 2(x^2-x)+(x^2-1)-2(x^2+x) & 1 < x\end{cases} \\ &= \begin{cases} x^2-4x-1 & x \leq -1 \\ 3x^2+1& -1 < x \leq 0 \\ -5x^2+1& 0 < x \leq 1 \\ x^2-4x-1 & 1 < x\end{cases} \\ \\ f'(x) &= \begin{cases} 2x-4 & x <-1 \\ 6x & -1 < x < 0 \\ -10x & 0 < x < 1 \\ 2x-4 & 1 < x\end{cases} \\ \end{align*} Therefore \(f'(x) = 0 \Rightarrow x = 0, 2\) and so we should check all the turning points. Therefore the minimum is \(x = 2, y = -5\), maximum is \(x = -2, y = 11\) (assuming the range is actually \(|x| < 2\). There is a point of inflection at \(x = 0, y = 1\).
The point \(P\) moves on a straight line in three-dimensional space. The position of \(P\) is observed from the points \(O_{1}(0,0,0)\) and \(O_{2}(8a,0,0).\) At times \(t=t_{1}\) and \(t=t_{1}'\), the lines of sight from \(O_{1}\) are along the lines \[ \frac{x}{2}=\frac{z}{3},y=0\quad\mbox{ and }\quad x=0,\frac{y}{3}=\frac{z}{4} \] respectively. At times \(t=t_{2}\) and \(t=t_{2}'\), the lines of sight from \(O_{2}\) are \[ \frac{x-8a}{-3}=\frac{y}{1}=\frac{z}{3}\quad\mbox{ and }\quad\frac{x-8a}{-4}=\frac{y}{2}=\frac{z}{5} \] respectively. Find an equation or equations for the path of \(P\).
The curve \(C\) has the differential equation in polar coordinates \[ \frac{\mathrm{d}^{2}r}{\mathrm{d}\theta^{2}}+4r=5\sin3\theta,\qquad\text{for }\quad\frac{\pi}{5}\leqslant\theta\leqslant\frac{3\pi}{5}, \] and, when \(\theta=\dfrac{\pi}{2},\) \(r=1\) and \(\dfrac{\mathrm{d}r}{\mathrm{d}\theta}=-2.\) Show that \(C\) forms a closed loop and that the area of the region enclosed by \(C\) is \[ \frac{\pi}{5}+\frac{25}{48}\left[\sin\left(\frac{\pi}{5}\right)-\sin\left(\frac{2\pi}{5}\right)\right]. \]
Solution: First we seek the complementary function. \begin{align*} && \frac{\mathrm{d}^{2}r}{\mathrm{d}\theta^{2}}+4r &= 0 \\ \Rightarrow && r &= A \sin 2\theta + B \cos 2 \theta \end{align*} Next we seek a particular integral, of the form \(r = C \sin 3 \theta\). \begin{align*} && \frac{\mathrm{d}^{2}r}{\mathrm{d}\theta^{2}}+4r &= 5 \sin 3 \theta \\ \Rightarrow && -9C \sin 3 \theta + 4C \sin 3 \theta &= 5 \sin 3 \theta \\ \Rightarrow && C &= -1 \\ \end{align*} So our general solution is \(A \sin 2\theta + B \cos 2 \theta -\sin 3 \theta\). Plugging in boundary conditions we obtain: \begin{align*} \theta = \frac{\pi}{2}, r = 1: &&1 &= -B +1 \\ \Rightarrow && B &= 0 \\ \theta = \frac{\pi}{2}, \frac{\d r}{\d \theta} = -2: && -2 &= -2A \\ \Rightarrow && A &= 1 \end{align*} So the general solution is \(r = \sin 2 \theta - \sin 3 \theta = 2 \sin \left ( \frac{-\theta}{2} \right) \cos \left (\frac{5 \theta}{2} \right)\) First notice that for \(\theta \in \left [\frac{\pi}{5}, \frac{3 \pi}{5} \right]\) this is positive, and it is zero on the end points, therefore we are tracing out a a loop. The area of the loop will be: \begin{align*} A &= \int_{\pi/5}^{3\pi/5} \frac12 \left ( \sin 2 \theta - \sin 3 \theta \right)^2 \d \theta \\ &= \frac12\int_{\pi/5}^{3\pi/5} \sin^2 2\theta + \sin^2 3 \theta - 2 \sin 2 \theta \cos 3 \theta \d \theta \\ &= \frac12\int_{\pi/5}^{3\pi/5} \frac{1-2 \cos 4 \theta}{2} + \frac{1-2 \cos6 \theta}{2} - \sin5 \theta-\cos\theta \d \theta \\ &= \frac12 \left [\theta - \frac14 \sin 4 \theta-\frac16 \sin 6 \theta + \frac15 \cos 5 \theta - \sin \theta \right]_{\pi/5}^{3\pi/5} \\ &= \frac{\pi}{5} +\frac{25}{48}\left [ \sin\left(\frac{\pi}{5}\right)-\sin\left(\frac{2\pi}{5}\right) \right] \end{align*}
