2018 HCI Promo Q11

The curve \({{C}_{1}}\) has equation \(y=\frac{2\left( {{x}^{2}}-8 \right)}{x-3}\).

1. Find algebraically the set of values that \(y\) can take.[3]
2. Sketch \({{C}_{1}}\), stating the equations of any asymptotes, the coordinates of any turning points and the points where \({{C}_{1}}\) crosses the axes.[3]
3. Find the range of values of \(k\) such that the equation
   \(2{{x}^{2}}-16=k{{\left( x-3 \right)}^{2}}+12\left( x-3 \right)\), where \(x\ne 3\),
   has no real roots.[2]
4. Solve the inequality \(\frac{2\left( {{x}^{2}}-8 \right)}{x-3}<\ln x\). Hence solve \(\frac{2\left( {{x}^{2}}-8 \right)}{\left| x \right|-3}<\ln \left| x \right|\).[3]
5. Describe a sequence of transformations that will map \({{C}_{1}}\) onto the curve that has equation \(y=2x+\frac{2}{x}\).[2]