OCR MEI C3 (Core Mathematics 3) 2007 January

1 Fig. 1 shows the graphs of \(y = | x |\) and \(y = | x - 2 | + 1\). The point P is the minimum point of \(y = | x - 2 | + 1\), and Q is the point of intersection of the two graphs. \begin{figure}[h] \end{figure}

1. Write down the coordinates of P .
2. Verify that the \(y\)-coordinate of Q is \(1 \frac { 1 } { 2 }\).

2 Evaluate \(\int _ { 1 } ^ { 2 } x ^ { 2 } \ln x \mathrm {~d} x\), giving your answer in an exact form.

3 The value \(\pounds V\) of a car is modelled by the equation \(V = A \mathrm { e } ^ { - k t }\), where \(t\) is the age of the car in years and \(A\) and \(k\) are constants. Its value when new is \(\pounds 10000\), and after 3 years its value is \(\pounds 6000\).

1. Find the values of \(A\) and \(k\).
2. Find the age of the car when its value is \(\pounds 2000\).

4 Use the method of exhaustion to prove the following result.
No 1 - or 2 -digit perfect square ends in \(2,3,7\) or 8
State a generalisation of this result.

5 The equation of a curve is \(y = \frac { x ^ { 2 } } { 2 x + 1 }\).

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 x ( x + 1 ) } { ( 2 x + 1 ) ^ { 2 } }\).
2. Find the coordinates of the stationary points of the curve. You need not determine their nature.

6 Fig. 6 shows the triangle OAP , where O is the origin and A is the point \(( 0,3 )\). The point \(\mathrm { P } ( x , 0 )\) moves on the positive \(x\)-axis. The point \(\mathrm { Q } ( 0 , y )\) moves between O and A in such a way that \(\mathrm { AQ } + \mathrm { AP } = 6\). \begin{figure}[h] \end{figure}

1. Write down the length AQ in terms of \(y\). Hence find AP in terms of \(y\), and show that $$( y + 3 ) ^ { 2 } = x ^ { 2 } + 9 .$$
2. Use this result to show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { x } { y + 3 }\).
3. When \(x = 4\) and \(y = 2 , \frac { \mathrm {~d} x } { \mathrm {~d} t } = 2\). Calculate \(\frac { \mathrm { d } y } { \mathrm {~d} t }\) at this time. Section B (36 marks)

7 Fig. 7 shows part of the curve \(y = \mathrm { f } ( x )\), where \(\mathrm { f } ( x ) = x \sqrt { 1 + x }\). The curve meets the \(x\)-axis at the origin and at the point P . \begin{figure}[h] \end{figure}

1. Verify that the point P has coordinates \(( - 1,0 )\). Hence state the domain of the function \(\mathrm { f } ( x )\).
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 + 3 x } { 2 \sqrt { 1 + x } }\).
3. Find the exact coordinates of the turning point of the curve. Hence write down the range of the function.
4. Use the substitution \(u = 1 + x\) to show that $$\int _ { - 1 } ^ { 0 } x \sqrt { 1 + x } \mathrm {~d} x = \int _ { 0 } ^ { 1 } \left( u ^ { \frac { 3 } { 2 } } - u ^ { \frac { 1 } { 2 } } \right) \mathrm { d } u$$ Hence find the area of the region enclosed by the curve and the \(x\)-axis.

8 Fig. 8 shows part of the curve \(y = \mathrm { f } ( x )\), where $$\mathrm { f } ( x ) = \left( \mathrm { e } ^ { x } - 1 \right) ^ { 2 } \text { for } x \geqslant 0 .$$ \begin{figure}[h] \end{figure}

1. Find \(\mathrm { f } ^ { \prime } ( x )\), and hence calculate the gradient of the curve \(y = \mathrm { f } ( x )\) at the origin and at the point \(( \ln 2,1 )\). The function \(\mathrm { g } ( x )\) is defined by \(\mathrm { g } ( x ) = \ln ( 1 + \sqrt { x } )\) for \(x \geqslant 0\).
2. Show that \(\mathrm { f } ( x )\) and \(\mathrm { g } ( x )\) are inverse functions. Hence sketch the graph of \(y = \mathrm { g } ( x )\). Write down the gradient of the curve \(y = \mathrm { g } ( x )\) at the point \(( 1 , \ln 2 )\).
3. Show that \(\int \left( \mathrm { e } ^ { x } - 1 \right) ^ { 2 } \mathrm {~d} x = \frac { 1 } { 2 } \mathrm { e } ^ { 2 x } - 2 \mathrm { e } ^ { x } + x + c\). Hence evaluate \(\int _ { 0 } ^ { \ln 2 } \left( \mathrm { e } ^ { x } - 1 \right) ^ { 2 } \mathrm {~d} x\), giving your answer in an exact form.
4. Using your answer to part (iii), calculate the area of the region enclosed by the curve \(y = \mathrm { g } ( x )\), the \(x\)-axis and the line \(x = 1\).