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2024 Paper 2 Q1
D: 1500.0 B: 1500.0

In the equality \[ 4 + 5 + 6 + 7 + 8 = 9 + 10 + 11, \] the sum of the five consecutive integers from 4 upwards is equal to the sum of the next three consecutive integers. Throughout this question, the variables \(n\), \(k\) and \(c\) represent positive integers.

  1. Show that the sum of the \(n + k\) consecutive integers from \(c\) upwards is equal to the sum of the next \(n\) consecutive integers if and only if \[ 2n^2 + k = 2ck + k^2. \]
  2. Find the set of possible values of \(n\), and the corresponding values of \(c\), in each of the cases
    1. \(k = 1\)
    2. \(k = 2\).
  3. Show that there are no solutions for \(c\) and \(n\) if \(k = 4\).
  4. Consider now the case where \(c = 1\).
    1. Find two possible values of \(k\) and the corresponding values of \(n\).
    2. Show, given a possible value \(N\) of \(n\), and the corresponding value \(K\) of \(k\), that \[ N' = 3N + 2K + 1 \] will also be a possible value of \(n\), with \[ K' = 4N + 3K + 1 \] as the corresponding value of \(k\).
    3. Find two further possible values of \(k\) and the corresponding values of \(n\).


Solution:

  1. Suppose the sum of the \(n + k\) consecutive integers from \(c\) upwards is equal to the sum of the next \(n\) consecutive integers then \begin{align*} && \sum_{i=c}^{i=c+n+k-1} i &= \sum_{i=c+n+k}^{c+2n+k-1} i \\ \Leftrightarrow && \frac{(c+n+k-1)(c+n+k)}{2} - \frac{(c-1)c}{2} &= \frac{(c+2n+k-1)(c+2n+k)}{2} - \frac{(c+n+k-1)(c+n+k)}{2} \\ \Leftrightarrow && 2(c+n+k-1)(c+n+k) &= (c+2n+k-1)(c+2n+k) + c(c-1) \\ \Leftrightarrow && 2c^2+4cn+4ck+2n^2+4kn+2k^2-2c-2n-2k&=2c^2+4cn+2ck+4n^2+4nk+k^2-2c-2n-k \\ \Leftrightarrow && 2ck+k^2&=2n^2+k \\ \end{align*}
    1. If \(k=1\) then \begin{align*} && 2n^2 + 1 &= 2c + 1 \\ \Rightarrow && c &= n^2 \end{align*} So \(n\) can take any value and \(c = n^2\)
    2. If \(k=2\) then \begin{align*} && 2n^2+2&= 4c+4 \\ \Rightarrow && n^2-1 &=2c \end{align*} So \(n\) must be odd, and \(c = \frac12(n^2-1)\)
  2. Suppose \(k=4\) then \(2n^2+4 = 8c+16\) or \(n^2-6 = 4c\) but then the left hand side is \(2, 3 \pmod{4}\) which is a contradiction.
  3. Suppose \(c =1\)
    1. Since \(2n^2+k = 2k + k^2\) or \(2n^2 = k^2+k\) we can have \(k = 1, n = 1\) or \(k = 8, n = 6\)
    2. Suppose \(2N^2 = K^2 + K\) then consider \begin{align*} && 2(N')^2 &= 2(3N+2K+1)^2 \\ &&&= 2(9N^2+4K^2+1+12NK+6N+4K) \\ &&&= 18N^2+8K^2+24NK+12N+8K+2 \\ && (K')^2+K' &= (4N+3K+1)^2 + (4N+3K+1) \\ &&&= 16N^2 + 9K^2+1+24NK+12N+9K+1 \\ &&&= 16N^2+9K^2+24NK+12N+9K+2 \\ \Rightarrow && 2(N')^2-(K')^2-K' &= 2N^2-K^2-K \\ &&&= 0 \end{align*} as required.
    3. So consider \((k,n) = (1,1), (8,6), (49, 35), (288,204)\)

2024 Paper 2 Q2
D: 1500.0 B: 1500.0

In this question, you need not consider issues of convergence.

  1. Find the binomial series expansion of \((8 + x^3)^{-1}\), valid for \(|x| < 2\). Hence show that, for each integer \(m \geqslant 0\), \[ \int_0^1 \frac{x^m}{8 + x^3}\,\mathrm{d}x = \sum_{k=0}^{\infty} \left( \frac{(-1)^k}{2^{3(k+1)}} \cdot \frac{1}{3k + m + 1} \right). \]
  2. Show that \[ \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} \left( \frac{1}{3k+3} - \frac{2}{3k+2} + \frac{4}{3k+1} \right) = \int_0^1 \frac{1}{x+2}\,\mathrm{d}x\,, \] and evaluate the integral.
  3. Show that \[ \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} \left( \frac{72(2k+1)}{(3k+1)(3k+2)} \right) = \pi\sqrt{a} - \ln b\,, \] where \(a\) and \(b\) are integers which you should determine.


Solution:

