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 7 - 2014 - Paper 6

To find the inverse function, we start with the equation: $y = e^x + k^2$ Rearranging gives: $e^x = y - k^2$ Taking the natural logarithm leads to: $x = \ln(y - k^2)$ Thus, the inverse function is: $f^{-1}(y) = \ln(y - k^2)$ The domain is determined by requiring the argument of the logarithm to be positive: $y - k^2 > 0 \\ \Rightarrow y > k^2$ Therefore, the domain of $f^{-1}$ is: $\text{Domain of } f^{-1} = (k^2, \, +\infty)$

Given the function $g(x) = \ln(2x)$, we can rewrite the equation: $\ln(2y) + \ln(2y^2) + \ln(2x) = 6$ This simplifies to: $\ln(2y) + \ln(4y) + \ln(2x) = 6$ Combining the logarithms yields: $\ln(8y^2x) = 6$ Exponentiating both sides gives: $8y^2x = e^6$ Therefore, solving for $y$ gives: $y^2 = \frac{e^6}{8x}\ \\ \Rightarrow y = \sqrt{\frac{e^6}{8x}} = \frac{e^3}{2\sqrt{2x}}$

To find $fg(y)$, we first compute: $g(y) = \ln(2y)$ Then substituting into $f$ results in: $fg(y) = f(g(y)) = f(\ln(2y))$ Thus: $fg(y) = e^{\ln(2y)} + k^2 = 2y + k^2$

Setting up the equation: $fg(x) = 2x + k^2 = 2k^2$ We need to isolate $x$: $2x + k^2 = 2k^2$ Rearranging gives: $2x = 2k^2 - k^2 \Rightarrow 2x = k^2$ Thus, solving for $x$ results in: $x = \frac{k^2}{2}$