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2020 Paper 3 Q2
D: 1500.0 B: 1500.0
hyperbolic functions implicit differentiation inflection points curve sketching stationary points quadratic in exponential

The curve \(C\) has equation \(\sinh x + \sinh y = 2k\), where \(k\) is a positive constant.

  1. Show that the curve \(C\) has no stationary points and that \(\dfrac{\mathrm{d}^2 y}{\mathrm{d}x^2} = 0\) at the point \((x,y)\) on the curve if and only if \[ 1 + \sinh x \sinh y = 0. \] Find the co-ordinates of the points of inflection on the curve \(C\), leaving your answers in terms of inverse hyperbolic functions.
  2. Show that if \((x,y)\) lies on the curve \(C\) and on the line \(x + y = a\), then \[ \mathrm{e}^{2x}(1 - \mathrm{e}^{-a}) - 4k\mathrm{e}^x + (\mathrm{e}^a - 1) = 0 \] and deduce that \(1 < \cosh a \leqslant 2k^2 + 1\).
  3. Sketch the curve \(C\).

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2010 Paper 1 Q13
D: 1484.0 B: 1516.0
Poisson distribution exponential waiting time quadratic in exponential probability independent random variables inequality roots of equations

The number of texts that George receives on his mobile phone can be modelled by a Poisson random variable with mean \(\lambda\) texts per hour. Given that the probability George waits between 1 and 2 hours in the morning before he receives his first text is \(p\), show that \[ p\e^{2\lambda}-\e^{\lambda}+1=0. \] Given that \(4p<1\), show that there are two positive values of \(\lambda\) that satisfy this equation. The number of texts that Mildred receives on each of her two mobile phones can be modelled by independent Poisson random variables with different means \(\lambda_{1}\) and \(\lambda_{2}\) texts per hour. Given that, for each phone, the probability that Mildred waits between 1 and 2 hours in the morning before she receives her first text is also \(p\), find an expression for \(\lambda_{1}+\lambda_{2}\) in terms of \(p\). Find the probability, in terms of \(p\), that she waits between 1 and 2 hours in the morning to receive her first text.

Solution: Let \(X_t\) be the number of texts he recieves before \(t\) hours. So \(X_t \sim P(t\lambda)\) \begin{align*} &&\mathbb{P}(X_1 = 0 \, \cap \, X_2 > 0) &= e^{-\lambda} \cdot \left ( 1-e^{-\lambda}\right) = p \\ \Rightarrow && e^{2\lambda}p &= e^{\lambda} - 1 \\ \Rightarrow && 0 &= pe^{2\lambda}-e^{\lambda} + 1 \\ \Rightarrow && e^{\lambda} &= \frac{1 \pm \sqrt{1-4p}}{2p} \end{align*} Which clearly has two positive roots if \(4p < 1\). We need to show both roots are \(>1\). So considering the smaller one we are looking at: \begin{align*} && \frac{1-\sqrt{1-4p}}{2p} & > 1 \\ \Leftrightarrow && 1-\sqrt{1-4p} &> 2p \\ \Leftrightarrow && 1-2p&> \sqrt{1-4p} \\ \Leftrightarrow && (1-2p)^2&> 1-4p \\ \Leftrightarrow && 1-4p+4p^2&> 1-4p \\ \end{align*} which is clearly true. We must have \(e^{\lambda_1}\cdot e^{\lambda_2} = \frac{1}{p}\), so \(\lambda_1 + \lambda_2 = -\ln p\) by considering the product of the roots in our quadratic. (Vieta). Therefore the probability she waits between 1 and 2 hours in the morning is \(e^{-(\lambda_1 + \lambda_2)} \cdot ( 1- e^{-(\lambda_1+\lambda_2)}) = p \cdot (1-p)\)

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