Problems
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2005 Paper 1 Q1
\(47231\) is a five-digit number whose digits sum to \(4+7+2+3+1 = 17\,\).
- Show that there are \(15\) five-digit numbers whose digits sum to \(43\). You should explain your reasoning clearly.
- How many five-digit numbers are there whose digits sum to \(39\)?
Solution:
- The largest a five-digit number can have for its digit sum is \(45 = 9+9+9+9+9\). To achieve \(43\) we can either have 4 9s and a 7 or 3 9s and 2 8s. The former can be achieved in \(5\) ways and the latter can be achieved in \(\binom{5}{2} = 10\) ways. (2 places to choose to put the 2 8s). In total this is \(15\) ways.
- To achieve \(39\) we can have: \begin{array}{c|l|c} \text{numbers} & \text{logic} & \text{count} \\ \hline 99993 & \binom{5}{1} & 5 \\ 99984 & 5 \cdot 4 & 20 \\ 99974 & 5 \cdot 4 & 20 \\ 99965 & 5 \cdot 4 & 20 \\ 99884 & \binom{5}{2} \binom{3}{2} & 30 \\ 99875 & \binom{5}{2} 3! & 60 \\ 99866 & \binom{5}{2} \binom{3}{2} & 30 \\ 98886 & 5 \cdot 4 & 20 \\ 98877 & \binom{5}{2} \binom{3}{2} & 30 \\ 88887 & \binom{5}{1} & 5 \\ \hline && 240 \end{array}
2005 Paper 1 Q2
The point \(P\) has coordinates \(\l p^2 , 2p \r\) and the point \(Q\) has coordinates \(\l q^2 , 2q \r\), where \(p\) and~\(q\) are non-zero and \(p \neq q\). The curve \(C\) is given by \(y^2 = 4x\,\). The point \(R\) is the intersection of the tangent to \(C\) at \(P\) and the tangent to \(C\) at \(Q\). Show that \(R\) has coordinates \(\l pq , p+q \r\). The point \(S\) is the intersection of the normal to \(C\) at \(P\) and the normal to \(C\) at \(Q\). If \(p\) and \(q\) are such that \(\l 1 , 0 \r\) lies on the line \(PQ\), show that \(S\) has coordinates \(\l p^2 + q^2 + 1 , \, p+q \r\), and that the quadrilateral \(PSQR\) is a rectangle.
2005 Paper 1 Q3
In this question \(a\) and \(b\) are distinct, non-zero real numbers, and \(c\) is a real number.
- Show that, if \(a\) and \(b\) are either both positive or both negative, then the equation \[ \displaystyle \frac {x }{ x-a} + \frac{x }{ x-b} = 1 \] has two distinct real solutions.
- Show that, if \(c\ne1\), the equation \[\displaystyle \frac x { x-a} + \frac{x}{ x-b} = 1 + c\] has exactly one real solution if $ \displaystyle c^2 = - \frac {4ab}{\l a - b \r ^2}\;. \;\; $ Show that this condition can be written $\displaystyle c^2= 1 - \l \frac {a+b}{a-b} \r ^2 $ and deduce that it can only hold if \(0 < c^2 \le 1\,\).
2005 Paper 1 Q4
- Given that \(\displaystyle \cos \theta = \frac35\) and that \(\displaystyle \frac{3\pi }{ 2} \le \theta \le 2\pi\), show that \(\displaystyle \sin 2 \theta = -\frac{24}{25}\), and evaluate \(\cos 3 \theta\).
- Prove the identity \(\displaystyle \tan 3\theta \equiv \frac {3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta}\). Hence evaluate \(\tan \theta\), given that \(\displaystyle \tan 3\theta = \frac{11}{ 2}\) and that \(\displaystyle \frac{\pi}{ 4} \le \theta \le \frac{\pi}{2}\).
