Problems

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Solution:

1. \begin{align*} && 0 &= x + 3\sqrt{x} - \frac12 \\ \sqrt{x} = y: && 0&= y^2 + 3y - \frac12 \\ \Rightarrow && y &= \frac{-3\pm\sqrt{3^2+2}}{2} \\ &&&= \frac{-3 \pm \sqrt{11}}{2} \\ y > 0: && x &= \left ( \frac{\sqrt{11}-3}{2} \right)^2 \end{align*}
2. • \begin{align*} && 0 &= x + 10\sqrt{x+2} - 22 \\ y = \sqrt{x+2}: && 0 &= y^2 - 2 + 10y - 22 \\ &&&= y^2 + 10y - 24 \\ &&&= (y-2)(y+12) \\ \Rightarrow && y &= 2, -12 \\ y > 0: && x &= 2 \end{align*}
   • Let \(y = \sqrt{2x^2-8x-3}\), so \begin{align*} && 0 &= x^2 - 4x +\sqrt{2x^2-8x-3} - 9 \\ && 0 &= \frac{y^2+3}{2} + y - 9 \\ &&&= \frac12 y^2 +y - \frac{15}{2} \\ &&&= \frac12 (y-3)(y+5) \\ \Rightarrow && y &= 3,-5 \\ y > 0: && 9 &= 2x^2-8x-3 \\ \Rightarrow && 0 &= 2x^2-8x-12 \\ &&&= 2(x^2-4x-6) \\ \Rightarrow && x &= 2 \pm \sqrt{10} \end{align*}

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Solution:

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2. Notice that there are no solutions when \(x < 0\) since \(f(x) \geq 1\) in that region. Suppose \(x = n + \epsilon, 0 < \epsilon < 1\), then \(f(x) = \frac{n}{n+\epsilon}\), ie \(12n = 7n + 7 \epsilon \Rightarrow 5 n = 7\epsilon \Rightarrow \epsilon = \frac{5}{7}n \Rightarrow n < \frac75\), so \(n = 1 ,\epsilon = \frac57, x = \frac{12}5\). \begin{align*} && \frac{17}{24} &= f(x) \\ \Rightarrow && 17n + 17 \epsilon &= 24 n \\ \Rightarrow && 17 \epsilon &= 7 n \\ \Rightarrow && n &< \frac{17}{7} \\ \Rightarrow && n &= 1, 2 \\ \Rightarrow && x &= \frac{24}{17}, \frac{48}{17} \end{align*}. For \(f(x) = \frac{4}{3}\) we notice that \(x < 0\), so let \(x = -n +\epsilon\), ie \begin{align*} && \frac43 &= f(x) \\ \Rightarrow && \frac43 &= \frac{-n}{-n+\epsilon} \\ \Rightarrow && 4\epsilon &= n \\ \Rightarrow && n &= 1,2,3 \\ \Rightarrow && x &= -\frac{5}{4}, -\frac{3}{2}, -\frac{9}{4} \end{align*}
3. \begin{align*} && \frac9{10} &= f(x) \\ \Rightarrow && 9n + 9 \epsilon &= 10 n \\ \Rightarrow && 9 \epsilon &= n \\ \Rightarrow && n < 9 \end{align} So largest will be when \(n = 8, \epsilon = \frac{8}{9}\), ie \(\frac{80}{9}\)

If \(c < 1\) \begin{align*} && c &= \frac{k}{k + \epsilon} \\ \Rightarrow && \frac{c}{1-c} \epsilon &= k \end{align*} For this to have exactly \(n\) solutions, we need \(n < \frac{c}{1-c} \leq n+1\). If \(c > 1\) \begin{align*} && c &= \frac{-k}{-k+\epsilon} \\ \Rightarrow && c \epsilon &= (c-1) k \\ \Rightarrow && \frac{c}{c-1} \epsilon &= k \end{align*} Therefore for there to be exactly \(n\) solutions we need \(n < \frac{c}{c-1} \leq n+1\)

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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*}

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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*}

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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 \(B_n = F_{n+2}\)

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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*}

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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, \((0, 1]\)

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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*}

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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.