Problems

---

Solution:

1. \(y = e^x(2x^2-5x+2) = e^x(2x-1)(x-2)\), we also have \(y' = e^x(2x^2-5x+2 + 4x-5) = e^x(2x^2-x-3) = e^x(2x-3)(x+1)\) \(y(-1) = \frac{9}{e}\), \(y(\frac32) = -e^{3/2}\)
   

   + - ↺
   If \(k < -e^{3/2}\) there are no solutions. If \(k = -e^{3/2}\) there is a unique solution. If \(-e^{3/2} < k \leq 0\) there are two solutions. If \(0 < k < \frac{9}{e}\) there are three solutions. Otherwise there is a unique solution.
2. 

   + - ↺

---

Solution:

1. \begin{align*} \cos 15^{\circ} &= \cos (45^{\circ} - 30^{\circ}) \\ &= \cos 45^{\circ} \cos 30^{\circ} + \sin 45^{\circ} \sin 30^{\circ} \\ &= \frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}}\frac{1}{2} \\ &= \frac{\sqrt{3}+1}{2\sqrt{2}} \\ \\ \sin15^{\circ} &= \sin(45^{\circ} - 30^{\circ}) \\ &= \sin45^{\circ} \cos 30^{\circ} - \cos 45^{\circ} \sin 30^{\circ} \\ &= \frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2} - \frac{1}{\sqrt{2}}\frac{1}{2} \\ &= \frac{\sqrt{3}-1}{2\sqrt{2}} \\ \end{align*}
2. \begin{align*} \cos 3 \alpha &= \cos 2\alpha \cos \alpha - \sin2\alpha \sin \alpha \\ &= (2\cos^2 \alpha -1)\cos \alpha - 2 \cos \alpha \sin^2 \alpha \\ &= 2\cos^3 \alpha - \cos \alpha - 2\cos \alpha (1-\cos^2 \alpha) \\ &= 4\cos^2 \alpha - 3\cos \alpha \end{align*} Therefore if \(x = \cos \alpha\) then \(4x^3 - 3x-\cos3\alpha = 0\). \begin{align*} 0 &= 4x^3 - 3x-\cos3\alpha \\ &= 4x^3 - 3x - 4\cos^3\alpha+ 3\cos \alpha \\ &= 4(x-\cos\alpha)(x^2+x\cos\alpha+\cos^2\alpha)-3(x-\cos\alpha)\\ &= (x - \cos \alpha)(4x^2+4x\cos\alpha+4\cos^2\alpha-3) \end{align*} Therefore the other roots will be solutions to the second quadratic which are: \begin{align*} \frac{-4\cos \alpha \pm \sqrt{16\cos^2\alpha - 16(4\cos^2\alpha-3)}}{8} &= \frac{-\cos \alpha \pm \sqrt{3(1-\cos^2\alpha)}}{2} \\ &= \frac{-\cos \alpha \pm \sqrt{3} \sin \alpha}{2} \end{align*}
3. Suppose \(y^3-3y-\sqrt{2} = 0\) then \(4\l \frac{y}{2} \r ^3-3(\frac{y}{2}) -\frac{\sqrt{2}}{2} = 0\) or alternatively, if \(x = \frac{y}{2}\), \(4x^3-3x-\cos 45^{\circ} = 0\). Therefore \(x = \cos 15^{\circ}, \frac{-\cos 15^{\circ} \pm \sqrt{3} \sin 15^{\circ}}{2}\) Therefore \(y =2\cos 15^{\circ}, -\cos 15^{\circ} \pm \sqrt{3} \sin 15^{\circ}\) or \(y = \frac{\sqrt{6}+\sqrt{2}}{2}\), \begin{align*} y &= -\frac{\sqrt{3}+1}{2\sqrt{2}} \pm \frac{3-\sqrt{3}}{2\sqrt{2}} \\ &= \frac{-4}{2\sqrt{2}}, \frac{-2\sqrt{3}}{2\sqrt{2}} \\ &= -\sqrt{2}, -\frac{\sqrt{6}-\sqrt{2}}{2} \end{align*}

---

Solution:

1. \(\,\) \begin{align*} && f(x) &= \int_1^3 (t-1)^{x-1} \d t \\ &&&= \left [ \frac1x(t-1)^{x} \right]_1^3 \\ &&&= \frac{2^x}{x} \end{align*}
   

   + - ↺
2. \(\,\) \begin{align*} && g(x) &= \int_{-1}^1 \frac{1}{\sqrt{1-2xt+x^2}} \d t \\ &&&= \left [ -\frac{1}{x}(1 +x^2 - 2xt)^{\frac12} \right]_{-1}^1 \\ &&&= \frac1x \left ( \sqrt{1+x^2+2x}-\sqrt{1+x^2-2x}\right) \\ &&&= \frac1x \left ( |1+x|-|1-x| \right) \end{align*}
   

