OCR MEI C3 (Core Mathematics 3)

1 Fig. 1 shows part of the curve \(y = \mathrm { e } ^ { 2 x } \cos x\). \begin{figure}[h] \end{figure} Find the coordinates of the turning point P .

2 Find the exact gradient of the curve \(y = \ln ( 1 - \cos 2 x )\) at the point with \(x\)-coordinate \(\frac { 1 } { 6 } \pi\).

3

1. Given that \(y = \mathrm { e } ^ { - x } \sin 2 x\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. Hence show that the curve \(y = \mathrm { e } ^ { - x } \sin 2 x\) has a stationary point when \(x = \frac { 1 } { 2 } \arctan 2\).

4 Fig. 8 shows parts of the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\), where \(\mathrm { f } ( x ) = \tan x\) and \(\mathrm { g } ( x ) = 1 + \mathrm { f } \left( x - \frac { 1 } { 4 } \pi \right)\). \begin{figure}[h] \end{figure}

1. Describe a sequence of two transformations which maps the curve \(y = \mathrm { f } ( x )\) to the curve \(y = \mathrm { g } ( x )\). [4] It can be shown that \(\mathrm { g } ( x ) = \frac { 2 \sin x } { \sin x + \cos x }\).
2. Show that \(\mathrm { g } ^ { \prime } ( x ) = \frac { 2 } { ( \sin x + \cos x ) ^ { 2 } }\). Hence verify that the gradient of \(y = \mathrm { g } ( x )\) at the point \(\left( \frac { 1 } { 4 } \pi , 1 \right)\) is the same as that of \(y = \mathrm { f } ( x )\) at the origin.
3. By writing \(\tan x = \frac { \sin x } { \cos x }\) and using the substitution \(u = \cos x\), show that \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \mathrm { f } ( x ) \mathrm { d } x = \int _ { \frac { 1 } { \sqrt { 2 } } } ^ { 1 } \frac { 1 } { u } \mathrm {~d} u\). Evaluate this integral exactly.
4. Hence find the exact area of the region enclosed by the curve \(y = \mathrm { g } ( x )\), the \(x\)-axis and the lines \(x = \frac { 1 } { 4 } \pi\) and \(x = \frac { 1 } { 2 } \pi\).

5 Differentiate \(x ^ { 2 } \tan 2 x\).

6 Given that \(y = \sqrt [ 3 ] { 1 + x ^ { 2 } }\), find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).

7 Given that \(y = x ^ { 2 } \sqrt { 1 + 4 x }\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 x ( 5 x + 1 ) } { \sqrt { 1 + 4 x } }\).