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1999 Paper 2 Q1
Let \(x=10^{100}\), \(y=10^{x}\), \(z=10^{y}\), and let $$ a_1=x!, \quad a_2=x^y,\quad a_3=y^x,\quad a_4=z^x,\quad a_5=\e^{xyz},\quad a_6=z^{1/y},\quad a_7 = y^{z/x}. $$
- Use Stirling's approximation \(n! \approx \sqrt{2 \pi}\, {n^{n+{1\over2}}\e^{-n}}\), which is valid for large \(n\), to show that \(\log_{10}\left(\log_{10} a_1 \right) \approx 102\).
- Arrange the seven numbers \(a_1\), \(\ldots\) , \(a_7\) in ascending order of magnitude, justifying your result.
Solution:
- \begin{align*} \log_{10}(\log_{10} a_1) &= \log_{10} (\log_{10} (x!) \\ &\approx \log_{10} (\log_{10} \sqrt{2 \pi} x^{x+\frac12} e^{-x}) \\ &= \log_{10} \l \log_{10} \sqrt{2 \pi} + (x+\frac12) \log_{10} x-x \r \\ &= \log_{10} \l \log_{10} \sqrt{2 \pi} + (100x+50)-x \r \\ &= \log_{10} \l 99x + \epsilon \r \\ &\approx \log_{10} 99 + \log_{10} x \\ &\approx 2 + 100 = 102 \end{align*}
- \begin{align*} \log_{10}(\log_{10} a_2) &= \log_{10}(\log_{10} x^y) \\ &= \log_{10} y + \log_{10} \log_{10} x \\ &= x + 2 \end{align*} \begin{align*} \log_{10}(\log_{10} a_3) &= \log_{10}(\log_{10} y^x) \\ &= \log_{10} x + \log_{10} \log_{10} y \\ &= 100 + \log_{10} x \\ &= 200 \end{align*} \begin{align*} \log_{10}(\log_{10} a_4) &= \log_{10}(\log_{10} z^x) \\ &= \log_{10} x + \log_{10} \log_{10} z \\ &= 100 + \log_{10} y \\ &= 100+x \end{align*} \begin{align*} \log_{10}(\log_{10} a_5) &= \log_{10}(\log_{10} e^{xyz}) \\ &= \log_{10} x + \log_{10}y+\log_{10} z+ \log_{10} \log_{10} e \\ &\approx 100 + x + y \end{align*} \begin{align*} \log_{10}(\log_{10} a_6) &= \log_{10}(\log_{10} z^{1/y}) \\ &= \log_{10}(\log_{10} 10) \\ &= 0 \end{align*} \begin{align*} \log_{10}(\log_{10} a_7) &= \log_{10}(\log_{10} y^{z/x}) \\ &= \log_{10}z-\log_{10} x + \log_{10} \log_{10} y \\ &= y \end{align*} Since \(0 < 102 < 200 < x+2 < x+100 < y < y+x+100\) we must have \(a_6 < a_1 < a_3 < a_2 < a_4 < a_7 < a_5\)
1999 Paper 2 Q2
Consider the quadratic equation $$ nx^2+2x \sqrt{pn^2+q} + rn + s = 0, \tag{*} $$ where \(p>0\), \(p\neq r\) and \(n=1\), \(2\), \(3\), \(\ldots\) .
- For the case where \(p=3\), \(q=50\), \(r=2\), \(s=15\), find the set of values of \(n\) for which equation \((*)\) has no real roots.
- Prove that if \(p < r\) and \(4q(p-r) > s^2\), then \((*)\) has no real roots for any value of \(n\).
- If \(n=1\), \(p-r=1\) and \(q={s^2}/8\), show that \((*)\) has real roots if, and only if, \(s \le 4-2\sqrt{2}\ \) or \(\ s \ge 4+2\sqrt{2}\).