  1. Note that \(\,\) \begin{align*} && (8+x^3)^{-1} &= \tfrac18(1 + \tfrac18x^3)^{-1} \\ &&&= \tfrac18( 1 - \left (\tfrac{x}{2} \right)^3 + \left (\tfrac{x}{2} \right)^6 -\left (\tfrac{x}{2} \right)^9 + \cdots ) \end{align*} So \begin{align*} && \int_0^1 \frac{x^m}{8+x^3} \d x &= \int_0^1 x^m \sum_{k=0}^{\infty} \frac{1}{2^3} \left ( - \frac{x}{2} \right)^{3k} \d x \\ &&&= \sum_{k=0}^{\infty} (-1)^k \frac{1}{2^{3(k+1)}} \int_0^1 x^{m+3k} \d x \\ &&&= \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} \frac{1}{3k+m+1} \\ \end{align*}
  2. Notice that \begin{align*} && S &= \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} \left( \frac{1}{3k+3} - \frac{2}{3k+2} + \frac{4}{3k+1} \right) \\ &&&= \int_0^1 \frac{x^2}{8+x^3} \d x - 2\int_0^1 \frac{x^1}{8+x^3}+4\int_0^1 \frac{x^0}{8+x^3} \d x \\ &&&= \int_0^1 \frac{x^2-2x+4}{x^3+8} \d x \\ &&&= \int_0^1 \frac{1}{x+2} \d x = \left[ \ln(x+2)\right]_0^1 \\ &&&= \ln 3 - \ln 2 = \ln \tfrac32 \end{align*}
  3. Firstly, note that \begin{align*} && \frac{2k+1}{(3k+1)(3k+2)} &= \frac13 \left ( \frac{1}{3k+1} +\frac{1}{3k+2} \right) \end{align*} so \begin{align*} && S_2 &= \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} \left( \frac{72(2k+1)}{(3k+1)(3k+2)} \right) \\ &&&= \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{3(k+1)}} 24 \left ( \frac{1}{3k+1} +\frac{1}{3k+2} \right) \\ &&&= 24 \left [ \int_0^1 \frac{x^0}{8+x^3} \d x + \int_0^1 \frac{x^1}{8+x^3} \d x \right] \\ &&&= 24 \left [ \int_0^1 \frac{1+x}{(x+2)(x^2-2x+4)} \d x \right] \\ &&&= 24 \left [ \int_0^1 \frac{x+8}{12(x^2-2x+4)}-\frac{1}{12(x+2)} \d x \right] \\ &&&= 2\left [ \int_0^1 \frac{x+8}{x^2-2x+4}-\frac{1}{x+2} \d x \right] \\ &&&= 2\left [ \int_0^1 \frac{x-1}{x^2-2x+4}+ \frac{9}{(x-1)^2+3}-\frac{1}{x+2} \d x \right] \\ &&&=2 \left [ \int_0^1 \frac12\ln(x^2-2x+4)+3\sqrt{3}\arctan \frac{x-1}{\sqrt{3}} -\ln(x+2) \right]_0^1 \\ &&&=2 \left ( \frac12 \ln3+3\sqrt3 \arctan 0 - \ln 3 \right) - 2\left (\frac12 \ln 4-3\sqrt3 \arctan \frac{1}{\sqrt3} - \ln 2 \right) \\ &&&=\sqrt3 \pi - \ln3 \end{align*}

2024 Paper 2 Q3
D: 1500.0 B: 1500.0

The unit circle is the circle with radius 1 and centre the origin, \(O\). \(N\) and \(P\) are distinct points on the unit circle. \(N\) has coordinates \((-1, 0)\), and \(P\) has coordinates \((\cos\theta, \sin\theta)\), where \(-\pi < \theta < \pi\). The line \(NP\) intersects the \(y\)-axis at \(Q\), which has coordinates \((0, q)\).

  1. Show that \(q = \tan\frac{1}{2}\theta\).
  2. In this part, \(q \neq 1\).
    1. Let \(\mathrm{f}_1(q) = \dfrac{1+q}{1-q}\). Show that \(\mathrm{f}_1(q) = \tan\frac{1}{2}\!\left(\theta + \frac{1}{2}\pi\right)\).
    2. Let \(Q_1\) be the point with coordinates \((0, \mathrm{f}_1(q))\) and \(P_1\) be the point of intersection (other than \(N\)) of the line \(NQ_1\) and the unit circle. Describe geometrically the relationship between \(P\) and \(P_1\).
    1. \(P_2\) is the image of \(P\) under an anti-clockwise rotation about \(O\) through angle \(\frac{1}{3}\pi\). The line \(NP_2\) intersects the \(y\)-axis at the point \(Q_2\) with co-ordinates \((0, \mathrm{f}_2(q))\). Find \(\mathrm{f}_2(q)\) in terms of \(q\), for \(q \neq \sqrt{3}\).
    2. In this part, \(q \neq -1\). Let \(\mathrm{f}_3(q) = \dfrac{1-q}{1+q}\), let \(Q_3\) be the point with coordinates \((0, \mathrm{f}_3(q))\) and let \(P_3\) be the point of intersection (other than \(N\)) of the line \(NQ_3\) and the unit circle. Describe geometrically the relationship between \(P\) and \(P_3\).
    3. In this part, \(0 < q < 1\). Let \(\mathrm{f}_4(q) = \mathrm{f}_2^{-1}\!\Big(\mathrm{f}_3\!\big(\mathrm{f}_2(q)\big)\Big)\), let \(Q_4\) be the point with coordinates \((0, \mathrm{f}_4(q))\) and let \(P_4\) be the point of intersection (other than \(N\)) of the line \(NQ_4\) and the unit circle. Describe geometrically the relationship between \(P\) and \(P_4\).


Solution:

