Problems

---

Solution: Suppose we have height \(h\) and radius \(r\), then: \(V = \pi r^2 h\) and \(S = 2\pi r^2 + 2\pi r h\). \(h = \frac{V}{\pi r^2}\), so \begin{align*} S &= 2 \pi r^2 + 2 \pi r\frac{V}{\pi r^2} \\ &= 2\pi r^2 +V \frac1{r}+V \frac1{r} \\ &\underbrace{ \geq }_{\text{AM-GM}} 3 \sqrt[3]{2\pi r^2 \frac{V^2}{r^2} } = 3 (2 \pi V^2)^{1/3} \end{align*} Equality holds when \(r = \sqrt[3]{\frac{V}{2 \pi}}, h = \frac{V}{\pi (V/2\pi)^{2/3}} = \sqrt[3]{\frac{4V}{\pi}}\) Since \(h > r\) the sphere has a maximum radius of \(r\) and so it's largest volume is \(\frac43 \pi r^3 = \frac43 \pi \frac{V}{2 \pi} = \frac23 V\).

The radius of the sphere is \(\sqrt{\left (\frac{r}{2} \right)^2 + \left (\frac{h}{2} \right)^2 } = \frac12 \sqrt{r^2+h^2}\) \begin{align*} V_{sphere} &= \frac43 \pi (r^2+h^2)^{3/2} \\ &= \frac43 \pi \left (\left( \frac{V}{2 \pi} \right)^{2/3}+\left( \frac{4V}{ \pi} \right)^{2/3} \right)^{3/2} \\ &= \frac43 \pi \frac{V}{ \pi} \left ( 2^{-2/3}+4^{2/3}\right)^{3/2} \\ &= \frac 43 V \left ( \frac{1+4}{2^{2/3}} \right)^{3/2} \\ &= \frac43 \frac{5^{3/2}}{2} V \\ &= \frac{2 \cdot \sqrt{125}}{3} V \end{align*}

---

Solution:

1. \begin{align*} u = 1+(\alpha-1)x: && \int_0^1 (1 + (\alpha - 1)x)^n \d x &= \int_{u=1}^{u=\alpha} u^n \frac{1}{\alpha - 1} \d u \\ &&&= \left [\frac{u^{n+1}}{(n+1)(\alpha-1)} \right]_1^\alpha \\ &&&= \frac{\alpha^{n+1}-1}{(n+1)(\alpha-1)} \end{align*}
2. \begin{align*} && \int_0^1 (\alpha x + (1-x))^n \d x &= \int_0^1 \sum_{k=0}^n \binom{n}{k} \alpha^k x^k (1-x)^{n-k} \d x \\ &&&= \sum_{k=0}^n \alpha^k \int_0^1 \binom{n}{k} x^k (1-x)^{n-k} \d x \end{align*} Therefore the coefficient of \(\alpha^k\) is \(\displaystyle \int_0^1 \binom{n}{k} x^k (1-x)^{n-k} \d x\)
3. The coefficient of \(\alpha^{k}\) in \(\displaystyle \frac{\alpha^{n+1}-1}{(n+1)(\alpha-1)}\) is \(\displaystyle \frac1{n+1}\). Therefore \begin{align*} && \frac1{n+1} &= \binom{n}{k} \int_0^1 x^k(1-x)^{n-k} \d x \\ \Rightarrow && \int_0^1 x^k (1-x)^{n-k} \d x &= \frac{k!(n-k)!}{(n+1)n!} \\ &&&= \frac{k!(n-k)!}{(n+1)!} \end{align*}

---

Solution:

1. \(n^5 -n^3 = n^3(n-1)(n+1)\). If \(n\) is even then \(8 \mid n^3\). if \(n\) is odd then both of \(n-1\) and \(n+1\) are divisible by \(2\) and one is divisible by \(4\), so regardless \(8\) divides our expression. We can write \(n = 3k, 3k+1, 3k+2\) and in all cases our expression is divisible by \(3\). \(n = 3k \Rightarrow 3 \mid n\), \(n = 3k+1 \Rightarrow 3 \mid n-1\), \(n = 3k+2 \Rightarrow 3 \mid n+1\). Therefore \(3\) and \(8\) both divide our expression, and they are coprime so their product (24) divides our expression.
2. \(2^{2n}-1 = (2^2-1) \cdot (1+2^2+\cdots + 2^{2n-2}) = 3 \cdot N\) therefore \(3\) divides our number.
3. Suppose \(n-1 = 3k\) then \(n^3-1 = (3k+1)^3-1 = 27k^3 + 27k^2 + 9k\) which is clearly divisible by 9

