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2016 Paper 3 Q1
D: 1700.0 B: 1500.0
integration reduction formula proof by induction trigonometric substitution improper integral completing the square binomial coefficient integration by parts

Let \[ \displaystyle I_n= \int_{-\infty}^\infty \frac 1 {(x^2+2ax+b)^n} \, \d x \] where \(a\) and \(b\) are constants with \(b > a^2\), and \(n\) is a positive integer.

  1. By using the substitution \(x + a = \sqrt{b- a^2} \, \tan u\,\), or otherwise, show that \[ I_1 = \dfrac \pi {\sqrt{b-a^2}}\, . \]
  2. Show that \(2n(b - a^2)\, I_{n+1} =(2n - 1) \, I_n\,\).
  3. Hence prove by induction that \[ I_n =\frac{\pi}{2^{2n-2}( b - a^2)^{n-\frac12}} \, \binom {2n-2}{n-1} \]

Solution:

  1. \(\,\) \begin{align*} && I_1 &= \int_{-\infty}^{\infty} \frac{1}{x^2+2ax+b} \d x \\ &&&= \int_{-\infty}^{\infty} \frac{1}{b-a^2 +(x+a)^2} \d x \\ &&&= \left [ \frac{1}{\sqrt{b-a^2}} \tan^{-1} \frac{x+a}{\sqrt{b-a^2}} \right]_{-\infty}^{\infty} \\ &&&= \frac{\pi}{\sqrt{b-a^2}} \end{align*}
  2. \(\,\) Here is the corrected LaTeX code for the second part, maintaining your exact styling and notation.
  3. \(\,\) \begin{align*} && I_{n} &= \int_{-\infty}^{\infty} \frac{1}{(x^2+2ax+b)^{n}} \d x \\ &&&= \left[ \frac{x}{(x^2+2ax+b)^n} \right]_{-\infty}^{\infty} - \int_{-\infty}^\infty x \cdot \frac{-n(2x+2a)}{(x^2+2ax+b)^{n+1}} \d x \\ &&&= 0 + n \int_{-\infty}^\infty \frac{2x^2+2ax}{(x^2+2ax+b)^{n+1}} \d x \\ &&&= n \int_{-\infty}^\infty \frac{2(x^2+2ax+b) - (2ax+2b)}{(x^2+2ax+b)^{n+1}} \d x \\ &&&= 2n I_n - n \int_{-\infty}^\infty \frac{2ax+2b}{(x^2+2ax+b)^{n+1}} \d x \\ &&&= 2n I_n - n \int_{-\infty}^\infty \frac{a(2x+2a) + 2(b-a^2)}{(x^2+2ax+b)^{n+1}} \d x \\ &&&= 2n I_n - n \int_{-\infty}^\infty \frac{a(2x+2a)}{(x^2+2ax+b)^{n+1}} \d x - 2n(b-a^2) I_{n+1} \\ &&&= 2n I_n - n \left[ \frac{-a}{n(x^2+2ax+b)^n} \right]_{-\infty}^\infty - 2n(b-a^2) I{n+1} \\ &&&= 2n I_n - 0 - 2n(b-a^2) I_{n+1} \\ \Rightarrow && 2n(b-a^2)I_{n+1} &= (2n-1)I_n \end{align*}
  4. \(\,\) \begin{align*} && I_{n+1} &= \frac{2n-1}{2n(b-a^2)} I_n \\ &&&= \frac{2n-1}{2n(b-a^2)} \cdot \frac{2n-3}{2(n-1)(b-a^2)} I_{n-1} \\ &&&= \frac{2n-1}{2n(b-a^2)} \cdot \frac{2n-3}{2(n-1)(b-a^2)} \cdots I_{1} \\ &&&= \frac{(2n-1)(2n-3) \cdots 1}{2n \cdot 2(n-1) \cdots 2 (b-a^2)^n} \frac{\pi}{\sqrt{b-a^2}} \\ &&&= \frac{(2n-1)(2n-3) \cdots 1}{2^n n!} \frac{\pi}{(b-a^2)^{n+\frac12}} \\ &&&= \frac{(2n-1)(2n-3) \cdots 1 \cdot 2n \cdot 2(n-1) \cdots 2}{2^{2n} n!n!} \frac{\pi}{(b-a^2)^{n+\frac12}} \\ &&&= \frac{(2n)!}{2^{2n}n!n!}\frac{\pi}{(b-a^2)^{n+\frac12}} \\ &&&= \frac{\pi}{2^{2n}(b-a^2)^{n+\frac12}} \binom{2n}{n} \\ \Rightarrow && I_n &= \frac{\pi}{2^{2n-2}(b-a^2)^{n-\frac12}} \binom{2n-2}{n-1} \\ \end{align*}

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2016 Paper 3 Q2
D: 1700.0 B: 1484.0
conics parabola normal parametric locus intersection inequality geometric proof

The distinct points \(P(ap^2 , 2ap)\), \(Q(aq^2 , 2aq)\) and \(R(ar^2,2ar)\) lie on the parabola \(y^2 = 4ax\), where \(a>0\). The points are such that the normal to the parabola at \(Q\) and the normal to the parabola at \(R\) both pass through \(P\).

  1. Show that \(q^2 +qp + 2 = 0\).
  2. Show that \(QR\) passes through a certain point that is independent of the choice of \(P\).
  3. Let \(T\) be the point of intersection of \(OP\) and \(QR\), where \(O\) is the coordinate origin. Show that \(T\) lies on a line that is independent of the choice of \(P\). Show further that the distance from the \(x\)-axis to \(T\) is less than \(\dfrac {\;a}{\sqrt2}\,\).

