Quod Erat Demonstrandum

2008/06/25

Marker’s report on SFXC F.2 Mathematics (I) Final Exam 2007-2008

Filed under: Junior Form Mathematics — johnmayhk @ 11:40 上午

Marker’s report on SFXC F.2 Mathematics (I) Final Exam 2007-2008

Passing rate : 63% (with project score considered)
Max : 86
Min : 9
Mean : 51.41
SD : 17.08

Question 1(a)

Students did not understand the word ‘solve’. To solve $6x^2 + 7x - 3 = 0$, many of them just stopped at $(2x + 3)(3x - 1) = 0$.

Students should be reminded of using the conjunction ‘OR’ in expressing the answer like $x = \frac{1}{3}$ or $x = -\frac{3}{2}$.

Question 1(b)

Not many students knew the correct way of “using the result in (a)". Unexpectedly, to solve $6(2x - \frac{1}{3})^2 + 7(2x - \frac{1}{3}) - 3 = 0$, some students replaced $\frac{1}{3}$ by $x$ (note that the root $x$ in (a) is different from that in (b)) and wrote $6(2x - x)^2 + 7(2x - x) - 3 = 0$. Actually, using previous results to solve further equation should be a common technique in further studies of mathematics, students should be reminded the importance of the technique.

There was a poor presentation. Many students knew the way of applying previous results, however, they wrote $x = 2x - \frac{1}{3}$ to make some confusion in the role of $x$. May be the question should be posted as $6(2y - \frac{1}{3})^2 + 7(2y - \frac{1}{3}) - 3 = 0$ to help students next time.

Surprisingly, students performed poorly in simple calculation of simple fractions. Many gave: $-\frac{2}{3} + 3 = -\frac{11}{3}$

Question 2(a)

Not many students could obtain full mark in simplifying $(\frac{a^{-3}}{-b^2})^{-4} \div (\frac{a^2}{b^{-3}})$. Many of them mistook that $(\frac{a^{-3}}{-b^2})^{-4} = \frac{a^{12}}{-b^{-8}}$ [instead of $\frac{a^{12}}{b^{-8}}$]. Some even further the calculation like $\frac{a^{12}}{-b^{-8}} = \frac{-b^{8}}{a^{12}}$.

The missing of bracket made wrong presentation like $\frac{a^{12}}{-b^{-8}} = a^{12} - b^8$.

Few students even made confusion in using index law in writing something like $(a^{-3})^{-4} = a^{-7}$.

Question 2(b)

This question: solve $2^{x + 1} = 4^{x - 1}$, seems to be very easy, however, students made mistakes mainly in simple algebraic computations.

The first main problem was about the distributive law. They wrote
$2^{x + 1} = 4^{x - 1}$
$\Rightarrow 2^{x + 1} = 2^{2x - 1}$
Students should be reminded of the proper use of brackets, like
$4^{x - 1} = (2^2)^{x - 1} = 2^{2(x - 1)} = 2^{2x - 2}$

Some students had difficulty in solving simple equation by writing:
$x + 1 = 2x - 2$
$\Rightarrow 3 = 3x$

The misuse of index law still prevailed, mistakes like
$2^{x + 1} = 4^{x - 1} \Rightarrow (4^2)^{x + 1} = 4^{x - 1}$
$2^{x + 1} = 4^{x - 1} \Rightarrow (2^{x + 1})2 = 4^{x - 1} \Rightarrow 4^{x + 1} = 4^{x - 1}$
$2^{x + 1} = 4^{x - 1} \Rightarrow 2^{x + 1} = 2^{(x - 1)^2} \Rightarrow x + 1 = (x - 1)^2$
$2^{x + 1} = 4^{x - 1}$
$\Rightarrow 2^x \times 2 = 4^x \times 4$
$\Rightarrow 2^x = 2^x \times 2$
$\Rightarrow 1^x = 2^x$
$\Rightarrow \frac{1^x}{2^x} = 1$
$\Rightarrow x = 1$

Question 3(a)

Students knew the identity $(a - b)^2 = a^2 - 2ab + b^2$, however, the following mistake was still common: $(\sqrt{2} - 2\sqrt{3})^2 = \sqrt{2}^2 - (2\sqrt{3})^2$

Still, students could not distinguish the difference between $(2\sqrt{3})^2$ and $2\sqrt{3}^2$; thus many wrote $(2\sqrt{3})^2 = 2(3) = 6$.

