Problems

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Solution:

Clearly the centre of mass will lie on the perpendicular from \(A\). We can also consider each side's wire as equivalent to a point mass at the centre of the side with mass proportional to the length of the side. Recalling that \(b = c\) (the triangle is isoceles we must have (for the \(y\)-coordinate \begin{align*} && a \cdot 0 + b \cdot \frac12 b \cos \theta + c \cdot \frac12 c \cos \theta &= (a+b+c) \overline{y} \\ \Rightarrow && b^2 \cos \theta &= (2b + 2b\sin \theta) \overline{y} \\ \Rightarrow && \overline{y} &= \frac{b \cos \theta}{2(1+\sin \theta)} \end{align*} \begin{align*} \text{N2}(\nearrow): && R - mg \cos \phi &= 0 \\ \text{N2}(\nwarrow): && F -mg \sin \phi &= 0 \\ \Rightarrow && F &\leq \mu R \\ \Rightarrow && \sin \phi &\leq \mu \cos \phi \\ \Rightarrow && \tan \phi &\leq \mu \end{align*} When the peg is at \(C\) \begin{align*} \tan \phi &= \frac{CM}{MG} \\ &= \frac{b\sin \theta}{\frac{b \cos \theta}{2(1+\sin \theta)}} \\ &= 2 \tan \theta(1+\sin \theta) \end{align*} Therefore \(2 \tan \theta(1+\sin \theta) \leq \mu\) as required.

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Solution: \begin{align*} && \mathbb{P}(A \cup B \cup C) &= \mathbb{P}(A \cup B) + \mathbb{P}(C) - \mathbb{P}((A \cup B) \cap C) \tag{applying with \(A\cup B\) and \(C\)} \\ &&&= \mathbb{P}(A \cup B) + \mathbb{P}(C) - \mathbb{P}((A \cap C) \cup (B \cap C)) \\ &&&= \mathbb{P}(A)+\mathbb{P}(B) - \mathbb{P}(A\cap B) + \mathbb{P}(C) - \mathbb{P}((A \cap C) \cup (B \cap C)) \tag{applying with \(A\) and \(B\)}\\ &&&= \mathbb{P}(A)+\mathbb{P}(B) - \mathbb{P}(A\cap B) + \mathbb{P}(C) - \left ( \mathbb{P}(A \cap C) +\mathbb{P}(B \cap C) - \mathbb{P}( (A \cap C) \cap (B \cap C) )\right) \\ &&&= \mathbb{P}(A)+\mathbb{P}(B) +\mathbb{P}(C)- \mathbb{P}(A\cap B)- \mathbb{P}(A \cap C) -\mathbb{P}(B \cap C)+\mathbb{P}( A \cap B \cap C) \end{align*} \[ \mathbb{P}(A_1 \cup A_2 \cup A_3 \cup A_4) = \sum_i \mathbb{P}(A_i) - \sum_{i \neq j} \mathbb{P}(A_i \cap A_j) + \sum_{i \neq j \neq j} \mathbb{P}(A_i \cap A_j \cap A_k) - \mathbb{P}(A_1 \cap A_2 \cap A_3 \cap A_4) \]

1. \(\mathbb{P}(E_i) = \frac{1}{n}\)
2. \(\mathbb{P}(E_i \cap E_j) = \frac{1}{n} \cdot \frac{1}{n-1} = \frac{1}{n(n-1)}\)
3. \(\mathbb{P})(E_i \cap E_j \cap E_k) = \frac{1}{n(n-1)(n-2)}\)

