OCR MEI C3 (Core Mathematics 3) 2015 June

1 Fig. 1 shows part of the curve \(y = \mathrm { e } ^ { 2 x } \cos x\). \begin{figure}[h] \end{figure} Find the coordinates of the turning point P .

2 Find \(\int \sqrt [ 3 ] { 2 x - 1 } \mathrm {~d} x\).

3 Find the exact value of \(\int _ { 1 } ^ { 2 } x ^ { 3 } \ln x \mathrm {~d} x\).

4 Fig. 4 shows a cone with its axis vertical. The angle between the axis and the slant edge is \(45 ^ { \circ }\). Water is poured into the cone at a constant rate of \(5 \mathrm {~cm} ^ { 3 }\) per second. At time \(t\) seconds, the height of the water surface above the vertex O of the cone is \(h \mathrm {~cm}\), and the volume of water in the cone is \(V \mathrm {~cm} ^ { 3 }\). \begin{figure}[h] \end{figure} Find \(V\) in terms of \(h\).
Hence find the rate at which the height of water is increasing when the height is 10 cm .
[0pt] [You are given that the volume \(V\) of a cone of height \(h\) and radius \(r\) is \(V = \frac { 1 } { 3 } \pi r ^ { 2 } h\) ].

5 A curve has implicit equation \(y ^ { 2 } + 2 x \ln y = x ^ { 2 }\).
Verify that the point \(( 1,1 )\) lies on the curve, and find the gradient of the curve at this point.

6 Solve each of the following equations, giving your answers in exact form.

1. \(6 \arcsin x - \pi = 0\).
2. \(\arcsin x = \arccos x\).

7

1. The function \(\mathrm { f } ( x )\) is defined by $$f ( x ) = \frac { 1 - x } { 1 + x } , x \neq - 1$$ Show that \(\mathrm { f } ( \mathrm { f } ( x ) ) = x\).
   Hence write down \(\mathrm { f } ^ { - 1 } ( x )\).
2. The function \(\mathrm { g } ( x )\) is defined for all real \(x\) by $$\mathrm { g } ( x ) = \frac { 1 - x ^ { 2 } } { 1 + x ^ { 2 } }$$ Prove that \(\mathrm { g } ( x )\) is even. Interpret this result in terms of the graph of \(y = \mathrm { g } ( x )\).

8 Fig. 8 shows the line \(y = 1\) and the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = \frac { ( x - 2 ) ^ { 2 } } { x }\). The curve touches the \(x\)-axis at \(\mathrm { P } ( 2,0 )\) and has another turning point at the point Q . \begin{figure}[h] \end{figure}

1. Show that \(\mathrm { f } ^ { \prime } ( x ) = 1 - \frac { 4 } { x ^ { 2 } }\), and find \(\mathrm { f } ^ { \prime \prime } ( x )\). Hence find the coordinates of Q and, using \(\mathrm { f } ^ { \prime \prime } ( x )\), verify that it is a maximum point.
2. Verify that the line \(y = 1\) meets the curve \(y = \mathrm { f } ( x )\) at the points with \(x\)-coordinates 1 and 4 . Hence find the exact area of the shaded region enclosed by the line and the curve. The curve \(y = \mathrm { f } ( x )\) is now transformed by a translation with vector \(\binom { - 1 } { - 1 }\). The resulting curve has equation \(y = \mathrm { g } ( x )\).
3. Show that \(\mathrm { g } ( x ) = \frac { x ^ { 2 } - 3 x } { x + 1 }\).
4. Without further calculation, write down the value of \(\int _ { 0 } ^ { 3 } \mathrm {~g} ( x ) \mathrm { d } x\), justifying your answer.

9 Fig. 9 shows the curve \(y = \mathrm { f } ( x )\), where $$\mathrm { f } ( x ) = \left( \mathrm { e } ^ { x } - 2 \right) ^ { 2 } - 1 , x \in \mathbb { R } .$$ The curve crosses the \(x\)-axis at O and P , and has a turning point at Q . \begin{figure}[h] \end{figure}

1. Find the exact \(x\)-coordinate of P .
2. Show that the \(x\)-coordinate of Q is \(\ln 2\) and find its \(y\)-coordinate.
3. Find the exact area of the region enclosed by the curve and the \(x\)-axis. The domain of \(\mathrm { f } ( x )\) is now restricted to \(x \geqslant \ln 2\).
4. Find the inverse function \(\mathrm { f } ^ { - 1 } ( x )\). Write down its domain and range, and sketch its graph on the copy of Fig. 9.