Problems

Filters
Clear Filters

137 problems found

1993 Paper 2 Q1
D: 1600.0 B: 1500.0
proof game theory combinatorics optimisation strategy integer partition inequality case analysis

In the game of ``Colonel Blotto'' there are two players, Adam and Betty. First Adam chooses three non-negative integers \(a_{1},a_{2}\) and \(a_{3},\) such that \(a_{1}+a_{2}+a_{3}=9,\) and then Betty chooses non-negative integers \(b_{1},b_{2}\) and \(b_{3}\), such that \(b_{1}+b_{2}+b_{3}=9.\) If \(a_{1} > b_{1}\) then Adam scores one point; if \(a_{1} < b_{1}\) then Betty scores one point; and if \(a_{1}=b_{1}\) no points are scored. Similarly for \(a_{2},b_{2}\) and \(a_{3},b_{3}.\) The winner is the player who scores the greater number of points: if the socres are equal then the game is drawn. Show that, if Betty knows the numbers \(a_{1},a_{2}\) and \(a_{3},\) she can always choose her numbers so that she wins. Show that Adam can choose \(a_{1},a_{2}\) and \(a_{3}\) in such a way that he will never win no matter what Betty does. Now suppose that Adam is allowed to write down two triples of numbers and that Adam wins unless Betty can find one triple that beats both of Adam's choices (knowing what they are). Confirm that Adam wins by writing down \((5,3,1)\) and \((3,1,5).\)

View
1993 Paper 2 Q6
D: 1600.0 B: 1516.0
matrices 2x2 matrix matrix multiplication determinant counterexample proof true or false commutativity singular matrix

In this question, \(\mathbf{A,\mathbf{B\) }}and \(\mathbf{X\) are non-zero \(2\times2\) real matrices.} Are the following assertions true or false? You must provide a proof or a counterexample in each case.

  1. If \(\mathbf{AB=0}\) then \(\mathbf{BA=0}.\)
  2. \((\mathbf{A-B)(A+B)=}\mathbf{A}^{2}-\mathbf{B}^{2}.\)
  3. The equation \(\mathbf{AX=0}\) has a non-zero solution \(\mathbf{X}\) if and only if \(\det\mathbf{A}=0.\)
  4. For any \(\mathbf{A}\) and \(\mathbf{B}\) there are at most two matrices \(\mathbf{X}\) such that \(\mathbf{X}^{2}+\mathbf{AX}+\mathbf{B}=\mathbf{0}.\)

Solution:

  1. This is false, for example let \(\mathbf{A} = \begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix}\) and \(\mathbf{B} = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\), then \begin{align*} \mathbf{AB} &= \begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \\ &= \begin{pmatrix}0 & 0 \\ 0 & 0\end{pmatrix} \\ \mathbf{BA} &= \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}\begin{pmatrix} 0 & 1 \\ 0 & 1 \end{pmatrix} \\ &= \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix} \\ \end{align*}
  2. This is also false, using the same matrices from part (i), we find: \begin{align*} (\mathbf{A - B})(\mathbf{A + B}) &= \mathbf{A}^2-\mathbf{BA}+\mathbf{AB}-\mathbf{B}^2 \\ &= \mathbf{A}^2-\mathbf{B}^2+\begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix} \\ &\neq \mathbf{A}^2-\mathbf{B}^2 \end{align*}
  3. This is true. Claim: The equation \(\mathbf{AX=0}\) has a non-zero solution \(\mathbf{X}\) if and only if \(\det\mathbf{A}=0.\) Proof: \((\Rightarrow)\) Suppose \(\det\mathbf{A} \neq 0\) then \(\mathbf{A}\) has an inverse, and so we must have \(\mathbf{A}^{-1}\mathbf{AX} = \mathbf{0} \Rightarrow \mathbf{X} = \mathbf{0}\). \((\Leftarrow)\) Suppose \(\det \mathbf{A} = 0\) then \(ad-bc=0\), so consider the matrix \(\mathbf{X} = \begin{pmatrix} d & d\\ -c & -c\end{pmatrix}\) (or if this is zero, \(\mathbf{X} = \begin{pmatrix} a & a\\ -b & -b\end{pmatrix}\))
  4. This is false. Consider \(\mathbf{A} = \mathbf{B} = \mathbf{0}\), then \(\mathbf{X} = \begin{pmatrix} 0 & x \\ 0 & 0\end{pmatrix}\) has the property that \(\mathbf{X}^2 = \mathbf{0}\) for all \(x\), so at least more than 2 values

View
1993 Paper 3 Q8
D: 1700.0 B: 1484.0
vectors 3D geometry plane equation square pyramid parametric proof parallelogram inequality

