Problems

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Solution:

Consider the function \(g(t) = f(tx_1 + (1-t)x_2) - tf(x_1) - (1-t)f(x_2)\), notice that \(g(0) = g(1) = 0\). Since \(g''(x) < 0\) over the whole interval, we must have two things: 1. \(g'(x)\) is increasing. 2. It \(g'(x) = 0\) can have at most one solution. Therefore \(g'(x)\) is initially \(0\), we have exactly one turning point. Therefore the function is initially decreasing and then increasing, therefore it is always negative and our inequality holds.

1. \(\,\) \begin{align*} && f \left ( \frac{u+v+w}{3} \right) &= f \left ( \frac{2}{3}\cdot \frac{u+v}2+\frac{1}{3}w \right) \\ &&&\geq \frac23 f \left ( \frac{u+v}{2} \right) + \frac13 f(w) \\ &&&\geq \frac23 \left (\frac12 f(u) + \frac12 f(v) \right) + \frac13 f(w) \\ &&&= \frac{f(u)+f(v)+f(w)}{3} \end{align*}
2. Notice that if \(A, B, C\) are angles in a triangle then they add to \(\pi\) \(0 < A,B,C < \pi\). We also have \(f(x) = \sin x \Rightarrow f''(x) = - \sin x < 0\) on this interval. Therefore \(\sin A + \sin B + \sin C \leq 3 \sin \frac{A+B+C}{3} = 3 \sin \frac{\pi}{3} = \frac{3\sqrt{3}}2\)
3. Also notice that \begin{align*} && f(x) &= \ln ( \sin x) \\ \Rightarrow && f'(x) &= \frac{\cos x}{ \sin x} \\ && f''(x) &= -\textrm{cosec}^2 x < 0 \\ \\ \Rightarrow && \ln( \sin A) + \ln (\sin B) + \ln (\sin C) &\leq 3 \ln \left (\sin \left ( \frac{A + B+ C}{3} \right) \right) \\ &&&= 3 \ln \left ( \frac{\sqrt{3}}{2} \right) = \ln \frac{3\sqrt{3}}{8} \\ \Rightarrow && \sin A \sin B \sin C &\leq \frac{3\sqrt{3}}8 \end{align*}

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Solution:

1. Let \(n! + 5 = p\). If \(n \geq 5\) then \(5\) divides the LHS and so must also divide the RHS. Since \(n!+5 > 5\) this means the RHS cannot be prime. Therefore consider \(n = 1, 2, 3, 4\). \begin{align*} n = 1: && 1! + 5 = 6 &&\text{ nope} \\ n=2: && 2! + 5 = 7 && \checkmark \\ n=3: && 3! + 5 = 11 && \checkmark \\ n=4: && 4! + 5 = 29 && \checkmark \end{align*} Therefore the solutions are \((2,7), (3,11), (4,29)\).
2. Suppose \(1! \times 3! \times \cdots \times (2n-1)! = m!\). If \(n \geq 7\) then \(m! > (4n)!\) (by the first theorem) in particular \(m > 4n\). Therefore (by the second theorem) the RHS is divisible by some prime which cannot divide the LHS. Therefore consider \(n = 1,2,3,4,5,6\) \begin{align*} n = 1: && 1! = 1 = 1! && \checkmark \\ n = 2: && 1! \times 3! = 6 = 3! && \checkmark \\ n = 3: && 1! \times 3! \times 5! = 6! && \checkmark \\ n = 4: && 1! \times 3! \times 5! \times 7! = 6! \times 7! = 10! && \checkmark \\ n = 5: && 1! \times 3! \times 5! \times 7! \times 9! = 10! 9! > 11! && \text{would need a factor of } 11\text{ so no} \\ n = 6: && 1! \times 3! \times 5! \times 7! \times 9! \times 11! = 10! 11! 9! > 13! && \text{would need a factor of } 13\text{ so no} \\ \end{align*} Therefore all solutions are \((1,1), (2,3), (3,6), (4,10)\)

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Solution:

\begin{align*} % \text{N2}(\uparrow): && R_B+ R_F &= mg \\ \overset{\curvearrowright}{G}: && -R_Fd - F_B h + R_B (b-d) &= 0 \\ \Rightarrow && -d R_F - \mu h R_B +R_B(b-d) &= 0 \\ \Rightarrow && R_B(b-d-\mu h) &= d R_F \\ \underbrace{\Rightarrow}_{R_F > 0 \text{ if not leaving ground}} && R_B(b-d-\mu h) & > 0 \\ \Rightarrow && \frac{b-d}{h} > \mu \end{align*} The acceleration is \(\frac{F_B}{m}\), so we wish to maximize \(F_B\) which is the same as maximising \(R_B\). Since the bike will slip before the front wheel lifts, we want the bike to be on the point of slipping, ie $$ \begin{align*} && R_B(b-d-\mu h) &= d R_F \\ \text{N2}(\uparrow): && R_B + R_F &= mg \\ \Rightarrow && R_B(b-d-\mu h) &= d(mg - R_B) \\ \Rightarrow && R_B(b-\mu h) &= dmg \\ \Rightarrow && R_B &= \frac{dmg}{b-\mu h} \\ \Rightarrow && a &= \frac{F_B}{m} \\ &&&= \frac{\mu R_B}{m} \\ &&&= \frac{\mu dg}{b-\mu h} \\ \end{align*} If the inequality doesn't hold, we want to be at the point just before \(R_F = 0\), since that gives us maximum friction at \(F_B\), ie \begin{align*} && R_B &= mg \\ \Rightarrow && a &= \frac{F_B}{m} \\ &&&= \frac{\mu mg}{m} \\ &&&= \mu g \end{align*}