Solution:

  1. \begin{align*} && 2y \frac{\d y}{\d x} &= 4a \\ \Rightarrow && \frac{\d y}{\d x} &= \frac{2a}{y} \end{align*} Therefore we must have \begin{align*} && \underbrace{-\frac{2aq}{2a}}_{\text{gradient of normal}} &= \underbrace{\frac{2ap-2aq}{ap^2-aq^2}}_{\Delta y / \Delta x} \\ \Rightarrow && -q &= \frac{2}{p+q} \\ && 0 &= 2 + pq+q^2 \end{align*}
  2. We must have that \(q,r\) are the two roots of \(x^2+px+2 = 0\) \(QR\) has the equation: \begin{align*} && \frac{y-2aq}{x-aq^2} &= \frac{2ar-2aq}{ar^2-aq^2} \\ \Rightarrow && \frac{y-2aq}{x-aq^2} &= \frac{2}{r+q} \\ \Rightarrow && y &= \frac{2}{q+r}(x-aq^2) +2aq \\ && y &= -\frac{2}{p}x+2a\left(q-\frac{q^2}{q+r} \right) \\ &&y&= -\frac{2}{p}x+2a \frac{qr}{q+r} \\ && y &= -\frac{2}{p}x - 2a \frac{2}{p} \\ && y & = -\frac{2}{p}(x+2a) \end{align*} Therefore the point \((-2a,0)\) lies on all such lines.
  3. \(OP\) has equation \(y = \frac{2}{p} x\) \begin{align*} && y &= \frac{2}{p} x \\ && y & = -\frac{2}{p}(x+2a) \\ && 2y &= -\frac{4a}{p} \\ \Rightarrow && y &= -\frac{2a}{p} \\ && x &= -a \end{align*} Therefore \(T\left (-a, -\frac{2a}{p} \right)\) always lies on the line \(x = -a\) The distance to the \(x\)-axis from \(T\) is \(\frac{2a}{|p|}\). We need to show that \(p\) can't be too small. Specifically \(x^2+px+2 = 0\) must have \(2\) real roots, ie \(\Delta = p^2-8 \geq 0 \Rightarrow |p| \geq 2\sqrt{2}\), ie \(\frac{2a}{|p|} \leq \frac{2a}{2\sqrt{2}} = \frac{a}{\sqrt{2}}\) as required.

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2016 Paper 3 Q3
D: 1700.0 B: 1484.0
integration partial fractions polynomial division proof by contradiction rational function exponential function differentiation

  1. Given that \[ \int \frac {x^3-2}{(x+1)^2}\, \e ^x \d x = \frac{\P(x)}{Q(x)}\,\e^x + \text{constant} \,, \] where \(\P(x)\)and \(Q(x)\) are polynomials, show that \(Q(x)\) has a factor of \(x + 1\). Show also that the degree of \(\P(x)\) is exactly one more than the degree of \(Q(x)\), and find \(\P(x)\) in the case \(Q(x) =x+1\).
  2. Show that there are no polynomials \(\P(x)\) and \(Q(x)\) such that \[ \int \frac 1 {x+1} \, \, \e^x \d x = \frac{\P(x)}{Q(x)}\,\e^x +\text{constant} \,. \] You need consider only the case when \(\P(x)\) and \(Q(x)\) have no common factors.

Solution:

  1. \begin{align*} && \int \frac {x^3-2}{(x+1)^2}\, \e ^x \d x &= \frac{\P(x)}{Q(x)}\,\e^x + \text{constant} \\ \underbrace{\Rightarrow}_{\frac{\d}{\d x}} && \frac{x^3-2}{(x+1)^2}e^x &= \frac{P'(x)Q(x)-Q'(x)P(x)}{Q(x)^2}e^x + \frac{P(x)}{Q(x)}e^x \\ \Rightarrow && \frac{x^3-2}{(x+1)^2} &= \frac{(P(x)+P'(x))Q(x)-Q'(x)P(x)}{Q(x)^2} \\ \Rightarrow && Q(x)^2(x^3-2) &= ((P(x)+P'(x))Q(x)-Q'(x)P(x))(x+1)^2 \\ \Rightarrow && Q(-1) &= 0 \\ \Rightarrow && x+1 &\mid Q(x) \end{align*} We have \(\frac{x^3-2}{(x+1)^2}\) has degree \(1\) (plus some remainder term). Therefore \begin{align*} 1 &= \deg \l (P(x)+P'(x))Q(x)-Q'(x)P(x)\r - 2 \deg Q(x) \\ &= \deg P(x) + \deg Q(x) - 2 \deg Q(x) \\ &= \deg P(x) - \deg Q(x) \end{align*} as required. Suppose \(Q(x) = x+1, P(x) = ax^2+bx+c\) then \begin{align*} && \frac{x^3-2}{(x+1)^2} &= \frac{(P(x)+P'(x))(x+1)-P(x)}{(x+1)^2} \\ \Rightarrow && x^3-2 &= (P(x)+P'(x))(x+1) - P(x) \\ \Rightarrow && x^3-2 &= (ax^2+bx+c+2ax+b)(x+1) - (ax^2+bx+c) \\ &&&= a x^3+ x^2 (2 a + b) + x (2 a + b + c)+b \\ \Rightarrow && a &= 1 \\ && b &= -2 \\ && c &= 0 \end{align*} So \(P(x) = x^2-2x\)
  2. \begin{align*} && \int \frac1{x+1}e^x \d x &= \frac{P(x)}{Q(x)}e^x + c \\ \Rightarrow && \frac{1}{x+1} e^x &= \frac{P'(x)Q(x)-Q'(x)P(x)}{Q(x)^2}e^x + \frac{P(x)}{Q(x)}e^x \\ \Rightarrow && \frac{1}{x+1} &= \frac{(P(x)+P'(x))Q(x)-Q'(x)P(x)}{Q(x)^2} \end{align*} Therefore \(Q(-1) = 0\) and so \(x +1 \mid Q(x)\). Considering degrees, we must have that \(P(x)\) has degree \(1\) less than \(Q(x)\). Consider also the number of factors of \(x+1\) in the numerator and denominator. Since \(P(x)\) and \(Q(x)\) have no common factors, the \(Q(x)\) could have \(q\) factors and \(P(x)\) must have none. The denominator therefore has \(2q\) factors and the numerator must have \(q-1\) factors (coming from \(Q'(x)\)), we must have \(2q = (q-1) + 1\), but that implies \(q = 0\). Contradiction! \end{align*}

