Problems
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2001 Paper 1 Q1
The points \(A\), \(B\) and \(C\) lie on the sides of a square of side 1 cm and no two points lie on the same side. Show that the length of at least one side of the triangle \(ABC\) must be less than or equal to \((\sqrt6 -\sqrt2)\) cm.
2001 Paper 1 Q2
Solve the inequalities
- \(1+2x-x^2 >2/x \quad (x\ne0)\) ,
- \(\sqrt{3x+10} > 2+\sqrt{x+4} \quad (x\ge -10/3)\).
Solution:
- \(\,\)
\begin{align*} && 1+2x-x^2 = 2/x \\ \Rightarrow && 0 &= x^3-2x^2-x+2 \\ &&&= (x+1)(x^2-3x+2) \\ &&&= (x+1)(x-1)(x-2) \end{align*} Therefore the inequality is satisfied on \((1,2)\) and \((-1,0)\)
- \(\,\)
\begin{align*} && \sqrt{3x+10} &= 2+\sqrt{x+4} \\ && 3x+10 &= x+8 + 4\sqrt{x+4} \\ && 16(x+4) &= 4(x+1)^2 \\ && 4x+16 &= x^2+2x+1 \\ \Rightarrow && 0 &= x^2-2x-15 \\ &&&= (x-5)(x+3) \end{align*} Therefore \(x > 5\)
2001 Paper 1 Q3
Sketch, without calculating the stationary points, the graph of the function \(\f(x)\) given by \[ \f(x) = (x-p)(x-q)(x-r)\;, \] where \(p < q < r\). By considering the quadratic equation \(\f'(x)=0\), or otherwise, show that \[ (p+q+r)^2 > 3(qr+rp+pq)\;. \] By considering \((x^2+gx+h)(x-k)\), or otherwise, show that \(g^2>4h\,\) is a sufficient condition but not a necessary condition for the inequality \[ (g-k)^2>3(h-gk) \] to hold.
Solution:
2001 Paper 1 Q4
Show that \(\displaystyle \tan 3\theta = \frac{3\tan\theta -\tan^3\theta}{1-3\tan^2\theta}\) . Given that \(\theta= \cos^{-1} (2/\sqrt5)\) and \(0<\theta<\pi/2\), show that \(\tan 3\theta =11/2\) Hence, or otherwise, find all solutions of the equations
- \(\tan(3\cos^{-1} x) =11/2\) ,
- \(\cos ({\frac13}\tan^{-1} y) = 2/\sqrt5\) .
Solution: Let \(\tan \theta = t\) \begin{align*} \tan 3 \theta &\equiv \tan (2 \theta + \theta) \\ &\equiv \frac{\tan 2 \theta +\tan \theta}{1 - \tan 2 \theta \tan \theta} \\ &\equiv \frac{\frac{2t}{1-t^2}+t}{1-\frac{2t^2}{1-t^2}} \\ &\equiv \frac{2t+t-t^3}{1-t^2-2t^2} \\ &\equiv \frac{3t-t^3}{1-3t^3} \\ &\equiv \frac{3\tan \theta - \tan^3 \theta}{1 - 3 \tan^3 \theta} \end{align*} If \(\theta = \cos^{-1} (2/\sqrt{5})\), then \(\sin \theta = 1/\sqrt{5}\) and \(\tan \theta = 1/2\). Hence \begin{align*} \tan 3 \theta &= \frac{3 \cdot \frac12 - \frac18}{1 - \frac34} \\ &= \frac{11}{2} \end{align*}
