1 Fig. 9 shows the curve $y = \mathrm { f } ( x )$, where $\mathrm { f } ( x ) = \frac { 2 x ^ { 2 } - 1 } { x ^ { 2 } + 1 }$ for the domain $0 \leqslant x \leqslant 2$.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{b7588524-8a5e-42af-8b52-29cdddc09eeb-1_976_1208_450_514}
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\caption{Fig. 9}
\end{center}
\end{figure}

(i) Show that $\mathrm { f } ^ { \prime } ( x ) = \frac { 6 x } { \left( x ^ { 2 } + 1 \right) ^ { 2 } }$, and hence that $\mathrm { f } ( x )$ is an increasing function for $x > 0$.\\
(ii) Find the range of $\mathrm { f } ( x )$.\\
(iii) Given that $\mathrm { f } ^ { \prime \prime } ( x ) = \frac { 6 - 18 x ^ { 2 } } { \left( x ^ { 2 } + 1 \right) ^ { 3 } }$, find the maximum value of $\mathrm { f } ^ { \prime } ( x )$.

The function $\mathrm { g } ( x )$ is the inverse function of $\mathrm { f } ( x )$.\\
(iv) Write down the domain and range of $\mathrm { g } ( x )$. Add a sketch of the curve $y = \mathrm { g } ( x )$ to a copy of Fig. 9 .\\
(v) Show that $\mathrm { g } ( x ) = \sqrt { \frac { x + 1 } { 2 - x } }$.

\hfill \mbox{\textit{OCR MEI C3  Q1 [19]}}