Problems

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Solution:

The curves touch when \(\sin x = 1\) ie \(x = \frac{4n+1}{2} \pi\). therefore \begin{align*} && I &= \int e^{-x} \sin x \d x \\ &&&= -e^{-x} \sin x + \int e^{-x} \cos x \d x \\ &&&= -e^{-x} \sin x -e^{-x} \cos x - I \\ && I &= -\frac{e^{-x}}2 ( \sin x + \cos x)\\ \\ && A_n &= \int_{\frac{4n+3}{2}\pi}^{\frac{4n-1}{2}\pi} e^{-x} (1-\sin x) \d x \\ &&&= \left [ -e^{-x} - \frac{e^{-x}}2(\sin x + \cos x) \right]_{\frac{4n+1}{2}\pi}^{\frac{4n+5}{2}\pi} \\ &&&= -\frac12\exp \left ( -\frac{4n+1}{2}\pi\right)+ \frac12\exp \left ( -\frac{4n-3}{2}\pi\right) \\ &&&= \frac12 (e^{2 \pi}-1) \left ( -\frac{4n+1}{2}\pi\right) \\ \\ && \sum_{n=1}^\infty A_n &= \frac12(e^{2\pi}-1)e^{-\pi/2} \sum_{n=1}^\infty e^{-2n\pi} \\ &&&= \frac12(e^{2\pi}-1)e^{-\pi/2} \frac{e^{-2\pi}}{1-e^{-2\pi}} \\ &&&= \frac12e^{-\pi/2} \end{align*}

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Solution:

1. 

   + - ↺
   Since the collision is perfectly elastic, the speed of approach and separation are equal, ie \(v_B = v_A - u\) \begin{align*} \text{COM}: && bmu &= bm(v_A - u) + mv_A \\ \Rightarrow && (b+1)v_A &= 2bu \\ \Rightarrow && v_A &= \frac{2b}{b+1} u \end{align*} Since \(0 < \frac{b}{b+1} < 1\), the largest \(0 < v_A < 2u\)
2. After the first collision with each \(B_i\) we will have \(\displaystyle v_{i+1} = \frac{2\lambda}{\lambda + 1}v_i\), ie \(\displaystyle v_{i+1} = \left (\frac{2\lambda}{\lambda + 1} \right)^i u\) and so \(\displaystyle v_A = \left (\frac{2\lambda}{\lambda + 1} \right)^n u\) which can be arbitrarily large. Suppose \(\lambda = 4\), then \begin{align*} && 20u &< v_A \\ &&&= \left (\frac{8}{5} \right)^n u \\ \Rightarrow && \log_{10} 20 < n \log_{10}(16/10) \\ && \log_{10} 2 + 1 < n 4\log_{10} 2 - n \\ \Rightarrow && n &> \frac{ \log_{10} 2 + 1}{ 4\log_{10} 2 - 1} \\ &&&\approx \frac{0.30103+1}{4 \times 0.30103 -1}\\ &&&= \frac{1.30103}{0.20412} \\ &&&>6 \end{align*} So \(n =7\) is the smallest possible

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Solution: \begin{align*} && 1 &= \int_0^1 f(x) \d x \\ &&&= ak + b(1-k) \\ &&&= b + (a-b)k \\ \Rightarrow && k &= \frac{1-b}{a-b} \\ \Rightarrow && b & < 1 \tag{\(0 < k, \,a > b\)} \\ && k &> 1 \\ \Rightarrow && a-b & > 1-b \\ \Rightarrow && a > 1 \end{align*}