The transformation \(T\) from \(\binom{x}{y}\) to \(\binom{x'}{y'}\) in two-dimensional space is given by \[ \begin{pmatrix}x'\\ y' \end{pmatrix}=\begin{pmatrix}\cosh u & \sinh u\\ \sinh u & \cosh u \end{pmatrix}\begin{pmatrix}x\\ y \end{pmatrix}, \] where \(u\) is a positive real constant. Show that the curve with equation \(x^{2}-y^{2}=1\) is transformed into itself. Find the equations of two straight lines through the origin which transform into themselves. A line, not necessary through the origin, which has gradient \(\tanh v\) transforms under \(T\) into a line with gradient \(\tanh v'\). Show that \(v'=v+u\). The lines \(\ell_{1}\) and \(\ell_{2}\) with gradients \(\tanh v_{1}\) and \(\tanh v_{2}\) transform under \(T\) into lines with gradients \(\tanh v_{1}'\) and \(\tanh v_{2}'\) respectively. Find the relation satisfied by \(v_{1}\) and \(v_{2}\) that is the necessary and sufficient for \(\ell_{1}\) and \(\ell_{2}\) to intersect at the same angle as their transforms. In the case when \(\ell_{1}\) and \(\ell_{2}\) meet at the origin, illustrate in a diagram the relation between \(\ell_{1}\), \(\ell_{2}\) and their transforms.
Solution:
Solution:
The parametric equations \(E_{1}\) and \(E_{2}\) define the same ellipse, in terms of the parameters \(\theta_{1}\) and \(\theta_{2}\), (though not referred to the same coordinate axes). \begin{alignat*}{2} E_{1}:\qquad & x=a\cos\theta_{1}, & \quad & y=b\sin\theta_{1},\\ E_{2}:\qquad & x=\dfrac{k\cos\theta_{2}}{1+e\cos\theta_{2}}, & \quad & y=\dfrac{k\sin\theta_{2}}{1+e\cos\theta_{2}}, \end{alignat*} where \(0< b< a,\) \(0< e< 1\) and \(0< k\). Find the position of the axes for \(E_{2}\) relative to the axes for \(E_{1}\) and show that \(k=a(1-e^{2})\) and \(b^{2}=a^{2}(1-e^{2}).\) {[}The standard polar equation of an ellipse is \(r=\dfrac{\ell}{1+e\cos\theta}.]\) By considering expressions for the length of the perimeter of the ellipse, or otherwise, prove that \[ \int_{0}^{\pi}\sqrt{1-e^{2}\cos^{2}\theta}\,\mathrm{d}\theta=\int_{0}^{\pi}\frac{1-e^{2}}{(1+e\cos\theta)^{2}}\sqrt{1+e^{2}+2e\cos\theta}\,\mathrm{d}\theta. \] Given that \(e\) is so small that \(e^{6}\) may be neglected, show that the value of either integral is \[ \tfrac{1}{64}\pi(64-16e^{2}-3e^{4}). \]
The equation \[ x^{n}-qx^{n-1}+r=0, \] where \(n\geqslant5\) and \(q\) and \(r\) are real constants, has roots \(\alpha_{1},\alpha_{2},\ldots,\alpha_{n}.\) The sum of the products of \(m\) distinct roots is denoted by \(\Sigma_{m}\) (so that, for example, \(\Sigma_{3}=\sum\alpha_{i}\alpha_{j}\alpha_{k}\) where the sum runs over the values of \(i,j\) and \(k\) with \(n\geqslant i>j>k\geqslant1\)). The sum of \(m\)th powers of the roots is denoted by \(S_{m}\) (so that, for example, \(S_{3}=\sum\limits_{i=1}^{n}\alpha_{i}^{3}\)). Prove that \(S_{p}=q^{p}\) for \(1\leqslant p\leqslant n-1.\) You may assume that for any \(n\)th degree equation and \(1\leqslant p\leqslant n\) \[ S_{p}-S_{p-1}\Sigma_{1}+S_{p-2}\Sigma_{2}-\cdots+(-1)^{p-1}S_{1}\Sigma_{p-1}+(-1)^{p}p\Sigma_{p}=0.] \] Find expressions for \(S_{n},\) \(S_{n+1}\) and \(S_{n+2}\) in terms of \(q,r\) and \(n\). Suggest an expression for \(S_{n+m},\) where \(m < n\), and prove its validity by induction.