Solution:
- Since \(\cos^2 \theta + \sin^2 \theta \equiv 1\), \(\sin \theta = \pm \frac45\) and since \(\displaystyle \frac{3\pi }{ 2} \le \theta \le 2\pi\) it must be the case that \(\sin\) is negative, ie \(\sin \theta = -\frac45\). Therefore \(\sin 2 \theta = 2 \sin \theta \cos \theta = 2 \cdot \frac35 \cdot (-\frac45) = -\frac{24}{25}\). \begin{align*} \cos 3 \theta &= \cos 2 \theta \cos \theta - \sin 2\theta \sin \theta \\ &= (\cos^2 \theta - \sin^2 \theta) \cos \theta - \sin 2 \theta \sin \theta \\ &= (\frac{9}{25} - \frac{16}{25}) \frac35 + \frac{24}{25} \cdot (-\frac{4}{5}) \\ &= -\frac{21}{125} - \frac{96}{125} \\ &= -\frac{117}{125} \end{align*}
- \begin{align*} \tan 3 \theta &\equiv \frac{\tan 2 \theta + \tan \theta}{1 - \tan 2 \theta \tan \theta} \\ &\equiv \frac{\frac{2 \tan \theta}{1- \tan^2 \theta} + \tan \theta}{1 - \frac{2 \tan^2 \theta}{1- \tan^2 \theta}} \\ &\equiv \frac{2\tan \theta + \tan \theta -\tan^3 \theta}{1 - \tan^2 \theta - 2 \tan^2 \theta} \\ &\equiv \frac {3 \tan \theta - \tan^3 \theta}{1 - 3 \tan^2 \theta} \end{align*} Let \(t = \tan \theta\), then \begin{align*} && \frac{11}{2} &= \frac{3t - t^3}{1-3t^2} \\ \Leftrightarrow && 11 - 33t^2 &= 6t -2t^3 \\ \Leftrightarrow && 0 &= 2t^3-33t^2-6t+11 \\ \Leftrightarrow && 0 &= (2t-1)(t^2-16t-11) \end{align*} Therefore \(\tan \theta = \frac12, \tan \theta = \frac{16 \pm \sqrt{16^2+4 \cdot 1 \cdot 11}}{2} = \frac{16\pm10\sqrt{3}}{2} = 8 \pm 5 \sqrt{3}\). Since \(\displaystyle \frac{\pi}{ 4} \le \theta \le \frac{\pi}{2}\) we must have that \(\tan\) is both positive and \(\geq 1\), therefore \(\tan \theta = 8 + 5 \sqrt{3}\)
2005 Paper 1 Q5
- Evaluate the integral \[ \int_0^1 \l x + 1 \r ^{k-1} \; \mathrm{d}x \] in the cases \(k\ne0\) and \(k = 0\,\). Deduce that \(\displaystyle \frac{2^k - 1}{k} \approx \ln 2\) when \(k \approx 0\,\).
- Evaluate the integral \[ \int_0^1 x \l x + 1 \r ^m \; \mathrm{d}x \; \] in the different cases that arise according to the value of \(m\).
Solution:
- Case \(k \neq 0\): \begin{align*} && \int_0^1 (x+1)^{k-1} \d x &= \left [\frac{1}{k}(x+1)^k \right]_0^1 \\ &&&= \frac{2^k-1}{k} \\ \end{align*} Case \(k = 0\): \begin{align*} && \int_0^1 (x+1)^{k-1} \d x &= \int_0^1 (x+1)^{-1} \d x \\ &&&= \left [\ln(x+1) \right]_0^1 \\ &&&= \ln 2 \end{align*} Therefore for \(k \approx 0\), we must have both integrals being close to each other, since the function is nice on this interval, ie \(\frac{2^k-1}{k} \approx \ln 2\)
- Case \(m = 0\). \(I = \frac12\) Case \(m \neq 0, -1, -2\) \begin{align*} u = x+1, \d u = \d x && \int_0^1 x(x+1)^m \d x &= \int_{u=1}^{u=2} (u-1)u^m \d u \\ &&&=\left[ \frac{u^{m+2}}{m+2} - \frac{u^{m+1}}{m+1} \right]_1^2 \\ &&&= 2^{m+1}\left ( \frac{2}{m+2} - \frac1{m+1} \right) - \frac{1}{m+2} + \frac{1}{m+1} \\ &&&= 2^{m+1} \frac{m}{(m+1)(m+2)} + \frac{1}{(m+1)(m+2)} \\ &&&= \frac{m2^{m+1}+1}{(m+1)(m+2)} \\ \end{align*} Case \(m = -1\). \begin{align*} && \int_0^1 \frac{x}{x+1} \d x &= \int_0^1 1 - \frac{1}{x+1} \d x \\ &&&= 1 - \ln2 \\ \end{align*} Case \(m = -2\): \begin{align*} && \int_0^1 \frac{x}{(x+1)^2} \d x &= \int_0^1\frac{x+1-1}{(x+1)^2} \d x \\ &&&= \left [ \ln (x+1) +(1+x)^{-1} \right]_0^1 \\ &&&= \ln 2 + \frac12 - 1 \\ &&&= \ln 2 - \frac12 \end{align*}
2005 Paper 1 Q6
- The point \(A\) has coordinates \(\l 5 \, , 16 \r\) and the point \(B\) has coordinates \(\l -4 \, , 4 \r\). The variable point \(P\) has coordinates \(\l x \, , y \r\,\) and moves on a path such that \(AP=2BP\). Show that the Cartesian equation of the path of \(P\) is \[ \displaystyle \l x+7 \r^2 + y^2 =100 \;. \]
- The point \(C\) has coordinates \(\l a \, , 0 \r\) and the point \(D\) has coordinates \(\l b \, , 0 \r\), where \(a\ne b\). The variable point \(Q\) moves on a path such that \[ QC = k \times QD\;, \] where \(k>1\,\). Given that the path of \(Q\) is the same as the path of \(P\), show that \[ \frac{a+7}{b+7}=\frac{a^2+51}{b^2+51}\;. \] Show further that \((a+7)(b+7)=100\,\).
