34 problems found
Solution:
The distinct points \(P(2ap,\, ap^2)\) and \(Q(2aq,\, aq^2)\) lie on the curve \(x^2 = 4ay\), where \(a > 0\).
Let \(\mathbf{M} = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\) be a real matrix with \(a \neq d\). The transformation represented by \(\mathbf{M}\) has exactly two distinct invariant lines through the origin.
In the triangle \(ABC\), the base \(AB\) is of length 1 unit and the angles at~\(A\) and~\(B\) are \(\alpha\) and~\(\beta\) respectively, where \(0<\alpha\le\beta\). The points \(P\) and~\(Q\) lie on the sides \(AC\) and \(BC\) respectively, with \(AP=PQ=QB=x\). The line \(PQ\) makes an angle of~\(\theta\) with the line through~\(P\) parallel to~\(AB\).
The line passing through the point \((a,0)\) with gradient \(b\) intersects the circle of unit radius centred at the origin at \(P\) and \(Q\), and \(M\) is the midpoint of the chord \(PQ\). Find the coordinates of \(M\) in terms of \(a\) and \(b\).
Solution: \begin{align*} && y &= bx-ba \\ && 1 &= x^2 + y^2 \\ \Rightarrow && 1 &= x^2 + b^2(x-a)^2 \\ \Rightarrow && 0 &= (1+b^2)x^2-2ab^2x+b^2a^2-1 \end{align*} This will have roots which sum to \(\frac{2ab^2}{1+b^2}\), therefore \(M = \left ( \frac{ab^2}{1+b^2}, \frac{ab^3}{1+b^2}-ba \right)=\left ( \frac{ab^2}{1+b^2}, \frac{-ba}{1+b^2} \right)\)
A tall shot-putter projects a small shot from a point \(2.5\,\)m above the ground, which is horizontal. The speed of projection is \(10\,\text{ms}^{- 1}\) and the angle of projection is \(\theta\) above the horizontal. Taking the acceleration due to gravity to be \(10\,\text{ms}^{-2}\), show that the time, in seconds, that elapses before the shot hits the ground is \[ \frac1{\sqrt2}\left ( \sqrt{1-c}+ \sqrt{2-c}\right), \] where \(c = \cos2\theta\). Find an expression for the range in terms of \(c\) and show that it is greatest when \(c= \frac15\,\). Show that the extra distance attained by projecting the shot at this angle rather than at an angle of \(45^\circ\) is \(5(\sqrt6 -\sqrt2 -1)\,\)m.
Solution: \begin{align*} && s &= ut + \frac12 gt^2 \\ \Rightarrow && -2.5 &= 10 \sin \theta \, T - 5 T^2 \\ \Rightarrow && T &= \frac{10\sin \theta \pm \sqrt{100\sin^2 \theta - 4 \cdot 5 \cdot (-2.5)}}{10} \\ &&&= \sin \theta +\sqrt{\sin^2 \theta + \frac12} \\ &&&= \frac1{\sqrt{2}} \left ( \sqrt{2} \sin \theta +\sqrt{2 \sin^2 \theta +1} \right) \\ &&&= \frac1{\sqrt{2}} \left ( \sqrt{2 (1-\cos^2 \theta)} + \sqrt{2-\cos 2\theta} \right) \\ &&&= \frac1{\sqrt{2}} \left ( \sqrt{1-\cos2 \theta} + \sqrt{2-\cos 2\theta} \right) \\ &&&= \frac{1}{\sqrt{2}}\left ( \sqrt{1-c}+\sqrt{2-c} \right)\\ \\ && s &= 10 \cos \theta T \\ &&&= 10 \sqrt{\frac{\cos 2 \theta +1}{2}}\frac{1}{\sqrt{2}}\left ( \sqrt{1-c}+\sqrt{2-c} \right) \\ &&&= 5 \sqrt{c+1}\left ( \sqrt{1-c}+\sqrt{2-c} \right) \\ \\ && \frac15\frac{\d s}{\d c} &= \frac12(c+1)^{-1/2}((1-c)^{1/2} + (2-c)^{1/2}) - \frac12(c+1)^{1/2}\left ((1-c)^{-1/2}+(2-c)^{-1/2} \right) \\ &&&= \frac{((1-c)(2-c)^{1/2}+(2-c)(1-c)^{1/2})-((c+1)(2-c)^{1/2}+(c+1)(1-c)^{1/2})}{2\sqrt{c+1}\sqrt{1-c}\sqrt{2-c}} \\ &&&= \frac{\sqrt{2-c}\left (1-c-c-1 \right)+\sqrt{1-c}\left(2-c-c-1) \right)}{2\sqrt{c+1}\sqrt{1-c}\sqrt{2-c}} \\ &&&= \frac{\sqrt{1-c}\left(1-2c\right)-2c\sqrt{2-c}}{2\sqrt{c+1}\sqrt{1-c}\sqrt{2-c}} \\ \\ \frac{\d s}{\d c} =0: && \sqrt{1-c}\left(1-2c\right)&=2c\sqrt{2-c} \\ \Rightarrow && (1-c)(1-2c)^2&=4c^2(2-c) \\ \Rightarrow && 1-5c+8c^2-4c^3 &= 8c^2-4c^3 \\ \Rightarrow && 0 &= -5c+1 \\ \Rightarrow && c &= \frac15 \end{align*} When \(\theta = 45^{\circ}, c = 0\), so \(s_{45^{\circ}} = 5(1+\sqrt{2})\) When \(c = \frac15\), \begin{align*} s &= 5 \sqrt{\frac15+1}\left ( \sqrt{1-\frac15}+\sqrt{2-\frac15} \right) \\ &= 5 \sqrt{\frac65} \left ( \sqrt{\frac45} + \sqrt{\frac95} \right) \\ &= 2\sqrt{6}+3\sqrt{6} = 5\sqrt{6} \end{align*} Therefore the additional distance is \(5(\sqrt{6}-\sqrt{2}-1)\)
Let \(x+{\rm i} y\) be a root of the quadratic equation \(z^2 + pz +1=0\), where \(p\) is a real number. Show that \(x^2-y^2 +px+1=0\) and \((2x+p)y=0\). Show further that either \(p=-2x\) or \(p=-(x^2+1)/x\) with \(x\ne0\). Hence show that the set of points in the Argand diagram that can (as \(p\) varies) represent roots of the quadratic equation consists of the real axis with one point missing and a circle. This set of points is called the root locus of the quadratic equation. Obtain and sketch in the Argand diagram the root locus of the equation \[ pz^2 +z+1=0\, \] and the root locus of the equation \[ pz^2 + p^2z +2=0\,.\]
Solution: \begin{align*} && 0 &= z^2 + pz + 1\\ &&&= (x+iy)^2 + (x+iy)p + 1 \\ &&& = (x^2-y^2+px+1) + (2xy+py)i \\ \Rightarrow && 0 &= x^2 - y^2 + px + 1 \\ && 0 &= (2x+p)y \\ \Rightarrow && p &= -2x \\ \text{ or } && y &= 0 \\ \Rightarrow && p &= -(x^2+1)/x \end{align*} Therefore as \(p\) varies with either have \(y = 0\) and \(x\) taking any real value except \(0\) ie the real axis minus the origin. Or \(p = -2x\) and \(-y^2-x^2+1 = 0 \Rightarrow x^2 + y^2 = 1\) which is a circle. Suppose \(pz^2 + z + 1 = 0\) \begin{align*} && 0 &= pz^2 + z +1\\ &&&= p(x+iy)^2 + (x+iy) + 1\\ &&&= (px^2-py^2+x+1) + (2xyp + y) i \\ \Rightarrow && 0 &= (2xp+1)y \\ \Rightarrow && y & = 0, p = \frac{-(x+1)}{x^2}, x \neq 0 \\ \text{ or } && p &= -\frac{1}{2x}\\ \Rightarrow && 0 &= -\frac{1}{2}x + \frac{y^2}{2x} + x + 1 \\ &&&= \frac{y^2 - x^2 +2x^2 + 2x}{2x} \\ &&&= \frac{(x+1)^2+y^2-1}{2x} \end{align*} So we either have the real axis (except \(0\)) or a circle radius \(1\) centre \((-1, 0)\) (excluding \(x = 0\)).
Show that, provided \(q^2\ne 4p^3\), the polynomial \[ \hphantom{(p\ne0, \ q\ne0)\hspace{2cm}} x^3-3px +q \hspace {2cm} (p\ne0, \ q\ne0) \] can be written in the form \[ a(x-\alpha)^3 + b(x-\beta)^3\,, \] where \(\alpha\) and \(\beta\) are the roots of the quadratic equation \(pt^2 -qt +p^2=0\), and \(a\) and \(b\) are constants which you should express in terms of \(\alpha\) and \(\beta\). Hence show that one solution of the equation \(x^3-24x+48=0\,\) is \[ x= \frac{2 (2-2^{\frac13})}{1-2^{\frac13}} \] and obtain similar expressions for the other two solutions in terms of \(\omega\), where \(\omega = \mathrm{e}^{2\pi\mathrm{i}/3}\,\). Find also the roots of \(x^3-3px +q=0\) when \(p=r^2\) and \(q= 2r^3\) for some non-zero constant \(r\).
