Problems
Filters
32 problems found
1989 Paper 2 Q4
The function \(\mathrm{f}\) is defined by \[ \mathrm{f}(x)=\frac{\left(x-a\right)\left(x-b\right)}{\left(x-c\right)\left(x-d\right)}\qquad\left(x\neq c,\ x\neq d\right), \] where \(a,b,c\) and \(d\) are real and distinct, and \(a+b\neq c+d\). Show that \[ \frac{x\mathrm{f}'(x)}{\mathrm{f}(x)}=\left(1-\frac{a}{x}\right)^{-1}+\left(1-\frac{b}{x}\right)^{-1}-\left(1-\frac{c}{x}\right)^{-1}-\left(1-\frac{d}{x}\right)^{-1}, \] \((x\neq0,x\neq a,x\neq b)\) and deduce that when \(\left|x\right|\) is much larger than each of \(\left|a\right|,\left|b\right|,\left|c\right|\) and \(\left|d\right|,\) the gradient of \(\mathrm{f}(x)\) has the same sign as \((a+b-c-d).\) It is given that there is a real value of real value of \(x\) for which \(\mathrm{f}(x)\) takes the real value \(z\) if and only if \[ [\left(c-d\right)^{2}z+\left(a-c\right)\left(b-d\right)+\left(a-d\right)\left(b-c\right)]^{2}\geqslant4\left(a-c\right)\left(b-d\right)\left(a-d\right)\left(b-c\right). \] Describe briefly a method by which this result could be proved, but do not attempt to prove it. Given that \(a < b\) and \(a < c < d\), make sketches of the graph of \(\mathrm{f}\) in the four distinct cases which arise, indicating the cases for which the range of \(\mathrm{f}\) is not the whole of \(\mathbb{R}.\)
Solution: Notice that \(\ln f(x) = \ln (x - a) + \ln (x-b) - \ln (x-c) - \ln (x-d)\) therefore: \begin{align*} \frac{\d}{\d x}: && \frac{f'(x)}{f(x)} &= (x-a)^{-1}+(x-b)^{-1}-(x-c)^{-1} - (x-d)^{-1} \\ &&&= \frac{1}{x} \left ( (1-\frac{a}{x})^{-1}+(1-\frac{b}{x})^{-1}-(1-\frac{c}{x})^{-1} - (1-\frac{d}{x})^{-1}\right) \end{align*} Multiplying by \(x\) gives the desired result. When \(|x|\) is very large then: \begin{align*} \frac{x f'(x)}{f(x)} &\approx 1 + \frac{a}{x} + o(\frac{1}{x^2})+ 1 + \frac{b}{x} + o(\frac{1}{x^2})-(1 + \frac{c}{x} + o(\frac{1}{x^2}))-(1 + \frac{d}{x} + o(\frac{1}{x^2})) \\ &= \frac{a+b-c-d}{x} + o(x^{-2}) \end{align*} Dividing by \(x\) we obtain \(\frac{f'(x)}{f(x)} \approx \frac{a+b-c-d}{x^2} + o(x^{-3})\) if \(|x|\) is sufficiently large this will be dominated by the \(\frac{a+b-c-d}{x^2}\) term which will have the same sign as \((a+b-c-d)\). When \(|x|\) is very large all of the brackets will have the same sign, and therefore \(f(x)\) will be positive, and so \(f'(x)\) must have the same sign as \(a+b-c-d\). To prove this result, we could set \(f(x) = k\) and rearrange to form a quadratic in \(x\). We could then check the discriminant is non-zero. Case 1: \(a < c < d < b\) and \(a+b > c+d \Rightarrow\) not all values reached and approx asymtope from below on the right and above on the left.
1988 Paper 2 Q16
Find the probability that the quadratic equation \[ X^{2}+2BX+1=0 \] has real roots when \(B\) is normally distributed with zero mean and unit variance. Given that the two roots \(X_{1}\) and \(X_{2}\) are real, find:
- the probability that both \(X_{1}\) and \(X_{2}\) are greater than \(\frac{1}{5}\);
- the expected value of \(\left|X_{1}+X_{2}\right|\);
Solution: The roots are \(X_1, X_2 = -B \pm \sqrt{B^2-1}\)
- The smallest root will be \(-B - \sqrt{B^2-1}\). For this to be larger than \(\frac15\) we must have, \begin{align*} && -B -\sqrt{B^2-1} &\geq \frac15 \\ \Rightarrow && -B - \frac15 &\geq \sqrt{B^2 - 1} \\ \Rightarrow && B^2 + \frac25 B + \frac1{25} &\geq B^2 - 1 \\ \Rightarrow && \frac25 B \geq -\frac{26}{25} \\ \Rightarrow && B \geq -\frac{13}{5} \end{align*} Therefore \(-\frac{13}5 \leq B \leq -1\). Therefore we want: \begin{align*} \frac{\P(-\frac{13}5 \leq B \leq -1)}{\P(B < -1) + \P(B > 1)} &= \frac{\Phi(-1) - \Phi(-\frac{13}{5})}{\Phi(-1)+1-\Phi(1)} \\ &= \frac{0.1586\ldots - 0.0046\ldots}{0.1586\ldots + 1- 0.8413\ldots} \\ &= 0.4853\ldots \\ &= 0.485 \,\,(3 \text{ s.f.}) \end{align*}
- \(X_1 + X_2 = -2B\). Therefore we want: \begin{align*} \mathbb{E}(|X_1 + X_2| &= \mathbb{E}(|2B|) \\ &= 2 \l\frac{1}{2\Phi(-1)} \int_1^{\infty} B \frac{1}{\sqrt{2 \pi}} e^{-\frac12 B^2} \, \d B+\frac{1}{2\Phi(-1)} \int_{-\infty}^{-1} B \frac{1}{\sqrt{2 \pi}} e^{-\frac12 B^2} \, \d B \r \\ &= \frac{4}{2\Phi(-1)} \int_1^{\infty} B \frac{1}{\sqrt{2 \pi}} e^{-\frac12 B^2} \, \d B \\ &=\frac{4}{2\sqrt{2 \pi} \Phi(-1)} \left [ -e^{-\frac12 B^2}\right]_1^{\infty} \\ &= \frac{4}{2\sqrt{2 \pi} \Phi(-1) \sqrt{e}} \\ &= 3.0502\ldots \\ &= 3.05\,\, (3\text{ s.f.}) \end{align*}