- Since \(\tan 3 y = 11/2\) has the solution \(y = \cos^{-1} (2/\sqrt{5})\) it will also have the solutions \(y = \cos^{-1}(2/\sqrt{5}) + \frac{\pi}{3}\) and \(y = \cos^{-1}(2/\sqrt{5})+\frac{2\pi}{3}\), therefore \begin{align*} && \cos^{-1} x &= \cos^{-1} (2/\sqrt{5})\\ \Rightarrow && x &= 2/\sqrt{5} \\ && \cos^{-1} x &= \cos^{-1} (2/\sqrt{5}) + \frac{\pi}{3}\\ \Rightarrow && x &= \frac{2}{\sqrt{5}} \frac{1}{2} - \frac{1}{\sqrt{5}} \frac{\sqrt{3}}{2} \\ &&&= \frac{2-\sqrt{3}}{2\sqrt{5}} \\ && \cos^{-1} x &= \cos^{-1} (2/\sqrt{5}) + \frac{2\pi}{3}\\ \Rightarrow && x &= \frac{2}{\sqrt{5}} \left (-\frac{1}{2} \right)- \frac{1}{\sqrt{5}} \frac{\sqrt{3}}{2} \\ &&&= \frac{-\sqrt{3}-2}{2\sqrt{5}} \\ \end{align*}
- Since \(\cos \frac13 x = \frac{2}{\sqrt{5}}\) has the solution \(x = \tan^{-1} \frac{11}{2}\) it will also have the solutions \(x = \tan^{-1} \frac{11}{2} + 2n \pi\) and \(x = -\tan^{-1} \frac{11}{2} + 2n \pi\). \begin{align*} && \tan^{-1} y &= \tan^{-1} \frac{11}{2} \\ \Rightarrow && y &= \frac{11}{2} \\ && \tan^{-1} y &= \tan^{-1} \frac{11}{2} + 2n \pi \\ \Rightarrow && y &= \frac{\frac{11}{2} + 0}{1-0} \\ &&&= \frac{11}{2} \\ && \tan^{-1} y &= -\tan^{-1} \frac{11}{2} + 2n \pi \\ \Rightarrow && y &= \frac{-\frac{11}{2} + 0}{1-0} \\ &&&= -\frac{11}{2} \\ \end{align*} So our two solutions are \(y = \pm \frac{11}{2}\)
2001 Paper 1 Q5
Show that (for \(t>0\))
- \[ \int_0^1 \frac1{(1+tx)^2} \d x = \frac1{(1+t)} \]
- \[ \int_0^1 \frac{-2x}{(1+tx)^3} \d x = -\frac1{(1+t)^2} \]
Solution:
- For the first one, consider \begin{align*} && \int_0^1 \frac{1}{(1+tx)^2} \d x &= \left [ -\frac{1}{t}(1+tx)^{-1} \right]_0^1 \\ &&&= \frac{1}{t} - \frac{1}{t(1+t)} \\ &&&= \frac{t+1-1}{t(t+1)} = \frac{1}{t+1} \end{align*}
- Consider \begin{align*} && \int_0^1 \frac{-2x}{(1+tx)^3} \d x &= \int_0^1 \frac{\frac{2}{t}(1+tx) -\frac{2}{t}}{(1+tx)^3} \d x \\ &&&= -\frac{2}{t} \int_0^1 \left (\frac{1}{(1+tx)^2}- \frac{1}{(1+tx)^3} \right) \d x \\ &&&= -\frac{2}{t} \frac{1}{t+1} + \frac{2}{t} \left [ \frac{1}{-2t}(1+tx)^{-2}\right]_0^1 \\ &&&= -\frac{2}{t(t+1)} + \frac2t\left (\frac{1}{2t} - \frac{1}{2t(1+t)^2} \right) \\ &&&= -\frac{2}{t} \left ( \frac{1}{t+1} + \frac{1}{2t(1+t)^2} - \frac{1}{2t}\right) \\ &&&= -\frac{2}{t} \frac{2t(1+t)+1-(1+t)^2}{2t(1+t)^2} \\ &&&= -\frac{2}{t} \frac{2t^2+2t+1-1-2t-t^2}{2t(1+t)^2} \\ &&&= -\frac{1}{(1+t)^2} \end{align*}
2001 Paper 1 Q6
A spherical loaf of bread is cut into parallel slices of equal thickness. Show that, after any number of the slices have been eaten, the area of crust remaining is proportional to the number of slices remaining. A European ruling decrees that a parallel-sliced spherical loaf can only be referred to as `crusty' if the ratio of volume \(V\) (in cubic metres) of bread remaining to area \(A\) (in square metres) of crust remaining after any number of slices have been eaten satisfies \(V/A<1\). Show that the