5. (a) Show that $g(x) = \frac{x + 1}{x - 2},\, x > 3$ - Edexcel - A-Level Maths Pure - Question 7 - 2014 - Paper 5

To find the range of $g(x)$, we analyze the expression:

1. The function $g(x) = \frac{x + 1}{x - 2}$ is a rational function.
2. The domain is restricted to $x > 3$.
3. Calculate the limits as $x$ approaches 3 and infinity:
   • When $x \to 3$: $g(3) = \frac{3 + 1}{3 - 2} = 4$.
   • As $x \to \infty$: $g(x) \to 1$.
4. Therefore, the range is $1 < g < 4$ or $(1, 4)$.

To find $\alpha$, first find the inverse of $g$:

1. Set $y = g(x)$: $y = \frac{x + 1}{x - 2}$
2. Solve for $x$: $y(x - 2) = x + 1$ $yx - 2y = x + 1$ $yx - x = 2y + 1$ $x(y - 1) = 2y + 1$ $x = \frac{2y + 1}{y - 1}$
3. Therefore, the inverse is: $g^{-1}(y) = \frac{2y + 1}{y - 1}$
4. Set $g(\alpha) = g^{-1}(\alpha)$: $\frac{\alpha + 1}{\alpha - 2} = \frac{2\alpha + 1}{\alpha - 1}$
5. Cross-multiply and solve: $(\alpha + 1)(\alpha - 1) = (2\alpha + 1)(\alpha - 2)$
6. Simplify this equation: $\alpha^2 - 1 = 2\alpha^2 - 4\alpha + \alpha - 2$
7. Rearranging leads to: $\alpha^2 + 3\alpha - 1 = 0$
8. Using the quadratic formula: $\alpha = \frac{-3 \pm \sqrt{9 + 4}}{2} = -\frac{3}{2} \pm \frac{\sqrt{13}}{2}$