A square pyramid has its base vertices at the points \(A\) \((a,0,0)\), \(B\) \((0,a,0)\), \(C\) \((-a,0,0)\) and \(D\) \((0,-a,0)\), and its vertex at \(E\) \((0,0,a)\). The point \(P\) lies on \(AE\) with \(x\)-coordinate \(\lambda a\), where \(0<\lambda<1\), and the point \(Q\) lies on \(CE\) with \(x\)-coordinate \(-\mu a\), where \(0<\mu<1\). The plane \(BPQ\) cuts \(DE\) at \(R\) and the \(y\)-coordinate of \(R\) is \(-\gamma a\). Prove that $$ \gamma = {\lambda \mu \over \lambda + \mu - \lambda \mu}. $$ Show that the quadrilateral \(BPRQ\) cannot be a parallelogram.

View
1993 Paper 3 Q9
D: 1700.0 B: 1485.7
inequality proof AM-GM inequality sum of squares generalisation algebraic manipulation quadratic forms

For the real numbers \(a_1\), \(a_2\), \(a_3\), \(\ldots\),

  1. prove that \(a_1^2+a_2^2 \ge 2a_1a_2\),
  2. prove that \(a_1^2+a_2^2 +a_3^2 \ge a_2a_3 + a_3a_1 +a_1a_2\),
  3. prove that $3(a_1^2+a_2^2 +a_3^2 +a_4^2) \ge 2(a_1a_2+a_1a_3 + a_1a_4 +a_2a_3 + a_2a_4 +a_3a_4)\(,
  4. state and prove a generalisation of (iii) to the case of \)n$ real numbers,
  5. prove that $$ \left(\sum_{i=1}^n a_i \right)^2 \ge {2n\over n-1} \sum_{i,j} a_ia_j, $$ where the latter sum is taken over all pairs \((i,j)\) with $1\le i

View
1992 Paper 1 Q1
D: 1484.0 B: 1500.0
proof modular arithmetic calendar calculation divisibility leap year systematic reasoning

Today's date is June 26th 1992 and the day of the week is Friday. Find which day of the week was April 3rd 1905, explaining your method carefully. {[}30 days hath September, April, June and November. All the rest have 31, excepting February alone which has 28 days clear and 29 in each leap year.{]}

Solution: There are \(87\) years between 1905 and 1992. Of those years, every 4th is a leap years, starting with 1908, and ending with 1992, so there are 22 leap years. There are 30 days between Apr 3 and May 3, 31 days between May 3 and Jun 3 and a further 23 days to the 26th. Therefore \(87 \times 365 + 22 \cdot 1 + 61 + 23\) total days. \(\pmod{7}\) this is \(3 \times 1 + 1 + 5 + 2 = 3\), therefore it is \(4\) week days before Friday, ie Monday.

View
1992 Paper 1 Q2
D: 1500.0 B: 1500.0
combinatorics magic square proof enumeration constraint satisfaction systematic counting

A \(3\times3\) magic square is a \(3\times3\) array \[ \begin{array}{ccc} a & b & c\\ d & e & f\\ g & h & k \end{array} \] whose entries are the nine distinct integers \(1,2,3,4,5,6,7,8,9\) and which has the property that all its rows, columns and main diagonals add up to the same number \(n\). (Thus \(a+b+c=d+e+f=g+h+k=a+d+g=b+e+h=c+f+k=a+e+k=c+e+g=n.)\)

  1. Show that \(n=15.\)
  2. Show that \(e=5.\)
  3. Show that one of \(b,d,h\) or \(f\) must have value \(9\).
  4. Find all \(3\times3\) magic squares with \(b=9.\)
  5. How many different \(3\times3\) magic squares are there? Why?
{[}Two magic squares are different if they have different entries in any place of the array.{]}

Solution:

  1. \((a+b+c)+(d+e+f)+(g+h+k) = 3n = 1 + 2 + \cdots + 9 = 45 \Rightarrow n = 15\).
  2. Summing all rows, columns, diagonals through \(e\) we have \((a+e+k)+(b+e+h)+(c+e+g)+(d+e+f) = 45 + 3e = 60 \Rightarrow e = 5\).
  3. Suppose that one of the corners is \(9\), then we need to find \(2\) ways to make \(6\) not using \(5\) and \(1\) (as \(5\) is in the middle and \(1\) diagonally opposite). Clearly this is not possible as the only remaining numbers are \(2,3,4\) and only \(2+4 = 6\). Therefore \(9\) cannot be in the corner or central squares, ie it's one of \(b,d,h,f\)
  4. We must have \begin{array}{ccc} a & 9 & c\\ d & 5 & f\\ g & 1 & k \end{array} and so \(a\) or \(c = 4\). Once we place \(4\) by symmetry there will be another solution with \(a = 2\). So: \begin{array}{ccc} 4 & 9 & 2\\ d & 5 & f\\ g & 1 & k \end{array} we now see \(k\), then \(f\), then \(d\) then \(g\) must be determined, ie: \begin{array}{ccc} 4 & 9 & 2\\ 3 & 5 & 7\\ 8 & 1 & 6 \end{array} so our two solutions must be this and \begin{array}{ccc} 2 & 9 & 4\\ 7 & 5 & 3\\ 6 & 1 & 8 \end{array}
  5. For each of the \(4\) possible placements of \(9\) there are two magic squares, so there are \(8\) possible magic squares, all related by reflection and rotation.

View
1992 Paper 1 Q5
D: 1484.0 B: 1500.0
polynomials factorization differentiation power sums product expansion combinatorial identity proof

Let \(\mathrm{p}_{0}(x)=(1-x)(1-x^{2})(1-x^{4}).\) Show that \((1-x)^{3}\) is a factor of \(\mathrm{p}_{0}(x).\) If \(\mathrm{p}_{1}(x)=x\mathrm{p}_{0}'(x)\) show, by considering factors of the polynomials involved, that \(\mathrm{p}_{0}'(1)=0\) and \(\mathrm{p}_{1}'(1)=0.\) By writing \(\mathrm{p}_{0}(x)\) in the form \[ \mathrm{p}_{0}(x)=c_{0}+c_{1}x+c_{2}x^{2}+c_{3}x^{3}+c_{4}x^{4}+c_{5}x^{5}+c_{6}x^{6}+c_{7}x^{7}, \] deduce that \begin{alignat*}{2} 1+2+4+7 & \quad=\quad & & 3+5+6\\ 1^{2}+2^{2}+4^{2}+7^{2} & \quad=\quad & & 3^{2}+5^{2}+6^{2}. \end{alignat*} Show that we can write the integers \(1,2,\ldots,15\) in some order as \(a_{1},a_{2},\ldots,a_{15}\) in such a way that \[ a_{1}^{r}+a_{2}^{r}+\cdots+a_{8}^{r}=a_{9}^{r}+a_{10}^{r}+\cdots+a_{15}^{r} \] for \(r=1,2,3.\)

Solution: \begin{align*} && p_0(x) &= (1-x)(1-x^2)(1-x^4) \\ &&&= (1-x)(1-x)(1+x)(1-x^2)(1+x^2) \\ &&&= (1-x)^2 (1+x)(1-x)(1+x)(1+x^2) \\ &&&= (1-x)^3 (1+x)^2 (1+x^2) \end{align*} \begin{align*} && p_0'(x) &= 3(1-x)^2(1+x)^2(1+x^2) + (1-x)^3 q(x) \\ \Rightarrow && p_0'(1) &= 3 \cdot 0 \cdots + 0 \cdots \\ &&&= 0 \\ && p_1'(x) &= p_0(x) + xp'_0(x) \\ \Rightarrow && p_1'(1) &= p_0(1) + 1\cdot p_0'(1) \\ &&&= 0 + 1 \cdot 0 \\ &&&= 0 \end{align*} Notice that \(p_0(x) = (1-x-x^2+x^3)(1-x^4) = 1-x-x^2+x^3-x^4+x^5+x^6-x^7\), so: \(p'_0(x) = -1-2x+3x^2-4x^3+5x^4+6x^5-7x^6 \Rightarrow p'_0(1) = 0 = -1 -2 -4 -7 + 3 + 5+6\). \((xp'_1(1))' = 0 = -1^2-2^2-4^2-7^2 + 3^2 + 5^2 + 6^2\). Consider \(q_0(x) = (1-x)(1-x^2)(1-x^4)(1-x^8)\), then \((1-x)^4\) is a factor, so in particular we know \(q_0(1), (xq_0(x))'|_{x=1} = 0,(x(xq_0(x))')'|_{x=1} = 0\), and so: \(q_0(x) = 1-x-x^2+x^3-x^4+x^5+x^6-x^7 - x^8+x^9+x^{10}-x^{11}+x^{12}-x^{13}-x^{14}+x^{15}\), and so: \(1^r+2^r+4^r+7^r+8^r+11^r+13^r+14^r = 3^r+5^r+6^r+9^r+10^r+12^r+15^r\) for \(r = 1,2,3\)

View
1992 Paper 2 Q5
D: 1577.1 B: 1470.1
groups matrix order of element inverse determinant group axioms proof