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2016 Paper 3 Q4
D: 1700.0 B: 1484.0
sequences telescoping series partial fractions convergence hyperbolic functions sech cosech infinite series substitution

  1. By considering \(\displaystyle \frac1{1+ x^r} - \frac1{1+ x^{r +1}}\) for \(\vert x \vert \ne 1\), simplify \[ \sum_{r=1}^N \frac{x^r}{(1+x^r)(1+x^{r+1})} \] Show that, for \(\vert x \vert <1\), \[ \sum_{r=1}^\infty \frac{x^r}{(1+x^r)(1+x^{r+1})} = \frac x {1-x^2} \]
  2. Deduce that \[ \sum_{r=1}^\infty \textrm{sech}(ry)\textrm{sech}((r + 1)y) = 2\e^{-y} \textrm{cosech}(2 y) \] for \(y > 0\). Hence simplify \[ \sum_{r=-\infty}^\infty \textrm{sech}(ry) \textrm{sech}((r + 1)y) \] for \(y>0\).

Solution:

  1. \(\,\) \begin{align*} && \frac{1}{1+x^r} - \frac{1}{1+x^{r+1}} &= \frac{1+x^{r+1}-1-x^r}{(1+x^r)(1+x^{r+1})} \\ &&&= \frac{x^r(x-1)}{(1+x^r)(1+x^{r+1})} \\ \\ && \sum_{r=1}^N \frac{x^r}{(1+x^r)(1+x^{r+1})} &= \sum_{r=1}^N \frac{1}{x-1} \left ( \frac{1}{1+x^r} - \frac{1}{1+x^{r+1}}\right) \\ &&&= \frac{1}{x-1} \Bigg ( \frac{1}{1+x} + \cdots \\ &&& \qquad \qquad \quad - \frac{1}{1+x^2} + \frac{1}{1+x^2} + \cdots \\ &&& \qquad \qquad \quad - \frac{1}{1+x^3} + \frac{1}{1+x^3} + \cdots \\ &&& \qquad \qquad \quad - \cdots \\ &&& \qquad \qquad \quad - \frac{1}{1+x^{N+1}} \Bigg ) \\ &&&= \frac{1}{x-1} \left (\frac{1}{1+x} - \frac{1}{1+x^{N+1}} \right) \\ \\ && \sum_{r=1}^{\infty} \frac{x^r}{(1+x^r)(1+x^{r+1})} &= \lim_{N\to \infty} \frac{1}{x-1} \left (\frac{1}{1+x} - \frac{1}{1+x^{N+1}} \right) \\ &&&= \frac{1}{x-1} \left ( \frac{1}{1+x} - 1\right) \\ &&&= \frac{1}{x-1} \left ( \frac{-x}{1+x} \right) \\ &&&= \frac{x}{1-x^2} \end{align*}
  2. \(\,\) \begin{align*} && \sum_{r=1}^\infty \textrm{sech}(ry)\textrm{sech}((r + 1)y) &= \sum_{r=1}^\infty \frac{4}{(e^{ry}+e^{-ry})(e^{(r+1)y}+e^{-(r+1)y})} \\ &&&=\sum_{r=1}^\infty \frac{4e^{-(2r+1)y}}{(1+e^{-2ry})(1+e^{-2(r+1)y})} \\ x = e^{-2y}: &&&= \frac{4e^{-y}e^{-2y}}{1-e^{-4y}} \\ &&&= \frac{4e^{-y}e^{-2y}}{e^{-2y}(e^{2y}-e^{-2y})} \\ &&&=2e^{-y}\textrm{cosech}(2y) \end{align*} \begin{align*} && \sum_{r=-\infty}^\infty \textrm{sech}(ry) \textrm{sech}((r + 1)y) &= \sum_{r=1}^\infty \textrm{sech}(ry) \textrm{sech}((r + 1)y) + \sum_{r=-\infty}^0 \textrm{sech}(ry) \textrm{sech}((r + 1)y) \\ &&&= 2e^{-y}\textrm{cosech}(2y) + \sum_{r=0}^\infty \textrm{sech}(-ry) \textrm{sech}(-(r-1)y) \\ &&&= 2e^{-y}\textrm{cosech}(2y) + \sum_{r=0}^\infty \textrm{sech}((r-1)y) \textrm{sech}(ry) \\ &&&= 4e^{-y}\textrm{cosech}(2y) + \textrm{sech}(y) + \textrm{sech}(-y) \\ &&&= 4e^{-y}\textrm{cosech}(2y)+2\textrm{sech}(y) \\ &&&= 4e^{-y} \frac12 \textrm{sech}(y) \textrm{cosech}(y) + 2 \textrm{sech}(y) \\ &&&= 2\textrm{sech}(y) \left ( e^{-y} \textrm{cosech}(y)+1 \right) \\ &&&= 2\textrm{sech}(y) \left ( \frac{2}{e^{2y}-1} + 1 \right) \\ &&&= 2\textrm{sech}(y) \left ( \frac{e^{2y}+1}{e^{2y}-1} \right) \\ &&&= 2 \textrm{cosech}(y) \end{align*}

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2016 Paper 3 Q5
D: 1700.0 B: 1500.0
binomial theorem proof by induction prime numbers inequality divisibility binomial coefficient number theory

  1. By considering the binomial expansion of \((1+x)^{2m+1}\), prove that \[ \binom{ 2m \! +\! 1}{ m} < 2^{2m}\,, \] for any positive integer \(m\).
  2. For any positive integers \(r\) and \(s\) with \(r< s\), \(P_{r,s}\) is defined as follows: \(P_{r,s}\) is the product of all the prime numbers greater than \(r\) and less than or equal to \(s\), if there are any such primes numbers; if there are no such primes numbers, then \(P_{r,s}=1\,\). For example, \(P_{3,7}=35\), \(P_{7,10}=1\) and \(P_{14,18}=17\). Show that, for any positive integer \(m\), \(P_{m+1\,,\, 2m+1} \) divides \(\displaystyle \binom{ 2m \! +\! 1}{ m} \,,\) and deduce that \[ P_{m+1\,,\, 2m+1} < 2^{2m} \,. \]
  3. Show that, if \(P_{1,k} < 4^k\) for \(k = 2\), \(3\), \(\ldots\), \(2m\), then \( P_{1,2m+1} < 4^{2m+1}\,\).
  4. Prove that \(\P_{1,n} < 4^n\) for \(n\ge2\).

