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2015 Paper 1 Q11
D: 1500.0 B: 1484.0

Two long circular cylinders of equal radius lie in equilibrium on an inclined plane, in \mbox{contact} with one another and with their axes horizontal. The weights of the upper and lower \mbox{cylinders} are \(W_1\) and \(W_2\), respectively, where \(W_1>W_2\)\,. The coefficients of friction \mbox{between} the \mbox{inclined} plane and the upper and lower cylinders are \(\mu_1\) and \(\mu_2\), respectively, and the \mbox{coefficient} of friction \mbox{between} the two cylinders is \(\mu\). The angle of inclination of the plane is~\(\alpha\) (which is positive).

  1. Let \(F\) be the magnitude of the frictional force between the two cylinders, and let \(F_1\) and \(F_2\) be the magnitudes of the frictional forces between the upper cylinder and the plane, and the lower cylinder and the plane, respectively. Show that \(F=F_1=F_2\,\).
  2. Show that \[ \mu \ge \dfrac{W_1+W_2}{W_1-W_2} \,,\] and that \[ \tan\alpha \le \frac{ 2 \mu_1 W_1}{(1+\mu_1)(W_1+ W_2)}\,. \]

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2015 Paper 1 Q12
D: 1500.0 B: 1461.6

The number \(X\) of casualties arriving at a hospital each day follows a Poisson distribution with mean 8; that is, \[ \P(X=n) = \frac{ \e^{-8}8^n}{n!}\,, \ \ \ \ n=0, \ 1, \ 2, \ \ldots \ . \] Casualties require surgery with probability \(\frac14\). The number of casualties arriving on any given day is independent of the number arriving on any other day and the casualties require surgery independently of one another.

  1. What is the probability that, on a day when exactly \(n\) casualties arrive, exactly \(r\) of them require surgery?
  2. Prove (algebraically) that the number requiring surgery each day also follows a Poisson distribution, and state its mean.
  3. Given that in a particular randomly chosen week a total of 12 casualties require surgery on Monday and Tuesday, what is the probability that 8 casualties require surgery on Monday? You should give your answer as a fraction in its lowest terms.

Solution:

  1. \(\mathbb{P}(r \text{ need surgery}|n \text{ casualties}) = \binom{n}{r} \left ( \frac14\right)^r \left ( \frac34\right)^{n-r}\)
  2. \(\,\) \begin{align*} && \mathbb{P}(r \text{ need surgery}) &= \sum_{n=r}^{\infty} \mathbb{P}(r \text{ need surgery} |n \text{ casualties}) \mathbb{P}(n \text{ casualties}) \\ &&&= \sum_{n=r}^{\infty} \binom{n}{r}\left ( \frac14\right)^r \left ( \frac34\right)^{n-r} \frac{e^{-8} 8^n}{n!} \\ &&&= \sum_{n=r}^{\infty} \frac{n!}{(n-r)!r!}\left ( \frac14\right)^r \left ( \frac34\right)^{n-r} \frac{e^{-8} 8^n}{n!} \\ &&&= \frac{e^{-8}8^r}{r!}\left ( \frac14\right)^r \sum_{n=r}^{\infty} \frac{8^{n-r}}{(n-r)} \left ( \frac34\right)^{n-r} \\ &&&= \frac{e^{-8}8^r}{r!}\left ( \frac14\right)^r \sum_{n=r}^{\infty} \frac{6^{n-r}}{(n-r)} \\ &&&= \frac{e^{-8}2^r}{r!} e^6 \\ &&&= \frac{e^{-2}2^r}{r!} \end{align*} Therefore the number requiring surgery is \(Po(2)\) with mean \(2\).
  3. \(\,\) \begin{align*} && \mathbb{P}(X_1 = 8| X_1 + X_2 =12) &= \frac{\mathbb{P}(X_1 = 8,X_2 =4)} {\mathbb{P}(X_1+X_2 = 12)}\\ &&&= \frac{\frac{e^{-2}2^8}{8!} \cdot \frac{e^{-2}2^4}{4!}}{\frac{e^{-4}4^{12}}{12!}} \\ &&&= \frac{12!}{8!4!} \frac{1}{2^{12}} \\ &&&= \binom{12}4 \left ( \frac12 \right)^4\left ( \frac12 \right)^8 \\ &&&= \frac{495}{4096} \end{align*}

