Problems

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Solution:

1. It is at the origin when both \(\sin 2t\) and \(\cos t = 0\), but this \(\sin 2t = 2 \sin t \cos t\) so this happens precisely when \(\cos t = 0\), ie when \(t = \frac{\pi}{2}, \frac{3\pi}{2}\)
2. \(\,\) \begin{align*} && \dot{\mathbf{r}} &= 2 \cos 2t \mathbf{i} - 2 \sin t \mathbf{j} \\ && \mathbf{r} \cdot \dot{\mathbf{r}} &= 2\cos 2t \sin 2t - 2 \sin t 2 \cos t \\ &&&= \sin 2t \left (2\cos 2t - 2 \right) \end{align*} Therefore they are perpendicular when \(\sin 2t = 0 \Rightarrow t = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi\) and when \(\cos 2t = 1 \Rightarrow 2t = 0, 2\pi, 4\pi \Rightarrow t = 0, \pi, 2\pi\), therefore all solutions are \( t = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi\)
3. For \(\mathbf{r}\) and \(\dot{\mathbf{r}}\) to be parallel, we would need \begin{align*} && \frac{2 \cos 2t}{\sin 2t} &= \frac{-2 \sin t}{2 \cos t}\\ && 2 \cos 2t \cos t &= - \sin t \sin 2t \\ && 0 &= 2\cos t (\cos 2t + \sin ^2 t) \\ &&&= 2 \cos t (\cos^2 t) \\ &&&= 2 \cos^3 t \end{align*} Therefore the only time we can be parallel is when \(\cos t = 0\), which is when we are at the origin.
4. \(\frac{\d }{\d t} (\mathbf{r} \cdot \mathbf{r}) = 2 \mathbf{r} \cdot \mathbf{\dot{r}}\) so we should check the values when velocity and displacement are perpendicular, ie \( t = 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi\) which have values \(\mathbf{r} = \binom{0}{2}, \binom{0}{0}, \binom{0}{-2}, \binom{0}{0}, \binom{0}{2}\). Therefore the maximum distance is \(2\).
   

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Solution:

1. \(\,\)
   

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   The overlap can be at most 0.4, which would mean \(p =0.7-0.4 = 0.3\) It must be at least 0.1, which would mean \(p =0.7-0.1 = 0.6\) so \(0.3 \leq p \leq 0.6\)
2. 

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   Notice that: \begin{align*} && 1 &= 0.05 + 0.7 -(hc+hr+0.1) + \\ &&&\quad\quad 0.8 - (hc+cr + 0.1) + \\ &&&\quad \quad \quad 0.4 - (hr+cr+0.1) +\\ &&&\quad \quad \quad \quad hc+hr+cr+0.1 \\ && &= 0.05 +0.7+0.8+0.4 - (hc+hr+cr)-2\cdot 0.1 \\ \Rightarrow && hc+hr+cr &=0.05 +0.7 + 0.8 + 0.4 - 0.2-1 \\ \Rightarrow && \mathbb{P}(\text{exactly 2}) &= 0.75 \end{align*} Notice \(q = \frac{hc}{0.8}\) Notice that we must have: \(hc, hr cr \geq 0\) as well as \(hc+hr+cr = 0.75\) \begin{align*} && \mathbb{P}(\text{only hat}) &= 0.7 -(hc+hr+0.1) \geq 0 \\ \Rightarrow && hc+hr & \leq 0.6 \\ && \mathbb{P}(\text{only cloak}) &= 0.8 - (hc+cr + 0.1)\geq 0 \\ \Rightarrow &&hc+cr & \leq 0.7 \\ && \mathbb{P}(\text{only ring}) &= 0.4 - (hr+cr+0.1) \geq 0 \\ \Rightarrow && hc+hr & \leq 0.3 \\ \end{align*} To find the minimum for \(hc\) we want to maximise \(hr+cr = 0.3\), so \(hc = 0.75 - 0.3 = 0.45\). To find the maximum for \(hc\) we want to minimise \(hr\) and \(cr\) \(cr \leq 0.7 - hc\) and \(hr \leq 0.6 - hc\) so \(0.75 \leq hc + (0.6 - hc) + (0.7 - hc) = 1.3-hc\) so \(hc \leq 1.3 - 0.75 = 0.55\) Therefore the range for \(q\) is \(\frac{.45}{.8}\) to \(\frac{.55}{.8}\) or \(\frac9{16} \leq q \leq \frac{11}{16}\)

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Solution:

