Problems

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Solution:

1. All factors of \(3^2 \times 5^3\) have factors of the form \(3^k \times 5^l\) where \(0 \leq k \leq 2\) and \(0 \leq l \leq 3\) therefore there are \(3\) possible values for \(k\) and \(4\) possible values for \(l\), which gives \(3 \times 4 = 12\) factors, which includes \(2\) factors we aren't counting, so \(10\) proper factors. By the same argument \(3^m \times 5^n\) has \((m+1) \times (n+1) - 2\) proper factors, so we want \((m+1) \times (n+1) = 12\), so we could have \begin{array}{cccc} \text{factor} & m+1 & n + 1 & m & n \\ 12 = 12 \times 1 & 12 & 1 & 11 & 0 \\ 12 = 6 \times 2 & 6& 2 & 5 & 1 \\ 12 = 4 \times 3 & 4& 3 & 3 & 2 \\ 12 = 3 \times 4 & 3& 4 & 2 & 3 \\ 12 = 2 \times 6 & 2& 6 & 1 & 5 \\ 12 = 1 \times 12 & 1& 12 & 0 & 11 \\ \end{array} So we could have \(3^{11}, 3^{5} \times 5^1 3^3 \times 5^2, 3^2 \times 5^3, 3^1 \times 5^5, 5^{11}\)
2. Suppose \(N\) has \(426\) proper factors, then it has \(428 = 2^2 \times 107\) factors, so it will either factor as \(p^{427}\) or \(p_1^{106} p_2^{3}\) or \(p_1^{106} p_2 p_3\). Clearly the first will be very large, and we should have \(p_1 < p_2 < p_3\), so lets consider \(2^{106}\) with either \(3^3 = 27\) or \(3 \times = 15 < 27\). Therefore we should take \(2^{106} \times 3 \times 5\)

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Solution: \begin{align*} && y^3 &= x^3 + a^3 + b^3 \\ \Rightarrow && 3y^2 \frac{\d y}{\d x} &= 3x^2 \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{x^2}{y^2} \end{align*} Therefore the tangent at the point \((-a,b)\) has gradient \(\frac{a^2}{b^2}\), ie \begin{align*} && \frac{y-b}{x+a} &= \frac{a^2}{b^2} \\ \Rightarrow && b^2y - b^3 &= a^2 x + a^3 \\ \Rightarrow && b^2 y-a^2 x &= a^3 + b^3 \end{align*} Notice that tangent will be, \(4y-x = 9\) so substituting this we obtain: \begin{align*} && \left (\frac{9+x}{4} \right)^3 &= x^3 + 9 \\ \Rightarrow && 9^3 + 3 \cdot 9^2 x + 3 \cdot 9x^2 + x^3 &= 64x^3 + 64 \cdot 9 \\ \Rightarrow && 9 \cdot (9^2 - 8^2) + 9 \cdot (3 \cdot 9) + 9 \cdot 3x^2 -9 \cdot 7x^3 &= 0 \\ \Rightarrow && 7x^3-3x^2-27x-17 &= 0 \\ \Rightarrow && (x+1)^2(7x-17) &= 0 \tag{repeated root since tangent} \end{align*} So we have another point on the curve \(y^3 = x^3 + 2^3 + 1^3\), namely \((\frac{17}7, \frac{17+9 \cdot 7}{28}) = (\frac{17}7, \frac{20}{7})\), so \begin{align*} 20^3 &= 17^3 + 14^3 + 7^3 \end{align*}

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Solution:

1. \(\,\) \begin{align*} && x &= (a-x)^{\frac12} \\ \Rightarrow && x^2 &= a - x \\ \Rightarrow && 0 &= x^2 + x - a \end{align*} This has a roots if \(\Delta = 1 + 4a \geq 0 \Rightarrow a \geq -\frac14\). These roots also need to be positive (since \(x \geq 0\)). Since \(f(0) = -a\) we have one positive root if \(a \geq 0\). If \(a \leq 0\) then since the roots are symmetric about \(x = -\frac12\), both roots are negative and there are no positive roots. Therefore we have on real solution if \(a \geq 0\) and non otherwise. \begin{align*} && x & = \left ( (1+a)x - a \right)^{\frac13} \\ \Leftrightarrow && x^ 3 &= (1+a)x - a \\ \Leftrightarrow && 0 &= x^3- (1+a)x + a \\ \Leftrightarrow && 0 &= (x-1)(x^2+x-a) \\ \end{align*} Since every solution to the first equation is a solution to the second, we have \(x = 1\) always works, and there is an additional two solutions if \(a > -\frac14\) and a single extra solution if \(a = -\frac14\). We can also repeat solutions if \(1\) is a root of \(x^2+x -a\), ie when \(a = 2\) Therefore: One solution if \(a < -\frac14\) Two solutions if \(a = -\frac14, 2\) Three solutions if \(a > -\frac14, a \neq 2\)
2. \(\,\) \begin{align*} && x &= (b+x)^{\frac12} \\ \Rightarrow && x^2 &= b + x \\ \Rightarrow && 0 &= x^2 - x - b \end{align*} This has a positive root if \(\frac14 - \frac12 - b \leq 0 \rightarrow b \geq \frac14\). It has two positive roots if \(b \geq 0\). Therefore two solutions if \(b > \frac14\) and one solution if \(b = \frac14\)

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Solution: The largest angle will be opposite the side with length \(p+q\). Similarly the smallest angle will be opposite the side with length \(p-q\). The cosine rule tells us that: \begin{align*} && (p+q)^2 &= p^2 + (p-q)^2 - 2p(p-q) \cos \alpha \\ && 0 &= p(p-4q-2(p-q)\cos \alpha)\\ && 0 &= p(1-2\cos \alpha) + q(2\cos \alpha - 4)\\ \Rightarrow && \frac{p}{q} & = \frac{4-2 \cos \alpha}{1-2 \cos \alpha} \\ && (p-q)^2 &= p^2 + (p+q)^2 - 2p(p+q) \cos \beta \\ && 0 &= p(p+4q-2(p+q) \cos \beta) \\ && 0 &= p(1-2\cos \beta)+q(4-2\cos \beta) \\ \Rightarrow && \frac{p}{q} &= \frac{2\cos \beta - 4}{1-2\cos \beta} \\ \Rightarrow && \frac{4-2 \cos \alpha}{1-2 \cos \alpha} &= \frac{2\cos \beta - 4}{1-2\cos \beta} \\ \Rightarrow && (2-\cos \alpha)(1-2\cos \beta) &= (\cos \beta - 2)(1 - 2 \cos \alpha) \\ \Rightarrow && 2 - \cos \alpha -4\cos \beta+2\cos \alpha \cos \beta &= \cos \beta - 2-2\cos \alpha \cos \beta + 4 \cos \alpha \\ \Rightarrow && 4-4\cos \alpha - 4\cos \beta+4\cos \alpha\cos \beta &= \cos \alpha + \cos \beta \\ \Rightarrow && 4(1-\cos \alpha)(1-\cos \beta) &= \cos \alpha + \cos \beta \end{align*} If \(\alpha = 2 \beta\), and let \(c = \cos \beta\) \begin{align*} && 4 (1- \cos 2 \beta)(1-\cos \beta) &= \cos 2 \beta + \cos \beta \\ \Rightarrow && 4(1-(2c^2-1))(1-c) &= 2c^2-1+c\\ \Rightarrow && 8(1+c)(1-c)^2 &= (2c-1)(c+1) \\ \Rightarrow && 0 &= (c+1)(8(1-c)^2-(2c-1)) \\ &&&= (c+1)(8c^2-18c+9) \\ &&&= (c+1)(4c-3)(2c-3) \\ \end{align*} Therefore \(c = -1, \frac32, \frac34\). Clearly \(\cos \beta \neq -1, \frac32\), since they are not valid angles in a triangle (or valid values of \(\cos \beta\)). \(\frac{p}{q} = \frac{2 \cdot \frac34-4 }{1 - 2\cdot \frac34} = \frac{3-8}{2-3} = 5\) so \(4:5:6\)

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Solution:

1. Given \(A\) is fixed, and \(h^2 + r^2 = \ell^2\), we can look at \begin{align*} && V^2 &= \frac19 \pi^2 r^4 h^2 \\ &&&= \frac19\pi^2r^4(\ell^2 - r^2) \\ &&&= \frac19\pi^2 r^4\left (\frac{A^2}{\pi^2r^2} - r^2 \right) \\ &&&= \frac{A^2r^2 - \pi^2r^6}{9} \end{align*} Differentiating wrt to \(r\) we find that \(2rA^2-6\pi^2 r^5 = 0\) or hence \(A^2 = 3\pi^2 r^4 \Rightarrow A = \sqrt{3}\pi r^2\). Therefore \(\sqrt{3}\pi r^2 = \pi r \ell \Rightarrow \ell = \sqrt{3}r\).
2. Supposing \(V\) is fixed, then \begin{align*} && A^2 &= \pi^2 r^2\ell^2 \\ &&&= \pi^2 r^2 (h^2+r^2) \\ &&&= \pi^2 r^2 \left ( \frac{9V^2}{\pi^2r^4} + r^2 \right) \\ &&&= 9V^2r^{-2} + \pi^2r^4 \\ \end{align*} Differentiating wrt to \(r\) we find \(-18V^2r^{-3} + 4\pi^2 r^3 = 0\) so \(V^2 = \frac{2\pi^2}{9}r^6\) or \(V = \frac{\sqrt{2}\pi}{3}r^3\), from which it follows: \(\frac{\sqrt{2}\pi}{3}r^3 = \frac13\pi r^2 h \Rightarrow h = \sqrt{2}r\)

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Solution:

1. \(\,\) \begin{align*} && \int_{1/m}^m \frac{x^2}{x+1} \d x &= \int_{1/m}^m \left ( x- 1 + \frac{1}{x+1} \right) \d x \\ &&&= \left [ \frac{x^2}{2} - x + \ln (x+1) \right]_{1/m}^m \\ &&&= \left ( m^2/2 - m + \ln(m+1) \right)- \left ( \frac{1}{2m^2} - \frac{1}{m} + \ln\left(\frac1m+1\right) \right) \\ &&&= \frac{m^4-2m^3-1+2m}{2m^2} + \ln (m+1) - \ln(m+1) + \ln m \\ &&&= \frac{(m-1)^3(m+1)}{2m^2} + \ln m \end{align*}
2. \(\,\) \begin{align*} u = \frac{1}x, \d x = -\frac{1}{u^2} \d u:&& \int_{1/m}^m \frac1 {x^n(x+1)}\,\d x &= \int_{u=m}^{u=1/m} \frac{1}{u^{-n}(u^{-1}+1)} \frac{-1}{u^2} \d u \\ &&&= \int_{1/m}^m \frac{u^{n-1}}{u+1} \d u \end{align*}
3. • \(\bf (a)\) \(\,\) \begin{align*} && I &= \int_{1/2}^2 \frac {x^5+3}{x^3(x+1)}\,\d x \\ &&&= \int_{1/2}^2 \left ( \frac{x^2}{x+1} + \frac{3}{x^3(x+1)} \right) \d x \\ &&&= \int_{1/2}^2 \frac{x^2}{x+1} \d x + 3 \int_{1/2}^2 \frac{x^2}{x+1} \d x \\ &&&= 4 \left ( \frac{(2-1)^3(2+1)}{2 \cdot 2^2} + \ln 2 \right) \\ &&&= \frac32+4 \ln 2 \end{align*}
   • \(\bf (b)\) \(\,\) \begin{align*} && J &= \int_1^2 \frac{x^5+x^3 +1}{x^3(x+1)}\, \d x \\ && K &= \int_1^2 \frac{x^5 +1}{x^3(x+1)}\, \d x\\ u = 1/x, \d x = -1/u^2 \d u: &&&= \int_{u=1}^{u=1/2} \frac{u^{-5}+1}{u^{-3}(u^{-1}+1)} \frac{-1}{u^2} \d u \\ &&&= \int_{1/2}^1 \frac{1 + u^5}{u^3(u+1)} \d u \\ \Rightarrow && K &= \frac12 \int_{1/2}^2 \frac{x^5+1}{x^3(x+1)} \d x \\ &&&= \frac{(2-1)^3(2+1)}{2 \cdot 2^2} + \ln 2 \\ &&&= \frac38 + \ln 2 \\ && L &= \int_1^2 \frac{x^3}{x^3(x+1)} \d x \\ &&&= \ln (3) - \ln 2 \\ \Rightarrow && J &= \frac38 + \ln 3 \end{align*}

