AQA C3 (Core Mathematics 3) 2009 June

1

1. The curve with equation $$y = \frac { \cos x } { 2 x + 1 } , \quad x > - \frac { 1 } { 2 }$$ intersects the line \(y = \frac { 1 } { 2 }\) at the point where \(x = \alpha\).
   1. Show that \(\alpha\) lies between 0 and \(\frac { \pi } { 2 }\).
   2. Show that the equation \(\frac { \cos x } { 2 x + 1 } = \frac { 1 } { 2 }\) can be rearranged into the form $$x = \cos x - \frac { 1 } { 2 }$$
   3. Use the iteration \(x _ { n + 1 } = \cos x _ { n } - \frac { 1 } { 2 }\) with \(x _ { 1 } = 0\) to find \(x _ { 3 }\), giving your answer to three decimal places.
2. 1. Given that \(y = \frac { \cos x } { 2 x + 1 }\), use the quotient rule to find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
   2. Hence find the gradient of the normal to the curve \(y = \frac { \cos x } { 2 x + 1 }\) at the point on the curve where \(x = 0\).

2 The functions \(f\) and \(g\) are defined with their respective domains by $$\begin{array} { l l } f ( x ) = \sqrt { 2 x + 5 } , & \text { for real values of } x , x \geqslant - 2.5
g ( x ) = \frac { 1 } { 4 x + 1 } , & \text { for real values of } x , x \neq - 0.25 \end{array}$$

1. Find the range of f.
2. The inverse of f is \(\mathrm { f } ^ { - 1 }\).
   1. Find \(\mathrm { f } ^ { - 1 } ( x )\).
   2. State the domain of \(\mathrm { f } ^ { - 1 }\).
3. The composite function fg is denoted by h .
   1. Find an expression for \(\mathrm { h } ( x )\).
   2. Solve the equation \(\mathrm { h } ( x ) = 3\).

3

1. Solve the equation \(\tan x = - \frac { 1 } { 3 }\), giving all the values of \(x\) in the interval \(0 < x < 2 \pi\) in radians to two decimal places.
2. Show that the equation $$3 \sec ^ { 2 } x = 5 ( \tan x + 1 )$$ can be written in the form \(3 \tan ^ { 2 } x - 5 \tan x - 2 = 0\).
3. Hence, or otherwise, solve the equation $$3 \sec ^ { 2 } x = 5 ( \tan x + 1 )$$ giving all the values of \(x\) in the interval \(0 < x < 2 \pi\) in radians to two decimal places.
   (4 marks)

4

1. Sketch the graph of \(y = \left| 50 - x ^ { 2 } \right|\), indicating the coordinates of the point where the graph crosses the \(y\)-axis.
2. Solve the equation \(\left| 50 - x ^ { 2 } \right| = 14\).
3. Hence, or otherwise, solve the inequality \(\left| 50 - x ^ { 2 } \right| > 14\).
4. Describe a sequence of two geometrical transformations that maps the graph of \(y = x ^ { 2 }\) onto the graph of \(y = 50 - x ^ { 2 }\).

5

1. Given that \(2 \ln x = 5\), find the exact value of \(x\).
2. Solve the equation $$2 \ln x + \frac { 15 } { \ln x } = 11$$ giving your answers as exact values of \(x\).

6 The diagram shows the curve with equation \(y = \sqrt { 100 - 4 x ^ { 2 } }\), where \(x \geqslant 0\).
\includegraphics[max width=\textwidth, alt={}, center]{a596af76-9680-4ccb-a512-5b2575414429-5_518_494_367_758}

1. Calculate the volume of the solid generated when the region bounded by the curve shown above and the coordinate axes is rotated through \(360 ^ { \circ }\) about the \(\boldsymbol { y }\)-axis, giving your answer in terms of \(\pi\).
2. Use the mid-ordinate rule with five strips of equal width to find an estimate for \(\int _ { 0 } ^ { 5 } \sqrt { 100 - 4 x ^ { 2 } } \mathrm {~d} x\), giving your answer to three significant figures.
3. The point \(P\) on the curve has coordinates \(( 3,8 )\).
   1. Find the gradient of the curve \(y = \sqrt { 100 - 4 x ^ { 2 } }\) at the point \(P\).
   2. Hence show that the equation of the tangent to the curve at the point \(P\) can be written as \(2 y + 3 x = 25\).
4. The shaded regions on the diagram below are bounded by the curve, the tangent at \(P\) and the coordinate axes.
   \includegraphics[max width=\textwidth, alt={}, center]{a596af76-9680-4ccb-a512-5b2575414429-5_642_546_1800_731} Use your answers to part (b) and part (c)(ii) to find an approximate value for the total area of the shaded regions. Give your answer to three significant figures.

7

1. Use integration by parts to find \(\int ( t - 1 ) \ln t \mathrm {~d} t\).
2. Use the substitution \(t = 2 x + 1\) to show that \(\int 4 x \ln ( 2 x + 1 ) \mathrm { d } x\) can be written as \(\int ( t - 1 ) \ln t \mathrm {~d} t\).
3. Hence find the exact value of \(\int _ { 0 } ^ { 1 } 4 x \ln ( 2 x + 1 ) \mathrm { d } x\).