TikZ diagram
  1. \(\,\) \begin{align*} && \frac{y-0}{x-(-1)} &= \frac{\sin \theta }{\cos \theta + 1} \\ \Rightarrow && y_0 &= \frac{\sin \theta}{\cos \theta + 1} \\ &&&= \frac{\frac{2t}{1+t^2}}{\frac{1-t^2}{1+t^2}+1} \\ &&&= t = \tan \tfrac{\theta}{2} \end{align*} Alternatively, it is straightforward to see from the angles.
    1. \(f_1(q) = \frac{1+q}{1-q}\) so \begin{align*} && f_1(\tan\tfrac12\theta) &= \frac{1+\tan\tfrac12\theta}{1-\tan\tfrac12\theta} \\ &&&= \frac{\cos \tfrac12 \theta + \sin \tfrac12 \theta}{\cos \tfrac12 \theta - \sin \tfrac12 \theta} \\ &&&= \frac{\sin(\tfrac14 \pi + \tfrac12 \theta)}{\cos(\tfrac14 \pi + \tfrac12 \theta)} \\ &&&= \tan \tfrac12(\theta + \tfrac{\pi}{2}) \end{align*}
    2. \(Q_1\) is the point \((0, f_1(q))\) so \(P_1\) will be the point \((\cos (\theta + \tfrac{\pi}{2}), \sin (\theta + \tfrac{\pi}{2}))\) which is a rotation anticlockwise by \(\frac{\pi}{2}\)
    1. \(P_2 = (\cos(\theta + \tfrac{\pi}{3}), \sin( \theta + \tfrac{\pi}{3})\) and so \(f_2(q) = \tan (\tfrac12(\theta + \tfrac{\pi}{3}))\) so \begin{align*} && f_2(q) &= \tan (\tfrac12(\theta + \tfrac{\pi}{3})) \\ &&&= \frac{q + \tan \frac{\pi}{3}}{1 - \tan \frac{\pi}{3} \cdot q} \\ &&&= \frac{q + \frac{1}{\sqrt3}}{1 - \frac{q}{\sqrt{3}}} \\ &&&= \frac{\sqrt3 q + 1}{\sqrt3-q} \end{align*}
    2. Since \(q \to -q\) reflects \((0,q)\) in the \(x\)-axis, \(f_3(q) = f_1(-q)\) so \(P_3\) is the reflection of \(P_1\) so it's rotation by \(\frac{\pi}{2}\) followed by reflection in the \(x\)-axis, which is reflection in \(y=x\). [ie \(\theta \to -\theta + \frac{\pi}{2} \to \frac{\pi}{2}-\theta\)]
    3. We are rotating by \(\frac{\pi}{3}\) then reflecting in \(y=x\) and then rotating by \(-\frac{\pi}{3}\), ie \(\theta \to \theta + \frac{\pi}{3} \to \frac{\pi}{6}-\theta \to -\theta -\frac{\pi}{6} \)

2024 Paper 2 Q4
D: 1500.0 B: 1500.0

In this question, if \(O\), \(C\) and \(D\) are non-collinear points in three dimensional space, we will call the non-zero vector \(\mathbf{v}\) a \emph{bisecting vector} for angle \(COD\) if \(\mathbf{v}\) lies in the plane \(COD\), the angle between \(\mathbf{v}\) and \(\overrightarrow{OC}\) is equal to the angle between \(\mathbf{v}\) and \(\overrightarrow{OD}\), and both angles are less than \(90^\circ\).

  1. Let \(O\), \(X\) and \(Y\) be non-collinear points in three-dimensional space, and define \(\mathbf{x} = \overrightarrow{OX}\) and \(\mathbf{y} = \overrightarrow{OY}\). Let \(\mathbf{b} = |\mathbf{x}|\mathbf{y} + |\mathbf{y}|\mathbf{x}\).
    1. Show that \(\mathbf{b}\) is a bisecting vector for angle \(XOY\). Explain, using a diagram, why any other bisecting vector for angle \(XOY\) is a positive multiple of \(\mathbf{b}\).
    2. Find the value of \(\lambda\) such that the point \(B\), defined by \(\overrightarrow{OB} = \lambda\mathbf{b}\), lies on the line \(XY\). Find also the ratio in which the point \(B\) divides \(XY\).
    3. Show, in the case when \(OB\) is perpendicular to \(XY\), that the triangle \(XOY\) is isosceles.
  2. Let \(O\), \(P\), \(Q\) and \(R\) be points in three-dimensional space, no three of which are collinear. A bisecting vector is chosen for each of the angles \(POQ\), \(QOR\) and \(ROP\). Show that the three angles between them are either all acute, all obtuse or all right angles.

2024 Paper 2 Q5
D: 1500.0 B: 1500.0

  1. The functions \(\mathrm{f}_1\) and \(\mathrm{F}_1\), each with domain \(\mathbb{Z}\), are defined by \[ \mathrm{f}_1(n) = n^2 + 6n + 11, \] \[ \mathrm{F}_1(n) = n^2 + 2. \] Show that \(\mathrm{F}_1\) has the same range as \(\mathrm{f}_1\).
  2. The function \(\mathrm{g}_1\), with domain \(\mathbb{Z}\), is defined by \[ \mathrm{g}_1(n) = n^2 - 2n + 5. \] Show that the ranges of \(\mathrm{f}_1\) and \(\mathrm{g}_1\) have empty intersection.
  3. The functions \(\mathrm{f}_2\) and \(\mathrm{g}_2\), each with domain \(\mathbb{Z}\), are defined by \[ \mathrm{f}_2(n) = n^2 - 2n - 6, \] \[ \mathrm{g}_2(n) = n^2 - 4n + 2. \] Find any integers that lie in the intersection of the ranges of the two functions.
  4. Show that \(p^2 + pq + q^2 \geqslant 0\) for all real \(p\) and \(q\). The functions \(\mathrm{f}_3\) and \(\mathrm{g}_3\), each with domain \(\mathbb{Z}\), are defined by \[ \mathrm{f}_3(n) = n^3 - 3n^2 + 7n, \] \[ \mathrm{g}_3(n) = n^3 + 4n - 6. \] Find any integers that lie in the intersection of the ranges of the two functions.


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.

2024 Paper 2 Q6
D: 1500.0 B: 1500.0

In this question, you need not consider issues of convergence.

