Problems

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Solution:

1. Step 1: There are only three possibilities for the remainder of \(a\) when divided by \(3\), (\(0\), \(1\), \(2\)). \(a = 3m+r\). If \(r = 0\) we are done, if \(r = 1\) take \(k = m\), and \(r=2\) take \(k=(m+1)\) and we have \(a = 3k-1\) as required. Then \(a^2 = (3k\pm1)^2 =9k^2\pm6k+1 = 3(3k^32\pm2k)+1\) which is clearly \(1\) more than a square. Step 3: \begin{align*} && \frac{a}{b} &= \sqrt{2}+\sqrt{3} \\ \Rightarrow && \frac{a^2}{b^2} &= 5+2\sqrt{6} \\ \Rightarrow && \frac{a^2}{b^2}-5 &= 2\sqrt{6} \\ \Rightarrow && 24 &= \left ( \frac{a^2}{b^2}-5 \right)^2 \\ &&&= 25 + \frac{a^4}{b^4}-10\frac{a^2}{b^2} \\ \Rightarrow && -b^4 &= a^4-10a^2b^2 \\ \Rightarrow && a^4+b^4 &= 10a^2b^2 \end{align*} Step 4: If \(a\) is divisible by \(3\) then \(b^4 = 10a^2b^2-a^4\) is a multiple of \(3\), but if \(b\) was not a multiple of \(3\) then \(b^2\) would be \(1\) more than a multiple of \(3\) (by Step 3) and \(b^4\) would also be \(1\) more than a multiple of \(3\), and we would have a contradiction. Step 5: Follows since either \(a,b\) are both divisible by \(3\) (contradicting Step 2), or neither is, but then \(a^2\) and \(b^2\) are both one more than a multiple of \(3\) and the RHS is one more than a multiple of \(3\) but the LHS is \(2\) more than a multiple of \(3\) which is a contradiction.
2. Step 1: If \(a\) is not divisible by \(5\) then \(a^2 \equiv \pm 1 \pmod{5}\) Step 2: Suppose \(\frac{a}{b} = \sqrt{6}+\sqrt{7}\) Step 3: \begin{align*} && \frac{a}{b} &= \sqrt{6}+\sqrt{7} \\ \Rightarrow && \frac{a^2}{b^2} &= 13 + 2\sqrt{42} \\ \Rightarrow && 168 &= \left (\frac{a^2}{b^2} - 13 \right)^2 \\ &&&= 169 - 26 \frac{a^2}{b^2} + \frac{a^4}{b^4} \\ \Rightarrow && a^4+b^4 &= 26a^2b^2 \end{align*} Step 4: If \(a\) is a multiple of \(5\) then so is \(b^4\) and hence so is \(b^2\) and \(b\). Step 5: But the left hand side is always \(2 \pmod{5}\) and the right hand side is never \(2 \pmod{5}\) contradiction. Divisibility by \(3\) doesn't work here since mod \(3\) we can have \(a = 1, b = 1\) and have a valid solution.

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Solution:

1. Let \(2t = u\), \begin{align*} \int_2^x \sqrt{\frac{u-2}{u+2}} du &= \int_{t=1}^{t=x/2} \sqrt{\frac{2t-2}{2t+2}}2 \d t \\ &= 2\int_{t=1}^{x/2} \sqrt{\frac{t-1}{t+1}} \d t \\ &= 2f\l\frac{x}{2}\r \end{align*}
2. Let \(v = u-2\), \begin{align*} \int_0^x \sqrt{\frac{u}{u+4}} du &= \int_{v = 2}^{x+2} \sqrt{\frac{v-2}{v+2}} \d v \\ &= 2 f \l \frac{x+2}{2} \r \end{align*}
3. Let \(v = u-2, \d v = \d u\) \begin{align*} \int_5^x \frac{u-5}{u+1} du &= \int_3^{x-2} \frac{v-3}{v+3} \d v \\ &= \int_1^{\frac{x-2}{3}} \frac{3t - 3}{3t+3} 3 \d t \\ &= 3 f \l \frac{x-2}{3} \r \end{align*}
4. Let \(v = u^2, \d v = 2u \d u\)\begin{align*}\int_1^2 \frac{u^2}{\sqrt{u^2+4}} du &= \int_1^2 \sqrt{\frac{u^2}{u^2+4}} u \d u \\ &= \int_1^4 \sqrt{\frac{v}{v+4}} \frac12 \d v \\ &= f \l \frac{4+2}{2} \r - f \l \frac{3}{2} \r \\ &= f(3) - f(\frac32) \end{align*}

