OCR MEI C2 (Core Mathematics 2) 2012 June

1 Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) when \(y = \sqrt { x } + \frac { 3 } { x }\).

2 Find the second and third terms in the sequence given by $$\begin{aligned} & u _ { 1 } = 5
& u _ { n + 1 } = u _ { n } + 3 . \end{aligned}$$ Find also the sum of the first 50 terms of this sequence.

3 \begin{figure}[h] \end{figure} \section*{Not to scale} In Fig. 3, BCD is a straight line. \(\mathrm { AC } = 9.8 \mathrm {~cm} , \mathrm { BC } = 7.3 \mathrm {~cm}\) and \(\mathrm { CD } = 6.4 \mathrm {~cm}\); angle \(\mathrm { ACD } = 53.4 ^ { \circ }\).

1. Calculate the length AD .
2. Calculate the area of triangle ABC .

4 The point \(\mathrm { P } ( 6,3 )\) lies on the curve \(y = \mathrm { f } ( x )\). State the coordinates of the image of P after the transformation which maps \(y = \mathrm { f } ( x )\) onto

1. \(y = 3 \mathrm { f } ( x )\),
2. \(y = \mathrm { f } ( 4 x )\).

5 A sector of a circle has angle 1.6 radians and area \(45 \mathrm {~cm} ^ { 2 }\). Find the radius and perimeter of the sector.

6 Fig. 6 shows the relationship between \(\log _ { 10 } x\) and \(\log _ { 10 } y\). \begin{figure}[h] \end{figure} Find \(y\) in terms of \(x\).

7 The gradient of a curve is given by \(\frac { \mathrm { d } y } { \mathrm {~d} x } = 6 x ^ { \frac { 1 } { 2 } } - 5\). Given also that the curve passes through the point (4, 20), find the equation of the curve.

8 Solve the equation \(\sin 2 \theta = 0.7\) for values of \(\theta\) between 0 and \(2 \pi\), giving your answers in radians correct to 3 significant figures.

9 A farmer digs ditches for flood relief. He experiments with different cross-sections. Assume that the surface of the ground is horizontal.

1. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{8f7413d8-2814-4d5c-bec0-ce118fec80eb-4_437_640_470_715} \captionsetup{labelformat=empty} \caption{Fig. 9.1}
   \end{figure} Fig. 9.1 shows the cross-section of one ditch, with measurements in metres. The width of the ditch is 1.2 m and Fig. 9.1 shows the depth every 0.2 m across the ditch. Use the trapezium rule with six intervals to estimate the area of cross-section. Hence estimate the volume of water that can be contained in a 50-metre length of this ditch.
2. Another ditch is 0.9 m wide, with cross-section as shown in Fig. 9.2. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{8f7413d8-2814-4d5c-bec0-ce118fec80eb-4_574_808_1402_632} \captionsetup{labelformat=empty} \caption{Fig. 9.2}
   \end{figure} With \(x\) - and \(y\)-axes as shown in Fig. 9.2, the curve of the ditch may be modelled closely by \(y = 3.8 x ^ { 4 } - 6.8 x ^ { 3 } + 7.7 x ^ { 2 } - 4.2 x\).
   (A) The actual ditch is 0.6 m deep when \(x = 0.2\). Calculate the difference between the depth given by the model and the true depth for this value of \(x\).
   (B) Find \(\int \left( 3.8 x ^ { 4 } - 6.8 x ^ { 3 } + 7.7 x ^ { 2 } - 4.2 x \right) \mathrm { d } x\). Hence estimate the volume of water that can be contained in a 50 -metre length of this ditch.

10

1. Use calculus to find, correct to 1 decimal place, the coordinates of the turning points of the curve \(y = x ^ { 3 } - 5 x\). [You need not determine the nature of the turning points.]
2. Find the coordinates of the points where the curve \(y = x ^ { 3 } - 5 x\) meets the axes and sketch the curve.
3. Find the equation of the tangent to the curve \(y = x ^ { 3 } - 5 x\) at the point \(( 1 , - 4 )\). Show that, where this tangent meets the curve again, the \(x\)-coordinate satisfies the equation $$x ^ { 3 } - 3 x + 2 = 0$$ Hence find the \(x\)-coordinate of the point where this tangent meets the curve again.

11 A geometric progression has first term \(a\) and common ratio \(r\). The second term is 6 and the sum to infinity is 25 .

1. Write down two equations in \(a\) and \(r\). Show that one possible value of \(a\) is 10 and find the other possible value of \(a\). Write down the corresponding values of \(r\).
2. Show that the ratio of the \(n\)th terms of the two geometric progressions found in part (i) can be written as \(2 ^ { n - 2 } : 3 ^ { n - 2 }\).