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1999 Paper 3 Q1
D: 1700.0 B: 1500.0

Consider the cubic equation \[ x^3-px^2+qx-r=0\;, \] where \(p\ne0\) and \(r\ne 0\).

  1. If the three roots can be written in the form \(ak^{-1}\), \(a\) and \(ak\) for some constants \(a\) and \(k\), show that one root is \(q/p\) and that \(q^3 -rp^3=0\;.\)
  2. If \(r=q^3/p^3\;\), show that \(q/p\) is a root and that the product of the other two roots is \((q/p)^2\). Deduce that the roots are in geometric progression.
  3. Find a necessary and sufficient condition involving \(p\), \(q\) and \(r\) for the roots to be in arithmetic progression.


Solution:

  1. If the roots are \(ak^{-1}, a, ak\) then we must have that \(p = a(k^{-1}+1+k)\), \(q = a^2(k^{-1}+k+1)\) and \(r = a^3\), therefore \(a = \frac{q}{p}\) (ie one of the roots is \(\frac q p\) and \(r = \left ( \frac{q}{p} \right)^3 \Rightarrow q^3 =rp^3 \Rightarrow q^3-rp^3 = 0\)
  2. Suppose \(r = q^3/p^3\) then \(\left (\frac{q}{p} \right)^3 - p\left (\frac{q}{p} \right)^2+q\left (\frac{q}{p} \right) - r = \frac{--pq^2+pq^2}{p^2} =0 \), therefore \(q/p\) is a root by the factor theorem. We must also have the product of the three roots is \(q^3/p^3\) but one of the roots is \(q/p\) therefore the product of the other two roots is \(q^2/p^2\), but the condition \(ac = b^2\) is precisely the condition that \(a,b,c\) is a geometric progression.
  3. If the three roots are \(a-d, a, a+d\) then \(p = 3a\), \(q = a^2-da+a^2+da+a^2-d^2 = 3a^2-d^2\), \(r = a(a^2-q^2)\), therefore \(\frac{p}{3}\left (q-\frac{2p^2}9 \right) = r\) Similarly, suppose \(\frac{p}{3}\) is a root, then the other two roots must sum to twice this and therefore they are in arithmetic progression. The condition \(\frac{p}{3}\) is a root is equivalent to: \(\frac{p^3}{27} - \frac{p^3}{9} + \frac{qp}{3} - r = 0\), ie exactly \(\frac{p}{3}\left (q-\frac{2p^2}9 \right) = r\), therefore this condition is both necessary and sufficient.

1999 Paper 3 Q2
D: 1700.0 B: 1486.1

  1. Let \(\f(x)=(1+x^2)\e^x\). Show that \(\f'(x)\ge 0\) and sketch the graph of \(\f(x)\). Hence, or otherwise, show that the equation \[ (1+x^2)\e^x = k, \] where \(k\) is a constant, has exactly one real root if \(k>0\) and no real roots if \(k\le 0\).
  2. Determine the number of real roots of the equation $$ (\e^x-1) - k \tan^{-1} x=0 $$ in the cases (a) \(0< k\le 2/\pi\) and (b) \(2/\pi < k < 1\).

1999 Paper 3 Q3
D: 1700.0 B: 1518.8

Justify, by means of a sketch, the formula $$ \lim_{n\rightarrow\infty}\left\{{1\over n}\sum_{m=1}^n \f(1+m/n)\right\} = \int_1^2 \f(x)\,\d x \,. $$ Show that $$ \lim_{n\rightarrow\infty}\left\{{1\over n+1} + {1\over n+2} + \cdots + {1\over n+n}\right\} = \ln 2 \,. $$ Evaluate $$ \lim_{n\rightarrow\infty}\left\{{n\over n^2+1} + {n\over n^2+4} + \cdots + {n\over n^2+n^2}\right\}\,. $$


Solution:

TikZ diagram
\begin{align*} && V &= \lim_{n\rightarrow\infty}\left\{{1\over n+1} + {1\over n+2} + \cdots + {1\over n+n}\right\} \\ && &= \lim_{n\rightarrow\infty}\left\{\sum_{m=1}^n \frac{1}{n+m}\right\} \\ && &= \lim_{n\rightarrow\infty}\left\{\frac1n\sum_{m=1}^n \frac{1}{1+\frac{m}{n}}\right\} \\ &&&=\int_1^2 \frac{1}{x} \d x \\ &&&= \left [\ln x \right]_1^2 = \ln 2 \end{align*} \begin{align*} V &= \lim_{n\rightarrow\infty}\left\{{n\over n^2+1} + {n\over n^2+4} + \cdots + {n\over n^2+n^2}\right\} \\ &= \lim_{n\rightarrow\infty}\left\{\sum_{m=1}^n \frac{n}{n^2+m^2}\right\} \\ &= \lim_{n\rightarrow\infty}\left\{\frac{1}{n}\sum_{m=1}^n \frac{1}{1+\left (\frac{m}{n} \right)^2}\right\} \\ &= \int_0^1 \frac{1}{1+x^2} \d x \\ &= \left [\tan^{-1} x \right]_0^1 \\ &= \frac{\pi}4 \end{align*}

1999 Paper 3 Q4
D: 1700.0 B: 1516.0

A polyhedron is a solid bounded by \(F\) plane faces, which meet in \(E\) edges and \(V\) vertices. You may assume {\em Euler's formula}, that \(V-E+F=2\). In a regular polyhedron the faces are equal regular \(m\)-sided polygons, \(n\) of which meet at each vertex. Show that $$ F={4n\over h} \,, $$ where \(h=4-(n-2)(m-2)\). By considering the possible values of \(h\), or otherwise, prove that there are only five regular polyhedra, and find \(V\), \(E\) and \(F\) for each.