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Solution:

1. \begin{align*} && f(\beta) &= \beta - \frac1{\beta}-\frac1{\beta^2} \\ \Rightarrow && f'(\beta) &= 1 +\frac{1}{\beta^2}+\frac{2}{\beta^3} \\ \Rightarrow && 0 &= f'(\beta) \\ &&&= 1 + \frac1{\beta^2} + \frac{2}{\beta^3} \\ \Rightarrow && 0 &= \beta^3 + \beta + 2 \\ &&&= (\beta+1)(\beta^2-\beta+2) \end{align*} Therefore the only stationary point is at \(\beta = -1, f(-1) = -1\)
   

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   \begin{align*} && g(\beta) &= \beta + \frac3{\beta}-\frac1{\beta^2} \\ \Rightarrow && g'(\beta) &= 1 -\frac{3}{\beta^2}+\frac{2}{\beta^3} \\ \Rightarrow && 0 &= f'(\beta) \\ &&&= 1 - \frac3{\beta^2} + \frac{2}{\beta^3} \\ \Rightarrow && 0 &= \beta^3 - 3\beta + 2 \\ &&&= (\beta-1)^2(\beta+2) \end{align*} Therefore there are stationary points at \(\beta=1,f(1) = 3, \beta=-2, f(-2) = \frac14\)
   

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2. Let \(u,v\) be the roots of \(x^2 + \alpha x + \beta = 0\), then since \((x-u)(x-v) = 0\) we must have \(\alpha = -(u+v), \beta = uv\). Therefore: \begin{align*} && u+v +\frac{1}{uv} &= -\alpha + \frac{1}{\beta} \\ && \frac1u+\frac1v + uv &= \frac{u+v}{uv} + uv \\ &&&= -\frac{\alpha}{\beta} + \beta \end{align*} Given \(u+v + \frac 1 {uv} = -1\), ie \(-\alpha + \frac{1}{\beta} = -1\). Since the roots are real, we must also have that \(\alpha^2 - 4\beta \geq 0\), so \begin{align*} && -\alpha + \frac1\beta &= -1 \\ \Rightarrow && \alpha &= 1 +\frac1\beta \\ \Rightarrow && -\frac{\alpha}{\beta}+\beta &= -\frac{1}{\beta} \l1+\frac1{\beta}\r + \beta \\ &&&=\beta - \frac{1}{\beta}-\frac{1}{\beta^2} \end{align*} So we want to maximise \(f(\beta)\) subject to \(\alpha ^2 - 4\beta \geq 0\) \begin{align*} && 0 &\leq \alpha^2 -4 \beta \\ &&&= \l 1 + \frac1{\beta} \r^2 - 4\beta \\ &&&= 1+ \frac2{\beta} + \frac1{\beta^2} - 4\beta \\ \Leftrightarrow && 0 &\leq -4\beta^3+\beta^2 + 2\beta + 1 \\ &&&=-(\beta-1)(4\beta^2+3\beta+1)\\ \Leftrightarrow && \beta &\leq 1 \end{align*} But we know \(f(\beta) \leq -1\) on \((-\infty,1]\) so we're done.
3. Given that \(-\alpha + \frac{1}{\beta} = 3\) we have \begin{align*} && -\alpha + \frac1\beta &= 3 \\ \Rightarrow && \alpha &= -3 +\frac1\beta \\ \Rightarrow && -\frac{\alpha}{\beta}+\beta &= -\frac{1}{\beta} \l-3+\frac1{\beta}\r + \beta \\ &&&=\beta + \frac{3}{\beta}-\frac{1}{\beta^2} \end{align*} which we want to maximise, subject to: \begin{align*} && 0 &\leq \alpha^2 -4 \beta \\ &&&= \l -3 + \frac1{\beta} \r^2 - 4\beta \\ &&&= 9- \frac6{\beta} + \frac1{\beta^2} - 4\beta \\ \Leftrightarrow && 0 &\leq -4\beta^3+9\beta^2 - 6\beta + 1 \\ &&&=-(\beta-1)^2(4\beta-1)\\ \Leftrightarrow && \beta &\leq \frac14 \end{align*} Therefore the maximum will either be \(f(-2) = \frac14\) or \(f(\frac14) = -\frac{15}4\). Therefore the maximum is \(\frac14\)

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Solution:

1. \begin{align*} \frac{\d y_n}{\d x} &= \frac{\d}{\d x} \l (-1)^n e^{x^2} \frac{\d^n}{\d x^{n}} \l e^{-x^2}\r \r \\ &= (-1)^n 2xe^{x^2} \frac{\d^n}{\d x^{n}} \l e^{-x^2}\r + (-1)^n e^{x^2} \frac{\d^{n+1}}{\d x^{n+1}} \l e^{-x^2}\r \\ &= 2xy_n - (-1)^{n+1} e^{x^2} \frac{\d^{n+1}}{\d x^{n+1}} \l e^{-x^2}\r \\ &= 2xy_n - y_{n+1} \end{align*}
2. \(y_0 = 1\), \(y_1 = (-1) e^{x^2} \cdot (-2x) \cdot e^{-x^2} = 2x\), \(y_2 = e^{x^2} \frac{\d^2}{\d x^2} \l e^{-x^2}\r = e^{x^2} \frac{\d }{\d x}\l -2xe^{-x^2} \r = e^{x^2} \l -2e^{-x^2}+4x^2e^{-x^2}\r = 4x^2-2\). Therefore \(2xy_1 - 2y_0 = 2x \cdot 2x - 2\cdot1 = 4x^2-2 = y_2\) so our statement is true for \(n=1\). Assume the statement is true for \(n=k\), then \begin{align*} && y_{k+1} &= 2xy_k - 2ky_{k-1} \\ \frac{\d }{\d x}: && \frac{\d y_{k+1}}{\d x} &= 2\frac{\d}{\d x}\l xy_k \r - 2k\frac{\d y_{k-1}}{\d x} \\ \Rightarrow && 2xy_{k+1}-y_{k+2} &= 2y_k+2x \l 2xy_k-y_{k+1}\r - 2k \l 2xy_{k-1}-y_k \r \\ \Rightarrow && y_{k+2} &=2y_k+ 4x \cdot y_{k+1}-(4x^2+2k)y_k+2x \cdot 2k y_{k-1} \\ &&&= 4x \cdot y_{k+1}-(4x^2+2(k+1))y_k+2x \l2xy_k - y_{k+1} \r \\ &&&= 2x \cdot y_{k+1} -2(k+1)y_k \end{align*} Therefore since our statement is true for \(n=1\) and if it is true for \(n=k\) it is true for \(n=k=1\), therefore by the principle of mathematical induction it is true for all \(n \geq 1\). Since \(2x = \frac{y_{n+1}+2ny_{n-1}}{y_n}\) for all \(n\), we must have \begin{align*} && \frac{y_{n+1}+2ny_{n-1}}{y_n} &= \frac{y_{n+2}+2(n+1)y_{n}}{y_{n+1}} \\ \Leftrightarrow && y_{n+1}^2+2ny_{n-1}y_{n+1} &= y_ny_{n+2}+2ny_n^2+2y_n^2 \\ \Leftrightarrow && y_{n+1}^2-y_ny_{n+2} &= 2n(y_n^2-y_{n-1}y_{n+1})+2y_n^2 \end{align*}
3. Consider the functions \(f_n(x) = y_{n}^2-y_{n-1}y_{n+1}\) then clearly \(f_{n+1} = 2nf_{n} + 2y_n^2 \geq f_{n}\) so to prove \(f_n(x) > 0\) for \(n \geq 1\) it suffices to prove it for \(n = 1\). But \(f_1 = y_1^2 - y_0y_{2} = (2x)^2-(4x^2-2) = 2 > 0\) so we are done.

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Solution:

1. \begin{align*} && (k+1) (A_{k+1} - G_{k+1}) & \geq k(A_k - G_k) \\ \Leftrightarrow && \sum_{i=1}^{k+1} a_i - (k+1)G_{k+1} &\geq \sum_{i=1}^k a_i - kG_k \\ \Leftrightarrow && a_{k+1} -(k+1)G_k^{k/(k+1)}a_{k+1}^{1/(k+1)} & \geq - k G_k \\ \Leftrightarrow && a_{k+1} -(k+1)G_k^{k/(k+1)}a_{k+1}^{1/(k+1)} + k G_k& \geq 0\\ \Leftrightarrow && \frac{a_{k+1}}{G_k} -(k+1)G_k^{k/(k+1)-1}a_{k+1}^{1/(k+1)} + k & \geq 0\\ \Leftrightarrow && \lambda_k^{k+1} -(k+1)\lambda_k+ k & \geq 0\\ \end{align*} as required.
2. \begin{align*} && f'(x) &= (k+1)x^k - (k+1) \\ &&&= (k+1)(x^k-1) \end{align*} Therefore \(f(x)\) is strictly decreasing on \((0,1)\) and strictly increasing on \((1,\infty)\) and so the minimum will be \(f(1) = 1 - (k+1) + k = 0\), so \(f(x) \geq 0\) with equality only at \(x = 1\).
3. 1. We can proceed by induction to show since the inequality holds for \(n=1\) and since if it holds for \(n=k\) it will hold for \(n=k+1\) as \(A_{k+1}-G_{k+1}\) must have the same sign as \(A_k - G_k\).
   2. The only way for equality to hold is if \(\lambda_k = 1\) for \(k = 1, \cdots n\), ie \(a_{k+1} = G_k\), but this means \(a_2 = a_1, a_3 = a_1\) etc. Therefore all values are equal.