  1. The sequence \(T_n\), for \(n = 0, 1, 2, \ldots\), is defined by \(T_0 = 1\) and, for \(n \geqslant 1\), by \[ T_n = \frac{2n-1}{2n}\,T_{n-1}. \] Prove by induction that \[ T_n = \frac{1}{2^{2n}}\binom{2n}{n}, \] for \(n = 0, 1, 2, \ldots\). [Note that \(\dbinom{0}{0} = 1\).]
  2. Show that in the binomial series for \((1-x)^{-\frac{1}{2}}\), \[ (1-x)^{-\frac{1}{2}} = \sum_{r=0}^{\infty} a_r x^r, \] successive coefficients are related by \[ a_r = \frac{2r-1}{2r}\,a_{r-1} \] for \(r = 1, 2, \ldots\)\,. Hence prove that \(a_r = T_r\) for all \(r = 0, 1, 2, \ldots\)\,.
  3. Let \(b_r\) be the coefficient of \(x^r\) in the binomial series for \((1-x)^{-\frac{3}{2}}\), so that \[ (1-x)^{-\frac{3}{2}} = \sum_{r=0}^{\infty} b_r x^r. \] By considering \(\dfrac{b_r}{a_r}\), find an expression involving a binomial coefficient for \(b_r\), for \(r = 0, 1, 2, \ldots\)\,.
  4. By considering the product of the binomial series for \((1-x)^{-\frac{1}{2}}\) and \((1-x)^{-1}\), prove that \[ \frac{(2n+1)}{2^{2n}}\binom{2n}{n} = \sum_{r=0}^{n} \frac{1}{2^{2r}}\binom{2r}{r}, \] for \(n = 1, 2, \ldots\)\,.


Solution:

  1. Claim: \(\displaystyle T_n = \frac{1}{2^{2n}}\binom{2n}{n}\) Proof: (By Induction) Base case: \(n=0\). Note that \(T_0 = 1\) and \(\frac{1}{2^0}\binom{0}{0} = 1\) so the base case is true. Assume true for some \(n=k\), ie \(T_k = \frac{1}{2^{2k}} \binom{2k}{k}\) so \begin{align*} && T_{k+1} &= \frac{2(k+1)-1}{2(k+1)} \frac{1}{2^{2k}} \binom{2k}{k} \\ &&&= \frac{2k+1}{k+1} \frac{1}{2^{2k+1}} \frac{(2k)!}{k!k!} \\ &&&= \frac{2(k+1)(2k+1)}{(k+1)(k+1)} \frac{1}{2^{2(k+1)}} \frac{(2k)!}{k!k!} \\ &&&= \frac{1}{2^{2(k+1)}} \frac{(2k+2)!}{(k+1)!(k+1)!} \\ &&&= \frac{1}{2^{2(k+1)}} \binom{2(k+1)}{k+1} \end{align*} and therefore it's true for all \(n\).
  2. Notice that \((1-x)^{-\frac12} = 1 + (-\tfrac12)(-x) + \frac{(-\frac12)(-\frac32)}{2!}(-x)^2+\cdots\) in particular \(a_r = \frac{(-\frac12 - r)}{r}(-1)a_{r-1} = \frac{2r-1}{2r}a_{r-1}\). Since \(a_0 = 1\) we have \(a_r = T_r\) for all \(r\).
  3. Notice that \begin{align*} && (1-x)^{-\frac32} &= \sum_{r=0}^\infty b_r x^r \\ &&&= \sum_{r=0}^\infty \frac{(-\frac32)\cdot(-\frac32-1)\cdots (-\frac32-(r-1))}{r!}(-x)^r \\ &&&= \sum_{r=0}^\infty \frac{(-\frac12-1)\cdot(-\frac12-2)\cdots (-\frac12-r)}{r!}(-x)^r \\ \end{align*} Therefore \(\frac{b_r}{a_r} = \frac{r+\frac12}{\frac12} = 2r+1\) so \(b_r = \frac{2r+1}{2^{2r}} \binom{2r}{r}\)
  4. Notice that \begin{align*} && (1-x)^{-\frac32} &= (1-x)^{-\frac12}(1-x)^{-1} \\ &&&= (1 + x+ x^2 + \cdots) \sum_{r=0}^{\infty} a_r x^r \\ &&&= \sum_{i=0}^{\infty} \sum_{k=0}^n a_r x^i \end{align*} So we must have \(b_r = \sum_{i=0}^ra_i\) which is the required result

2024 Paper 2 Q7
D: 1500.0 B: 1500.0

  1. Sketch the curve \(C_1\) with equation \[ \left(y^2 + (x-1)^2 - 1\right)\left(y^2 + (x+1)^2 - 1\right) = 0. \]
  2. Consider the curve \(C_2\) with equation \[ \left(y^2 + (x-1)^2 - 1\right)\left(y^2 + (x+1)^2 - 1\right) = \tfrac{1}{16}. \]
    1. Show that the line \(y = k\) meets the curve \(C_2\) at points for which \[ x^4 + 2(k^2 - 2)x^2 + \left(k^4 - \tfrac{1}{16}\right) = 0. \] Hence determine the number of intersections between curve \(C_2\) and the line \(y = k\) for each positive value of \(k\).
    2. Determine whether the points on curve \(C_2\) with the greatest possible \(y\)-coordinate are further from, or closer to, the \(y\)-axis than those on curve \(C_1\).
    3. Show that it is not possible for both \(y^2 + (x-1)^2 - 1\) and \(y^2 + (x+1)^2 - 1\) to be negative, and deduce that curve \(C_2\) lies entirely outside curve \(C_1\).
    4. Sketch the curves \(C_1\) and \(C_2\) on the same axes.


