Problems

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Solution:

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Solution:

1. Case \(b > 0\). Then \(bx+a\) is increasing and the greatest value is \(10b+a\), and the least value \(a-10b\) Case \(b=0\), then \(a\) is constant and the greatest and least value is \(a\) Case \(b < 0\), then \(bx+a\) is decreasing and the greatest value is \(-10b+a\) and the least value is \(10b+a\)
2. If \(c = 0\) we have the same cases as above. If \( c > 0\) the consider \(2cx+b\). if \(b-20c > 0\) then our function is increasing on our interval and the greatest value is \(100c+10b+a\) and the least value is \(100c-10b+a\) If \(20c+b < 0\) then our function is decreasing and that calculation is reversed. If neither of these are true, then the minimum will be when \(x = - \frac{b}{2c}\) and the max at one end point.

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Solution: \begin{align*} && y &= \sin(k \sin^{-1} x ) \\ &&y' &= \cos (k \sin^{-1} x) \cdot k \frac{1}{\sqrt{1-x^2}} \\ && y'' &= -\sin (k \sin^{-1} x) \cdot k^2 \frac{1}{(1-x^2)} - \cos(k \sin^{-1} x) \cdot k \frac{x}{(1-x^2)\sqrt{1-x^2}} \\ && (1-x^2)y'' &= -k^2y -xy' \\ \Rightarrow && 0 &= (1-x^2)y''+xy' + k^2y \end{align*} \begin{align*} && y &= Ax^3 + Bx^2 + Cx + D \\ && y' &= 3Ax^2 + 2Bx + C \\ && y'' &= 6Ax+2B \\ && 0 &= (1-x^2)(6Ax+2B) - x( 3Ax^2 + 2Bx + C) + 9(Ax^3 + Bx^2 + Cx + D ) \\ &&&= x^3(-6A-3A+9A) + x^2(-2B-2B+9B) + x(6A-C+9C) + (2B +9D) \\ \Rightarrow && B &= 0 \\ \Rightarrow && D &= 0 \\ \Rightarrow && C &= -\frac34 A \\ \\ x = 0, y = 0, y' = 0: && y &= 3x-4x^3 \\ \end{align*} And so \(\sin 3 x = 3 \sin x - 4\sin^3 x\)

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Solution: If \(T_1\) is the time they meet after \(\frac12\)km from the starting points and \(T_2\) is the time they meet a second time, then \begin{align*} && \frac12 &= vT_1 \\ &&&= V(T_1-\tfrac12) \\ && X - \frac54 &= vT_2 \\ && X + \frac54 &= V(T_2 - 1) \\ && \frac{T_2}{T_1} &= \frac{4X-5}{2} \\ && X + \frac54 + V &= VT_2 \\ && \frac12 + \frac12 V &= VT_1 \\ \Rightarrow && \frac{T_2}{T_1} &= \frac{4X + 5 + 4V}{2(1+V)} \\ \Rightarrow && \frac{4X-5}{2}&=\frac{4X + 5 + 4V}{2(1+V)} \\ \Rightarrow && V(4X-9) &= 10 \\ \Rightarrow && V &= \frac{10}{4X-9} \\ \\ && T_1 &= \frac{1}{2V} + \frac12 \\ &&&= \frac{4X+1}{20} \\ && v &= \frac{1}{2T_1} \\ &&&= \frac{10}{4X+1} \end{align*} \begin{align*} && 2X &= \frac12 V \\ \Rightarrow && 2X(4X-9) &= 5 \\ \Rightarrow && 0 &= 8X^2-18X-5 \\ &&&= (4X+1)(2X-5) \\ \Rightarrow && X &= -\frac14, \frac52 \end{align*} Since \(X\) is positive, we must have \(X = \frac52\)km

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Solution: The net force on the planets will always be towards the centre of mass (by symmetry or similar arguments). Therefore it suffices to check whether we can find a speed where the planets follow uniform circular motion, ie \(F = mr \omega^2\). (But clearly this is possible, we just need to find the speed)

\begin{align*} && F &= m r \omega^2 \\ && 2\frac{\gamma m^2}{a^2} \cos 30^{\circ} &= m \frac{a}{\sqrt{3}} \omega^2 \\ \Rightarrow && \frac{\sqrt{3}\gamma m^2}{a^2} &= \frac{ma \omega^2}{\sqrt{3}} \\ \Rightarrow && \omega^2 &= \frac{3\gamma m}{a^3} \\ \Rightarrow && \omega &= \left ( \frac{3\gamma m}{a^3}\right)^{1/2} \end{align*} In the second scenario, we are interested in when: \begin{align*} && F &= m r \omega^2 \\ && \underbrace{2\frac{\gamma m^2}{a^2} \cos 30^{\circ}}_{\text{to other symmetric planets}} + \underbrace{\frac{\gamma \lambda m^2}{a^2}}_{\text{central planet}} &= m \frac{a}{\sqrt{3}} \omega^2 \\ \Rightarrow && \frac{(\sqrt{3}+\lambda)\gamma m^2}{a^2} &= \frac{ma \omega^2}{\sqrt{3}} \\ \Rightarrow && \omega^2 &= \frac{(3+\sqrt{3}\lambda)\gamma m}{a^3} \\ \Rightarrow && \omega &= \left ( \frac{(3+\sqrt{3}\lambda)\gamma m}{a^3}\right)^{1/2} \end{align*}

