Problems

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Solution:

Since we're at limiting equilibrium and about to slip \(Fr_B = \mu R_B\) and \(Fr_A = \mu R_A\) \begin{align*} \text{N2}(\parallel OB): && \mu R_B + R_A - W \cos \alpha &= 0 \\ \text{N2}(\parallel OA): && R_B - \mu R_A - W \sin \alpha &= 0 \\ \\ \Rightarrow && \sin\alpha \l \mu R_B + R_A \r - \cos \alpha \l R_B - \mu R_A \r &= 0 \\ \Leftrightarrow && R_A(\sin \alpha + \mu \cos \alpha) - R_B(\cos \alpha - \mu \sin \alpha) &= 0 \\ \Rightarrow && \frac{\tan \alpha + \mu}{1 - \mu \tan \alpha} R_A &= R_B\\ && \tan (\alpha + \gamma) R_A &= R_B \\ \\ \\ \overset{\curvearrowleft}{\text{midpoint}}: && R_A \sin \beta - \mu R_A \cos \beta - R_B \cos \beta - \mu R_B \sin \beta &= 0\\ \Rightarrow && \tan \beta - \mu - \tan (\alpha + \gamma) - \mu \tan (\alpha + \gamma) \tan \beta &= 0\\ \Rightarrow && \tan \beta \l 1 - \mu \tan (\alpha + \gamma) \r - \mu - \tan (\alpha + \gamma) &= 0\\ \Rightarrow && \frac{\mu + \tan (\alpha + \gamma)}{1 - \mu \tan (\alpha + \gamma)} &= \tan \beta \\ \Rightarrow && \tan (\alpha + 2\gamma) &= \tan \beta \end{align*} Since \(\alpha < \beta\) and \(\gamma < \frac{\pi}{4}\) we must have \(\alpha + 2\gamma = \beta\)

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Solution:

1. The probability no-one picks the winning number is \(\left ( 1 - \frac{1}{N}\right)^N \approx \frac1e\). \begin{align*} && \mathbb{E}(\text{profit}) &= Nc - (1-e^{-1})J \\ &&& < Nc -(1- \tfrac12 )J \\ &&& < Nc - \frac12 J \\ &&&= \frac{2Nc-J}{2} \end{align*} Therefore if \(J = 2Nc\) the expected profit is negative.
2. \(\,\) \begin{align*} && 1 &= \sum_{\text{all numbers}} \mathbb{P}(\text{pick }i) \\ &&&= \sum_{\text{popular numbers}} \mathbb{P}(\text{pick }i)+\sum_{\text{unpopular numbers}} \mathbb{P}(\text{pick }i) \\ &&&=\gamma N \frac{a}{N} + (1-\gamma)N \frac{b}{N} \\ &&&= \gamma a + (1-\gamma)b \end{align*} \begin{align*} && \mathbb{P}(\text{no-one picks winning number}) &= \mathbb{P}(\text{no-one picks winning number} | \text{winning number is popular})\mathbb{P})(\text{winning number is popular}) + \\ &&&\quad + \mathbb{P}(\text{no-one picks} | \text{unpopular})\mathbb{P}(\text{unpopular}) \\ &&&= \left (1 - \frac{a}{N} \right)^N \gamma + \left (1 - \frac{b}{N} \right)^N (1-\gamma) \\ &&&\approx \gamma e^{-a} + (1-\gamma)e^{-b} \\ \\ && \mathbb{E}(\text{profit}) &= Nc - (1-\gamma e^{-a} - (1-\gamma)e^{-b})J \\ &&&= Nc-J+J\gamma e^{-a} +J(1-\gamma)e^{-b} \end{align*} If \(\gamma = \frac18\) and \(a=9b\), then \(1=\frac18 a + \frac78b = 2b \Rightarrow b = \frac12, a = \frac92\) and \begin{align*} && \mathbb{E}(\text{profit}) &= Nc-J +J\tfrac18e^{-9/2}+J\tfrac78e^{-1/2} \\ &&&= Nc-J+\tfrac18Je^{-1/2}(e^{-4}+7) \end{align*} If we can show \(e^{-1/2}\frac{e^{-4}+7}{8} > \frac12\) we'd be done, so \begin{align*} && e^{-1/2}\frac{e^{-4}+7}{8} &> \frac12 \\ \Leftrightarrow && e^{-4}+7 &>4e^{1/2} \\ \Leftrightarrow && 49+14e^{-4}+e^{-8} &>16e \\ \end{align*} But clearly the LHS \(>49\) and the RHS \(<48\) so we're done