Solution:

  1. Notice that \((1+x)^{2m+1} = 1+\binom{2m+1}{1}x+\cdots + \binom{2m+1}{m}x^{m} + \binom{2m+1}{m+1} + \cdots\). Notice also that \(\binom{2m+1}{m} = \binom{2m+1}{m+1}\). Therefore evaluating at \(x = 1\), we see \(2^{2m+1} > \binom{2m+1}{m} + \binom{2m+1}{m+1} = 2 \binom{2m+1}{m} \Rightarrow \binom{2m+1}{m} < 2^{2m}\)
  2. Each prime dividing \(P_{m+1, 2m+1}\) divides the numerator of \(\binom{2m+1}{m}\) since it appears in \((2m+1)!\), but not the denominator, since they wont appear in \(m!\) or \((m+1)!\), and since they are prime they have to appear to divide it. Therefore the must divide \(\binom{2m+1}{m}\) and therefore \(P_{m+1,2m+1}\) must divide that binomail coefficient. Since \(a \mid b \Rightarrow a \leq b\) we must have \(P_{m+1, 2m+1} \leq \binom{2m+1}{m} < 2^{2m}\)
  3. Since \begin{align*} P_{1,2m+1} &= P_{1,m+1}P_{m+1, 2m+1} \tag{split into primes below \(m+1\) and abvoe} \\ &< 4^{m+1}P_{m+1,2m+1} \tag{use the condition from the question}\\ &<4^{m+1}2^{2m} \tag{use our inequality} \\ &= 4^{2m+1} \end{align*}
  4. We proceed by (strong) induction. Base case: (\(n = 2\)): \(P_{1,2} = 2 < 4^2 =16\) Suppose it is true for all \(k=2,3,\cdots,2m\) then it is true for \(k=2m+1\) by the previous part of the question. However it is also true for \(k=2m+2\), since that can never be prime (as n is now an even number bigger than 2). Therefore by the principle of mathematical induction it is true for all \(n\).

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2016 Paper 3 Q6
D: 1700.0 B: 1484.0
hyperbolic functions sech tanh curve intersection arcosh artanh identity manipulation necessary and sufficient conditions tangency optimisation

Show, by finding \(R\) and \(\gamma\), that \(A \sinh x + B\cosh x \) can be written in the form \(R\cosh (x+\gamma)\) if \(B>A>0\). Determine the corresponding forms in the other cases that arise, for \(A>0\), according to the value of \(B\). Two curves have equations \(y = \textrm{sech} x\) and \(y = a\tanh x + b\,\), where \(a>0\).

  1. In the case \(b>a\), show that if the curves intersect then the \(x\)-coordinates of the points of intersection can be written in the form \[ \pm\textrm{arcosh} \left( \frac 1 {\sqrt{b^2-a^2}}\right) - {\rm artanh \,} \frac a b .\]
  2. Find the corresponding result in the case \(a>b>0\,\).
  3. Find necessary and sufficient conditions on \(a\) and \(b\) for the curves to intersect at two distinct points.
  4. Find necessary and sufficient conditions on \(a\) and \(b\) for the curves to touch and, given that they touch, express the \(y\)-coordinate of the point of contact in terms of \(a\).

Solution: \begin{align*} && R\cosh(x + \gamma) &=R \cosh x \cosh \gamma + R \sinh x \sinh \gamma \\ \Rightarrow && R \cosh \gamma &= B \\ && R \sinh \gamma &= A \\ \Rightarrow && R^2 &= B^2 - A^2 \\ \Rightarrow && \tanh \gamma &= \frac{A}{B} \\ \end{align*} Therefore it is possible, by writing \(R = \sqrt{B^2-A^2}\) and \(\gamma = \textrm{artanh} \left ( \frac{A}{B} \right)\). This works as long as \(|B| > A > 0\). Supposing \(A >|B| \), try \(S \sinh (x + \delta) = S \sinh x \cosh \delta +S \cosh x \sinh \delta\) \begin{align*} && S \cosh \delta &= A \\ && -S \sinh \delta &= B \\ \Rightarrow && S^2 &= A^2 - B^2 \\ \Rightarrow && \tanh \delta &= \frac{B}{A} \\ \end{align*} Therefore in this case we can write \(\sqrt{A^2-B^2} \sinh \left (x + \tanh^{-1} \left ( \frac{B}{A} \right) \right)\) If \(A = \pm B > 0\) we can we have \(A \sinh x + B \cosh x = \pm Ae^{\pm x}\)