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Solution: \begin{align*} && I &= \int_0^{2 \pi} e^{x} \cos m x \d x \\ &&&= \left [e^x \cos m x \right]_0^{2 \pi}-\int_0^{2 \pi} e^x m (-\sin mx) \d x\\ &&&= e^{2\pi}-1 + m\int_0^{2\pi}e^x \sin m x \d x \\ &&&= e^{2\pi}-1 + m\left [e^x \sin m x \right]_0^{2\pi} - m \int_0^{2\pi} e^x m \cos x \d x \\ &&&= e^{2\pi}-1+0 - m^2 I\\ \Rightarrow && (m^2+1)I &= e^{2\pi}-1 \\ \Rightarrow && I &= \frac{1}{m^2+1} (e^{2\pi}-1) \end{align*}

1. \(\,\) \begin{align*} && \cos(A+B) + \cos(A-B) &= \cos A\cos B - \sin A \sin B + \cos A \cos B + \sin A \sin B \\ &&&= 2 \cos A \cos B \end{align*} Therefore \begin{align*} && I &= \int_0^{2\pi} e^x \cos x \cos 6x \d x \\ &&&= \int_0^{2\pi} e^x \frac12\left (\cos 7x + \cos 5x \right) \d x\\ &&&= \left ( \frac{1}{2(1+7^2)} + \frac1{2(1+5^2)}\right)(e^{2\pi}-1) \\ &&&= \left (\frac{1}{100}+\frac{1}{52} \right) (e^{2\pi}-1) \\ &&&= \frac{19}{650}(e^{2\pi}-1) \end{align*}
2. \(\,\) \begin{align*} && I &= \int_0^{2\pi} e^x \sin 2x \sin 4x \cos x \d x\\ &&&= \int_0^{2\pi} e^x \tfrac12(\cos2x-\cos 6x) \cos x \d x\\ &&&= \frac12 \int_0^{2\pi} e^x \left (\cos 2x \cos x -\cos 6x \cos x \right) \d x \\ &&&= \frac14 \int_0^{2\pi} e^x \left (\cos 3x + \cos x-\cos 7x-\cos 5x \right) \d x \\ &&&= \frac14 \left (\frac{1}{1+3^2}+\frac{1}{1+1^2}-\frac{1}{1+7^2} - \frac{1}{1+5^2} \right)(e^{2\pi}-1) \\ &&&= \frac14 \left (\frac{1}{10}+\frac{1}{2}-\frac{1}{50} - \frac{1}{26} \right)(e^{2\pi}-1) \\ &&&= \frac{44}{325}(e^{2\pi}-1) \end{align*}

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Solution:

1. This is a circle centre \((2t,t)\) with radius \(t\). Therefore it is exactly \(t\) away from the line \(y = 0\) so just touches that line. Not that \(\tan \alpha = \frac{t}{2t} = \frac12\) so \(\tan 2\alpha = \frac{2\tan \alpha}{1-\tan^2\alpha} = \frac{1}{1-\frac14} = \frac43\). Therefore the line \(y = \frac43x\) or \(3y = 4x\) is tangent to \(C\).
2. Note that \(3y=4x\) and \(4y+3x=15\) are perpendicular, so this is a right-angled triangle with incenter \((2t,t)\) for some \(t\) and hypotenuse \(15\) We can find the third coordinate when \(3y-4x = 0\) and \(4y+3x = 15\) meet, ie \((\frac{9}{5}, \frac{12}5)\) The incentre lies on the bisector of the right angle at this point, which is the line through \((\frac{9}{5}, \frac{12}5)\) and \((\frac{15}{2}, 0)\), so \begin{align*} && \frac{2t-\frac{12}{5}}{t - \frac{9}{5}} &= \frac{-\frac{12}{5}}{\frac{15}2-\frac95} \\ \Rightarrow && \frac{10t-12}{5t-9} &= \frac{-24}{57} = -\frac{8}{19} \\ \Rightarrow && 190t - 12 \cdot 19 &= -40t + 72 \\ \Rightarrow && t &= 2 \end{align*} Therefore the center is \((4, 2)\) and the equation is \((x-4)^2+(y-2)^2=2^2\)

