OCR MEI C3 (Core Mathematics 3)

1 The function \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = \sqrt { 4 - x ^ { 2 } }\) for \(- 2 \leqslant x \leqslant 2\).

1. Show that the curve \(y = \sqrt { 4 - x ^ { 2 } }\) is a semicircle of radius 2 , and explain why it is not the whole of this circle. Fig. 9 shows a point \(\mathrm { P } ( a , b )\) on the semicircle. The tangent at P is shown. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{9e68f5e0-3394-4962-acd9-25bb31f09f2b-1_628_935_728_657} \captionsetup{labelformat=empty} \caption{Fig. 9}
   \end{figure}
2. (A) Use the gradient of OP to find the gradient of the tangent at P in terms of \(a\) and \(b\).
   (B) Differentiate \(\sqrt { 4 - x ^ { 2 } }\) and deduce the value of \(\mathrm { f } ^ { \prime } ( a )\).
   (C) Show that your answers to parts (A) and (B) are equivalent. The function \(\mathrm { g } ( x )\) is defined by \(\mathrm { g } ( x ) = 3 \mathrm { f } ( x - 2 )\), for \(0 \leqslant x \leqslant 4\).
3. Describe a sequence of two transformations that would map the curve \(y = \mathrm { f } ( x )\) onto the curve \(y = \mathrm { g } ( x )\). Hence sketch the curve \(y = \mathrm { g } ( x )\).
4. Show that if \(y = \mathrm { g } ( x )\) then \(9 x ^ { 2 } + y ^ { 2 } = 36 x\).

2 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 $$\int _ { - 1 } ^ { 0 } x \sqrt { 1 + x } \mathrm {~d} x = \int _ { 0 } ^ { 1 } \left( \begin{array} { l l } u ^ { \frac { 3 } { 2 } } & u ^ { \frac { 1 } { 2 } } \end{array} \right) \mathrm { d } u .$$ Hence find the area of the region enclosed by the curve and the \(x\)-axis.

3 Fig. 7 shows the curve \(y = \frac { x ^ { 2 } } { 1 + 2 x ^ { 3 } }\). It is undefined at \(x = a\); the line \(x = a\) is a vertical asymptote. \begin{figure}[h] \end{figure}

1. Calculate the value of \(a\), giving your answer correct to 3 significant figures.
2. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 x - 2 x ^ { 4 } } { \left( 1 + 2 x ^ { 3 } \right) ^ { 2 } }\). Hence determine the coordinates of the turning points of the curve.
3. Show that the area of the region between the curve and the \(x\)-axis from \(x = 0\) to \(x = 1\) is \(\frac { 1 } { 6 } \ln 3\).