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Solution:

1. \begin{align*} \frac{\left(\cot \theta + i\right)^{2n+1} -\left(\cot \theta - i\right)^{2n+1}}{2i} &= \frac{1}{\sin^{2n+1} \theta}\frac{\left(\cos \theta + i \sin \theta \right)^{2n+1} -\left(\cos\theta - i\sin \theta\right)^{2n+1}}{2i} \\ &= \frac{1}{\sin^{2n+1} \theta} \frac{e^{i(2n+1) \theta} - e^{-i(2n+1) \theta} }{2i} \\ &=\frac{\sin (2n+1) \theta}{\sin^{2n+1} \theta} \end{align*} Notice that: \begin{align*} (\cot \theta + i)^{2n+1} - (\cot \theta - i)^{2n+1} &= \sum_{k=0}^{2n+1} \binom{2n+1}{k}(i)^k \cdot \cot^{2n+1-k} \theta - \sum_{k=0}^{2n+1} \binom{2n+1}{k}(-i)^k \cdot \cot^{2n+1-k} \theta \\ &= \sum_{k=0}^{2n+1} \binom{2n+1}{k} \l i^k - (-i)^k \r \cdot \cot^{2n+1-k} \theta \\ &= \sum_{l=0}^{n} \binom{2n+1}{2l+1} \l i^{2l+1} - (-i)^{2l+1} \r \cdot \cot^{2n+1-(2l+1)} \theta \\ &= \sum_{l=0}^{n} \binom{2n+1}{2l+1} 2i \cdot \cot^{2(n-l)} \theta \\ &= 2i\sum_{l=0}^{n} \binom{2n+1}{2l+1} \cot^{2(n-l)} \theta \\ \end{align*} Therefore if \(\theta\) satisfies \(\frac{\sin (2n+1) \theta}{\sin^{2n+1} \theta} = 0\) then \(z = \cot^2 \theta\) satisfies the equation. But \(\theta = \frac{m \pi}{2n+1}, m = 1, 2, \ldots, n\) are \(n\) distinct all the roots must be \(\cot^2 \frac{m \pi}{2n+1}\).
2. Notice that the sum of the roots will be \(\displaystyle \frac{\binom{2n+1}{3}}{\binom{2n+1}{1}} = \frac{(2n+1)\cdot 2n \cdot (2n-1)}{3! \cdot (2n+1)} = \frac{n \cdot (2n-1)}{3}\) and so \[ \sum_{m=1}^n \cot^{2}\left(\frac{m\pi}{2n+1}\right) =\frac{n\left(2n-1\right)}{3}. \]
3. For \(0 < \theta < \frac{1}{2}\pi\), \begin{align*} && 0 < \sin \theta < \theta < \tan \theta \\ \Leftrightarrow && 0 < \cot \theta < \frac{1}{\theta} < \frac{1}{\sin \theta} \\ \Leftrightarrow && 0 < \cot^2 \theta < \frac{1}{\theta^2} < \cosec^2 \theta = 1 + \cot^2 \theta\\ \end{align*} Therefore \begin{align*} && \sum_{n=1}^N \cot^2 \frac{n \pi}{2N+1} < \sum_{n=1}^N \frac{(2N+1)^2}{n^2 \pi^2} < N + \sum_{n=1}^N \cot^2 \frac{n \pi}{2N+1} \\ \Rightarrow && \frac{1}{(2N+1)^2} \frac{N(2N-1)}{3} < \sum_{n=1}^N \frac{1}{n^2 \pi^2} < \frac{1}{(2N+1)^2} \l \frac{N(2N-1)}{3} + 1 \r \\ \Rightarrow && \lim_{N \to \infty}\frac{1}{(2N+1)^2} \frac{N(2N-1)}{3} < \lim_{N \to \infty}\sum_{n=1}^N \frac{1}{n^2 \pi^2} < \lim_{N \to \infty}\frac{1}{(2N+1)^2} \l \frac{N(2N-1)}{3} + 1 \r \\ \Rightarrow && \frac{1}{6} \leq \lim_{N \to \infty}\sum_{n=1}^N \frac{1}{n^2 \pi^2} \leq \frac16 \\ \Rightarrow && \sum_{n=1}^N \frac{1}{n^2} = \frac{\pi^2}{6} \end{align*}

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Solution:

1. Just before collision \(n-1\): Velocity of \(P\) is \(u_{n-2}\) Velocity of \(Q\) is \(-v_{n-2}\) \begin{align*} COM: && mu_{n-2}+km(-v_{n-2}) &= mu_{n-1}+kmv_{n-1} \\ \Rightarrow && u_{n-2}-kv_{n-2} &= u_{n-1}+kv_{n-1} \\ NEL: && v_{n-1}-u_{n-1} &= -e((-v_{n-2})-u_{n-2}) \\ \Rightarrow && v_{n-1}-u_{n-1} &= e(v_{n-2}+u_{n-2}) \end{align*} \begin{align*} &&kv_{n-1} &= u_{n-2} - kv_{n-2}-u_{n-1} \\ &&kv_{n-1}&= ku_{n-1}+kev_{n-2}+keu_{n-2} \\ \Rightarrow && kv_{n-2}(1+e) &= u_{n-2}(1-ke)-u_{n-1}(1+k) \\ \Rightarrow && kv_{n-1}(1+e) &= u_{n-1}(1-ke)-u_{n}(1+k) \\ && k(1+e)v_{n-1}-k(1+e)u_{n-1} &= k(1+e)e(v_{n-2}+u_{n-2}) \\ \Rightarrow && u_{n-1}(1-ke)-u_{n}(1+k)-k(1+e)u_{n-1} &= e(u_{n-2}(1-ke)-u_{n-1}(1+k))+k(1+e)eu_{n-2} \\ \Rightarrow && 0 &= (1+k)u_n + ((ke-1)+k(1+e)-e(1+k))u_{n-1} \\ &&& \quad \quad + (e(1-ke)+k(1+e)e)u_{n-2} \\ \Rightarrow && 0 &= (1+k)u_n- (1-k)(1+e)u_{n-1} +e(1+k)u_{n-2} \end{align*}
2. \(u_0 = A + B\) \begin{align*} &&& \begin{cases}u_0 - kv_0 &= kv_1 + u_1 \\ \frac12 (u_0+v_0) &= v_1 - u_1 \\ \end{cases} \\ \Rightarrow && (1+k)u_1 &= u_0 - kv_0 - \frac{k}{2}(u_0 + v_0) \\ \Rightarrow && u_1 &= \frac{1}{k+1} \l u_0 (1-\frac{k}{2}) - \frac32 k v_0 \r \\ &&&= \frac{67}{70} u_0 - \frac{3}{70} v_0 \end{align*} Therefore \(A+B = u_0, \frac{49A+50B}{70} = \frac{67}{70} u_0 - \frac{3}{70} v_0\) \begin{align*} && A+B &= u_0 \\ && 49A+50B &= 67u_0 - 3v_0 \\ \Rightarrow && 50u_0 - A &= 67u_0 - 3v_0 \\ \Rightarrow && A &= -17u_0 + 3v_0 \\ && B &= 18u_0 - 3v_0 \end{align*} If \(0 < 6u_0 < v_0\), then \(B < 0\) and as \(n \to \infty\) we will find that \(\l \frac57 \r^n\) dominates \(\l \frac7{10} \r^n\) and so our velocity will be negative and the particle will change direction

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Solution:

1. \(\,\) \begin{align*} (x \geq 1): && \int_1^x \frac{1}{t} \d t &\leq \int_1^x 1 \d t \\ \Rightarrow && \ln x - \ln 1 &\leq x - 1 \\ \Rightarrow && \ln x & \leq x - 1 \\ \\ (0 < x \leq 1):&& \int_x^1 1\d t &\leq \int_x^1 \frac{1}{t} \d t \\ \Rightarrow&& 1- x &\leq \ln 1 - \ln x \\ \Rightarrow&& \ln x &\leq x - 1 \end{align*}
2. \(\,\) \begin{align*} (x \geq 1): && \int_1^x \frac{1}{t^2} \d t &\leq \int_1^x \frac{1}{t} \d t \\ \Rightarrow && -\frac1x+1 &\leq \ln x - \ln 1 \\ \Rightarrow && 1 - \frac1x &\leq \ln x \\ \\ (0 < x \leq 1): && \int_x^1 \frac{1}{t} \d t &\leq \int_x^1 \frac{1}{t^2} \d t \\ \Rightarrow && \ln 1 - \ln x & \leq -1 + \frac{1}{x} \\ \Rightarrow && 1 - \frac1x &\leq \ln x \\ \end{align*}
3. \(\,\) \begin{align*} (1 < y): && \int_1^y \left (1 - \frac1{x} \right)\d x &\leq \int_1^y \ln x \d x \\ \Rightarrow && \left [x - \ln x \right]_1^y & \leq \left [ x \ln x - x\right]_1^y \\ \Rightarrow && y - \ln y - 1 &\leq y \ln y - y +1 \\ \Rightarrow && 2y-2 & \leq (y+1) \ln y \\ \Rightarrow && \frac{2}{y+1} & \leq \frac{\ln y}{y-1} \\ (0 < y < 1): && \int_y^1 \left (1 - \frac1{x} \right)\d x &\leq \int_y^1 \ln x \d x \\ \Rightarrow && \left [x - \ln x \right]_y^1 & \leq \left [ x \ln x - x\right]_y^1 \\ \Rightarrow && 1 - (y - \ln y) &\leq -1-(y \ln y-y) \\ \Rightarrow && 2-2y &\leq -(y+1)\ln y \\ \Rightarrow && \frac{2}{y+1} &\leq \frac{-\ln y}{1-y} \tag{\(1-y > 0\)} \\ \Rightarrow && \frac{2}{y+1} &\leq \frac{\ln y}{y-1} \\ \\ (1 < y): && \int_1^y \ln x \d x &\leq \int_1^y (x-1) \d x \\ \Rightarrow && \left [x \ln x -x \right]_1^y &\leq \left[ \frac12 x^2 - x \right]_1^y\\ \Rightarrow && y \ln y - y +1 &\leq \frac12y^2 - y+\frac12 \\ \Rightarrow && y \ln y &\leq \frac12 \left (y^2-1 \right) \\ \Rightarrow && \frac{\ln y}{y-1} &\leq \frac{y+1}{2y} \\ \\ (0 < y < 1) && \int_y^1 \ln x \d x &\leq \int_y^1 (x-1) \d x \\ \Rightarrow && \left [x \ln x -x \right]_y^1&\leq \left[ \frac12 x^2 - x \right]_y^1\\ \Rightarrow && -1-(y \ln y - y +1) &\leq-\frac12 - \left ( \frac12y^2 - y\right)\\ \Rightarrow && \frac12 \left (y^2-1 \right) &\leq y \ln y \\ \Rightarrow && \frac{\ln y}{y-1} & \leq \frac{y+1}{2y} \tag{\(y-1 < 0\)} \end{align*}