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2015 Paper 1 Q13
D: 1500.0 B: 1501.1

A fair die with faces numbered \(1, \ldots, 6\) is thrown repeatedly. The events \(A\), \(B\), \(C\), \(D\) and \(E\) are defined as follows. \begin{align*} A: && \text{the first 6 arises on the \(n\)th throw.}\\ B: && \text{at least one 5 arises before the first 6.} \\ C: && \text{at least one 4 arises before the first 6.}\\ D: && \text{exactly one 5 arises before the first 6.}\\ E: && \text{exactly one 4 arises before the first 6.} \end{align*} Evaluate the following probabilities:

  1. \(\P(A)\)
  2. \(\P(B)\)
  3. \(\P(B\cap C)\)
  4. \(\P(D)\)
  5. \(\P(D\cup E)\)
For some parts of this question, you may want to make use of the binomial expansion in the form: \[ (1-x)^{-n} = 1 +nx +\frac {n(n+1)}2 x^2 + \cdots + \frac {(n+r-1)!}{r! (n-1)!}x^r +\cdots\ .\]

Solution:

  1. \(\,\) \begin{align*} \mathbb{P}(A) &= \mathbb{P}(\text{the first 6 arises on the \(n\)th throw.}) \\ &= \mathbb{P}(\text{\(n-1\) not 6s, followed by a 6.})\\ &= \left ( \frac56\right)^{n-1} \cdot \frac16 = \frac{5^{n-1}}{6^n} \end{align*}
  2. There is nothing special about \(5\) or \(6\), so which comes first is \(50:50\), therefore this probability is \(\frac12\)
  3. There is nothing special about \(4\), \(5\) or \(6\) so this is the probability that \(6\) appears last out of these three numbers, hence \(\frac13\)
  4. \(\,\) \begin{align*} \mathbb{P}(D) &= \mathbb{P}(\text{exactly one 5 arises before the first 6.}) \\ &=\sum_{n=2}^{\infty} \mathbb{P}(\text{exactly one 5 arises before the first 6 which appears on the \(n\)th roll.}) \\ &= \sum_{n=2}^{\infty} \binom{n-1}{1} \left ( \frac46 \right)^{n-2} \frac16 \cdot \frac16 \\ &= \frac1{36} \sum_{n=2}^{\infty} (n-1) \left ( \frac23 \right)^{n-2} \\ &= \frac1{36} \sum_{n=1}^{\infty} n \left ( \frac23 \right)^{n-1} \\ &= \frac1{36} \frac{1}{\left ( 1- \frac23 \right)^2} = \frac14 \end{align*}
  5. \(\,\) \begin{align*} \mathbb{P}(D \cup E) &= \mathbb{P}(D) + \mathbb{P}(E) - \mathbb{P}(D \cap E) \\ &= \frac12 - \mathbb{P}(D \cap E) \\ &=\frac12 - \sum_{n=3}^{\infty} \mathbb{P}(\text{exactly one 5 and one 4 arises before the first 6 which appears on the \(n\)th roll.}) \\ &=\frac12 - \sum_{n=3}^{\infty} 2\binom{n-1}{2} \left ( \frac36 \right)^{n-3}\cdot \frac16 \cdot \frac16 \cdot \frac16 \\ &=\frac12 - \frac2{6^3}\sum_{n=3}^{\infty} \frac{(n-1)(n-2)}{2} \left ( \frac12 \right)^{n-3} \\ &=\frac12 - \frac2{6^3}\frac{1}{(1-\tfrac12)^3}\\ &= \frac12 - \frac{2}{27} \\ &= \frac{23}{54} \end{align*}

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2015 Paper 2 Q1
D: 1600.0 B: 1516.0

  1. By use of calculus, show that \(x- \ln(1+x)\) is positive for all positive \(x\). Use this result to show that \[ \sum_{k=1}^n \frac 1 k > \ln (n+1) \,. \]
  2. By considering \( x+\ln (1-x)\), show that \[ \sum_{k=1}^\infty \frac 1 {k^2} <1+ \ln 2 \,. \]

Solution:

  1. Consider \(f(x) = x - \ln (1+ x)\), then \(f'(x) = 1 - \frac{1}{1+x} = \frac{x}{1+x} > 0\) if \(x >0\). Therefore \(f(x)\) is strictly increasing on the positive reals. Since \(f(0) = 0\) we must have \(f(x) > 0\) for all positive \(x\), ie \(x - \ln(1+x)\) is positive for all positive \(x\). \begin{align*} \sum_{k=1}^n \frac1k &\underbrace{>}_{x > \ln(1+x)} \sum_{k=1}^n \ln \left (1 + \frac1k \right ) \\ &= \sum_{k=1}^n \ln \left ( \frac{k+1}{k} \right ) \\ &= \sum_{k=1}^n \left ( \ln (k+1) - \ln (k) \right) \\ &= \ln (n+1) - \ln 1 \\ &= \ln (n+1) \end{align*}
  2. Let \(g(x) = x + \ln (1-x)\) ,then \(g'(x) = 1 - \frac{1}{1-x} = \frac{-x}{1-x} < 0\) if \(0 < x < 1\) and \(g(0) = 0\). Therefore \(g(x)\) is decreasing and hence negative on \(0 < x < 1\), in particular \(x < -\ln(1-x) \) \begin{align*} \sum_{k=2}^n \frac1{k^2} &\underbrace{<}_{x < -\ln(1+x)} \sum_{k=2}^n - \ln \left (1-\frac1{k^2} \right) \\ &= -\sum_{k=2}^n \ln \left ( \frac{k^2-1}{k^2}\right) \\ &= \sum_{k=2}^n \l 2 \ln k - \ln(k-1) - \ln(k+1) \r \\ &= \ln n - \ln(n+1) - \ln 0+\ln 2 \\ &= \ln 2 + \ln \frac{n}{n+1} \end{align*} as \(n \to \infty\) we must have \(\displaystyle \sum_{k=2}^{\infty} \frac1{k^2} < \ln 2\) ie \[ \sum_{k=1}^\infty \frac 1 {k^2} <1+ \ln 2\]

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2015 Paper 2 Q2
D: 1600.0 B: 1484.0

In the triangle \(ABC\), angle \(BAC = \alpha\) and angle \(CBA= 2\alpha\), where \(2\alpha\) is acute, and \(BC= x\). Show that \(AB = (3-4 \sin^2\alpha)x\). The point \(D\) is the midpoint of \(AB\) and the point \(E\) is the foot of the perpendicular from \(C\) to \(AB\). Find an expression for \(DE\) in terms of \(x\). The point \(F\) lies on the perpendicular bisector of \(AB\) and is a distance \(x\) from \(C\). The points \(F\) and \(B\) lie on the same side of the line through \(A\) and \(C\). Show that the line \(FC\) trisects the angle \(ACB\).

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2015 Paper 2 Q3
D: 1600.0 B: 1483.4

Three rods have lengths \(a\), \(b\) and \(c\), where \(a< b< c\). The three rods can be made into a triangle (possibly of zero area) if \(a+b\ge c\). Let \(T_{n}\) be the number of triangles that can be made with three rods chosen from \(n\) rods of lengths \(1\), \(2\), \(3\), \(\ldots\) , \(n\) (where \(n\ge3\)). Show that \(T_8-T_7 = 2+4+6\) and evaluate \(T_8 -T_6\). Write down expressions for \(T_{2m}-T_{2m-1}\) and \(T_{2m} - T_{2m-2}\). Prove by induction that \(T_{2m}=\frac 16 m (m-1)(4m+1)\,\), and find the corresponding result for an odd number of rods.