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Solution: The \(1\)st slice of bread can only be the first slice in a sandwich or a slice of toast. Therefore \(s_1 = 0\) \begin{align*} && t_r &= \underbrace{s_{r-1}}_{r-1\text{th is the end of a sandwich}} \cdot \underbrace{p}_{\text{and we make toast}} + \underbrace{t_{r-1}}_{r-1\text{th is toast}} \cdot \underbrace{p}_{\text{and we make toast}} \\ &&&= (s_{r-1}+t_{r-1})p \\ \\ && s_r &= 1-\mathbb{P}(\text{previous slice is not the first of a sandwich}) \\ &&&= 1-(s_{r-1} + t_{r-1}) \\ \\ \Rightarrow && s_r &= 1 - \frac{t_r}{p} \\ \Rightarrow && t_r &= p - ps_r \\ \Rightarrow && s_r &= 1 - s_{r-1} - (p-ps_{r-1}) \\ &&&= 1 -p -(1-p)s_{r-1} \\ &&&= q(1-s_{r-1}) \end{align*} Therefore since \(s_r + qs_{r-1} = q\) we should look for a solution of the form \(s_r = A(-q)^r + B\). The particular solution will have \((1+q)B = q \Rightarrow B = \frac{q}{1+q}\), the initial condition will have \(s_1 = \frac{q}{1+q} +A(-q) = 0 \Rightarrow q = \frac{1}{1+q}\), so we must have \begin{align*} && s_r &= \frac{q+(-q)^r}{1+q}\\ \Rightarrow && t_r &= p(1-s_r) \\ &&&= p \frac{1+q-q-(-q)^r}{1+q} \\ &&&= \frac{(1-q)(1-(-q)^r)}{1+q} \\ && s_n &= 1-\frac{q+(-q)^{n-1}}{1+q} - \frac{p(1-(-q)^{n-1})}{1+q} \\ &&&= 1-\frac{1+(1-p)(-q)^{n-1}}{1+q}\\ &&&= 1-\frac{1-(-q)^n}{1+q}\\ &&&= \frac{q+(-q)^n}{1+q}\\ && t_n &=1-s_n \\ &&&=\frac{1-(-q)^n}{1+q} \end{align*}

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Solution:

1. \(\,\) \begin{align*} && I_n &= \int_0^1 x^n \arctan x \d x \\ &&&= \left [ \frac{x^{n+1}}{n+1} \arctan x\right]_0^1 - \int_0^1 \frac{x^{n+1}}{n+1} \frac{1}{1+x^2} \d x \\ &&&= \frac{1}{n+1} \frac{\pi}{4} - \frac{1}{n+1} \int_0^1 \frac{x^{n+1}}{1+x^2}\d x \\ \Rightarrow && (n+1)I_n &= \frac{\pi}{4} - \int_0^1 \frac{x^{n+1}}{1+x^2}\d x \\ && I_0 &= \frac{\pi}{4} - \int_0^1 \frac{x}{1+x^2} \d x \\ &&&= \frac{\pi}{4} - \left [\frac12 \ln(1+x^2) \right]_0^1 \\ &&&= \frac{\pi}{4} - \frac12 \ln 2 \end{align*}
2. \(\,\) \begin{align*} && (n+3)I_{n+2} + (n+1)I_n &=\left ( \frac{\pi}{4} - \int_0^1 \frac{x^{n+3}}{1+x^2} \d x \right)+ \left (\frac{\pi}{4} - \int_0^1 \frac{x^{n+1}}{1+x^2} \d x \right) \\ &&&=\frac{\pi}{2}+ \int_0^1 \frac{x^{n+1}+x^{n+3}}{1+x^2} \d x \\ &&&=\frac{\pi}{2}+ \int_0^1 x^{n+1} \d x \\ &&&= \frac{\pi}{2} + \frac{1}{n+2} \\ && 3I_2 + I_0 &= \frac{\pi}{2} + \frac{1}{2} \\ \Rightarrow && 3I_2 &=\frac{\pi}{4} + \frac12 \ln 2 + \frac12 \\ && 5I_4 + 3I_2 &= \frac{\pi}{2} + \frac14 \\ \Rightarrow && 5I_4 &= \frac{\pi}{2} + \frac14 - \left ( \frac{\pi}{4} + \frac12 \ln 2 + \frac12\right) \\ &&&= \frac{\pi}4-\frac12 \ln 2-\frac14 \\ \Rightarrow && I_4 &= \frac15 \left (\frac{\pi}4-\frac12 \ln 2-\frac14 \right) \\ &&&= \frac1{20} \left (\pi - 2\ln 2 -1 \right) \end{align*}
3. Claim: \[ (4n+1) I_{4n} =\frac{\pi}4-\frac12 \ln 2 - \frac12 \sum_{r=1}^{2n} (-1)^r \frac 1 {r} \,, \] Proof: Base case we have just shown above Assume true for \(n = k\), consider \(n = k+1\), then \begin{align*} && (4(k+1)+1) I_{4(k+1)} &= \frac{\pi}{2} + \frac{1}{4(k+1)} - (4k+3)I_{4k+2} \\ &&&= \frac{\pi}{2} + \frac{1}{4(k+1)} - \left (\frac{\pi}{2} + \frac{1}{2(2k+1)} - (4k+1)I_{4k} \right)\\ &&&= (4k+1)I_{4k} - \frac12 \left ( \frac{1}{2k+2} - \frac{1}{2k+1}\right) \\ &&&= \frac{\pi}4-\frac12 \ln 2 - \frac12 \sum_{r=1}^{2k} (-1)^r \frac 1 {r} - \frac12 \left ( \frac{1}{2k+2} - \frac{1}{2k+1}\right)\\ &&&= \frac{\pi}4-\frac12 \ln 2 - \frac12 \sum_{r=1}^{2(k+1)} (-1)^r \frac 1 {r} \\ \end{align*} as required.