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Solution:

1. \begin{align*} \text{COM}: && mu + Mv &= mp + Mq \\ \Rightarrow && m(u-p) &= M(q-v) \\ \text{NEL}: && q-p &= e(u-v) \\ && q +ev &= p+eu \\ && \Delta \text{ k.e.} &= \frac12 m u^2 + \frac12 M v^2 -\frac12 m p^2 - \frac12 M q^2 \\ &&&= \frac12m (u^2 - p^2)+\frac12M(v^2-q^2) \\ &&&= \frac12m (u^2 - p^2)+\frac12M(v-q)(v+q) \\ &&&= \frac12m(u^2-p^2) - \frac12 m(u-p)(v+q) \\ &&&= \frac12 m(u-p) \left ( u+p-v-q\right) \\ &&&= \frac12 m(u-p) \left (u-v+(p-q)\right) \\ &&&= \frac12 m(u-p) \left (u-v-e(u-v)\right) \\ &&&= \frac12m(u-p)(u-v)(1-e) \end{align*} Since the loss in energy is positive, and \(m\), \(u-v\) and \(1-e\) are all positive, so is \(u-p\), ie \(u \geq p\)
2. \begin{align*} && \frac12 m u^2 - \frac12mp^2 &= \frac12Mv^2 - \frac12Mq^2 \\ && \frac12 m(u-p)(u+p) &= \frac12 M (v-q)(v+q) \\ && \frac12 m (u-p)(u+p) &= -\frac12 m(u-p)(v+q) \\ \Rightarrow && u+p+v+q &= 0 \end{align*} \begin{align*} && p+q &= -(u+v)\\ &&mp+Mq &= mu+Mv \\ \Rightarrow && (M-m)q &= mu+Mv+mu+mv\\ \Rightarrow && q &= \frac{(M+m)v+2mu}{M-m} \\ \Rightarrow && (m-M)p &= mu+Mv+Mu+Mv \\ \Rightarrow && p &= -\frac{(M+m)u+2Mv}{M-m} \\ \\ && e &= \frac{q-p}{u-v} \\ &&&= \frac{(M+m)v+2mu+(M+m)u+2Mv}{(u-v)(M-m)} \\ &&&= \frac{(3M+m)v+(3m+M)u}{(u-v)(M-m)} \end{align*}

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Solution:

1. If all the girls are say together there are \(n+1\) ways to place the block of 3 girls. There are \(\binom{n+3}{3}\) ways to choose where to place the girls in total, therefore: \begin{align*} && \mathbb{P}(K =3) &= \frac{n+1}{\binom{n+3}3} \\ &&&= \frac{6(n+1)}{(n+3)(n+2)(n+1)} \\ &&&= \frac{6}{(n+3)(n+2)} \end{align*}
2. If \(K= 1\) then all of the girls are separated. We can place three girls and two boys separating them, then we are allocating \(N-2\) boys to \(4\) gaps, ie \(\binom{N-2+3}{3} = \binom{N+1}{3}\). \begin{align*} && \mathbb{P}(K=3) &= \frac{\binom{n+1}{3}}{\binom{n+3}{3}} \\ &&&= \frac{(n+1)n(n-1)}{(n+3)(n+2)(n+1)} \\ &&&= \frac{n(n-1)}{(n+3)(n+2)} \end{align*}
3. \(\,\) \begin{align*} \mathbb{E}(K) &= \sum_{k=1}^3 k \mathbb{P}(K=k) \\ &= \frac{6}{(n+3)(n+2)} + 2 \left (1 - \frac{6}{(n+3)(n+2)} - \frac{n(n-1)}{(n+3)(n+2)} \right) + 3\frac{n(n-1)}{(n+3)(n+2)} \\ &= 2+\frac{6-12+n(n-1)}{(n+3)(n+2)} \\ &= 2 + \frac{n^2-n-6}{(n+2)(n+3)}\\ &= 2 + \frac{(n-3)(n+2)}{(n+2)(n+3)} \\ &= 2 + \frac{n-3}{n+3} \\ &= \frac{2n}{n+3} \end{align*}

