Problems

Solution:

1. \(\,\) \begin{align*} && f(x) &= \frac{x+\sqrt3}{1-\sqrt3x} \\ \Rightarrow && f(f(x)) &= \frac{f(x)+\sqrt3}{1-\sqrt3f(x)} \\ &&&= \frac{\frac{x+\sqrt3}{1-\sqrt3x}+\sqrt3}{1-\sqrt3 \frac{x+\sqrt3}{1-\sqrt3x}} \\ &&&= \frac{x+\sqrt{3}+\sqrt3(1-\sqrt3x)}{1-\sqrt3x-\sqrt3(x+\sqrt3)} \\ &&&= \frac{-2x+2\sqrt3}{-2-2\sqrt3x} \\ &&&= \frac{x-\sqrt3}{1+\sqrt3 x} \\ \\ && f^3(x) &= f^2(f(x)) \\ &&&= \frac{f(x)-\sqrt3}{1+\sqrt3 f(x)} \\ &&&=\frac{\frac{x+\sqrt3}{1-\sqrt3x}-\sqrt3}{1+\sqrt3 \frac{x+\sqrt3}{1-\sqrt3x}} \\ &&&= \frac{(x+\sqrt3)-\sqrt3(1-\sqrt3 x)}{(1-\sqrt3x)+\sqrt3 (x+\sqrt3)} \\ &&&= \frac{-2x}{-2} = x \\ \\ && f^{2007}(x) &= x \end{align*}
2. If \(x = \tan \theta\) then \(f(x) = \frac{\tan \theta + \tan \frac{\pi}{3}}{1 - \tan \frac{\pi}{3} \tan \theta} = \tan (\theta + \frac{\pi}{3})\) and hence \(f^n(x) = \tan (\theta + \frac{n \pi}{3})\)
3. Note that if \(t = \sin \theta\) then \(g(t) = \cos \frac{\pi}{6} t\sin \theta + \frac12 \cos \theta = \sin(\theta + \frac{\pi}6)\) therefore \(g^n(t) = \sin(\sin^{-1}(t) + \frac{n\pi}{6})\)

Solution:

1. \begin{align*} && t(x) &= \int_0^x \frac{1}{1+u^2} \d u \\ u = v^{-1}, \d u = -v^{-2} \d v&&&= \int_{v = \infty}^{v = 1/x} \frac{1}{1+v^{-2}} \frac{-1}{v^2} \d v \\ &&&= \int_{1/x}^\infty \frac{1}{1+v^2} \d v \\ \\ \Rightarrow && t(x) + t(1/x) &= \int_0^x \frac{1}{1+u^2} \d u + \int_0^{1/x} \frac{1}{1+u^2} \d u \\ &&&= \int_{1/x}^{\infty} \frac{1}{1+u^2} \d u + \int_0^{1/x} \frac{1}{1+u^2} \d u \\ &&&= \int_0^{\infty} \frac{1}{1+u^2} \d u \\ &&&= \frac12 p \\ \\ \Rightarrow && t(1) +t(1/1) = 2t(1) &= \frac12 p \end{align*}
2. \(\,\) \begin{align*} && y &= \frac{u}{\sqrt{1+u^2}} \\ \Rightarrow && y^2 &= \frac{u^2}{1+u^2} \\ &&&= 1-\frac{1}{1+u^2} \\ \Rightarrow && 1+u^2 &= \frac{1}{1-y^2} \\ \Rightarrow && u &= \frac{y}{\sqrt{1-y^2}} \\ \\ && \frac{\d u}{\d y} &= \frac{\sqrt{1-y^2} + y^2(1-y^2)^{-1/2}}{1-y^2} \\ &&&= \frac{1}{(1-y^2)^{3/2}} \\ \\ && t(x) &= \int_0^x \frac{1}{1+u^2} \d u \\ &&&= \int_0^{y = x/\sqrt{1+x^2}} \frac{1}{1 + \frac{y^2}{1-y^2}} \frac{1}{(1-y^2)^{3/2}} \d y \\ &&&= \int_0^{x/\sqrt{1+x^2}} \frac{1-y^2}{(1-y^2)^{3/2}} \d y \\ &&&= \int_0^{x/\sqrt{1+x^2}} \frac{1}{(1-y^2)^{1/2}} \d y \\ &&&= s\left ( \frac{x}{\sqrt{1+x^2}} \right) \\ \\ \Rightarrow && s\left ( \frac{1}{\sqrt{2}} \right) &= t(1) = \frac14p \end{align*}
3. \(\,\) \begin{align*} && z &= \frac{u + \frac1{\sqrt{3}}}{1- \frac{1}{\sqrt{3}} u}\\ \Rightarrow && z - \frac{z}{\sqrt{3}}u &= u + \frac{1}{\sqrt{3}} \\ \Rightarrow && u &= \frac{z-\frac{1}{\sqrt{3}}}{1 + \frac{z}{\sqrt{3}}} \\ \\ \Rightarrow && \frac{\d u}{\d z} &= \frac{\sqrt{3}(\sqrt{3}+z ) -(\sqrt{3}z-1)}{\left (\sqrt{3}+z \right)^2} \\ &&&= \frac{4}{(\sqrt{3}+z)^2} \\ \\ \Rightarrow && t \left ( \frac{1}{\sqrt{3}} \right) &= \int_0^{1/\sqrt{3}} \frac{1}{1+u^2} \d u \\ &&&= \int_{z=1/\sqrt{3}}^{z=\sqrt{3}} \frac{1}{1 + \left ( \frac{\sqrt{3}z-1}{\sqrt{3}+z}\right)^2} \frac{4}{(\sqrt{3}+z)^2} \d z\\ &&&= \int_{1/\sqrt{3}}^{\sqrt{3}} \frac{4}{(\sqrt{3}+z)^2+(\sqrt{3}z-1)^2} \d z \\ &&&= \int_{1/\sqrt{3}}^{\sqrt{3}} \frac{4}{4+4z^2} \d z \\ &&&= \int_{1/\sqrt{3}}^{\sqrt{3}} \frac{1}{1+z^2} \d z \end{align*} Notice that \(t(1/\sqrt{3})+t(\sqrt{3}) = \frac12p\) and also notice that \(t(1/\sqrt{3}) + t(1/\sqrt{3}) =t(\sqrt{3})\) so \(3t(1/\sqrt{3}) = \frac12p\)

Solution:

1. \begin{align*} && I &= \int \frac{3}{(x-4)\sqrt{2x+1}}\, \d x \\ u^2 =2x+1, 2u \frac{\d u}{\d x}=2: && &= \int \frac{3}{\left(\frac{u^2-1}{2}-4\right)u} u \d u \\ &&&= \int \frac{6}{u^2-9} \d u \\ &&&= \int \frac{6}{(u-3)(u+3)} \d u\\ &&&= \int \left ( \frac{1}{u-3} - \frac{1}{u+3} \right )\d u \\ &&&= \ln (u-3) - \ln (u+3) + C \\ &&&= \ln \frac{u-3}{u+3} + C \\ &&&= \ln \left (\frac{\sqrt{2x+1}-3}{\sqrt{2x+1}+3} \right) + C \end{align*}
2. \begin{align*} && I &= \int_{\ln 3}^{\ln 8} \frac{2}{e^x\sqrt{e^x+1}} \d x \\ u = e^x, \frac{\d u}{\d x} = e^x: &&&= \int_{u=3}^{u=8} \frac{2}{u\sqrt{u+1}} \frac{1}{u} \d u \\ v^2=u+1, 2v \frac{\d v}{\d u} = 1: &&&= \int_{v=2}^{v=3} \frac{2}{v(v^2-1)^2} \d v \\ &&&= \int_2^3 \left ( \frac{2}{v} - \frac{}{v-1} - \frac{1}{2(v-1)^2} - \frac{1}{v+1} - \frac{1}{2(v+1)^2}\right) \d v \\ &&&= \left [2\ln v - \ln(v^2-1)+\frac12(v-1)^{-1}+\frac12(v+1)^{-1} \right]_2^3 \\ &&&= \left ( 2\ln3-\ln 8+\frac14+\frac18\right)-\left ( 2\ln2-\ln 3+\frac12+\frac16\right) \\ &&&= \end{align*}

Solution:

1. Suppose \(a^2-2b^2=1\) then \begin{align*} (3a+4b)^2-2(2a+3b)^2 &= 9a^2+24ab+16b^2-2\cdot(4a^2+12ab+9b^2) \\ &=a^2-2b^2 \\ &= 1 \end{align*} Therefore \((3a+4b,2a+3b)\) also lies on the curve.
2. Suppose \(Ma^2-Nb^2 = 1\) then \begin{align*} M(5a+6b)^2-N(4a+5b)^2 &= M\cdot(25a^2+60ab+36b^2) - N\cdot(16a^2+40ab+25b^2) \\ &= (25M-16N)a^2+20\cdot(3M-2N)ab+(36M-25N)b^2 \end{align*} Therefore we need \(3M = 2N\) so the smallest possible value would have to be \(M = 2, N = 3\), which does work
3. Consider \(x + \sqrt{3}y\), then consider \((x+\sqrt{3}y)(2+\sqrt{3}) = (2x+3y)+(x+2y)\sqrt{3}\). Notice that \((x+\sqrt{3}y)(x-\sqrt{3}y) = 1\) and \((2+\sqrt{3})(2-\sqrt{3}) = 1\) so \(((2x+3y)+(x+2y)\sqrt{3})((2x+3y)-(x+2y)\sqrt{3}) = 1\), so we can take \(P=2,Q=3,R=1,S=2\)

Solution:

1. \begin{align*} 1 &= \int_{1/4}^{1/2} \frac{\lambda}{x\ln x} \, dx \\ &= \lambda\left [ \ln |\ln x| \right ]_{1/4}^{1/2} \\ &= \lambda \l \ln |-\ln 2| - \ln |-2 \ln 2| \r \\ &= \lambda (-\ln 2) \end{align*} So \(\lambda = -\frac{1}{\ln 2} = \frac{1}{\ln \frac12}\)
2. \begin{align*} 1 &= \int_{a}^{b} \frac{1}{x\ln x} \, dx \\ &= \left [ \ln |\ln x| \right ]_{a}^{b} \\ &= \l \ln \ln b - \ln \ln a \r \\ &= \ln \l \frac{\ln b}{\ln a} \r \\ \end{align*} So \(b = e^{a}\)
3. If \(\lambda = 1, a = e, b = e^e\), then \begin{align*} \P(\e^{3/2}\le X \le \e^2) &= \int_{e^{3/2}}^{e^2} \frac{1}{x \ln x} \, dx \\ &= \left [ \ln \ln x \right]_{e^{3/2}}^{e^2} \\ &= \ln 2 - \ln \frac{3}{2} \\ &= \ln \frac{4}{3} \\ &= \ln \l 1 + \frac{1}{3} \r \\ &\approx \frac{1}{3} - \frac{1}{2 \cdot 3^2} + \frac{1}{3 \cdot 3^3} - \frac{1}{4 \cdot 3^4} \\ &= \frac{31}{108} \end{align*}
4. Note that \(2 > e^{\frac12} > 1\) so \(a = e^{\frac12}, b = e^{\frac{e}2}\). Since \(3 > e \Rightarrow e^{3/2} > e^{\frac{e}{2}}\) this probability is out of range, therefore \(\P(\e^{3/2}\le X \le \e^2) = 0\)

Solution:

1. We must have the total probability summing to \(1\), therefore \(1 =m + k\) (as \(x \to \infty\)) therefore \(k = 1-m\).
2. \(\,\) \begin{align*} && \E[X] &= \mathbb{P}(X=-1) \cdot (-1) + \int_0^{\infty} kx e^{-x} \d x \\ &&&= -m + (1-m) = 1-2m \end{align*}
3. \(\,\) \begin{align*} && \var[X] &= \E[X^2]-\E[X]^2 \\ &&&= \mathbb{P}(X=-1)\cdot(-1)^2 + (1-m)\int_0^{\infty} x^2e^{-x} \d x - (1-2m)^2 \\ &&&= m + (1-m)(1+1^2) - (1-2m)^2 \\ &&&= 3-4m - 1+4m -4m^2 \\ &&&= 2(1-m^2) \end{align*} To find the median \(q\), we need \begin{align*} && \frac12 &= \mathbb{P}(X \leq q) \\ &&&= m + (1-m)(1-e^{-q}) \\ \Rightarrow && e^{-q} &= 1-\frac{\frac12-m}{1-m} \\ &&&= \frac{1-m - \frac12+m}{1-m} \\ &&&= \frac{1}{2(1-m)} \\ \Rightarrow && q &= \ln 2(1-m) \end{align*}
4. \(\,\) \begin{align*} && \E\left [|X|^{\frac12}\right] &= \mathbb{P}(X=-1) \cdot 1 + \int_0^{\infty} \sqrt{x} (1-m)e^{-x} \d x \\ &&&= m + (1-m)\int_0^\infty \sqrt{x} e^{-x} \d x \\ u^2 = x, \d x = 2u \d u : &&&= m + (1-m) \int_{u=0}^{u=\infty} u e^{-u^2} \cdot 2u \d u \\ &&&= m + 2(1-m) \int_0^{\infty} u^2 e^{-u^2} \d u \\ &&&= m + (1-m)\frac{\sqrt{\pi}}2 \end{align*}