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Solution: \begin{align*} x_{n+2} &= \frac{ax_{n+1}-1}{x_{n+1}+b} \\ &= \frac{a \frac{ax_n - 1}{x_n+b}-1}{\frac{ax_n - 1}{x_n+b}+b} \\ &= \frac{a(ax_n-1)-(x_n+b)}{ax_n-1+b(x_n+b)} \\ &= \frac{(a^2-1)x_n-(a+b)}{(a+b)x_n+b^2-1} \end{align*}

1. If \(x_{n+2} = x_n\) then \begin{align*} && x_n &= \frac{(a^2-1)x_n-(a+b)}{(a+b)x_n+b^2-1} \\ \Rightarrow && 0 &=(a+b)x_n^2+(b^2-a^2)x_n+(a+b) \\ &&&= (a+b)(x_n^2+(a-b)x_n + 1) \end{align*} If \(x_{n+1} = x_n\) then \(x_n^2+(a-b)x_n + 1\) and since our sequence has period \(2\) rather than \(1\) it must be the case this is non-zero. Therefore \(a+b =0\).
2. \begin{align*} x_{n+4} &= \frac{(a^2-1)x_{n+2}-(a+b)}{(a+b)x_{n+2}+b^2-1} \\ &= \frac{(a^2-1)\frac{(a^2-1)x_{n}-(a+b)}{(a+b)x_{n}+b^2-1} -(a+b)}{(a+b)\frac{(a^2-1)x_{n}-(a+b)}{(a+b)x_{n}+b^2-1} +b^2-1} \\ &= \frac{((a^2-1)^2-(a+b)^2)x_n -(a^2+b^2-2)(a+b)}{(a^2+b^2-2)(a+b)x_n + (b^2-1)^2-(a+b)^2} \end{align*} If \(x_{n+4} = x_n\) then \begin{align*} x_n &=\frac{((a^2-1)^2-(a+b)^2)x_n -(a^2+b^2-2)(a+b)}{(a^2+b^2-2)(a+b)x_n + (b^2-1)^2-(a+b)^2} \\ 0 &= (a^2+b^2-2)(a+b)x_n^2 + \l (b^2-1)^2-(a^2-1)^2 \r x_n+(a^2+b^2-2)(a+b) \\ &= (a^2+b^2-2)(a+b)x_n^2+(b^2-a^2)(a^2+b^2-2)x_n + (a^2+b^2-2)(a+b) \\ &= (a^2+b^2-2)(a+b)(x_n^2+(b-a)x_n + 1) \end{align*} Since we do not want \(x_n\) to be periodic with period \(1\) we must have the quadratic in \(x_n\) \(\neq 0\). If \(a+b = 0\) then \(x_n\) is periodic with period \(2\) since \(x_{n+2} = \frac{(a^2-1)x_n}{((-a)^2-1)} = x_n\). Therefore it is necessary that \(a^2+b^2-2 = 0\). If \(a^2+b^2-2= 0\) then \begin{align*} x_{n+4} &= \frac{((a^2-1)^2-(a+b)^2)x_n}{(b^2-1)^2-(a+b)^2} \\ &=\frac{((a^2-1)^2-(a+b)^2)x_n}{((2-a^2)-1)^2-(a+b)^2} \\ &=\frac{((a^2-1)^2-(a+b)^2)x_n}{((1-a^2)^2-(a+b)^2} \\ &= x_n \end{align*} Therefore it is sufficient too. So our conditions are \(a+b \neq 0, \, \, x_n^2+(a-b)x_n + 1 \neq 0\) and \(a^2+b^2-2 = 0\)