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Solution: The curve \(x^4 + y^4 = u\) has lines of symmetry:

• \(y = 0\)
• \(x = 0\)
• \(y = x\)
• \(y = -x\)

The curve \(xy = v\) has lines of symmetry:

• \(y = x\)
• \(y = -x\)

The points are \(A = (\alpha, \beta), B = (\beta, \alpha), C = (-\alpha, -\beta), D = (-\beta, -\alpha)\) \(AD\) has gradient \(\frac{\beta+\alpha}{\alpha+\beta} = 1\), \(BC\) has the same gradient. \(AB\) has gradient \(\frac{\alpha-\beta}{\beta-\alpha} = -1\), as does \(CD\). Therefore it has two sets of perpendicular and parallel sides, hence a rectangle. The area is \(|AD||AB| = \sqrt{2(\alpha+\beta)^2}\sqrt{2(\alpha-\beta)^2} = 2(\alpha^2-\beta^2)\) The squared area is \(4(\alpha^4+\beta^4 - 2 \alpha^2\beta^2) = 4(u - 2v^2)\) ie the area is \(2\sqrt{u-2v^2}\) When \(u = 81, v = 4\) we have the area is \(2 \sqrt{81 - 2 \cdot 16} = 14\) as required.

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Solution:

1. \(\,\) \begin{align*} && y & = a^{\sin(\pi e^x)} \\ \Rightarrow && \frac{\d y}{\d x} &= a^{\sin(\pi e^x)} \cdot ( \ln a) \cdot (\cos (\pi e^x)) \cdot \pi e^x \\ \frac{\d y}{\d x} = 0: && 0 &= \cos(\pi e^x) \\ \Rightarrow && \pi e^x &= \left ( \frac{2n+1}{2} \right) \pi \\ \Rightarrow && x &= \ln \left ( \frac{2n+1}{2} \right) \\ && y &= a^{(-1)^n} \\ &&(x,y) &= \left (\ln \left ( \frac{2n+1}{2} \right), a^{(-1)^n} \right) \end{align*}
2. \(\,\) \begin{align*} && y &= a^{\sin(\pi e^x)} \\ &&&= e^{\ln a \cdot \sin(\pi e^x)} \\ &&&\approx e^{\ln a \cdot \sin(\pi (1+x))} \\ &&&\approx e^{-\ln a \cdot \sin(\pi x)} \\ &&&\approx e^{-\ln a \cdot \pi x} \\ &&&\approx 1-( \pi\ln a) x \end{align*}
3. 

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4. The \(k\)th maxima is at \(\ln \left ( \frac{4(k-1)+1}{2}\right)\) and \(a\) ,and the \((k+1)\)th is at \(\ln \left ( \frac{4k+1}{2}\right)\). They have a minima between at \(\ln \left ( \frac{4k-3}{2}\right)\). \begin{align*} && \text{Area} &\approx \frac12 \left (\ln \left ( \frac{4k-1}{2}\right)- \ln \left ( \frac{4k-3}{2}\right)\right) \left ( a + \frac1a \right) + \frac12 \left ( \ln \left ( \frac{4k+1}{2}\right)-\ln \left ( \frac{4k-1}{2}\right)\right) \left ( a + \frac1a \right) \\ &&&= \frac{a^2+1}{2a} \ln \left (\frac{4k+1}{4k-3} \right) \\ &&&= \frac{a^2+1}{2a} \ln \left (1 + \frac{4}{4k-3} \right) \\ &&&= \frac{a^2+1}{2a} \ln \left (1 + \frac{1}{k-\tfrac34} \right) \\ &&&= \frac{a^2+1}{2a} \ln \left (1 + (k-\tfrac34)^{-1} \right) \\ \end{align*}