Explain what is meant by the order of an element \(g\) of a group \(G\). The set \(S\) consists of all \(2\times2\) matrices whose determinant is \(1\). Find the inverse of the element \(\mathbf{A}\) of \(S\), where \[ \mathbf{A}=\begin{pmatrix}w & x\\ y & z \end{pmatrix}. \] Show that \(S\) is a group under matrix multiplication (you may assume that matrix multiplication is associative). For which elements \(\mathbf{A}\) is \(\mathbf{A}^{-1}=\mathbf{A}\)? Which element or elements have order 2? Show that the element \(\mathbf{A}\) of \(S\) has order 3 if, and only if, \(w+z+1=0.\) Write down one such element.

Solution: The order of an element \(g\) is the smallest positive number \(k\) such that \(g^k = e\). $\mathbf{A}^{-1} = \begin{pmatrix}z & -x\\ -y & w \end{pmatrix}$. Claim, \(S\) is a group. \begin{enumerate} \item (Closure) The product of two \(2\times2\) matrices is always a \(2\times 2\) matrix so we only need to check the determinant. Suppose \(\det(\mathbf{A}) = \det (\mathbf{B}) = 1\), then \(\det(AB) = \det(A)\det(B) = 1\), so our operation is closed \item (Associativity) Inherited from matrix multiplication \item (Identity) $\mathbf{I} =\begin{pmatrix}1 & 0\\ 1 & 1 \end{pmatrix}\( has determinant \)1$. \item (Inverses) The inverse is always fine since the matrix of cofactors always contains integers and the determinant is one, so we never end up with anything which isn't an integer. \end{itemize} If \(\mathbf{A}^-1 = \mathbf{A}\) then assuming $\mathbf{A} = \begin{pmatrix}a & b\\ c & d \end{pmatrix}\( then \)\mathbf{A}^{-1} = \begin{pmatrix}d & - b\\ -c & a \end{pmatrix}\( so we must have \)a=d, -b=b, -c=c\(, so \)b = c = 0\( and \)a = d\(. For the determinant to be \)1\( we must have \)ad = a^2 = 1\(, ie \)a = \pm 1\(. Therefore we must have \)\mathbf{A} = \begin{pmatrix}1 & 0\\ 0 & 1 \end{pmatrix}\( or \)\mathbf{A} = \begin{pmatrix}-1 & 0\\ 0 & -1 \end{pmatrix}$. For an element to have order \(2\) then \(\mathbf{A}^2 = \mathbf{I}\) ie, \(\mathbf{A} = \mathbf{A}^{-1}\) and \(\mathbf{A} \neq \mathbf{I}\) therefore the only element of order \(2\) is $\begin{pmatrix}-1 & 0\\ 0 & -1 \end{pmatrix}$. For an element to have order \(3\) we must have \(\mathbf{A}^2 = \mathbf{A}^{-1}\), ie $\begin{pmatrix}w^2 + xy & x(w+z)\\ y(w+z) & z^2 + xy \end{pmatrix} = \begin{pmatrix}z & -x\\ -y & w \end{pmatrix}$. Therefore \(w^2 + xy = z, x(w+z) = -x, y(w+z) = -y, z^2+xy = w\). The second and third equations are satisfied iff \(w+z+1 = 0\) or \(x = 0\) and \(y = 0\), but if \(x = 0\) and \(y = 0\) then we aren't order \(3\), so we just need to check this is sufficient for the first and last equations. Since \(\det(\mathbf{A}) = 1\) we have \(wz =xy +1\), so the first and last equations are equivalent to \(w^2 + wz - 1 = z\) and \(x^2 + wz-1 = w\) which are equivalent to \(w(w+z) = z+1\) or \(w + z+ 1 = 0\) as required

View
1991 Paper 1 Q1
D: 1484.0 B: 1513.2
trigonometry trig identity proof sin²θ identity cubic equation angle sum constraint double angle formula

If \(\theta+\phi+\psi=\tfrac{1}{2}\pi,\) show that \[ \sin^{2}\theta+\sin^{2}\phi+\sin^{2}\psi+2\sin\theta\sin\phi\sin\psi=1. \] By taking \(\theta=\phi=\tfrac{1}{5}\pi\) in this equation, or otherwise, show that \(\sin\tfrac{1}{10}\pi\) satisfies the equation \[ 8x^{3}+8x^{2}-1=0. \]