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Solution:

1. \(\,\)
   

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2. \(\,\) \begin{align*} && \sin y &= \tfrac12 \sin x \\ \Rightarrow && \frac{\d y}{\d x} \cos y &= \tfrac12 \cos x \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{\cos x}{2 \cos y} \\ &&&= \frac{\cos x}{2 \sqrt{1-\sin^2 y}} \\ &&&= \frac{\cos x}{2 \sqrt{1-\frac14 \sin^2 x}} \\ &&&= \frac{\cos x}{\sqrt{4-\sin^2 x}} \\ \\ && y'' &= \frac{-\sin x \cdot (4-\sin^2 x)^{\frac12} - \cos x \cdot (4-\sin^2 x)^{-\frac12} \cdot 2 \sin x \cos x}{(4-\sin^2 x)} \\ &&&= \frac{-\sin x \cdot (4-\sin^2 x) - \cos x \cdot 2 \sin x \cos x}{(4-\sin^2x)^{\frac32}} \\ &&&= \frac{-\sin x \cdot (4-\sin^2 x) - \sin x (1-\sin^2x)}{(4-\sin^2x)^{\frac32}} \\ &&&= \frac{-3\sin x }{(4-\sin^2x)^{\frac32}} \\ \end{align*}
   

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3. \(\,\)
   

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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. \(\,\) \begin{align*} && \frac{\d x}{\d t} &= 2at \\ && \frac{\d y}{\d t} &= 2a \\ \Rightarrow && \frac{\d y}{\d x} &= \frac1t \\ && -p &= \text{grad of normal} \\ &&&= \frac{y-2ap}{x-ap^2} \\ \Rightarrow && y &= -px + ap^3+2ap \\ && 2an &= -pan^2 + ap^3 + 2ap \\ \Rightarrow && 0 &= pan^2+2an-ap(2+p^2) \\ \Rightarrow && n &= p, -\left ( p + \frac2{p}\right) \\ \Rightarrow && n &= -\left ( p + \frac2{p}\right) \end{align*}
2. \(\,\) \begin{align*} && |PN|^2 &= (ap^2-an^2)^2 +(2ap-2an)^2 \\ &&&= a^2(p-n)^2(p+n)^2+4a^2(p-n)^2 \\ &&&= a^2(p-n)^2((p+n)^2+4) \\ &&&= a^2\left(p+p+\frac2p \right)^2 \left ( \left ( -\frac2p\right)^2+4\right)\\ &&&= a^2\left(\frac{2p^2+2}p \right)^2 \left ( \frac{4}{p^2}+4\right)\\ &&&= 16a^2 \frac{(p^2+1)^3}{p^4} \\ \\ && \frac{\d |PN|^2}{\d p^2} &= 16a^2\frac{3(p^2+1)^2p^4-2(p^2+1)^3p^2}{p^8} \\ &&&= 16a^2(p^2+1)^2 \frac{3p^2-2(p^2+1)}{p^6} \\ &&&= 16a^2(p^2+1)^2 \frac{p^2-2}{p^6} \end{align*} Therefore minimized when \(p^2=2\) (clearly a minimum by considering behaviour as \(p^2 \to 0, \infty\))
3. If \(PN\) is the diameter of \(PNQ\) then \(QP\) and \(QN\) are perpendicular, ie \begin{align*} && -1 &= \frac{2ap-2aq}{ap^2-aq^2} \cdot \frac{2aq-2an}{aq^2-an^2} \\ &&&= \frac{2}{p+q} \cdot \frac{2}{q+n} \\ &&&= \frac{2}{p+q} \cdot \frac{2}{q - p -\frac{2}{p}} \\ \Rightarrow && 4 &= (p+q)(p+\frac2{p}-q) \\ &&&= p^2-q^2 + \frac{2q}{p} + 2 \\ \Rightarrow && 2 &= p^2 - q^2 + \frac{2q}{p} \end{align*} Therefore \(q = 0 \Rightarrow p^2 = 2 \Rightarrow |PN|\) is at it's minimum.

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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*}