  1. Suppose \(y \cosh x = 1\) and \(y \cosh x = a \sinh x +b \cosh x\) so \begin{align*} && 1 & = a \sinh x + b \cosh x \\ &&&= \sqrt{b^2-a^2} \cosh(x + \textrm{artanh} \frac{a}{b} ) \\ \Rightarrow && x + \textrm{artanh} \frac{a}{b} &= \pm \textrm{arcosh} \left ( \frac{1}{\sqrt{b^2-a^2}} \right) \\ \Rightarrow && x &= \pm \textrm{arcosh} \left ( \frac{1}{\sqrt{b^2-a^2}} \right) -\textrm{artanh} \frac{a}{b} \end{align*}
  2. If \( a > b > 0\) we have \begin{align*} && 1 & = \sqrt{a^2-b^2} \sinh \left ( x - \textrm{artanh} \frac{b}{a} \right) \\ \Rightarrow && x &= \textrm{arsinh} \left ( \frac{1}{\sqrt{a^2-b^2}} \right) + \textrm{artanh} \left ( \frac{b}{a} \right) \end{align*}
  3. To intersect at distinct points we must have \(b > a\) and \(\textrm{arcosh} \left ( \frac{1}{\sqrt{b^2-a^2}} \right) \neq 0\) which is always true.
  4. For the curves to touch, we need them to intersect and have matching derivatives, ie \begin{align*} && -\tanh x \cdot \textrm{sech}x &= a\textrm{sech}^2 x \\ \Rightarrow && 0 &= \textrm{sech}^2 x (a + \sinh x) \\ \Rightarrow && x &= -\textrm{arsinh} \, a\\ \Rightarrow && \sinh x &= - a\\ \Rightarrow &&\cosh x &= \sqrt{1 + a^2} \\ \end{align*} So if the curves touch, we must have \(1 = -a^2+b\sqrt{1+a^2} \Rightarrow b = \sqrt{1+a^2}\) and since this does work it is a necessary and sufficient condition. We will also have the \(y\) coordinate is \(\frac{1}{\sqrt{1+a^2}}\)

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2016 Paper 3 Q7
D: 1700.0 B: 1516.0
complex numbers roots of unity De Moivre regular polygon modulus factorization geometric proof product of distances

Let \(\omega = \e^{2\pi {\rm i}/n}\), where \(n\) is a positive integer. Show that, for any complex number \(z\), \[ (z-1)(z-\omega) \cdots (z - \omega^{n-1}) = z^n -1\,. \] The points \(X_0, X_1, \ldots\, X_{n-1}\) lie on a circle with centre \(O\) and radius 1, and are the vertices of a regular polygon.

  1. The point \(P\) is equidistant from \(X_0\) and \(X_1\). Show that, if \(n\) is even, \[ |PX_0| \times |PX_1 |\times \,\cdots\, \times |PX_{n-1}| = |OP|^n +1\, ,\] where \(|PX_ k|\) denotes the distance from \(P\) to \(X_k\). Give the corresponding result when \(n\) is odd. (There are two cases to consider.)
  2. Show that \[ |X_0 X_1|\times |X_0 X_2|\times \,\cdots\, \times |X_0 X_{n-1}| =n\,. \]

Solution: Notice that \(\omega^n = e^{2\pi i} = 1\), so \(\omega\) is a root of \(z^n - 1\), notice also that \((\omega^k)^n =1\) so therefore the \(n\) roots are \(1, \omega, \omega^2, \cdots, \omega^{n-1}\) and so \((z-1)(z-\omega) \cdots (z-\omega^{n-1}) = C(z^n-1)\). By considering the coefficient of \(z^n\) we can see that \(C = 1\).

  1. \(P\) lies on the perpendicular bisect of \(1\) and \(\omega\), so \(p = re^{\pi i/n}\), where \(r\) can be positive or negative, but \(|r| = |OP|\). \begin{align*} && |PX_0| \times |PX_1| \times \cdots \times |PX_{n-1}| &= |(p-1)(p-\omega) \cdots (p-\omega^{n-1})| \\ &&&= |p^n - 1| \\ &&&= |r^ne^{\pi i} - 1| \\ &&&= |-|OP|^n - 1| \tag{since \(n\) even} \\ &&&= |OP|^n+1 \end{align*} If \(n\) is odd, depending on the sign of \(r\) we get \(|OP|^n+1\) or \(||OP|^n-1|\).
  2. \(\,\) \begin{align*} && (z-\omega) \cdots(z-\omega^{n-1}) &= \frac{z^n-1}{z-1} \\ &&&= 1 + z +\cdots + z^{n-1} \\ && |X_0X_1| \times |X_0X_2| \times \cdots \times |X_0X_{n-1}| &= |(1 - \omega)\cdots(1-\omega^{n-1})| \\ &&&= 1+1+1^2+\cdots + 1^{n-1} \\ &&&= n \end{align*}

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2016 Paper 3 Q8
D: 1700.0 B: 1484.0
simultaneous equations functional equations substitution involution rational function algebraic manipulation self-inverse function

  1. The function f satisfies, for all \(x\), the equation \[ \f(x) + (1- x)\f(-x) = x^2\, . \] Show that \(\f(-x) + (1 + x)\f(x) = x^2\,\). Hence find \(\f(x)\) in terms of \(x\). You should verify that your function satisfies the original equation.
  2. The function \({\rm K}\) is defined, for \(x\ne 1\), by \[{\rm K}(x) = \dfrac{x+1}{x-1}\,.\] Show that, for \(x\ne1\), \({\rm K(K(}x)) =x\,\). The function g satisfies the equation \[ \g(x)+ x\, \g\Big(\frac{ x+1 }{x-1}\Big) = x \ \ \ \ \ \ \ \ \ \ \ ( x\ne 1) \,. \] Show that, for \(x\ne1\), \(\g(x)= \dfrac{2x}{x^2+1}\,\).
  3. Find \(\h(x)\), for \(x\ne0\), \(x\ne1\), given that \[ \h(x)+ \h\Big(\frac 1 {1-x}\Big)= 1-x -\frac1{1-x} \ \ \ \ \ \ ( x\ne0, \ \ x\ne1 ) \,. \]

Solution:

  1. \(\,\) Let \(P(x)\) mean the proposition that \(f(x) + (1-x)f(-x) = x^2\) so \begin{align*} P(x): && f(x) + (1-x)f(-x) &= x^2 \\ P(-x): && f(-x)+(1+x)f(x) &= (-x)^2 = x^2 \\ \Rightarrow && f(x)+(1-x)\left (x^2-(1+x)f(x) \right) &= x^2 \\ \Rightarrow && f(x) \left (1 -(1-x^2) \right) &= x^2 + (x-1)x^2 \\ \Rightarrow && f(x)x^2 &= x^3 \\ \Rightarrow && f(x) &= x \end{align*} Notice that \(x + (1-x)(-x) = x^2\) so it does satisfy the functional equation.
  2. Let \(K(x) = \frac{x+1}{x-1}\) if \(x \neq 1\) so \begin{align*} && K(K(x)) &= \frac{K(x)+1}{K(x)-1} \\ &&&= \frac{\frac{x+1}{x-1}+1}{\frac{x+1}{x-1}-1} \\ &&&= \frac{\frac{2x}{x-1}}{\frac{2}{x-1}} \\ &&&= x \end{align*} Let \(Q(x)\) denote the proposition that \(g(x) + xg(K(x)) = x\) so \begin{align*} Q(x): && g(x) + xg(K(x)) &= x \\ Q(K(x)): && g(K(x)) + K(x)g(x) &= K(x) \\ \Rightarrow && g(x) +xK(x)[1-g(x)] &= x \\ \Rightarrow && g(x)[1-xK(x)] &= x(1-K(x)) \\ \Rightarrow && g(x) \frac{x-1-x^2-x}{x-1} &= \frac{-2x}{x-1} \\ \Rightarrow && g(x) &= \frac{2x}{x^2+1} \end{align*}. And notice that \(\frac{2x}{x^2+1} + x \frac{2\frac{x+1}{x-1}}{\left( \frac{x+1}{x-1}\right)^2+1} = \frac{2x}{x^2+1} + \frac{2x(x^2-1)}{2x^2+2} = x\)
  3. Consider \(H(x) = \frac{1}{1-x}\) then notice that \(H(H(x)) = \frac{1}{1-\frac{1}{1-x}} = \frac{x-1}{x}\) and \(H^3(x) = \frac{\frac{1}{1-x}-1}{\frac{1}{1-x}} = 1-(1-x) = x\). So So letting \(S(x)\) be the statement that \(h(x) + h(H(x)) = 1 - x - \frac{1}{1-x}\) we have \begin{align*} S(x): && h(x) + h(H(x)) &= 1 - x - H(x) \\ S(H(x)): && h(H(x)) + h(H^2(x)) &= 1 - H(x) - H^2(x) \\ S(H^2(x)): && h(H^2(x)) + h(x) &= 1 - H^2(x) - x \\ S(x) - S(H(x)) + S(H^2(x)): && 2h(x) &= 1 - 2x \\ \Rightarrow && h(x)& = \frac12 - x \end{align*} and notice that \(\frac12 -x +\frac12 - \frac{1}{1-x} = 1 - x - \frac{1}{1-x}\) so it does satisfy the equation.

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2016 Paper 3 Q9
D: 1700.0 B: 1475.6
SHM elastic spring Hooke's law equilateral triangle equilibrium small displacement approximation energy tension modulus of elasticity natural length

Three pegs \(P\), \(Q\) and \(R\) are fixed on a smooth horizontal table in such a way that they form the vertices of an equilateral triangle of side \(2a\). A particle \(X\) of mass \(m\) lies on the table. It is attached to the pegs by three springs, \(PX\), \(QX\) and \(RX\), each of modulus of elasticity \(\lambda\) and natural length \(l\), where \(l < \frac{ \ 2 }{\sqrt3}\, a\). Initially the particle is in equilibrium. Show that the extension in each spring is \(\frac{\ 2}{\sqrt3}\,a -l\,\). The particle is then pulled a small distance directly towards \(P\) and released. Show that the tension \(T\) in the spring \(RX\) is given by \[ T= \frac {\lambda} l \left( \sqrt{\frac {4a^2}3 + \frac{2ax}{\sqrt3} +x^2\; }\; -l\right) , \] where \(x\) is the displacement of \(X\) from its equilibrium position. Show further that the particle performs approximate simple harmonic motion with period \[ 2\pi \sqrt{ \frac{4mla}{3 (4a-\sqrt3 \, l)\lambda } \; }\,. \]

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2016 Paper 3 Q10
D: 1700.0 B: 1484.0
circular motion conical pendulum inclined plane string tension normal reaction inequality energy conservation constraint condition

A smooth plane is inclined at an angle \(\alpha\) to the horizontal. A particle \(P\) of mass \(m\) is attached to a fixed point \(A\) above the plane by a light inextensible string of length \(a\). The particle rests in equilibrium on the plane, and the string makes an angle \(\beta\) with the plane. The particle is given a horizontal impulse parallel to the plane so that it has an initial speed of \(u\). Show that the particle will not immediately leave the plane if \(ag\cos(\alpha + \beta)> u^2 \tan\beta\). Show further that a necessary condition for the particle to perform a complete circle whilst in contact with the plane is \(6\tan\alpha \tan \beta < 1\).

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2016 Paper 3 Q11
D: 1700.0 B: 1484.0
variable force power resistance Newton's second law integration logarithm partial fractions acceleration car motion inequality comparison

A car of mass \(m\) travels along a straight horizontal road with its engine working at a constant rate \(P\). The resistance to its motion is such that the acceleration of the car is zero when it is moving with speed \(4U\).

  1. Given that the resistance is proportional to the car's speed, show that the distance \(X_1\) travelled by the car while it accelerates from speed \(U\) to speed \(2U\), is given by \[ \lambda X_1 = 2\ln \tfrac 9 5 - 1 \,, \] where \(\lambda= P/(16mU^3)\).
  2. Given instead that the resistance is proportional to the square of the car's speed, show that the distance \(X_2\) travelled by the car while it accelerates from speed \(U\) to speed \(2U\) is given by \[ \lambda X_2 = \tfrac43 \ln \tfrac 98 \,. \]
  3. Given that \(3.17<\ln 24 < 3.18\) and \(1.60<\ln 5 < 1.61\), determine which is the larger of \(X_1\) and \(X_2\).