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Solution:

1. \(\,\)
   

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   Notice that \(1+r+r^2\) ranges from \(\frac34\) to \(3\) over \((-1,1)\) therefore \(\frac34 \leq p < 3\) attaining its minimum but not its maximum. If \(p > 1\) we know we must be on the right branch and hence we can determine \(r\) and \(S\) uniquely. If \(\frac34 < p < 1\) then we must have two possible values for \(r\), satisfying \(r_i^2 + r+1 = p\) and so our two possible values for \(S_i = \frac{1}{1-r_i}\) or \(r_i = 1-\frac{1}{S_i}\) and so \begin{align*} && p &= \left ( 1 - \frac{1}{S} \right)^2 + 1 - \frac1S + 1 \\ \Rightarrow && pS^2 &= (S-1)^2 + S^2-S + S^2 \\ \Rightarrow && 0 &= (3-p)S^2 -3S + 1 \end{align*}
2. \(\,\)
   

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   So \(\frac23 \leq q < 6\) and \(r\) and hence \(T\) is uniquely determined if \(2 \leq q < 6\). if \(\frac23

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Solution:

1. Claim: If \(f(x)\) non-zero for some \(x \in [0,1]\) and \(\int_0^1 f(x) \d x =0\) then \(f\) takes both positive and negative values in the interval \([0,1]\). Proof: Suppose not, then WLOG suppose \(f(x) > 0\) for some \(x \in [0,1]\). Then notice (since \(f\) is continuous) that there is some interval where \(f(x) > 0\) around the \(x\) we have already shown exists. But then \(\int_{\text{interval}} f(x) \d x > 0\) and since \(f(x) \geq 0\) everywhere \(\int_0^1 f(x) \d x > 0\), which is a contradiction.
2. \(\,\) \begin{align*} && \int_0^1 (x - \alpha)^2 g(x) \d x &= \int_0^1 x^2g(x) \d x - 2 \alpha \int_0^1 x g(x) \d x + \alpha^2 \int_0^1 g(x) \d x \\ &&&= \alpha^2 - 2\alpha \cdot \alpha + \alpha^2 \cdot 1 \\ &&&= 0 \end{align*} Therefore \(g(x)(x-\alpha)^2\) is a continuous function which is either exactly \(0\) (in which case we've already found our \(0\)) or it is a continuous function which is both positive somewhere and has \(0\) integral, and therefore by part (i) must take both positive and negative values (and therefore takes \(0\) in between those points by continuity). \begin{align*} &&1 &= \int_0^1 a+bx \d x \\ &&&= a + \frac12 b \\ && \alpha &= \int_0^1 ax + bx^2 \d x \\ &&&= \frac12 a + \frac13 b \\ && \alpha^2 &= \int_0^1 ax^2 + bx^3 \d x\\ &&&= \frac13 a + \frac14 b \\ \Rightarrow && \frac1{36}(3a+2b)^2 &= \frac1{12}(4a+3b) \\ \Rightarrow && \frac1{36}(3a+2(2-2a))^2 &= \frac1{12}(4a+3(2-2a)) \\ \Rightarrow && (4-a)^2 &= 3(6-2a) \\ \Rightarrow && 16-8a+a^2 &= 18-6a \\ \Rightarrow && a^2-2a-2 &= 0 \\ \Rightarrow && (a-1)^2 - 3 &= 0 \\ \Rightarrow && a &= \pm \sqrt{3}+1 \\ && b &= \mp 2\sqrt{3} \end{align*} So say \(a = \sqrt{3}+1, b = -2\sqrt{3}\) This has a root at \(-\frac{a}{b} = \frac{1+\sqrt{3}}{2\sqrt{3}} = \frac{\sqrt{3}+3}{6} < 1\) so we have met our condition.
3. Consider \(h'\), we must have \begin{align*} && \int_0^1 h'(x)\d x &= h(1)-h(0) =1\\ && \int_0^1 xh'(x) \d x &= \left [x h(x) \right]_0^1 - \int_0^1 h(x) \d x \\ &&&= 1 - \beta \\ && \int_0^1 x^2 h'(x) \d x &= \left [ x^2h(x) \right]_0^1 - \int_0^1 2xh(x) \d x \\ &&&= 1 - 2\tfrac12 \beta(2-\beta) \\ &&&= (1-\beta)^2 \end{align*} Therefore \(h'\) satisfies the conditions with \(\alpha = 1-\beta\), so \(h'\) must have a root in our interval.