Two particles \(P\) and \(Q\) are projected simultaneously from points \(O\) and \(D\), respectively, where~\(D\) is a distance \(d\) directly above \(O\). The initial speed of \(P\) is \(V\) and its angle of projection {\em above} the horizontal is \(\alpha\). The initial speed of \(Q\) is \(kV\), where \(k>1\), and its angle of projection {\em below} the horizontal is \(\beta\). The particles collide at time \(T\) after projection. Show that \(\cos\alpha = k\cos\beta\) and that \(T\) satisfies the equation \[ (k^2-1)V^2T^2 +2dVT\sin\alpha -d^2 =0\,. \] Given that the particles collide when \(P\) reaches its maximum height, find an expression for~\(\sin^2\alpha\) in terms of \(g\), \(d\), \(k\) and \(V\), and deduce that \[ gd\le (1+k)V^2\,. \]
The points \(S\), \(T\), \(U\) and \(V\) have coordinates \((s,ms)\), \((t,mt)\), \((u,nu)\) and \((v,nv)\), respectively. The lines \(SV\) and \(UT\) meet the line \(y=0\) at the points with coordinates \((p,0)\) and \((q,0)\), respectively. Show that \[ p = \frac{(m-n)sv}{ms-nv}\,, \] and write down a similar expression for \(q\). Given that \(S\) and \(T\) lie on the circle \(x^2 + (y-c)^2 = r^2\), find a quadratic equation satisfied by \(s\) and by \(t\), and hence determine \(st\) and \(s+t\) in terms of \(m\), \(c\) and \(r\). Given that \(S\), \(T\), \(U\) and \(V\) lie on the above circle, show that \(p+q=0\).
A particle is projected under gravity from a point \(P\) and passes through a point \(Q\). The angles of the trajectory with the positive horizontal direction at \(P\) and at \(Q\) are \(\theta\) and \(\phi\), respectively. The angle of elevation of \(Q\) from \(P\) is \(\alpha\).
In this question, use \(g=10\,\)m\,s\(^{-2}\). In cricket, a fast bowler projects a ball at \(40\,\)m\,s\(^{-1}\) from a point \(h\,\)m above the ground, which is horizontal, and at an angle \(\alpha\) above the horizontal. The trajectory is such that the ball will strike the stumps at ground level a horizontal distance of \(20\,\)m from the point of projection.
In the High Court of Farnia, the outcome of each case is determined by three judges: the ass, the beaver and the centaur. Each judge decides its verdict independently. Being simple creatures, they make their decisions entirely at random. Past verdicts show that the ass gives a guilty verdict with probability \(p\), the beaver gives a guilty verdict with probability \(p/3\) and the centaur gives a guilty verdict with probability \(p^2\). Let \(X\) be the number of guilty verdicts given by the three judges in a case. Given that \(\E(X)= 4/3\), find the value of \(p\). The probability that a defendant brought to trial is guilty is \(t\). The King pronounces that the defendant is guilty if at least two of the judges give a guilty verdict; otherwise, he pronounces the defendant not guilty. Find the value of \(t\) such that the probability that the King pronounces correctly is \(1/2\).
Solution: \begin{align*} && \mathbb{E}(X) &= p + \frac{p}{3} + p^2 = \frac43p+p^2 \\ \Rightarrow && \frac43 &= \frac43p+p^2 \\ \Rightarrow && 0 &= 3p^2+4p-4 \\ &&&= (3p-2)(p+2) \\ \Rightarrow && p &= \frac23, -2 \end{align*} Since \(p \in [0,1]\) we must have \(p = \frac23\). \begin{align*} && \mathbb{P}(\text{correct verdict}) &= t p+ (1-t) (1-p) \\ &&&= t(2p-1)+(1-p)\\ \Rightarrow && \frac12 &= t(2p-1)+(1-p) \\ \Rightarrow && t &= \frac{\frac12-(1-p)}{2p-1} \\ &&&= \frac{2p-1}{2(2p-1)} = \frac12 \end{align*} (so it doesn't depend at all on what the judges are doing, the only way to be fair is if the trials happen at random!)
The functions \({\rm T}_n(x)\), for \(n=0\), 1, 2, \(\ldots\,\), satisfy the recurrence relation \[ {\rm T}_{n+1}(x) -2x {\rm T}_n(x) + {\rm T}_{n-1}(x) =0\, \ \ \ \ \ \ \ (n\ge1). \tag{\(*\)} \] Show by induction that \[ \left({\rm T}_n(x)\right)^2 - {\rm T}_{n-1}(x) {\rm T}_{n+1}(x) = \f(x)\,, \] where \(\f(x) = \left({\rm T}_1(x)\right)^2 - {\rm T}_0(x){\rm T}_2(x)\,\). In the case \(\f(x)\equiv 0\), determine (with proof) an expression for \({\rm T}_n(x)\) in terms of \({\rm T}_0(x)\) (assumed to be non-zero) and \({\rm r}(x)\), where \({\rm r}(x) = {\rm T}_1(x)/ {\rm T}_0(x)\). Find the two possible expressions for \({\rm r}(x)\) in terms of \(x\). %Conjecture (without proof) the general form of the solution of \((*)\).