Problems

Solution:

1. Suppose \(X*Y = Y*X\), then \begin{align*} && X * Y &= \lambda \mathbf{x} + (1-\lambda) \mathbf{y} \\ && Y * X &= \lambda \mathbf{y} + (1-\lambda) \mathbf{x}\\ \Rightarrow && 0 &= (2\lambda - 1)(\mathbf{x} -\mathbf{y}) \end{align*} Therefore, either \(\mathbf{x} = \mathbf{y}\) or \(\lambda = \frac12\). Since we assumed \(X,Y\) were distinct, \(\mathbf{x} \neq \mathbf{y}\) and so \(X*Y\) and \(Y*X\) are distinct unless \(\lambda = \frac12\)
2. Suppose \((X*Y)*Z = X*(Y*Z)\) \begin{align*} &&(X*Y)*Z &= (\lambda \mathbf{x} + (1-\lambda) \mathbf{y}) * \mathbf{z} \\ &&&= (\lambda^2 \mathbf{x} + \lambda(1-\lambda)\mathbf{y} + (1-\lambda)\mathbf{z}\\ &&X*(Y*Z) &=\mathbf{x}* (\lambda \mathbf{y} + (1-\lambda) \mathbf{z}) \\ &&&= (\lambda \mathbf{x} + \lambda(1-\lambda)\mathbf{y} + (1-\lambda)^2\mathbf{z}\\ \Rightarrow && 0 &= (\lambda^2 - \lambda)\mathbf{x} + ((1-\lambda) - (1-\lambda)^2)\mathbf{z} \\ &&&=(1-\lambda)(-\lambda \mathbf{x} +\lambda \mathbf{z}) \\ &&&= \lambda(1-\lambda)(\mathbf{z}-\mathbf{x}) \end{align*} Therefore they are distinct unless \(\lambda = 1, 0\) or \(\mathbf{x} = \mathbf{z}\).
3. Claim: \((X*Y)*Z = (X*Z)*(Y*Z)\) Proof: \begin{align*} && (X*Y)*Z &= (\lambda \mathbf{x} + \lambda(1-\lambda)\mathbf{y} + (1-\lambda)^2\mathbf{z} \\ && (X*Z)*(Y*Z) &= (\lambda \mathbf{x} + (1-\lambda)\mathbf{z}) * (\lambda \mathbf{y} + (1-\lambda)\mathbf{z}) \\ &&&= \lambda(\lambda \mathbf{x} + (1-\lambda)\mathbf{z}) + (1-\lambda)(\lambda \mathbf{y} + (1-\lambda)\mathbf{z}) \\ &&&= \lambda^2 \mathbf{x} + \lambda(1-\lambda)\mathbf{y} + (1-\lambda) \mathbf{z} \end{align*} Claim: \(X*(Y*Z) = (X*Y)*(X*Z)\) Proof: \begin{align*} X*(Y*Z) &= \lambda \mathbf{x} + \lambda(1-\lambda)\mathbf{y} + (1-\lambda)^2\mathbf{z} \\ (X*Y)*(X*Z) &= (\lambda \mathbf{x} + (1-\lambda)\mathbf{y})*(\lambda \mathbf{x} + (1-\lambda)\mathbf{z}) \\ &= \lambda (\lambda \mathbf{x} + (1-\lambda)\mathbf{y}) + (1-\lambda)(\lambda \mathbf{x} + (1-\lambda)\mathbf{z}) \\ &= \lambda \mathbf{x} + \lambda(1-\lambda)\mathbf{y} + (1-\lambda)^2\mathbf{z} \end{align*}
4. \(P_1 = X*Y\) divides the line segment into the ratio \(\lambda:(1-\lambda)\). \(P_n\) divides the line segment \(P_{n-1}Y\) into the ratio \(\lambda:(1-\lambda)\), therefore it divides the line segment \(XY\) in the ratio \(\lambda^n : 1- \lambda^n\) Alternatively, \begin{align*} P_1 &= \lambda \mathbf{x} + (1-\lambda)\mathbf{y} \\ P_2 &= (\lambda \mathbf{x} + (1-\lambda)\mathbf{y} )*\mathbf{y} \\ &= \lambda^2 \mathbf{x} + (1-\lambda^2) \mathbf{y} \end{align*} Suppose \(P_k = \lambda^k\mathbf{x} + (1-\lambda^k)\mathbf{y}\) then \begin{align*}P_{k+1} &= (\lambda^k\mathbf{x} + (1-\lambda^k)\mathbf{y}) * \mathbf{y} \\ &= \lambda^{k+1}\mathbf{x} + \lambda(1-\lambda^k)\mathbf{y} + (1-\lambda)\mathbf{y}\\ & = \lambda^{k+1}\mathbf{x} + (1-\lambda^{k+1})\mathbf{y}\end{align*}

Solution:

1. \begin{align*} u = \tan x, \d u = \sec^2 x \d x: &&\int_0^{\pi/4} \tan^n x \sec^2 x \d x &= \int_0^1 u^n \d u \\ &&&= \frac{1}{n+1} \end{align*} \begin{align*} u = \sec x, \d u = \sec x \tan x \d x: &&\int_0^{\pi/4} \sec^n x \tan x \d x &= \int_{u=1}^{u=\sqrt{2}} u^{n-1} \d u \\ &&&= \left [ \frac{u^n}{n}\right] \\ &&&= \frac{(\sqrt{2})^n - 1}n \end{align*}
2. \begin{align*} &&\int_0^{\frac14\pi} x \sec ^4 x \tan x \d x &= \left [x \frac{1}{4} \sec^4 x \right]_0^{\frac14\pi} - \frac14 \int_0^{\frac14\pi} \sec^4 x \d x \\ &&&= \frac{\pi}{4} - \frac14 \int_0^{\frac14\pi} \sec^2 x(1+ \tan^2 x) \d x \\ &&&= \frac{\pi}{4} - \frac14 \left [ \tan x+ \frac13 \tan^3 x \right] _0^{\frac14\pi} \\ &&&= \frac{\pi}{4} - \frac{1}{3} \end{align*} \begin{align*} \int_0^{\frac14\pi} \!\! x^2 \sec ^2 \! x\, \tan x \, \d x &= \left [x^2 \frac12 \tan^2 x \right]_0^{\frac14\pi} - \int_0^{\frac14\pi} x \tan^2 x \d x\\ &= \frac{\pi^2}{32} - \int_0^{\frac14\pi} x (\sec^2 x - 1) \d x\\ &= \frac{\pi^2}{32} - \left [x (-x + \tan x) \right]_0^{\frac14\pi} + \int_0^{\frac14\pi}-x + \tan x \d x \\ &= \frac{\pi^2}{32} - \frac{\pi}{4} (-\frac{\pi}{4} + 1) -\frac{\pi^2}{32} + \left [ -\ln \cos x \right]_0^{\pi/4} \\ &= \frac{\pi^2}{16}- \frac{\pi}{4} + \frac12 \ln 2 \end{align*}

Solution:

1. \(k = 0\), we have \(x^2 + y^2 + 3x + y = 0\), ie \((x+\tfrac32)^2+(y+\tfrac12)^2 = \frac{10}{4}\).
   

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2. \(3x^2 + 3y^2 +10xy = (3x+y)(x+3y)\) so \(x^2 + y^2 + \tfrac{10}3xy + 3x+y = (3x+y)(\frac{x+3y}{3}+1) = 0\) so we have the line pair \(3x +y =0\), \(x+3y + 3 = 0\)
   

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3. If \(k = 2\) then \((x+y)^2 + (x+y)+2x = 0\). If \(\theta = 45^\circ\) then \( X = \frac1{\sqrt{2}}(x+y), Y = \frac{1}{\sqrt{2}}(y-x)\), ie \(x+y = \sqrt{2}X\) and \(x = \frac{1}{\sqrt2}(X-Y)\), so our equation is: \begin{align*} 0 &= 2X^2 + \sqrt{2}X + \sqrt{2}(X-Y) \\ &= (\sqrt{2}X + 1)^2 - 1 - \sqrt{2} Y \end{align*} which would be a parabola with line of symmetry \(X = -\frac{1}{\sqrt{2}}\). However, we are actually looking at that parabola rotated by \(45^\circ\) anticlockwise.
   

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Solution: The coefficient of \(x^{r-1}\) in \((1+x)^n\) is \(\binom{n}{r-1}\) and the coefficient of \(x^r\) in \((1+x)^n\) is \(\binom{n}{r}\). The only ways to get \(x^r\) in the expansion of \((1+x)(1+x)^n\) is to either multiply the \(x^r\) term from the second expansion by \(1\) or the \(x^{r-1}\) term by \(x\). This is \(\binom{n}{r-1} + \binom{n}{r}\). However, the coefficient of \(x^r\) in \((1+x)^{n+1}\) is \(\binom{n+1}r\), so \(\binom{n}{r} + \binom{n}{n-1} = \binom{n+1}r\). Claim: \(B_{n+2} - B_{n+1} = B_{n}\). Proof: Consider \(n\) even, ie \(n = 2m\) \begin{align*} B_{n+2} - B_{n+1} &= \sum_{j=0}^{m+1} \binom{2(m+1)-j}{j} - \sum_{j=0}^m \binom{2m+1-j}{j} \\ &= \binom{2(m+1)-(m+1)}{m+1} +\sum_{j=0}^m \left ( \binom{2(m+1)-j}{j} - \binom{2m+1-j}{j} \right) \\ &= 1 + \sum_{j=1}^m \binom{2m+1-j}{j-1} \\ &= 1 + \sum_{j=0}^{m-1} \binom{2m-j}{j} \\ &= \binom{m}{m} + \sum_{j=0}^{m-1} \binom{2m-j}{j} \\ &= \sum_{j=0}^{m} \binom{2m-j}{j} \\ &= B_n \end{align*} Consider \(n\) even, ie \(n = 2m+1\) \begin{align*} B_{n+2} - B_{n+1} &= \sum_{j=0}^{m+1} \binom{2(m+1)+1-j}{j} - \sum_{j=0}^{m+1} \binom{2(m+1)-j}{j} \\ &= \sum_{j=0}^{m+1} \left (\binom{2(m+1)+1-j}{j} - \binom{2(m+1)-j}{j}\right)\\ &= \sum_{j=1}^{m+1} \binom{2(m+1)-j}{j-1} \\ &= \sum_{j=0}^{m} \binom{2m+1-j}{j} \\ &= B_n \end{align*} as required. \(B_0 = 1, B_1 = 2\), therefore \(B_n = F_{n+2}\)