Also, may be affected by the format of the identity, many students stopped at
$(\sqrt{2} - 2\sqrt{3})^2 = 2 - 4\sqrt{6} + 12$ without adding up 2 and 12.

Even, students failed to perform simple arithmetic, as for example, they wrote $2 - 4\sqrt{6} + 12 = -10 - 4\sqrt{6}$.

The confusion among identities was also found. Some students wrote
$(\sqrt{2} - 2\sqrt{3})^2 = (\sqrt{2} - 2\sqrt{3})(\sqrt{2} + 2\sqrt{3})$

Question 3(b)

The computation on surd forms was not satisfactory, here is one of the cases: $\frac{\sqrt{12}}{6} = \sqrt{2}$.

Some students could not figure out that $\sqrt{6}$ and $\sqrt{150}$ are like surds, that was, the following calculation could go further: $\frac{\sqrt{6}}{3} - \frac{\sqrt{150}}{2}$

Students should be reminded that it may be easier to perform rationalization of denominators first and then combine the fractions next.

Question 4(a)

Not bad. Few students could not solve simple equation $\frac{x}{x + 12} = \frac{5}{11}$ properly. They wrote $\frac{x}{x + 12} = \frac{5}{11} \Rightarrow 11x = 5x + 12$. Students should be reminded of using bracket whenever necessary, like $\frac{x}{(x + 12)} = \frac{5}{11} \Rightarrow 11x = 5(x + 12) = 5x + 60$

Question 4(b)(i)

Some students did not understand what is the meaning of the question. The phrase “ratio of the number of John’s cards to the number of May’s cards is not greater than $\frac{y}{10}$" was wrongly interpreted by students as: $10 - 3y < \frac{y}{10}$, instead of the correct one: $\frac{10 - 3y}{22} \le \frac{y}{10}$.

Some set up equation instead of inequality.

Question 4(b)(ii)

The following was a common mistake in solving inequality:
$\frac{10 - 3y}{22} \le \frac{y}{10} \Rightarrow 22y \le 100 - 3y$ [instead of $22y \ge 100 - 3y$]
Students did not pay attention on the difference between solving equation and inequality.

Only one or two students could figure out the ‘hidden condition’: the number of John’s cards remaining should be non-negative, that is $10 - 3y \ge 0$; therefore it is incorrect to simply write $y \ge 1.92$, thus the possible values of $y$ are 2 and 3.

Question 5(a)

Not bad, but some students wrote $0.0000000000000123 = 1.23^{-14}$.

Question 5(b)

Some mistook the place value of $1_{(2)}$ as $2^1$ (instead of $2^0$).

Question 5(c)

The conversion of the unit $km/h$ into $m/s$ was not performed well. Students wrote
$36 km/h = 36 \times 1000 \times 60 \times 60 m/s$ or $36 km/h = 36 \times 1000 \div 60 m/s$. Students may be reminded of expressing $36 km/h$ as $\frac{36 km}{1 h}$ and the calculation should be proceeded as $\frac{36 km}{1 h} = \frac{36 \times 1000 m}{60 \times 60 s}$.

Question 6(a)

The purpose of setting up this question is about a common technique in trigonometry, that is, given a trigonometric ratio, as for example, $\sin \theta = \frac{1}{3}$, without finding the value of the angle $\theta$, we can find exact values of other trigonometric ratio like $\tan \theta$.