1. \(\,\) \begin{align*} && \mathbb{P}\left (\mu - \tfrac12 \sigma \leq X \right) &= \mathbb{P}\left (X \leq \mu + \tfrac12 \sigma \right) \tag{by symmetry} \\ &&&= b \\ \Rightarrow && \mathbb{P} \left (\mu - \tfrac12 \sigma \leq X \leq \mu + \sigma \right) &= a - (1-b) = a+b - 1\\ \\ && \mathbb{P} \left ( X \le \mu +\tfrac12 \sigma \vert X \ge \mu - \tfrac12 \sigma \right ) &= \frac{ \mathbb{P} \left (\mu - \tfrac12 \sigma \leq X \leq \mu + \tfrac12 \sigma \right)}{\mathbb{P} \left ( X \ge \mu - \tfrac12 \sigma \right )} \\ &&&= \frac{b-(1-b)}{1-(1-b)} \\ &&&= \frac{2b-1}{b} \end{align*}
2. 1. Let \(Y\) be the volume of milk in the carton I bring home, we are interested in: \begin{align*} && \mathbb{P}(Y \geq 500 | Y \leq 505) &= \frac{\mathbb{P}(500 \leq Y \leq 505)}{\mathbb{P}(Y \leq 505)} \\ &&&=\frac{\mathbb{P}(500 \leq Y \leq 505|\text{skimmed})\mathbb{P}(\text{skimmed})+\mathbb{P}(500 \leq Y \leq 505|\text{full fat})\mathbb{P}(\text{full fat})}{\mathbb{P}(Y \leq 505|\text{skimmed})\mathbb{P}(\text{skimmed})+\mathbb{P}(Y \leq 505|\text{full fat})\mathbb{P}(\text{full fat})} \\ &&&= \frac{\frac35 \cdot \mathbb{P}(\mu \leq X \leq \mu + \tfrac12 \sigma) + \frac25 \cdot \mathbb{P}(\mu+\tfrac12 \sigma \leq X \leq \mu +\sigma)}{\frac35 \cdot \mathbb{P}(X \leq \mu + \tfrac12 \sigma) + \frac25 \cdot \mathbb{P}(X \leq \mu +\sigma)} \\ &&&= \frac{\frac35 \cdot(b-\tfrac12) + \frac25 \cdot (a-b)}{\frac35 \cdot b + \frac25 \cdot a} \\ &&&= \frac{b+2a-\frac32}{3b+2a} \\ &&&= \frac{4a+2b-3}{4a+6b} \end{align*}
   2. \(70\%\) of cartons have contained at most 505 ml, so: \begin{align*} && \tfrac7{10} &= \mathbb{P}(Y \leq 505) \\ &&&= \mathbb{P}(Y \leq 505 | \text{ skimmed}) \mathbb{P}(\text{skimmed}) + \mathbb{P}(Y \leq 505 | \text{ full fat}) \mathbb{P}(\text{full fat}) \\ &&&= \mathbb{P}(X \leq \mu + \tfrac12 \sigma) \cdot \tfrac35 + \mathbb{P}(X\leq \mu + \sigma ) \cdot \tfrac25 \\ \Rightarrow && 7 &= 6b+ 4a \end{align*} \(\tfrac13\) of cartons containing 495 ml contained full fat milk: \begin{align*} && \tfrac13 &= \mathbb{P}(\text{full fat} | Y \geq 495) \\ &&&= \frac{\mathbb{P}(\text{full fat and} Y \geq 495) }{\mathbb{P}(Y \geq 495)} \\ &&&= \frac{\mathbb{P}(X \geq \mu)\frac25}{\mathbb{P}(X \geq \mu)\cdot \frac25+\mathbb{P}(X \geq \mu-\tfrac12 \sigma)\cdot \frac35} \\ &&&= \frac{\frac15}{\frac12 \cdot \frac25 + b\frac35}\\ &&&= \frac{1}{1+ 3b }\\ \Rightarrow && 3b+1 &= 3 \\ \Rightarrow && b &= \frac23 \\ && a &= \frac34 \end{align*}

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Solution: \begin{align*} && y &= x^2e^{-x^2} \\ \Rightarrow && y' &= 2xe^{-x^2} +x^2 \cdot (-2x)e^{-x^2} \\ &&&= e^{-x^2}(2x-2x^3) \\ &&&= 2e^{-x^2}x(1-x^2) \end{align*} Therefore the derivative is zero iff \(x = 0, \pm 1\) \begin{align*} && y &= \P(x) e^{-x^2} \\ \Rightarrow && y' &= e^{-x^2} (\P'(x)-2x\P(x)) \end{align*} Therefore we want \(\P'(x) - 2x\P(x) = Kx(x^2-a^2)(x^2-b^2)\) Since this has degree \(5\), we should look at polynomials degree \(4\) for \(\P\). We can also immediately see that \(0\) is a root of \(\P'(x)\), so \(\P(x) = a_4x^4+a_3x^3+a_2x^2+a_0\). WLOG \(a_4 = 1\) and \(K = -2\), so \begin{align*} && -2(x^5-(a^2+b^2)x^3+a^2b^2x) &= 4x^3+3a_3x^2+2a_2x- 2x(x^4+a_3x^3+a_2x^2+a_0) \\ &&&= -2x^5-2a_3 x^4+(4-2a_2)x^3+(2a_2-2a_0)x \\ \Rightarrow && a_3 &= 0 \\ && a^2+b^2 &= 2-a_2 \\ \Rightarrow && a_2 &= 2-a^2-b^2 \\ && a^2b^2 &= a_0-a_2 \\ \Rightarrow && a_0 &= a^2b^2 + 2-a^2-b^2 \\ \Rightarrow && \P(x) &= x^4+(2-a^2-b^2)x^2+(a^2-1)(b^2-1)x \end{align*}

