Problems

Solution:

1. \(\,\) \begin{align*} && 3 &= 2^2 - 1^2 \\ && 5 &= 3^2 - 2^2 \\ && 8 &= 3^2 - 1^2 \\ && 16 &= 5^2 - 3^2 \end{align*}
2. Suppose \(n = 2k+1\), then \(n = (k+1)^2 - k^2\)
3. Suppose \(n = 4k\) then \(n = (2k+1)^2 - (2k-1)^2\)
4. All squares leave a remainder of \(0\) or \(1\) on division by \(4\). Therefore the difference can leave a remainder of \(0\), \(1\), \(-1 \equiv 3\), none of which are \(2\).
5. Suppose \(n = pq = a^2 - b^2\) with \(a > b\) ie \((a-b)(a+b) = pq\). Since \(p\) is prime, \(p \mid (a-b)\) or \(p \mid (a+b)\). Similarly for \(q\). Suppose also (wlog) that \(p > q\) Since the factors of \(pq\) are \(1, p, q, pq\) then \(a-b = 1, p\) (which are two possibilities) and \(a+b = pq, q\), ie \(a = \frac{1+pq}{2}, \frac{p+q}{2}\) and \(b = \frac{pq-1}{2}, \frac{p-q}{2}\) \begin{align*} && pq &= \left ( \frac{1+pq}{2} \right)^2- \left ( \frac{1-pq}{2} \right)^2 \\ &&&= \left ( \frac{p+q}{2} \right)^2- \left ( \frac{p-q}{2} \right)^2 \\ \end{align*} Where everything is an integer since \(p\) and \(q\) are odd. If we have \(p > 2\) and \(q = 2\) then \(p\) is odd and the number has the form \(4k+2\) which cannot be expressed as the difference of two squares.
6. \(675 = 3^3 \cdot 5^2\), each factor pair of \(675\) will lead to a different solution of \(675 = a^2-b^2\), since we will have an equation \(a-b = X, a+b = Y\) where \(X, Y\) are both odd. Therefore there are as many solution as (half) the number of factors, ie \(4 \times 3 = 12\)

Solution:

1. \(\,\) \begin{align*} && \int \ln (2- x) \d x &= \left [ (x-2) \ln (2- x) \right] - \int -\frac{x-2}{2-x} \d x \\ && \int \ln (2- x) \d x &= \left [ (x-2) \ln (2- x) \right] + \int 1 \d x \\ &&&= -(2-x) \ln (2-x) +(2-x) + C \end{align*}
2. \(\,\)
   

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3. \begin{align*} && \text{Area} &= \int_0^{\sqrt{3}} \ln | x^2 - 4 | \d x \\ &&&= \int_0^\sqrt{3} \ln(4-x^2) \d x \\ &&&= \int_0^\sqrt{3} \left ( \ln(2-x) + \ln (2+x) \right) \d x \\ &&&= \left [ -(2-x) \ln (2-x) +(2-x) +(2+x)\ln(2+x)-(2+x)\right]_0^{\sqrt{3}} \\ &&&= \left ( -(2-\sqrt{3}) \ln (2-\sqrt{3}) +(2-\sqrt{3}) +(2+\sqrt{3})\ln(2+\sqrt{3})-(2+\sqrt{3}) \right) - \\ &&&\quad \quad \left (- 2\ln (2)+2 +2\ln(2)-2 \right) \\ &&&=\left ( -(2-\sqrt{3}) \ln \left ( \frac{1}{2+\sqrt{3}} \right) -2\sqrt{3} +(2+\sqrt{3})\ln(2+\sqrt{3}) \right) \\ &&&= 4\ln(2 + \sqrt{3}) - 2 \sqrt{3} \end{align*}
4. 

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   \begin{align*} && \text{Area} &= 2 \left ( \int_0^\sqrt{3} \ln (4-x^2) \d x - \lim_{t \to 2}\int_{\sqrt{3}}^t \ln(4-x^2) \d x \right) \\ &&&= 8\ln(2 + \sqrt{3})-4\sqrt{3} - 2 \lim_{t \to 2}\int_{\sqrt{3}}^t \left ( \ln (2-x) + \ln (2+x) \right) \d x \\ &&&= 8\ln(2 + \sqrt{3})-4\sqrt{3} - 2 \lim_{t \to 2} \left [ -(2-x) \ln (2-x) +(2-x) +(2+x)\ln(2+x)-(2+x)\right]_{\sqrt{3}}^{t} \\ &&&= 16 \ln(2+\sqrt{3})-8\sqrt{3} - 2 \lim_{t \to 2} \left(-(2-t) \ln (2-t) +(2-t) +(2+t)\ln(2+t)-(2+x) \right) \\ &&&= 16 \ln(2+\sqrt{3})-8\sqrt{3} -2(4 \ln4-4) \\ &&&= 16 \ln(2 + \sqrt{3}) - 16 \ln 2 +8(1-\sqrt{3}) \end{align*}

Solution:

1. \(\,\) \begin{align*} && \int_0^b x^2 \d x &= \left ( \int_0^b x \d x \right)^2 \\ \Rightarrow && \frac{b^3}{3} &= \left ( \frac{b^2}{2} \right)^2 \\ \Rightarrow && b &= \frac{4}{3} \end{align*}
2. \(\,\) \begin{align*} && \int_1^b x^2 \d x &= \left ( \int_1^b x \d x \right)^2 \\ \Rightarrow && \frac{b^3}{3} - \frac{1}{3} &= \left ( \frac{b^2}{2} - \frac{1}{2} \right)^2 \\ \Rightarrow && 4(b^3 - 1) &= 3(b^2-1)^2 \\ \Rightarrow && 4(b^3-1) &= 3(b^4-2b^2+1) \\ \Rightarrow && 0 &= 3b^4-4b^3-6b^2+7 \\ &&&= (b-1)(3b^3-b^2-7b-7) \\ \Rightarrow && 0 &= 3b^3-b^2-7b-7 \end{align*}
   

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   Let \(f(x) = 3x^3-x^2-7x-7\) then \(f(2) = 3 \cdot 8 - 4 - 14 - 7 = -1 < 0\), \(f(3) = 3 \cdot 27 - 9 - 21 - 7 = 44 > 0\) therefore the root must lie between \(2\) and \(3\).
3. \(,\) \begin{align*} && \int_a^b x^2 \d x &= \left ( \int_a^b x \d x \right)^2 \\ \Rightarrow && \frac{b^3}{3} - \frac{a^3}{3} &= \left ( \frac{b^2}{2} - \frac{a^2}{2} \right)^2 \\ \Rightarrow && 4(b^3 - a^3) &= 3(b^2-a^2)^2 \\ \Rightarrow && 4(b^2+ab+a^2) &= 3(b-a)(b+a)^2 \\ \Rightarrow && 4 \left ( \left ( \frac{p+q}{2}\right)^2+\left ( \frac{p+q}{2}\right)\left ( \frac{p-q}{2}\right)+\left ( \frac{p-q}{2}\right)^2\right) &= 3qp^2 \\ \Rightarrow && 3p^2 + q^2 &= 3qp^2 \\ \Rightarrow && 3p^2(q-1) &= q^2 \\ \Rightarrow && p^2 &= \frac{q^2}{3(q-1)} \\ \Rightarrow && 1 &\leq \frac{1}{3(q-1)} \\ \Rightarrow && 3(q-1) &\leq 1 \\ \Rightarrow && q & \leq \frac{4}{3} \\ \end{align*}

Solution: The position of the hands are \(\begin{pmatrix} a\sin(-t) \\ a \cos(-t) \end{pmatrix}\) and \(\begin{pmatrix} b\sin(-60t) \\ b \cos(-60t) \end{pmatrix}\), the distance between the hands is \begin{align*} x &= \sqrt{\left ( a \sin t - b \sin 60t\right)^2+\left ( a \cos t - b \cos 60t\right)^2} \\ &= \sqrt{a^2+b^2-2ab\left (\sin t \sin 60t+\cos t \cos 60t \right)} \\ &= \sqrt{a^2+b^2-2ab \cos(59t)} = \sqrt{a^2+b^2-2ab \cos \theta} \\ \\ \frac{\d x}{\d \theta} &= \frac{ab \sin \theta}{ \sqrt{a^2+b^2-2ab \cos \theta}} \\ \frac{\d^2 x}{\d \theta^2} &= \frac{ab \cos \theta\sqrt{a^2+b^2-2ab \cos \theta} - \frac{a^2b^2 \sin^2 \theta}{\sqrt{a^2+b^2-2ab \cos \theta}} }{a^2+b^2-2ab \cos \theta} \\ &= \frac{ab \cos \theta(a^2+b^2-2ab \cos \theta) - a^2b^2 \sin^2 \theta }{(a^2+b^2-2ab \cos \theta)^{3/2}} \\ &= \frac{ab \cos \theta(a^2+b^2-2ab \cos \theta) - a^2b^2(1-\cos^2 \theta)}{(a^2+b^2-2ab \cos \theta)^{3/2}} \\ &= \frac{ab(a^2+b^2) \cos \theta-a^2b^2 \cos \theta- a^2b^2}{(a^2+b^2-2ab \cos \theta)^{3/2}} \\ &= \frac{-ab(a\cos \theta -b)(b \cos \theta - a)}{(a^2+b^2-2ab \cos \theta)^{3/2}} \\ \end{align*} So the rate of increase is largest when \(\cos \theta = \frac{a}{b}\) (since \(\frac{b}{a}\) is impossible. Therefore when \(x = \sqrt{a^2+b^2-2ab \frac{a}{b}} = \sqrt{a^2+b^2-2a^2} = \sqrt{b^2-a^2}\) If \(b = 2a\) then \(\cos \theta = \frac{a}{2a} = \frac12 = \frac{\pi}{3} = 60^\circ\) The relative speed of the hands is \(5.5^\circ\) per minute, so \(\frac{60}{5.5} = \frac{120}{11} \approx 11\) but clearly also less than since \(121 = 11^2\).