Solution:

  1. \(\,\)
    TikZ diagram
    1. Suppose \(y=k\) meets the curve \(C_2\) then \begin{align*} && \tfrac1{16} &= (k^2+(x-1)^2-1)(k^2+(x+1)^2-1) \\ &&&= (k^2+x^2-2x)(k^2+x^2+2x) \\ &&&= k^4+2k^2x^2+x^4-4x^2 \\ &&&= x^4+(2k^2-4)x^2+k^4 \\ \Rightarrow && 0 &= x^4+(2k^2-4)x^2+(k^4-\tfrac1{16}) \\ \\ && \Delta &= 4(k^2-2)^2 - 4 \cdot 1 \cdot (k^4-\tfrac1{16}) \\ &&&= 4(k^4-4k^2+4 - k^4 +\tfrac{1}{16}) \\ &&&= 16(1+\tfrac{1}{64} - k^2) \\ &&&= 16(\tfrac{65}{64} - k^2) \end{align*} Therefore if \(|k| < \frac{\sqrt{65}}{8}\) there are \(4\) intersections. If \(|k| = \frac{\sqrt{65}}{8}\) there are \(2\) intersections, otherwise there are \(0\).
    2. The greatest possible \(y\) value is \( \frac{\sqrt{65}}{8}\) and at this point \(x^2 = \frac{2(2-\frac{65}{64}}{2} =1 - \frac{1}{64} < 1\) so they are close to the \(y\)-axis.
    3. The regions where \(y^2+(x-1)^2-1 < 0\) and \(y^2+(x+1)^2-1 < 0\) is the interior of the two circles from the first part. However, since they don't overlap they can never both be negative. Therefore in our equation both are positive and therefore \(C_2\) is entirely outside \(C_1\)
    4. \(\,\)
      TikZ diagram

2024 Paper 2 Q8
D: 1500.0 B: 1500.0

In this question, the following theorem may be used without proof. Let \(u_1, u_2, \ldots\) be a sequence of real numbers. If the sequence is

  • bounded above, so \(u_n \leqslant b\) for all \(n\), where \(b\) is some fixed number
  • and increasing, so \(u_n \leqslant u_{n+1}\) for all \(n\)
then there is a number \(L \leqslant b\) such that \(u_n \to L\) as \(n \to \infty\). For positive real numbers \(x\) and \(y\), define \(\mathrm{a}(x,y) = \frac{1}{2}(x+y)\) and \(\mathrm{g}(x,y) = \sqrt{xy}\). Let \(x_0\) and \(y_0\) be two positive real numbers with \(y_0 < x_0\) and define, for \(n \geqslant 0\) \[ x_{n+1} = \mathrm{a}(x_n, y_n)\,, \] \[ y_{n+1} = \mathrm{g}(x_n, y_n)\,. \]
  1. By considering \((\sqrt{x_n} - \sqrt{y_n})^2\), show that \(y_{n+1} < x_{n+1}\), for \(n \geqslant 0\). Show further that, for \(n \geqslant 0\)
    • \(x_{n+1} < x_n\)
    • \(y_n < y_{n+1}\).
    Deduce that there is a value \(M\) such that \(y_n \to M\) as \(n \to \infty\). Show that \(0 < x_{n+1} - y_{n+1} < \frac{1}{2}(x_n - y_n)\) and hence that \(x_n - y_n \to 0\) as \(n \to \infty\). Explain why \(x_n\) also tends to \(M\) as \(n \to \infty\).
  2. Let \[ \mathrm{I}(p,q) = \int_0^{\infty} \frac{1}{\sqrt{(p^2 + x^2)(q^2 + x^2)}}\,\mathrm{d}x, \] where \(p\) and \(q\) are positive real numbers with \(q < p\). Show, using the substitution \(t = \frac{1}{2}\!\left(x - \dfrac{pq}{x}\right)\) in the integral \[ \int_{-\infty}^{\infty} \frac{1}{\sqrt{\left(\frac{1}{4}(p+q)^2 + t^2\right)(pq + t^2)}}\,\mathrm{d}t, \] that \[ \mathrm{I}(p,q) = \mathrm{I}\!\left(\mathrm{a}(p,q),\, \mathrm{g}(p,q)\right). \] Hence evaluate \(\mathrm{I}(x_0, y_0)\) in terms of \(M\).

2024 Paper 2 Q9
D: 1500.0 B: 1500.0

A long straight trench, with rectangular cross section, has been dug in otherwise horizontal ground. The width of the trench is \(d\) and its depth \(2d\). A particle is projected at speed \(v\), where \(v^2 = \lambda dg\), at an angle \(\alpha\) to the horizontal, from a point on the ground a distance \(d\) from the nearer edge of the trench. The vertical plane in which it moves is perpendicular to the trench.

  1. The particle lands on the base of the trench without first touching either of its sides.
    1. By considering the vertical displacement of the particle when its horizontal displacement is \(d\), show that \((\tan\alpha - \lambda)^2 < \lambda^2 - 1\) and deduce that \(\lambda > 1\).
    2. Show also that \((2\tan\alpha - \lambda)^2 > \lambda^2 + 4(\lambda - 1)\) and deduce that \(\alpha > 45^\circ\).
  2. Show that, provided \(\lambda > 1\), \(\alpha\) can always be chosen so that the particle lands on the base of the trench without first touching either of its sides.

2024 Paper 2 Q10
D: 1500.0 B: 1500.0

A triangular prism lies on a horizontal plane. One of the rectangular faces of the prism is vertical; the second is horizontal and in contact with the plane; the third, oblique rectangular face makes an angle \(\alpha\) with the horizontal. The two triangular faces of the prism are right angled triangles and are vertical. The prism has mass \(M\) and it can move without friction across the plane. A particle of mass \(m\) lies on the oblique surface of the prism. The contact between the particle and the plane is rough, with coefficient of friction \(\mu\).