   + - ↺

---

Solution: Let \(O\) be the origin, and let \(\mathbf{a}, \mathbf{b}, \mathbf{a}', \mathbf{b}'\) be the four points. The conditions give us \begin{align*} && \mathbf{a} \cdot \mathbf{b} &= 0 \\ && |\mathbf{a}| &= |\mathbf{b}| \\ && \mathbf{a}' \cdot \mathbf{b}' &= 0 \\ && |\mathbf{a}'| &= |\mathbf{b}'| \\ \end{align*} So \begin{align*} \text{midpoint }AB \text{ to midpoint } BA' &= (\tfrac12(\mathbf{a}+\mathbf{b}) - \tfrac12(\mathbf{b}+\mathbf{a}'))\cdot (\tfrac12(\mathbf{a}+\mathbf{b}) - \tfrac12(\mathbf{b}+\mathbf{a}')) \\ &= \frac12(\mathbf{a}-\mathbf{a}')\cdot \frac12(\mathbf{a} - \mathbf{a}') \\ \text{midpoint }BA' \text{ to midpoint } A'B' &= (\tfrac12(\mathbf{b}+\mathbf{a}') - \tfrac12(\mathbf{a}'+\mathbf{b}')) \cdot (\tfrac12(\mathbf{b}+\mathbf{a}') - \tfrac12(\mathbf{a}'+\mathbf{b}'))\\ &= \frac12(\mathbf{b}-\mathbf{b}')\cdot \frac12(\mathbf{b} - \mathbf{b}') \\ &= \frac14 (|\mathbf{b}|^2 + |\mathbf{b}'|^2 - 2\mathbf{b}\cdot\mathbf{b}')\\ &= \frac14(|\mathbf{a}|^2 + |\mathbf{a}'|^2 - 2\mathbf{b}\cdot\mathbf{b}') \\ \text{midpoint }A'B' \text{ to midpoint } B'A &= (\tfrac12(\mathbf{a}'+\mathbf{b}') - \tfrac12(\mathbf{b}'+\mathbf{a})) \cdot (\tfrac12(\mathbf{a}'+\mathbf{b}') - \tfrac12(\mathbf{b}'+\mathbf{a}))\\ &= \frac12(\mathbf{a}'-\mathbf{a})\cdot \frac12(\mathbf{a}' - \mathbf{a}) \\ \text{midpoint }B'A \text{ to midpoint } AB &= (\tfrac12(\mathbf{b}'+\mathbf{a}) - \tfrac12(\mathbf{a}+\mathbf{b})) \cdot (\tfrac12(\mathbf{b}'+\mathbf{a}) - \tfrac12(\mathbf{a}+\mathbf{b}))\\ &= \frac12(\mathbf{b}'-\mathbf{b})\cdot \frac12(\mathbf{b}' - \mathbf{b}) \\ \end{align*} So it's sufficient to prove \(\mathbf{a}\cdot \mathbf{a}' = \mathbf{b}\cdot \mathbf{b}'\) but this is clear from looking at a diagram for 1 second. Given the length of the square is what it is, we want to minimise \(\mathbf{b}\cdot \mathbf{b}'\) which is when they are vertically opposite each other, ie \(\angle BOA' = 90^\circ\)

---

Solution: \(f'(x) = 6ax-18x^2\), therefore \(f\) has turning points at \(0\) and \(\frac{a}3\) (ie decreasing for \(x \leq 0\) and \(x \geq \frac{a}{3}\) and increasing otherwise). Therefore possible maxima are \(f(-\tfrac13), f(\frac{a}{3}), f(1)\) where we consider \(\frac{a}{3}\) if \(a \leq 3\) and \(1\) otherwise. \(f(-\frac13) = \frac{a}{3} + \frac{2}{9} = \frac{3a+2}{9}\) \(f(\frac{a}{3}) = \frac{a^3}{3} - \frac{2a^3}{9} = \frac{a^3}{9}\) \(f(1) = 3(a-2)\) Comparing \(\frac{a^3}{9}\) to \(\frac{3a+2}{9}\) we have a double root at \(a = -1\) and a single root at \(a = 2\), therefore \(\frac{a^3}9\) is larger if \(a \geq 2\) Comparing \(3(a-2)\) to \(\frac{3a+2}9\) we have a cross-over at \(a = \frac{7}3\). Therefore we have: \begin{align*} M(a) &= \begin{cases} \frac{3a+2}{9} & 0 \leq a \leq 2 \\ \frac{a^3}{9} & 2 \leq a \leq 3 \\ 3(a-2) & 3 \leq a \end{cases} \end{align*}

---

Solution:

1. \(\,\) \begin{align*} && S & = 1 +\quad 2\quad \;\;+ \quad 3 \quad+ \cdots + \quad n \\ && S &= n + (n-1) + (n-2) + \cdots + 1 \\ && 2S &= (n+1) + (n+1) + \cdots + (n+1) \\ \Rightarrow && S &= \frac12n(n+1) \end{align*}
2. \(\,\) \begin{align*} && (N-m)^{k} + m^k&= \sum_{i=0}^k \binom{k}{i} N^{k-i} (-m)^{i} + m^k \\ &&&= N\sum_{i=0}^{k-1} \binom{k}{i}N^{k-i-1}(-m)^i -m^k+m^k \\ &&&= N\sum_{i=0}^{k-1} \binom{k}{i}N^{k-i-1}(-m)^i \end{align*} which is clearly divisible by \(N\).

\begin{align*} 2S &= 2\sum_{i=1}^n i^k \\ &= \sum_{i=0}^n (\underbrace{(n-i)^k + i^k}_{\text{divisible by }n}) \\ \end{align*} Therefore \(2S\) is divisible by \(n\) and so if \(n\) is odd, \(n\) divides \(S\) and if \(n\) is even, \(\frac{n}{2}\) divides \(S\). Also notice \begin{align*} 2S &= 2\sum_{i=1}^n i^k \\ &= \sum_{i=1}^{n} (\underbrace{(n+1-i)^k + i^k}_{\text{divisible by }n+1}) \\ \end{align*} Therefore if \(n+1\) is odd, \(n+1 \mid S\) otherwise \(\frac{n+1}{2} \mid S\), and in either case \(\frac{n(n+1)}{2} \mid S\) (since they are both coprime) but this is the same as \(1 + 2 + \cdots + n \mid S\)