Solution: \begin{align*} u = x^m, \d u = m x^{m-1} && \int \frac{m}{x f(x^m)} \d x &= \int \frac{m x^{m-1}}{uf(u)} \d x \\ &&&= \int \frac{1}{u f(u)} \d u \\ &&&= F(u) + c \\ &&&= F(x^m) + c \end{align*}

1. \begin{align*} && \int \frac{1}{u(u-1)} \d u &= \int \left ( \frac{1}{u-1}-\frac{1}{u} \right ) \d u \\ &&&= \ln \left ( \frac{u-1}{u} \right) + c \\ &&&= \ln \left ( 1 - \frac{1}{u} \right) + c \\ && \int \frac{1}{x^n - x} \d x &= \int \frac{1}{x (x^{n-1}-1)} \d x \\ f(u) = u - 1: && &= \frac{1}{n-1} \ln \left ( 1 - \frac{1}{x^{n-1}} \right) + c \end{align*}
2. \begin{align*} v = \sqrt{u+1}, \d v = \tfrac12 (u+1)^{-1/2} \d u && \int \frac{1}{u\sqrt{u+1}} \d u &= \int \frac{1}{(v^2-1)} (u+1)^{-1/2} \d u \\ &&&= \int \frac{2}{v^2-1} \d v \\ &&&=\ln \frac{1-v}{1+v} + c \\ &&&= \ln \left (\frac{1-\sqrt{u+1}}{1+\sqrt{u+1}} \right)+ c \\ f(u) = \sqrt{x+1}:&& \int \frac{1}{\sqrt{x^n + x^2}} \d x &= \int \frac{1}{x\sqrt{x^{n-2}+1}} \d x \\ &&&= \frac{1}{n-2} \ln \left ( \frac{1-\sqrt{x^{n-2}+1}}{1+\sqrt{x^{n-2}+1}} \right)+c \end{align*}