Solution: Claim: \(S_p = q^p\) for \(1 \leq p \leq n-1\) Proof: When \(p = 1\), \(S_p = \Sigma_1 = q\) as expected. Note that \(\Sigma_i = 0\) for \(i = 2, \cdots, n-1\). Using \(S_p = S_{p-1}\Sigma_{1}-S_{p-2}\Sigma_{2}+\cdots+(-1)^{p-1+1}S_{1}\Sigma_{p-1}+(-1)^{p+1}p\Sigma_{p}\), we can see that \(S_p = qS_{p-q}\) when \(1 \leq p \leq n-1\), ie \(S_p = q^p\). Note that \begin{align*} S_n &= \sum \alpha_i^n \\ &= q\sum \alpha_i^{n-1} - \sum r \\ &= qS_{n-1} - nr \\ &= q^n - nr \\ \\ S_{n+1} &= \sum \alpha_i^{n+1} \\ &= q \sum \alpha_i^{n} - r \sum \alpha_i \\ &= q^{n+1} - rq \\ \\ S_{n+2} &= \sum \alpha_i^{n+2} \\ &= q \sum \alpha_i^{n+1} - r \sum \alpha_i^2 \\ &= q^{n+2} - rq^2 \\ \end{align*} Claim: \(S_{n+m} = q^{n+m} - rq^{m}\) Proof: The obvious
A smooth tube whose axis is horizontal has an elliptic cross-section in the form of the curve with parametric equations \[ x=a\cos\theta\qquad y=b\sin\theta \] where the \(x\)-axis is horizontal and the \(y\)-axis is vertically upwards. A particle moves freely under gravity on the inside of the tube in the plane of this cross-section. By first finding \(\ddot{x}\) and \(\ddot{y},\) or otherwise, show that the acceleration along the inward normal at the point with parameter \(\theta\) is \[ \frac{ab\dot{\theta}^{2}}{\sqrt{a^{2}\sin^{2}\theta+b^{2}\cos^{2}\theta}}. \] The particle is projected along the surface in the vertical cross-section plane, with speed \(2\sqrt{bg},\) from the lowest point. Given that \(2a=3b,\) show that it will leave the surface at the point with parameter \(\theta\) where \[ 5\sin^{3}\theta+12\sin\theta-8=0. \]
A smooth particle \(P_{1}\) is projected from a point \(O\) on the horizontal floor of a room with has a horizontal ceiling at a height \(h\) above the floor. The speed of projection is \(\sqrt{8gh}\) and the direction of projection makes an acute angle \(\alpha\) with the horizontal. The particle strikes the ceiling and rebounds, the impact being perfectly elastic. Show that for this to happen \(\alpha\) must be at least \(\frac{1}{6}\pi\) and that the range on the floor is then \[ 8h\cos\alpha\left(2\sin\alpha-\sqrt{4\sin^{2}\alpha-1}\right). \] Another particle \(P_{2}\) is projected from \(O\) with the same velocity as \(P_{1}\) but its impact with the ceiling is perfectly inelastic. Find the difference \(D\) between the ranges of \(P_{1}\) and \(P_{2}\) on the floor and show that, as \(\alpha\) varies, \(D\) has a maximum value when \(\alpha=\frac{1}{4}\pi.\)
A pack of \(2n\) (where \(n\geqslant4\)) cards consists of two each of \(n\) different sorts. If four cards are drawn from the pack without replacement show that the probability that no pairs of identical cards have been drawn is \[ \frac{4(n-2)(n-3)}{(2n-1)(2n-3)}. \] Find the probability that exactly one pair of identical cards is included in the four. If \(k\) cards are drawn without replacement and \(2 < k < 2n,\) find an expression for the probability that there are exactly \(r\) pairs of identical cards included when \(r < \frac{1}{2}k.\) For even values of \(k\) show that the probability that the drawn cards consist of \(\frac{1}{2}k\) pairs is \[ \frac{1\times3\times5\times\cdots\times(k-1)}{(2n-1)(2n-3)\cdots(2n-k+1)}. \]