Problems

Solution:

1. \(\,\) \begin{align*} && f_1(n) &= n^2 + 6n + 11 \\ &&&= (n+3)^2 + 2 \\ &&&=F_1(n+3) \end{align*} Since \(n \mapsto n+3\) is a bijection on \(\mathbb{Z}\) both functions must have exactly the same range.
2. \(g_1(n) = n^2-2n+5 = (n-1)^2 + 4\). Since squares are always \(0, 1 \pmod{4}\) it's impossible for \(f_1\) and \(g_1\) to take the same value therefore the ranges have empty intersection.
3. \(\,\) \begin{align*} && f_2(n) &= n^2-2n - 6 \\ &&&= (n-1)^2-7 \\ && g_2(n) &= n^2-4n+2 \\ &&&= (n-2)^2 - 2 \end{align*} so suppose \(x^2 - 7 = y^2 - 2\) then \begin{align*} && x^2 - 7 &= y^2 -2 \\ \Rightarrow && 5 &= y^2 - x^2 \\ &&&= (y-x)(y+x) \end{align*} So we have cases: \(y-x = -5, y + x = -1 \Rightarrow y = -3\) and the output is \(7\) \(y-x=-1, y+x = -5 \Rightarrow y = -3\) same output \(y-x=1, y+x = 5 \Rightarrow y = 3\) same output \(y-x=5, y-x = 1 \Rightarrow y = 3\) same ouput.
4. \begin{align*} && 0 &\leq \frac12(p^2+q^2)+\frac12(p+q)^2 \\ &&&= p^2 + q^2 + pq \end{align*} Looking at \(f_3\) we see \begin{align*} && f_3(n) &= n^3 - 3n^2 + 7n \\ &&&= (n-1)^3 -3n + 7n +1 \\ &&&= (n-1)^3 +4(n-1) -3 \\ &&&= g_3(n-1) + 3 \end{align*} So suppose we have two values which are equal, ie \begin{align*} && x^3 + 4x -3 &= y^3 +4y -6 \\ \Rightarrow && 3 &= y^3-x^3+4y-4x \\ &&&= (y-x)(y^2+xy+x^2+4) \end{align*} Since \(x^2+xy+y^2 \geq 0\) then the right hand factor is always a positive integer bigger than \(3\) and in particular there will be no solutions and hence no integers in the intersection of the ranges.

Solution:

1. The curves meet when they have the same radius for a given \(\theta\) ie \begin{align*} && a + 2 \cos \theta &= 2 + \cos 2 \theta \\ &&&= 2 + 2\cos^2 \theta - 1 \\ \Rightarrow && 0 &= 2 \cos ^2 \theta - 2 \cos \theta + 1 - a \end{align*} The curves touch if this has a repeated root, ie \(0 = \Delta = 4 - 8(1-a) \Rightarrow a = \frac12\). The second way the curves can touch is if there is a single root, but it's at an extreme value of \(\cos \theta = \pm 1\) ie \(0 = 2 - 2\cdot(\pm1) + 1 - a \Rightarrow a = 3 \pm 2 = 1, 5\)
2. Suppose \(a = \frac12\) then the curves touch when \(0 = 2\cos^2 \theta - 2 \cos \theta + \frac12 = (2 \cos \theta-1 )(\cos \theta -\frac12) \Rightarrow \theta = \pm \frac{\pi}{3}\)
   

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3. \(a = 1\)
   

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   \(a = 5\)
   

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Solution: \begin{align*} && v_{\uparrow} &= U\sin \theta - g t \\ \Rightarrow && T_H &= \frac{U \sin \theta}{g} \\ \\ && s_{\uparrow} &= U \sin \theta t - \frac12 g t^2 \\ \Rightarrow && 0 &= U\sin \theta T_L - \frac12 g T_L^2 \\ && T_L &= \frac{2 U \sin \theta}{g} \end{align*} \(T = U\cos \theta / (kg)\) is the point when the particle's horizontal motion is reversed.

