Problems

---

Solution:

1. 

   + - ↺
2. \begin{enumerate}
3. since \(\frac{\ln x}{x}\) is decreasing on \((e, \infty)\) we must have that \(\frac{\ln 3}{3} > \frac{\ln \pi}{\pi} \Rightarrow e^\pi > \pi^3\)
4. similarly, since \(\frac{\ln x}{x}\) is increasing on \((0, e)\) we must have that \(\frac{\ln \sqrt{5}}{\sqrt{5}} < \frac{\ln 9/4}{9/4} \Rightarrow \left(\frac{9}{4}\right)^{\sqrt{5}} > \sqrt{5}^{\frac{9}{4}}\)
5. Since \(2^4 = 4^2\) notice also that:
   

   + - ↺
   from the graph we must have the green area between \(1\) and \(2\) mapping to the (higher) green area between \(3\) and \(4\). Therefore \((x+2)^x > x^{x+2}\) for \(1 < x < 2\)
6. \begin{align*} && 9^{\sqrt 2} & \stackrel{?}{>} \sqrt{2}^9 \\ \Leftrightarrow && (3^2)^{\sqrt2} &\stackrel{?}{>} (\sqrt{2}^3)^3 \\ \Leftrightarrow && 3^{2 \sqrt2} &\stackrel{?}{>} (2\sqrt2)^3 \end{align*} Since \(e^2 < 8 < 9\Rightarrow e < 2\sqrt2 < 3\) therefore: \begin{align*} && \frac{\ln 2 \sqrt2}{2 \sqrt 2} &> \frac{\ln 3}{3} \\ \Leftrightarrow && (2 \sqrt{2})^3 &> 3^{2 \sqrt{2}} \\ \Leftrightarrow && \sqrt{2}^9 &> 9^{\sqrt 2} \\ \end{align*}
7. \begin{align*} && 8^{\sqrt[3]{3}} & \stackrel{?}{>} \sqrt[3]{3}^8 \\ \Leftrightarrow && 2^{3 \sqrt[3] 3} & \stackrel{?}{>} (\sqrt[3]{3}^4)^2 \\ \Leftrightarrow && 2^{3 \sqrt[3] 3} & \stackrel{?}{>} (3\sqrt[3]{3})^2 \\ \end{align*} Since \(3\sqrt[3]{3} > 4\) we have \begin{align*} && \frac{\ln (3 \sqrt[3]3)}{3 \sqrt[3]3} &< \frac{\ln 4}{4} \\ &&&= \frac{\ln 2}{2}\\ \Rightarrow && (3 \sqrt[3]{3})^2 &< 2^{3 \sqrt[3]{3}} \\ \Rightarrow && \sqrt[3]3^8 &< 8^{\sqrt[3]3} \end{align*}

---

Solution:

1. \begin{align*} && a^{\frac{1}{y} \int_{0}^{y} \log_a f(x) \, dx} &= e^{\ln a \cdot \frac{1}{y} \int_{0}^{y} \log_a f(x) \, dx} \\ &&&= e^{\ln a \cdot \frac{1}{y} \int_{0}^{y} \frac{\ln f(x)}{\ln a} \, dx} \\ &&&= e^{ \frac{1}{y} \int_{0}^{y} \ln f(x) \, dx} \\ &&&= F(y) \end{align*}
2. \(\,\) \begin{align*} && H(y) &= e^{\frac1y \int_0^y \ln h(x) \d x} \\ &&&= e^{\frac1y \int_0^y \ln (f(x)g(x))\d x} \\ &&&= e^{\frac1y \int_0^y \left ( \ln f(x)+\ln g(x) \right)\d x} \\ &&&= e^{\frac1y \int_0^y \ln f(x) \d x +\frac1y \int_0^y \ln g(x) \d x} \\ &&&= e^{\frac1y \int_0^y \ln f(x) \d x }e^{\frac1y \int_0^y \ln g(x) \d x} \\ &&&= F(y)G(y) \end{align*}
3. Suppose \(f(x) = b^x\), then \begin{align*} && F(y) &= b^{\frac1y \int_0^y \log_b f(x) \d x} \\ &&&= b^{\frac1y \int_0^y x \d x}\\ &&&= b^{\frac1y \frac{y^2}{2}} \\ &&&= b^{\frac{y}2} = \sqrt{b^y} \end{align*}
4. Suppose the geometric mean of \(f(x)\) is \(\sqrt{f(y)}\) then the geometric mean of \(f(x)^2\) is \(f(y)\) by the the second part.