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Solution:

At the point we are about to topple, reaction and friction forces will be acting at \(C\)

1. \(\,\) \begin{align*} \overset{\curvearrowright}{X}:&& kW \cdot \lambda h\cos \alpha - R h &= 0 \\ \Rightarrow && R &= k\lambda W \cos \alpha \\ \end{align*}
2. \(\,\) \begin{align*} \overset{\curvearrowright}{C}:&& R \sin \alpha \cdot h-R\cos \alpha \cdot b-W\frac{b}{2} &= 0 \\ && k\lambda W \cos \alpha \sin \alpha \cdot b \tan \alpha- k\lambda W \cos \alpha\cos \alpha \cdot h-W\frac{b}{2} &= 0 \\ && k \lambda (\cos^2 \alpha - \sin^2 \alpha) +\frac12 &= 0 \\ \Rightarrow && 2k \lambda \cos 2\alpha + 1 &= 0 \end{align*}
3. \(\,\) \begin{align*} \text{N2}(\uparrow): && R_b -W-R\cos \alpha &= 0 \\ \Rightarrow && R_b &= W + k\lambda W \cos^2 \alpha\\ \text{N2}(\rightarrow): && R\sin \alpha - F_b &= 0 \\ \Rightarrow && F_b &= R \sin \alpha \\ \\ && F_b &\leq \mu R \\ \Rightarrow && k\lambda W \cos \alpha \sin \alpha &= \mu (W + k\lambda W \cos^2 \alpha) \\ \Rightarrow && \mu &\geq \frac{k\lambda \cos \alpha \sin \alpha}{1 + k\lambda \cos^2 \alpha} \\ &&&= \frac{k\lambda \sin 2\alpha}{2 + 2k\lambda cos^2 \alpha} \\ &&&= \frac{k\lambda \sin 2\alpha}{2 + k\lambda (\cos 2 \alpha+1)} \\ &&&= \frac{k\lambda \sin 2\alpha}{-4k\lambda \cos 2 \alpha + k\lambda (\cos 2 \alpha+1)} \\ &&&= \frac{\sin 2 \alpha}{1 -3 \cos 2\alpha} \end{align*}

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Solution:

1. The horizontal position of the particle at time \(t\) is \(u \sin\alpha t\), so \(T = \frac{h \tan \beta}{u \sin \alpha}\) The vertical position of the particle at this time \(T\) satisifes: \begin{align*} && h &= u \cos\alpha \frac{h \tan \beta}{u \sin\alpha} - \frac12 g \left ( \frac{h \tan \beta}{u \sin\alpha} \right)^2 \\ &&&= h\cot \alpha \tan \beta - \frac{gh^2}{2u^2} \tan^2 \beta \cosec^2 \beta \\ \Rightarrow && 1 &= c \tan \beta - \frac{1}{k} \tan^2 \beta (1 + c^2) \\ \Rightarrow && k \cot^2 \beta &= kc\cot \beta -1-c^2 \\ \Rightarrow && 0 &= c^2 -ck \cot \beta + 1 + k \cot^2 \beta \end{align*}
   1. As a quadratic in \(c\) the sum of the roots is \(k \cot \beta\), therefore \(\cot \alpha_1 + \cot \alpha_2 = k \cot \beta\). We also have that \(\cot \alpha_1 \cot \alpha_2 = 1 + k \cot^2 \beta\), so \begin{align*} && \cot (\alpha_1 + \alpha_2) &= \frac{\cot \alpha_1 \cot \alpha_2-1}{\cot \alpha_1 + \cot \alpha_2} \\ &&&= \frac{1 + k \cot^2 \beta - 1}{k \cot \beta} \\ &&&= \cot \beta \\ \Rightarrow && \beta &= \alpha_1 + \alpha_2 \pmod{\pi} \end{align*} but since \(\alpha_i \in (0, \frac{\pi}{2})\) the equation must hold exactly.
   2. Since it has two real roots we must have \begin{align*} && 0 &<\Delta = k^2 \cot^2 \beta - 4 (1 + k \cot^2 \beta) \\ &&&= k^2 \cot^2 \beta-4k \cot^2 \beta -4 \\ &&&= \cot^2 \beta (k^2 - 4k - 4(\sec^2 \beta - 1)) \\ &&&= \cot^2 \beta ( (k-2)^2 -4\sec^2 \beta) \\ \Rightarrow && k &> 2 + 2\sec \beta = 2(1+\sec \beta) \end{align*}
2. The greatest height will satisfy \(v^2 = u^2 + 2as\) so \(0 = u^2 \cos^2 \alpha - 2gh_{max} \Rightarrow 4\sec^2 \alpha = \frac{2u^2}{gh_{max}} = k_{max}\), but this decreases with \(h\), so the smallest \(k\) can be is \(4\sec^2 \alpha\), ie \(k \geq 4 \sec^2 \alpha\)

