OCR MEI C3 (Core Mathematics 3)

1 Given that \(\mathrm { f } ( x ) = \frac { x + 1 } { x - 1 }\), show that \(\mathrm { ff } ( x ) = x\).
Hence write down the inverse function \(\mathrm { f } ^ { - 1 } ( x )\). What can you deduce about the symmetry of the curve \(y = \mathrm { f } ( x ) ?\)

2 The functions \(\mathrm { f } ( x )\) and \(\mathrm { g } ( x )\) are defined for all real numbers \(x\) by $$\mathrm { f } ( x ) = x ^ { 2 } , \quad \mathrm {~g} ( x ) = x - 2 .$$

1. Find the composite functions \(\mathrm { fg } ( x )\) and \(\mathrm { gf } ( x )\).
2. Sketch the curves \(y = \mathrm { f } ( x ) , y = \mathrm { fg } ( x )\) and \(y = \mathrm { gf } ( x )\), indicating clearly which is which.

3 Given that \(\mathrm { f } ( x ) = 1 - x\) and \(\mathrm { g } ( x ) = | x |\), write down the composite function \(\mathrm { gf } ( x )\). On separate diagrams, sketch the graphs of \(y = \mathrm { f } ( x )\) and \(y = \mathrm { gf } ( x )\).

4 The function \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = 1 + 2 \sin x\) for \(- \frac { 1 } { 2 } \pi \leqslant x \leqslant \frac { 1 } { 2 } \pi\).

1. Show that \(\mathrm { f } ^ { - 1 } ( x ) = \arcsin \left( \frac { x \mathrm { r } } { 2 } \right)\) and state the domain of this function. Fig. 6 shows a sketch of the graphs of \(y = \mathrm { f } ( x )\) and \(y = \mathrm { f } ^ { - 1 } ( x )\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{d4aa92fb-d21b-4387-b711-b0a6b0d57baa-2_499_562_779_785} \captionsetup{labelformat=empty} \caption{Fig. 6}
   \end{figure}
2. Write down the coordinates of the points \(\mathrm { A } , \mathrm { B }\) and C .

5 Given that \(\arcsin x = \frac { 1 } { 6 } \pi\), find \(x\). Find \(\arccos x\) in terms of \(\pi\).

6 The functions \(\mathrm { f } ( x )\) and \(\mathrm { g } ( x )\) are defined for the domain \(x > 0\) as follows: $$\mathrm { f } ( x ) = \ln x , \quad \mathrm {~g} ( x ) = x ^ { 3 } .$$ Express the composite function \(\mathrm { fg } ( x )\) in terms of \(\ln x\).
State the transformation which maps the curve \(y = \mathrm { f } ( x )\) onto the curve \(y = \mathrm { fg } ( x )\).

7 Fig. 9 shows the line \(y = x\) and the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { 1 } { 2 } \left( \mathrm { e } ^ { x } - 1 \right)\). The line and the curve intersect at the origin and at the point \(\mathrm { P } ( a , a )\). \begin{figure}[h] \end{figure}

1. Show that \(\mathrm { e } ^ { a } = 1 + 2 a\).
2. Show that the area of the region enclosed by the curve, the \(x\)-axis and the line \(x = a\) is \(\frac { 1 } { 2 } a\). Hence find, in terms of \(a\), the area enclosed by the curve and the line \(y = x\).
3. Show that the inverse function of \(\mathrm { f } ( x )\) is \(\mathrm { g } ( x )\), where \(\mathrm { g } ( x ) = \ln ( 1 + 2 x )\). Add a sketch of \(y = \mathrm { g } ( x )\) to the copy of Fig. 9.
4. Find the derivatives of \(\mathrm { f } ( x )\) and \(\mathrm { g } ( x )\). Hence verify that \(\mathrm { g } ^ { \prime } ( a ) = \frac { 1 } { \mathrm { f } ^ { \prime } ( a ) }\). Give a geometrical interpretation of this result.

8 The function \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = 1 - 2 \sin x\) for \(- \frac { 1 } { 2 } \pi \leqslant x \leqslant \frac { 1 } { 2 } \pi\). Fig. 3 shows the curve \(y = \mathrm { f } ( x )\). \begin{figure}[h] \end{figure}

1. Write down the range of the function \(\mathrm { f } ( x )\).
2. Find the inverse function \(\mathrm { f } ^ { - 1 } ( x )\).
3. Find \(\mathrm { f } ^ { \prime } ( 0 )\). Hence write down the gradient of \(y = \mathrm { f } ^ { - 1 } ( x )\) at the point \(( 1,0 )\).