OCR MEI C3 (Core Mathematics 3) 2009 January

1 Solve the inequality \(| x - 1 | < 3\).

2

1. Differentiate \(x \cos 2 x\) with respect to \(x\).
2. Integrate \(x \cos 2 x\) with respect to \(x\).

3 Given that \(\mathrm { f } ( x ) = \frac { 1 } { 2 } \ln ( x - 1 )\) and \(\mathrm { g } ( x ) = 1 + \mathrm { e } ^ { 2 x }\), show that \(\mathrm { g } ( x )\) is the inverse of \(\mathrm { f } ( x )\).

4 Find the exact value of \(\int _ { 0 } ^ { 2 } \sqrt { 1 + 4 x } \mathrm {~d} x\), showing your working.

5

1. State the period of the function \(\mathrm { f } ( x ) = 1 + \cos 2 x\), where \(x\) is in degrees.
2. State a sequence of two geometrical transformations which maps the curve \(y = \cos x\) onto the curve \(y = \mathrm { f } ( x )\).
3. Sketch the graph of \(y = \mathrm { f } ( x )\) for \(- 180 ^ { \circ } < x < 180 ^ { \circ }\).

6

1. Disprove the following statement. $$\text { 'If } p > q \text {, then } \frac { 1 } { p } < \frac { 1 } { q } \text {. }$$
2. State a condition on \(p\) and \(q\) so that the statement is true.

7 The variables \(x\) and \(y\) satisfy the equation \(x ^ { \frac { 2 } { 3 } } + y ^ { \frac { 2 } { 3 } } = 5\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \left( \frac { y } { x } \right) ^ { \frac { 1 } { 3 } }\). Both \(x\) and \(y\) are functions of \(t\).
2. Find the value of \(\frac { \mathrm { d } y } { \mathrm {~d} t }\) when \(x = 1 , y = 8\) and \(\frac { \mathrm { d } x } { \mathrm {~d} t } = 6\). Section B (36 marks)

8 Fig. 8 shows the curve \(y = x ^ { 2 } - \frac { 1 } { 8 } \ln x\). P is the point on this curve with \(x\)-coordinate 1 , and R is the point \(\left( 0 , - \frac { 7 } { 8 } \right)\). \begin{figure}[h] \end{figure}

1. Find the gradient of PR.
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\). Hence show that PR is a tangent to the curve.
3. Find the exact coordinates of the turning point Q .
4. Differentiate \(x \ln x - x\). Hence, or otherwise, show that the area of the region enclosed by the curve \(y = x ^ { 2 } - \frac { 1 } { 8 } \ln x\), the \(x\)-axis and the lines \(x = 1\) and \(x = 2\) is \(\frac { 59 } { 24 } - \frac { 1 } { 4 } \ln 2\).

9 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] \end{figure}

1. Find \(a\). Hence write down the domain of the function.
2. 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 \(\mathrm { g } ( x ) = \frac { 1 } { \sqrt { 1 - x ^ { 2 } } }\).
3. (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.