Problems

Solution:

1. \begin{align*} cb(x) &= c(b(x)) \\ &= 2 \ln x \quad (x > 0) \\ ab(x) &= (b(x))^2 \\ &= (\ln x)^2 \quad (x > 0) \\ da(x) &= \sqrt{a(x)} \\ &= \sqrt{x^2} \\ &= |x| \quad (-\infty < x < \infty) \\ ad(x) &= (d(x))^2 \\ &= (\sqrt{x})^2 \\ &= x \quad (x \geq 0) \end{align*} The domains are specified above. The ranges are \(\mathbb{R}, \mathbb{R}_{\geq 0}, \mathbb{R}_{\geq 0}, \mathbb{R}_{\geq 0}\) respectively.
2. \begin{align*} fg(x) &= \sqrt{g(x)^2-1} \quad (|g(x)| \geq 1) \\ &= \sqrt{x^2+1-1} \\ &= |x| \end{align*} So \(fg: \mathbb{R} \to \mathbb{R}_{\geq 0}\). \begin{align*} gf(x) &= \sqrt{f(x)^2 + 1} \\ &= \sqrt{\left ( \sqrt{x^2-1} \right)^2+1} \quad (|x| \geq 1) \\ &= |x| \end{align*} So \(gf: \mathbb{R}_{\geq 0} \to \mathbb{R}_{\geq 0}\)
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   \begin{align*} kh(x) &= h(x) - \sqrt{h(x)^2 -1} \quad (|h(x)| \geq 1)\\ &= x + \sqrt{x^2+1} - \sqrt{(x + \sqrt{x^2+1})^2 - 1} \\ &= x + \sqrt{x^2+1} - \sqrt{x^2 + x^2 - 2x} \quad (x \geq 1) \\ &= x + \sqrt{x^2+1} - \sqrt{2x^2-2x} \quad (x \geq 1) \end{align*} This has domain \(x \geq 1\) and range, \((0, 1]\)

Solution: Since \(A\) and \(B\) return to their starting points, and at their highest points there is no vertical component to their velocities, their horizontal must perfectly reverse, ie \begin{align*} && m u \cos \alpha - M v \cos \beta &= -m u \cos \alpha + M v \cos \beta \\ \Rightarrow && mu \cos \alpha &= Mv \cos \beta \end{align*} Since they reach their highest points at the same time, they must have the same initial vertical speed, ie \(u \sin \alpha = v \sin \beta\), so \begin{align*} && m v \frac{\sin \beta}{\sin \alpha} \cos \alpha &= M v \cos \beta \\ \Rightarrow && m \cot \alpha &= M \cot \beta \end{align*} The horizontal distance travelled by \(A\) & \(B\) will be: \begin{align*} && d_A &= u \cos \alpha t \\ && d_B &= v \cos \beta t \\ \Rightarrow && \frac{d_A}{d_A+d_B} &= \frac{u \cos \alpha}{u \cos \alpha + v \cos \beta} \\ &&&= \frac{\frac{M}{m}v \cos \beta}{\frac{M}{m}v \cos \beta + v \cos \beta} \\ &&&= \frac{M}{M+m} \\ \Rightarrow && d_A = b &= \frac{Md}{m+M} \end{align*} Applying \(v^2 = u^2 + 2as\) we see that \begin{align*} && 0 &= u \sin \alpha - gt \\ \Rightarrow && t &= \frac{u \sin \alpha}{g} \\ && b &=u \cos \alpha \frac{u \sin \alpha}{g} \\ \Rightarrow && u^2 &= \frac{2bg}{\sin 2 \alpha} \\ && 0 &= u^2 \sin^2 \alpha - 2g h \\ \Rightarrow && h &= \frac{u^2 \sin^2 \alpha}{2g} \\ &&&= \frac{2bg}{\sin 2 \alpha} \frac{ \sin^2 \alpha}{2g} \\ &&&= \frac12 b \tan \alpha \end{align*}