1. \(\,\) \begin{align*} && \E[X] &= \int_0^1 x \cdot f(x) \d x \\ &&&= \int_0^k ax \d x + \int_k^1 b x \d x \\ &&&= a \frac{k^2}{2} + b \frac{1-k^2}{2} \\ &&&= \frac12b + (a-b) \frac{(1-b)^2}{2(a-b)^2} \\ &&&= \frac{(1-b)^2+b(a-b)}{2(a-b)} \\ &&&= \frac{1-2b+ab}{2(a-b)} \end{align*}
2. \(\,\) The median \(M\) satisfies \[\frac12 = \int_0^M f(x) \d x \] If \(ka = \frac{a-ab}{a-b} \leq \frac12 \Leftrightarrow 2a-2ab \leq a-b \Leftrightarrow a+b \leq 2ab\) then \(M > k\) otherwise \(M < k\). In the latter case: \begin{align*} && \frac12 &= Ma \\ \Rightarrow && M &= \frac{1}{2M} \end{align*} In the former case \begin{align*} && \frac12 &= ka + (M-k)b \\ &&&= k(a-b) + Mb \\ &&&= 1-b + M b \\ \Rightarrow && M &= 1-\frac1{2b} \end{align*}

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Solution:

1. If she plays \(P\) twice her probability is \(q(p^2+2p(1-p)) = qp(2-p)\). If she plays \(Q\) twice her probability is \(pq(2-q)\). Since \(p < q\) she should play \(P\) twice.
2. Under strategy 1, her probability is \(q(p^3+3p^2(1-p)+3p(1-p)^2) = qp(p^2+3p-3p^2+3-6p+3p^2) = qp(3-3p+p^2)\) Under strategy 2 her probability is \((p^2+2p(1-p))(q^2+2q(1-q)) = pq(2-p)(2-q)\). Under strategy 3 her probability is \(qp(3-3q+q^2)\) \begin{align*} && q - p &> \frac12 \\ \Rightarrow && (2-p)(2-q) & < (2-p)(\frac32 - p) \\ &&&= 3 - \frac72p + p^2 \\ &&&< 3- 3p + p^2 \end{align*} Therefore Strategy 1 dominates if \(q-p > \frac12\). If \(p = \frac14, q = \frac12\) then \((2-p)(2-q) =\frac74 \cdot \frac32 = \frac{21}{8}\) and \(3-3p + p^2 = 3 - \frac34 + \frac1{16} = \frac{48-12+1}{16} = \frac{37}{16} < \frac{42}{16}\) so strategy 2 dominates. Notice that strategy 1 always dominates strategy 3 since \(f(x) = 3-3x+x^2\) is decreasing for \(x < 1.5\). If \(p = \frac14, q = \frac12\) then \((2-p)(2-q) =\frac74 \cdot \frac32 = \frac{21}{8}\) and \(3-3p + p^2 = 3 - \frac34 + \frac1{16} = \frac{48-12+1}{16} = \frac{37}{16} < \frac{42}{16}\) so strategy 2 dominates. For strategy 1 to dominate, we need \(3-3p+p^2 > (2-q)(2-p)\) or \(\frac{3-3p+p^2}{2-p} > 2-q\). When \(p = \frac12\) this is \(\frac{3-\frac32 + \frac14}{2 - \frac12} = \frac{\frac{7}{4}}{\frac{3}{2}} = \frac76 = 2-\frac{5}{6}\) so take any value of \(q\) larger than \(\frac56\).

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Solution:

1. \(\cosh a = \frac12 (e^a + e^{-a})\) \begin{align*} \int_0^1 \frac 1{ x^2 +2x\cosh a +1} \, \d x &= \int_0^1 \frac{1}{x^2+(e^a+e^{-a})x+e^ae^{-a}} \d x \\ &= \int_0^1 \frac{1}{e^a-e^{-a}}\left (\frac{1}{x+e^{-a}}-\frac{1}{x+e^a} \right)\d x \\ &= \frac{1}{2 \sinh a} \int_0^1 \left (\frac{1}{x+e^{-a}}-\frac{1}{x+e^a} \right)\d x \\ &= \frac{1}{2 \sinh a}\left [\ln(x+e^{-a})-\ln(x+e^a) \right]_0^1 \\ &= \frac{1}{2 \sinh a} \left (\ln(1+e^a)-\ln(1+e^{-a}) - (\ln e^{-a}-\ln e^a) \right) \\ &= \frac{1}{2\sinh a}\left (2a + \ln \frac{1+e^a}{1+e^{-a}}\right) \\ &= \frac1{2\sinh a} \left ( 2a -a \right) \\ &= \frac{a}{2 \sinh a} \end{align*}
2. \begin{align*} \int_1^\infty \frac 1 {x^2 +2x \sinh a -1} \,\d x &= \int_1^{\infty} \frac{1}{(x+e^a)(x-e^{-a})} \d x \\ &= \int_1^{\infty} \frac{1}{e^a+e^{-a}} \left ( \frac{1}{x-e^{-a}} - \frac{1}{x+e^{a}} \right)\d x \\ &= \frac{1}{2\cosh a} \int_1^{\infty} \left ( \frac{1}{x-e^{-a}} - \frac{1}{x+e^{a}} \right)\d x \\ &= \frac{1}{2\cosh a} \left [\ln(x-e^{-a}) - \ln (x + e^{a} ) \right]_1^{\infty} \\ &= \frac1{2\cosh a} \left [ \ln \frac{x-e^{-a}}{x+e^{a}} \right]_1^{\infty} \\ &= \frac{1}{2\cosh a} \left ( 0 - \ln \frac{1-e^{-a}}{1+e^a}{}\right) \\ &= \frac{1}{2\cosh a} \ln \frac{1+e^a}{1-e^{-a}}\\ &= \frac{1}{2\cosh a} \left ( a + \ln \coth \frac{a}{2} \right) \end{align*} and \begin{align*} \int_0^\infty \frac 1 {x^4 +2x^2\cosh a +1} \,\d x &= \int_0^\infty\frac{1}{(x^2+e^a)(x^2+e^{-a})} \d x \\ &= \int_0^\infty \frac{1}{e^a-e^{-a}} \left ( \frac{1}{x^2+e^{-a}} - \frac{1}{x^2+e^{a}} \right) \d x \\ &= \frac{1}{2\sinh a} \left [ \frac{1}{e^{-a/2}} \tan^{-1} \frac{x}{e^{-a/2}} - \frac{1}{e^{a/2}}\tan^{-1} \frac{x}{e^{a/2}} \right]_0^{\infty} \\ &= \frac{1}{2\sinh a} \left (e^{a/2}\frac{\pi}{2}-e^{-a/2}\frac{\pi}{2} - 0 \right) \\ &= \frac{1}{2\sinh a} \pi \sinh \frac{a}{2} \\ &= \frac{\pi \sinh \tfrac{a}{2}}{2\sinh a} \\ &= \frac{\pi \sinh \tfrac{a}{2}}{4\sinh \tfrac{a}{2} \cosh \tfrac{a}{2}} \\ &= \frac{\pi}{4\cosh \tfrac{a}{2}} \end{align*}

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Solution: The primitive 4th roots of unity are \(i\) and \(-i\). (Since the other two roots of \(x^4-1\) are also roots of \(x^2-1\) \({\rm C}_4(x) = (x-i)(x+i) = x^2+1\) as required.

1. \(\,\) \begin{align*} && {\rm C}_1 (x) &= x-1 \\ && {\rm C}_2 (x) &= x+1 \\ && {\rm C}_3 (x) &= x^2+x+1 \\ && {\rm C}_5 (x) &= x^4+x^3+x^2+x+1 \\ && {\rm C}_6 (x) &= x^2-x+1 \\ \end{align*}
2. Since \((x^4+1)(x^4-1) = x^8-1\) we must have \(n \mid 8\). But \(n \neq 1,2,4\) so \(n = 8\).
3. \({\rm C}_p(x) = x^{p-1} +x^{p-2}+\cdots+x+1\)
4. Suppose \({\rm C_q}(x) \equiv {\rm C}_r(x){\rm C}_s(x)\), then if \(\omega\) is a primitive \(q\)th root of unity we must \({\rm C}_q(\omega) = 0\), but that means that one of \({\rm C}_r(\omega)\), \({\rm C}_s(\omega)\) is \(0\). But that's only possible if \(r\) or \(s\) \(=q\). If this were the case, then what would the other value be? There are no possible values, hence it's not possible.