  1. Show that if \(\mu < \tan\alpha\), then the system cannot be in equilibrium.
Let \(\mu = \tan\lambda\), with \(0 < \lambda < \alpha < \frac{1}{4}\pi\). A force \(P\) is exerted on the vertical rectangular face of the prism, perpendicular to that face and directed towards the interior of the prism. The particle and prism accelerate, but the particle remains in the same position relative to the prism.
  1. Show that the magnitude, \(F\), of the frictional force between the particle and the prism is \[ F = \frac{m}{M+m}\left|(M+m)g\sin\alpha - P\cos\alpha\right|. \] Find a similar expression for the magnitude, \(N\), of the normal reaction between the particle and the prism.
  2. Hence show that the force \(P\) must satisfy \[ (M+m)g\tan(\alpha - \lambda) \leqslant P \leqslant (M+m)g\tan(\alpha + \lambda). \]

2024 Paper 2 Q11
D: 1500.0 B: 1500.0

  1. Sketch a graph of \(y = x^{\frac{1}{x}}\) for \(x > 0\), showing the location of any turning points. Find the maximum value of \(n^{\frac{1}{n}}\), where \(n\) is a positive integer.
\(N\) people are to have their blood tested for the presence or absence of an enzyme. Each person, independently of the others, has a probability \(p\) of having the enzyme present in a sample of their blood, where \(0 < p < 1\). The blood test always correctly determines whether the enzyme is present or absent in a sample. The following method is used.
  • The people to be tested are split into \(r\) groups of size \(k\), with \(k > 1\) and \(rk = N\).
  • In every group, a sample from each person in that group is mixed into one large sample, which is then tested.
  • If the enzyme is not present in the combined sample from a group, no further testing of the people in that group is needed.
  • If the enzyme is present in the combined sample from a group, a second sample from each person in that group is tested separately.
  1. Find, in terms of \(N\), \(k\) and \(p\), the expected number of tests.
  2. Given that \(N\) is a multiple of \(3\), find the largest value of \(p\) for which it is possible to find an integer value of \(k\) such that \(k > 1\) and the expected number of tests is at most \(N\). Show that this value of \(p\) is greater than \(\frac{1}{4}\).
  3. Show that, if \(pk\) is sufficiently small, the expected number of tests is approximately \[ N\!\left(\frac{1}{k} + pk\right). \] In the case where \(p = 0.01\), show that choosing \(k = 10\) gives an expected number of tests which is only about \(20\%\) of \(N\).


Solution:

  1. \(\,\)
    TikZ diagram
    \begin{align*} && y & = x^{1/x} = \exp \left ( \tfrac1x \ln x \right ) \\ \Rightarrow && y' &= \exp \left ( \tfrac1x \ln x \right ) \cdot \left ( \frac{1}{x^2} - \frac{\ln x}{x^2} \right ) \\ &&&= \frac{\exp \left ( \tfrac1x \ln x \right ) }{x^2}(1 - \ln x) \\ y' =0: && x &= e \end{align*} Therefore the largest integer values will be \(2\) or \(3\). Comparing \((2^{\frac12})^6 = 8 < 9 = (3^{1/3})^6\) we see the maximum value of \(n^{1/n}\) where \(n\) is an integer is \(\sqrt[3]{3}\)
  2. The number of tests is \(r\) plus however many groups fail times \(k\). The probability of a group failing is \(g = 1-(1-p)^k\) and the number of failing groups is \(\sim B(r, g)\) so the expected number of additional groups is \(rg\) and the expected total number of tests is \[ \frac{N}{k} + N(1-(1-p)^k) \]
  3. \(\,\) \begin{align*} && N &\geq E = \frac{N}{k} + N(1-(1-p)^k) \\ \Rightarrow && 1 &\geq \frac{1}{k} + 1-(1-p)^k \\ \Rightarrow && (1-p)^k &\geq \frac1k \\ \Rightarrow && k \ln(1-p) &\geq - \ln k \\ \Rightarrow && \ln(1 - p) &\geq -\frac{1}{k} \ln k \geq -\frac13 \ln 3 \\ \Rightarrow && 1-p &\geq \frac{1}{\sqrt[3]{3}} \\ \Rightarrow && p &\leq 1-\frac{1}{\sqrt[3]{3}} \end{align*} (taking \(k=3\)) Claim: \(1 - \frac{1}{\sqrt[3]{3}} > \frac14\) Proof: This is equivalent to \(\sqrt[3]{3} > \frac43\) or \(81 > 4^3 = 64\) which is clearly true.
  4. If \(pk\) is small then \((1-p)^k \approx 1 - pk\) and so we obtain \(N \left ( \frac1k +pk \right)\) as required. If \(p = 0.01\) and \(k = 10\) then \(\frac{1}{10} + 0.01 \cdot 10 = 0.2\) so the expected number of tests is \(\sim 20\%\) of \(N\)

2024 Paper 2 Q12
D: 1500.0 B: 1500.0

In this question, you may use without proof the results \[ \sum_{i=1}^{n} i^2 = \tfrac{1}{6}n(n+1)(2n+1) \quad \text{and} \quad \sum_{i=1}^{n} i^3 = \tfrac{1}{4}n^2(n+1)^2. \] Throughout the question, \(n\) and \(k\) are integers with \(n \geqslant 3\) and \(k \geqslant 2\).

  1. In a game, \(k\) players, including Ada, are each given a random whole number from \(1\) to \(n\) (that is, for each player, each of these numbers is equally likely and assigned independently of all the others). A player wins the game if they are given a smaller number than all the other players, so there may be no winner in this game. Find an expression, in terms of \(n\), \(k\) and \(a\), for the probability that Ada is given number \(a\), where \(1 \leqslant a \leqslant n-1\), and all the other players are given larger numbers. Hence show that, if \(k = 4\), the probability that there is a winner in this game is \[ \frac{(n-1)^2}{n^2}\,. \]
  2. In a second game, \(k\) players, including Ada and Bob, are each given a random whole number from \(1\) to \(n\). A player wins the game if they are given a smaller number than all the other players or if they are given a larger number than all the other players, so it is possible for there to be zero, one or two winners in this game. Find an expression, in terms of \(n\), \(k\) and \(d\), for the probability that Ada is given number \(a\) and Bob is given number \(a + d + 1\), where \(1 \leqslant d \leqslant n-2\) and \(1 \leqslant a \leqslant n - d - 1\), and all the other players are given numbers greater than \(a\) and less than \(a + d + 1\). Hence show that, if \(k = 4\), the probability that there are two winners in this game is \[ \frac{(n-2)(n-1)^2}{n^3}\,. \] If \(k = 4\), what is the minimum value of \(n\) for which there are more likely to be exactly two winners than exactly one winner in this game?