Solution: \begin{align*} \text{W.E.P.}: && \text{change in energy} &= \text{work done on particle} \\ \Rightarrow && \underbrace{\frac12mv^2}_{\text{speed before hitting barrier}} - \underbrace{\frac12mu^2}_{\text{speed leaving first barrier}} &= \underbrace{\left( -\mu mg \right)}_{F} \cdot \underbrace{d}_{d} \\ \Rightarrow && v^2 &= v_i^2-2\mu gd \end{align*} Newton's experimental law tells us that the speed leaving the barrier will be \(r\) times the speed approaching, ie \begin{align*} && v_{i+1} &= rv \\ \Rightarrow && v_{i+1}^2 &= r^2 v^2 \\ &&&= r^2v_i^2 - 2r^2\mu gd \\ \Rightarrow && v_{i+1}^2 - r^2v_i^2 &= - 2r^2\mu gd \end{align*} It must be the case that after \(n+1\) collisions the speed is zero, ie \(v_{n+1}^2 = 0\). Not that we can consider \(w_i = \frac{v_i^2}{2\mu gd}\) and we have the recurrence: \begin{align*} && w_{i+1} &=r^2w_i -r^2 \\ \end{align*} Looking at this we have a linear recurrence with a constant term, so let's try \(w_i = C\), then \begin{align*} && C &= r^2 C - r^2 \\ \Rightarrow && C &= \frac{-r^2}{1-r^2} \\ \end{align*} So \(w_i = Ar^{2i} - \frac{r^2}{1-r^2}\). \(w_0 = k \Rightarrow A = k+\frac{r^2}{1-r^2}\) Therefore \(w_n = \left (k+\frac{r^2}{1-r^2} \right)r^{2n} - \frac{r^2}{1-r^2}\) Suppose \(w_n = 0\) then, \begin{align*} && 0 &= \left (k+\frac{r^2}{1-r^2} \right)r^{2n} - \frac{r^2}{1-r^2} \\ \Rightarrow && r^{2n} &= \frac{r^2}{1-r^2} \frac{1}{k+\frac{r^2}{1-r^2}} \\ &&&= \frac{r^2}{k(1-r^2)+r^2} \\ \Rightarrow && 2n \ln r &= 2\ln r - \ln[k(1-r^2)+r^2] \\ \Rightarrow && n &= 1 - \frac1{2\ln r} \ln[k(1-r^2)+r^2)] \end{align*} If \(r = e^{-1}\) then \(\ln r = -1\) \begin{align*} && n &= 1 + \frac12 \ln [k(1-e^{-2}) + e^{-2}] \\ &&&= 1 + \frac12 \ln [e^{-2}(k(e^2-1)+1)] \\ &&&= 1 + \frac12 \ln e^{-2} + \frac12 \ln [1+k(e^2-1)] \\ &&&= \frac12 \ln [1+k(e^2-1)] \end{align*} If \(r = 1\) the recurrence becomes: \(w_{i+1} = w_i - 1\), so \(w_i = k-n\), so we have \(k\) collisions.

Solution:

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1. Assuming the block is at rest we must have: \begin{align*} \text{N2}(\uparrow): && 0 &= T \sin \beta\cos \alpha + T \cos \beta \sin \alpha +R -W \\ \Rightarrow && W &> T \sin \beta\cos \alpha + T \cos \beta \sin \alpha \\ &&&= T\sin(\alpha+\beta) \\ \Rightarrow && R &= W-T\sin(\alpha+\beta)\\ \\ \text{N2}(\rightarrow): && 0 &= T \cos \beta \cos \alpha - T \sin \beta \sin \alpha - F \\ \Rightarrow && T \cos(\alpha+\beta) &= F \\ &&&\leq \mu (W-T\sin(\alpha+\beta)) \\ \Rightarrow && W \sin \lambda &\geq T \cos (\alpha+\beta)\cos \lambda +T \sin (\alpha+\beta) \sin \lambda \\ &&&= T\cos(\alpha+\beta-\lambda) \end{align*}
2. If \(W = T\tan \phi\) where \(2\phi = \alpha + \beta\) then \begin{align*} \text{N2}(\uparrow): && ma_y &= T\sin(\alpha+\beta) - W \\ &&&= T \sin(\alpha+\beta) - T \tan \left ( \frac{\alpha+\beta}{2} \right ) \\ &&&= T \tan \left ( \frac{\alpha+\beta}{2} \right ) \left ( 2 \cos^2 \left ( \frac{\alpha+\beta}{2} \right ) -1\right) \\ &&&= T \tan \phi \cos \left ( \alpha+\beta\right ) \tag{notice this is positive so \(R=F=0\)} \\ \text{N2}(\rightarrow): && ma_x &= T \cos(\alpha+\beta) \\ \Rightarrow && \frac{a_y}{a_x} &= \tan \phi \end{align*} Therefore we are accelerating at an angle \(\phi\) to the horizontal