Solution:

1. Suppose \(u_0 = u = \sin^2 \theta\) then \begin{align*} && u_1 &= 4 u_0 (1-u_0) \\ &&&= 4 \sin^2 \theta ( 1- \sin^2 \theta) \\ &&&= 4 \sin^2 \theta \cos^2 \theta \\ &&&= (2 \sin \theta \cos \theta)^2 \\ &&&= (\sin 2 \theta)^2 = \sin^2 2 \theta \\ \\ && u_2 & = 4u_1 (1-u_1) \\ &&&= 4 \sin^2 2\theta \cos^2 2 \theta \\ &&&= \sin^2 4 \theta \end{align*} Claim: \(u_n = \sin^2 2^n \theta\). Proof: (By Induction) Base case is clear, suppose it's true for \(n=k\), then \begin{align*} && u_{k+1} &= 4u_k(1-u_k) \\ &&&= 4 \sin^2 2^k \theta(1-\sin^2 2^k \theta) \\ &&&= (2 \sin 2^k \theta \cos 2^k \theta)^2 \\ &&&= (\sin 2^{k+1} \theta)^2 \\ &&&= \sin^2 2^{k+1} \theta \end{align*} Therefore since it is true for \(n = 1\) and if it's true for \(n = k\) it is true for \(n=k+1\) it must be true for all \(k\).
2. Suppose \(v_n = \alpha u_n + \beta\) then \begin{align*} && (\alpha u_{n+1}+\beta) &= -p(\alpha u_n + \beta)^2 + q(\alpha u_n + \beta) + r \\ &&&= -p\alpha^2u_n^2+\alpha(q-2p\beta) u_n -p \beta^2 +q \beta+r \\ \Rightarrow && u_{n+1} &= u_n(q-2p\beta -p \alpha u_n) -(p\beta^2-(q-1)\beta-r) \end{align*} So if \(\alpha = \frac{4}{p}\) and \(q-2p\beta = 4\) ie \(\beta = \frac{q-4}{2p}\) then we also need the constant term to vanish, ie \begin{align*} 0 &&&= p\beta^2-(q-1)\beta+r \\ &&&= p \left (\frac{q-4}{2p} \right)^2 - (q-1) \frac{q-4}{2p} - r \\ \Rightarrow && 0 &= p(q-4)^2 -(q-1)(q-4)2p - 4p^2r \\ \Rightarrow && 0 &= (q-4)^2-2(q-1)(q-4)-4pr \\ &&&= q^2-8q+16-2q^2+10q-8-4pr \\ \Rightarrow && 4pr &= -q^2+2q+8 \end{align*} Suppose \(v_{n+1} = -v_n^2 + 2v_n +2\) then since \(4\cdot 1 \cdot 2 = 8\) and \(8 + 4 -4 = 8\) we can apply our method. \(v_n = 4u_n + \frac{-2}{2} = 4u_n -1 = 4\sin^2 (2^{n-1} \pi)-1\)

Solution:

1. The story for first ball (before it collides with the second ball) will be exactly the same, ie it rebounds with speed \(eV\). For the second ball, it will also have fallen a distance \(H\) and will be travelling with the same speed \(V\). Their speed of approach therefore will be \((1+e)V\), and the speed of separating therefore must be \(e(1+e)V\) Given the centre of the second ball reaches a height of \(h\) (from a position of height) \(2R+r\), we must have: \begin{align*} && v^2 &= u^2 + 2as \\ && 0 &= w^2 - 2g(h - 2R-r) \\ \Rightarrow && w^2 &= 2g(h-2R-r) \end{align*} Taking upwards to be positive, then we have:
   

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   \begin{align*} \text{COM}: && MeV + m(-V) &= M(w-e(1+e)V) + mw \\ \Rightarrow && V(Me-m)+e(1+e)MV &= w(M+m) \\ \Rightarrow && w &= \frac{2Me+e^2M-m}{M+m} V \\ \Rightarrow && w^2 &= \left ( \frac{M/m(2e+e^2)-1}{M/m+1}\right)^2 V^2 \\ \Rightarrow && 2g(h-2R-r) &= \left ( \frac{M/m(2e+e^2)-1}{M/m+1}\right)^2 2gH \\ \Rightarrow && \left ( \frac{M/m(2e+e^2)-1}{M/m+1}\right)^2 &= \frac{h-2R-r}{H} \\ &&&= \frac{4.5-0.4-0.05}{1.8} \\ &&&= \frac{9}{4} \\ \Rightarrow && \frac{M/m(\frac43+\frac49)-1}{M/m+1} &= \frac32 \\ \Rightarrow && \frac{16}9M/m-1 &= \frac32 M/m+\frac32 \\ \Rightarrow && \frac{5}{18}M/m &= \frac{5}{2} \\ \Rightarrow && M/m &= 9 \end{align*}