Problems

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Solution:

1. Let \(f(x,y) = x^3-5x^2+2x^2y+xy^2-8xy-3y^2+6x+6y\), then \begin{align*} f(3,y) &= 27 - 5 \cdot 9 +18y + 3y^2-24y-3y^2+18 + 6y \\ &= 0 \end{align*}, therefore \(x-3\) is a factor of \(f(x,y)\). \begin{align*} f(x,y) &= x^3-5x^2+6x+y(2x^2-8x+6) + y^2(x-3) \\ &= (x-3)(x^2-2x)+y(x-3)(2x-2)+y^2(x-3) \\ &= (x-3)(x^2-2x+2y(x-1)+y^2) \\ &= (x-3)(x+y)(x+y-2) \end{align*}
2. Let \(g(x,y) = 6y^3-y^2-21y+2x^2+12x-4xy+x^2y-5xy^2+10\), notice that \(g(x,-2) = 0\), so \(y+2\) is a factor, \begin{align*} g(x,y) &= 6y^3-y^2-21y+2x^2+12x-4xy+x^2y-5xy^2+10 \\ &= x^2(2+y) + x(12-4y-5y^2) + 6y^3-y^2-21y+10 \\ &= x^2(y+2) + x(y+2)(6-5y) + (y+2)(6y^2-13y+5) \\ &= (y+2)(x^2+(6-5y)x+(6y^2-13y+5)) \\ &= (y+2)(x-2y +1)(x-3y+5) \end{align*}

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Solution: \[\frac{\d}{\d t} \left ( \sec t \right) = \frac{\sin t }{\cos^2 t} = \sec t \tan t \]

1. \(\,\) \begin{align*} && I_1 &= \int_{\sqrt{2}}^2 \frac{1}{x^3 \sqrt{x^2-1}} \\ x = \sec t, \d x = \sec t \tan t:&&&= \int_{t=\pi/4}^{t=\pi/3} \frac{1}{\sec^3 t \tan t} \sec t \tan t \d t \\ &&&= \int_{t=\pi/4}^{t=\pi/3} \cos^2 t \d t \\ &&&= \int_{t=\pi/4}^{t=\pi/3} \frac{1+\cos 2t}{2} \d t \\ &&&= \frac12 \frac{\pi}{12} + \frac12 \left (\sin \frac{\pi}{3} - \sin \frac{\pi}{4} \right) \\ &&&= \frac{\pi}{24} + \frac{\sqrt{3}-2}{8} \\ \end{align*}
2. \(\,\) \begin{align*} && I_2 &= \int \frac{1}{(x+2)\sqrt{(x+1)(x+3)}} \d x \\ &&&= \int \frac{1}{(x+2)\sqrt{(x+2)^2-1}} \d x \\ &&&= \int \frac{1}{u\sqrt{u^2-1}} \d u \\ &&&= \sec^{-1} u + C \\ &&&= \sec^{1} (x+2) + C \end{align*}
3. \(\,\) \begin{align*} && I_3 &= \int \frac{1}{(x+2)\sqrt{(x+2)^2 - 9}} \d x \\ &&&= \int\frac{1}{9(\frac{x+2}{3})\sqrt{(\frac{x+2}3)^2 - 1}} \d x \\ u = \frac{x+2}{3}, 3\d u =\d x &&&= \frac19 \int \frac{1}{u\sqrt{u^2-1}} 3 \d u \\ &&&= \frac13 \sec^{-1} u + C \\ &&&= \frac13 \sec^{-1} \frac{x+2}{3} + C \end{align*}

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Solution:

1. \begin{align*} && f(x) -7xf(x)+10x^2f(x) &= \sum_{n=0}^\infty a_n x^n - 7x \sum_{n=0}^{\infty} a_n x^n + 10x^2 \sum_{n=0}^{\infty} a_nx^n \\ &&&= \sum_{n=2}^\infty (a_n-7a_{n-1}+10a_{n-2})x^n + a_0+a_1x-7a_0x \\ &&&= 0 + 2-7x \\ \\ \Rightarrow && f(x) &= \frac{2-7x}{1-7x+10x^2} \\ &&&= \frac{2-7x}{(1-5x)(1-2x)} \\ &&&= \frac{1}{1-2x} + \frac{1}{1-5x} \\ &&&= \sum_{n=0}^{\infty} (2^n + 5^n)x^n \end{align*} Therefore \(a_n = 2^n +5^n\)
2. \(\,\) \begin{align*} && 40 &= 10p + 5 q \\ && 100 &= 40p+10q \\ && 10 &= 4p + q \\ \Rightarrow && (p,q) &= (1,6) \\ \\ && g(x) -xg(x)-6x^2g(x) &= 5+5x \\ \Rightarrow && g(x) &= \frac{5+5x}{1-x-6x^2} \\ &&&= \frac{5+5x}{(1-3x)(1+2x)} \\ &&&= \frac{4}{1-3x} + \frac{1}{1+2x} \\ &&&= \sum_{n=0}^{\infty} (4 \cdot 3^n + (-2)^n)x^n \\ \Rightarrow && b_n &= 4 \cdot 3^n + (-2)^n \end{align*}