Solution: \begin{align*} && f(x) &= \frac{C \, k^{a+1} \, x^a}{(x+k)^{2a+2}} \\ \Rightarrow && 1 &= \int_0^{\infty} f(x) \d x \\ &&&= \int_0^{\infty} \frac{C \, k^{a+1} \, x^a}{(x+k)^{2a+2}} \d x \\ &&&= Ck^{a+1} \int_0^{\infty} \frac{x^a}{(x+k)^{2a+2} }\d x \\ &&&= Ck^{a+1} \frac{a!(2a-a)!}{(2a+1)!k^{2a-a+1}} \\ &&&= C \frac{a!a!}{(2a+1)!} \\ \Rightarrow && C &= \frac{(2a+1)!}{a!a!} \end{align*} \begin{align*} && I &= \int_0^v \frac{x^a}{(x+k)^{2a+2}} \d x\\ u = k^2/x, \d x = -k^2u^{-2} \d u: &&&= \int_{u = +\infty}^{u = k^2/v} \frac{k^{2a}u^{-a}}{(k^2u^{-1} +k)^{2a+2}}(-k^2u^{-2}) \d u \\ &&&= \int_{u = +\infty}^{u = k^2/v} \frac{k^{2a-2a-2}u^{2a+2-a}}{(k +u)^{2a+2}}(-k^2u^{-2}) \d u \\ &&&= \int_{ k^2/v}^{\infty} \frac{u^{a}}{(k +u)^{2a+2}} \d u \\ \end{align*} At the median we want a value \(M\) such that \(M = k^2/M\) ie \(M = k\) \begin{align*} && \mathbb{E}(V) &= \int_0^{\infty} x f(x) \d x \\ &&&= \frac{(2a+1)!k^{a+1}}{a!a!} \int_0^{\infty} \frac{x^{a+1}}{(x+k)^{2a+2}} \d x \\ &&&= \frac{(2a+1)!k^{a+1}}{a!a!} \frac{(a+1)!(2a-(a+1))!}{(2a+1)!k^{2a-(a+1)+1}}\\ &&&= \frac{k^{a+1}}{a!} \frac{(a+1)(a-1)!}{k^{a}} \\ &&&= \frac{k(a+1)}{a} = \frac{a+1}a k \end{align*} \begin{align*} && \mathbb{P}(T < t) &= \mathbb{P}(\frac{s}{V} < t) \\ &&&= \mathbb{P}(V > \frac{s}{t}) \\ &&&= \int_{s/t}^{\infty} f(x) \d x \\ &&&= \int_{s/t}^{\infty} \frac{C \, k^{a+1} \, x^a}{(x+k)^{2a+2}} \d x \\ \\ \Rightarrow && f_T(t) &= \frac{\d}{\d t} \left ( \mathbb{P}(T < t)\right) \\ &&&= \frac{\d}{\d t} \left ( \int_{s/t}^{\infty} \frac{C \, k^{a+1} \, x^a}{(x+k)^{2a+2}} \d x \right) \\ &&&= - \frac{C \, k^{a+1} \, \left ( \frac{s}{t} \right)^a}{(\frac{s}{t}+k)^{2a+2}} \cdot \left (-\frac{s}{t^2} \right) \\ &&&= \frac{Ck^{a+1}s^{a+1}t^{2a+2}}{t^{a+2}(s+kt)^{2a+2}} \\ &&&= \frac{C(ks)^{a+1}t^a}{(s+kt)^{2a+2}} \\ &&&= \frac{C(\frac{s}{k})^{a+1}t^a}{(\frac{s}{k}+t)^{2a+2}} \end{align*} Therefore \(T\) follows the same distribution, but with parameter \(s/k\) rather than \(k\). In particular it has median \(s/k\) (and the product of the medians is \(s\)). However, the product of the expected time and expected speed is \(\frac{a+1}{a} k \frac{a+1}{a} \frac{s}{k} = \left ( \frac{a+1}{a} \right)^2s > s\)

Solution:

1. \begin{align*} (3+2\sqrt{5})^3 &= 3^3 + 3 \cdot 3^2 \cdot 2\sqrt{5} + 3 \cdot 3 \cdot (2 \sqrt{5})^2 + (2\sqrt{5})^3 \\ &= 27 + 180 + (54+40)\sqrt{5} \\ &= 207 + 94\sqrt{5} \end{align*}
2. \begin{align*} && (c-d\sqrt{2})^3 &= c^3+6cd -(3c^2d+2d^3)\sqrt{2} \\ \Rightarrow && 99 &= c(c^2+6d^2) \\ && 70 &= d(3c^2+2d^2) \\ \Rightarrow && c & \mid 99, d \mid 70 \\ && c &= 3, d = 2 \end{align*}
3. \begin{align*} && 0 &= x^6 - 198x^3 + 1 \\ \Rightarrow && 0 &= (x^3-99)^2+1-99^2 \\ \Rightarrow && x^3 &= 99 \pm \sqrt{99^2-1} \\ &&&= 99 \pm 10 \sqrt{98} \\ &&&= 99 \pm 70 \sqrt{2} \\ \Rightarrow && x &= 3 \pm 2 \sqrt{2} \end{align*}

Solution:

1. \(\,\) \begin{align*} && 0 &= \sqrt{3x^2+1} + \sqrt{x} -2x-1 \\ \Rightarrow && 2x+1 &= \sqrt{3x^2+1} + \sqrt{x} \\ \Rightarrow && 4x^2+4x+1 &= 3x^2+1+x+2\sqrt{x(3x^2+1)} \\ \Rightarrow && x^2+3x &= 2\sqrt{x(3x^2+1)} \\ \Rightarrow && x^2(x^2+6x+9) &= 4x(3x^2+1) \\ \Rightarrow && 0 &= x^4-6x^3+9x^2-4x \\ &&&= x(x-4)(x-1)^2 \end{align*} So clearly we have \(x = 0, x = 1, x = 4\). \(x = 0\) works, \(x = 1\) works, \(x = 4\) works.
2. \(\,\) \begin{align*} && 0 &= \sqrt{3x^2+1} - 2\sqrt{x} +x-1 \\ \Rightarrow && 1-x &= \sqrt{3x^2+1}-2\sqrt{x} \\ \Rightarrow && x^2-2x+1 &= 3x^2+1+4x-4\sqrt{x(3x^2+1)} \\ \Rightarrow && x^2+3x &= 2\sqrt{x(3x^2+1)} \\ \Rightarrow && 0 &= x(x-4)(x-1)^2 \end{align*} Again we must check \(x = 0, x = 1, x = 4\). \(x = 0,1\) work, but \(x = 4\) is not a solution.
3. \(\,\) \begin{align*} && 0 &= \sqrt{3x^2+1} - 2\sqrt{x} -x+1 \\ \Rightarrow && x-1 &= \sqrt{3x^2+1} - 2\sqrt{x} \\ \Rightarrow && x^2-2x+1 &= 3x^2+1+4x-4\sqrt{x(3x^2+1)} \\ \Rightarrow && x^2+3x &= 2\sqrt{x(3x^2+1)}\\ \Rightarrow && 0 &= x(x-4)(x-1)^2 \end{align*} So again, we need to check \(x = 0, 1, 4\). \(x = 0, 4\) work, but \(x = 1\) fails.

Solution: \begin{align*} && y &= x(x+1)(x-2)^4 \\ \Rightarrow && y' &= (x+1)(x-2)^4+x(x-2)^4+4x(x+1)(x-2)^3 \\ &&&= (x-2)^3 \left ( (2x+1)(x-2)+4x(x+1) \right) \\ &&&= (x-2)^3 \left (2x^2-3x-2+4x^2+4x \right) \\ &&&=(x-2)^3(6x^2+x-2) \\ &&&=(x-2)^3(2x-1)(3x+2) \end{align*} Therefore there are stationary points at \((2,0), (\frac12, -\frac{625}{64}), (-\frac23, -\frac{4078}{81})\) \((0,2)\) is a minimum by considering the sign of \(y'\) either side. \( (-\frac23, \frac{2560}{729})\) is a minimum, since it's the first stationary point. \( (\frac12, \frac{243}{64})\) is a maximum since you can't have consecutive minima and the second derivative is clearly non-zero.

+ - ↺

1. \(\,\)
   

   + - ↺
2. \(\,\)
   

   + - ↺

Solution:

1. \(\,\) \begin{align*} && \dot{x} &= (1-x) \\ \Rightarrow &&\int \frac{1}{1-x} \d x &= \int \d t \\ \Rightarrow && -\ln |1-x| &= t + C \\ t=0, x = 0: && -\ln 1 &= C \Rightarrow C = 0\\ \Rightarrow && -\ln|1-x| &= t \\ \Rightarrow && 1-x&= e^{-t} \\ \Rightarrow && x &= 1-e^{-t} \end{align*}
   

   + - ↺
2. Notice that \((1-x^2)^{1/2} = (1-x)^{1/2}(1+x)^{1/2} > (1-x)^{1/2} > 1-x\)
   

   + - ↺
   \begin{align*} && \dot{x} &= \sqrt{1-x^2} \\ \Rightarrow && \int \frac{1}{\sqrt{1-x^2}} \d x &= t + C \\ x = \sin y, \d x = \cos y && \int \frac{\cos y}{\cos y} \d y &= t + C \\ \Rightarrow && y &= t + C \\ \Rightarrow && \sin^{-1} x &= t + C \\ t = 0, x = 0: && x &= \sin t \end{align*}
3. \(\,\)
   

   + - ↺
   We know the gradient is steeper, so the solution must always be above \(\sin t\), which means it reaches \(1\) before \(\frac{\pi}{2}\)