OCR MEI C3 (Core Mathematics 3)

1

1. Differentiate \(\sqrt { 1 + 3 x ^ { 2 } }\).
2. Hence show that the derivative of \(x \sqrt { 1 + 3 x ^ { 2 } }\) is \(\frac { 1 + 6 x ^ { 2 } } { \sqrt { 1 + 3 x ^ { 2 } } }\).

2 Given that \(y ^ { 3 } = x y - x ^ { 2 }\), show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { y - 2 x } { 3 y ^ { 2 } - x }\).
Hence show that the curve \(y ^ { 3 } = x y - x ^ { 2 }\) has a stationary point when \(x = \frac { 1 } { 8 }\).

3 Fig. 8 shows the curve \(y = x ^ { 2 } - \frac { 1 } { 8 } \ln x\). P is the point on this curve with \(x\)-coordinate 1 , and R is the point \(\left( 0 , - \frac { 7 } { 8 } \right)\). \begin{figure}[h] \end{figure}

1. Find the gradient of PR.
2. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\). Hence show that PR is a tangent to the curve.
3. Find the exact coordinates of the turning point Q .
4. Differentiate \(x \ln x - x\). Hence, or otherwise, show that the area of the region enclosed by the curve \(y = x ^ { 2 } - \frac { 1 } { 8 } \ln x\), the \(x\)-axis and the lines \(x = 1\) and \(x = 2\) is \(\frac { 59 } { 24 } - \frac { 1 } { 4 } \ln 2\).

4 The equation of a curve is given by \(\mathrm { e } ^ { 2 y } = 1 + \sin x\).

1. By differentiating implicitly, find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) in terms of \(x\) and \(y\).
2. Find an expression for \(y\) in terms of \(x\), and differentiate it to verify the result in part (i).

5 Fig. 6 shows the curve \(\mathrm { e } ^ { 2 y } = x ^ { 2 } + y\). \begin{figure}[h] \end{figure}

1. Show that \(\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 2 x } { 2 \mathrm { e } ^ { 2 y } - 1 }\).
2. Hence find to 3 significant figures the coordinates of the point P , shown in Fig. 6, where the curve has infinite gradient.