Problems

Solution:

1. \begin{align*} && 0 &= x + 3\sqrt{x} - \frac12 \\ \sqrt{x} = y: && 0&= y^2 + 3y - \frac12 \\ \Rightarrow && y &= \frac{-3\pm\sqrt{3^2+2}}{2} \\ &&&= \frac{-3 \pm \sqrt{11}}{2} \\ y > 0: && x &= \left ( \frac{\sqrt{11}-3}{2} \right)^2 \end{align*}
2. • \begin{align*} && 0 &= x + 10\sqrt{x+2} - 22 \\ y = \sqrt{x+2}: && 0 &= y^2 - 2 + 10y - 22 \\ &&&= y^2 + 10y - 24 \\ &&&= (y-2)(y+12) \\ \Rightarrow && y &= 2, -12 \\ y > 0: && x &= 2 \end{align*}
   • Let \(y = \sqrt{2x^2-8x-3}\), so \begin{align*} && 0 &= x^2 - 4x +\sqrt{2x^2-8x-3} - 9 \\ && 0 &= \frac{y^2+3}{2} + y - 9 \\ &&&= \frac12 y^2 +y - \frac{15}{2} \\ &&&= \frac12 (y-3)(y+5) \\ \Rightarrow && y &= 3,-5 \\ y > 0: && 9 &= 2x^2-8x-3 \\ \Rightarrow && 0 &= 2x^2-8x-12 \\ &&&= 2(x^2-4x-6) \\ \Rightarrow && x &= 2 \pm \sqrt{10} \end{align*}

Solution:

1. \(\,\) \begin{align*} && \mathbb{P}(\text{both roots real}) &= \mathbb{P}(\Delta \geq 0) \\ &&&= \mathbb{P}(U^2 \geq 4V) \\ &&&= \mathbb{P}(V = -1) \\ &&&= \tfrac13 ( \tfrac23 + \tfrac12 + \tfrac13) \\ &&&= \tfrac13 \cdot \tfrac 32 = \frac12 \end{align*}
2. Our equations will be: \(x^2+x-1 = 0\) with larger root \(\frac{-1 + \sqrt{5}}{2}\) \(x^2-1 = 0\) with larger root \(1\) \(x^2-x-1 = 0\) with larger root \(\frac{1 + \sqrt5}{2}\) and the expected value is \begin{align*} && \E[\text{larger root}|\text{both real}] &= \frac23 \left ( \frac23 \cdot \frac{-1+\sqrt5}{2} + \frac12 \cdot 1 + \frac13 \cdot \frac{1+\sqrt5}{2} \right) \\ &&&= \frac23 \left ( \frac{2+3\sqrt5}{6} \right) \\ &&&= \frac{2+3\sqrt5}{9} \end{align*}
3. Suppose we have \(x^3+(U-2V)x^2+(1-2UV)x + U = 0\), then for all roots to be positive, we need \(U < 0 \Rightarrow U = -1\) (otherwise there is a root at or below zero). Therefore our two possible cubics are: \(x^3 -3x^2+3x-1 = (x-1)^3\) (all roots positive) \(x^3+x^2-x-1 = (x-1)(x+1)^2\) (not all roots positive!) Therefore the probability is \(\frac13 \cdot \frac23 = \frac29\)

Solution: If \(p \geq 0\) then the derivative is \(x^2+p \geq 0\) and in particular the function is increasing. Therefore it will have exactly \(1\) real root (as for very large negative \(x\) it is negative, and vice-versa fo positive \(x\)). \begin{align*} && \frac{\d y}{\d x} &= \frac{\dot{y}}{\dot{x}} \\ &&&= \frac{2a}{2at} \\ &&&= \frac{1}{t} \\ \text{eq of normal} && \frac{k-2at}{h-at^2} &= -t \\ \Rightarrow && k-2at &= at^3-th \\ && 0 &= at^3+(2a-h)t-k \end{align*} Since \(a > 0\) this is the same constraint as the first part, in particular \(2a-h \geq 0 \Leftrightarrow 2a \geq h\). If exactly two normals can be drawn to \(C\) we must have that our equation has a repeated root, ie \begin{align*} && 0 &= at^3+(2a-h)t-k\\ && 0 &= 3at^2+2a-h\\ \Rightarrow && 0 &= 3at^3+ 3(2a-h)t-3k \\ && 0 &= 3at^3+(2a-h)t \\ \Rightarrow && 0 &= 2(2a-h)t-3k \\ \Rightarrow && t &= \frac{3k}{2(2a-h)} \\ \Rightarrow && 0 &= 3a \left (\frac{3k}{2(2a-h)} \right)^2+2a-h \\ && 0 &= 27ak^2+4(2a-h)^3 \end{align*}

+ - ↺

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*}