However, some students still did not familiarize with trigonometric ratios. The given in question was $\sin \theta = \frac{1}{3}$, but they wrote $\sin \frac{1}{3} = \frac{AC}{9}$.

Students should be reminded of writing proper unit in the answer whenever necessary.

Question 6(b)

Not bad. It is just a simple application of Pythagorean theorem. Students may be reminded of simplying surd form $\sqrt{72}$ as $6\sqrt{2}$.

Question 6(c)

The question is asking to find the value of $\tan \theta$, however, some students gave $\theta$.

There was a serious mistake, some students thought that $\tan \theta$ means $\tan \times \theta$. They gave:
$\tan \theta = \frac{3}{\sqrt{72}}$
$\Rightarrow \tan \theta \sqrt{72} = 3$
$\Rightarrow \theta = \frac{3}{\tan \sqrt{72}}$

Question 7

To find the total surface area of the solid above, many students ignored the area of the rectangular base $6 \times 8 cm^2$.

Also, students overlooked the semi-circle in calculating the curved surface area, they wrote $2\pi(3)(8)$ instead of $2\pi(3)(8) \div 2$.

Some considered $\pi(3)^2 \times 8$. Few made confusion between the formulae of circle area and circumference.

Finally, students should read the question carefully that the answer should be in terms of $\pi$.

Question 8(a)

Students may not understand the meaning of incomplete frequency polygon as shown.

The question asked about “the frequency of the class interval whose class mark is 95.5″, some students gave interval : 90.5 – 100.5, while some gave zero as the required frequency without paying attention on the total frequency of 100.

Question 8(b)

Not bad, provided that the reading and counting of frequencies in the graph is correct.

Question 8(c)

To find the passing mark provided that there are 22 students failed, we need to consider cumulative frequencies, however, some just frequency = 22 and read the corresponding mark as the required answer.

Question 9(a)(i)

Not bad. Just few students mixed up the formula of arc length with the formula of sector area.

Some students mistook the radius of the sector AFG as 5 cm (instead of 6 cm).

Question 9(a)(ii)

There was a common mistake, many students regarded the length of arc ED is the same as the length of arc FHG. And hence they could not calculate the perimeter of the figure correctly. This may reflect the weak spatial sense of some students.

Question 9(b)

In this part, the misunderstanding of measurement in figure was shown. The following are common mistakes given by students.

Some students mistook the length of $AK$ as 5 cm.
Some students mistook the length of $ED$ as 5 cm.

Many students mistook the length of $AJ$ as 2.5 cm. Students should be reminded that ‘perpendicular’ does not imply ‘bisecting’.

Question 10(a)

To find the vertical distance from A to the ground, many students could apply Pythagorean theorem and gave

$(2x)^2 + x^2 = (3\sqrt{5})^2$, however, many students regarded $(2x)^2$ wrongly as $2x^2$.

Some just gave $x = 3$ without finding $AC = 2 \times 3 = 6 m$.

Few misunderstood the question by writing $2AC = 3\sqrt{5}$.

Algebraic computation is still a hard problem for some students, here shows a ‘terrible’ presentation:
$(2x)^2 + x^2 = 3\sqrt{5}^2$
$\Rightarrow 2x^2 + x^2 = 3 \times 5$
$\Rightarrow 3x^4 = 15$
$\Rightarrow x^4 = 5$
$\Rightarrow x = \sqrt{2.5}$

Question 10(b)

Students knew what is the meaning of shortest distance from the man and the point C, however, they applied Pythagorean theorem and obtained nothing. Students should be reminded of a common technique of considering different ways of calculating triangle area.

Also, students did not use surd form to express the final answer.

Question 10(c)

Some students did not understand the phrase “2 m below its original position", they wrote $\sin \angle CBA = \frac{2}{3\sqrt{2}}$ instead of $\sin \angle CBA = \frac{6 - 2}{3\sqrt{2}}$.

The question asked $\angle CBA$, but few students gave $BC$.