1999 Paper 3 Q5
D: 1700.0 B: 1516.0

The sequence \(u_0\), \(u_1\), \(u_2\), ... is defined by $$ u_0=1,\hspace{0.2in} u_1=1, \quad u_{n+1}=u_n+u_{n-1} \hspace{0.2in}{\rm for}\hspace{0.1in}n \ge 1\,. $$ Prove that $$ u^2_{n+2} + u^2_{n-1} = 2( u^2_{n+1} + u^2_n ) \,. $$ Using induction, or otherwise, prove the following result: \[ u_{2n} = u^2_n + u^2_{n-1} \quad \mbox{ and }\quad u_{2n+1} = u^2_{n+1} - u^2_{n-1} \] for any positive integer \(n\).


Solution: Claim: \(u^2_{n+2} + u^2_{n-1} = 2( u^2_{n+1} + u^2_n )\) Proof: (By Induction). (Base Case): \(n = 1\) \begin{align*} && LHS &= u_{n+2}^2 + u_{n-1}^2 \\ &&&= u_3^2 + u_0^2 \\ &&&= 3^2 + 1^2 = 10\\ && RHS &= 2(u_{n+1}^2+u_n^2) \\ &&&= 2(2^2 + 1^2) \\ &&&= 10 \end{align*} Therefore the base case is true. (Inductive hypothesis) Suppose \(u^2_{n+2} + u^2_{n-1} = 2( u^2_{n+1} + u^2_n )\) is true for some \(n = k\), ie \(u^2_{k+2} + u^2_{k-1} = 2( u^2_{k+1} + u^2_k )\), the consider \(n = k+1\) \begin{align*} && LHS &= u_{k+1+2}^2 + u_{k+1-1}^2 \\ &&&= (u_{k+1}+u_{k+2})^2+u_k^2 \\ &&&= u_{k+2}^2+u_{k+1}^2+ u_k^2 + 2u_{k+1}u_{k+2} \\ &&&= u_{k+2}^2+u_{k+1}^2+ u_k^2 + 2u_{k+1}(u_{k+1}+u_k) \\ &&&= u_{k+2}^2 + u_{k+1}^2+u_k^2+2u_{k+1}^2+2u_{k+1}u_k \\ &&&= u_{k+1}^2+2u_{k+1}^2+ u_{k+1}^2+u_k^2+2u_{k+1}u_k \\ &&&= u_{k+2}^2+2u_{k+1}^2+ (u_{k+1}+u_k)^2 \\ &&&= u_{k+2}^2+2u_{k+1}^2+ u_{k+2}^2 \\ &&&=2(u_{k+2}^2+u_{k+1}^2) \\ &&&= RHS \end{align*} Therefore it is true for \(n = k+1\). Therefore by the principle of mathematical induction it is true for all \(n \geq 1\) Claim: $ u_{2n} = u^2_n + u^2_{n-1} \quad \mbox{ and }\quad u_{2n+1} = u^2_{n+1} - u^2_{n-1} $ Proof: Notice that \(\begin{pmatrix}u_{n+1} \\ u_n \end{pmatrix}= \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^n \begin{pmatrix}1 \\1 \end{pmatrix}\), in particular \begin{align*} && \begin{pmatrix}u_{n}& u_{n-1} \\ u_{n-1} & u_{n-2} \end{pmatrix}&= \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^{n} \\ \Rightarrow && \begin{pmatrix}u_{2n}& u_{2n-1} \\ u_{2n-1} & u_{2n-2} \end{pmatrix}&= \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^{2n} \\ &&&= \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^{n} \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}^{n} \\ &&&=\begin{pmatrix}u_{n}& u_{n-1} \\ u_{n-1} & u_{n-2} \end{pmatrix}\begin{pmatrix}u_{n}& u_{n-1} \\ u_{n-1} & u_{n-2} \end{pmatrix}\\ &&&= \begin{pmatrix}u_{n}^2+u_{n-1}^2& u_{n-1}(u_n+u_{n-2}) \\ u_{n-1}(u_n+u_{n-2}) & u_{n-1}^2+u_{n-2}^2 \end{pmatrix} \end{align*} Therefore \(u_{2n} = u_{n}^2+u_{n-1}^2\) and \(u_{2n+1} = u_n(u_{n+1}+u_{n-1}) =(u_{n+1}-u_{n-1})(u_{n+1}-u_{n-1}) = u_{n+1}^2-u_{n-1}^2\)

