Problems

Filters
Clear Filters

7 problems found

2022 Paper 2 Q7
D: 1500.0 B: 1500.0

Let \(\mathrm{h}(z) = nz^6 + z^5 + z + n\), where \(z\) is a complex number and \(n \geqslant 2\) is an integer.

  1. Let \(w\) be a root of the equation \(\mathrm{h}(z) = 0\).
    1. Show that \(|w^5| = \sqrt{\dfrac{\mathrm{f}(w)}{\mathrm{g}(w)}}\), where \[\mathrm{f}(z) = n^2 + 2n\operatorname{Re}(z) + |z|^2 \quad \text{and} \quad \mathrm{g}(z) = n^2|z|^2 + 2n\operatorname{Re}(z) + 1.\]
    2. By considering \(\mathrm{f}(w) - \mathrm{g}(w)\), prove by contradiction that \(|w| \geqslant 1\).
    3. Show that \(|w| = 1\).
  2. It is given that the equation \(\mathrm{h}(z) = 0\) has six distinct roots, none of which is purely real.
    1. Show that \(\mathrm{h}(z)\) can be written in the form \[\mathrm{h}(z) = n(z^2 - a_1 z + 1)(z^2 - a_2 z + 1)(z^2 - a_3 z + 1),\] where \(a_1\), \(a_2\) and \(a_3\) are real constants.
    2. Find \(a_1 + a_2 + a_3\) in terms of \(n\).
    3. By considering the coefficient of \(z^3\) in \(\mathrm{h}(z)\), find \(a_1 a_2 a_3\) in terms of \(n\).
    4. How many of the six roots of the equation \(\mathrm{h}(z) = 0\) have a negative real part? Justify your answer.

2012 Paper 3 Q3
D: 1700.0 B: 1468.7

It is given that the two curves \[ y=4-x^2 \text{ and } m x = k-y^2\,, \] where \(m > 0\), touch exactly once.

  1. In each of the following four cases, sketch the two curves on a single diagram, noting the coordinates of any intersections with the axes:
    1. \(k < 0\, \);
    2. \(0 < k < 16\), \(k/m < 2\,\);
    3. \(k > 16\), \(k/m > 2\,\);
    4. \(k > 16\), \(k/m < 2\,\).
  2. Now set \(m=12\). Show that the \(x\)-coordinate of any point at which the two curves meet satisfies \[ x^4-8x^2 +12x +16-k=0\,. \] Let \(a\) be the value of \(x\) at the point where the curves touch. Show that \(a\) satisfies \[ a^3 -4a +3 =0 \] and hence find the three possible values of \(a\). Derive also the equation \[ k= -4a^2 +9a +16\,. \] Which of the four sketches in part (i) arise?


Solution:

    1. \(\,\)
      TikZ diagram
    2. \(\,\)
      TikZ diagram
    3. \(\,\)
      TikZ diagram
    4. \(\,\)
      TikZ diagram
  1. Suppose \(m = 12\) \begin{align*} && y &= 4-x^2 \\ && 12x &= k-y^2 \\ \Rightarrow && 12 x&=k-(4-x^2)^2 \\ &&&= k-16+8x^2-x^4 \\ \Rightarrow && 0 &= x^4- 8x^2+12x+16-k \end{align*} When the curves touch, we will have repeated root, ie \(a\) is a root of \(4x^3-16x+12 \Rightarrow a^3-4a+3 =0\). \begin{align*} &&0 &= a^3-4a+3 \\ &&&= (a-1)(a^2+a-3) \\ \Rightarrow &&a &= 1, \frac{-1 \pm \sqrt{13}}{2} \end{align*} \begin{align*} && 0 &= a^4-8a^2+12a+16-k \\ \Rightarrow && k &= a(a^3-8a+12)+16 \\ &&&= a(4a-3-8a+12)+16 \\ &&&= -4a^2+9a+16 \\ \\ \Rightarrow && a = 1& \quad k = 21 \\ && k &= -4(3-a)+9a+16 = 13a+4\\ && a = \frac{-1-\sqrt{13}}2& \quad k = \frac{-5 - 13\sqrt{13}}{2} < 0 \\ && a = \frac{-1+\sqrt{13}}2& \quad k = \frac{-5 + 13\sqrt{13}}{2} \\ \end{align*} So we have type (a), and (d).

