OCR MEI C3 (Core Mathematics 3)

1 Fig. 8 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = ( 1 - x ) \mathrm { e } ^ { 2 x }\), with its turning point P . \begin{figure}[h] \end{figure}

1. Write down the coordinates of the intercepts of \(y = \mathrm { f } ( x )\) with the \(x\) - and \(y\)-axes.
2. Find the exact coordinates of the turning point P .
3. Show that the exact area of the region enclosed by the curve and the \(x\) - and \(y\)-axes is \(\frac { 1 } { 4 } \left( \mathrm { e } ^ { 2 } - 3 \right)\). The function \(\mathrm { g } ( x )\) is defined by \(\mathrm { g } ( x ) = 3 \mathrm { f } \left( \frac { 1 } { 2 } x \right)\).
4. Express \(\mathrm { g } ( x )\) in terms of \(x\). Sketch the curve \(y = \mathrm { g } ( x )\) on the copy of Fig. 8, indicating the coordinates of its intercepts with the \(x\) - and \(y\)-axes and of its turning point.
5. Write down the exact area of the region enclosed by the curve \(y = \mathrm { g } ( x )\) and the \(x\) - and \(y\)-axes.

2 Fig. 9 shows the curve with equation \(y ^ { 3 } = \frac { x ^ { 3 } } { 2 x - 1 }\). It has an asymptote \(x = a\) and turning point P . \begin{figure}[h] \end{figure}

1. Write down the value of \(a\).
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 4 x ^ { 3 } - 3 x ^ { 2 } } { 3 y ^ { 2 } ( 2 x - 1 ) ^ { 2 } }\). Hence find the coordinates of the turning point P , giving the \(y\)-coordinate to 3 significant figures.
3. Show that the substitution \(u = 2 x - 1\) transforms \(\int \frac { x } { \sqrt [ 3 ] { 2 x - 1 } } \mathrm {~d} x\) to \(\frac { 1 } { 4 } \int \left( u ^ { \frac { 2 } { 3 } } + u ^ { - \frac { 1 } { 3 } } \right) \mathrm { d } u\). Hence find the exact area of the region enclosed by the curve \(y ^ { 3 } = \frac { x ^ { 3 } } { 2 x - 1 }\), the \(x\)-axis and the lines \(x = 1\) and \(x = 4.5\).

3 Fig. 9 shows the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\). The function \(y = \mathrm { f } ( x )\) is given by $$f ( x ) = \ln \left( \frac { 2 x } { 1 + x } \right) , x > 0$$ The curve \(y = \mathrm { f } ( x )\) crosses the \(x\)-axis at P , and the line \(x = 2\) at Q . \begin{figure}[h] \end{figure}

1. Verify that the \(x\)-coordinate of P is 1 . Find the exact \(y\)-coordinate of Q .
2. Find the gradient of the curve at P. [Hint: use \(\ln \frac { a } { b } = \ln a - \ln b\).] The function \(\mathrm { g } ( x )\) is given by $$\mathrm { g } ( x ) = \frac { \mathrm { e } ^ { x } } { 2 - \mathrm { e } ^ { x } } , \quad x < \ln 2 .$$ The curve \(y = \mathrm { g } ( x )\) crosses the \(y\)-axis at the point R .
3. Show that \(\mathrm { g } ( x )\) is the inverse function of \(\mathrm { f } ( x )\). Write down the gradient of \(y = \mathrm { g } ( x )\) at R .
4. Show, using the substitution \(u = 2 - \mathrm { e } ^ { x }\) or otherwise, that \(\int _ { 0 } ^ { \ln \frac { 4 } { 3 } } \mathrm {~g} ( x ) \mathrm { d } x = \ln \frac { 3 } { 2 }\). Using this result, show that the exact area of the shaded region shown in Fig. 9 is \(\ln \frac { 32 } { 27 }\). [Hint: consider its reflection in \(y = x\).]