OCR MEI C3 (Core Mathematics 3) 2005 June

1 Solve the equation \(| 3 x + 2 | = 1\).

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

3 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 \(y = \mathrm { fg } ( x )\).

4 The temperature \(T ^ { \circ } \mathrm { C }\) of a liquid at time \(t\) minutes is given by the equation $$T = 30 + 20 \mathrm { e } ^ { - 0.05 t } , \quad \text { for } t \geqslant 0 .$$ Write down the initial temperature of the liquid, and find the initial rate of change of temperature.
Find the time at which the temperature is \(40 ^ { \circ } \mathrm { C }\).

5 Using the substitution \(u = 2 x + 1\), show that \(\int _ { 0 } ^ { 1 } \frac { x } { 2 x + 1 } \mathrm {~d} x = \frac { 1 } { 4 } ( 2 - \ln 3 )\).

6 A curve has equation \(y = \frac { x } { 2 + 3 \ln x }\). Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\). Hence find the exact coordinates of the stationary point of the curve.

7 Fig. 7 shows the curve defined implicitly by the equation $$y ^ { 2 } + y = x ^ { 3 } + 2 x ,$$ together with the line \(x = 2\). \begin{figure}[h] \end{figure} Fig. 7 Find the coordinates of the points of intersection of the line and the curve.
Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\). Hence find the gradient of the curve at each of these two points.

8 Fig. 8 shows part of the curve \(y = x \sin 3 x\). It crosses the \(x\)-axis at P . The point on the curve with \(x\)-coordinate \(\frac { 1 } { 6 } \pi\) is Q . \begin{figure}[h] \end{figure}

1. Find the \(x\)-coordinate of P .
2. Show that Q lies on the line \(y = x\).
3. Differentiate \(x \sin 3 x\). Hence prove that the line \(y = x\) touches the curve at Q .
4. Show that the area of the region bounded by the curve and the line \(y = x\) is \(\frac { 1 } { 72 } \left( \pi ^ { 2 } - 8 \right)\).

9 The function \(\mathrm { f } ( x ) = \ln \left( 1 + x ^ { 2 } \right)\) has domain \(- 3 \leqslant x \leqslant 3\).
Fig. 9 shows the graph of \(y = \mathrm { f } ( x )\). \begin{figure}[h] \end{figure}

1. Show algebraically that the function is even. State how this property relates to the shape of the curve.
2. Find the gradient of the curve at the point \(\mathrm { P } ( 2 , \ln 5 )\).
3. Explain why the function does not have an inverse for the domain \(- 3 \leqslant x \leqslant 3\). The domain of \(\mathrm { f } ( x )\) is now restricted to \(0 \leqslant x \leqslant 3\). The inverse of \(\mathrm { f } ( x )\) is the function \(\mathrm { g } ( x )\).
4. Sketch the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\) on the same axes. State the domain of the function \(\mathrm { g } ( x )\). Show that \(\mathrm { g } ( x ) = \sqrt { \mathrm { e } ^ { x } - 1 }\).
5. Differentiate \(\mathrm { g } ( x )\). Hence verify that \(\mathrm { g } ^ { \prime } ( \ln 5 ) = 1 \frac { 1 } { 4 }\). Explain the connection between this result and your answer to part (ii).