Solution:

1. The probability the first is different to the second is \(\frac68\), the probability the third is different to both of the first two is \(\frac37\) therefore the probability is \(\frac{6}{8} \cdot \frac37 = \frac9{28}\) Whatever pills we are left with on the last day is essentially the same random choice as we make on the first day, therefore \(\frac9{28}\)
2. The probability the first is different to the second is \(\frac{2n}{3n-1}\), the probability the third is different to both of the first two is \(\frac{n}{3n-2}\) therefore the probability is \(\frac{2n^2}{(3n-1)(3n-2)}\). [We can also view this as \(\frac{(3n) \cdot (2n) \cdot n}{(3n) \cdot (3n-1) \cdot (3n-2)}\)]
3. Suppose describe the pills as \(B\) and \(R\) and also number them, then we must have a sequence of the form: \[ B_1R_1 \, B_2R_2 \, B_3R_3 \, \cdots \, B_{n}R_n \] However, we can also rearrange the order of the \(B\) and \(R\) pills in \(n!\) ways each, and also the order of the pairs in \(2^n\) ways. There are \((2n)!\) orders we could have taken the pills out therefore the probability is \begin{align*} && P &= \frac{2^n (n!)^2}{(2n)!} = \frac{2^n}{\binom{2n}{n}} \\ &&&\approx \frac{2^n \cdot 2n \pi \left ( \frac{n}{e} \right)^{2n}}{\sqrt{2 \cdot 2n \cdot \pi} \left ( \frac{2n}{e} \right)^{2n}} \\ &&&= \frac{2^n \sqrt{n \pi} \cdot n^{2n} \cdot e^{-2n}}{2^{2n} \cdot n^{2n} \cdot e^{-2n}} \\ &&&= 2^{-n} \sqrt{n \pi} \end{align*} There is a nice way to think about this question using conditional probability. Suppose we are drawing out of an infinitely supply of \(R\) and \(B\) pills, then each day there is a \(\frac12\) chance of getting different pills. Therefore over \(n\) days there is a \(2^{-n}\) chance of getting different pills. Conditional on the balanced total we see that \begin{align*} && \mathbb{P}(\text{balanced every day} |\text{balanced total}) &= \frac{\mathbb{P}(\text{balanced every day})}{\mathbb{P}(\text{balanced total})} \end{align*} We have already seen the term that is balanced total is \(\frac{1}{2^{2n}}\binom{2n}{n}\), but we can also approximate the balanced total using a normal approximation. \(B(2n, \tfrac12) \approx N(n, \frac{n}{2})\) and so: \begin{align*} \mathbb{P}(X = n) &\approx \mathbb{P}\left (n-0.5 \leq \sqrt{\tfrac{n}{2}} Z + n \leq n+0.5 \right) \\ &= \mathbb{P}\left (- \frac1{\sqrt{2n}} \leq Z \leq \frac{1}{\sqrt{2n}} \right) \\ &= \int_{- \frac1{\sqrt{2n}}}^{\frac1{\sqrt{2n}}} \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \d x \approx \frac{2}{\sqrt{2n}} \frac{1}{\sqrt{2\pi}} \\ &\approx \frac{1}{\sqrt{n\pi}} \end{align*}