Solution: Every \(T_7\) triangle is valid, so we are interested in new triangles which have \(8\) has a longest side. We can have: \begin{array}{c|c|c} \text{longest} & \text{middle} & \text{shortest} \\ \hline 8 & 7 & 1-6 \\ 8 & 6 & 2-5 \\ 8 & 5 & 3-4 \end{array} which is \(6+4+2\) extra triangles. The new ones excluding all the sixes are: \begin{array}{c|c|c} \text{longest} & \text{middle} & \text{shortest} \\ \hline 8 & 7 & 1-6 \\ 8 & 6 & 2-5 \\ 8 & 5 & 3-4 \\ 7 & 6 & 1-5 \\ 7 & 5 & 2-4 \\ 7 & 4 & 3 \\ \end{array} Ie \(2+4+6 + 1 + 3+5\) \(T_{2m}-T_{2m-1} = 2 \frac{(m-1)m}{2} = m(m-1)\) and \(T_{2m}-T_{2m-2} = \frac{(2m-2)(2m-1)}{2}\) \(T_4 = 3\) (\(1,2,3\), \(1,3,4\), \(2,3,4\)) and \(\frac16 \cdot 2 \cdot 1 \cdot 9 = 3\) so the base case holds. Suppose it's true for some \(m = k\), then \begin{align*} && T_{2(k+1)} &= T_{2k} + \frac{2m(2m+1)}{2} \\ &&&= \frac{m(m-1)(4m+1)}{6} + \frac{6m(2m+1)}{6}\\ &&&= \frac{m(4m^2-3m-1+12m+6)}{6} \\ &&&= \frac{m(4m^2+9m+5)}{6}\\ &&&= \frac{m(4m+5)(m+1)}{6}\\ &&&= \frac{(m+1-1)(4(m+1)+5)(m+1)}{6}\\ \end{align*} as required, therefore it is true by induction. For odd numbers, we can see that \(T_{2m-1} = \frac{m(m-1)(4m+1)}{6} - m(m-1) = \frac{m(m-1)(4m-5)}{6}\)

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2015 Paper 2 Q4
D: 1600.0 B: 1516.0

  1. The continuous function \(\f\) is defined by \[ \tan \f(x) = x \ \ \ \ \ (-\infty < x <\infty) \] and \(\f(0)=\pi\). Sketch the curve \(y=\f(x)\)\,.
  2. The continuous function \(\g\) is defined by \[ \tan \g(x) = \frac x {1+x^2} \ \ \ \ \ \ (-\infty < x <\infty) \] and \(\g(0)=\pi\). Sketch the curves \(y= \dfrac x {1+x^2} \ \) and \(y=\g(x)\)\,.
  3. The continuous function \(\h \) is defined by \(\h (0)=\pi\) and \[ \tan \h (x)= \frac x {1-x^2}\, \ \ \ \ \ (x \ne \pm 1) \,. \] (The values of \(\h (x)\) at \(x=\pm1\) are such that \(\h (x)\) is continuous at these points.) Sketch the curves \(y= \dfrac x {1-x^2} \ \) and \(y=\h (x)\). %
  4. The continuous functions \(\h_1\) and \(\h_2\) are % defined by: \(\h_1(0)=\h_2(0)=\pi \), %\[ %\tan \h_1(x) = \frac {x+x^4} {1+x^2+x^4} %\ \ \ \ \ \text{and} \ \ \ \ \ \ %\tan \h_2(x) = \frac {4x-x^3} {1-x^4} %\,. %\] %for values of \(x\) at which the right hand sides are defined. %Find \(\lim\limits_{x\to\infty}\h_1(x)\) and \(\lim\limits_{x\to\infty}\h_2(x)\,\).

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2015 Paper 2 Q5
D: 1600.0 B: 1484.9

In this question, the \(\mathrm{arctan}\) function satisfies \(0\le \arctan x <\frac12 \pi\) for \(x\ge0\,\).

  1. Let \[ S_n= \sum_{m=1}^n \arctan \left(\frac1 {2m^2}\right) \,, \] for \(n=1, 2, 3, \ldots\) . Prove by induction that \[ \tan S_n = \frac n {n+1} \,. \] Prove also that \[ S_n = \arctan \frac n {n+1} \,. \]
  2. In a triangle \(ABC\), the lengths of the sides \(AB\) and \(BC\) are \(4n^2\) and \(4n^4-1\), respectively, and the angle at \(B\) is a right angle. Let \(\angle BCA = 2\alpha_n\). Show that \[ \sum_{n=1}^\infty \alpha_n = \tfrac14\pi \,. \]

Solution:

  1. Claim: \(\tan S_n = \frac n {n+1}\) Proof: (By Induction) Base case: (\(n=1\)): \begin{align*} && \tan \left ( \sum_{m=1}^1 \arctan \left ( \frac{1}{2m^2} \right) \right) &= \tan \left ( \arctan \left ( \frac{1}{2} \right) \right) \\ &&&= \frac12 = \frac{1}{1+1} \end{align*} Therefore the base case is true. Inductive step: Suppose our statement is true for some \(n = k\), ie \begin{align*} && \frac{k}{k+1} &= \tan \left ( \sum_{m=1}^k \arctan \left ( \frac{1}{2m^2} \right) \right) \\ \Rightarrow && \tan S_{k+1} &= \tan \left ( \sum_{m=1}^k \arctan \left ( \frac{1}{2m^2} \right) + \arctan \left ( \frac{1}{2 (k+1)^2} \right) \right) \\ &&&= \frac{\tan S_k + \tan \left ( \arctan \left ( \frac{1}{2 (k+1)^2} \right) \right)}{1-\tan S_k \tan \left ( \arctan \left ( \frac{1}{2 (k+1)^2} \right) \right)} \\ &&&= \frac{\frac{k}{k+1} + \frac{1}{2(k+1)^2}}{1-\frac{k}{k+1} \frac{1}{2(k+1)^2}} \\ &&&= \frac{2k(k+1)^2+(k+1)}{2(k+1)^3-k} \\ &&&= \frac{k+1}{(k+1)+1} \end{align*} Therefore it is true for \(n=k+1\). Conclusion: Therefore by the principle of mathematical induction since our statement is true for \(n=1\) and if it is true for \(n=k\) it is true for \(n=k+1\) it is true for all \(n\geq1\) Since \(S_n < \frac12 \pi\) for all \(n\), we must have \(\arctan \frac{n}{n+1} = S_n\)
  2. \(\tan (2\alpha_n) = \frac{4n^2}{4n^4-1} = \frac{2n^2+2n^2}{(2n^2)(2n^2)-1} = \frac{\frac{1}{2n^2}+\frac{1}{2n^2}}{1-\frac{1}{2n^2}\frac{1}{2n^2}} \Rightarrow \tan (\alpha_n) = \arctan \frac{1}{2n^2}\). In particular \(\displaystyle \sum_{n=1}^{N} \alpha_n = \arctan \frac{n}{n+1} \Rightarrow \sum_{n=1}^{\infty} \alpha_n \to \arctan 1 = \frac{\pi}{4} \)

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2015 Paper 2 Q6
D: 1600.0 B: 1504.2

  1. Show that \[ \mathrm{sec}^2\left(\tfrac14\pi-\tfrac12 x\right)=\frac{2}{1+\sin x} \,. \] Hence integrate \(\dfrac{1}{1+\sin x}\) with respect to \(x\).
  2. By means of the substitution \(y=\pi -x\), show that \[ \int_0^\pi x \f (\sin x)\, \d x = \frac \pi 2 \int_0^\pi \f(\sin x) \, \d x ,\] where \(\mathrm{f}\) is any function for which these integrals exist. Hence evaluate \[ \int_0^\pi \frac x {1+\sin x} \, \d x \,. \]
  3. Evaluate \[ \int_0^\pi\frac{ 2x^3 -3\pi x^2}{(1+\sin x)^2}\, \d x .\]

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2015 Paper 2 Q7
D: 1600.0 B: 1484.0

A circle \(C\) is said to be {\em bisected} by a curve \(X\) if \(X\) meets \(C\) in exactly two points and these points are diametrically opposite each other on \(C\).

  1. Let \(C\) be the circle of radius \(a\) in the \(x\)-\(y\) plane with centre at the origin. Show, by giving its equation, that it is possible to find a circle of given radius \(r\) that bisects \(C\) provided \(r>a\). Show that no circle of radius \(r\) bisects \(C\) if \(r\le a\,\).
  2. Let \(C_1\) and \(C_2\) be circles with centres at \((-d,0)\) and \((d,0)\) and radii \(a_1\) and \(a_2\), respectively, where \(d>a_1\) and \(d>a_2\). Let \(D\) be a circle of radius~\(r\) that bisects both \(C_1\) and \(C_2\). Show that the \(x\)-coordinate of the centre of \(D\) is \(\dfrac{a_2^2 - a_1^2}{4d}\). Obtain an expression in terms of \(d\), \(r\), \(a_1\) and \(a_2\) for the \(y\)-coordinate of the centre of~\(D\), and deduce that \(r\) must satisfy \[ 16r^2d^2 \ge \big (4d^2 +(a_2-a_1)^2\big) \, \big (4d^2 +(a_2+a_1)^2\big) \,. \]

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