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 $f^{-1}(y)$, we start from the equation:

$y = e^x + k^2$

Rearranging gives:

$e^x = y - k^2$

Taking the natural logarithm of both sides, we have:

$x = ext{ln}(y - k^2)$

Thus, the inverse function is:

$f^{-1}(y) = ext{ln}(y - k^2)\, , y > k^2.$

Therefore, the domain of $f^{-1}$ is $(k^2, + ext{∞})$.

We start by expressing $g(x)$:

$g(x) = ext{ln}(2x)$

Then we substitute into the equation:

$g(y) + g(x^2) + g(x) = \ln(2y) + \ln(2x^2) + \ln(2x) = 6$

Using properties of logarithms:

$\ln(2y) + \ln(4x^2) + \ln(2x) = 6\n$

Combining:

$\ln(8xy^2) = 6\n$

Exponentiating both sides:

$8xy^2 = e^6\n$

Solving for $y^2$ gives:

$y^2 = \frac{e^6}{8x}$

Therefore:

$y = \sqrt{\frac{e^6}{8x}} = \frac{e^3}{2\sqrt{x}}$

To find $fg(g)$, we first compute $g(g)$:

From earlier, $g(g) = g(y) = \ln(2y)$. Now we need to substitute this result into $f$:

$fg(g) = f(g(y)) = f(\ln(2y)) = e^{\ln(2y)} + k^2 = 2y + k^2$.

Setting $fg(x) = 2k^2$, we use the previous result:

$2x + k^2 = 2k^2$.

Rearranging gives:

$2x = 2k^2 - k^2\n$

Therefore:

$2x = k^2\n$

Dividing both sides by 2:

$x = \frac{k^2}{2}.$

Thus, the solution is $x = \frac{k^2}{2}$.