2003 Paper 3 Q14
D: 1700.0 B: 1599.8

Write down the probability generating function for the score on a standard, fair six-faced die whose faces are labelled \(1, 2, 3, 4, 5, 6\). Hence show that the probability generating function for the sum of the scores on two standard, fair six-faced dice, rolled independently, can be written as \[ \frac1{36} t^2 \l 1 + t \r^2 \l 1 - t + t^2 \r^2 \l 1 + t + t^2 \r^2 \;. \] Write down, in factorised form, the probability generating functions for the scores on two fair six-faced dice whose faces are labelled with the numbers \(1, 2, 2, 3, 3, 4\) and \(1, 3, 4, 5, 6, 8,\) and hence show that when these dice are rolled independently, the probability of any given sum of the scores is the same as for the two standard fair six-faced dice. Standard, fair four-faced dice are tetrahedra whose faces are labelled \(1, 2, 3, 4,\) the score being taken from the face which is not visible after throwing, and each score being equally likely. Find all the ways in which two fair four-faced dice can have their faces labelled with positive integers if the probability of any given sum of the scores is to be the same as for the two standard fair four-faced dice.

2002 Paper 3 Q5
D: 1700.0 B: 1516.0

Give a condition that must be satisfied by \(p\), \(q\) and \(r\) for it to be possible to write the quadratic polynomial \(px^2 + qx + r\) in the form \(p \l x + h \r^2\), for some \(h\). Obtain an equation, which you need not simplify, that must be satisfied by \(t\) if it is possible to write \[ \l x^2 + \textstyle{{1 \over 2}} bx + t \r^2 - \l x^4 + bx^3 + cx^2 +dx +e \r \] in the form \(k \l x + h \r^2\), for some \(k\) and \(h\). Hence, or otherwise, write \(x^4 + 6x^3 + 9x^2 -2x -7\) as a product of two quadratic factors.

2000 Paper 1 Q8
D: 1484.0 B: 1484.0

Show that \[ \int_{-1}^1 \vert \, x\e^x \,\vert \d x =- \int_{-1}^0 x\e^x \d x + \int_0^1 x\e^x \d x \] and hence evaluate the integral. Evaluate the following integrals:

  1. \(\displaystyle \int_0^4 \vert\, x^3-2x^2-x+2 \,\vert \, \d x\,;\)
  2. \(\displaystyle \int_{-\pi}^\pi \vert\, \sin x +\cos x \,\vert \; \d x\,.\)


Solution: \begin{align*} && \int_{-1}^1 |x e^x |\d x &= \int_{-1}^0 |xe^x| \d x + \int_0^1 |xe^x| \d x \\ &&&= \int_{-1}^0 -xe^x \d x + \int_0^1 x \e^x \d x \\ &&&= -\int_{-1}^0 xe^x \d x + \int_0^1 x \e^x \d x \\ \\ && \int xe^x \d x &= xe^x - \int e^x \d x \\ &&&= xe^x - e^x \\ \\ \Rightarrow && \int_{-1}^1 |x e^x |\d x &= \left [ xe^x - e^x \right]_0^{-1}+ \left [ xe^x - e^x \right]_0^{1} \\ &&&= -e^{-1}-e^{-1} +e^{0} + e^1 - e^1 +e^0 \\ &&&= 2-2e^{-1} \end{align*}

  1. \(\,\) \begin{align*} && I &= \int_0^4 | x^3-2x^2-x+2| \d x \\ &&&= \int_0^4 |(x-2)(x-1)(x+1)| \d x\\ &&&= \int_0^1( x^3-2x^2-x+2) \d x- \int_1^2 ( x^3-2x^2-x+2) \d x + \int_2^4 ( x^3-2x^2-x+2) \d x \\ &&&= \left [\frac14 x^4-\frac23 x^3- \frac12 x^2 +2x \right]_0^1 - \left [\frac14 x^4-\frac23 x^3- \frac12 x^2 +2x \right]_1^2 + \left [\frac14 x^4-\frac23 x^3- \frac12 x^2 +2x \right]_2^4 \\ &&&= 2 \left ( \frac14 - \frac23 -\frac12 + 2\right) - 2 \left ( \frac14 2^4 - \frac23 2^3 -\frac12 2^2 + 2 \cdot 2\right)+ \left ( \frac14 4^4 - \frac23 4^3 -\frac12 4^2 + 2 \cdot 4\right) \\ &&&= \frac{133}{6} \end{align*}
  2. \(\,\) \begin{align*} && J &= \int_{-\pi}^\pi | \sin x + \cos x | \d x \\ &&&= \int_{-\pi}^{\pi} | \sqrt{2} \sin(x + \tfrac{\pi}{4})| \d x \\ &&&= 2\sqrt{2}\int_0^\pi \sin x \d x \\ &&&= 4\sqrt{2} \end{align*}