---

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

---

Solution: \begin{align*} && I &= \int_0^{\frac{1}{2}\pi} \frac{\sin 2x}{1+\sin^2x} \d x \\ &&&= \int_0^{\frac{1}{2}\pi} \frac{2 \sin x \cos x}{1+\sin^2x} \d x \\ &&&= \left [\ln (1 + \sin^2 x) \right]_0^{\pi/2} \\ &&&= \ln 2 \\ \\ && J &= \int_0^{\frac{1}{2}\pi} \frac{\sin x}{1+\sin^2x} \d x \\ &&&= \int_0^{\frac{1}{2}\pi} \frac{\sin x}{2-\cos^2x} \d x \\ &&&= \frac{1}{2\sqrt{2}}\int_0^{\frac{1}{2}\pi} \left ( \frac{\sin x}{\sqrt{2}-\cos x}+ \frac{\sin x}{\sqrt{2}+\cos x} \right) \d x \\ &&&= \frac{1}{2\sqrt{2}} \left [\ln (\sqrt{2}-\cos x) - \ln (\sqrt{2}+\cos x) \right]_0^{\pi/2} \\ &&&= \frac{1}{2\sqrt{2}} \left (-\ln(\sqrt{2}-1)+\ln(\sqrt{2}+1) \right) \\ &&&= \frac1{2\sqrt{2}} \ln \left (\frac{\sqrt{2}+1}{\sqrt{2}-1} \right)\\ &&&= \frac1{\sqrt{2}} \ln (\sqrt{2}+1) \end{align*} \begin{align*} && (1+\sqrt{2})^5 + (1-\sqrt{2})^5 &= 2(1+10\cdot2+5\cdot2^2) \\ &&&= 82 \\ && |(1-\sqrt{2})^5| & < 1 \\ && (1+\sqrt{2})^5 &< 83 < 99 \\ \\ && 1.4^2 &= 1.96 \\ &&&< 2 \\ \Rightarrow && 1.4 &<\sqrt{2} \\ \\ \Rightarrow && 2^{\sqrt{2}} &> 2^{1.4} \\ &&&=2^{7/5} \\ &&&= {128}^{1/5} \\ &&&>99^{1/5} \\ &&&>1+\sqrt{2} \end{align*} \begin{align*} && \ln 2 & > \frac{1}{\sqrt{2}} \ln(\sqrt{2}+1) \\ \Leftrightarrow && \sqrt{2} \ln 2 &> \ln(\sqrt{2}+1) \\ \Leftrightarrow && 2^{\sqrt{2}} &> 1+\sqrt{2} \end{align*} which we have already shown, so the first integral is larger.

---

Solution: \begin{align*} && f(x) &= \sin x \\ \Rightarrow && f''(x) &= -\sin x \end{align*} which is clearly negative on \((0,\pi)\) since \(\sin\) is positive on this interval. \begin{align*} && f(x) &= \ln x \\ \Rightarrow && f''(x) &= -1/x^2 \end{align*} which is clearly negative for \(x > 0\)

