Write down the coordinates of point \({\text{E}}\).

\({\text{C}}\) is a point on the line \({\text{DE}}\). \({\text{B}}\) is a point on the \(x\)-axis such that \({\text{BC}}\) is parallel to the \(y\)-axis. The \(x\)-coordinate of \({\text{C}}\) is \(t\).

Show that the \(y\)-coordinate of \({\text{C}}\) is \(4 - \frac{1}{2}t\).

\({\text{OBCD}}\) is a trapezium. The \(y\)-coordinate of point \({\text{D}}\) is \(4\).

Show that the area of \({\text{OBCD}}\) is \(4t - \frac{1}{4}{t^2}\).

The area of \({\text{OBCD}}\) is \(9.75\) square units. Write down a quadratic equation that expresses this information.

(i) Using your graphic display calculator, or otherwise, find the two solutions to the quadratic equation written in part (d).

(ii) Hence find the correct value for \(t\). Give a reason for your answer.

\({\text{E}}(8{\text{, }}0)\)    (A1)(A1)

Notes: Brackets required but do not penalize again if mark lost in Q4 (i)(d). If missing award (A1)(A0).
Accept \(x = 8\), \(y = 0\)
Award (A1) for \(x = 8\)

\(y + \frac{1}{2}t = 4\)     (M1)(M1)

Note: (M1) for the equation of the line seen. (M1) for substituting \(t\).

\(y = 4 - \frac{1}{2}t\)     (AG)

Note: Final line must be seen or previous (M1) mark is lost.

[2 marks]

\({\text{Area}} = \frac{1}{2} \times (4 + 4 - \frac{1}{2}t) \times t\)     (M1)(A1)

Note: (M1) for substituting in correct formula, (A1) for correct substitution.

\( = \frac{1}{2} \times (8 - \frac{1}{2}t) \times t = \frac{1}{2}(8t - \frac{1}{2}{t^2})\)     (A1)
\( = 4t - \frac{1}{4}{t^2}\)     (AG)

Note: Final line must be seen or previous (A1) mark is lost.

[3 marks]

\(4t - \frac{1}{4}{t^2} = 9.75\) or any equivalent form.     (A1)

[1 mark]

(i) \(t = 3\) or \(t =13\)     (A1)(ft)(A1)(ft)(G2)

Note: Follow through from candidate’s equation to part (d). Award (A0)(A1)(ft) for \((3{\text{, }}0)\) and \((13{\text{, }}0)\).

(ii) \(t\) must be a value between \(0\) and \(8\) then \(t = 3\)

Note: Accept \({\text{B}}\) is between \({\text{O}}\) and \({\text{E}}\). Do not award (R0)(A1).

A number of candidates did not attempt this question worth 12 marks but the majority answered this question partially and were able to gain some marks. Parts (a) and (b) were mostly well done. Very few candidates managed to answer part (c) well; this part of the question required good algebra along with a clear understanding of the situation given in the diagram. Many recovered then in (d) when they were asked to write down the quadratic equation. Solving the equation was not always found to be easy. Use of the GDC was expected but many used the formula. The correct solution, \(t = 3\), was chosen in the last part of the question. However, their justification was often false causing them to lose both the reasoning and the answer mark.

A number of candidates did not attempt this question worth 12 marks but the majority answered this question partially and were able to gain some marks. Parts (a) and (b) were mostly well done. Very few candidates managed to answer part (c) well; this part of the question required good algebra along with a clear understanding of the situation given in the diagram. Many recovered then in (d) when they were asked to write down the quadratic equation. Solving the equation was not always found to be easy. Use of the GDC was expected but many used the formula. The correct solution, \(t = 3\), was chosen in the last part of the question. However, their justification was often false causing them to lose both the reasoning and the answer mark.

A number of candidates did not attempt this question worth 12 marks but the majority answered this question partially and were able to gain some marks. Parts (a) and (b) were mostly well done. Very few candidates managed to answer part (c) well; this part of the question required good algebra along with a clear understanding of the situation given in the diagram. Many recovered then in (d) when they were asked to write down the quadratic equation. Solving the equation was not always found to be easy. Use of the GDC was expected but many used the formula. The correct solution, \(t = 3\), was chosen in the last part of the question. However, their justification was often false causing them to lose both the reasoning and the answer mark.

A number of candidates did not attempt this question worth 12 marks but the majority answered this question partially and were able to gain some marks. Parts (a) and (b) were mostly well done. Very few candidates managed to answer part (c) well; this part of the question required good algebra along with a clear understanding of the situation given in the diagram. Many recovered then in (d) when they were asked to write down the quadratic equation. Solving the equation was not always found to be easy. Use of the GDC was expected but many used the formula. The correct solution, \(t = 3\), was chosen in the last part of the question. However, their justification was often false causing them to lose both the reasoning and the answer mark.

A number of candidates did not attempt this question worth 12 marks but the majority answered this question partially and were able to gain some marks. Parts (a) and (b) were mostly well done. Very few candidates managed to answer part (c) well; this part of the question required good algebra along with a clear understanding of the situation given in the diagram. Many recovered then in (d) when they were asked to write down the quadratic equation. Solving the equation was not always found to be easy. Use of the GDC was expected but many used the formula. The correct solution, \(t = 3\), was chosen in the last part of the question. However, their justification was often false causing them to lose both the reasoning and the answer mark.