---

Solution: First suppose \(|a| < 1\), then \begin{align*} && I &= \int_0^1 \frac{1}{x^2+2ax+1} \d x \\ &&&= \int_0^1 \frac{1}{(x+a)^2 +1-a^2} \d x \\ &&&= \int_0^1 \frac{1}{(x+a)^2 +(\sqrt{1-a^2})^2} \d x \tag{\(1-a^2 > 0\)}\\ &&&= \left [\frac{1}{\sqrt{1-a^2}} \tan^{-1} \frac{x+a}{\sqrt{1-a^2}} \right]_0^1 \\ &&&= \frac{1}{\sqrt{1-a^2}} \left ( \tan^{-1} \frac{a+1}{\sqrt{1-a^2}} - \tan^{-1} \frac{a}{\sqrt{1-a^2}} \right) \\ &&&= \frac{1}{\sqrt{1-a^2}} \tan^{-1} \left ( \frac{\frac{a+1}{\sqrt{1-a^2}}-\frac{a}{\sqrt{1-a^2}}}{1+\frac{(a+1)a}{1-a^2}} \right) \\ &&&= \frac{1}{\sqrt{1-a^2}} \tan^{-1} \left ( \frac{\frac{1}{\sqrt{1-a^2}}}{\frac{1+a}{1-a^2}} \right) \\ &&&= \frac{1}{\sqrt{1-a^2}} \tan^{-1} \left ( \frac{1-a}{\sqrt{1-a^2}}\right) \\ &&&= \frac{1}{\sqrt{1-a^2}} \tan^{-1} \sqrt { \frac{1-a}{1+a}} \\ \end{align*} Second, suppose \(|a| > 1\), then \begin{align*} && I &= \int_0^1 \frac{1}{x^2+2ax+1} \d x \\ &&&= \int_0^1 \frac{1}{(x+a)^2-(a^2-1)} \d x \\ &&&= \int_0^1 \frac{1}{(x+a-\sqrt{a^2-1})(x+a+\sqrt{a^2-1})} \d x \tag{\(a^2-1 > 0\)} \\ &&&= \frac{1}{2\sqrt{a^2-1}}\int_0^1 \left ( \frac{1}{x+a-\sqrt{a^2-1}} - \frac{1}{x+a+\sqrt{a^2-1}} \right) \d x \\ &&&= \frac{1}{2\sqrt{a^2-1}} \left [ \ln |x+a-\sqrt{a^2-1}|- \ln |x+a+\sqrt{a^2-1}| \right]_0^1 \\ &&&= \frac{1}{2\sqrt{a^2-1}} \left ( \ln |1+a-\sqrt{a^2-1}| - \ln|1+a+\sqrt{a^2-1}| - \ln|a-\sqrt{a^2-1}| +\ln|a + \sqrt{a^2-1}| \right) \\ &&&= \frac{1}{2\sqrt{a^2-1}} \ln | \frac{(1+a-\sqrt{a^2-1})(a+\sqrt{a^2-1})}{(1+a+\sqrt{a^2-1})(a-\sqrt{a^2-1})}|\\ &&&= \frac{1}{2\sqrt{a^2-1}} \ln |\frac{a+a^2-(a^2-1) +\sqrt{a^2-1}}{1+a-\sqrt{a^2-1}}| \\ &&&= \frac{1}{2\sqrt{a^2-1}} \ln |\frac{(1+a +\sqrt{a^2-1})^2}{(1+a)^2-(a^2-1)}| \\ &&&= \frac{1}{2\sqrt{a^2-1}} \ln |\frac{1+2a+a^2+a^2-1+2(1+a)\sqrt{a^2-1}}{2+2a}| \\ &&&= \frac{1}{2\sqrt{a^2-1}} \ln |a+\sqrt{a^2-1}| \\ \end{align*}