1995 Paper 1 Q1
D: 1484.0 B: 1484.0

  1. Find the real values of \(x\) for which \[ x^{3}-4x^{2}-x+4\geqslant0. \]
  2. Find the three lines in the \((x,y)\) plane on which \[ x^{3}-4x^{2}y-xy^{2}+4y^{3}=0. \]
  3. On a sketch shade the regions of the \((x,y)\) plane for which \[ x^{3}-4x^{2}y-xy^{2}+4y^{3}\geqslant0. \]


Solution:

  1. \(\,\) \begin{align*} && 0 & \leq x^3 - 4x^2 - x + 4 \\ &&&= (x-1)(x^2-3x-4) \\ &&&= (x-1)(x-4)(x+1) \\ \Leftrightarrow && x &\in [-1, 1] \cup [4, \infty) \end{align*}
  2. \(\,\) \begin{align*} && 0 &= x^{3}-4x^{2}y-xy^{2}+4y^{3} \\ && 0 &= (x-y)(x-4y)(x+y) \end{align*} Therefore the lines are \(y = x, 4y = x, y=-x\).
  3. TikZ diagram
    (quickest way to see this is to check the \(x\) or \(y\)-axis)

1990 Paper 2 Q10
D: 1600.0 B: 1496.1

Two square matrices \(\mathbf{A}\) and \(\mathbf{B}\) satisfies \(\mathbf{AB=0}.\) Show that either \(\det\mathbf{A}=0\) or \(\det\mathbf{B}=0\) or \(\det\mathbf{A}=\det\mathbf{B}=0\). If \(\det\mathbf{B}\neq0\), what must \(\mathbf{A}\) be? Give an example to show that the condition \(\det\mathbf{A}=\det\mathbf{B}=0\) is not sufficient for the equation \(\mathbf{AB=0}\) to hold. Find real numbers \(p,q\) and \(r\) such that \[ \mathbf{M}^{3}+2\mathbf{M}^{2}-5\mathbf{M}-6\mathbf{I}=(\mathbf{M}+p\mathbf{I})(\mathbf{M}+q\mathbf{I})(\mathbf{M}+r\mathbf{I}), \] where \(\mathbf{M}\) is any square matrix and \(\mathbf{I}\) is the appropriate identity matrix. Hence, or otherwise, find all matrices \(\mathbf{M}\) of the form $\begin{pmatrix}a & c\\ 0 & b \end{pmatrix}$ which satisfy the equation \[ \mathbf{M}^{3}+2\mathbf{M}^{2}-5\mathbf{M}-6\mathbf{I}=\mathbf{0}. \]


Solution: Since \(0 = \det \mathbf{0} = \det \mathbf{AB} = \det \mathbf{A} \det\mathbf{B}\) at least one of \(\det \mathbf{A}\) or \(\det \mathbf{B}\) is zero. If \(\det \mathbf{B} \neq 0\) then \(\mathbf{B}\) is invertible, and multiplying on the right by \(\mathbf{B}^{-1}\) gives us \(\mathbf{A} = \mathbf{0}\). If \(\mathbf{A} = \begin{pmatrix} 1 & 1 \\ 0 & 0 \end{pmatrix}\) and \(\mathbf{B} = \begin{pmatrix} 1 & 0 \\1 & 0 \end{pmatrix}\), then \(\det \mathbf{A} = \det \mathbf{B} = 0\), but \(\mathbf{AB} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} \neq \mathbf{0}\) Since \(\mathbf{M}\) commutes with itself and the identity matrix, this is equivalent to factorising the polynomial over the reals. Therefore $$\mathbf{M}^{3}+2\mathbf{M}^{2}-5\mathbf{M}-6\mathbf{I}=(\mathbf{M}-2\mathbf{I})(\mathbf{M}+\mathbf{I})(\mathbf{M}+3\mathbf{I}),$$ Since we now know at least one of \(\det (\mathbf{M}-2\mathbf{I})\), \(\det (\mathbf{M}+\mathbf{I})\), \(\det (\mathbf{M}+3\mathbf{I})\), we should look at cases: Since at least one of those must be non-zero, we must have the following cases: \((a,b) = (2,-1), (-1,2), (-1,-3), (-3,-1), (2,-3), (-3,2)\) In each of those cases, we will have: \(\begin{pmatrix} 0 & c \\ 0 & b+k \end{pmatrix}\begin{pmatrix} a+l & c \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0\end{pmatrix}\) and so all of those solutions are valid. So \(c\) can be anything as long as \((a,b)\) are in that set of solutions