1. Since \(A,B,C\) are angles in a triangle, we must have \(0 < A,B,C< \pi\) and so we can apply Jensen with \(f = \sin\) to obtain: \begin{align*} &&\frac13( \sin A + \sin B + \sin C) &\leq \sin \left ( \frac{A+B+C}{3}\right) \\ &&&= \sin \frac{\pi}{3} = \frac{\sqrt{3}}2 \\ \Rightarrow && \sin A + \sin B + \sin C &\leq\frac{3\sqrt{3}}2 \end{align*}
2. Suppose \(f(x) = \ln x\), then applying Jensen on the positive numbers \(t_1, \ldots, t_n\) we obtain \begin{align*} && \frac1n \left ( \sum_{i=1}^n \ln t_n \right) &\leq \ln \left ( \frac1n\sum_{i=1}^n t_n \right) \\ \Rightarrow && \frac1n \ln\left (\prod_{i=1} t_n\right)&\leq \ln \left ( \frac1n\sum_{i=1}^n t_n \right) \\ \Rightarrow && \ln\left (\left (\prod_{i=1} t_n\right)^{1/n}\right)&\leq \ln \left ( \frac1n\sum_{i=1}^n t_n \right) \\ \Rightarrow && \left (\prod_{i=1} t_n\right)^{1/n}&\leq\frac1n\sum_{i=1}^n t_n \\ \Rightarrow && \sqrt[n]{t_1t_2 \cdots t_n}&\leq\frac1n(t_1 + t_2 + \cdots + t_n) \tag{AM-GM}\\ \end{align*}
   1. Applying AM-GM with \(t_1 = x^4, t_2 = y^4, t_3 = z^4, t_4 = 2^4\) we have \begin{align*} && \frac{x^4+y^4+z^4+16}{4} & \geq \sqrt[4]{x^4y^4z^42^4} \\ \Rightarrow && x^4+y^4+z^4+16 &\geq 8xyz \end{align*}
   2. Applying AM-GM with \(t_1 = x^5, t_2 = y^5, t_3 = z^5, t_4 = 1^5, t_5 = 1^5\) we have \begin{align*} && \frac{x^5+y^5+z^5+1+1}{5} & \geq \sqrt[5]{x^5y^5z^5} \\ \Rightarrow && x^5+y^5+z^5+2 &\geq 5xyz \\ \Rightarrow && x^5+y^5+z^5 - 5xyz &\geq -2 \end{align*} Therefore the minimum is \(-2\)

---

Solution:

1. \begin{align*} \log_{10}5 &= \log_{10} 10 - \log_{10}2 \\ &= 1- \log_{10} 2 \\ &= 0.699\\ \\ \log_{10} 6 &= \log_{10} 2 + \log_{10} 3 \\ &= 0.301029996+0.477121255 \\ &= 0.778 \end{align*} \begin{align*} && 5 \times 10^{47} < 3^{100} < 6 \times 10^{47} \\ \Leftrightarrow && 47 + \log_{10} 5 < 100 \log_{10} 3 < \log_{10} 6 + 47 \\ \Leftrightarrow &&47.699< 47.71 < 47.778 \\ \end{align*} Which is true. Therefore the first digit of \(3^{100}\) is 5.
2. \(\log_{10} 2^{1000} = 1000 \log_{10} 2 = 301.02\cdots\). Therefore it starts with a \(1\). \(\log_{10}2^{10\, 000} = 10\,000 \log_{10} 2 = 3010.2\) therefore this also starts with a \(1\). \(\log_{10} 2^{100\, 000} = 100\,000 \log_{10} 2 = 30102.9996\) therefore it starts with a \(9\)

---

Solution:

1. \begin{align*} \log_{10}(\log_{10} a_1) &= \log_{10} (\log_{10} (x!) \\ &\approx \log_{10} (\log_{10} \sqrt{2 \pi} x^{x+\frac12} e^{-x}) \\ &= \log_{10} \l \log_{10} \sqrt{2 \pi} + (x+\frac12) \log_{10} x-x \r \\ &= \log_{10} \l \log_{10} \sqrt{2 \pi} + (100x+50)-x \r \\ &= \log_{10} \l 99x + \epsilon \r \\ &\approx \log_{10} 99 + \log_{10} x \\ &\approx 2 + 100 = 102 \end{align*}
2. \begin{align*} \log_{10}(\log_{10} a_2) &= \log_{10}(\log_{10} x^y) \\ &= \log_{10} y + \log_{10} \log_{10} x \\ &= x + 2 \end{align*} \begin{align*} \log_{10}(\log_{10} a_3) &= \log_{10}(\log_{10} y^x) \\ &= \log_{10} x + \log_{10} \log_{10} y \\ &= 100 + \log_{10} x \\ &= 200 \end{align*} \begin{align*} \log_{10}(\log_{10} a_4) &= \log_{10}(\log_{10} z^x) \\ &= \log_{10} x + \log_{10} \log_{10} z \\ &= 100 + \log_{10} y \\ &= 100+x \end{align*} \begin{align*} \log_{10}(\log_{10} a_5) &= \log_{10}(\log_{10} e^{xyz}) \\ &= \log_{10} x + \log_{10}y+\log_{10} z+ \log_{10} \log_{10} e \\ &\approx 100 + x + y \end{align*} \begin{align*} \log_{10}(\log_{10} a_6) &= \log_{10}(\log_{10} z^{1/y}) \\ &= \log_{10}(\log_{10} 10) \\ &= 0 \end{align*} \begin{align*} \log_{10}(\log_{10} a_7) &= \log_{10}(\log_{10} y^{z/x}) \\ &= \log_{10}z-\log_{10} x + \log_{10} \log_{10} y \\ &= y \end{align*} Since \(0 < 102 < 200 < x+2 < x+100 < y < y+x+100\) we must have \(a_6 < a_1 < a_3 < a_2 < a_4 < a_7 < a_5\)