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Solution:

1. \begin{align*} && 0 &= \alpha^2 + a \alpha + b \tag{1} \\ && 0 &= \alpha^2 + c \alpha + d \tag{2} \\ \\ (1) - (2): && 0 & =\alpha ( a-c) + (b-d) \\ \Rightarrow && \alpha &= - \frac{b-d}{a-c} \tag{\(a\neq c\)} \end{align*} (\(\Rightarrow\)) Suppose they have a common root, then given we know it's form, we must have: \begin{align*} && 0 &= \left ( - \frac{b-d}{a-c} \right)^2 +a\left ( - \frac{b-d}{a-c} \right) + b \\ \Rightarrow && 0 &= (b-d)^2 - a(b-d)(a-c) + b(a-c)^2 \end{align*} (\(\Leftarrow\)) Suppose the equation holds, then \begin{align*} && 0 &= (b-d)^2 - a(b-d)(a-c) + b(a-c)^2 \\ \Rightarrow && 0 &= \left ( - \frac{b-d}{a-c} \right)^2 +a\left ( - \frac{b-d}{a-c} \right) + b \\ \end{align*} So \(\alpha\) is a root of the first equation. Considering \((1) - (2)\) we must have that \(\alpha(a-c) +(b-d) = t\) (whatever the second equation is), but that value is clearly \(0\), therefore \(\alpha\) is a root of both equations. If \(a = c\) then the equation becomes \(0 = (b-d)^2\), ie the two equations are the same, therefore they must have common roots!
2. \begin{align*} && 0 &= x^2+ax+b \tag{1} \\ && 0 &= x^3+(a+1)x^2+qx+r \tag{2} \\ \\ (2) - x(1) && 0 &= x^2 + (q-b)x + r \tag{3} \end{align*} Therefore if the equations have a common root, \((1)\) and \((3)\) have a common root, ie \((b-r)^2-a(b-r)(a-(q-b))+b(a-(q-b))^2 = 0\) which is exactly our condition. \(a = \frac52, q = \frac52, r = \frac12\) \begin{align*} && 0 &= \left (b-\frac12 \right)^2 - \frac52\left (b-\frac12\right) b + b^3 \\ &&&= b^2 -b + \frac14 - \frac52 b^2+\frac54b + b^3 \\ &&&= b^3 -\frac32 b^2 +\frac14 b + \frac14 \\\Rightarrow && 0 &= 4b^3 - 6b^2+b + 1 \\ &&&= (b-1)(4b^2-2b-1) \\ \Rightarrow && b &= 1, \frac{1 \pm \sqrt{5}}{4}\end{align*}

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Solution:

Note that \(CP\) has equation \(\frac{y-0}{x-an} = \frac{a-0}{a-an} = \frac{1}{1-n} \Rightarrow y = \frac{x-an}{1-n}\) Therefore \(R = (0, -\frac{an}{1-n})\) Note that \(CQ\) has equation \(\frac{y-am}{x} = \frac{a-am}{a} = 1-m \Rightarrow y = (1-m)x + am\) Therefore \(S = (-\frac{am}{1-m}, 0)\) \(PQ\) has equation \(\frac{y}{x-an} = \frac{am-0}{0-an} \Rightarrow y = -\frac{m}{n}x +am\) \(SR\) has equation \(\frac{y}{x+\frac{am}{1-m}} = \frac{-\frac{an}{1-n}}{\frac{am}{1-m}} = -\frac{n(1-m)}{m(1-n)} \Rightarrow y =-\frac{n(1-m)}{m(1-n)} x -a\frac{n}{1-n}\) So \(PQ \cap SR\) has \begin{align*} && -\frac{m}{n}x +am &= -\frac{n(1-m)}{m(1-n)} x -a\frac{n}{1-n} \\ && x \left (\frac{n(1-m)}{m(1-n)} - \frac{m}{n} \right) &= -am - \frac{an}{1-n} \\ \Rightarrow && x \left ( \frac{n^2(1-m)-m^2(1-n)}{nm(1-n)} \right) &= -\frac{a(m(1-n)+n)}{1-n} \\ \Rightarrow && x \left ( \frac{(m-n)(mn-n-m)}{mn(1-n)} \right) &= \frac{a(mn-m-n)}{1-n} \\ \Rightarrow && x &= \frac{amn}{m-n} \\ && y &= -\frac{amn}{m-n} \end{align*} Therefore clearly \(TA\) is perpendicular to \(AC\) since they are the lines \(y = -x\) and \(y = x\) Given this method we can construct the perpendicular to the diagonal through the vertex. Doing this at \(A\) we can construct \(C'\) the reflection of \(C\) in \(AB\). We can do the same to find the reflection of \(A\) and so we have a square with sidelengths \(\sqrt{2}a\) and hence area \(2a^2\)

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Solution: \begin{align*} && y &= \cos(m \arcsin x) \\ && y' &= -m \sin (m \arcsin x) \cdot (1-x^2)^{-\frac12} \\ && y'' &= -m^2 \cos(m \arcsin x) \cdot (1-x^2)^{-1} -m \sin(m \arcsin x) \cdot (1-x^2)^{-\frac32} \cdot (-x) \\ &&&= -m^2 y (1-x^2)^{-1} + x(1-x^2)^{-1} y' \\ \Rightarrow && 0 &= (1-x^2)y'' - x y' + m^2y \\ \\ && 0 &= (1-x^2)y^{(3)} -2xy'' - xy''-y' + m^2y' \\ &&&= (1-x^2)y^{(3)} - 3xy'' + (m^2-1)y' \\ \\ && 0 &= (1-x^2)y^{(4)} - 2xy^{(3)} - 3xy^{(3)} - 3y^{(2)} + (m^2-1)y^{(2)} \\ &&&= (1-x^2)y^{(4)}- 5xy^{(3)} - (m^2-4)y^{(2)} \end{align*} Claim: \(0 = (1-x^2)y^{(n+2)} - (2n+1)y^{(n+1)} + (m^2-n^2)y^{(n)}\) Proof: (By induction) Clearly the first few base cases are true. Suppose it is true for some \(n\), then \begin{align*} && 0 &= (1-x^2)y^{(n+2)} - (2n+1)xy^{(n+1)} + (m^2-n^2)y^{(n)} \\ \Rightarrow && 0 &= (1-x^2)y^{(n+3)} - 2xy^{(n+2)} - (2n+1)xy^{(n+2)} - (2n+1)y^{(n+1)} + (m^2-n^2)y^{(n+1)} \\ &&&= (1-x^2)y^{(n+3)} - (2n+3)xy^{(n+2)} + (m^2-n^2-2n-1)y^{(n+1)} \\ &&&= (1-x^2)y^{(n+1+2)} - (2(n+1)+1)xy^{(n+1+1)} +(m^2-(n+1)^2)y^{(n)} \end{align*} And so we can conclude the result by induction. Notice that \begin{align*} && y(0) &= \cos(m 0) = 1 \\ && y'(0) &= -m\sin(m 0) = 0 \\ && y''(0) &= -m^2 y(0) = -m^2\\ \end{align*} Notice that \(y^{(n+2)}(0) + (m^2-n^2)y^{(n)} = 0\) so in particular all the odd terms will be \(0\) and the even terms will be \(1, -m^2, m^2(m^2-2^2), \cdots\), therefore \begin{align*} && \cos (m \arcsin x) &= 1 -\frac{m^2}{2!} x^2 + \frac{m^2(m^2-2^2)}{4!}x^4 - \cdots \\ \Rightarrow && \cos(m \theta) &= 1 - \frac{m^2}{2!} \sin^2 \theta + \frac{m^2(m^2-2^2)}{4!} \sin^4 \theta \end{align*} Notice that if \(m\) is even, then at some point we will have \(m^2-m^2\) appearing in our expansion and all remaining terms will be zero. Therefore we will end up with a polynomial series, of degree \(m\) in \(\sin \theta\)

