UNIVERSITY OF MALTA
THE MATRICULATION CERTIFICATE EXAMINATION
INTERMEDIATE LEVEL
PURE MATHEMATICS
May 2009
EXAMINERS’ REPORT
MATRICULATION AND SECONDARY EDUCATION
CERTIFICATE EXAMINATIONS BOARD
IM EXAMINERS’ REPORT MAY 2009
IM Pure Mathematics
May 2009 Session
Examiners’ Report
Part 1: Statistical Information
Table 1 shows the distribution of grades for the May 2009 session.
Table 1: Distribution of Grades awarded in May 2009
GRADE
A
B
C
D
14
37
86
72
Number
2.67
7.06
16.41
13.74
% of Total
E
112
21.37
F
188
35.88
Abs
15
2.86
Total
524
100
Part 2: Comments regarding candidates’ performance
Q1: a) This was very poorly attempted. Only three candidates managed to obtain the correct answer.
Many students do not know what a rational number is. Most got as far as
1
10
6 10
log 2 10 − log 0.8 − log = log 3
, and then evaluated the expression using a
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3
10 0.8
calculator.
b) This part of the question was well attempted. Some candidates showed correctly that
x log 2 = (1 − x) log 3 , but could not find x algebraically, and so they again referred to their
calculator to show that
x=
log 3
.
log 6
In c), many students found no difficulty in simplifying the expression, but some found the common
denominator of the expression incorrectly. The main error was that
2 2 − 1 was taken equal to
2( 2 − 1) .
Q2: Part (a) was very poorly attempted. Almost all students found the derivative of the function by
integrating f (x) instead of differentiating it. Others found f (0) instead of f / (0).
Part (b) was fairly well done, the main errors being algebraic mistakes like
(−a ) 2 = − a 2 and
(2a ) 2 = 2a 2 .
In (c), some candidates showed a lack of understanding of the method of completing the squares, and
wrote expressions like ( x + 2)
= −3 . Most candidates showed correctly that q ( x) = ( x + 2) 2 − 3 ,
but did not conclude that the minimum of q ( x ) is (-2, 3). Instead, they went on to find q / (x) = 0. The
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graph of y = q(x) was well sketched, but some candidates took the unnecessary trouble of finding
where the curve cuts the x-axis.
Q3:
Part (a) was poorly answered. Very few recognised that the three numbers were;
d = y and
x − y, x, x + y . Most candidates wrote down the numbers as; a, a + d , a + 2d , with
x = a+d . This lead to, many difficulties in finding the values of y and a.
Part (b) was generally well attempted. Some took 5(3) = (15) . Some calculated this part by finding
n
n
S1 , S 2 ,..., S9 , explicitly, rather than by using the quadratic equation for Sn .
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IM EXAMINERS’ REPORT MAY 2009
Q4: Knowledge of the domain and range of a function was not clear, and this led to an incorrect
sketch. Some candidates used the wrong base to solve 1 + ln x = 0. Few candidates identified
differentiation as the inverse of integration to calculate the required area after identifying the
appropriate limits.
Q5: In (a), candidates found it difficult to set up the equations for n and a from the coefficients of the
second and third terms of the binomial expansion; others found it difficult to solve for n and a from
these equations. Some candidates resorted to finding n and a using trial and error.
In (b), only a handful of candidates used Venn diagrams to evaluate the probability. Marks were still
awarded for any correct method.
In (c) was generally well attempted. Some candidates could not distinguish between the number of
combinations of an event, and the probability of an event.
Q6: In part (a) some candidates expressed their result in degrees, even though it was clearly indicated
that the answer was to be expressed in radians.
Many students obtained the formula for the area of the segment in part (b), but very few were able to
draw a sketch properly for the situation discussed in the problem. As a result the angle of the segment
was not evaluated correctly. Hardly any students identified the required area as twice the area of a
segment.
Q7: Part (a) was well attempted by a good number of students.
In part (b) about a third of the candidates answered this part correctly. Those who did not do it
correctly made various mistakes, including using an incorrect formula for the volume of a sphere,
differentiating it with respect to t incorrectly, not even differentiating the volume, and applying the
chain rule wrongly.
Q8: (a) The differential equation was solved correctly by about half the students. Common mistakes
by the others included failure to integrate the separated parts, or omission of the constant of integration.
(b) Few students got this part completely correct. Some failed to find the values of x for which the
graphs intersect, whilst others did not find the corresponding y coordinates. Very few candidates drew
the required sketch correctly, whilst many did not attempt it at all.
Q9: Most candidates found the matrices A, B, BA and (BA)2 correctly. However, few candidates could
explain convincingly that (BA)2 should be the identity matrix.
Q10: (a) A good number of candidates did this part correctly although some could not interpret the
matrix multiplication as an equation which could be solved.
In (b), some students did not use the inverse matrix method, or found the inverse incorrectly, or
multiplied it incorrectly with the vector on the right hand side.
Chairperson
Board of Examiners
July 2009
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