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

To find the inverse function $f^{-1}(y)$, we start by letting $y = e^{x} + k^{2}$.

Rearranging the equation gives: $e^{x} = y - k^{2}$

Taking the natural logarithm of both sides: $x = \ln(y - k^{2})$

Thus, the inverse function is: $f^{-1}(y) = \ln(y - k^{2})$

The domain of $f^{-1}$ is determined where $y - k^{2} > 0$, i.e., $y > k^{2}$.

Substituting into the equation:

$g(y) = \ln(2y), \, g(y^{*}) = \ln(2y^{*}), \, g(x) = \ln(2x)$

Thus, we have: $\ln(2y) + \ln(2y^{*}) + \ln(2x) = 6$

Using the properties of logarithms, this collapses to: $\ln(2y imes 2y^{*} imes 2x) = 6$

This simplifies to: $2y imes 2y^{*} imes 2x = e^{6}$

This means: $8xyz = e^{6}$
Using this, we can express the relationship between y, y*, and x.

To find $fg(x)$, we compute:

$fg(x) = f(g(x))$

Substituting:

$g(x) = \ln(2x)$

Hence, we find: $f(g(x)) = f(\ln(2x)) = e^{\ln(2x)} + k^{2}$

The property of exponential functions gives: $= 2x + k^{2}$

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

We set this equal to $2k^{2}$: $2x + k^{2} = 2k^{2}$

Rearranging the equation leads us to: $2x = 2k^{2} - k^{2}$ $2x = k^{2}$

Thus, we find: $x = \frac{k^{2}}{2}$