Solution:

  1. \(\,\) \begin{align*} && F_{res} &= kv \\ && P &= Fv \\ v = 4U: && 0 &= F-F_{res} \\ \Rightarrow && 0 &= \frac{P}{4U} - 4Uk \\ \Rightarrow && k &= \frac{P}{16U^2} \\ \\ &&m v \frac{\d v}{\d x}&= \frac{P}{v} - \frac{P}{16U^2}v \\ \Rightarrow && X_1 &= \int_{v=U}^{v=2U} \frac{16U^2mv^2}{P(16U^2-v^2)} \d v \\ v = Ut&& &= \frac{16mU^2}{P} \int_{t=1}^{t=2}\left ( \frac{t^2}{16-t^2} \right)U\d t \\ &&&= \frac{16mU^3}{P} \int_1^2 \left ( -1 + \frac{16}{16-t^2} \right) \d t \\ &&&= \frac{16mU^3}{P} \int_1^2 \left ( -1 +\frac{2}{4+t} +\frac{2}{4-t} \right) \d t \\ &&&= \frac{1}{\lambda}\left (-1 + 2\ln(6)-2\ln(2)-2\ln(5)+2\ln(3) \right) \\ \Rightarrow && \lambda X_1 &= 2\ln \tfrac95-1 \end{align*}
  2. \(\,\) \begin{align*} && F_{res} = kv^2 \\ v = 4U: && 0 &= \frac{P}{4U} - 16U^2k \\ \Rightarrow && k &= \frac{P}{64U^3} \\ \\ && mv \frac{\d v}{\d x} &= \frac{P}{v} - \frac{P}{64U^3}v^2 \\ \Rightarrow && X_2 &= \int_{v=U}^{v=2U} \frac{64U^3mv^2}{P(64U^3-v^3)} \d v \\ &&&= \frac{64U^3m}{P} \int_{v=U}^{v=2U} \frac{v^2}{64U^3-v^3} \d v\\ v = Ut &&&= \frac{64U^3m}{P} \int_{t=1}^{t=2} \frac{U^2t^2}{64U^3-U^3v^3} U \d t\\ &&&= \frac{4}{\lambda} \int_1^2 \frac{t^2}{64-t^3} \d t \\ &&&= \frac{4}{\lambda} \left [ -\frac13\ln(64-t^3) \right]_1^2 \\ &&&= \frac{4}{3\lambda} \ln (63/56) \\ \Rightarrow && \lambda X_2 &= \tfrac43 \ln \tfrac98 \end{align*}
  3. \(\,\) \begin{align*} && 2\ln \tfrac95 - 1 &\overset{?}{>} \frac43 \ln \frac98 \\ \Leftrightarrow && 4 \ln 3 - 2\ln 5 - 1 &\overset{?}{>} \frac83\ln 3 -4 \ln 2 \\ \Leftrightarrow && \frac43(3\ln 3 + 3\ln 2 - 2 \ln 3) &\overset{?}{>} 2 \ln 5 + 1\\ \Leftrightarrow && \frac43\ln 24 &\overset{?}{>} 2 \ln 5 + 1\\ \end{align*} The \(LHS\) is \(>4.22\). The \(RHS\) is \(< 4.22\), and therefore our inequality holds, in particular, \(X_1 > X_2\).

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2016 Paper 3 Q12
D: 1700.0 B: 1516.0
Chebyshev's inequality Poisson distribution binomial distribution probability bounds inequality mean and variance summation

Let \(X\) be a random variable with mean \(\mu\) and standard deviation \(\sigma\). Chebyshev's inequality, which you may use without proof, is \[ \P\left(\vert X-\mu\vert > k\sigma\right) \le \frac 1 {k^2} \,, \] where \(k\) is any positive number.

  1. The probability of a biased coin landing heads up is \(0.2\). It is thrown \(100n\) times, where \(n\) is an integer greater than 1. Let \(\alpha \) be the probability that the coin lands heads up \(N\) times, where \(16n \le N \le 24n\). Use Chebyshev's inequality to show that \[ \alpha \ge 1-\frac 1n \,. \]
  2. Use Chebyshev's inequality to show that \[ 1+ n + \frac{n^2}{ 2!} + \cdots + \frac {n^{2n}}{(2n)!} \ge \left(1-\frac1n\right) \e^n \,. \]

Solution:

  1. Let \(N\) be the number of times the coin lands heads up, ie \(N \sim Binomial(100n, 0.2)\), then \(\mathbb{E}(N) = \mu = 20n, \mathrm{Var}(N) = \sigma^2 = 100n \cdot 0.2 \cdot 0.8 = 16n \Rightarrow \sigma = 4\sqrt{n}\). \begin{align*} && \mathbb{P}(|X - \mu| > k\sigma) &\leq \frac{1}{k^2} \\ \Rightarrow && 1 - \mathbb{P}(|X - \mu| \leq k\sigma) &\leq \frac1{k^2} \\ \Rightarrow && 1 - \mathbb{P}(|X - 20n| \leq \sqrt{n} \cdot 4\sqrt{n}) &\leq \frac1{{\sqrt{n}}^2} \\ \Rightarrow && 1 - \mathbb{P}(16n \leq N \leq 24n) &\leq \frac{1}{n} \\ \Rightarrow && 1 - \frac1n &\leq \alpha \end{align*}
  2. Suppose \(X \sim Pois(n)\), then \(\mathbb{E}(X) = n, \mathrm{Var}(X) = n\). Therefore \begin{align*} && \mathbb{P}(|X - \mu| > k\sigma) &\leq \frac{1}{k^2} \\ \Rightarrow && 1-\mathbb{P}(|X - n| \leq \sqrt{n} \cdot \sqrt{n}) &> \frac{1}{\sqrt{n}^2} \\ \Rightarrow && 1 - \sum_{i=0}^{2n} \mathbb{P}(X = i) & \leq \frac{1}{n} \\ \Rightarrow && \sum_{i=0}^{2n} e^{-n} \frac{n^i}{i!} \geq 1 - \frac{1}{n} \\ \Rightarrow && \sum_{i=0}^{2n} \frac{n^i}{i!} \geq \left ( 1 - \frac1n \right)e^n \end{align*}

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2016 Paper 3 Q13
D: 1700.0 B: 1500.0
kurtosis normal distribution integration expectation summation independent random variables fourth moment proof

Given a random variable \(X\) with mean \(\mu\) and standard deviation \(\sigma\), we define the kurtosis, \(\kappa\), of \(X\) by \[ \kappa = \frac{ \E\big((X-\mu)^4\big)}{\sigma^4} -3 \,. \] Show that the random variable \(X-a\), where \(a\) is a constant, has the same kurtosis as \(X\).