Solution: There are \(\binom{26}3\binom{23}{1}2\) ways to obtain \(3\) pairs There are \(\binom{26}2 \binom{24}3 \cdot 2^3\) ways to obtain \(2\) pairs There are \(\binom{26}1 \binom{25}5 \cdot 2^5\) ways to obtain \(1\) pairs There are \(\binom{26}7 \cdot 2^7\) ways to obtain \(0\) pairs There are \(\binom{52}{7}\) ways to choose our integers, so \begin{align*} && \mathbb{P}(X = 3) &= \frac{\binom{26}{3} \cdot \binom{23}{1} \cdot 2}{\binom{52}{7}} \\ &&&= \frac{7! \cdot 26 \cdot 25 \cdot 24 \cdot 23 \cdot 2}{3! \cdot 52 \cdot 51 \cdot 50 \cdot 49 \cdot 48 \cdot 47 \cdot 46} \\ &&&= \frac{7 \cdot 6 \cdot 5 \cdot 4 }{51 \cdot 2\cdot 49 \cdot 2\cdot 47 \cdot 2} \\ &&&= \frac{ 5 }{17\cdot 7\cdot 47} = \frac{5}{5593} \\ \\ && \mathbb{P}(X = 2) &= \frac{\binom{26}2 \binom{24}3 \cdot 2^3}{\binom{52}{7}} \\ &&&= \frac{220}{5593} \\ \\ && \mathbb{P}(X = 1) &= \frac{\binom{26}1 \binom{25}5 \cdot 2^5}{\binom{52}{7}} \\ &&&= \frac{1848}{5593} \\ \\ && \mathbb{P}(X = 0) &= \frac{\binom{26}7 \cdot 2^7}{\binom{52}{7}} \\ &&&= \frac{3520}{5593} \\ \\ && \mathbb{E}(X) &= \frac{1848}{5593} + 2 \cdot \frac{220}{5593} + 3 \cdot \frac{5}{5593} \\ &&&= \frac{2303}{5593} = \frac{7}{17} \end{align*} Notice we can find the expected value directly: Let \(X_i\) be the random variable the \(i\)th number is discarded. Notice that \(\mathbb{E}(X) = \mathbb{E}\left (\frac12 \left (X_1 +X_2 +X_3 +X_4 +X_5 +X_6 +X_7\right) \right)\) and also notice that each \(X_i\) has the same distribution (although not independent!). Then \begin{align*} &&\mathbb{E}(X) &= \frac72 \mathbb{E}(X_i) \\ &&&= \frac72 \cdot \left (1 - \frac{50}{51} \cdot \frac{49}{50} \cdots \frac{45}{46} \right) = \frac74 \left ( 1 - \frac{45}{51}\right) \\ &&&= \frac72 \cdot \frac{6}{51} \\ &&&= \frac7{17} \end{align*}

Solution: \begin{questionparts} \item The tangent to \(y = \ln x\) is \begin{align*} && \frac{y - \ln x_1}{x - x_1} &= \frac{1}{x_1} \\ \Rightarrow && \frac{x_1y -x_1 \ln x_1}{ x- x_1} &= 1 \\ \Rightarrow && x_1 y - x_1 \ln x_1 &= x - x_1 \end{align*} So to run through the origin, we need \(\ln x_1 = 1 \Rightarrow x_1 = e\) so the line will be \(y = \frac1{e} x\) If \(ma = \ln a \Rightarrow m = \frac{\ln a}{a} = \frac{\ln b}{b} \Rightarrow b \ln a = a \ln b \Rightarrow a^b = b^a\). \item

Solution:

1. \(\,\) \begin{align*} u = 1-x, \d u = -\d x && \int_0^1 x^{n-1}(1-x)^n \d x &= \int_{u=1}^{u=0} (1-u)^{n-1}u^n (-1) \d u \\ &&&= \int_0^1 u^n (1-u)^{n-1} \d u \\ &&&= \int_0^1 x^n (1-x)^{n-1} \d x \\ \\ \Rightarrow && 2\int_0^1 x^{n-1}(1-x)^n \d x &= \int_0^1 \left ( x^{n-1}(1-x)^n + x^{n}(1-x)^{n-1} \right)\d x \\ &&&= \int_0^1x^{n-1}(1-x)^{n-1} \left ( (1-x) + x \right) \d x\\ &&&= I_{n-1} \\ \\ && I_n &= \left [x^n \cdot (-1)\frac{(1-x)^{n+1}}{n+1}\right]_0^1 + \int_0^1 n x^{n-1} \frac{(1-x)^{n+1}}{n+1} \d x\\ &&&= \frac{n}{n+1} \int_0^1 x^{n-1} (1-x)^{n+1} \d x \\ \\ && I_n &= \frac{n}{n+1} \int_0^1 x^{n-1} (1-x)^{n+1} \d x \\ &&&= \frac{n}{n+1} \int_0^1 \left ( x^{n-1} (1-x)^{n} - x^n(1-x)^n \right) \d x \\ &&&= \frac{n}{n+1} \left (\frac12 I_{n-1} - I_n \right) \\ \Rightarrow && I_n \cdot \left ( \frac{2n+1}{n+1} \right) &= \frac{n}{2(n+1)} I_{n-1}\\ \Rightarrow && I_n &= \frac{n}{2(2n+1)} I_{n-1} \end{align*}
2. \(\,\) \begin{align*} && I_0 &= \int_0^1 1 \d x = 1 \\ \Rightarrow && I_1 &= \frac{1}{2 \cdot 3} \\ && I_n &= \frac{n}{2(2n+1)} \cdot \frac{n-1}{2(2n-1)}\cdot \frac{n-2}{2(2n-3)} \cdots \frac{1}{2 \cdot 3} \\ &&&= \frac{n!}{2^n (2n+1)(2n-1)(2n-3) \cdots 3} \\ &&&= \frac{n! (2n)(2n-2)\cdots 2}{2^n (2n+1)!} \\ &&&= \frac{(n!)^2 2^n}{2^n(2n+1)!} \\ &&&= \frac{(n!)^2}{(2n+1)^2} \end{align*}
3. \(\,\) \begin{align*} && I_{\frac12} &= \int_0^1 \sqrt{x(1-x)} \d x\\ x = \sin^2 \theta, \d x = 2 \sin \theta \cos \theta \d \theta: &&&= \int_{\theta =0}^{\theta = \frac{\pi}{2}} \sin \theta \cos \theta 2 \sin \theta \cos \theta \d \theta \\ &&&= \frac12 \int_0^{\pi/2} \sin^2 2 \theta \d \theta \\ &&&= \frac12 \int_0^{\pi/2} \frac{1-\cos 2 \theta}{2} \d \theta \\ &&&= \frac14 \left [\theta - \frac12 \sin 2 \theta \right]_0^{\pi/2} \\ &&&= \frac{\pi}{8} \\ \\ && I_{\frac32} &= \frac{3/2}{2 \cdot ( 2 \cdot \frac32 + 1)} I_{\frac12} \\ &&&= \frac{3}{4 \cdot 4} \frac{\pi}{8} \\ &&&= \frac{3 \pi}{128} \end{align*}

Solution: \begin{align*} && y &= bx-ba \\ && 1 &= x^2 + y^2 \\ \Rightarrow && 1 &= x^2 + b^2(x-a)^2 \\ \Rightarrow && 0 &= (1+b^2)x^2-2ab^2x+b^2a^2-1 \end{align*} This will have roots which sum to \(\frac{2ab^2}{1+b^2}\), therefore \(M = \left ( \frac{ab^2}{1+b^2}, \frac{ab^3}{1+b^2}-ba \right)=\left ( \frac{ab^2}{1+b^2}, \frac{-ba}{1+b^2} \right)\)

1. Since \(b\) is fixed so is \(\frac{b}{1+b^2} = t\) and all the points are \((bta, -ta)\), ie \(x = -by\). If \(a \in [-1,1]\) we are ranging on the points \((bt, -t)\) to \((-bt, t)\) which is a distance of \begin{align*} && d &= \sqrt{(bt+bt)^2+(-2t)^2} \\ &&&= \sqrt{4(b^2+1)t^2} \\ &&&=2 \sqrt{(b^2+1)\frac{b^2}{(b^2+1)^2}} \\ &&&= \frac{2b}{\sqrt{b^2+1}} \end{align*}
   

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2. • If \(a\) is fixed we have \(\left ( \frac{ab^2}{1+b^2}, -\frac{ba}{1+b^2} \right)\) \begin{align*} && \frac{x}{y} &= - b \\ \Rightarrow && y &= \frac{a\frac{x}{y}}{1 + \frac{x^2}{y^2}} \\ \Rightarrow && y^2 \left ( 1 + \frac{x^2}{y^2} \right) &= ax \\ \Rightarrow && x^2-ax + y^2 &= 0 \\ \Rightarrow && \left (x - \frac{a}{2} \right)^2 + y^2 &= \frac{a^2}{4} \end{align*} Therefore we will end up with a circle centre \((\tfrac{a}{2}, 0)\) going through the origin.
     

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   • If \(ab = 1\), we have \(\left ( \frac{b}{1+b^2}, -\frac{1}{1+b^2} \right)\) \begin{align*} && \frac{x}{y} &= -b \\ \Rightarrow && y &= -\frac{1}{1+\frac{x^2}{y^2}} \\ \Rightarrow && y + \frac{x^2}{y} &= - 1 \\ \Rightarrow && y^2 +y+ x^2 &= 0 \\ \Rightarrow && \left ( y + \frac12 \right)^2 + x^2 &= \frac14 \end{align*}
     

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