It is given that $\mathrm{f}(x) = Ax(e^{kx})$, where $A$ and $k$ are constants.
Find the exact values of $A$ and $k$ such that $\mathrm{f}'(x) + 2ke^{kx} + 6\mathrm{f}(x) = 0$.

1. Find the derivative of $\mathrm{f}(x)$ using the product rule.

$\mathrm{f}(x) = Ax e^{kx}$
$\mathrm{f}'(x) = A(e^{kx}) + Ax(ke^{kx}) = Ae^{kx} + Akxe^{kx}$

2. Substitute $\mathrm{f}(x)$ and $\mathrm{f}'(x)$ into the given equation.

$\mathrm{f}'(x) + 2ke^{kx} + 6\mathrm{f}(x) = 0$
$(Ae^{kx} + Akxe^{kx}) + 2ke^{kx} + 6(Axe^{kx}) = 0$

3. Factorise and group terms by powers of $x$.

$e^{kx}(A + Akx + 2k + 6Ax) = 0$
Since $e^{kx} \neq 0$ for all $x$, we have:
$(A + 2k) + x(Ak + 6A) = 0$

4. Compare coefficients to form simultaneous equations.

For the equation to hold for all $x$, the coefficients must be zero:
$A + 2k = 0 \implies A = -2k$ --- (1)
$Ak + 6A = 0 \implies A(k + 6) = 0$ --- (2)

5. Solve the simultaneous equations for $A$ and $k$.

From (2), since $A \neq 0$ (for a non-trivial function), we have:
$k + 6 = 0 \implies k = -6$
Substitute $k = -6$ into (1):
$A = -2(-6) = 12$