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Solution:

1. The probability they don't make a decision in a round is \(qp + pq = 2qp\) (TH and HT). The probability they make a decision in a round is \(q^2+p^2\) (TT and HH). Therefore the probability they make a decision in the \(n\)th round is: \[ (q^2+p^2)(2qp)^{n-1} \] by having \(n-1\) failures and one success. The probability they make a decision on or before the \(n\)th round is the \(1-\) the probability they don't, ie \(1 - (2qp)^n\). Notice that \(\sqrt{qp} \leq \frac{p+1}{2} = \frac12 \Rightarrow qp \leq \frac14\) so \(1-(2pq)^n \leq 1 - \frac1{2^n}\)
2. The probability it's decided in the first round is \(p^3 + q^3\) (HHH, TTT). The probability it's decided in the second round is \(3p^2q \cdot p^2 + 3qq^2 \cdot q^2 = 3pq(p^3+q^3)\) (HHT -> HHH) and (TTH -> TTT) with reorderings). Therefore the probability of making a decision in the first or second round is \((p^3+q^3)(1 + 3pq)\) which is minimised when \(p = q\) by Muirhead (or whatever your favourite inequality is). So \(\frac{2}{8} \cdot \left ( 1 + \frac{3}{4} \right) = \frac{7}{16}\)

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Solution: The tangent to the curve \(y = f(x)\) at \(x = a\) has the equation \(\frac{y-f(a)}{x-a} = f'(a) = g(a)+(a-p)g'(a)\). This passes through \((p,0)\) iff \begin{align*} && \frac{-f(a)}{p-a} &= g(a)+(a-p)g'(a) \\ \Leftrightarrow && -f(a) &= (p-a)g(a) -(a-p)^2g'(a) \\ \Leftrightarrow && -f(a) &= -f(a) -(a-p)^2g'(a) \\ \Leftrightarrow && 0 &= g'(a) \\ \end{align*}

1. In this case \(g(x) = A(x-q)(x-r) = A(x^2-(q+r)x+qr)\) and so we must have that \(g'(a) = 0\), ie \(A(2a-(q+r)) = 0 \Rightarrow 2a = q+r\) The gradient is \(g(a) +(a-p)g'(a) = g(a) = A(a-q)(a-r)\)
2. By the same reasoning, but with \(g(x) = A(x-p)(x-q)\) we have the gradient is \(A(c-p)(c-r)\). This is parallel iff \begin{align*} && (c-p)(c-r) &= (a-q)(a-r) \end{align*} The tangent at \(x = q\) is \(\frac{y-0}{x-q} = A(q-p)(q-r)\) or \( y = A(q-p)(q-r)(x-q)\)

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Solution:

Notice the total area is \(xf(x)\) and it is made up of the sum of the two integrals.

1. Suppose \((g(t))^3 + g(t) = t\). Notice that \((g(0))^3 + g(0) =0 \Rightarrow g(0)((g(0))^2 + 1) = 0 \Rightarrow g(0) = 0\). \begin{align*} && t &= (g(t))^3 + g(t) \\ \Rightarrow && 1 &= 3(g(t))^2 g'(t) + g'(t) \\ \Rightarrow && g'(t) &= \frac{1}{1 + 3(g(t))^2} > 0 \end{align*}
   

   + - ↺
   From our sketch, we can see we are interested in: \begin{align*} && \int_0^2 g(t) \d t &= 2 - \int_0^1 (x^3 + x) \d x \\ &&&= 2 - \frac14 - \frac12 = \frac54 \end{align*}
2. \(\,\)
   

   + - ↺
   From our second sketch, we can see that: \begin{align*} && \int_0^8 h(t) \d t &= 16 - \int_1^2 (x^3+x-2) \d x \\ &&&= 16 - \left ( \frac{8}{4} + \frac{2^2}{2} - 2 \cdot 2 \right)+ \left ( \frac{1}{4} + \frac{1}{2} - 2 \right) \\ &&&= \frac{59}{4} \end{align*}

