OCR MEI C3 (Core Mathematics 3) 2012 January

Question 1
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1 Differentiate \(x ^ { 2 } \tan 2 x\).
Question 2
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2 The functions \(\mathrm { f } ( x )\) and \(\mathrm { g } ( x )\) are defined as follows. $$\begin{array} { l l } \mathrm { f } ( x ) = \ln x , & x > 0
\mathrm {~g} ( x ) = 1 + x ^ { 2 } , & x \in \mathbb { R } \end{array}$$ Write down the functions \(\mathrm { fg } ( x )\) and \(\mathrm { gf } ( x )\), and state whether these functions are odd, even or neither.
Question 3
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3 Show that \(\int _ { 0 } ^ { \frac { \pi } { 2 } } x \cos \frac { 1 } { 2 } x \mathrm {~d} x = \frac { \sqrt { 2 } } { 2 } \pi + 2 \sqrt { 2 } - 4\).
Question 5
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5 Each of the graphs of \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\) below is obtained using a sequence of two transformations applied to the corresponding dashed graph. In each case, state suitable transformations, and hence find expressions for \(\mathrm { f } ( x )\) and \(\mathrm { g } ( x )\).

  1. \includegraphics[max width=\textwidth, alt={}, center]{8b8958be-0ebc-4f72-ac3f-c16a8ec9e4ab-2_430_712_1366_680}

  2. \includegraphics[max width=\textwidth, alt={}, center]{8b8958be-0ebc-4f72-ac3f-c16a8ec9e4ab-2_394_608_1925_731}
Question 6
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6 Oil is leaking into the sea from a pipeline, creating a circular oil slick. The radius \(r\) metres of the oil slick \(t\) hours after the start of the leak is modelled by the equation $$r = 20 \left( 1 - \mathrm { e } ^ { - 0.2 t } \right) .$$
  1. Find the radius of the slick when \(t = 2\), and the rate at which the radius is increasing at this time.
  2. Find the rate at which the area of the slick is increasing when \(t = 2\).
Question 7
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7 Fig. 7 shows the curve \(x ^ { 3 } + y ^ { 3 } = 3 x y\). The point P is a turning point of the curve. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{8b8958be-0ebc-4f72-ac3f-c16a8ec9e4ab-3_583_513_708_776} \captionsetup{labelformat=empty} \caption{Fig. 7}
\end{figure}
  1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y - x ^ { 2 } } { y ^ { 2 } - x }\).
  2. Hence find the exact \(x\)-coordinate of P . Section B (36 marks)
Question 8
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8 Fig. 8 shows the curve \(y = \frac { x } { \sqrt { x - 2 } }\), together with the lines \(y = x\) and \(x = 11\). The curve meets these lines at P and Q respectively. R is the point \(( 11,11 )\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{8b8958be-0ebc-4f72-ac3f-c16a8ec9e4ab-4_609_736_440_667} \captionsetup{labelformat=empty} \caption{Fig. 8}
\end{figure}
  1. Verify that the \(x\)-coordinate of P is 3 .
  2. Show that, for the curve, \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x - 4 } { 2 ( x - 2 ) ^ { \frac { 3 } { 2 } } }\). Hence find the gradient of the curve at P . Use the result to show that the curve is not symmetrical about \(y = x\).
  3. Using the substitution \(u = x - 2\), show that \(\int _ { 3 } ^ { 11 } \frac { x } { \sqrt { x - 2 } } \mathrm {~d} x = 25 \frac { 1 } { 3 }\). Hence find the area of the region PQR bounded by the curve and the lines \(y = x\) and \(x = 11\).
Question 9
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9 Fig. 9 shows the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\). The function \(y = \mathrm { f } ( x )\) is given by $$f ( x ) = \ln \left( \frac { 2 x } { 1 + x } \right) , x > 0$$ The curve \(y = \mathrm { f } ( x )\) crosses the \(x\)-axis at P , and the line \(x = 2\) at Q . \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{8b8958be-0ebc-4f72-ac3f-c16a8ec9e4ab-5_552_636_470_715} \captionsetup{labelformat=empty} \caption{Fig. 9}
\end{figure}
  1. Verify that the \(x\)-coordinate of P is 1 . Find the exact \(y\)-coordinate of Q .
  2. Find the gradient of the curve at P. [Hint: use \(\ln \frac { a } { b } = \ln a - \ln b\).] The function \(\mathrm { g } ( x )\) is given by $$\mathrm { g } ( x ) = \frac { \mathrm { e } ^ { x } } { 2 - \mathrm { e } ^ { x } } , \quad x < \ln 2 .$$ The curve \(y = \mathrm { g } ( x )\) crosses the \(y\)-axis at the point R .
  3. Show that \(\mathrm { g } ( x )\) is the inverse function of \(\mathrm { f } ( x )\). Write down the gradient of \(y = \mathrm { g } ( x )\) at R .
  4. Show, using the substitution \(u = 2 - \mathrm { e } ^ { x }\) or otherwise, that \(\int _ { 0 } ^ { \ln \frac { 4 } { 3 } } \mathrm {~g} ( x ) \mathrm { d } x = \ln \frac { 3 } { 2 }\). Using this result, show that the exact area of the shaded region shown in Fig. 9 is \(\ln \frac { 32 } { 27 }\).
    [0pt] [Hint: consider its reflection in \(y = x\).]