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Solution:

1. Suppose \(x_n> 3\) then \begin{align*} && x_{n+1} &= \frac{x_n^2+9-3}{5} \\ &&& \geq \frac{2\sqrt{x_n^2 \cdot 9} - 3}{5} \\ &&&= \frac{6x_n -3}{5} = x_n + \frac{x_n-3}{5} \\ &&&> x_n > 3 \end{align*} Therefore if \(x_i > 3 \Rightarrow x_{i+1} > x_i\) and \(x_{i+1} > 3\) so by induction \(x_n\) strictly increasing for all \(n\).
2. Suppose \(2 < y_n < 3\) then \begin{align*} && y_{n+1} &= 5 - \frac6{y_n} \\ &&&< 5 - \frac63 = 3 \\ \\ && y_{n+1} &= 5 - \frac4{y_n} - \frac{2}{y_n} \\ \\ &&&= y_n + 5 - \frac{2}{y_n} - \left ( y_n + \frac4{y_n} \right) \\ &&&\geq y_n + 5 - \frac{2}{y_n} - 2\sqrt{y_n \frac{4}{y_n}} \\ &&&= y_n + 1 - \frac{2}{y_n} \\ &&&> y_n \end{align*} Therefore if \(y_n \in (2,3)\) we have \(y_{n+1} \in ( y_n, 3)\) and so \(y_n\) is strictly increasing and bounded.

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Solution: \begin{align*} && v^2 &= u^2 + 2as \\ (\uparrow): && 0 &= (u \sin \theta)^2 - 2gh \\ \Rightarrow && h &= \frac{u^2 \sin^2 \theta}{2g} \end{align*} To avoid hitting the ceiling \begin{align*} && \frac9{10}d &>\frac{u^2 \sin^2 \theta}{2g} \\ \Rightarrow && \frac{gd}{u^2}&>\frac{5\sin^2 \theta}{9} \\ \end{align*} In order to pass through \(P\) we need \begin{align*} && d &= u \cos \theta t \\ && \frac12 d &= u \sin \theta t - \frac12 g t^2 \\ \Rightarrow && \frac12 &= \tan \theta - \frac12 \frac{g d}{u^2} \sec^2 \theta \\ &&&<\tan \theta - \frac12 \frac{5 \sin^2 \theta}{9\cos^2 \theta} \\ \Rightarrow && 0 & >5 \tan^2 \theta -18\tan \theta +9 \\ &&&= (5\tan \theta - 3)(\tan \theta - 3) \\ \\ \Rightarrow && \tan \theta &\in \left ( \frac35, 3\right) \\ && \theta &= \left (\arctan \tfrac35, \arctan 3 \right) \end{align*}

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Solution:

\begin{align*} \text{N2}(\rightarrow): && F_A - F_C &= 0\\ && F_A &= F_C \\ \text{N2}(\uparrow): && R_A + R_C - 7W - W - 4W &= 0\\ && R_A + R_C &= 12W \\ \overset{\curvearrowright}{A}: && \frac{1}{6}7W + \frac{1}{4}W + \frac{3}{4}4W - R_C &= 0 \\ \Rightarrow && R_C &= \frac{53}{12}W\\ \Rightarrow && R_A = 12W - \frac{53}{12}W &= \frac{91}{12}W \\ \overset{\curvearrowleft}{B}(AB): && \frac{1}{2}W + \frac{2}{3}7W - R_A+\sqrt{3}F_A &= 0 \\ \Rightarrow && F_A = \frac{1}{\sqrt{3}} \l \frac{91}{12}-\frac12-\frac{14}3\r W &= \frac{29}{12\sqrt{3}}W \end{align*} We know that the system is about to slip, so equality holds in one of \(F_A \leq \mu R_A\) or \(F_C \leq \mu R_C\). Since \(F_A = F_C\), we know it must occur for whichever of \(\mu R_A\) and \(\mu R_C\) is smaller. Since \(R_C\) is much smaller, this must be the ladder about to slip \(BC\) and \[ \mu = \frac{F_C}{R_C} = \frac{\frac{29}{12\sqrt{3}}W}{\frac{53}{12}W} = \boxed{\frac{29}{53\sqrt{3}}}\]