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Solution: Claim: \(|x_1 + x_2| \leq |x_1| + |x_2|\) Proof: Case 1: \(x_1, x_2 \geq 0\). The inequality is equivalent to \(|x_1 + x_2| = x_1 + x_2 = |x_1|+|x_2|\) so it's an equality. Case 2: \(x_1, x_2 \leq 0\). The inequality is equivalent to \(|x_1+x_2| = -x_1-x_2 = |x_1|+|x_2\), so it's also an equality in this case. Case 3: (wlog) \(|x_1| \geq |x_2| > 0\) and \(x_1x_2 < 0\) then \(|x_1+x_2| = x_1-x_2 \leq x_1 \leq |x_1|+|x_2|\) We can prove this by induction, we've already proven the base case and: \(|x_1+x_2 + \cdots + x_n| \leq |x_1 + x_2 + \cdots x_{n-1}| + |x_n| \leq |x_1| + |x_2| + \cdots + |x_n|\)

1. \(\,\) \begin{align*} && |f(x) - 1| &= |a_1 x + a_2x^2 + \cdots + a_{n-1}x^{n-1} + x^n| \\ &&&\leq |a_1x| + |a_2x^2| + \cdots + |a_{n-1}x^{n-1}| + |x^n| \\ &&&\leq |a_1||x| + |a_2||x|^2 + \cdots + |a_{n-1}||x|^{n-1} + |x|^n \\ &&&\leq A|x| + A|x|^2 + \cdots + A|x|^{n-1} + |x|^n \\ &&&=A|x| \frac{1-|x|^{n-1}}{1-|x|} + |x|^n \\ &&&= \frac{A|x|-A|x|^{n}+|x|^{n+1}-|x|^n}{1-|x|} \\ &&&= \frac{A|x|-|x|^n(\underbrace{A-|x|+1}_{\geq0})}{1-|x|} \\ &&&\leq \frac{A|x|}{1-|x|} \end{align*}
2. If \(f(\omega) = 0\) then \begin{align*} && 1 & \leq \frac{A|\omega|}{1-|\omega|} \\ \Leftrightarrow && 1-|\omega| &\leq A |\omega| \\ \Leftrightarrow && 1 &\leq (1+A) |\omega| \\ \Leftrightarrow && \frac{1}{1+A} &\leq |\omega| \\ \end{align*} We also know \(\omega \leq 1 < 1 + A\)
3. If \(\omega\) is a root of \(f(x)\) then \(1/\omega\) is a root of \(1 + a_{n-1}x + a_{n-2}x^2 + \cdots + a_1x^{n-1}+x^n\) and so \(1/\omega\) satisfies that inequality, ie \begin{align*} && \frac{1}{1+A} && \leq &&|1/\omega| && \leq &&1 + A \\ \Leftrightarrow &&1+A && \geq&& |\omega| && \geq&& \frac{1}{1 + A} \end{align*}
4. First notice that it's equivalent to: \(0 = x^5 - 1x^4 - \frac{100}{135}x^3-\frac{91}{135}x^2-\frac{126}{135} + 1\) therefore all integer roots must be between \(-2,-1\) and \(1\) and \(2\). \(1\) doesn't work. \(-1\) works. Clearly \(2\) cannot work by parity argument, therefore the only integer root is \(-1\).

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Solution:

1. \begin{align*}\cos(\frac{\pi}{9}) \cos(\frac{2\pi}{9}) \cos(\frac{4\pi}{9}) &= \frac{\sin(\frac{2\pi}{9}) \cos(\frac{2\pi}{9}) \cos(\frac{4\pi}{9})}{2 \sin \frac{\pi}{9}} \\ &= \frac{\sin(\frac{4\pi}{9})\cos(\frac{4\pi}{9})}{4 \sin \frac{\pi}{9}} \\ &= \frac{\sin(\frac{8\pi}{9})}{8 \sin \frac{\pi}{9}} \\ &= \frac{1}{8} \end{align*}
2. Let \(\displaystyle P_n = \prod_{k=0}^{n} \cos\left(\frac{x}{2^k}\right)\). Claim: \(P_n = \frac{\sin 2x}{2^{n+1} \sin \l \frac{x}{2^n} \r}\). Proof: This is true for \(n = 0\), assume true for \(n-1\) \begin{align*} \sin\l \frac{x}{2^{n}} \r P_n &= P_{n-1} \cos\l \frac{x}{2^{n}} \r \sin\l \frac{x}{2^{n}} \r \\ &= P_{n-1} \frac{1}{2} \sin\l \frac{x}{2^{n-1}} \r \\ &= \frac{\sin 2x}{2^{n} \sin \l \frac{x}{2^{n-1}}\r} \frac{1}{2} \sin\l \frac{x}{2^{n}} \r \\ &= \frac{\sin 2x}{2^{n+1}} \end{align*} Hence \(P_n = \frac{\sin 2x}{2^{n+1} \sin \l \frac{x}{2^n}\r}\) Taking logs, we determine that: \begin{align*} && \sum_{k=0}^n \ln \cos \l \frac{x}{2^k} \r &= \ln \sin 2x - \ln \sin \l \frac{x}{2^n} \r - (n+1) \ln 2 \\ \Rightarrow && \sum_{k=0}^n \frac{1}{2^k} \tan \l \frac{x}{2^k} \r &= -2 \cot 2x + \frac{1}{2^n} \cot \l \frac{x}{2^n} \r - 0 \\ \end{align*} as required.
3. As \(n \to \infty\) \(\frac{x}{2^n} \to 0\), so \(\frac{\sin \frac{x}{2^n}}{\frac{x}{2^n}} = \frac{2^n \sin \frac{x}{2^n}}{x} \to 1\) \begin{align*}\prod_{k=1}^{\infty} \cos\left(\frac{x}{2^k}\right) &= \lim_{n \to \infty} \frac{\sin x}{2^n \sin \l \frac{x}{2^n} \r} \\ &= \lim_{n \to \infty} \frac{\sin x}{x \frac{2^n \sin \l \frac{x}{2^n} \r}{x} } \\ &= \lim_{n \to \infty} \frac{\sin x}{x} \\ \end{align*} \begin{align*} \sum_{j=2}^{\infty} \frac{1}{2^{j-2}} \tan\left(\frac{\pi}{2^j}\right) &= \sum_{j=0}^{\infty} \frac{1}{2^{j}} \tan\left(\frac{1}{2^j}\frac{\pi}{4}\right) \\ &= \lim_{n \to \infty} \l -2 \cot \frac{\pi}{2} + \frac{1}{2^n} \cot \l \frac{\pi}{4 \cdot 2^n} \r\r \\ &= \frac{4}{\pi} \lim_{n \to \infty} \l \frac{1}{2^n} \frac{\pi}{4} \cot \l \frac{\pi}{4 \cdot 2^n} \r\r \\ &\to \frac{\pi}{4} \end{align*}

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Solution:

1. If \(f(a) = a\) then the sequence is constant, ie \(a = p+a^2-pa \Rightarrow 0 = (a-p)(a-1)\). Therefore \(a = 1, p\) If there sequence has period \(2\) then there must be a solution to \(f(f(x)) = x\), ie \begin{align*} && x &= p+(f(x)-p)f(x) \\ &&&= p+(p+(x-p)x-p)(p+(x-p)x) \\ &&&= p + (x-p)x(p+(x-p)x) \\ &&&= p+(x^2-px)(x^2-px+p) \\ \Rightarrow && 0 &= x^4-2px^3+(p+p^2)x^2-(p^2+1)x+p \\ &&&= (x-1)(x-p)(x^2-(p-1)x+1) \end{align*} The first two roots (\(x = 1, p\)) are constant sequences, so we need the second quadratic to have a root, ie \((p-1)^2-4 \geq 0 \Rightarrow p \geq 3 , p \leq -1\). We also need this root not to be \(1\) or \(p\), ie \(1-(p-1)+1 = 3-p \neq 0\) and \(p^2-(p-1)p + 1 = 1+p \neq 0\) so \(p \neq -1, 3\). Therefore \(p > 3\) or \(p < -1\).
2. There exists a constant sequence iff there is a solution to \(f(x) = x\), ie \begin{align*} && x &= f(x) \\ &&&= q + (x-p)x \\ \Leftrightarrow && 0 &= x^2-(p+1)x + q \tag{has a solution} \\ \end{align*} But if it doesn't have a solution, clearly the RHS is always larger, and if it does have a solution then there is some point where the inequality doesn't hold. Suppose \(f(x) > x\) for all \(x\) then \(f(f(x)) > f(x) > x\) therefore there is no value where \(f(f(x)) = x\) which is required for any sequence of period 2. No, consider \(p = q = 0\) so \(f(x) = x^2\) then there cannot be a period \(2\) sequence by the first part, but also clearly \(u_n = 1\) is a valid constant sequence.