1. (i) Step 1: Use the Factor Theorem. Given that $(x-1)$ is a factor of $f(x) = ax^4 + x^3 - 12x^2 - x + 2$, by the Factor Theorem, $f(1) = 0$. Substitute $x=1$ into $f(x)$: $$f(1) = a(1)^4 + (1)^3 - 12(1)^2 - (1) + 2 = 0$$ $$a + 1 - 12 - 1 + 2 = 0$$ $$a - 10 = 0$$ $$a = 10$$ The value of the constant $a$ is $\boxed{10}$. Step 2: Verify that $f(-1) = 0$. Now, substitute $a=10$ into $f(x)$, so $f(x) = 10x^4 + x^3 - 12x^2 - x + 2$. Substitute $x=-1$ into $f(x)$: $$f(-1) = 10(-1)^4 + (-1)^3 - 12(-1)^2 - (-1) + 2$$ $$f(-1) = 10(1) + (-1) - 12(1) + 1 + 2$$ $$f(-1) = 10 - 1 - 12 + 1 + 2$$ $$f(-1) = 9 - 12 + 1 + 2$$ $$f(-1) = -3 + 1 + 2$$ $$f(-1) = 0$$ This verifies that $f(-1) = 0$. 1. (ii) Step 1: Define the roots. Let the roots of the equation $x^2 + (k+1)x + k = 0$ be $\alpha$ and $2\alpha$. Step 2: Apply Vieta's formulas. For a quadratic equation $Ax^2 + Bx + C = 0$, the sum of roots is $-\frac{B}{A}$ and the product of roots is $\frac{C}{A}$. In this equation, $A=1$, $B=(k+1)$, $C=k$. Sum of roots: $$\alpha + 2\alpha = -(k+1)$$ $$3\alpha = -(k+1) \quad (1)$$ Product of roots: $$\alpha \cdot 2\alpha = k$$ $$2\alpha^2 = k \quad (2)$$ Step 3: Solve the system of equations for $k$. From equation (1), express $\alpha$ in terms of $k$: $$\alpha = -\frac{k+1}{3}$$ Substitute this expression for $\alpha$ into equation (2): $$2\left(-\frac{k+1}{3}\right)^2 = k$$ $$2\frac{(k+1)^2}{9} = k$$ $$2(k^2 + 2k + 1) = 9k$$ $$2k^2 + 4k + 2 = 9k$$ $$2k^2 + 4k - 9k + 2 = 0$$ $$2k^2 - 5k + 2 = 0$$ Step 4: Solve the quadratic equation for $k$. Factor the quadratic equation: $$2k^2 - 4k - k + 2 = 0$$ $$2k(k - 2) - 1(k - 2) = 0$$ $$(2k - 1)(k - 2) = 0$$ This gives two possible values for $k$: $$2k - 1 = 0 \implies k = \frac{1}{2}$$ $$k - 2 = 0 \implies k = 2$$ The values of the constant $k$ are $\boxed{\frac{1}{2} \text{ and } 2}$.