Solution:

1. Let \(y = ux\), then \(\frac{\d y}{\d x} = x\frac{\d u}{\d x} = u\) and the differential equation becomes, \begin{align*} && xu' + u &= \frac{1}{u} +u \\ \Rightarrow && u' &= \frac{1}{ux} \\ \Rightarrow && u u' &= \frac1{x} \\ \Rightarrow && \frac12 u^2 &= \ln x + C \\ (x,y) = (1,2): && \frac12 4 &= C \\ \Rightarrow && \frac12 \frac{y^2}{x^2} &= \ln x + 2 \\ \Rightarrow && y^2 &= x^2 \l 2\ln x + 4 \r \\ \Rightarrow && y &= x \sqrt{4 + 2 \ln x} \quad (x > e^{-2}) \end{align*}
2. Let \(y = ux^2\) then \begin{align*} && \frac{\d y}{\d x} &= \frac{x^2}{y} + \frac{2y}{x} \\ \Rightarrow && u' x^2 + 2x u &= \frac{1}{u} + 2x u \\ \Rightarrow && u' u &= \frac{1}{x^2} \\ \Rightarrow && \frac12 u^2 &= -\frac{1}{x} + C \\ (x,y) = (1,2): && 2 &= C - 1 \\ \Rightarrow && \frac12 \frac{y^2}{x^4} &= 3 - \frac{1}{x} \\ \Rightarrow && y &= x\sqrt{2(3x^2-x)}, \quad (x > \frac13) \end{align*}

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:

1. First notice that this cubic has leading first term \(1\) and three real roots, so it must have the shape:
   

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   With the \(x\)-axis running somewhere between the dashed lines. Since \(c < 0\), the \(y\)-axis must meet the curve below the \(x\)-axis, ie somewhere on the blue section of this curve:
   

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   Therefore there will be either \(1\) (if it meets it in the \(\cup\) area) or \(3\) (if it meets it on the far left) positive roots.
2. First notice that if \(c > 0\) we cannot have three positive real roots since the function would need to pass \(0\) between \(0\) and \(-\infty\). Secondly, notice both turning points must be larger than zero, ie \begin{align*} && 0 &= 3x^2 + 6ax + 3b \\ \Leftrightarrow && 0 &= (x+a)^2 + b - a^2 \end{align*} has both roots larger than zero, (and it needs to have two roots, so \(a^2 > b\) and \(-a > 0\), ie \(a < 0\). If \(b < 0\), then just looking at \(x^2+2ax+b\) we can see that it is \(<0\) at \(0\) and one of the roots will be negative, therefore \(c < 0\), \(a^2 > b > 0\) and \(a < 0\)
3. Since \(c > 0\) we can see that at least one root is negative.
   

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   ie the \(y\)-axis passes through an orange section of this curve. What now matters is where the larger turning point is. Considering \(x^2 + 2ax + b\), we notice that \(ab < 0\) means that \((x-\alpha)(x-\beta)\) we must have \((\alpha + \beta)\alpha \beta > 0\) which isn't possible if both roots are negative. Therefore the \(y\)-axis passes through the orange \(\cap\) and there are \(2\) positive real roots.
4. If we take \(a = 1, b = -1, c = 1\) then we have \(x^3 + 3x^2-3x+1\). This has turning points when \(x^2+2x-1 = 0\), ie \(x = -1 \pm \sqrt{2}\) Notice that \begin{align*} && y(-1\pm \sqrt2) &= (-1 \pm \sqrt{2})^3 + 3(-1 \pm \sqrt{2})^2-3(-1 \pm \sqrt{2}) + 1 \\ &&&= (-1\pm \sqrt{2}) \cdot (3 \mp 2\sqrt2) + 3(3 \mp \sqrt2) -3(-1\pm \sqrt2) + 1 \\ &&&= (-7 \pm 5 \sqrt2) + (9 \mp 3\sqrt2) +(3 \mp 3\sqrt2) + 1 \\ &&&= 24 \mp 16\sqrt2 = 8(3 \mp 2 \sqrt2) >0 \end{align*} ie both turning points are above zero and hence only one real root

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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