---

Solution:

1. True. \begin{align*} && \ln a \cdot \ln b &= \ln b \cdot \ln a \\ \Leftrightarrow && \exp ( \ln a \cdot \ln b) &= \exp ( \ln b \cdot \ln a) \\ \Leftrightarrow && \exp ( \ln a )^{\ln b} &= \exp ( \ln b )^{\ln a} \\ \Leftrightarrow && a^{\ln b} &= b^{\ln a} \\ \end{align*}
2. False. Consider \(\theta = 0\). We'd need \(\cos 0 = 1 = \sin 1\), but \(0 < 1 < \frac{\pi}{2}\) so \(\sin 1 \neq 1\)
3. False. If the polynomial has positive degree, then as \(n \to \infty\), \(\P(x) \to \pm \infty\), in particular it must be well outside the interval \([-1,1]\). Therefore it can't be within \(10^{-6}\) of \(\cos \theta\) which is confined to that interval. The only polynomial which is restricted to that range are constants, but then \(|\cos 0 - c| \leq 10^{-6}\) and \(|\cos \pi - c| \leq 10^{-6}\) \(2 = |1-(-1)| \leq |1-c| + |-1-c| \leq 2\cdot 10^{-6}\) contradiction.
4. True. \begin{align*} && (x^2-x^{-2})^2 &\geq 0 \\ \Leftrightarrow && x^4-2+x^{-4} &\geq0 \\ \Leftrightarrow && x^4+3+x^{-4} &\geq 5 \\ \end{align*}

---

Solution: \begin{align*} && y &= \ln x - x \ln a \\ \Rightarrow && y' &= \frac1x - \ln a \\ && y'' &= -\frac{1}{x^2} \end{align*} Therefore the maximum is when \(x = \frac{1}{\ln a}\) and \(y_{max} = -\ln \ln a - 1\). If \(y_{max} < 0\) then \(y \neq 0\). But that's equivalent to \(a > e^{1/e}\). \begin{align*} && 0 &> -\ln \ln a - 1 \\ \Leftrightarrow && 1 &> - \ln \ln a \\ \Leftrightarrow && \ln \ln a &>-1 \\ \Leftrightarrow && \ln a &> e^{-1} \\ \Leftrightarrow && a & > e^{1/e} \end{align*} If \(0 < a < 1\) then, when \(x\) is small, \(\ln x - x \ln a\) is large and negative. When \(x\) is large and positive \(\ln x\) is positive and \(-x \ln a\) is positive. We also notice there is no turning point. Hence exactly one solution

---

Solution:

If \(n^m = m^n \Rightarrow \frac{\ln m}{m} = \frac{\ln n}{n}\) so if \(n < m\) we must have that \(n < e \Rightarrow n = 1,2\). \(n = 1\) has no solutions, so consider \(n = 2\) which obviously has the corresponding solution \(m=4\). Since \(h(x)\) is decreasing for \(x > e\) we know this is the only solution. Hence the only solution for distinct \(m,n\) are \((m,n) = (2,4), (4,2)\)

---

Solution: Let \(S = \log_{2}\mathrm{e}-\log_{4}\mathrm{e}+\log_{16}\mathrm{e}-\ldots+(-1)^{n}\log_{2^{2^{n}}}\mathrm{e}+\ldots\) then \begin{align*} S &= \sum_{n=0}^{\infty} (-1)^n \log_{2^{2^n}} e \\ &= \sum_{n=0}^{\infty} (-1)^n \frac{\log e}{\log {2^{2^n}}} \\ &= \sum_{n=0}^{\infty} (-1)^n \frac{\log e}{2^n\log {2}} \\ &= \frac{\log e}{\log 2} \sum_{n=0}^{\infty} \frac{(-1)^n}{2^n} \\ &= \frac{1}{\log_e 2} \frac{1}{1+\frac12} \\ &= \frac{1}{\ln (2^{3/2})} \\ &= \frac{1}{\ln (2 \sqrt{2})} \end{align*}