OCR MEI C4 (Core Mathematics 4) 2014 June

Question 1
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1 Express \(\frac { 3 x } { ( 2 - x ) \left( 4 + x ^ { 2 } \right) }\) in partial fractions.
Question 3
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3 Fig. 3 shows the curve \(y = x ^ { 3 } + \sqrt { ( \sin x ) }\) for \(0 \leqslant x \leqslant \frac { \pi } { 4 }\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{413a0c52-b506-46d4-b1e4-fe13466abcc1-02_577_538_744_758} \captionsetup{labelformat=empty} \caption{Fig. 3}
\end{figure}
  1. Use the trapezium rule with 4 strips to estimate the area of the region bounded by the curve, the \(x\)-axis and the line \(x = \frac { \pi } { 4 }\), giving your answer to 3 decimal places.
  2. Suppose the number of strips in the trapezium rule is increased. Without doing further calculations, state, with a reason, whether the area estimate increases, decreases, or it is not possible to say.
Question 4
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4
  1. Show that \(\cos ( \alpha + \beta ) = \frac { 1 - \tan \alpha \tan \beta } { \sec \alpha \sec \beta }\).
  2. Hence show that \(\cos 2 \alpha = \frac { 1 - \tan ^ { 2 } \alpha } { 1 + \tan ^ { 2 } \alpha }\).
  3. Hence or otherwise solve the equation \(\frac { 1 - \tan ^ { 2 } \theta } { 1 + \tan ^ { 2 } \theta } = \frac { 1 } { 2 }\) for \(0 ^ { \circ } \leqslant \theta \leqslant 180 ^ { \circ }\).
Question 5
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5 A curve has parametric equations \(x = \mathrm { e } ^ { 3 t } , y = t \mathrm { e } ^ { 2 t }\).
  1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(t\). Hence find the exact gradient of the curve at the point with parameter \(t = 1\).
  2. Find the cartesian equation of the curve in the form \(y = a x ^ { b } \ln x\), where \(a\) and \(b\) are constants to be determined.
Question 6
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6 Fig. 6 shows the region enclosed by the curve \(y = \left( 1 + 2 x ^ { 2 } \right) ^ { \frac { 1 } { 3 } }\) and the line \(y = 2\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{413a0c52-b506-46d4-b1e4-fe13466abcc1-03_426_673_340_678} \captionsetup{labelformat=empty} \caption{Fig. 6}
\end{figure} This region is rotated about the \(y\)-axis. Find the volume of revolution formed, giving your answer as a multiple of \(\pi\). \section*{Question 7 begins on page 4.}
Question 7
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7 Fig. 7 shows a tetrahedron ABCD . The coordinates of the vertices, with respect to axes \(\mathrm { O } x y z\), are \(\mathrm { A } ( - 3,0,0 ) , \mathrm { B } ( 2,0 , - 2 ) , \mathrm { C } ( 0,4,0 )\) and \(\mathrm { D } ( 0,4,5 )\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{413a0c52-b506-46d4-b1e4-fe13466abcc1-04_794_844_456_589} \captionsetup{labelformat=empty} \caption{Fig. 7}
\end{figure}
  1. Find the lengths of the edges AB and AC , and the size of the angle CAB . Hence calculate the area of triangle ABC .
  2. (A) Verify that \(4 \mathbf { i } - 3 \mathbf { j } + 10 \mathbf { k }\) is normal to the plane ABC .
    (B) Hence find the equation of this plane.
  3. Write down a vector equation for the line through D perpendicular to the plane ABC . Hence find the point of intersection of this line with the plane ABC . The volume of a tetrahedron is \(\frac { 1 } { 3 } \times\) area of base × height.
  4. Find the volume of the tetrahedron ABCD .
Question 8
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8 Fig. 8.1 shows an upright cylindrical barrel containing water. The water is leaking out of a hole in the side of the barrel. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{413a0c52-b506-46d4-b1e4-fe13466abcc1-05_254_442_347_794} \captionsetup{labelformat=empty} \caption{Fig. 8.1}
\end{figure} The height of the water surface above the hole \(t\) seconds after opening the hole is \(h\) metres, where $$\frac { \mathrm { d } h } { \mathrm {~d} t } = - A \sqrt { h }$$ and where \(A\) is a positive constant. Initially the water surface is 1 metre above the hole.
  1. Verify that the solution to this differential equation is $$h = \left( 1 - \frac { 1 } { 2 } A t \right) ^ { 2 } .$$ The water stops leaking when \(h = 0\). This occurs after 20 seconds.
  2. Find the value of \(A\), and the time when the height of the water surface above the hole is 0.5 m . Fig. 8.2 shows a similar situation with a different barrel; \(h\) is in metres. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{413a0c52-b506-46d4-b1e4-fe13466abcc1-05_232_447_1489_794} \captionsetup{labelformat=empty} \caption{Fig. 8.2}
    \end{figure} For this barrel, $$\frac { \mathrm { d } h } { \mathrm {~d} t } = - B \frac { \sqrt { h } } { ( 1 + h ) ^ { 2 } } ,$$ where \(B\) is a positive constant. When \(t = 0 , h = 1\).
  3. Solve this differential equation, and hence show that $$h ^ { \frac { 1 } { 2 } } \left( 30 + 20 h + 6 h ^ { 2 } \right) = 56 - 15 B t .$$
  4. Given that \(h = 0\) when \(t = 20\), find \(B\). Find also the time when the height of the water surface above the hole is 0.5 m . \section*{END OF QUESTION PAPER}