The function $f$ is defined by $$f: x ightarrow e^{x} + k^{2}, \, x \in \mathbb{R}, \, k \text{ is a positive constant.}$$ (a) State the range of $f$ - Edexcel - A-Level Maths Pure - Question 6 - 2014 - Paper 6

$f(x) = e^{x} + k^{2}$ implies that we first need to solve for $x$:

1. Rearranging the equation: $y = e^{x} + k^{2} \Rightarrow e^{x} = y - k^{2}$
2. Taking natural logarithm: $x = \ln(y - k^{2})$
   • The function is defined for $y > k^{2}$.
     Thus, the inverse function is $f^{-1}(y) = \ln(y - k^{2}), \, y > k^{2}$.

Given that $g(x) = \ln(2x)$, we can substitute:

1. Substitute into the equation: $\ln(2y) + \ln(2y^{2}) + \ln(2x) = 6$
2. Combine the logarithms: $\ln(2y) + \ln(4y^{2}) + \ln(2x) = 6$ $\ln(8xy^{2}) = 6$
3. Exponentiating both sides gives: $8xy^{2} = e^{6}$
   or
   $y^{2} = \frac{e^{6}}{8x}$.
4. Thus, $y = \sqrt{\frac{e^{6}}{8x}} = \frac{e^{3}}{2\sqrt{2x}}$.

To find $fg(y)$, we first need to compute:

1. Calculate $g(y)$: $g(y) = \ln(2y)$
2. Then find: $f(g(y)) = f(\ln(2y)) = e^{\ln(2y)} + k^{2} = 2y + k^{2}.$ Thus, the answer is: $fg(y) = 2y + k^{2}.$

From part (d), we have: $fg(x) = 2x + k^{2}.$

1. Now solve: $2x + k^{2} = 2k^{2}$
   $2x = 2k^{2} - k^{2}$
   $2x = k^{2}$
   $x = \frac{k^{2}}{2}.$
   Thus, the solution is: $x = \frac{k^{2}}{2}.$