Majority of the candidates who attempted this question were reported to have performed better in part (a) then in part (b). Their performance in part (b) was described as poor. Majority of them could not apply the trapezium rule correctly.

In part (a) candidates were expected to show that if P = and Q = then

**PQ** = = and **QP** = = .

Since **PQ** = **QP**, then, 32 = 24 + 2*c* which gave *c* = 4. Similarly, 8 + 2*d* = 18, hence *d *= 5. i.e. *c* and *d* are 4 and 5 respectively.

In part (b), candidates were expected to obtain the following table:

x |
2 |
3 |
4 |
5 |
6 |

y = |
0.5 |
0.333 |
0.25 |
0.20 |
0.167 |

Using the trapezium rule, = [(0.5 + 0.167) + 2(0.333 + 0.25 + 0.2)] = 1.12, correct to 2 decimal places.