---

Solution:

1. Suppose \begin{align*} && 4 - 2\sqrt{3} &= (r+s\sqrt{3})^2 \\ &&&= r^2+3s^2+2sr \sqrt{3} \\ \Rightarrow && rs &= -1 \\ && r^2+3s^2 &= 4 \\ \Rightarrow && (r,s) &= (1,-1), (-1,1) \end{align*}
2. \begin{align*} && (3-2\sqrt{3})+2(1-\sqrt{3})i &= (p+qi)^2 \\ &&&= p^2-q^2 + 2pq i \\ \Rightarrow && pq &= (1-\sqrt{3}) \\ && p^2 - q^2 &= 3-2\sqrt{3} \\ \Rightarrow &&3-2\sqrt{3} &= p^2 - \frac{(1-\sqrt{3})^2}{p^2} \\ \Rightarrow && 0 &= p^4-(3-2\sqrt{3})p^2-(4-2\sqrt{3}) \\ &&&= (p^2-(4-2\sqrt{3}))(p^2+1) \\ \Rightarrow && p &= \pm (1-\sqrt{3}) \\ && q &=\mp \frac12(1+\sqrt{3}) \end{align*}
3. \begin{align*} && 0 &= (1+i)z^2 - 2z + 2(\sqrt{3}-1) \\ \Rightarrow && z &= \frac{2 \pm \sqrt{4-4(1+i)2(\sqrt{3}-1)}}{2(1+i)} \\ &&&= \frac{1 \pm \sqrt{1-(1+i)2(\sqrt{3}-1)}}{1+i} \\ &&&= \frac{1 \pm \sqrt{(3-2\sqrt{3})+(2-2\sqrt{3})i}}{1+i} \\ &&&= \frac{1 \pm (1 - \sqrt{3}) \mp \frac12 (1+\sqrt{3})i}{1+i} \\ &&&= \frac{5-\sqrt{3}}{4} + \frac{3-3\sqrt{3}}{4}i, \\ &&& \frac{\sqrt{3}-1}{4} + \frac{1+3\sqrt{3}}{4}i \end{align*}

---

Solution:

1. \(\,\) \begin{align*} && 2\sin \tfrac12 x \sum_{k=1}^n \cos kx &= \sum_{k=1}^n 2\sin \tfrac12 x \cos kx \\ &&&= \sum_{k=1}^n \left ( \sin(k+\tfrac12)x - \sin(k - \tfrac12)x \right) \\ &&&= \left ( \sin\tfrac32x - \sin\tfrac12x \right) + \\ &&&\quad \quad \left ( \sin\tfrac52x - \sin \tfrac32 x \right) + \\ &&&\quad \quad \quad +\cdots + \\ &&&\quad \quad \quad \quad +\left ( \sin(n+\tfrac12)x - \sin(n - \tfrac12)x \right) \\ &&&= \sin(n+\tfrac12)x - \sin \tfrac12 x \\ \Rightarrow && \sin(n+\tfrac12)x &= \sin \tfrac12 x + 2\sin \tfrac12 x \sum_{k=1}^n \cos kx \\ \Rightarrow && f(x) &= 1 + 2 \sum_{k=1}^n \cos kx \end{align*}
2. \(\,\) \begin{align*} && \int_0^{\pi} f(x) \d x &= \int_0^{\pi} \left (1 + 2 \sum_{k=1}^n \cos kx \right) \d x \\ &&&= \pi + 2 \left [ \sum_{k=1}^n \frac{1}{k} \sin k x\right]_0^\pi \\ &&&= \pi \\ \\ && \int_0^{\pi} f(x) \cos x \d x &= \int_0^{\pi} \left (\cos x + 2 \sum_{k=1}^n \cos kx \cos x \right) \d x \\ &&&= 0 + \sum_{k=1}^n \left ( \int_0^{\pi} 2 \cos k x \cos x \d x \right) \\ &&&= \sum_{k=1}^n \left ( \int_0^{\pi} (\cos (k+1)x - \cos (k-1) x)\d x\right) \\ &&&= -\pi \end{align*}

---

Solution:

1. Since water flows out a rate proportional to the water in the tank we must have \(\dot{y} = -ky\), ie \(y = Ae^{-k t}\). Since \(t = 0, y = 1\) we have \(y = e^{-kt} = (e^{-k})^t\), so call \(b = e^{-k}\) and we have the result. (Since \(k > 0 \Rightarrow b < 1\)
2. Notice that \begin{align*} && \dot{y} &= -\underbrace{ky}_{\text{flow out}} + \underbrace{a}_{\text{flow in}} \\ &&&= y\ln b + a \\ \Rightarrow && \dot{y} - y \ln b &= a \\ \\ \text{CF}: && y &= Ae^{\ln b t} = Ab^t\\ \text{PI}: && y &= -\frac{a}{\ln b} \\ t = 0, y = 1: && 1 &= A-\frac{a}{\ln b} \\ \Rightarrow && y &= \frac{a}{\ln b} \left ( b^t - 1 \right)+b^t \end{align*}

---

Solution:

1. \(\,\) \begin{align*} && 0.383838\ldots &= \frac{3}{10} + \frac{8}{100} + \frac{3}{1000} + \cdots \\ &&&= \frac{38}{100} + \frac{38}{10000} + \cdots \\ &&&= \frac{38}{100} \left (1 + \frac{1}{100} + \frac{1}{100^2} + \cdots \right) \\ &&&= \frac{\frac{38}{100}}{1 - \frac{1}{100}} \\ &&&= \frac{38}{99} \end{align*}
2. Let \(x = 0\cdot a_{1}a_{2}a_{2}\cdots\) such that there exists \(N\), \(k\) such that \(a_{n+k} = a_n\) for \(n > N\). First notice that \begin{align*} x &= \frac{a_1a_2\cdots a_N}{10\ldots000} + \frac{a_{N+1}}{10^{N+1}} + \cdots \\ &= \frac{b}{10^N} + \frac{a_{N+1}a_{N+2}\cdots a_{N+k}}{10^{N+k}} + \frac{a_{N+1}a_{N+2}\cdots a_{N+2k}}{10^{N+k}} + \cdots \\ &= \frac{b}{10^N} + \frac{c}{10^{N+k}} \left (1 + \frac{1}{10^k} + \cdots \right) \\ &= \frac{b}{10^N} + \frac{c}{10^{N+k}} \frac{1}{1- \frac{1}{10^k}} \\ &= \frac{b}{10^N} + \frac{c}{10^{N+k}} \frac{10^k}{10^k-1} \\ &= \frac{b}{10^N} + \frac{c}{10^N(10^k-1)} \end{align*} where \(b = a_1a_2\cdots a_N\) and \(c = a_{N+1}a_{N+2}\cdots a_{N+k}\). Clearly as the sum of two rational numbers, \(x\) is rational.

---

Solution: \begin{align*} && \text{Energy at the top} &= mgh \\ && \text{Energy at the bottom} &= \frac12\frac{\lambda (h-l)^2}{l} \\ \Rightarrow && mgh & = \frac{\frac{mg}{k}(h-l)^2}{2l} \\ \Rightarrow && 2hkl &= (h-l)^2 \\ \Rightarrow && 0 &= h^2-2lh-2hlk+l^2 \\ &&0&= h^2-2hl(k+1)+l^2 \\ \Rightarrow && \frac{h}{l} &= \frac{2(k+1)\pm \sqrt{4(k+1)^2-4}}{2} \\ &&&= (k+1) \pm \sqrt{k^2+2k} \\ \Rightarrow && h &= l \left ( (k+1) \pm \sqrt{k^2+2k} \right) \end{align*} Since the negative root is less than \(1\), she would have not fully extended the cord. Therefore \(h = l \left ( (k+1) + \sqrt{k^2+2k} \right)\) Her maximum speed will be when her acceleration is \(0\), ie \(g = \text{force from cord}\) ie \(mg = \frac{\lambda x}{l}\) or \(x = \frac{mgl}{\lambda} = \frac{mglk}{mg} = kl\). At this point by conservation of energy we will have \begin{align*} && mgh &= mg(h-l-x) + \frac12 m v^2+\frac{1}{2} \frac{mgx^2}{kl} \\ \Rightarrow && mg\left ( l + kl \right) &= \frac12 m v^2 + \frac12 \frac{mgl^2k^2}{kl} \\ \Rightarrow && 2g\left ( l + kl \right) &= v^2 + glk \\ \Rightarrow && v^2 &= gl(2+k) \end{align*}

