5. (a) Show that $g(x) = \frac{x + 1}{x - 2}, \; x > 3$ (4) (b) Find the range of $g$ - Edexcel - A-Level Maths Pure - Question 6 - 2014 - Paper 5

To find the range of ( g(x) = \frac{x + 1}{x - 2} ):

1. Set ( y = g(x) ), leading to ( y = \frac{x + 1}{x - 2} ).

2. Rearranging gives: [ y(x - 2) = x + 1 ] [ yx - 2y = x + 1 ] [ x(y - 1) = 2y + 1 ]

   Therefore, [ x = \frac{2y + 1}{y - 1} ]

3. Restrictions: since ( x > 3 ), we should find the constraints on ( y ):

   • Solve for you and understand that as ( x \to \infty ), ( y \to 1 ) and as ( x \to 3 ), ( y \to \frac{4}{1} = 4 ).

4. Thus, the range is ( 1 < y < 4 ) or in interval notation: ( (1, 4) ).

To find ( \alpha ) such that ( g(\alpha) = g^{-1}(\alpha) ):

1. Start by determining the inverse of ( g(x) ).
2. From earlier work, we know that: [ g^{-1}(y) = \frac{2y + 1}{y - 1} ]
3. Setting ( g(\alpha) = g^{-1}(\alpha) ): [ \frac{\alpha + 1}{\alpha - 2} = \frac{2\alpha + 1}{\alpha - 1} ]
4. Cross-multiplying leads to: [ (\alpha + 1)(\alpha - 1) = (2\alpha + 1)(\alpha - 2) ]
5. Expanding both sides:
   • Left: ( \alpha^2 - 1 )
   • Right: ( 2\alpha^2 - 4\alpha + \alpha - 2 = 2\alpha^2 - 3\alpha - 2 )
6. Setting up the equation: [ \alpha^2 - 1 = 2\alpha^2 - 3\alpha - 2 ]
7. Simplifying: [ 0 = \alpha^2 - 3\alpha + 1 ]
8. Solving with the quadratic formula: [ \alpha = \frac{3 \pm \sqrt{(3)^2 - 4(1)(1)}}{2(1)} = \frac{3 \pm \sqrt{5}}{2} ]
9. The exact values of ( \alpha ) are therefore: ( \alpha = \frac{3 + \sqrt{5}}{2} ) and ( \alpha = \frac{3 - \sqrt{5}}{2} ).