Solution: \begin{align*} S &= \sin^{2}\theta+\sin^{2}\phi+\sin^{2}\psi+2\sin\theta\sin\phi\sin\psi \\ &= \sin^{2}\theta+\sin^{2}\phi+\sin^{2}(\tfrac\pi2-\theta-\phi)+2\sin\theta\sin\phi\sin(\tfrac\pi2-\theta-\phi) \\ &= \sin^{2}\theta+\sin^{2}\phi+\cos^{2}(\theta+\phi)+2\sin\theta\sin\phi\cos(\theta+\phi) \\ &= \sin^{2}\theta+\sin^{2}\phi+\left ( \cos(\theta)\cos(\phi)-\sin(\theta)\sin(\phi)\right)^2+2\sin\theta\sin\phi\left ( \cos(\theta)\cos(\phi)-\sin(\theta)\sin(\phi)\right) \\ &= \sin^{2}\theta+\sin^{2}\phi+\cos^2 \theta\cos^2 \phi-\sin^2 \theta \sin^2 \phi \\ &= \sin^{2}\theta(1-\sin^2 \phi)+\sin^{2}\phi+\cos^2 \theta\cos^2 \phi \\ &= \sin^{2}\theta\cos^2 \phi+\sin^{2}\phi+\cos^2 \theta\cos^2 \phi \\ &= \sin^{2}\phi+\cos^2 \phi \\ &= 1 \end{align*} Suppose \(\theta = \phi = \tfrac15 \pi, \psi = \tfrac1{10}\pi\). Also let \(s = \sin \tfrac1{10}\) \begin{align*} 1 &= 2\sin^2 \tfrac15 \pi + \sin^2 \tfrac1{10} \pi + 2 \sin^2\tfrac15 \pi \sin \tfrac1{10} \pi \\ &= 8\sin^2 \tfrac1{10} \pi \cos^2 \tfrac1{10} \pi + \sin^2 \tfrac1{10} \pi + 8 \sin^2 \tfrac1{10} \pi \cos^2 \tfrac1{10} \pi \sin \tfrac1{10} \pi \\ &= 8\sin^2 \tfrac1{10} \pi(1- \sin^2 \tfrac1{10} \pi) + \sin^2 \tfrac1{10} \pi + 8 \sin^2 \tfrac1{10} \pi (1-\sin^2 \tfrac1{10} \pi) \sin \tfrac1{10} \pi \\ &= 8s^2(1-s^2)+s^2 + 8s^2(1-s^2)s \\ &= -8 s^5 - 8 s^4 + 8 s^3 + 9 s^2 \end{align*} Therefore \(s\) is a root of \(8s^5+8s^4-8s^3-9s^2+1 = 0\), but notice that \begin{align*} 8s^5+8s^4-8s^3-9s^2+1 &= (s-1)(8 s^4 + 16 s^3 + 8 s^2 - s - 1 ) \\ &= (s-1)(s+1)(8s^3+8s^2-1) \end{align*} Therefore since \(\sin \tfrac{1}{10} \pi \neq \pm 1\) it must be a root of \(8x^3+8x^2-1=0\)

View
1991 Paper 1 Q9
D: 1500.0 B: 1516.0
proof inequality exponentials harmonic series summation bounding arguments proof by contradiction

  1. Suppose that the real number \(x\) satisfies the \(n\) inequalities \begin{alignat*}{2} 1<\ & x & & < 2\\ 2<\ & x^{2} & & < 3\\ 3<\ & x^{3} & & < 4\\ & \vdots\\ n<\ & x^{n} & & < n+1 \end{alignat*} Prove without the use of a calculator that \(n\leqslant4\).
  2. If \(n\) is an integer strictly greater than 1, by considering how many terms there are in \[ \frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n^{2}}, \] or otherwise, show that \[ \frac{1}{n}+\frac{1}{n+1}+\cdots+\frac{1}{n^{2}}>1. \] Hence or otherwise find, with justification, an integer \(N\) such that \({\displaystyle {\displaystyle \sum_{n=1}^{N}\frac{1}{n}>10.}}\)

Solution:

  1. Suppose \(n > 4\) then the following inequalities are both true \begin{align*} 3 < x^3 < 4 & \Rightarrow 3^5 < x^{15} < 4^{5}\\ 5 < x^5 < 6 & \Rightarrow 5^{3} < x^{15} < 6^3 \end{align*} But \(3^5 = 243\) and \(6^3 = 216\) so \(243 < x^{15} < 216\) whichis a contradiction.
  2. This question is wrong. Consider \(n = 2\), then \(\frac{1}{2+1} + \frac{1}{2+2} = \frac13+\frac14 = \frac{7}{12} < 1\). The question should be about \(n \geq 4\). \begin{align*} \frac{1}{n+1}+\frac1{n+2}+\cdots + \frac{1}{2n} > \frac{n}{2n} &= \frac12 \\ \frac{1}{2n+1}+\frac1{2n+2}+\cdots + \frac{1}{3n} > \frac{n}{3n} &= \frac13 \\ \frac{1}{4n+1}+\frac1{4n+2}+\cdots + \frac{1}{4n} > \frac{n}{4n} &= \frac14 \\ \sum_{k=1}^{n^2-n} \frac{1}{n+k} > \frac{13}{12} &> 1 \end{align*} We have a stronger result, \(\frac1{n+1} + \cdots + \frac1{4n} > 1\) for \(n > 4\) so we can take \(N = 4^{10}\) since, since there will be \(9\) sequences from \(\frac{1}{4^{i}+1} \to \frac{1}{4^{i+1}}\) and we will have \(\frac1{1}\) at the start to give use the extra \(1\).

View
1991 Paper 2 Q3
D: 1600.0 B: 1516.0
simultaneous equations symmetric functions algebraic manipulation proof cyclic equations distinct values constraint

It is given that \(x,y\) and \(z\) are distinct and non-zero, and that they satisfy \[ x+\frac{1}{y}=y+\frac{1}{z}=z+\frac{1}{x}. \] Show that \(x^{2}y^{2}z^{2}=1\) and that the value of \(x+\dfrac{1}{y}\) is either \(+1\) or \(-1\).

Solution: \begin{align*} && x-y &= \frac1z - \frac1y \\ && x-z &= \frac1x - \frac1y \\ && y-z &= \frac1x - \frac1z \\ \Rightarrow && (x-y)(x-z)(y-z) &= \frac{(y-z)(y-x)(z-x)}{x^2y^2z^2} \\ \Rightarrow && x^2y^2 z^2 &= 1 \\ \end{align*} Suppose \(x + \frac1{y} =k \Rightarrow xy + 1 = ky\) Therefore \(y + \frac{1}{z} = y \pm xy = k\) Therefore \(1 \mp y = k(y \mp 1) \Rightarrow k = \pm 1\)

View
1991 Paper 2 Q6
D: 1600.0 B: 1485.5
integration inequality comparison of integrals trigonometric bounding integrals Taylor polynomial bounds proof estimation

Show by means of a sketch, or otherwise, that if \(0\leqslant\mathrm{f}(y)\leqslant\mathrm{g}(y)\) for \(0\leqslant y\leqslant x\) then \[ 0\leqslant\int_{0}^{x}\mathrm{f}(y)\,\mathrm{d}y\leqslant\int_{0}^{x}\mathrm{g}(y)\,\mathrm{d}y. \] Starting from the inequality \(0\leqslant\cos y\leqslant1,\) or otherwise, prove that if \(0\leqslant x\leqslant\frac{1}{2}\pi\) then \(0\leqslant\sin x\leqslant x\) and \(\cos x\geqslant1-\frac{1}{2}x^{2}.\) Deduce that \[ \frac{1}{1800}\leqslant\int_{0}^{\frac{1}{10}}\frac{x}{(2+\cos x)^{2}}\,\mathrm{d}x\leqslant\frac{1}{1797}. \] Show further that if \(0\leqslant x\leqslant\frac{1}{2}\pi\) then \(\sin x\geqslant x-\frac{1}{6}x^{3}.\) Hence prove that \[ \frac{1}{3000}\leqslant\int_{0}^{\frac{1}{10}}\frac{x^{2}}{(1-x+\sin x)^{2}}\,\mathrm{d}x\leqslant\frac{2}{5999}. \]

View
1991 Paper 2 Q7
D: 1600.0 B: 1500.0
implicit differentiation functional equation logarithm chain rule differential equation substitution proof

The function \(\mathrm{g}\) satisfies, for all positive \(x\) and \(y\), \[ \mathrm{g}(x)+\mathrm{g}(y)=\mathrm{g}(z),\tag{*} \] where \(z=xy/(x+y+1).\) By treating \(y\) as a constant, show that \[ \mathrm{g}'(x)=\frac{y^{2}+y}{(x+y+1)^{2}}\mathrm{g}'(z)=\frac{z(z+1)}{x(x+1)}\mathrm{g}'(z), \] and deduce that \(2\mathrm{g}'(1)=(u^{2}+u)\mathrm{g}'(u)\) for all \(u\) satisfying \(0 < u < 1.\) Now by treating \(u\) as a variable, show that \[ \mathrm{g}(u)=A\ln\left(\frac{u}{u+1}\right)+B, \] where \(A\) and \(B\) are constants. Verify that \(\mathrm{g}\) satisfies \((*)\) for a suitable value of \(B\). Can \(A\) be determined from \((*)\)? The function \(\mathrm{f}\) satisfies, for all positive \(x\) and \(y\), \[ \mathrm{f}(x)+\mathrm{f}(y)=\mathrm{f}(z) \] where \(z=xy.\) Show that \(\mathrm{f}(x)=C\ln x\) where \(C\) is a constant.