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Solution: Claim \(a_n^2+2a_nb_n - b_n^2 = 1\) for all \(n \geq 1\) Proof: (By induction) Base case: (\(n = 1\)). When \(n = 1\) we have \(a_1^2 + 2a_1 b_1-b_1^2 = 1^2+2\cdot1\cdot2-2^2 = 1\) as required. (Inductive step). Now we assume our result is true for some \(n =k\), ie \(a_k^2+2a_kb_k - b_k^2 = 1\), now consider \(n = k+1\) \begin{align*} && a_{k+1}^2+2a_{k+1}b_{k+1} - b_{k+1}^2 &= (a_k+2b_k)^2+2(a_k+2b_k)(2a_k+5b_k) - (2a_k+5b_k)^2 \\ &&&= a_k^2+4a_kb_k+4b_k^2 +4a_k^2+18a_kb_k+20b_k^2 - 4a_k^2-20a_kb_k-25b_k^2 \\ &&&= (1+4-4)a_k^2+(4+18-20)a_kb_k +(4+20-25)b_k^2 \\ &&&= a_k^2+2a_kb_k -b_k^2 = 1 \end{align*} Therefore since our statement is true for \(n = 1\) and when it is true for \(n=k\) it is true for \(n=k+1\) by the POMI it is true for \(n \geq 1\)

1. Notice that \(b_{n+1} \geq 5 b_n\) and therefore \(b_n \geq 5^{n-1} b_1 = 2\cdot 5^{n-1}\), so \begin{align*} && 1 &= a_n^2+2a_nb_n - b_n^2\\ \Rightarrow && \frac1{b_n^2} &= c_n^2 + 2c_n - 1 \\ \to && 0 &= c_n^2 + 2c_n - 1 \quad \text{ as } n\to \infty \\ \end{align*} This has roots \(c = -1 \pm \sqrt{2}\), and since \(c_n > 0\) it must tend to the positive value, ie \(c_n \to \sqrt{2}-1\)
2. Notice that \(c_n^2 + 2c_n - 1 > 0\) so either \(c_n > \sqrt{2}-1\) or \(c_n < -1-\sqrt{2}\), but again, since \(c_n > 0\) we must have \(c_n > \sqrt{2}-1\). Therefore \(\sqrt{2} < c_n + 1\) and \(1+c_n > \sqrt{2} \Rightarrow \frac{1}{1+c_n} < \frac{\sqrt{2}}2 \Rightarrow \frac{2}{1+c_n} < \sqrt{2}\) \begin{array}{c|c|c} n & a_n & b_n \\ \hline 1 & 1 & 2 \\ 2 & 5 & 12 \\ 3 & 29 & 70 \end{array} Therefore \(c_3 = \frac{29}{70}\) and so \(\frac{2}{1 + \frac{29}{70}} = \frac{140}{99} < \sqrt{2} < \frac{29}{70} + 1 = \frac{99}{70}\)

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Solution:

1. Let \(f(x) = 1, g(x) = e^x, a = 0, b = t\), so \begin{align*} && \left ( \int_0^t e^x \d x \right)^2 &\leq \left (\int_0^t 1^2 \d x \right) \cdot \left (\int_0^t (e^x)^2 \d x \right) \\ \Rightarrow && (e^t-1)^2 &\leq t \cdot (\frac12e^{2t} - \frac12) \\ \Rightarrow && \frac{e^t-1}{e^t+1} & \leq \frac{t}{2} \end{align*}
2. Let \(f(x) = x, g(x) = e^{-\frac14 x^2}, a = 0, b = 1\) \begin{align*} && \left ( \int_0^1 xe^{-\frac14 x^2} \d x \right)^2 &\leq \left (\int_0^1 x^2 \d x \right) \cdot \left (\int_0^1 (e^{-\frac14x^2})^2 \d x \right) \\ \Rightarrow && \left ( \left [-2e^{-\frac14x^2} \right]_0^1 \right)^2 & \leq \frac{1}{3} \int_0^1 e^{-\frac12 x^2} \d x \\ \Rightarrow && \int_0^1 e^{-\frac12 x^2} \d x & \geq 12(1-e^{-\frac14})^2 \end{align*}
3. Let \(f(x) = 1, g(x) = \sqrt{\sin x}, a = 0, b = \tfrac12 \pi\), then \begin{align*} && \left ( \int_0^{\frac12 \pi} \sqrt{\sin x} \d x \right)^2 &\leq \left (\int_0^{\frac12 \pi} 1^2 \d x \right) \cdot \left (\int_0^{\frac12 \pi}|\sin x| \d x \right) \\ &&&= \frac{\pi}{2} \cdot 1 \\ \Rightarrow && \int_0^{\frac12 \pi} \sqrt{\sin x} \d x & \leq \sqrt{\frac{\pi}{2}} \end{align*} Let \(f(x) =(\sin x)^{\frac14}, g(x) = \cos x, a = 0, b = \tfrac12 \pi\), so \begin{align*} && \left ( \int_0^{\frac12 \pi} (\sin x)^{\frac14} \cos x \d x \right)^2 &\leq \left (\int_0^{\frac12 \pi} \cos^2 x \d x \right) \cdot \left (\int_0^{\frac12 \pi}\sqrt{\sin x} \d x \right) \\ \Rightarrow &&\left ( \left [\frac45 (\sin x)^{\frac54} \right]_0^{\frac12 \pi} \right)^2 & \leq \frac{\pi}{4} \int_0^{\frac12 \pi}\sqrt{\sin x} \d x \\ \Rightarrow && \frac{64}{25\pi} &\leq \int_0^{\frac12 \pi}\sqrt{\sin x} \d x \end{align*}

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Solution:

1. Claim: \(S_n \leq 2\sqrt{n} -1\). Proof: (By induction) (Base case: \(n = 1\)). \(\frac{1}{\sqrt{1}} \leq 1 = 2 \cdot \sqrt1 - 1\). Therefore the base case is true. (Inductive step): Suppose our result is true for \(n = k\). Then consider \(n = k+1\). \begin{align*} && \sum_{r=1}^{k+1} \frac{1}{\sqrt{r}} &=\sum_{r=1}^{k} \frac{1}{\sqrt{r}} + \frac{1}{\sqrt{k+1}} \\ &&&\leq 2\sqrt{k} - 1 + \frac{1}{\sqrt{k+1}} \\ &&&= \frac{2 \sqrt{k}\sqrt{k+1}+1}{\sqrt{k+1}} - 1 \\ &&&\underbrace{\leq}_{AM-GM} \frac{(k+k+1)+1}{\sqrt{k+1}} - 1 \\ &&&=\frac{2(k+1)}{\sqrt{k+1}} - 1 \\ &&&= 2\sqrt{k+1}-1 \end{align*} Therefore, since if our statement is true for \(n = k\), it is also true for \(n = k+1\). By the principle of mathematical induction we can say that it is true for all \(n \geq 1, n \in \mathbb{Z}\)
2. Claim: \((4k+1)\sqrt{k+1} > (4k+3)\sqrt k\,\) for \(k\ge0\,\) Proof: \begin{align*} && (4k+1)\sqrt{k+1} &> (4k+3)\sqrt k \\ \Leftrightarrow && (4k+1)^2(k+1) &> (4k+3)^2k \\ \Leftrightarrow && (16k^2+8k+1)(k+1) &> (16k^2 + 24k+9)k \\ \Leftrightarrow && 16 k^3 + 24 k^2 + 9 k +1&> 16k^3 + 24k^2+9k \end{align*} But this last inequality is clearly true, hence our original inequality is true. Suppose \(S_n \geq 2\sqrt{n} + \frac{1}{2 \sqrt{n}} - C\), then adding \(\frac{1}{\sqrt{n+1}}\) to both sides we have: \begin{align*} S_{n+1} &\geq 2\sqrt{n} + \frac{1}{2 \sqrt{n}} - C + \frac{1}{\sqrt{n+1}} \\ &= 2\sqrt{n+1} + \frac{1}{2\sqrt{n+1}} - C + \frac{1}{2\sqrt{n+1}} +\frac{1}{2 \sqrt{n}} +2(\sqrt{n} - \sqrt{n+1})\\ &= 2\sqrt{n+1} + \frac{1}{2\sqrt{n+1}} - C + \frac{1}{2\sqrt{n+1}} +\frac{1}{2 \sqrt{n}} -\frac{2}{\sqrt{n+1} + \sqrt{n}}\\ \end{align*} Therefore as long as the inequality is satisfied for \(n=1\), ie \(1 \geq 2\sqrt{1} + \frac{1}{2 \sqrt{1}} - C = \frac52 - C \Rightarrow C \geq \frac32\)

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Solution:

1. \(\,\) \begin{align*} && \ln f(x) &= x \ln x \\ &&&> \ln x \quad (\text{if } 0 < x < 1)\\ \Rightarrow && f(x) &> x\quad\quad (\text{if } 0 < x < 1)\\ \Rightarrow && x^{f(x)} &< x^x \\ && g(x) &< f(x) \\ && 1&>f(x) \\ \Rightarrow && x &< x^{f(x)} = g(x) \end{align*}
2. \(\,\) \begin{align*} && f(x) &= e^{x \ln x} \\ \Rightarrow && f'(x) &= (\ln x + 1)e^{x \ln x} \\ \Rightarrow && f'(x) = 0 &\Leftrightarrow x = \frac1e \end{align*}
3. \(\,\) \begin{align*} && \lim_{x \to 0} f(x) &= \lim_{x \to 0} \exp \left ( x \ln x \right ) \\ &&&= \exp \left ( \lim_{x \to 0} \left ( x \ln x \right )\right) \\ &&&= \exp \left ( 0 \right) = 1\\ \\ && \lim_{x \to 0} g(x) &= \lim_{x \to 0} \exp \left ( f(x) \ln x\right) \\ &&&= \exp \left (\lim_{x \to 0} \ln x f(x)\right) \\ &&&= \exp \left (\lim_{x \to 0} \ln x \lim_{x \to 0}f(x)\right) \\ &&&= \exp \left (\lim_{x \to 0} \ln x\right) \\ &&&= 0 \end{align*}
4. \(y = x^{-1} + \ln x \Rightarrow y' = -x^{-2} + x^{-1}\) which has roots at \(x =1\), therefore the minimum value is \(1\). (We can see it's a minimum by considering \(x \to 0, x \to \infty\). So \begin{align*} && g'(x) &= x^{f(x)} \cdot (f'(x) \ln x + f(x) x^{-1})\\ &&&= x^{f(x)} \cdot f(x) \cdot ((1+\ln x) \ln x + x^{-1}) \\ &&&= x^{f(x)} \cdot f(x) \cdot (\ln x + x^{-1} + (\ln x)^2) \\ &&&\geq x^{f(x)} \cdot f(x) > 0 \end{align*}