OCR MEI C3 (Core Mathematics 3)

Question 1
View details
1 Fig. 8 shows part of the curve \(y = x \cos 2 x\), together with a point P at which the curve crosses the \(x\)-axis. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{aee8da6a-7d5c-442f-9729-55d81d9a606f-1_427_968_432_584} \captionsetup{labelformat=empty} \caption{Fig. 8}
\end{figure}
  1. Find the exact coordinates of P .
  2. Show algebraically that \(x \cos 2 x\) is an odd function, and interpret this result graphically.
  3. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
  4. Show that turning points occur on the curve for values of \(x\) which satisfy the equation \(x \tan 2 x = \frac { 1 } { 2 }\).
  5. Find the gradient of the curve at the origin. Show that the second derivative of \(x \cos 2 x\) is zero when \(x = 0\).
  6. Evaluate \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } x \cos 2 x \mathrm {~d} x\), giving your answer in terms of \(\pi\). Interpret this result graphically.
Question 2
View details
2 Fig. 8 shows part of the curve \(y = x \sin 3 x\). It crosses the \(x\)-axis at P . The point on the curve with \(x\)-coordinate \(\frac { 1 } { 6 } \pi\) is Q . \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{aee8da6a-7d5c-442f-9729-55d81d9a606f-2_418_769_516_673} \captionsetup{labelformat=empty} \caption{Fig. 8}
\end{figure}
  1. Find the \(x\)-coordinate of P .
  2. Show that Q lies on the line \(y = x\).
  3. Differentiate \(x \sin 3 x\). Hence prove that the line \(y = x\) touches the curve at Q .
  4. Show that the area of the region bounded by the curve and the line \(y = x\) is \(\frac { 1 } { 72 } \left( \pi ^ { 2 } - 8 \right)\).
Question 3
View details
3 The function \(f ( x ) = \ln \left( t + x ^ { 2 } \right)\) has domain \(- 3 \leqslant x \leqslant 3\).
Fig. 9 shows the graph of \(y = f ( x )\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{aee8da6a-7d5c-442f-9729-55d81d9a606f-3_510_895_523_604} \captionsetup{labelformat=empty} \caption{Fig. 9}
\end{figure}
  1. Show algebraically that the function is even. State how this property relates to the shape of the curve.
  2. Find the gradient of the curve at the point \(\mathrm { P } ( 2 , \ln 5 )\).
  3. Explain why the function does not have an inverse for the domain \(- 3 \leqslant x \leqslant 3\). The domain of \(f ( x )\) is now restricted to \(0 \leqslant x \leqslant 3\). The inverse of \(f ( x )\) is the function \(g ( x )\).
  4. Sketch the curves \(y = f ( x )\) and \(y = g ( x )\) on the same axes. State the domain of the function \(g ( x )\),
    Show that \(\mathrm { g } ( x ) = \sqrt { \mathrm { e } ^ { x } - 1 }\).
  5. Differentiate \(\mathrm { g } ( \mathrm { x } )\). Hence verify that \(\mathrm { g } ^ { \prime } ( \ln 5 ) = 1 \frac { 1 } { 4 }\). Explain the connection between this result and your answer to part (ii).