Find the coordinates of the points on C at which $\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$ .

The tangent to C at the point P(1, 2) cuts the x-axis at the point T. Determine the coordinates of T.

The normal to C at the point P cuts the y-axis at the point N. Find the area of triangle PTN.

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = 2x - \frac{1}{2}{x^3}$     A1

$x\left( {2 - \frac{1}{2}{x^2}} \right) = 0$

$x = 0,{\text{ }} \pm 2$

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$ at $\left( {0,\frac{9}{8}} \right),{\text{ }}\left( { - 2,\frac{{25}}{8}} \right),{\text{ }}\left( {2,\frac{{25}}{8}} \right)$     A1A1A1

Note: Award A2 for all three x-values correct with errors/omissions in y-values.

[4 marks]

at x =1, gradient of tangent $= \frac{3}{2}$     (A1)

Note: In the following, allow FT on incorrect gradient.

equation of tangent is $y - 2 = \frac{3}{2}(x - 1)\,\,\,\,\,\left( {y = \frac{3}{2}x + \frac{1}{2}} \right)$     (A1)

meets x-axis when y = 0 , $- 2 = \frac{3}{2}(x - 1)$     (M1)

$x = - \frac{1}{3}$

coordinates of T are $\left( { - \frac{1}{3},0} \right)$     A1

[4 marks]

gradient of normal $= - \frac{2}{3}$     (A1)

equation of normal is $y - 2 = - \frac{2}{3}(x - 1)\,\,\,\,\,\left( {y = - \frac{2}{3}x + \frac{8}{3}} \right)$     (M1)

at x = 0 , $y = \frac{8}{3}$     A1

Note: In the following, allow FT on incorrect coordinates of T and N.

lengths of ${\text{PN}} = \sqrt {\frac{{13}}{9}}$ , ${\text{PT}} = \sqrt {\frac{{52}}{9}}$     A1A1

area of triangle ${\text{PTN}} = \frac{1}{2} \times \sqrt {\frac{{13}}{9}}  \times \sqrt {\frac{{52}}{9}}$     M1

$= \frac{{13}}{9}$ (or equivalent e.g. $\frac{{\sqrt {676} }}{{18}}$)     A1

[7 marks]

The whole of this question seemed to prove accessible to a high proportion of candidates.

(a)     was well answered by most, although a number of candidates gave only the x-values of the points or omitted the value at 0.

(b)     was successfully solved by the majority of candidates, who also found the correct equation of the normal in (c).

The last section proved more difficult for many candidates, the most common error being to use the wrong perpendicular sides. There were a number of different approaches here all of which were potentially correct but errors abounded. 

The whole of this question seemed to prove accessible to a high proportion of candidates.

(a)     was well answered by most, although a number of candidates gave only the x-values of the points or omitted the value at 0.

(b)     was successfully solved by the majority of candidates, who also found the correct equation of the normal in (c).

The last section proved more difficult for many candidates, the most common error being to use the wrong perpendicular sides. There were a number of different approaches here all of which were potentially correct but errors abounded. 

The whole of this question seemed to prove accessible to a high proportion of candidates.

(a)     was well answered by most, although a number of candidates gave only the x-values of the points or omitted the value at 0.

(b)     was successfully solved by the majority of candidates, who also found the correct equation of the normal in (c).

The last section proved more difficult for many candidates, the most common error being to use the wrong perpendicular sides. There were a number of different approaches here all of which were potentially correct but errors abounded.