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1996 Paper 3 Q7
D: 1700.0 B: 1515.1
roots of polynomials symmetric functions Newton's sums Vieta's formulas power sums simultaneous equations Heron's formula cubic

  1. If \(x+y+z=\alpha,\) \(xy+yz+zx=\beta\) and \(xyz=\gamma,\) find numbers \(A,B\) and \(C\) such that \[ x^{3}+y^{3}+z^{3}=A\alpha^{3}+B\alpha\beta+C\gamma. \] Solve the equations \begin{alignat*}{1} x+y+z & =1\\ x^{2}+y^{2}+z^{2} & =3\\ x^{3}+y^{3}+z^{3} & =4. \end{alignat*}
  2. The area of a triangle whose sides are \(a,b\) and \(c\) is given by the formula \[ \mathrm{area}=\sqrt{s(s-a)(s-b)(s-c)} \] where \(s\) is the semi-perimeter \(\frac{1}{2}(a+b+c).\) If \(a,b\) and \(c\) are the roots of the equation \[ x^{3}-16x^{2}+81x-128=0, \] find the area of the triangle.

Solution:

  1. \begin{align*} (x+y+z)^3 &= x^3+y^3+z^3+ \\ &\quad 3xy^2 + 3xz^2 + 3yx^2 + \cdots + 3zy^2 \\ &\quad\quad + 6xyz \\ (x+y+z)(xy+yz+zx) &= x^2y+x^2z + \cdots + z^2 x + 3xyz \\ x^3+y^3+z^3 &= (x+y+z)^3 - 3(xy^2 + \cdots + zy^2) - 6xyz \\ &= \alpha^3 - 3(\alpha \beta - 3\gamma)-6\gamma \\ &= \alpha^3-3\alpha \beta+3\gamma \end{align*} Since \(4 = 1^3-3\cdot1\cdot(-1) + 3 \gamma \Rightarrow \gamma = 0\), therefore one of \(x,y,z = 0\). WLOG \(x = 0\), so \(y+z = 1, y^2 + z^2 = 3 \Rightarrow y^2 + (1-y)^2 = 3 \Rightarrow y^2 -y -1 = 0 \Rightarrow y = \frac{1 \pm \sqrt{5}}{2}\), so we have \((x,y,z) = (0, \frac{1 +\sqrt{5}}{2}, \frac{1 - \sqrt{5}}{2})\) and permutations.
  2. \begin{align*} A^2 &= s(s-a)(s-b)(s-c) \\ \end{align*} Notice the second part is the same as plugging \(s= 16/2 = 8\) into our polynomial Therefore \begin{align*} A^2 &= 8 \cdot (8^3 - 16 \cdot 8^2 + 81 \cdot 8 - 128) \\ &= 8 \cdot 8 (8^2 - 16 \cdot 8 + 81- 16) \\ &= 64 (-64+81-16) \\ &= 64 \end{align*} Therefore \(A = 8\)

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1995 Paper 2 Q6
D: 1600.0 B: 1516.0
complex numbers roots of unity De Moivre quadratic equation Vieta's formulas trigonometric identity 7th roots of unity geometric series

If \(u\) and \(v\) are the two roots of \(z^{2}+az+b=0,\) show that \(a=-u-v\) and \(b=uv.\) Let \(\alpha=\cos(2\pi/7)+\mathrm{i}\sin(2\pi/7).\) Show that \(\alpha\) is a root of \(z^{6}-1=0\) and express the roots in terms of \(\alpha.\) The number \(\alpha+\alpha^{2}+\alpha^{4}\) is a root of a quadratic equation \[ z^{2}+Az+B=0 \] where \(A\) and \(B\) are real. By guessing the other root, or otherwise, find the numerical values of \(A\) and \(B\). Show that \[ \cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{8\pi}{7}=-\frac{1}{2}, \] and evaluate \[ \sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}, \] making it clear how you determine the sign of your answer.

