1. **Problem statement:** (i)(a) Express the quadratic $2x^2 - 6x + 13$ in the form $A(x+B)^2 + C$ where $A$, $B$, and $C$ are constants. (i)(b) Using the form found, find the maximum value of $f(x) = \frac{1}{2x^2 - 6x + 13}$ as an exact fraction. (i)(c) Find the value of $x$ at which this maximum occurs. (ii)(a) Given roots $\alpha$ and $\beta$ of $2x^2 - 5x + p = 0$, find a quadratic equation with roots $1 + \frac{1}{\alpha}$ and $1 + \frac{1}{\beta}$. (ii)(b) Given the new quadratic has equal roots, find $p$. --- 2. **Step (i)(a): Complete the square for $2x^2 - 6x + 13$** Start with the quadratic: $$2x^2 - 6x + 13$$ Factor out 2 from the first two terms: $$2(x^2 - 3x) + 13$$ Complete the square inside the parentheses: Take half of $-3$ which is $-\frac{3}{2}$, square it: $\left(-\frac{3}{2}\right)^2 = \frac{9}{4}$. Add and subtract $\frac{9}{4}$ inside the parentheses: $$2\left(x^2 - 3x + \frac{9}{4} - \frac{9}{4}\right) + 13 = 2\left(\left(x - \frac{3}{2}\right)^2 - \frac{9}{4}\right) + 13$$ Distribute 2: $$2\left(x - \frac{3}{2}\right)^2 - 2 \times \frac{9}{4} + 13 = 2\left(x - \frac{3}{2}\right)^2 - \frac{9}{2} + 13$$ Simplify constants: $$-\frac{9}{2} + 13 = -\frac{9}{2} + \frac{26}{2} = \frac{17}{2}$$ So the expression is: $$2\left(x - \frac{3}{2}\right)^2 + \frac{17}{2}$$ Therefore, $$A = 2, \quad B = -\frac{3}{2}, \quad C = \frac{17}{2}$$ --- 3. **Step (i)(b): Find the maximum value of $f(x) = \frac{1}{2x^2 - 6x + 13}$** Since $2x^2 - 6x + 13 = 2\left(x - \frac{3}{2}\right)^2 + \frac{17}{2}$, the minimum value of the denominator occurs when the squared term is zero, i.e., at $x = \frac{3}{2}$. Minimum denominator value: $$2 \times 0 + \frac{17}{2} = \frac{17}{2}$$ Maximum value of $f(x)$ is the reciprocal of this minimum denominator: $$\max f(x) = \frac{1}{\frac{17}{2}} = \frac{2}{17}$$ --- 4. **Step (i)(c): Value of $x$ at maximum $f(x)$** From above, maximum occurs at: $$x = \frac{3}{2}$$ --- 5. **Step (ii)(a): Find quadratic with roots $1 + \frac{1}{\alpha}$ and $1 + \frac{1}{\beta}$** Given $\alpha, \beta$ roots of: $$2x^2 - 5x + p = 0$$ Sum and product of roots: $$\alpha + \beta = \frac{5}{2}, \quad \alpha \beta = \frac{p}{2}$$ New roots: $$r_1 = 1 + \frac{1}{\alpha}, \quad r_2 = 1 + \frac{1}{\beta}$$ Sum of new roots: $$r_1 + r_2 = \left(1 + \frac{1}{\alpha}\right) + \left(1 + \frac{1}{\beta}\right) = 2 + \frac{1}{\alpha} + \frac{1}{\beta} = 2 + \frac{\alpha + \beta}{\alpha \beta} = 2 + \frac{\frac{5}{2}}{\frac{p}{2}} = 2 + \frac{5}{p}$$ Product of new roots: $$r_1 r_2 = \left(1 + \frac{1}{\alpha}\right)\left(1 + \frac{1}{\beta}\right) = 1 + \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\alpha \beta} = 1 + \frac{\alpha + \beta}{\alpha \beta} + \frac{1}{\alpha \beta} = 1 + \frac{\frac{5}{2}}{\frac{p}{2}} + \frac{1}{\frac{p}{2}} = 1 + \frac{5}{p} + \frac{2}{p} = 1 + \frac{7}{p} = \frac{p + 7}{p}$$ The quadratic with roots $r_1$ and $r_2$ is: $$x^2 - (r_1 + r_2)x + r_1 r_2 = 0$$ Substitute sums and products: $$x^2 - \left(2 + \frac{5}{p}\right)x + \frac{p + 7}{p} = 0$$ Multiply through by $p$ to clear denominators: $$p x^2 - (2p + 5) x + p + 7 = 0$$ --- 6. **Step (ii)(b): Find $p$ such that the quadratic has equal roots** For equal roots, discriminant $D = 0$: $$D = b^2 - 4ac = 0$$ Here, $$a = p, \quad b = -(2p + 5), \quad c = p + 7$$ Calculate discriminant: $$D = (2p + 5)^2 - 4 p (p + 7) = 0$$ Expand: $$(2p + 5)^2 = 4p^2 + 20p + 25$$ So, $$4p^2 + 20p + 25 - 4p^2 - 28p = 0$$ Simplify: $$4p^2 - 4p^2 + 20p - 28p + 25 = 0 \implies -8p + 25 = 0$$ Solve for $p$: $$-8p = -25 \implies p = \frac{25}{8}$$ --- **Final answers:** (i)(a) $A = 2$, $B = -\frac{3}{2}$, $C = \frac{17}{2}$ (i)(b) Maximum value of $f(x) = \frac{2}{17}$ (i)(c) Maximum occurs at $x = \frac{3}{2}$ (ii)(a) Quadratic equation: $$p x^2 - (2p + 5) x + p + 7 = 0$$ (ii)(b) $p = \frac{25}{8}$