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Solution:

1. Notice that if \(x > 0\) we must have \begin{align*} && \left ( \sqrt{x} - \frac{1}{\sqrt{x}} \right)^2 &\geq 0 \\ \Leftrightarrow && x - 2 + x^{-1} & \geq 0 \\ \Leftrightarrow && x + x^{-1} & \geq 2 \end{align*} Let \(S\) be the sum, and assume all probabilities are equal \begin{align*} && \mathbb{P}(S = 2) &= q_1 r_1 \\ && \mathbb{P}(S = 12) &= q_6 r_6 \\ && \mathbb{P}(S = 7) &= \sum_{i=1}^6 q_i r_{7-i} \\ \Rightarrow && q_1r_1 &= q_6r_6 \\ \Rightarrow && q_1r_6+q_6r_1 &\leq q_1r_1 \\ \Rightarrow && \frac{r_6}{r_1} + \frac{q_6}{q_1} &\leq 1 \\ \Rightarrow && q_1r_6+q_6r_1 &\leq q_6r_6 \\ \Rightarrow && \frac{q_1}{q_6} + \frac{r_1}{r_6} &\leq 1 \\ \Rightarrow && \frac{r_6}{r_1} + \frac{q_6}{q_1}+\frac{q_1}{q_6} + \frac{r_1}{r_6} &\leq 2\\ \text{but} && \frac{r_6}{r_1} + \frac{q_6}{q_1}+\frac{q_1}{q_6} + \frac{r_1}{r_6} &\geq 4 \end{align*} Since we have a contradiction they cannot all be equal.
2. We would like \(\displaystyle \sum q_i^2 \leq 1/36\) (subject to \(\displaystyle \sum q_i = 1\), clearly this cannot be true since: \begin{align*} && 1 &= \left ( \sum_{i=1}^6 q_i \right)^2 \\ &&&= \sum_{i=1}^6 q_i^2 + \sum_{i \neq j} 2q_i q_j \\ &&&\leq \sum_{i=1}^6 q_i^2 + 5\sum_{i=1}^6 q_i^2 \\ &&&=6 \sum_{i=1}^6 q_i^2 \\ \Rightarrow && \sum_{i=1}^6 q_i^2 &\geq 1/6 > 1/36 \end{align*} [For a weaker solution to the last part, notice that the largest value of \(q_i\) is \(\geq 1/6\) and therefore \(q_{max}^2 \geq 1/36\), but if equality holds then the other values must also be non-zero, and therefore the inequality cannot hold]

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Solution: There are \(N-1\) gaps between the magnets which are independently gaps or not gaps. Therefore the total number of gaps is \(X \sim Binomial(N-1, \frac12)\) and \begin{align*} \mathbb{E}(X) &= \frac{N-1}{2} \\ \textrm{Var}(X) &= \frac{N-1}{4} \end{align*}

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Solution:

1. \begin{align*} \log_{10}(\log_{10} a_1) &= \log_{10} (\log_{10} (x!) \\ &\approx \log_{10} (\log_{10} \sqrt{2 \pi} x^{x+\frac12} e^{-x}) \\ &= \log_{10} \l \log_{10} \sqrt{2 \pi} + (x+\frac12) \log_{10} x-x \r \\ &= \log_{10} \l \log_{10} \sqrt{2 \pi} + (100x+50)-x \r \\ &= \log_{10} \l 99x + \epsilon \r \\ &\approx \log_{10} 99 + \log_{10} x \\ &\approx 2 + 100 = 102 \end{align*}
2. \begin{align*} \log_{10}(\log_{10} a_2) &= \log_{10}(\log_{10} x^y) \\ &= \log_{10} y + \log_{10} \log_{10} x \\ &= x + 2 \end{align*} \begin{align*} \log_{10}(\log_{10} a_3) &= \log_{10}(\log_{10} y^x) \\ &= \log_{10} x + \log_{10} \log_{10} y \\ &= 100 + \log_{10} x \\ &= 200 \end{align*} \begin{align*} \log_{10}(\log_{10} a_4) &= \log_{10}(\log_{10} z^x) \\ &= \log_{10} x + \log_{10} \log_{10} z \\ &= 100 + \log_{10} y \\ &= 100+x \end{align*} \begin{align*} \log_{10}(\log_{10} a_5) &= \log_{10}(\log_{10} e^{xyz}) \\ &= \log_{10} x + \log_{10}y+\log_{10} z+ \log_{10} \log_{10} e \\ &\approx 100 + x + y \end{align*} \begin{align*} \log_{10}(\log_{10} a_6) &= \log_{10}(\log_{10} z^{1/y}) \\ &= \log_{10}(\log_{10} 10) \\ &= 0 \end{align*} \begin{align*} \log_{10}(\log_{10} a_7) &= \log_{10}(\log_{10} y^{z/x}) \\ &= \log_{10}z-\log_{10} x + \log_{10} \log_{10} y \\ &= y \end{align*} Since \(0 < 102 < 200 < x+2 < x+100 < y < y+x+100\) we must have \(a_6 < a_1 < a_3 < a_2 < a_4 < a_7 < a_5\)