radius of a crusty parallel-sliced spherical loaf must be less than \(2\frac23\) metres. [{\sl The area \(A\) and volume \(V\) formed by rotating a curve in the \(x\)--\(y\) plane round the \(x\)-axis from \(x=-a\) to \(x=-a+t\) are given by \[ A= 2\pi\int_{-a}^{-a+t} { y}\left( 1+ \Big(\frac{\d {y}}{\d x}\Big)^2\right)^{\frac12} \d x\;, \ \ \ \ \ \ \ \ \ \ \ V= \pi \int_{-a}^{-a+t} {y}^2 \d x \;. \ \ ] \] }
2001 Paper 1 Q7
In a cosmological model, the radius \(\rm R\) of the universe is a function of the age \(t\) of the universe. The function \(\rm R\) satisfies the three conditions: $$ \mbox{\({\rm R}(0)=0\)}, \ \ \ \ \ \ \ \ \ \mbox{\({\rm R'}(t)>0\) for \(t>0\)}, \ \ \ \ \ \ \ \ \ \ \mbox{\({\rm R''}(t)<0\) for \(t>0\)}, \tag{*} $$ where \({\rm R''}\) denotes the second derivative of \(\rm R\). The function \({\rm H}\) is defined by \[ {\rm H} (t)= \frac{{\rm R}'(t)}{{\rm R}( t)}\;. \]
- Sketch a graph of \({\rm R} (t)\). By considering a tangent to the graph, show that \(t<1/{\rm H}(t)\).
- Observations reveal that \({\rm H}(t) = a/t\), where \(a\) is constant. Derive an expression for \({\rm R}(t)\). What range of values of \(a\) is consistent with the three conditions \((*)\)?
- Suppose, instead, that observations reveal that \({\rm H}(t)= b t^{-2}\), where \(b\) is constant. Show that this is not consistent with conditions \((*)\) for any value of \(b\).
Solution:
- \(\,\)
Notice the tangent must hit the \(y\)-axis above the origin, ie \begin{align*} && 0 &< R'(t)(0-t) + R(t) \\ \Rightarrow && R'(t) t &< R(t) \\ \Rightarrow && t &< \frac{R(t)}{R'(t)} = \frac{1}{H(t)} \end{align*}
- Suppose \(H(t) = a/t\) then \begin{align*} && \frac{R'}{R} &= \frac{a}{t} \\ \Rightarrow && \int \frac{1}{R} \d R &= \int \frac{a}{t} \d t \\ \Rightarrow && \ln R &= a \ln t + C \tag{t, R > 0} \\ \Rightarrow && R &= Kt^a \end{align*} Since we need \(R(t) > 0\), \(K > 0\), since \(R'(t) > 0\) we need \(a > 0\), since \(R''(t) < 0\) we need \(a(a-1) < 0\) ie \(0 < a < 1\)
- Suppose instead \(H(t) = bt^{-2}\) then \begin{align*} && \frac{R'}{R} &= \frac{b}{t^2} \\ \Rightarrow && \int \frac{1}{R} \d R &= \int \frac{b}{t^2} \d t \\ \Rightarrow && \ln R &= -bt^{-1} + C \tag{R > 0} \\ \Rightarrow && R &= Ke^{-b/t} \end{align*} Since \(R > 0\) we must have \(K > 0\). \begin{align*} R' > 0: && R' &= K(b/t^2)e^{-b/t} > 0 \\ \Rightarrow && b &> 0 \\ R'' < 0: && R'' &= K(b^2/t^4)e^{-b/t} -K2b/t^3 e^{-b/t} \\ &&&= Kb/t^4 (b-2t)e^{-b/t} < 0 \\ \Rightarrow && b &< 2t\\ \Rightarrow && b &< 2t \end{align*} which cannot be true for all \(t\), ie there is no \(b\) which satisfies this.
