1. **Problem statement:** Given the curve $$y = \frac{(2x^2 + 10)^{3/2}}{x - 1}$$ for $$x > 1$$, (a) Show that $$\frac{dy}{dx}$$ can be written as $$\frac{(2x^2 + 10)^{1/2}}{(x - 1)^2}(Ax^2 + Bx + C)$$ where $$A, B, C$$ are integers. (b) Show that for $$x > 1$$, the curve has exactly one stationary point and find its $$x$$-value. 2. **Step 1: Differentiate using the quotient rule.** Recall the quotient rule: $$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}$$. Let $$u = (2x^2 + 10)^{3/2}$$ and $$v = x - 1$$. 3. **Step 2: Compute $$\frac{du}{dx}$$.** Use the chain rule: $$\frac{du}{dx} = \frac{3}{2}(2x^2 + 10)^{1/2} \cdot 4x = 6x(2x^2 + 10)^{1/2}$$. 4. **Step 3: Compute $$\frac{dv}{dx}$$.** $$\frac{dv}{dx} = 1$$. 5. **Step 4: Apply quotient rule.** $$\frac{dy}{dx} = \frac{(x - 1) \cdot 6x (2x^2 + 10)^{1/2} - (2x^2 + 10)^{3/2} \cdot 1}{(x - 1)^2}$$. 6. **Step 5: Factor out $$ (2x^2 + 10)^{1/2} $$ in numerator.** $$\frac{dy}{dx} = \frac{(2x^2 + 10)^{1/2} \left[6x(x - 1) - (2x^2 + 10) \right]}{(x - 1)^2}$$. 7. **Step 6: Simplify the bracket.** Calculate: $$6x(x - 1) - (2x^2 + 10) = 6x^2 - 6x - 2x^2 - 10 = 4x^2 - 6x - 10$$. 8. **Step 7: Final form.** $$\frac{dy}{dx} = \frac{(2x^2 + 10)^{1/2}}{(x - 1)^2} (4x^2 - 6x - 10)$$. Here, $$A = 4$$, $$B = -6$$, $$C = -10$$, all integers. 9. **Step 8: Find stationary points where $$\frac{dy}{dx} = 0$$.** Since $$x > 1$$, denominator and $$ (2x^2 + 10)^{1/2} $$ are positive, so set numerator polynomial to zero: $$4x^2 - 6x - 10 = 0$$. 10. **Step 9: Solve quadratic equation.** Use quadratic formula: $$x = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 4 \cdot (-10)}}{2 \cdot 4} = \frac{6 \pm \sqrt{36 + 160}}{8} = \frac{6 \pm \sqrt{196}}{8} = \frac{6 \pm 14}{8}$$. 11. **Step 10: Calculate roots.** $$x_1 = \frac{6 + 14}{8} = \frac{20}{8} = 2.5$$ $$x_2 = \frac{6 - 14}{8} = \frac{-8}{8} = -1$$ Only $$x = 2.5$$ is valid since $$x > 1$$. 12. **Step 11: Conclusion.** There is exactly one stationary point for $$x > 1$$ at $$x = 2.5$$. **Final answers:** (a) $$\frac{dy}{dx} = \frac{(2x^2 + 10)^{1/2}}{(x - 1)^2} (4x^2 - 6x - 10)$$ with $$A=4$$, $$B=-6$$, $$C=-10$$. (b) The curve has exactly one stationary point for $$x > 1$$ at $$x = 2.5$$.