---

Solution: If the highest mark is \(k\), then there are \(n-1\) remaining marks to give, and they have to be chosen from the numbers \(1, 2, \ldots, k-1\), ie in \(\binom{k-1}{n-1}\) ways. There are \(n\) numbers to be chosen from \(1, 2, \ldots, m\) in total, therefore \(\displaystyle \mathbb{P}(K=k) = \left.\binom{k-1}{n-1} \right/ \binom{m}{n}\) Since \(K\) can take any of the values \(n, \cdots, m\), we must have \begin{align*} && 1 &= \sum_{k=n}^m \mathbb{P}(K=k) \\ &&&= \sum_{k=n}^m \left.\binom{k-1}{n-1} \right/ \binom{m}{n} \\ \Rightarrow && \binom{m}{n} &= \sum_{k=n}^m \binom{k-1}{n-1} \\ \\ && \mathbb{E}(K) &= \sum_{k=n}^m k \cdot \mathbb{P}(K=k) \\ &&&= \sum_{k=n}^m k \cdot \left.\binom{k-1}{n-1} \right/ \binom{m}{n} \\ &&&= n\binom{m}{n}^{-1} \sum_{k=n}^m \frac{k}{n} \cdot \binom{k-1}{n-1} \\ &&&= n\binom{m}{n}^{-1} \sum_{k=n}^m \binom{k}{n} \\ &&&= n\binom{m}{n}^{-1} \sum_{k=n+1}^{m+1} \binom{k-1}{n+1-1} \\ &&&= n\binom{m}{n}^{-1} \binom{m+1}{n+1} \\ &&&= n \cdot \frac{m+1}{n+1} \end{align*}

---

Solution: \begin{align*} && \mathbb{P}(N = k) &= \mathbb{P}(\text{guesses }k-1\text{ correctly then 1 wrong})\\ &&&= \frac12 \cdot \frac{1}{3} \cdots \frac{1}{k-1} \frac{k-1}{k} \\ &&&= \frac{k-1}{k!} \\ &&\mathbb{E}(N) &= \sum_{k=2}^\infty k \cdot \mathbb{P}(N=k) \\ &&&= \sum_{k=2}^{\infty} \frac{k(k-1)}{k!} \\ &&&= \sum_{k=0}^{\infty} \frac{1}{k!} = e \\ && \textrm{Var}(N) &= \mathbb{E}(N^2) - \mathbb{E}(N)^2 \\ && \mathbb{E}(N^2) &= \sum_{k=2}^{\infty} k^2 \mathbb{P}(N=k) \\ &&&= \sum_{k=2}^{\infty} \frac{k^2(k-1)}{k!} \\ &&&= \sum_{k=0}^{\infty} \frac{k+2}{k!} \\ &&&= \sum_{k=0}^{\infty} \frac{1}{k!} + 2 \sum_{k=0}^{\infty} \frac{1}{k!} = 3e \\ \Rightarrow && \textrm{Var}(N) &= 3e-e^2 \end{align*}

---

Solution: \begin{align*} && \P(A|H) &= \P(\text{achieve }r\text{ heads immediately}) + \P(\text{don't and then achieve it from having flipped a tail}) \\ &&&= p^{r-1} + (1-p^{r-1}) \cdot \P(A|T) \\ && \P(A|T) &= (1-q^{s-1})\P(A|H) \\ \\ &&\P(A|H) &= p^{r-1}+(1-p^{r-1})(1-q^{s-1})\P(A|H) \\ \Rightarrow && \P(A|H) &= \frac{p^{r-1}}{1-(1-p^{r-1})(1-q^{s-1})} \\ && \P(A|T) &= \frac{(1-q^{s-1})p^{r-1}}{1-(1-p^{r-1})(1-q^{s-1})} \\ && \P(A) &= \frac{(2-q^{s-1})p^{r-1}}{2(1-(1-p^{r-1})(1-q^{s-1}))} \end{align*}