2001 Paper 1 Q8
Given that \(y=x\) and \(y=1-x^2\) satisfy the differential equation $$ \frac{\d^2 {y}}{\d x^2} + \p(x) \frac{\d {y}}{\d x} + \q(x) {y}=0\;, \tag{*} $$ show that \(\p(x)= -2x(1+x^2)^{-1}\) and \(\q(x) = 2(1+x^2)^{-1}\). Show also that \(ax+b(1-x^2)\) satisfies the differential equation for any constants \(a\) and \(b\). Given instead that \(y=\cos^2(\frac{1}{2}x^2)\) and \(y=\sin^2(\frac{1}{2}x^2)\) satisfy the equation \((*)\), find \(\p(x)\) and \(\q(x)\).
Solution: \begin{align*} && y &= x \\ && y' &= 1 \\ && y'' &= 0 \\ \Rightarrow && 0 &= 0 + p(x) + xq(x) \tag{1} \\ \\ && y &= 1-x^2 \\ && y' &= -2x \\ && y'' &= -2 \\ \Rightarrow && 0 &= -2 -2x p(x)+(1-x^2)q(x) \tag{2}\\ \\ 2x*(1) +(2): && 2 &= (2x^2+1-x^2) q(x) \\ \Rightarrow && q(x) &= 2(1+x^2)^{-1} \\ \Rightarrow && p(x) &= -2x(1+x^2)^{-1} \tag{by (1)} \end{align*} \begin{align*} && \frac{\d^2}{\d x^2} \left (a x + b(1-x^2) \right) + p(x) \frac{\d}{\d x} \left (a x + b(1-x^2) \right)+q(x) \left (a x + b(1-x^2) \right) \\ &&= a \frac{\d^2 x}{\d x^2} + b \frac{\d^2}{\d x^2} \left ( 1- x^2 \right) + ap(x) \frac{\d x}{ \d x} + bp(x) \frac{\d }{\d x} \left ( 1- x^2 \right) + aq(x) x + bq(x)(1-x^2) \\ &&= a \left (\frac{\d^2 x}{\d x^2}+ p(x) \frac{\d x}{ \d x} +q(x)x\right)+b \left ( \frac{\d^2}{\d x^2} \left ( 1- x^2 \right)+ p(x) \frac{\d }{\d x} \left ( 1- x^2 \right)+q(x)(1-x^2)\right) &= 0 \end{align*} \begin{align*} && y &= \cos^2(\tfrac12 x^2) = \frac12 \left (1 + \cos(x^2) \right) \\ && y' &= -x \sin(x^2) \\ && y'' &= -2x^2 \cos(x^2)-\sin(x^2) \\ \Rightarrow && 0 &= -2x^2 \cos(x^2)-\sin(x^2)+p(x)(-x \sin(x^2)) +\frac12 \left (1 + \cos(x^2) \right)q(x) \\ \Rightarrow && 2x^2\cos(x^2)+\sin(x^2) &= -x \sin(x^2) p(x) + \frac12(1 + \cos(x^2)) q(x) \tag{3}\\ \\ && y &= \sin^2(\tfrac12 x^2) = \frac12 \left ( 1 - \cos (x^2) \right) \\ && y' &= x\sin(x^2) \\ && y'' &= 2x^2 \cos(x^2)+\sin(x^2) \\ \Rightarrow && 0 &= 2x^2 \cos(x^2)+\sin(x^2) +p(x) x \sin(x^2) + \frac12 \left ( 1 - \cos (x^2) \right)q(x)\\ \Rightarrow && -2x^2 \cos(x^2)-\sin(x^2) &= p(x) x \sin(x^2) + \frac12 \left ( 1 - \cos (x^2) \right)q(x) \tag{4}\\ (3)+(4): && 0 &= q(x) \\ \Rightarrow && p(x) &= -\frac{2x^2 \cos(x^2)+\sin(x^2)}{x \sin(x^2)} \end{align*}
2001 Paper 1 Q9
A ship sails at \(20\) kilometres/hour in a straight line which is, at its closest, 1 kilometre from a port. A tug-boat with maximum speed 12 kilometres/hour leaves the port and intercepts the ship, leaving the port at the latest possible time for which the interception is still possible. How far does the tug-boat travel?