Solution: \begin{align*} 0 &= z^2+az+b \\ &= (z-u)(z-v) \\ &= z^2-(u+v)z+uv \end{align*} Therefore by comparing coefficients, \(a = -u-v\) and \(b = uv\). Suppose \(\alpha = \cos(2\pi/7) + i \sin (2\pi/7)\), then by De Moivre, \(\alpha^7 = \cos(2\pi) + i \sin (2\pi) = 1\), ie \(\alpha^7-1 = 0\). Notice that \((\alpha+\alpha^2 + \alpha^4) + (\alpha^3+\alpha^5+\alpha^6) = -1\) and \begin{align*} P &= (\alpha+\alpha^2 + \alpha^4)(\alpha^3+\alpha^5+\alpha^6) \\ &= \alpha^4 + \alpha^6 + \alpha^7 + \alpha^5 + \alpha^7 + \alpha^8 + \alpha^{7}+\alpha^{9}+\alpha^{10} \\ &= 3 + \alpha+ \alpha^2 + \alpha^3 + \alpha^4 + \alpha^5 + \alpha^6 \\ &= 2 \end{align*} Therefore it is a root of \(x^2+x+2 = 0 \Rightarrow x = \frac{-1 \pm i\sqrt{7}}{2}\) Therefore $\cos\frac{2\pi}{7} + \cos \frac{4\pi}{7} + \cos \frac{8\pi}{7} = \textrm{Re}(\alpha + \alpha^2 + \alpha^4) = -\frac12$ And \(\sin\frac{2\pi}{7} + \sin\frac{4\pi}{7} + \sin\frac{8\pi}{7} = \textrm{Im}(\alpha + \alpha^2 + \alpha^4) = \pm\frac{\sqrt{7}}2\) since it is positive it is \(\frac{\sqrt{7}}{2}\)

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1992 Paper 2 Q7
D: 1600.0 B: 1500.0
roots of polynomials cubic symmetric functions Vieta's formulas repeated roots discriminant algebraic manipulation

The cubic equation \[ x^{3}-px^{2}+qx-r=0 \] has roots \(a,b\) and \(c\). Express \(p,q\) and \(r\) in terms of \(a,b\) and \(c\).

  1. If \(p=0\) and two of the roots are equal to each other, show that \[ 4q^{3}+27r^{2}=0. \]
  2. Show that, if two of the roots of the original equation are equal to each other, then \[ 4\left(q-\frac{p^{2}}{3}\right)^{3}+27\left(\frac{2p^{3}}{27}-\frac{pq}{3}+r\right)^{2}=0. \]

Solution: \(p = a+b+c, q = ab+bc+ca, r = abc\)

  1. Suppose two roots are equal to each other, this means that one of the roots is also a root of the derivative. ie \begin{align*} && 0 &= x^3+qx - r \\ && 0 &= 3x^2+q \end{align*} have a common root, but this root must satisfy \(x^2 = -\frac{q}{3}\). Then \begin{align*} &&0 &= x^3 + qx - r \\ &&&= x^3 -3x^3 - r \\ &&&= -2x^3 -r \\ \Rightarrow && r^2 &= 4x^6 \\ &&&= 4 \left ( -\frac{q}{3}\right)^3 \\ \Rightarrow && 0 &= 27r^2+4q^3 \end{align*}
  2. Consider \(x = z + \frac{p}{3}\), then the equation is: \begin{align*} x^{3}-px^{2}+qx-r &= (z + \frac{p}{3})^3 - p(z + \frac{p}{3})^2 + q(z + \frac{p}{3}) - r \\ &= z^3 + pz^2 + \frac{p^2}{3}z + \frac{p^3}{27} - \\ &\quad -pz^2-\frac{2p^2}{3}z-\frac{p^3}{9} + \\ &\quad\quad qz + \frac{pq}{3} - r \\ &= z^3+\left (\frac{p^2}{3}-\frac{2p^2}{3}+q \right)z + \left (\frac{p^3}{27}-\frac{p^3}{9}+\frac{pq}{3}-r \right) \\ &= z^3+\left (-\frac{p^2}{3}+q \right)z + \left (-\frac{2p^3}{27}+\frac{pq}{3}-r \right) \\ \end{align*} Since this equation must also have repeated roots we must have: \begin{align*} 4\left (-\frac{p^2}{3}+q \right)^3 + 27 \left (-\frac{2p^3}{27}+\frac{pq}{3}-r \right)^2 = 0 \end{align*} which is exactly our desired result

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1992 Paper 3 Q8
D: 1700.0 B: 1515.1
complex numbers De Moivre trig identity roots of polynomials binomial coefficients summation Vieta's formulas cot cosec