Solution:
- Since \(AP = 2BP\) we also have \(|AP|^2 = 4|BP|^2\) ie \begin{align*} && (x-5)^2 + (y-16)^2 &= 4(x+4)^2 + 4(y-4)^2 \\ \Rightarrow && x^2 - 10x+25 + y^2 -32y + 256 &= 4x^2+32x+64+4y^2-32y+64 \\ \Rightarrow && 281 &= 3x^2+42x+3y^2+128\\ && 281 &= 3(x+7)^2-147+3y^2+128 \\ \Rightarrow && 300 &= 3(x+7)^2 + 3y^2 \\ && 100 &= (x+7)^2 + y^2 \end{align*}
- Since \(|QC|^2 = k^2 |QD|^2\), \begin{align*} && (x-a)^2 + y^2 &= k^2 (x-b)^2 + k^2y^2 \\ \Rightarrow && x^2-2ax+a^2 &= k^2x^2-2k^2bx+k^2b^2 + (k^2-1)y^2 \\ && a^2-k^2b^2 &= (k^2-1)x^2-2(k^2b-a)x + (k^2-1)y^2 \\ && a^2-k^2b^2&= (k^2-1)\left(x-\frac{k^2b-a}{k^2-1}\right)^2-(k^2-1)\left(\frac{k^2b-a}{k^2-1}\right)^2+(k^2-1)y^2 \\ && \frac{a^2-k^2b^2}{k^2-1}+\left(\frac{k^2b-a}{k^2-1}\right)^2&= \left(x-\frac{k^2b-a}{k^2-1}\right)^2+y^2 \\ \Rightarrow && -7 &= \frac{k^2b-a}{k^2-1} \tag{*} \\ && 100 &= \frac{a^2-k^2b^2}{k^2-1}+\left(\frac{k^2b-a}{k^2-1}\right)^2 \\ &&&= \frac{a^2-k^2b^2}{k^2-1}+7^2 \\ \Rightarrow && 51 &= \frac{a^2-k^2b^2}{k^2-1} \tag{**} \\ (*) \Rightarrow && k^2(b+7)&= a+7 \\ (**) \Rightarrow && k^2(51+b^2)&= a^2+51 \\ \Rightarrow && \frac{a^2+51}{b^2+51} &= \frac{a+7}{b+7} \\ \\ \Rightarrow && a^2b+51b+7a^2 &= ab^2+51a+7b^2 \\ && 0 &= ab(b-a)-51(b-a)+7(b-a)(b+a) \\ &&&= (b-a)(ab+7(b+a)-51) \\ &&&= (b-a)((a+7)(b+7)-100) \\ \Rightarrow && 100 &= (a+7)(b+7) \end{align*} Since \(a \neq b\)
2005 Paper 1 Q7
The notation \(\displaystyle \prod^n_{r=1} \f (r)\) denotes the product $\f (1) \times \f (2) \times \f(3) \times \cdots \times \f(n)$. %For example, \(\displaystyle \prod_{r=1}^4 r = 24\). %Simplify \(\displaystyle \prod^n_{r=1} \frac{\g (r) }{ \g (r-1) }\). %You may assume that \(\g (r) \neq 0\) for any integer \(0 \le r \le n \). Simplify the following products as far as possible:
- \(\displaystyle \prod^n_{r=1} \l \frac{r+ 1 }{ r } \r\,\);
- \(\displaystyle \prod^n_{r=2} \l \frac{r^2 -1}{r^2 } \r\,\);
- $\displaystyle \prod^n_{r=1} \l {\cos \frac{2\pi }{ n} + \sin \frac{2\pi}{ n} \cot \frac{\l 2r-1 \r \pi }{ n} }\r\,$, where \(n\) is even.