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Solution:

1. \(\,\) \begin{align*} && 0 &= nx^2 + 2\sqrt{3n^2+50}x + 2n + 15 \\ && 0 &> \Delta = 4(3n^2+5) - 4\cdot n \cdot (2n + 15) \\ \Leftrightarrow && 0 &> n^2-15n+5\\ \text{cv}: && n &= \frac{15 \pm \sqrt{225 - 20}}{2} \\ &&&\approx \frac{15\pm14.x}{2}\\ \Leftrightarrow &&n &\in \{1, 2, \cdots, 14\} \end{align*}
2. \(\,\) \begin{align*} && 0 &> \Delta = 4(pn^2+q) - 4\cdot n \cdot (rn+s) \\ \Leftrightarrow && 0&>(p-r)n^2-sn+q \end{align*} Which is always true if \(r > p\) and \(s^2 < 4q(p-r)\)
3. \(\,\) \begin{align*} && 0 &= x^2 + 2\sqrt{p+q}x+ r+s \\ && 0 &\leq \Delta = 4(p+q) - 4(r+s) \\ && 0 &\leq 1 + s^2/8 - s \\ \text{c.v}: && s &= \frac{1 \pm \sqrt{1-4 \cdot \frac{1}{8}}}{2\cdot\frac18} \\ &&&= 4 \pm 4\sqrt{\frac12} \\ &&&= 4 \pm 2\sqrt{2} \\ \Rightarrow && s \leq 4 - 2\sqrt{2} &\text{ or } s \geq 4 + 2\sqrt{2} \end{align*}

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Solution: \begin{align*} && S_2(x) &= e^{x^3} \frac{d^2}{\d x^2} \left [e^{-x^3} \right] \\ &&&= e^{x^3} \frac{d}{\d x} \left [e^{-x^3}(-3x^2) \right] \\ &&&= e^{x^3} \left [e^{-x^3}(9x^4-6x) \right] \\ &&&=9x^4-6x \\ \\ && S_3(x) &= e^{x^3} \frac{\d^3}{\d x^3} \left [ e^{-x^3} \right]\\ &&&= e^{x^3} \frac{\d}{\d x} \left [ e^{-x^3}(9x^4-6x) \right ] \\ &&&= e^{x^3} e^{-x^3}\left [ (-3x^2)(9x^4-6x)+(36x^3-6) \right ] \\ &&&= -27x^6 +54x^3-6 \end{align*} Claim: \(S_n\) is a polynomial of degree \(2n\) with leading coefficient \((-3)^n\). Proof: Clearly this is true for \(n = 1, 2, 3\) by demonstration. Suppose it is true for some \(n = k\), then \begin{align*} && S_k(x) &= e^{x^2} \frac{\d^k}{\d x^k} \left [ e^{x^3}\right] \\ && (-3)^kx^{2k} +\cdots &= e^{x^3} \frac{\d^k}{\d x^k} \left [ e^{x^3}\right] \\ \Rightarrow && \frac{\d^k}{\d x^k} \left [ e^{x^3}\right] &= e^{-x^3} \left ( (-3)^kx^{2k} +\cdots\right) \\ \Rightarrow && \frac{\d^k}{\d x^k}\left [ e^{x^3}\right] &= e^{-x^3} (-3x^2)\left ( (-3)^kx^{2k} +\cdots\right) + e^{-x^3} S_k'(x) \\ &&&= e^{-x^3} \left (\underbrace{ (-3)^{k+1}x^{2k+2} + \cdots + S_k'(x)}_{\deg =2k+2}\right) \\ \Rightarrow && S_{k+1}(x) &= (-3)^{k+1}x^{2k+2} + \cdots + S_k'(x) \end{align*} And therefore \(S_{k+1}\) is a polynomial degree \(2(k+1)\) with leading coefficient \((-3)^{k+1}\) so by induction it's true for all \(n\). If \(S'_n(a) = 0\) then \(S_{n+1}(a) = (-3a^2)S_n(a) + S_n'(a) \Rightarrow S_{n+1}(a)S_n(a) = -3 (aS_n(a))^2 \leq 0\)