The curve $C$ has equation

$$y = \frac { x ^ { 2 } + q x + 1 } { 2 x + 3 } ,$$

where $q$ is a positive constant.\\
(i) Obtain the equations of the asymptotes of $C$.\\
(ii) Find the value of $q$ for which the $x$-axis is a tangent to $C$, and sketch $C$ in this case.\\
(iii) Sketch $C$ for the case $q = 3$, giving the exact coordinates of the points of intersection of $C$ with the $x$-axis.\\
(iv) It is given that, for all values of the constant $\lambda$, the line

$$y = \lambda x + \frac { 3 } { 2 } \lambda + \frac { 1 } { 2 } ( q - 3 )$$

passes through the point of intersection of the asymptotes of $C$. Use this result, with the diagrams you have drawn, to show that if $\lambda < \frac { 1 } { 2 }$ then the equation

$$\frac { x ^ { 2 } + q x + 1 } { 2 x + 3 } = \lambda x + \frac { 3 } { 2 } \lambda + \frac { 1 } { 2 } ( q - 3 )$$

has no real solution if $q$ has the value found in part (ii), but has 2 real distinct solutions if $q = 3$.

\hfill \mbox{\textit{CAIE FP1 2006 Q12 EITHER}}