Solution:
- \(\,\) \begin{align*} \prod^n_{r=1} \left ( \frac{r+ 1 }{ r } \right) &= \frac{2}{1} \cdot \frac{3}{2} \cdot \frac{4}{3} \cdots \frac{n-1}{n-2} \cdot \frac{n}{n-1} \cdot \frac{n+1}{n} \\ &= \frac{n+1}{1} = n+1 \end{align*}
- \(\,\) \begin{align*} \prod^n_{r=2} \left ( \frac{r^2 -1}{r^2 } \right) &= \prod^n_{r=2} \left ( \frac{(r -1)(r+1)}{r^2 } \right) \\ &= \left ( \frac{1}{2} \cdot \frac{3}{2} \right) \cdot \left ( \frac{2}{3} \cdot \frac{4}{3} \right) \cdots \left ( \frac{r-1}{r} \cdot \frac{r+1}{r}\right) \cdots \frac{n-1}{n} \cdot \frac{n+1}{n} \\ &= \frac{1}{n} \cdot \frac{n+1}{2} \\ &= \frac{n+1}{2n} \end{align*}
- When \(n\) is odd, the product is undefined, since we have a \(\cot \pi\) lurking in there. \begin{align*} \prod^n_{r=1} \left ( {\cos \frac{2\pi }{ n} + \sin \frac{2\pi}{ n} \cot \frac{ (2r-1 ) \pi }{ n} } \right) &= \prod^n_{r=1} \left ( {\cos \frac{2\pi }{ n} + \sin \frac{2\pi}{ n} \frac{\cos \frac{ (2r-1 ) \pi }{ n}}{\sin\frac{ (2r-1 ) \pi }{ n}} } \right) \\ &= \prod^n_{r=1} \frac{1}{\sin\frac{ (2r-1 ) \pi }{ n}} \left ( {\cos \frac{2\pi }{ n} \sin\frac{ (2r-1 ) \pi }{ n} + \sin \frac{2\pi}{ n} \cos \frac{ (2r-1 ) \pi }{ n} } \right) \\ &= \prod^n_{r=1} \frac{1}{\sin\frac{ (2r-1 ) \pi }{ n}} \sin \left ( \frac{2\pi}{n} + \frac{(2r-1)\pi}{n} \right) \\ &= \prod^n_{r=1} \frac{1}{\sin\frac{ (2r-1 ) \pi }{ n}} \sin \left ( \frac{(2r+1)\pi}{n} \right) \\ &= \frac{\sin \frac{3\pi}{n}}{\sin \frac{\pi}{n}} \cdot \frac{\sin \frac{5\pi}{n}}{\sin \frac{3\pi}{n}} \cdots \frac{\sin \frac{(2n+1)\pi}{n}}{\sin \frac{(2n-1)\pi}{n}} \\ &= \frac{\sin \frac{(2n+1)\pi}{n}}{\sin \frac{\pi}{n}} \\ &= 1 \end{align*}
2005 Paper 1 Q8
Show that, if \(y^2 = x^k \f(x)\), then $\displaystyle 2xy \frac{\mathrm{d}y }{ \mathrm{d}x} = ky^2 + x^{k+1} \frac{\mathrm{d}\f }{ \mathrm{d}x}$\,.
- By setting \(k=1\) in this result, find the solution of the differential equation \[ \displaystyle 2xy \frac{\mathrm{d}y }{ \mathrm{d}x} = y^2 + x^2 - 1 \] for which \(y=2\) when \(x=1\). Describe geometrically this solution.
- Find the solution of the differential equation \[ 2x^2y\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} = 2 \ln(x) - xy^2 \] for which \(y=1\) when \(x=1\,\).
2005 Paper 1 Q9
A non-uniform rod \(AB\) has weight \(W\) and length \(3l\). When the rod is suspended horizontally in equilibrium by vertical strings attached to the ends \(A\) and \(B\), the tension in the string attached to \(A\) is \(T\). When instead the rod is held in equilibrium in a horizontal position by means of a smooth pivot at a distance \(l\) from \(A\) and a vertical string attached to \(B\), the tension in the string is \(T\). Show that \(5T = 2W\). When instead the end \(B\) of the rod rests on rough horizontal ground and the rod is held in equilibrium at an angle \(\theta\) to the horizontal by means of a string that is perpendicular to the rod and attached to \(A\), the tension in the string is \(\frac12 T\). Calculate \(\theta\) and find the smallest value of the coefficient of friction between the rod and the ground that will prevent slipping.
Solution:
2005 Paper 1 Q10
Three collinear, non-touching particles \(A\), \(B\) and \(C\) have masses \(a\), \(b\) and \(c\), respectively, and are at rest on a smooth horizontal surface. The particle \(A\) is given an initial velocity \(u\) towards~\(B\). These particles collide, giving \(B\) a velocity \(v\) towards \(C\). These two particles then collide, giving \(C\) a velocity \(w\). The coefficient of restitution is \(e\) in both collisions. Determine an expression for \(v\), and show that \[ \displaystyle w = \frac {abu \l 1+e \r^2}{\l a + b \r \l b+c \r}\;. \] Determine the final velocities of each of the three particles in the cases:
- \(\displaystyle \frac ab = \frac bc = e\,\);
- \(\displaystyle \frac ba = \frac cb = e\,\).