Show that \[ \sin(2n+1)\theta=\sin^{2n+1}\theta\sum_{r=0}^{n}(-1)^{n-r}\binom{2n+1}{2r}\cot^{2r}\theta, \] where \(n\) is a positive integer. Deduce that the equation \[ \sum_{r=0}^{n}(-1)^{r}\binom{2n+1}{2r}x^{r}=0 \] has roots \(\cot^{2}(k\pi/(2n+1))\) for \(k=1,2,\ldots,n\). Show that

  • sep}{3mm}
  • \(\bf (i)\) \({\displaystyle \sum_{k=1}^{n}\cot^{2}\left(\frac{k\pi}{2n+1}\right)=\frac{n(2n-1)}{3}},\)
  • \(\bf (ii)\) \({\displaystyle \sum_{k=1}^{n}\tan^{2}\left(\frac{k\pi}{2n+1}\right)=n(2n+1)},\)
  • \(\bf (iii)\) \({\displaystyle \sum_{k=1}^{n}\mathrm{cosec}^{2}\left(\frac{k\pi}{2n+1}\right)=\frac{2n(n+1)}{3}}.\)

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1991 Paper 3 Q10
D: 1700.0 B: 1516.0
roots of polynomials symmetric functions Newton's sums proof by induction power sums Vieta's formulas recurrence relation

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

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1988 Paper 2 Q16
D: 1600.0 B: 1570.7
normal distribution probability quadratic equation real roots conditional probability expected value discriminant Vieta's formulas

Find the probability that the quadratic equation \[ X^{2}+2BX+1=0 \] has real roots when \(B\) is normally distributed with zero mean and unit variance. Given that the two roots \(X_{1}\) and \(X_{2}\) are real, find:

  1. the probability that both \(X_{1}\) and \(X_{2}\) are greater than \(\frac{1}{5}\);
  2. the expected value of \(\left|X_{1}+X_{2}\right|\);
giving your answers to three significant figures.

Solution: The roots are \(X_1, X_2 = -B \pm \sqrt{B^2-1}\)

  1. The smallest root will be \(-B - \sqrt{B^2-1}\). For this to be larger than \(\frac15\) we must have, \begin{align*} && -B -\sqrt{B^2-1} &\geq \frac15 \\ \Rightarrow && -B - \frac15 &\geq \sqrt{B^2 - 1} \\ \Rightarrow && B^2 + \frac25 B + \frac1{25} &\geq B^2 - 1 \\ \Rightarrow && \frac25 B \geq -\frac{26}{25} \\ \Rightarrow && B \geq -\frac{13}{5} \end{align*} Therefore \(-\frac{13}5 \leq B \leq -1\). Therefore we want: \begin{align*} \frac{\P(-\frac{13}5 \leq B \leq -1)}{\P(B < -1) + \P(B > 1)} &= \frac{\Phi(-1) - \Phi(-\frac{13}{5})}{\Phi(-1)+1-\Phi(1)} \\ &= \frac{0.1586\ldots - 0.0046\ldots}{0.1586\ldots + 1- 0.8413\ldots} \\ &= 0.4853\ldots \\ &= 0.485 \,\,(3 \text{ s.f.}) \end{align*}
  2. \(X_1 + X_2 = -2B\). Therefore we want: \begin{align*} \mathbb{E}(|X_1 + X_2| &= \mathbb{E}(|2B|) \\ &= 2 \l\frac{1}{2\Phi(-1)} \int_1^{\infty} B \frac{1}{\sqrt{2 \pi}} e^{-\frac12 B^2} \, \d B+\frac{1}{2\Phi(-1)} \int_{-\infty}^{-1} B \frac{1}{\sqrt{2 \pi}} e^{-\frac12 B^2} \, \d B \r \\ &= \frac{4}{2\Phi(-1)} \int_1^{\infty} B \frac{1}{\sqrt{2 \pi}} e^{-\frac12 B^2} \, \d B \\ &=\frac{4}{2\sqrt{2 \pi} \Phi(-1)} \left [ -e^{-\frac12 B^2}\right]_1^{\infty} \\ &= \frac{4}{2\sqrt{2 \pi} \Phi(-1) \sqrt{e}} \\ &= 3.0502\ldots \\ &= 3.05\,\, (3\text{ s.f.}) \end{align*}

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