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Solution:

1. \begin{align*} && 0 &= \alpha^2 + a \alpha + b \tag{1} \\ && 0 &= \alpha^2 + c \alpha + d \tag{2} \\ \\ (1) - (2): && 0 & =\alpha ( a-c) + (b-d) \\ \Rightarrow && \alpha &= - \frac{b-d}{a-c} \tag{\(a\neq c\)} \end{align*} (\(\Rightarrow\)) Suppose they have a common root, then given we know it's form, we must have: \begin{align*} && 0 &= \left ( - \frac{b-d}{a-c} \right)^2 +a\left ( - \frac{b-d}{a-c} \right) + b \\ \Rightarrow && 0 &= (b-d)^2 - a(b-d)(a-c) + b(a-c)^2 \end{align*} (\(\Leftarrow\)) Suppose the equation holds, then \begin{align*} && 0 &= (b-d)^2 - a(b-d)(a-c) + b(a-c)^2 \\ \Rightarrow && 0 &= \left ( - \frac{b-d}{a-c} \right)^2 +a\left ( - \frac{b-d}{a-c} \right) + b \\ \end{align*} So \(\alpha\) is a root of the first equation. Considering \((1) - (2)\) we must have that \(\alpha(a-c) +(b-d) = t\) (whatever the second equation is), but that value is clearly \(0\), therefore \(\alpha\) is a root of both equations. If \(a = c\) then the equation becomes \(0 = (b-d)^2\), ie the two equations are the same, therefore they must have common roots!
2. \begin{align*} && 0 &= x^2+ax+b \tag{1} \\ && 0 &= x^3+(a+1)x^2+qx+r \tag{2} \\ \\ (2) - x(1) && 0 &= x^2 + (q-b)x + r \tag{3} \end{align*} Therefore if the equations have a common root, \((1)\) and \((3)\) have a common root, ie \((b-r)^2-a(b-r)(a-(q-b))+b(a-(q-b))^2 = 0\) which is exactly our condition. \(a = \frac52, q = \frac52, r = \frac12\) \begin{align*} && 0 &= \left (b-\frac12 \right)^2 - \frac52\left (b-\frac12\right) b + b^3 \\ &&&= b^2 -b + \frac14 - \frac52 b^2+\frac54b + b^3 \\ &&&= b^3 -\frac32 b^2 +\frac14 b + \frac14 \\\Rightarrow && 0 &= 4b^3 - 6b^2+b + 1 \\ &&&= (b-1)(4b^2-2b-1) \\ \Rightarrow && b &= 1, \frac{1 \pm \sqrt{5}}{4}\end{align*}

Solution: We have \begin{align*} \rightarrow: && a &= v\cos \alpha t_{net} \\ \Rightarrow && t_{net} &= \frac{a}{v \cos \alpha} \\ \downarrow: && H-h &= v\sin \alpha t_{net} + \frac12 g t_{net}^2 \\ &&&= a \tan \alpha + \frac12 g \frac{a^2}{v^2} \sec^2 \alpha \\ &&&= a \tan \alpha + \frac{a^2g}{2v^2}(1 + \tan^2 \alpha) \tag{*}\\ \\ \rightarrow: && a+b &= v \cos \alpha t_{ground} \\ && t_{ground} &= \frac{a+b}{v \cos \alpha}\\ \downarrow: && H &= v\sin \alpha t_{ground} + \frac12 g t_{ground}^2 \\ &&&= (a+b)\tan \alpha + \frac{(a+b)^2g}{2v^2}(1+\tan^2\alpha) \tag{**} \\ \\ (**): && v^2 &= \frac{g(a+b)^2(1+\tan^2\alpha)}{2[H-(a+b)\tan \alpha]} \\ (a+b)^2(*) - a^2(**): && (a+b)^2(H-h) -a^2H &= [(a+b)^2a - a^2(a+b)]\tan \alpha \\ \Rightarrow && (2ab+b^2)H - (a+b)^2h &= ab(a+b) \tan \alpha \\ \Rightarrow && \tan \alpha &= \frac{2a+b}{a(a+b)}H - \frac{a+b}{ab} h \end{align*} Noting that \(v^2 \geq 0\) and the numerator is positive, we must have \begin{align*} && H &> (a+b)\tan \alpha \\ &&&= \frac{2a+b}{a}H - \frac{(a+b)^2}{ab} h \\ \Rightarrow && \frac{a+b}{a}H &< \frac{(a+b)^2}{ab} h \\ \Rightarrow && H &< \frac{a+b}{b} h \end{align*} Noting that \(\tan \alpha > 0\) we must have \begin{align*} && \frac{2a+b}{a(a+b)} H & > \frac{a+b}{ab} h \\ \Rightarrow && H &> \frac{(a+b)^2}{b(2a+b)}h \end{align*}