4 Fig. 9 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 1 } { \sqrt { 2 x - x ^ { 2 } } }\).
The curve has asymptotes \(x = 0\) and \(x = a\).
\begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{2437cecc-f084-4e49-ab36-1c132ba13267-2_652_795_876_717}
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\caption{Fig. 9}
\end{figure}
- Find \(a\). Hence write down the domain of the function.
- Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x - 1 } { \left( 2 x - x ^ { 2 } \right) ^ { \frac { 3 } { 2 } } }\).
Hence find the coordinates of the turning point of the curve, and write down the range of the function.
The function \(\mathrm { g } ( x )\) is defined by \(\mathrm { g } ( x ) = \frac { 1 } { \sqrt { 1 - x ^ { 2 } } }\).
- (A) Show algebraically that \(\mathrm { g } ( x )\) is an even function.
(B) Show that \(\mathrm { g } ( x - 1 ) = \mathrm { f } ( x )\).
(C) Hence prove that the curve \(y = \mathrm { f } ( x )\) is symmetrical, and state its line of symmetry.