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Solution: \begin{align*} && \int \frac{{\rm P} ( x)}{ \big( {\rm Q} ( x)\big )^ 2}\, \d x &= \int \frac{{\rm Q} (x){\rm R}'(x) - {\rm Q}'(x){\rm R}(x)}{ \big( {\rm Q} ( x)\big )^ 2}\, \d x \\ &&&= \int \frac{\d}{\d x} \left ( \frac{R(x)}{Q(x)} \right) \d x \\ &&&= \frac{R(x)}{Q(x)} + C \end{align*}

1. Suppose \(Q(x) = 1 + 2x + 3x^2, P(x) = 5x^2-4x-3\), and \(R(x) = a +bx + cx^2\), then \begin{align*} && 5x^2-4x-3 &= (1 + 2x + 3x^2)(2cx+b) - (6x+2)(a+bx+cx^2) \\ &&&= (6c-6c)x^3 + (3b+4c-6b-2c)x^2 + \\ &&&\quad+(2c+2b-6a-2b)x + (b-2a) \\ \Rightarrow && 2c-3b &= 5 \\ && 2c-6a &= -4 \\ && b - 2a &= -3 \\ \Rightarrow && b &= 2a - 3\\ && c &= 3a-2 \end{align*} So say \(a = 0, b = -3, c = -2\) we will have \begin{align*} \int \frac{5x^2 - 4x - 3} {(1 + 2x + 3x^2 )^2 } \, \d x &= \frac{-3x-2x^2}{1+2x+3x^2} + C \end{align*} Suppose we have a different value of \(a\), then we end up with: \begin{align*} \frac{a+(2a-3)x+(3a-2)x^2}{1+2x+3x^2} = a +\frac{-3x-2x^2}{1+2x+3x^2} \end{align*} So the different antiderivatives differ by a constant.
2. \(\,\) \begin{align*} && \frac{\d }{\d x} \left ( \frac{1}{1+\cos x + 2 \sin x } y\right) &= \frac{5-3\cos x + 4 \sin x }{(1+\cos x + 2 \sin x)^2} \\ \end{align*} We want to find \(R(x) = A \cos x + B\sin x + C\) such that \begin{align*} &&5-3\cos x + 4 \sin x &= (1+\cos x + 2 \sin x)R'(x) - R(x)(-\sin x + 2 \cos x) \\ &&&= (1+\cos x + 2 \sin x)(-A\sin x + B \cos x) \\ &&&\quad- (A\cos x + B \sin x + C)(-\sin x + 2 \cos x) \\ &&&=(-A+C) \sin x + (B-2C)\cos x +\\ &&&\quad\quad (2B-A+A-2B)\sin x \cos x \\ &&&\quad\quad (-2A+B)\sin^2x+(B-2A)\cos^2x \\ &&&= (-A+C)\sin x + (B-2C)\cos x +(B-2A) \\ \Rightarrow && B-2A &= 5\\ && C-A &= 4 \\ && B-2C &= -3 \\ \end{align*} There are many solutions so WLOG \(C=4, A = 0, B = 5\) and so \begin{align*} && \int \frac{5-3\cos x + 4 \sin x }{(1+\cos x + 2 \sin x)^2} \d x &= \frac{5\sin x +4}{1+\cos x + 2 \sin x} + K \\ \Rightarrow && y &= 5\sin x + 4 + K(1 + \cos x + 2 \sin x) \end{align*}