Solution:
- \(\,\) \begin{align*} && 0 &= nx^2 + 2\sqrt{3n^2+50}x + 2n + 15 \\ && 0 &> \Delta = 4(3n^2+5) - 4\cdot n \cdot (2n + 15) \\ \Leftrightarrow && 0 &> n^2-15n+5\\ \text{cv}: && n &= \frac{15 \pm \sqrt{225 - 20}}{2} \\ &&&\approx \frac{15\pm14.x}{2}\\ \Leftrightarrow &&n &\in \{1, 2, \cdots, 14\} \end{align*}
- \(\,\) \begin{align*} && 0 &> \Delta = 4(pn^2+q) - 4\cdot n \cdot (rn+s) \\ \Leftrightarrow && 0&>(p-r)n^2-sn+q \end{align*} Which is always true if \(r > p\) and \(s^2 < 4q(p-r)\)
- \(\,\) \begin{align*} && 0 &= x^2 + 2\sqrt{p+q}x+ r+s \\ && 0 &\leq \Delta = 4(p+q) - 4(r+s) \\ && 0 &\leq 1 + s^2/8 - s \\ \text{c.v}: && s &= \frac{1 \pm \sqrt{1-4 \cdot \frac{1}{8}}}{2\cdot\frac18} \\ &&&= 4 \pm 4\sqrt{\frac12} \\ &&&= 4 \pm 2\sqrt{2} \\ \Rightarrow && s \leq 4 - 2\sqrt{2} &\text{ or } s \geq 4 + 2\sqrt{2} \end{align*}
1999 Paper 2 Q3
Let $$ {\rm S}_n(x)=\mathrm{e}^{x^3}{{\d^n}\over{\d x^n}}{(\mathrm{e}^{-x^3})}. $$ Show that \({\rm S}_2(x)=9x^4-6x\) and find \({\rm S}_3(x)\). Prove by induction on \(n\) that \({\rm S}_n(x)\) is a polynomial. By means of your induction argument, determine the order of this polynomial and the coefficient of the highest power of~\(x\). Show also that if \ \(\displaystyle \frac{\d S_n}{\d x}=0\) \ for some value \(a\) of \(x\), then \( \ S_n(a)S_{n+1}(a)\le0\).
1999 Paper 2 Q4
By considering the expansions in powers of \(x\) of both sides of the identity $$ {(1+x)^n}{(1+x)^n}\equiv{(1+x)^{2n}}, $$ show that $$ \sum_{s=0}^n {n\choose s}^2 = {2n\choose n}, $$ where \(\displaystyle {n\choose s}= \frac{n!}{s!\,(n-s)!}\). By considering similar identities, or otherwise, show also that:
- if \(n\) is an even integer, then \(\displaystyle \sum_{s=0}^n {{(-1)}^s}{n \choose s}^2= (-1)^{n/2}{n \choose n/2};\)
- \(\displaystyle \sum\limits_{t=1}^ n 2t { n \choose t}^2 = n {2n\choose n} .\)
Solution: To obtain the coefficient of \(x^n\) on the RHS we clearly have \(\displaystyle \binom{2n}n\). To obtain the coefficient of \(x^n\) on the LHS we can obtain \(x^s\) from the first bracket and \(x^{n-s}\) from the second bracket, ie \(\displaystyle \sum_{s=0}^n \binom{n}{s}\binom{n}{n-s} = \sum_{s=0}^n \binom{n}{s}\binom{n}{s} = \sum_{s=0}^n \binom{n}{s}^2\)
- Consider \((1-x)^n(1+x)^n = (1-x^2)^n\), then the coefficient of \(x^n\) (if \(n\) is even) is for the RHS \(\displaystyle (-1)^{n/2} \binom{n}{n/2}\). For the LHS, we can obtain \(x^n\) via \(x^s\) and \(x^{n-s}\) which is \(\displaystyle \sum_{s=0}^n (-1)^s\binom{n}{s}\binom{n}{n-s} = \sum_{s=0}^n (-1)^s\binom{n}{s}^2\)
- Notice that \begin{align*} && \sum_{t=1}^ n 2t { n \choose t}^2 &= n {2n\choose n} \\ \Leftrightarrow && \sum_{t=1}^ n 2t \frac{n}{t} { n-1 \choose t-1}\binom{n}{t} &= n \frac{2n}{n}{2n-1\choose n-1} \\ \Leftrightarrow && \sum_{t=1}^ n { n-1 \choose t-1}\binom{n}{t} &= {2n-1\choose n-1} \\ \Leftrightarrow && \sum_{t=1}^ n { n-1 \choose t-1}\binom{n}{n-t} &= {2n-1\choose n-1} \\ \end{align*} but this is exactly what we would obtain by considering the coefficient of \(x^{n-1}\) in \((1+x)^{n-1}(1+x)^n \equiv (1+x)^{2n-1}\)
1999 Paper 2 Q5
Show that if \(\alpha\) is a solution of the equation $$ 5{\cos x} + 12{\sin x} = 7, $$ then either $$ {\cos }{\alpha} = \frac{35 -12\sqrt{120}}{169} $$ or \(\cos \alpha\) has one other value which you should find. Prove carefully that if \(\frac{1}{2}\pi< \alpha < \pi\), then \(\alpha < \frac{3}{4}\pi\).