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Solution: \begin{align*} && \frac25 &= \frac{\lambda p}{\lambda p + (1-\lambda) q} \\ \Rightarrow && 2(1-\lambda)q &= 3\lambda p \\ \Rightarrow && \lambda(3p+2q) &= 2q \\ \Rightarrow && \lambda &= \frac{2q}{3p+2q} \end{align*} \begin{align*} && \frac25 &= \frac{\lambda (p + \frac14(1-p))}{\lambda (p + \frac14(1-p))+(1-\lambda) (q + \frac14(1-q))} \\ &&&= \frac{\lambda(\frac34p + \frac14)}{\lambda(\frac34p + \frac14)+(1-\lambda)(\frac34q + \frac14)} \\ \Rightarrow && \lambda &= \frac{2(\frac34q+\frac14)}{3(\frac34p + \frac14)+2(\frac34q+\frac14)} \\ &&&= \frac{\frac32q + \frac12}{\frac94p + \frac32q + \frac54} \\ &&&= \frac{6q+2}{9p+6q+5} \end{align*}

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Solution: \begin{align*} \P(X \leq 0.8) &= \P(X_1 \leq 0.8,X_2 \leq 0.8,X_3 \leq 0.8) \\ &= 0.8^3 \\ &= 0.512 \end{align*} \begin{align*} && \P(X < c) &= c^3 \\ \Rightarrow && f_X(x) &= 3x^2 \\ \Rightarrow && \E[X] &= \int_0^1 x \cdot (3x^2) \, dx \\ && &= \left [ \frac{3}{4}x^4 \right]_0^1 \\ &&&= \frac{3}{4} \end{align*} \(X\) is distributed the maximum of \(N\) numbers on \([0,a]\). \begin{align*} H_0 : & x= 1 \\ H_1 : & x < 1 \end{align*} \begin{align*} &&\P(X < c) &= c^N \\ &&&= \frac1{20} \\ \Rightarrow && N &= -\frac{\log(20)}{\log(c)} \end{align*} where \(c = 0.8\), we have \begin{align*} N &= \frac{\log(20)}{\log(5/4)} \\ &= \frac{\log(5)+\log(4)}{\log(5)-\log(4)} \\ &= \frac{ \frac{\log(5)}{\log(4)}+1}{\frac{\log(5)}{\log(4)} - 1} \end{align*} \begin{align*} && 2^{10} &\approx 10^{3} \\ && 10\log(2) &\approx 3 (\log(5) + \log(2)) \\ && 7\log(2) &\approx 3 \log(5) \\ && \frac{\log(5)}{2\log(2)} &\approx \frac{7}{6} \end{align*} \begin{align*} &= \frac{ \frac{\log(5)}{\log(4)}+1}{\frac{\log(5)}{\log(4)} - 1} &= \frac{\frac{7}{6} + 1}{\frac{7}{6} -1} \\ &= 13 \end{align*} Since \(2^{10} > 10^3\) then \(N=14\) is the value we seek. \(\P(X < 0.8 | a= 0.8) = 1\) \(\P(X < 0.8 | a= 0.9, N=14) = \frac{8^{14}}{9^{14}}\)

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Solution:

1. The first pirate will have some gold coins as long as the very first coin drawn is a gold coin, ie \(\frac{n}{n+2}\).
2. The probability the second pirate will have some gold coins is the probability the two lead coins are separate. There are \((n+2)!\) ways to arrange the coins, and there are \((n+1)! \cdot 2\) ways to arrange the coins where they are together, therefore the probability is: \[ 1 - \frac{2(n+1)!}{(n+2)!} = 1 - \frac{2}{n+2} = \frac{n}{n+2} \]
3. For all three pirates to have some gold coins we need the lead coins to be separate and not first or last. If we line up all \(n\) gold coins, there are \(n-1\) gaps between them we could place the \(2\) lead coins in, therefore \(\binom{n-1}{2}\) ways to place the lead coins with the restriction. Without the restriction there are \(\binom{n+2}{2}\) ways to choose where to put the coins, therefore \begin{align*} && P &= \frac{\binom{n-1}{2}}{\binom{n+2}{2}} \\ &&&= \frac{(n-1)(n-2)}{(n+2)(n+1)} \end{align*} [Notice this is clearly \(0\) if there are only \(1\) or \(2\) gold coins]

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Solution:

1. \(\,\) \begin{align*} && 0 &= \sqrt{3x^2+1} + \sqrt{x} -2x-1 \\ \Rightarrow && 2x+1 &= \sqrt{3x^2+1} + \sqrt{x} \\ \Rightarrow && 4x^2+4x+1 &= 3x^2+1+x+2\sqrt{x(3x^2+1)} \\ \Rightarrow && x^2+3x &= 2\sqrt{x(3x^2+1)} \\ \Rightarrow && x^2(x^2+6x+9) &= 4x(3x^2+1) \\ \Rightarrow && 0 &= x^4-6x^3+9x^2-4x \\ &&&= x(x-4)(x-1)^2 \end{align*} So clearly we have \(x = 0, x = 1, x = 4\). \(x = 0\) works, \(x = 1\) works, \(x = 4\) works.
2. \(\,\) \begin{align*} && 0 &= \sqrt{3x^2+1} - 2\sqrt{x} +x-1 \\ \Rightarrow && 1-x &= \sqrt{3x^2+1}-2\sqrt{x} \\ \Rightarrow && x^2-2x+1 &= 3x^2+1+4x-4\sqrt{x(3x^2+1)} \\ \Rightarrow && x^2+3x &= 2\sqrt{x(3x^2+1)} \\ \Rightarrow && 0 &= x(x-4)(x-1)^2 \end{align*} Again we must check \(x = 0, x = 1, x = 4\). \(x = 0,1\) work, but \(x = 4\) is not a solution.
3. \(\,\) \begin{align*} && 0 &= \sqrt{3x^2+1} - 2\sqrt{x} -x+1 \\ \Rightarrow && x-1 &= \sqrt{3x^2+1} - 2\sqrt{x} \\ \Rightarrow && x^2-2x+1 &= 3x^2+1+4x-4\sqrt{x(3x^2+1)} \\ \Rightarrow && x^2+3x &= 2\sqrt{x(3x^2+1)}\\ \Rightarrow && 0 &= x(x-4)(x-1)^2 \end{align*} So again, we need to check \(x = 0, 1, 4\). \(x = 0, 4\) work, but \(x = 1\) fails.

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Solution: There are two cases to consider by they are equivalent to \(x^2 \pm \alpha x + 2 < 0\), which has no solution solutions if \(\Delta < 0\), ie if \(\alpha^2 - 4\cdot1\cdot2 < 0 \Leftrightarrow |\alpha| < 2\sqrt{2}\). If \(\alpha = 3\), we have \begin{align*} && 0 & > x^2-3x+2 \\ &&&= (x-2)(x-1) \\ \Rightarrow && x & \in (1,2) \\ \\ && 0 &> x^2+3x+2 \\ &&& = (x+2)(x+1) \\ \Rightarrow && x &\in (-2,-1) \end{align*} Both cases work here, so \(x \in (-2, -1) \cup (1,2)\). \begin{align*} && 0 &> x^2 \pm \alpha x + 2 \\ &&&= (x \pm \tfrac{\alpha}{2})^2 -\frac{\alpha^2-8}{4} \end{align*} The potential intervals therefore are \((\frac{\alpha -\sqrt{\alpha^2-8}}{2}, \frac{\alpha +\sqrt{\alpha^2-8}}{2})\) and \((\frac{-\alpha -\sqrt{\alpha^2-8}}{2}, \frac{-\alpha +\sqrt{\alpha^2-8}}{2})\). Neither of these intervals overlap with \(0\), since \(\alpha^2 > \alpha^2-8\), and their lengths are both \(\sqrt{\alpha^2-8}\), therefore \(S = 2\sqrt{\alpha^2-8} < 2\alpha\)

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Solution: \begin{align*} && y &= x(x+1)(x-2)^4 \\ \Rightarrow && y' &= (x+1)(x-2)^4+x(x-2)^4+4x(x+1)(x-2)^3 \\ &&&= (x-2)^3 \left ( (2x+1)(x-2)+4x(x+1) \right) \\ &&&= (x-2)^3 \left (2x^2-3x-2+4x^2+4x \right) \\ &&&=(x-2)^3(6x^2+x-2) \\ &&&=(x-2)^3(2x-1)(3x+2) \end{align*} Therefore there are stationary points at \((2,0), (\frac12, -\frac{625}{64}), (-\frac23, -\frac{4078}{81})\) \((0,2)\) is a minimum by considering the sign of \(y'\) either side. \( (-\frac23, \frac{2560}{729})\) is a minimum, since it's the first stationary point. \( (\frac12, \frac{243}{64})\) is a maximum since you can't have consecutive minima and the second derivative is clearly non-zero.

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