2025 DHS P1 Q7

The function \(\mathrm{f}\) is defined by

\[\mathrm{f} : x \mapsto \mathrm{e}^x + \frac{1}{2x + 2} \text{ for}\hspace{0.5em} x \in \mathbb{R}, x \neq -1.\]

A function \(\mathrm{g}\), defined for \(x \in \mathbb{R}, x \geq 1\), is such that \(y \to \infty\) as \(x\) increases. It is also given that \(\mathrm{g}(1) = -0.5\).

1. Explain why the composite function \(\mathrm{fg}\) exists and find the corresponding range of \(\mathrm{fg}\).[2]
2. Given that \(\mathrm{fg}(x) = \frac{x}{\sqrt{\mathrm{e}}} + \frac{1}{2 \ln x + 1}\), find an expression for \(\mathrm{g}(x)\) in terms of \(x\).[2]
3. The domain of \(\mathrm{f}\) is now further restricted to \(x > k\). State the least value of integer \(k\) for which the function \(\mathrm{f}^{-1}\) exists.[1]

For the rest of the question, use the value of \(k\) found in part (c).

4. Without finding \(\mathrm{f}^{-1}\),
   1. sketch, on the same diagram, the graphs of \(\mathrm{f}\), \(\mathrm{f}^{-1}\) and \(\mathrm{f}\mathrm{f}^{-1}\) showing clearly the relationships between the graphs,[2]
   2. find the gradient of the tangent to the graph of \(y = \mathrm{f}^{-1}(x)\) at \(x = \mathrm{e} + \frac{1}{4}\).[3]