Solution: The position of the ship is \(\mathbf{s} = \binom{20t}{1}\). Suppose the interception is at \(T\), then the ship leaves at \(T-\frac1{12}\underbrace{\sqrt{400T^2+1}}_{\text{distance to intercept}}\). We wish to maximise this, ie \begin{align*} && \frac{\d}{\d T} \left ( T - \frac1{12}\sqrt{400T^2+1}\right) &= 1 - \frac{1}{12} \cdot \frac12 \cdot 400 \cdot 2T \cdot \left (400T^2+1 \right)^{-1/2} \\ &&&= 1 - \frac{100}3 T(400T^2+1)^{-1/2} \\ \Rightarrow && \frac{T}{\sqrt{400T^2+1}} &= \frac{3}{100} \\ \Rightarrow && \frac{T^2}{400T^2+1} &= \frac{9}{10000} \\ \Rightarrow && 10000T^2 &= 3600T^2+9 \\ \Rightarrow && 6400T^2 &= 9 \\ \Rightarrow && T &= \pm \frac{3}{80} \quad \text{(T > 0)} \end{align*} Therefore the distance is \(\sqrt{400 \frac{9}{6400} + 1} = \sqrt{\frac{9}{16}+1} = \frac{5}{4} = 1.25 \text{ km}\)
2001 Paper 1 Q10
A gun is sited on a horizontal plain and can fire shells in any direction and at any elevation at speed \(v\). The gun is a distance \(d\) from a straight railway line which crosses the plain, where \(v^2>gd\). The gunner aims to hit the line, choosing the direction and elevation so as to maximize the time of flight of the shell. Show that the time of flight, \(T\), of the shell satisfies \[ %\frac{2v}{g} \sin \left( \frac12 \arccos \frac{gd}{v^2}\right)\,. g^2 T^2 = 2 v^2 +2 \left(v^4 -g^2d^2\right)^{\frac12}\,. \] Extension: (Not in original paper) Find the time of flight if the gun is constrained so that the angle of elevation \(\alpha \) is not greater than \( 45^\circ\).
Solution: If we fire the gun at an angle to the track, as long as we can travel a horizontal distance \(\geq d\) we can hit the track. Suppose we am at an elevatation \(\alpha\), then \begin{align*} (\uparrow): && s &= ut + \frac12 at^2 \\ && 0 &= v\sin \alpha T - \frac12 g T^2 \\ \Rightarrow &&T &= \frac{2v\sin \alpha}{g}\\ \\ (\rightarrow): && s &= ut \\ && s &= v \cos \alpha T \\ &&&= v\sqrt{1-\sin^2 \alpha} T \\ &&&= vT\sqrt{1 - \frac{g^2T^2}{4v^2}}\\ &&&= \frac{T}{2}\sqrt{4v^2-g^2T^2}\\ \Rightarrow && d & \leq \frac{T}{2}\sqrt{4v^2-g^2T^2} \\ \Rightarrow && 4g^2d^2&\leq g^2T^2(4v^2-g^2T^2) \\ \Rightarrow && 0 &\leq -(g^2T^2)^2 + 4v^2 (g^2T^2)-4g^2d^2 \\ &&&=4v^4-4g^2d^2 -\left (g^2T^2-2v^2 \right)^2 \\ \Rightarrow && \left (g^2T^2-2v^2 \right)^2 & \leq 4v^4-4g^2d^2 \\ \Rightarrow && g^2T^2 &\leq 2v^2+2\sqrt{v^4-g^2d^2} \end{align*} Therefore the maximum value for \(g^2T^2\) is \(2v^2+2\sqrt{v^4-g^2d^2}\) Notice that we are hitting the track directly at \(d\). This is because to maximise the time of flight (for a fixed speed) we want to maximise the angle of elevation. Therefore we want the highest angle where we still hit the track (which is clearly the shortest distance). If we are constraint to \(\alpha \leq 45^\circ\) we know that \(T\) is maximised when \(\alpha = 45^\circ\) (and we will reach the track since the range \(\frac{v^2 \sin 2 \alpha}{g}\) is increasing). Therefore the maximum time is \(T = \frac{\sqrt{2}v}{g}\)