Solution: Note that \(z = xy/(x+y+1) \Rightarrow y(x-z) = z(x+1)\) \begin{align*} && g(x) + g(y) &= g(z) \\ \Rightarrow && g'(x) &= g'(z) \cdot \frac{y(x+y+1) - xy \cdot 1} {(x+y+1)^2} \\ &&&= g'(z) \frac{y^2+y}{(x+y+1)^2} \\ &&&= g'(z) \frac{z^2(y^2+y)}{x^2y^2} \\ &&&= g'(z) \frac{z^2(y+1)}{x^2y} \\ &&&= g'(z) \frac{z^2}{x^2} \left (1 + \frac{x-z}{z(x+1)} \right) \\ &&&= g'(z) \frac{z}{x^2} \frac{zx+x}{x+1} \\ &&&= g'(z) \frac{z(z+1)}{x(x+1)} \end{align*} If \(x = 1\) then as \(y\) ranges from \(0\) to \(\infty\), \(z\) ranges from \(0\) to \(1\), so \(g'(1) = \frac{z(z+1)}{1(1+1)}g'(z)\), ie \(2g'(1) = (u^2+u)g'(u)\). \begin{align*} && g'(u) &= \frac{A}{u(u+1)} \\ \Rightarrow && g(u) &= A\int \left ( \frac{1}{u} - \frac{1}{u+1} \right) \d u \\ &&&= A \left ( \ln u - \ln(u+1) \right) + B \\ &&&= A \ln \left ( \frac{u}{u+1} \right) + B \\ \\ && A \ln \left ( \frac{x}{x+1} \right) + B+A \ln \left ( \frac{y}{y+1} \right) + B &=A \ln \left ( \frac{z}{z+1} \right) + B \\ \Rightarrow && B &= A \ln \left ( \frac{z}{z+1} \frac{y+1}{y} \frac{x+1}{x} \right) \\ &&&= A \ln \left ( \frac{1}{1+\frac{x+y+1}{xy}} \frac{(y+1)(x+1)}{xy} \right) \\ &&&= A \ln 1 \\ &&& = 0 \end{align*} Therefore \(B = 0\). \(A\) cannot be determined from \((*)\). Suppose \(f(x) + f(y) = f(z)\), then \(f'(x) = yf'(z)\). Letting \(x = 1\) we find \(f'(1) = uf'(u) \Rightarrow f(u) = C \ln u + D\), but \(D = 0\) so \(f(x) = C \ln x\)

View
1991 Paper 3 Q8
D: 1700.0 B: 1500.1
reduction formula integration by parts trigonometric binomial coefficients cos nθ summation proof

  1. The integral \(I_{k}\) is defined by \[ I_{k}=\int_{0}^{\theta}\cos^{k}x\,\cos kx\,\mathrm{d}x. \] Prove that \(2kI_{k}=kI_{k-1}+\cos^{k}\theta\sin k\theta.\)
  2. Prove that \[ 1+m\cos2\theta+\binom{m}{2}\cos4\theta+\cdots+\binom{m}{r}\cos2r\theta+\cdots+\cos2m\theta=2^{m}\cos^{m}\theta\cos m\theta. \]
  3. Using the results of (i) and (ii), show that \[ m\frac{\sin2\theta}{2}+\binom{m}{2}\frac{\sin4\theta}{4}+\cdots+\binom{m}{r}\frac{\sin2r\theta}{2r}+\cdots+\frac{\sin2m\theta}{2m} \] is equal to \[ \cos\theta\sin\theta+\cos^{2}\theta\sin2\theta+\cdots+\frac{1}{r}2^{r-1}\cos^{r}\theta\sin r\theta+\cdots+\frac{1}{m}2^{m-1}\cos^{m}\theta\sin m\theta. \]

Solution:

  1. \begin{align*} kI_k &= \int_0^\theta \cos^k x k\cos kx \d x \\ &= \left [\cos^k x \sin kx \right]_0^\theta - \int_0^\theta k \cos^{k-1} x \sin x \sin k x \d x \\ &= \cos^k \theta \sin k \theta - k\int_0^\theta \cos^{k-1} x \sin x \sin k x \d x \\ &= \cos^k \theta \sin k \theta + k\int_0^\theta \cos^{k-1} x (\cos (k-1) x - \cos k x \cos x) \d x \\ &= \cos^k\theta \sin k \theta + k I_{k-1} - kI_k \\ \end{align*} So \(2kI_k = kI_{k-1} + \cos^k \sin k \theta\)
  2. \begin{align*} && \sum_{r=0}^m \binom{m}{r} \cos 2r \theta &= \textrm{Re} \left ( \sum_{r=0}^m \binom{m}{r} \exp(i 2r \theta)\right) \\ &&&= \textrm{Re} \left ( \sum_{r=0}^m \binom{m}{r} \exp(i 2 \theta)^r\right) \\ &&&= \textrm{Re} \left ( \left (1+e^{2i \theta} \right)^m\right) \\ &&&= \textrm{Re} \left (\left (e^{i\theta}(e^{i \theta} +e^{-i\theta})\right)^m\right) \\ &&&= \textrm{Re} \left (\left (e^{i\theta}2\cos \theta\right)^m\right) \\ &&&= \textrm{Re} \left (2^m \cos^m \theta e^{im\theta}\right) \\ &&&= 2^m \cos^m \theta \cos m\theta \\ \end{align*}
  3. \begin{align*} \sum_{r=1}^m \binom{m}{r}\frac{\sin 2r \theta}{2r} &= \int_0^\theta \sum_{r=1}^m \binom{m}{r}\cos 2r x\d x \\ &= \int_0^\theta \left (2^m \cos^m x\cos mx- 1 \right ) \d x\\ &= 2^m I_m - \theta \\ &= \frac{2^{m-1}}{m}\cos^m \theta \sin m \theta + 2^{m-1} I_{m-1} - \theta \\ &= \frac{2^{m-1}}{m}\cos^m \theta \sin m \theta + \frac{2^{m-2}}{m-1}\cos^m \theta \sin m \theta+2^{m-2} I_{m-2} - \theta \\ &= \sum_{r=0}^{m} \frac{2^{m-1-r}}{m-r} \cos^{m-r} \theta \sin (m-r) \theta + I_0 - \theta \\ &= \sum_{r=0}^{m} \frac{2^{m-1-r}}{m-r} \cos^{m-r} \theta \sin (m-r) \theta + \int_0^\theta 1 \cdot 1 \d \theta - \theta \\ &= \sum_{r=0}^{m} \frac{2^{m-1-r}}{m-r} \cos^{m-r} \theta \sin (m-r) \theta \\ &= \sum_{r=0}^{m} \frac{2^{r-1}}{r} \cos^{r} \theta \sin r \theta \end{align*}

View
1990 Paper 2 Q1
D: 1600.0 B: 1516.0
inequality quadratics quartic completing the square proof no real roots case analysis polynomials

Prove that both \(x^{4}-2x^{3}+x^{2}\) and \(x^{2}-8x+17\) are non-negative for all real \(x\). By considering the intervals \(x\leqslant0\), \(0 < x\leqslant2\) and \(x > 2\) separately, or otherwise, prove that the equation \[ x^{4}-2x^{3}+x^{2}-8x+17=0 \] has no real roots. Prove that the equation \(x^{4}-x^{3}+x^{2}-4x+4=0\) has no real roots.

Solution: \begin{align*} x^4 - 2x^3+x^2 &= x^2(x^2-2x+1) \\ &= x^2(x-1)^2 > 0 \end{align*} Since \(x\)and \(x-1\) can't both be zero, and square cannot be negative. \begin{align*} x^2 - 8x+17 &= (x-4)^2 +1 \geq 1 > 0 \end{align*} If \(x \leq 2\) then \(x^4 - 2x^3+x^2 > 0\) and \(17-8x \geq 1\) so \(x^4-2x^3+x^2-8x+17 > 0\) If \(x > 2\) then \(x^4-2x^3 = x^3(x-2) \geq 0\) and \(x^2-8x+17 > 0\) so \(x^4-2x^3+x^2-8x+17 > 0\), so for all real numbers our polynomial is positive and therefore cannot have any roots. Note that: \(x^4-x^3+x^2 = x^2(x^2-x+1) > 0\) and \(x^2-4x+4 =(x-2)^2 \geq 0\) If \(x < 1\) then \(x^4-x^3+x^2 > 0\) and \(4(1-x) > 0\) so \(x^4-x^3+x^2-4x+4 > 0\). If \(x > 1\) then \(x^4-x^3 = x^3(x-1) > 0\) and \(x^2-4x+4 \geq 0\) therefore \(x^4-x^3+x^2-4x+4 > 0\). Therefore \(x^4-x^3+x^2-4x+4 > 0\) for all real \(x\) and hence there are no real roots.

View