General Mathematics Paper 2,Nov/Dec. 2014
 Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 Main
Weakness/Remedies
Strengths

Question 12

In the diagram,  is a tangent to the circle at S. If O is the centre of the circle, ∠TSP = 21o and RQP = 100o, find, with reasons:

• SPR;
• QSR.
• Given the relation T = ,
• Make g the subject of the relation;
• Find g when T = 3, f = 4 and U = 5.

Observation

The Chief Examiner reported that this question was quite unpopular among the candidates and the performance of those who attempted it was poor. Majority of the candidates were reported to display poor knowledge of circle theorems.
In part (a), candidates’ expected response was as follows:
Angle PRS = 21o (angle in alternate segment to ∠PST). Angle RSP = 80o (opposite angles of a cyclic quadrilateral). Therefore, ∠ SPR = 180o – 80o – 21o = 79o. In part (a(ii)), ∠PQS = ∠SRP = 21o (angle in the same segment). Angle QPS = 90o (angle on a semi-circle). Therefore, angle QSP = 180o – 90o – 21o = 69o. Angle QSR = ∠PSR – ∠QSP = 80o – 69o = 11o
In part (b), candidates were expected to square both sides of the equality to obtain T2= . By cross multiplying, we obtain  +  = U. Multiplying both sides by the LCM of f and g gave gT2 + fT2 = Ugf. Bringing like terms together gave Ufg – gT2 = fT2. Hence, g = . When T = 3, f = 4 and U = 5,
g =  =  = 3.