Solution: \begin{align*} && 5 \cos x + 12\sin x &= 7 \\ \Rightarrow && 5 \cos x - 7 &= -12 \sin x \\ \Rightarrow && 25 \cos^2 x - 70\cos x + 49 &= 144 \sin^2 x \\ \Rightarrow && 25 \cos^2 x - 70\cos x + 49 &= 144 (1-\cos^2 x) \\ \Rightarrow && 169 \cos^2 x - 70 \cos x -95 &= 0 \\ \Rightarrow && \cos \alpha &= \frac{70 \pm \sqrt{70^2 - 4 \cdot 169 \cdot (-95)}}{2 \cdot 169} \\ &&&= \frac{35 \pm \sqrt{35^2 + 169 \cdot 95} }{169} \\ &&&= \frac{35 \pm 12\sqrt{120}}{169} \end{align*} If \(\frac12 \pi < \alpha < \pi\) then \(\cos \alpha\) is negative, in particular \(\cos \alpha = \frac{35 -12\sqrt{120}}{169}\). Since \(\cos\) is decreasing over this range, if \(\cos \alpha > \cos \frac34 \pi = -\frac{\sqrt{2}}2\), then we will have shown \(\alpha < \frac34 \pi\) \begin{align*} && \cos \alpha &= \frac{35 - 12 \sqrt{120}}{169} \\ &&&> \frac{35 - 12 \cdot \sqrt{121}}{169} \\ &&&= \frac{35 - 12 \cdot 11}{169} \\ &&&= \frac{35 - 132}{169} \\ &&&= -\frac{97}{169} \\ &&&> -\frac{8}{13} \end{align*} but \(\left ( \frac{8}{13} \right)^2 = \frac{64}{169} < \frac12\), so we are done.
1999 Paper 2 Q6
Find \(\displaystyle \ \frac{\d y}{\d x} \ \) if $$ y = \frac{ax+b}{cx+d}. \eqno(*) $$ By using changes of variable of the form \((*)\), or otherwise, show that \[ \int_0^1 \frac{1}{(x+3)^2} \; \ln \left(\frac{x+1}{x+3}\right)\d x = {\frac16} \ln3 - {\frac14}\ln 2 - \frac 1{12}, \] and evaluate the integrals \[ \int_0^1 \frac{1}{(x+3)^2} \; \ln \left(\frac{x^2+3x+2}{(x+3)^2}\right)\d x \mbox{ \ \ and \ \ } \int_0^1 \frac{1}{(x+3)^2} \; \ln\left(\frac{x+1}{x+2}\right)\d x . \] %By changing to the variable \(y\) defined by %$$ %y=\frac{2x-3}{x+1}, %$$ % evaluate the integral %$$ %\int_2^4 \frac{2x-3}{(x+1)^3}\; %\ln\!\left(\frac{2x-3}{x+1}\right)\d x. %$$ %Evaluate the integral %$$ %\int_9^{25} {\big({2z^{-3/2} -5z^{-2}}\big)} %\ln{\big(2-5z^{-1/2}\big)}\; \d z. %$$
1999 Paper 2 Q7
The curve \(C\) has equation $$ y = \frac x {\sqrt{x^2-2x+a}}\; , $$ where the square root is positive. Show that, if \(a>1\), then \(C\) has exactly one stationary point. Sketch \(C\) when (i) \(a=2\) and (ii) \(a=1\).