Solution:

  1. Suppose Ada is given \(a\), then she wins if the other \(k-1\) players all get a number between \(a+1\) and \(n\). Since each of these choices are independent, this occurs with probability: \begin{align*} && \mathbb{P}(\text{Ada wins with }a) &= \left ( \frac{n-a}{n} \right)^{k-1} \\ \\ && \mathbb{P}(\text{Ada wins}) &= \sum_{a=1}^{n-1} \mathbb{P}(\text{Ada wins with }a) \mathbb{P}(\text{Ada has }a) \\ &&&= \sum_{a=1}^{n-1}\frac{1}{n}\left ( \frac{n-a}{n} \right)^{3}\\ &&&= \frac{1}{n^4} \sum_{a=1}^{n-1} (n-a)^3 \\ &&&= \frac{1}{n^4} \sum_{a=1}^{n-1} a^3 \\ &&&= \frac{1}{n^4} \tfrac14(n-1)^2n^2 \\ &&&= \frac{(n-1)^2}{4n^2} \end{align*} Since each the game is symmetric, each player is equally likely to win, therefore the probability anyone wins is \(\displaystyle \frac{(n-1)^2}{n^2}\)
  2. The probability that Ada gets \(a\), Bob gets \(a+d+1\) and the other players are in between is \begin{align*} && \mathbb{P}(\text{event}) &= \mathbb{P}(\text{Ada gets }a) \mathbb{P}(\text{Bob gets }a+d+1) \mathbb{P}(\text{everyone else between}) \\ &&&= \frac1{n^2} \cdot \left ( \frac{d}{n} \right) ^{k-2} \end{align*} Therefore the probability that Ada and Bob jointly win is \begin{align*} && \mathbb{P}(\text{Ada and Bob win}) &= \sum_{d=1}^{n-2} \sum_{a=1}^{n-d-1} \frac{1}{n^4} d^2 \\ &&&= \frac{1}{n^4} \sum_{d=1}^{n-2} (n-1-d) d^2 \\ &&&= \frac{n-1}{n^4} \frac{(n-2)(n-1)(2n-3)}{6} - \frac{1}{n^4} \frac{(n-2)^2(n-1)^2}{4} \\ &&&= \frac{(n-1)^2(n-2)}{12n^4} \left ( 2(2n-3)-3(n-2) \right) \\ &&&= \frac{(n-1)^2(n-2)}{12n^3} \\ \end{align*} There are \(4\) players so there are \(4\) ways to choose the lowest player and \(3\) remaining ways to choose the highest, so we get \(\displaystyle \frac{(n-2)(n-1)^2}{n^3}\) probability of a winner happening. The probability of there being a highest winner is the same as the probability of there being a lowest winner (both \(\frac{(n-1)^2}{n^2}\)) and the probability of there being exactly one winner is therefore \begin{align*} && P_1 &= P_{\geq1}-P_2+P_{\geq1}-P_2 \\ \end{align*} this is less than \(P_2\) iff \begin{align*} && P_{\geq1}+P_{\geq1}-2P_2 &< P_2 \\ \Leftrightarrow && 2P_{\geq 1} & < 3P_2 \\ \Leftrightarrow && \frac{2(n-1)^2}{n^2} &< \frac{3(n-1)^2(n-2)}{n^3} \\ \Leftrightarrow && 2n&<3(n-2) \\ \Leftrightarrow && 6 &< n \end{align*} So \(n = 7\)

2024 Paper 3 Q1
D: 1500.0 B: 1500.0

Throughout this question, \(N\) is an integer with \(N \geqslant 1\) and \(S_N = \displaystyle\sum_{r=1}^{N} \frac{1}{r^2}\). You may assume that \(\displaystyle\lim_{N\to\infty} S_N\) exists and is equal to \(\frac{1}{6}\pi^2\).

  1. Show that \[\frac{1}{r+1} - \frac{1}{r} + \frac{1}{r^2} = \frac{1}{r^2(r+1)}.\] Hence show that \[\sum_{r=1}^{N} \frac{1}{r^2(r+1)} = \sum_{r=1}^{N} \frac{1}{r^2} - 1 + \frac{1}{N+1}.\] Show further that \(\displaystyle\sum_{r=1}^{\infty} \frac{1}{r^2(r+1)} = \frac{1}{6}\pi^2 - 1\).
  2. Find \(\displaystyle\sum_{r=1}^{N} \frac{1}{r^2(r+1)(r+2)}\) in terms of \(S_N\), and hence evaluate \(\displaystyle\sum_{r=1}^{\infty} \frac{1}{r^2(r+1)(r+2)}\).
  3. Show that \[\sum_{r=1}^{\infty} \frac{1}{r^2(r+1)^2} = \sum_{r=1}^{\infty} \frac{2}{r^2(r+1)} - 1.\]


Solution:

  1. \(\,\) \begin{align*} && \frac1{r+1} - \frac1r + \frac1{r^2} &= \frac{-1}{r(r+1)} + \frac{1}{r^2} \\ &&&= \frac{r+1-r}{r^2(r+1)} \\ &&&= \frac{1}{r^2(r+1)} \end{align*} Therefore \begin{align*} && \sum_{r=1}^N \frac1{r^2(r+1)} &= \sum_{r=1}^N \left (\frac1{r+1} - \frac1r + \frac1{r^2} \right) \\ &&&= \sum_{r=1}^N \left (\frac1{r+1} - \frac1r\right) + \sum_{r=1}^N \frac1{r^2} \\ &&&=\frac{1}{N+1} - 1 + \sum_{r=1}^N \frac1{r^2} \\ \end{align*} therefore \begin{align*} && \sum_{r=1}^{\infty} \frac{1}{r^2(r+1)} &= \lim_{N \to \infty } \sum_{r=1}^{N} \frac{1}{r^2(r+1)} \\ &&&= \lim_{N \to \infty } \left (\frac{1}{N+1} - 1 + \sum_{r=1}^N \frac1{r^2} \right) \\ &&&= -1 +\lim_{N \to \infty } \sum_{r=1}^N \frac1{r^2} \\ &&&= -1 + \sum_{r=1}^\infty \frac1{r^2} \\ &&&= \frac{\pi^2}{6}-1 \end{align*}
  2. Note that \begin{align*} && \frac{1}{r^2(r+1)(r+2)} &= \frac{Ar+B}{r^2} + \frac{C}{r+1} + \frac{D}{r+2} \\ &&&= \frac{1}{2r^2} + \frac{1}{r+1} - \frac{1}{4(r+2)} - \frac{3}{4r} \end{align*} So \begin{align*} && \sum_{r=1}^N \frac{1}{r^2(r+1)(r+2)} &= \sum_{r=1}^N \left ( \frac{1}{2r^2} + \frac{1}{r+1} - \frac{1}{4(r+2)} - \frac{3}{4r} \right ) \\ &&&= \frac12 \sum_{r=1}^N \frac{1}{r^2} + \frac{1}{2} - \frac14 \cdot \frac1{3} - \frac34 \frac11 + \\ &&& \quad \quad \quad + \frac13 - \frac14\frac14 - \frac34\frac12 + \\ &&& \quad \quad \quad + \frac14 - \frac14\frac15 - \frac34\frac13 + \\ &&&\quad \quad \quad+ \cdots + \\ &&&\quad \quad \quad + \frac1{N+1} - \frac14\frac1{N+2} - \frac34\frac1N \\ &&&= \frac12 \sum_{r=1}^N \frac{1}{r^2} +\frac14\frac12 - \frac34\frac11-\frac14\frac1{N+2}+\frac34 \frac1{N+1} \\ \\ \Rightarrow && \sum_{r=1}^{\infty} \frac{1}{r^2(r+1)(r+2)} &= \lim_{N \to \infty} \left [ \frac12 \sum_{r=1}^N \frac{1}{r^2} +\frac14\frac12 - \frac34\frac11-\frac14\frac1{N+2}+\frac34 \frac1{N+1}\right] \\ &&&= \frac{\pi^2}{12} -\frac58 \end{align*}
  3. Notice that \(\frac{1}{r^2(r+1)^2} - \frac{2}{r^2(r+1)} = \frac{1-2(r+1)}{r^2(r+1)^2} = \frac{-1-2r}{r^2(r+1)^2} = \frac{1}{(r+1)^2} - \frac{1}{r^2}\) and so \begin{align*} && \sum_{r=1}^N \frac{1}{r^2(r+1)^2} &= \sum_{r=1}^N \left ( \frac{2}{r^2(r+1)} +\frac{1}{(r+1)^2} - \frac{1}{r^2} \right) \\ &&&= \sum_{r=1}^N \frac{2}{r^2(r+1)} +\frac{1}{(N+1)^2} - 1 \\ \end{align*} and the result follows as \(N \to \infty\)
[There is a beautiful paper by KConrad about this question: https://kconrad.math.uconn.edu/blurbs/analysis/series_acceleration.pdf]

2024 Paper 3 Q2
D: 1500.0 B: 1500.0

  1. Solve the inequalities
    1. \(\sqrt{4x^2 - 8x + 64} \leqslant |x+8|\,\),
    2. \(\sqrt{4x^2 - 8x + 64} \leqslant |3x-8|\,\).
    1. Let \(\mathrm{f}(x) = \sqrt{4x^2 - 8x + 64} - 2(x-1)\). Show, by considering \(\bigl(\sqrt{4x^2 - 8x + 64} + 2(x-1)\bigr)\mathrm{f}(x)\) or otherwise, that \(\mathrm{f}(x) \to 0\) as \(x \to \infty\).
    2. Sketch \(y = \sqrt{4x^2 - 8x + 64}\) and \(y = 2(x-1)\) on the same axes.
  2. Find a value of \(m\) and the corresponding value of \(c\) such that the solution set of the inequality \[\sqrt{4x^2 - 5x + 4} \leqslant |mx + c|\] is \(\{x : x \geqslant 3\}\).
  3. Find values of \(p\), \(q\), \(m\) and \(c\) such that the solution set of the inequality \[|x^2 + px + q| \leqslant mx + c\] is \(\{x : -5 \leqslant x \leqslant 1\} \cup \{x : 5 \leqslant x \leqslant 7\}\).

2024 Paper 3 Q3
D: 1500.0 B: 1500.0

Throughout this question, consider only \(x > 0\).

  1. Let \[\mathrm{g}(x) = \ln\left(1 + \frac{1}{x}\right) - \frac{x+c}{x(x+1)}\] where \(c \geqslant 0\).
    1. Show that \(y = \mathrm{g}(x)\) has positive gradient for all \(x > 0\) when \(c \geqslant \frac{1}{2}\).
    2. Find the values of \(x\) for which \(y = \mathrm{g}(x)\) has negative gradient when \(0 \leqslant c < \frac{1}{2}\).
  2. It is given that, for all \(c > 0\), \(\mathrm{g}(x) \to -\infty\) as \(x \to 0\). Sketch, for \(x > 0\), the graphs of \[y = \mathrm{g}(x)\] in the cases
    1. \(c = \frac{3}{4}\),
    2. \(c = \frac{1}{4}\).
  3. The function \(\mathrm{f}\) is defined as \[\mathrm{f}(x) = \left(1 + \frac{1}{x}\right)^{x+c}.\] Show that, for \(x > 0\),
    1. \(\mathrm{f}\) is a decreasing function when \(c \geqslant \frac{1}{2}\);
    2. \(\mathrm{f}\) has a turning point when \(0 < c < \frac{1}{2}\);
    3. \(\mathrm{f}\) is an increasing function when \(c = 0\).