1999 Paper 3 Q6
D: 1700.0 B: 1516.0

A closed curve is given by the equation $$ x^{2/n} + y^{2/n} = a^{2/n} \eqno(*) $$ where \(n\) is an odd integer and \(a\) is a positive constant. Find a parametrization \(x=x(t)\), \(y=y(t)\) which describes the curve anticlockwise as \(t\) ranges from \(0\) to \(2\pi\). Sketch the curve in the case \(n=3\), justifying the main features of your sketch. The area \(A\) enclosed by such a curve is given by the formula $$ A= {1\over 2} \int_0^{2\pi} \left[ x(t) {\d y(t)\over \d t} - y(t) {\d x(t)\over \d t} \right] \,\d t \,. $$ Use this result to find the area enclosed by (\(*\)) for \(n=3\).

1999 Paper 3 Q7
D: 1680.5 B: 1516.0

Let \(a\) be a non-zero real number and define a binary operation on the set of real numbers by $$ x*y = x+y+axy \,. $$ Show that the operation \(*\) is associative. Show that \((G,*)\) is a group, where \(G\) is the set of all real numbers except for one number which you should identify. Find a subgroup of \((G,*)\) which has exactly 2 elements.


Solution: Claim: \(*\) is associative. Proof: Then \(x*(y*z) = x*(y+z+ayz) = x + (y+z+ayz) + ax(y+z+ayz) = x + y + z + a(yz + xy + zx) + a^2xyz\) and \((x*y)*z = (x+y+axy)*z = (x+y+axy) + z+ a(x+y+axy)z = x + y + z + a(yz + xy + zx) + a^2xyz\) so \(x*(y*z) = (x*y)*z\) and we are done. Let \(G = \mathbb{R} \setminus \{-\frac1{a} \}\) In order to show that \((G, *)\) is a group we need to check:

  1. closure \(x*y = x + y + axy\) is a real number
  2. associativity we have already checked
  3. identity \(x*0 = x + 0 + 0 = x = 0*x\), so \(0\) is an identity
  4. inverses \(x*\left ( \frac{-x}{1+ax} \right ) = x - \frac{x}{1+ax} + ax \frac{-x}{1+ax} = \frac{x +ax^2 - x - ax^2}{1+ax} = 0\) so \(x\) has an identity, assuming \(1+ax \neq 0\) which is true for everything in our set
Consider the set \(\{0, \frac{-2}{a} \}\). Then \(\frac{-2}{a} * \frac{-2}{a} = \frac{-4}{a} + a \frac{4}{a^2} = 0\), so this is a group of order \(2\)

1999 Paper 3 Q8
D: 1700.0 B: 1516.0

The function \(y(x)\) is defined for \(x\ge0\) and satisfies the conditions \[ y=0 \mbox{ \ \ and \ \ } \frac{\d y}{\d x}=1 \mbox{ \ \ at \(x=0\)}. \] When \(x\) is in the range \(2(n-1)\pi< x <2n\pi\), where \(n\) is a positive integer, \(y(t)\) satisfies the differential equation $$ {\d^2y \over \d x^2} + n^2 y=0. $$ Both \(y\) and \(\displaystyle \frac{\d y}{\d x} \) are continuous at \(x=2n\pi\) for \(n=0,\; 1,\;2,\; \ldots\;\).

  1. Find \(y(x)\) for \(0\le x \le 2\pi\).
  2. Show that \(y(x) = \frac12 \sin 2x \) for \(2\pi\le x\le 4\pi\), and find \(y(x)\) for all \(x\ge0\).
  3. Show that $$ \int_0^\infty y^2 \,\d x = \pi \sum_{n=1}^\infty {1\over n^2} \,. $$

1999 Paper 3 Q9
D: 1700.0 B: 1470.3

The gravitational force between two point particles of masses \(m\) and \(m'\) is mutually attractive and has magnitude $$ {G m m' \over r^2}\,, $$ where \(G\) is a constant and \(r\) is the distance between them. A particle of unit mass lies on the axis of a thin uniform circular ring of radius \(r\) and mass \(m\), at a distance \(x\) from its centre. Explain why the net force on the particle is directed towards the centre of the ring and show that its magnitude is $$ {G m x \over (x^2 + r^2)^{3/2}} \,. $$ The particle now lies inside a thin hollow spherical shell of uniform density, mass \(M\) and radius \(a\), at a distance \(b\) from its centre. Show that the particle experiences no gravitational force due to the shell. %Explain without calculation the effect on this result if %the shell has finite thickness \(x\).

1999 Paper 3 Q10
D: 1700.0 B: 1484.0

A chain of mass \(m\) and length \(l\) is composed of \(n\) small smooth links. It is suspended vertically over a horizontal table with its end just touching the table, and released so that it collapses inelastically onto the table. Calculate the change in momentum of the \((k+1)\)th link from the bottom of the chain as it falls onto the table. Write down an expression for the total impulse sustained by the table in this way from the whole chain. By approximating the sum by an integral, show that this total impulse is approximately \[ {\textstyle \frac23} m \surd(2gl) \] when \(n\) is large.