OCR MEI C3 (Core Mathematics 3) 2016 June

1 Find the exact value of \(\int _ { 0 } ^ { \frac { 1 } { 2 } \pi } \left( 1 + \cos \frac { 1 } { 2 } x \right) \mathrm { d } x\).

2 The functions \(\mathrm { f } ( x )\) and \(\mathrm { g } ( x )\) are defined by \(\mathrm { f } ( x ) = \ln x\) and \(\mathrm { g } ( x ) = 2 + \mathrm { e } ^ { x }\), for \(x > 0\).
Find the exact value of \(x\), given that \(\mathrm { fg } ( x ) = 2 x\).

3 Find \(\int _ { 1 } ^ { 4 } x ^ { - \frac { 1 } { 2 } } \ln x \mathrm {~d} x\), giving your answer in an exact form.

4 By sketching the graphs of \(y = | 2 x + 1 |\) and \(y = - x\) on the same axes, show that the equation \(| 2 x + 1 | = - x\) has two roots. Find these roots.

5 The volume \(V \mathrm {~m} ^ { 3 }\) of a pile of grain of height \(h\) metres is modelled by the equation $$V = 4 \sqrt { h ^ { 3 } + 1 } - 4$$

1. Find \(\frac { \mathrm { d } V } { \mathrm {~d} h }\) when \(h = 2\). At a certain time, the height of the pile is 2 metres, and grain is being added so that the volume is increasing at a rate of \(0.4 \mathrm {~m} ^ { 3 }\) per minute.
2. Find the rate at which the height is increasing at this time.

6 Fig. 6 shows part of the curve \(\sin 2 y = x - 1\). P is the point with coordinates \(\left( 1.5 , \frac { 1 } { 12 } \pi \right)\) on the curve. \begin{figure}[h] \end{figure}

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(y\). Hence find the exact gradient of the curve \(\sin 2 y = x - 1\) at the point P . The part of the curve shown is the image of the curve \(y = \arcsin x\) under a sequence of two geometrical transformations.
2. Find \(y\) in terms of \(x\) for the curve \(\sin 2 y = x - 1\). Hence describe fully the sequence of transformations.
   \(7 \quad\) You are given that \(n\) is a positive integer.
   By expressing \(x ^ { 2 n } - 1\) as a product of two factors, prove that \(2 ^ { 2 n } - 1\) is divisible by 3 . Section B (36 marks)

8 Fig. 8 shows the curve \(y = \frac { x } { \sqrt { x + 4 } }\) and the line \(x = 5\). The curve has an asymptote \(l\).
The tangent to the curve at the origin O crosses the line \(l\) at P and the line \(x = 5\) at Q . \begin{figure}[h] \end{figure}

1. Show that for this curve \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x + 8 } { 2 ( x + 4 ) ^ { \frac { 3 } { 2 } } }\).
2. Find the coordinates of the point P .
3. Using integration by substitution, find the exact area of the region enclosed by the curve, the tangent OQ and the line \(x = 5\).

9 Fig. 9 shows the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \mathrm { e } ^ { 2 x } + k \mathrm { e } ^ { - 2 x }\) and \(k\) is a constant greater than 1 . The curve crosses the \(y\)-axis at P and has a turning point Q . \begin{figure}[h] \end{figure}

1. Find the \(y\)-coordinate of P in terms of \(k\).
2. Show that the \(x\)-coordinate of Q is \(\frac { 1 } { 4 } \ln k\), and find the \(y\)-coordinate in its simplest form.
3. Find, in terms of \(k\), the area of the region enclosed by the curve, the \(x\)-axis, the \(y\)-axis and the line \(x = \frac { 1 } { 2 } \ln k\). Give your answer in the form \(a k + b\). The function \(\mathrm { g } ( x )\) is defined by \(\mathrm { g } ( x ) = \mathrm { f } \left( x + \frac { 1 } { 4 } \ln k \right)\).
4. (A) Show that \(\mathrm { g } ( x ) = \sqrt { k } \left( \mathrm { e } ^ { 2 x } + \mathrm { e } ^ { - 2 x } \right)\).
   (B) Hence show that \(\mathrm { g } ( x )\) is an even function.
   (C) Deduce, with reasons, a geometrical property of the curve \(y = \mathrm { f } ( x )\). \section*{END OF QUESTION PAPER}