Separable variables

5．In this question u and v are functions of \(x\) ．Given that \(\int \mathrm { u } \mathrm { d } x , \int \mathrm { v } \mathrm { d } x\) and \(\int \mathrm { uv } \mathrm { d } x\) satisfy $$\int \text { uv } \mathrm { d } x = \left( \int \mathrm { u } \mathrm {~d} x \right) \times \left( \int \mathrm { v } \mathrm {~d} x \right) \quad \text { uv } \neq 0$$ （a）show that \(1 = \frac { \int \mathrm { u } \mathrm { d } x } { \mathrm { u } } + \frac { \int \mathrm { v } \mathrm { d } x } { \mathrm { v } }\) Given also that \(\frac { \int \mathrm { u } \mathrm { d } x } { \mathrm { u } } = \mathrm { sin } ^ { 2 } x\),
（b）use part（a）to write down an expression，in terms of \(x\) ，for \(\frac { \int \mathrm { v } \mathrm { d } x } { \mathrm { v } }\) ，
（c）show that $$\frac { 1 } { \mathrm { u } } \frac { \mathrm { du } } { \mathrm {~d} x } = \frac { 1 - 2 \sin x \cos x } { \sin ^ { 2 } x }$$ （d）hence use integration to show that \(\mathrm { u } = A \mathrm { e } ^ { - \cot x } \operatorname { cosec } ^ { 2 } x\) ，where \(A\) is an arbitrary constant．
（e）By differentiating \(\mathrm { e } ^ { \tan x }\) find a similar expression for v ．

9 The continuous random variable \(X\) has probability density function given by $$f ( x ) = \begin{cases} \frac { 1 } { 20 } \left( 3 - \frac { 1 } { \sqrt { } x } \right) & 1 \leqslant x \leqslant 9 \\ 0 & \text { otherwise } \end{cases}$$ The random variable \(Y\) is defined by \(Y = \sqrt { } X\).

1. Show that the probability density function of \(Y\) is given by $$\operatorname { g } ( y ) = \begin{cases} \frac { 1 } { 10 } ( 3 y - 1 ) & 1 \leqslant y \leqslant 3 \\ 0 & \text { otherwise } \end{cases}$$
2. Find the mean value of \(Y\).

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4 \end{array} \right) + \mu \left( \begin{array} { r } 1
- 2
1 \end{array} \right) \end{aligned}$$ Find, in exact form, the distance between \(l _ { 1 }\) and \(l _ { 2 }\).