\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)

[next] [tail] [up]

2018.2.1 Question 1

First, notice that \(x = 0\) must not be a root to this quartic equation. Therefore, we can divide both sides by \(x^2\), and the original equation is equivalent to \[ x^2 + \frac {1}{x^2} + a \left (x + \frac {1}{x}\right ) + b = 0, \] and this rearranges to \[ \left (x + \frac {1}{x}\right )^2 + a \left (x + \frac {1}{x}\right ) + (b - 2) = 0. \]

Notice that \[ k + \frac {1}{k} = \frac {1}{k^{-1}} + k^{-1} = k^{-1} + \frac {1}{k^{-1}}, \] so if \(x = k\) satisfies this equation, then \(x = k^{-1}\) also satisfies this equation.

Notice that the range of \(t = x + \frac {1}{x}\) for non-zero real \(x\) is \(t \in (-\infty , -2] \cup [2, \infty )\).

Since it is given that all the roots are real, it must be the case that the quadratic equation \[ t^2 + at + (b - 2) = 0 \] produces two real roots situated within \((-\infty , -2] \cup [2, \infty )\).

Notice that for \(t \in (-\infty , -2] \cup [2, \infty )\), the equation \[ x + \frac {1}{x} = t \] has precisely two real roots for \(t \neq \pm 2\), and precisely one \(x = \pm 1\) for \(t = \pm 2\).

  1. In this case, by the previous analysis, the only possibility is that \(x_1 = x_2 = x_3 = x_4 = \mp 1\). This means that \[ x^4 + ax^3 + bx^2 + ax + 1 = (x \pm 1)^4 = x^4 \pm 4x^3 + 6x^2 \pm 4x + 1, \] and hence \((a, b) = (\pm 4, 6)\).

  2. Since there are exactly three distinct roots for \(x\), this means that the one which repeated must be \(x_1 = x_2 = \pm 1\), which leads to \(t_1 = \pm 2\), and those two which does not leads to \(t_2 \neq \pm 2\).

    Putting \(t_1 = \pm 2\) into the quadratic equation in \(t\), we have \[ 4 \pm 2a + (b - 2) = 0, \] and hence \[ b = \mp 2a - 2, \] precisely as desired.

  3. When \(b = 2a - 2\), we have \[ t^2 + at + (2a - 4) = 0, \] which solves to \(t_1 = -2\), \(t_2 = - a + 2\).

    For \(x + \frac {1}{x} = t_1 = -2\), this solves to \(x_1 = x_2 = -1\).

    For \(x + \frac {1}{x} = t_2 = -a + 2\), this rearranges to \[ x^2 + (a - 2)x + 1 = 0, \] and hence the two roots are \[ x_{3, 4} = \frac {-(a - 2) \pm \sqrt {(a - 2)^2 - 4}}{2} = \frac {-a + 2 \pm \sqrt {a^2 - 4a}}{2} \]

  4. We first look at necessary condition. Given the equation has precisely two roots, we have \(b = \pm 2a - 2\), and hence the quadratic equation in \(t\) becomes \[ t^2 + at + (\pm 2a - 4) = 0. \]

    \(t_1 = \mp 2\) must be a root, and notice that this factorises to \[ t^2 + at + (\pm 2a - 4) = (t \pm 2) (t - (-a \pm 2)), \] and hence the other root is \(t_2 = -a \pm 2\).

    As discussed before, we must have that \(t_2 < -2\) or \(t_2 > 2\) to produce two distinct roots for \(x\), and hence \[ -a \pm 2 < -2 \text { or } -a \pm 2 > 2, \] and hence \[ a \mp 2 > 2 \text { or } a \mp 2 < -2, \] and hence \[ a > 2 \pm 2 \text { or } a < -2 \pm 2. \]

    Therefore, a necessary condition is \(b = \pm 2a - 2\), and \(a \in \left (-\infty , -2 \pm 2\right ) \cup \left (2 \pm 2, \infty \right )\).

    We would like to show that this is a sufficient condition as well. If \(b = \pm 2a - 2\) and \(a \in \left (-\infty , -2 \pm 2\right ) \cup \left (2 \pm 2, \infty \right )\), we have the quadratic in \(t\) simplifies to \[ t^2 + at + (\pm 2a - 4) = (t \pm 2) (t - (-a \pm 2)) = 0. \]

    This gives roots \(t_1 = \mp 2\) which in turn gives \(x_1 = x_2 = \mp 1\), and \(t_2 = -a \pm 2\). In the second case, since \[ a \in \left (-\infty , -2 \pm 2\right ) \cup \left (2 \pm 2, \infty \right ), \] we must have \[ a \mp 2 \in \left (-\infty , -2\right ) \cup \left (2, \infty \right ) \] and hence \[ -a \pm 2 \in \left (-\infty , -2\right ) \cup \left (2, \infty \right ). \]

    This shows that there are two distinct \(x\)s corresponding to \(t_2\), both of which are not equal to \(\pm 1\).

    Hence, in this case, the original equation has \(3\) distinct roots precisely, and \[ b = \pm 2a - 2, a \in \left (-\infty , -2 \pm 2\right ) \cup \left (2 \pm 2, \infty \right ) \] is a necessary and sufficient condition for the original equation to have precisely \(3\) distinct real roots.

    The following is to simplify this to what is written in the mark scheme. \(b = \pm 2a - 2\) is equivalent to \(b + 2 = \pm 2a\), and \((b + 2)^2 = 4a^2\).

    The second part is equivalent to \(a \mp 2 \in \left (-\infty , -2\right ) \cup \left (2, \infty \right )\), i.e. \[ (a \mp 2)^2 = a^2 \mp 4a + 4 > 4, \] i.e. \[ a^2 > \pm 4a = 2 \pm 2a = 2 (b + 2) = 2b + 4, \] precisely what is in the mark scheme.

[next] [front] [up]