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Solution:

1. \(f(12) = f(2^2 \cdot 3) = 12 \cdot (1-\frac12)\cdot(1-\frac13) = 12 \cdot \frac12 \cdot \frac 23 = 4\) \(f(180) = f(2^2 \cdot 3^2 \cdot 5) = 180 \cdot \frac12 \cdot \frac23 \cdot \frac 45 = 12 \cdot 4 = 48\) Suppose \(N\) has prime decomposition \(p_1^{a_1} \cdots p_k^{a_k}\), then \(f(N) = p_1^{a_1} \cdots p_k^{a_k} (1 - \frac{1}{p_1})\cdots ( 1- \frac{1}{p_k}) = p_1^{a_1-1} \cdots p_k^{a_k-1}(p_1-1) \cdots (p_k-1)\) which is clearly an integer.
2. \(f(2) = 1, f(4) = 2 \neq f(2) \cdot f(2)\) If \(p, q\) are distinct primes then \(f(p) = p \cdot \frac{p-1}{p} = p-1\) and \(f(q) = q-1\). \(f(pq) = pq \frac{p-1}{p} \cdot \frac{q-1}{q} = (p-1)(q-1) = f(p)f(q)\) \(f(12) = 4 = 2\cdot 2 = f(4) \cdot f(3)\)
3. \(f(p^m) = p^{m-1} (p-1)\) \(146410 = 10 \cdot 14641 = 10 \cdot 11^4\). Therefore \(p = 11, m = 5\)

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Solution:

Since \(y' = \cos x - \cos x + x \sin x = x \sin x > 0\) which is positive on \((0, \frac{\pi}{2})\), so \(y\) is increasing, and therefore will achieve it's highest value at \(\frac{\pi}{2}\) which is \(y(\frac{\pi}{2}) = 1\) and it's smallest value at \(y(0) = 0\). Therefore \(0 \leq y \leq 1\)

1. \(\,\) \begin{align*} \int_0^{\frac{1}{2}\pi}\,y\;\d x &= \int_0^{\frac{1}{2}\pi} (\sin x - x \cos x) \d x \\ &= \left [-\cos x \right]_0^{\frac{1}{2}\pi} +\left [ -x \sin x \right]_0^{\frac{1}{2}\pi} + \int_0^{\frac{1}{2}\pi} \sin x \d x \\ &= 1-\frac{\pi}{2} + 1 = 2 - \frac{\pi}{2} \end{align*}
2. \(\,\) \begin{align*} \int_0^{\frac{1}{2}\pi}y^2\d x &= \int_0^{\frac{1}{2}\pi} (\sin x - x \cos x)^2 \d x \\ &= \int_0^{\frac{1}{2}\pi} (\sin^2x - 2x\sin x \cos x+x^2\cos^2 x) \d x\\ &= \int_0^{\frac{1}{2}\pi} (\sin^2x -x \sin 2x+\tfrac12x^2(\cos 2 x + 1)) \d x\\ &= \frac{\pi}{4} + \frac{\pi^3}{48} + \int_0^{\frac{1}{2}\pi} (-x \sin 2x+\tfrac12x^2\cos 2 x) \d x \\ &= \frac{\pi}{4} + \frac{\pi^3}{48} + \left [\frac12 x \cos 2x +\frac14 x^2 \sin2x\right]_0^{\frac{1}{2}\pi}-\int_0^{\frac{1}{2}\pi}(\tfrac12 \cos 2x +\tfrac12 x \sin 2x) \d x\\ &= \frac{\pi}{4} + \frac{\pi^3}{48} - \frac{\pi}{4} - \left [ \frac14 \sin 2x \right]_0^{\frac{1}{2}\pi} - \int_0^{\frac{1}{2}\pi} \tfrac12 x \sin 2x \d x\\ &= \frac{\pi^3}{48} - \left( \left[ -\frac14 x \cos 2x \right]_0^{\frac{1}{2}\pi} - \int_0^{\frac{1}{2}\pi} -\frac14 \cos 2x \d x \right)\\ &= \frac{\pi^3}{48} - \left( \frac{\pi}{8} + \left[ \frac18 \sin 2x \right]_0^{\frac{1}{2}\pi} \right)\\ &= \frac{\pi^3}{48} - \frac{\pi}{8} \end{align*}

Since \(y^2 < y\) on this interval, we must have \( \frac{\pi^3}{48} - \frac{\pi}{8} < 2 - \frac{\pi}{2} \Rightarrow \pi^3 +18\pi < 96\) as required.