First notice that the probability that \(k\) (or more) cards are in the correct place is \(\frac{(n-k)!}{n!}\) (place the other \(n-k\) cards in any order. We are interested in: \begin{align*} \mathbb{P} \left ( \bigcup_{i=1}^n E_i \right) &= \sum_{i} \mathbb{P}(E_i) - \sum_{i \neq j} \mathbb{P}(E_i \cap E_j) + \sum_{i \neq j \neq k} \mathbb{P}(E_i \cap E_j \cap E_k) - \cdots \\ &= \sum_i \frac1n - \sum_{i\neq j} \frac{1}{n(n-1)} + \sum_{i \neq j \neq k} \frac{1}{n(n-1)(n-2)} - \cdots + (-1)^{k+1} \sum_{i_1 \neq i_2 \neq \cdots \neq i_k} \frac{(n-k)!}{n!} + \cdots\\ &= 1 - \binom{n}{2} \frac{1}{n(n-1)} + \binom{n}{3} \frac{1}{n(n-1)(n-2)} - \cdots + (-1)^{k+1} \binom{n}{k} \frac{(n-k)}{n!} + \cdots \\ &= 1 - \frac12 + \frac1{3!} - \cdots + (-1)^{k+1} \frac{n!}{k!(n-k)!} \frac{(n-k)!}{n!} + \cdots \\ &= 1 - \frac1{2!} + \frac1{3!} - \cdots + (-1)^{k+1} \frac{1}{k!} + \cdots + (-1)^{n+1} \frac{1}{n!} \end{align*} The probability exactly one card is in the right place is the probability none of the other \(n-1\) are in the right place, which is: \(\frac1n \left (1 - \left (1 - \frac1{2!} + \frac1{3!} - \cdots + (-1)^{k+1} \frac{1}{k!} + \cdots + (-1)^{n} \frac{1}{(n-1)!} \right) \right)\) but there are also \(n\) cards we can choose to be the card in the right place, hence \(\frac{1}{2!} - \frac{1}{3!} + \cdots +(-1)^n \frac{1}{(n-1)!}\)

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Solution:

1. \(X \sim B(16, \tfrac12)\), then \(X \approx N(8, 2^2)\), in particular \begin{align*} && \mathbb{P}(X = 8) &\approx \mathbb{P} \left ( 8 - \frac12 \leq 2Z + 8 \leq 8 + \frac12 \right) \\ &&&= \mathbb{P} \left (-\frac14 \leq Z \leq \frac14 \right) \\ &&&= \int_{-\frac14}^{\frac14} \frac{1}{\sqrt{2 \pi}}e^{-\frac12 x^2} \d x \\ &&&\approx \frac{1}{\sqrt{2\pi}} \int_{-\frac14}^{\frac14} 1\d x\\ &&&= \frac{1}{2 \sqrt{2\pi}} \end{align*}
2. Suppose \(X \sim B(2n, \frac12)\) then \(X \approx N(n, \frac{n}{2})\), and \begin{align*} && \mathbb{P}(X = n) &\approx \mathbb{P} \left ( n - \frac12 \leq \sqrt{\frac{n}{2}} Z + n \leq n + \frac12 \right) \\ &&&= \mathbb{P} \left ( - \frac1{\sqrt{2n}} \leq Z \leq \frac1{\sqrt{2n}}\right) \\ &&&= \int_{-\frac1{\sqrt{2n}}}^{\frac1{\sqrt{2n}}} \frac{1}{\sqrt{2 \pi}} e^{-\frac12 x^2} \d x \\ &&&\approx \frac{1}{\sqrt{n\pi}}\\ \Rightarrow && \binom{2n}{n}\frac1{2^n} \frac{1}{2^n} & \approx \frac{1}{\sqrt{n \pi}} \\ \Rightarrow && (2n)! &\approx \frac{2^{2n}(n!)^2}{\sqrt{n\pi}} \end{align*}
3. \(X \sim Po(n)\), then \(X \approx N(n, (\sqrt{n})^2)\), therefore \begin{align*} && \mathbb{P}(X = n) &\approx \mathbb{P} \left (-\frac12 \leq \sqrt{n} Z \leq \frac12 \right) \\ &&&= \int_{-\frac{1}{2 \sqrt{n}}}^{\frac{1}{2 \sqrt{n}}} \frac{1}{\sqrt{2\pi}}e^{-\frac12 x^2} \d x \\ &&&\approx \frac{1}{\sqrt{2 \pi n}} \\ \Rightarrow && e^{-n} \frac{n^n}{n!} & \approx \frac{1}{\sqrt{2 \pi n}} \\ \Rightarrow && n! &\approx \sqrt{2 \pi n} e^{-n}n^n \end{align*}

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Solution:

1. \(\,\) \begin{align*} && I_1 &= \int_{-\infty}^{\infty} \frac{1}{x^2+2ax+b} \d x \\ &&&= \int_{-\infty}^{\infty} \frac{1}{b-a^2 +(x+a)^2} \d x \\ &&&= \left [ \frac{1}{\sqrt{b-a^2}} \tan^{-1} \frac{x+a}{\sqrt{b-a^2}} \right]_{-\infty}^{\infty} \\ &&&= \frac{\pi}{\sqrt{b-a^2}} \end{align*}
2. \(\,\) Here is the corrected LaTeX code for the second part, maintaining your exact styling and notation.
3. \(\,\) \begin{align*} && I_{n} &= \int_{-\infty}^{\infty} \frac{1}{(x^2+2ax+b)^{n}} \d x \\ &&&= \left[ \frac{x}{(x^2+2ax+b)^n} \right]_{-\infty}^{\infty} - \int_{-\infty}^\infty x \cdot \frac{-n(2x+2a)}{(x^2+2ax+b)^{n+1}} \d x \\ &&&= 0 + n \int_{-\infty}^\infty \frac{2x^2+2ax}{(x^2+2ax+b)^{n+1}} \d x \\ &&&= n \int_{-\infty}^\infty \frac{2(x^2+2ax+b) - (2ax+2b)}{(x^2+2ax+b)^{n+1}} \d x \\ &&&= 2n I_n - n \int_{-\infty}^\infty \frac{2ax+2b}{(x^2+2ax+b)^{n+1}} \d x \\ &&&= 2n I_n - n \int_{-\infty}^\infty \frac{a(2x+2a) + 2(b-a^2)}{(x^2+2ax+b)^{n+1}} \d x \\ &&&= 2n I_n - n \int_{-\infty}^\infty \frac{a(2x+2a)}{(x^2+2ax+b)^{n+1}} \d x - 2n(b-a^2) I_{n+1} \\ &&&= 2n I_n - n \left[ \frac{-a}{n(x^2+2ax+b)^n} \right]_{-\infty}^\infty - 2n(b-a^2) I{n+1} \\ &&&= 2n I_n - 0 - 2n(b-a^2) I_{n+1} \\ \Rightarrow && 2n(b-a^2)I_{n+1} &= (2n-1)I_n \end{align*}
4. \(\,\) \begin{align*} && I_{n+1} &= \frac{2n-1}{2n(b-a^2)} I_n \\ &&&= \frac{2n-1}{2n(b-a^2)} \cdot \frac{2n-3}{2(n-1)(b-a^2)} I_{n-1} \\ &&&= \frac{2n-1}{2n(b-a^2)} \cdot \frac{2n-3}{2(n-1)(b-a^2)} \cdots I_{1} \\ &&&= \frac{(2n-1)(2n-3) \cdots 1}{2n \cdot 2(n-1) \cdots 2 (b-a^2)^n} \frac{\pi}{\sqrt{b-a^2}} \\ &&&= \frac{(2n-1)(2n-3) \cdots 1}{2^n n!} \frac{\pi}{(b-a^2)^{n+\frac12}} \\ &&&= \frac{(2n-1)(2n-3) \cdots 1 \cdot 2n \cdot 2(n-1) \cdots 2}{2^{2n} n!n!} \frac{\pi}{(b-a^2)^{n+\frac12}} \\ &&&= \frac{(2n)!}{2^{2n}n!n!}\frac{\pi}{(b-a^2)^{n+\frac12}} \\ &&&= \frac{\pi}{2^{2n}(b-a^2)^{n+\frac12}} \binom{2n}{n} \\ \Rightarrow && I_n &= \frac{\pi}{2^{2n-2}(b-a^2)^{n-\frac12}} \binom{2n-2}{n-1} \\ \end{align*}

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Solution:

1. \begin{align*} && 2y \frac{\d y}{\d x} &= 4a \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{2a}{y} \end{align*} Therefore we must have \begin{align*} && \underbrace{-\frac{2aq}{2a}}_{\text{gradient of normal}} &= \underbrace{\frac{2ap-2aq}{ap^2-aq^2}}_{\Delta y / \Delta x} \\ \Rightarrow && -q &= \frac{2}{p+q} \\ && 0 &= 2 + pq+q^2 \end{align*}
2. We must have that \(q,r\) are the two roots of \(x^2+px+2 = 0\) \(QR\) has the equation: \begin{align*} && \frac{y-2aq}{x-aq^2} &= \frac{2ar-2aq}{ar^2-aq^2} \\ \Rightarrow && \frac{y-2aq}{x-aq^2} &= \frac{2}{r+q} \\ \Rightarrow && y &= \frac{2}{q+r}(x-aq^2) +2aq \\ && y &= -\frac{2}{p}x+2a\left(q-\frac{q^2}{q+r} \right) \\ &&y&= -\frac{2}{p}x+2a \frac{qr}{q+r} \\ && y &= -\frac{2}{p}x - 2a \frac{2}{p} \\ && y & = -\frac{2}{p}(x+2a) \end{align*} Therefore the point \((-2a,0)\) lies on all such lines.
3. \(OP\) has equation \(y = \frac{2}{p} x\) \begin{align*} && y &= \frac{2}{p} x \\ && y & = -\frac{2}{p}(x+2a) \\ && 2y &= -\frac{4a}{p} \\ \Rightarrow && y &= -\frac{2a}{p} \\ && x &= -a \end{align*} Therefore \(T\left (-a, -\frac{2a}{p} \right)\) always lies on the line \(x = -a\) The distance to the \(x\)-axis from \(T\) is \(\frac{2a}{|p|}\). We need to show that \(p\) can't be too small. Specifically \(x^2+px+2 = 0\) must have \(2\) real roots, ie \(\Delta = p^2-8 \geq 0 \Rightarrow |p| \geq 2\sqrt{2}\), ie \(\frac{2a}{|p|} \leq \frac{2a}{2\sqrt{2}} = \frac{a}{\sqrt{2}}\) as required.

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Solution:

1. \begin{align*} && \int \frac {x^3-2}{(x+1)^2}\, \e ^x \d x &= \frac{\P(x)}{Q(x)}\,\e^x + \text{constant} \\ \underbrace{\Rightarrow}_{\frac{\d}{\d x}} && \frac{x^3-2}{(x+1)^2}e^x &= \frac{P'(x)Q(x)-Q'(x)P(x)}{Q(x)^2}e^x + \frac{P(x)}{Q(x)}e^x \\ \Rightarrow && \frac{x^3-2}{(x+1)^2} &= \frac{(P(x)+P'(x))Q(x)-Q'(x)P(x)}{Q(x)^2} \\ \Rightarrow && Q(x)^2(x^3-2) &= ((P(x)+P'(x))Q(x)-Q'(x)P(x))(x+1)^2 \\ \Rightarrow && Q(-1) &= 0 \\ \Rightarrow && x+1 &\mid Q(x) \end{align*} We have \(\frac{x^3-2}{(x+1)^2}\) has degree \(1\) (plus some remainder term). Therefore \begin{align*} 1 &= \deg \l (P(x)+P'(x))Q(x)-Q'(x)P(x)\r - 2 \deg Q(x) \\ &= \deg P(x) + \deg Q(x) - 2 \deg Q(x) \\ &= \deg P(x) - \deg Q(x) \end{align*} as required. Suppose \(Q(x) = x+1, P(x) = ax^2+bx+c\) then \begin{align*} && \frac{x^3-2}{(x+1)^2} &= \frac{(P(x)+P'(x))(x+1)-P(x)}{(x+1)^2} \\ \Rightarrow && x^3-2 &= (P(x)+P'(x))(x+1) - P(x) \\ \Rightarrow && x^3-2 &= (ax^2+bx+c+2ax+b)(x+1) - (ax^2+bx+c) \\ &&&= a x^3+ x^2 (2 a + b) + x (2 a + b + c)+b \\ \Rightarrow && a &= 1 \\ && b &= -2 \\ && c &= 0 \end{align*} So \(P(x) = x^2-2x\)
2. \begin{align*} && \int \frac1{x+1}e^x \d x &= \frac{P(x)}{Q(x)}e^x + c \\ \Rightarrow && \frac{1}{x+1} e^x &= \frac{P'(x)Q(x)-Q'(x)P(x)}{Q(x)^2}e^x + \frac{P(x)}{Q(x)}e^x \\ \Rightarrow && \frac{1}{x+1} &= \frac{(P(x)+P'(x))Q(x)-Q'(x)P(x)}{Q(x)^2} \end{align*} Therefore \(Q(-1) = 0\) and so \(x +1 \mid Q(x)\). Considering degrees, we must have that \(P(x)\) has degree \(1\) less than \(Q(x)\). Consider also the number of factors of \(x+1\) in the numerator and denominator. Since \(P(x)\) and \(Q(x)\) have no common factors, the \(Q(x)\) could have \(q\) factors and \(P(x)\) must have none. The denominator therefore has \(2q\) factors and the numerator must have \(q-1\) factors (coming from \(Q'(x)\)), we must have \(2q = (q-1) + 1\), but that implies \(q = 0\). Contradiction! \end{align*}