  1. Show by integration that a random variable which is Normally distributed with mean 0 has kurtosis 0.
  2. Let \(Y_1, Y_2, \ldots, Y_n\) be \(n\) independent, identically distributed, random variables with mean 0, and let \(T = \sum\limits_{r=1}^n Y_r\). Show that \[ \E(T^4) = \sum_{r=1}^n \E(Y_r^4) + 6 \sum_{r=1}^{n-1} \sum_{s=r+1}^{n} \E(Y^2_s) \E(Y^2_r) \,. \]
  3. Let \(X_1\), \(X_2\), \(\ldots\)\,, \(X_n\) be \(n\) independent, identically distributed, random variables each with kurtosis \(\kappa\). Show that the kurtosis of their sum is \(\dfrac\kappa n\,\).

Solution: \begin{align*} &&\kappa_{X-a} &= \frac{\mathbb{E}\left(\left(X-a-(\mu-a)\right)^4\right)}{\sigma_{X-a}^4}-3 \\ &&&= \frac{\mathbb{E}\left(\left(X-\mu\right)^4\right)}{\sigma_X^4}-3\\ &&&= \kappa_X \end{align*}

  1. \(\,\) \begin{align*} && \kappa &= \frac{\mathbb{E}((X-\mu)^4)}{\sigma^4} - 3 \\ &&&= \frac{\mathbb{E}((\mu+\sigma Z-\mu)^4)}{\sigma^4} - 3 \\ &&&= \frac{\mathbb{E}((\sigma Z)^4)}{\sigma^4} - 3 \\ &&&= \mathbb{E}(Z^4)-3\\ &&&= \int_{-\infty}^{\infty} x^4\frac{1}{\sqrt{2\pi}} \exp \left ( - \frac12x^2 \right)\d x -3 \\ &&&= \left [\frac{1}{\sqrt{2\pi}}x^{3} \cdot \left ( -\exp \left ( - \frac12x^2 \right)\right) \right]_{-\infty}^{\infty} + \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty 3x^2 \exp \left ( - \frac12x^2 \right) \d x - 3 \\ &&&= 0 + 3 \textrm{Var}(Z) - 3 =0 \end{align*}
  2. \(\,\) \begin{align*} && \mathbb{E}(T^4) &= \mathbb{E} \left [\left ( \sum\limits_{r=1}^n Y_r\right)^4\right] \\ &&&= \mathbb{E} \left [ \sum_{r=1}^n Y_r^4+\sum_{i\neq j} 4Y_iY_j^3+\sum_{i\neq j} 6Y_i^2Y_j^2+\sum_{i\neq j \neq k} 12Y_iY_jY_k^2 +\sum_{i\neq j\neq k \neq l}24 Y_iY_jY_kY_l\right] \\ &&&= \sum_{r=1}^n \mathbb{E} \left [ Y_r^4 \right]+\sum_{i\neq j} \mathbb{E} \left [ 4Y_iY_j^3\right]+\sum_{i\neq j} \mathbb{E} \left [ 6Y_i^2Y_j^2\right]+\sum_{i\neq j \neq k} \mathbb{E} \left [ 12Y_iY_jY_k^2\right] +\sum_{i\neq j\neq k \neq l} \mathbb{E} \left [ 24 Y_iY_jY_kY_l\right] \\ &&&= \sum_{r=1}^n \mathbb{E} \left [ Y_r^4 \right]+4\sum_{i\neq j} \mathbb{E} \left [ Y_i]\mathbb{E}[Y_j^3\right]+6\sum_{i\neq j} \mathbb{E} \left [ Y_i^2]\mathbb{E}[Y_j^2\right]+12\sum_{i\neq j \neq k} \mathbb{E} \left [ Y_i]\mathbb{E}[Y_j]\mathbb{E}[Y_k^2\right] +24\sum_{i\neq j\neq k \neq l} \mathbb{E} \left [ Y_i]\mathbb{E}[Y_j]\mathbb{E}[Y_k]\mathbb{E}[Y_l\right] \\ &&&= \sum_{r=1}^n \mathbb{E} \left [ Y_r^4 \right]+6\sum_{i\neq j} \mathbb{E} \left [ Y_i^2]\mathbb{E}[Y_j^2\right] \end{align*}
  3. Without loss of generality, we may assume they all have mean zero. Therefore we can consider the sitatuion as in the previous case with \(T\) and \(Y_i\)s. Note that \(\mathbb{E}(Y_i^4) = \sigma^4(\kappa + 3)\) and \(\textrm{Var}(T) = n \sigma^2\) \begin{align*} && \kappa_T &= \frac{\mathbb{E}(T^4)}{(\textrm{Var}(T))^2} - 3 \\ &&&= \frac{\sum_{r=1}^n \mathbb{E} \left [ Y_r^4 \right]+6\sum_{i\neq j} \mathbb{E} \left [ Y_i^2\right]\mathbb{E}\left[Y_j^2\right]}{n^2\sigma^4}-3 \\ &&&= \frac{n\sigma^4(\kappa+3)+6\binom{n}{2}\sigma^4}{n^2\sigma^4} -3\\ &&&= \frac{\kappa}{n} + \frac{3n + \frac{6n(n-1)}{2}}{n^2} - 3 \\ &&&= \frac{\kappa}{n} + \frac{3n^2}{n^2}-3 \\ &&&= \frac{\kappa}{n} \end{align*}

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