7 problems found
Let \(\mathrm{h}(z) = nz^6 + z^5 + z + n\), where \(z\) is a complex number and \(n \geqslant 2\) is an integer.
It is given that the two curves \[ y=4-x^2 \text{ and } m x = k-y^2\,, \] where \(m > 0\), touch exactly once.
Solution:
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.
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.
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:
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*}
Solution:
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