Solution: \begin{align*} && y &= \frac x {\sqrt{x^2-2x+a}} \\ && y' &= \frac{\sqrt{x^2-2x+a} - \frac{x(x-1)}{\sqrt{x^2-2x+a}}}{x^2-2x+a} \\ &&&= \frac{-x+a}{(x^2-2x+a)^{3/2}} \end{align*} Since the denominator is always positive, the only stationary point is when \(x = a\)
1999 Paper 2 Q8
Prove that $$ \sum_{k=0}^n \sin k\theta = \frac { \cos \half\theta - \cos (n+ \half) \theta} {2\sin \half\theta}\;. \eqno(*) $$
- Deduce that, when \(n\) is large, \[ \sum_{k=0}^n \sin \left(\frac{k\pi}{n}\right) \approx \frac{2n}\pi\;. \]
- By differentiating \((*)\) with respect to \(\theta\), or otherwise, show that, when \(n\) is large, \[ \sum_{k=0}^n k \sin^2 \left(\frac{k\pi}{2n}\right) \approx \left(\frac{1}4 +\frac{1}{\pi^2} \right)n^2\;. \]
1999 Paper 2 Q9
In the \(Z\)--universe, a star of mass \(M\) suddenly blows up, and the fragments, with various initial speeds, start to move away from the centre of mass \(G\) which may be regarded as a fixed point. In the subsequent motion the acceleration of each fragment is directed towards \(G\). Moreover, in accordance with the laws of physics of the \(Z\)--universe, there are positive constants \(k_1\), \(k_2\) and \(R\) such that when a fragment is at a distance \(x\) from \(G\), the magnitude of its acceleration is \(k_1x^3\) if \(x < R\) and is \(k_2x^{-4}\) if \(x \ge R\). The initial speed of a fragment is denoted by \(u\).
- For \(x < R\), write down a differential equation for the speed \(v\), and hence determine \(v\) in terms of \(u\), \(k_1\) and \(x\) for \( x < R\).
- Show that if \(u < a\), where \(2a^2=k_1 R^4\), then the fragment does not reach a distance \(R\) from \(G\).
- Show that if \(u \ge b\), where $ 6b^2= 3k_1R^4 + 4k_2 /R^3, $ then from the moment of the explosion the fragment is always moving away from \(G\).
- If \(a < u < b\), determine in terms of \(k_2\), \(b\) and \(u\) the maximum distance from \(G\) attained by the fragment.
1999 Paper 2 Q10
\(N\) particles \(P_1\), \(P_2\), \(P_3\), \(\ldots\), \(P_N\) with masses \(m\), \(qm\), \(q^2m\), \(\ldots\) , \({q^{N-1}}m\), respectively, are at rest at distinct points along a straight line in gravity-free space. The particle \(P_1\) is set in motion towards \(P_2\) with velocity \(V\) and in every subsequent impact the coefficient of restitution is \(e\), where \(0 < e < 1\). Show that after the first impact the velocities of \(P_1\) and \(P_2\) are $$ {\left({{1-eq}\over{1+q}}\right)}V \mbox{ \ \ \ and \ \ \ } {\left({{1+e}\over{1+q}}\right)}V, $$ respectively. Show that if \(q \le e\), then there are exactly \(N-1\) impacts and that if \(q=e\), then the total loss of kinetic energy after all impacts have occurred is equal to $$ {1\over 2}{me}{\left(1-e^{N-1}\right)}{V^2}. $$