2008 NANYANGJUNIOR COLLEGE H2MATH PRELIM PAPER 2 SOLUTIONS
1 / Differentiate wrt x, we have .Substituting into original d.e. gives
Let C = 1,
Let C = -1,
Comments on Question 1:
Ahandfulcandidates were not able to differentiate y=ux2 as they treated u as a constant. Most candidates were able to perform the integration to obtain the correct general solution.
To obtain the family of solution curves, some candidates did not simplify the general solution to only C as they are expected to. This resulted in erroneous sketches.
2 / (i)Since every horizontal line y = k cuts the curve of y = g(x) at most once, g is a one-one function.
Let
Since , we have
Thus
(ii)Since , thus fg exists
,
(iii).. Thus f2g does not exist.
Comments on Question 2:
Most students were able to apply horizontal line test and state the relevant points regarding when a function was 1-1. The usual mistakes were applying the vertical line test, using the wrong "range" for y = k (e.g. k was stated to be the set of integers) or failing to produce a graph or drawing an incorrect graph (e.g. straight line, n shape quadratic).
For the inverse, many students omitted the + - sign when taking square root, and/or ignored the need to consider which branch of thequadratic was relevant. Domain of the inverse was omitted by many students.
Showing the composite existed was generally fine. Problems included wrong range for g (i.e. including the value -3), and failure to state the rule for fg (which many interpreted as range).
The final part was well done when students obtained the correct range for fg; otherwise, many problems arosed when students simply used the GC to try and read off the range (wrongly in most cases; also, penalty was especially severe when the left limit was indicated as ln 0, which made comparison impossible for the existence of the composite function).
3 / (i) Given ------(1)
At (0 , 2), .
Equation of normal at point (0 , 2) is
(ii) Differentiating (1) with respect to x,
Differentiating again with respect to x,
When x = 0, y = 2 and, and
Hence
(iii)
Comments on Question 3:
The first 2 parts of the question were well-attempted by most students. However, some candidates still did not master the technique of implicit differentiation which leads to disastrous answers.
The last part of the question was poorly answered, with less than 10 candidates obtaining the correct answer. The part asked for the expansion of ey, using the expansion of y found in the previous part. Most candidates substituted the correct expansion of but merely used the standard expansion of ex with x replaced by . The problem with this method is we need to consider an infinite series in order to determine the value of the constant in the expansion. The correct method is as follows:
, where was factored out. Then the standard expansion of ex with x replaced by can be applied up to the term in x2.
4 / (i)As , thus
.
(ii)Since , thus . Accordingly, and .
(iii)If , using graphic calculator, we can see that . Thus . Hence .
(iv)Using the recurrence relation,
Since , thus . (Since both terms are negative)
Since , thus .
Since , thus . Accordingly, . Thus
.
(v)Using GC, if , then we can see that for all . Using (iv) recursively,
Since , . For , it suffices for . Thus
.
Thus a possible value of N is 30.
Comments on Question 4:
The first three parts were well attempted by the entire cohort. Except for some candidates who showed complete ignorance to this topic, they were able to give the correct method to finding the equation that the limit satisfies and the two roots to the equation. Majority of the candidates were not able to give the fact that L < 0 and resulted in loss of marks.
For (iv), candidates were able to give good response to the first portion of this part. Some candidates were unable to prove the given equation and give long and erroneous working. A fraction of them just simply write the given equality without justification. Candidates are reminded that for “show” questions, workings for each step are to be presented clearly. The inequality that was to be shown proved to be a tantamount to all candidates. Almost all candidates were unable to see that the expression is negative and fail to provide the correct bounds for this expression.
The entire cohort was unable to answer (v). Some of them attempted to use graphic calculator to get the answer but was baffled by the sudden change of value from 10-13 to 0.
5 / (i)(a)number of ways = = 768
(b)number of ways =
(ii)number of ways =
Comments on Question 5:
(i)(a) This is an easy question which many students manage to solve correctly. However there were several students who did not consider permutation of each of the 5 couples, and some students who strangely seemed to think that 25 is the same as 5 X 2.
(b) Solving this part entails a technique students ought to know well, that is to use the other people who are not in the subcommittee to separate those in the subcommittee. Some students tried to consider the complement instead, which is rather cumbersome and, except for a few students, most of them got it wrong.
(ii) This last part can be done directly quite easily. A number of students tried to use complement either without considering correct complementary set, or without being exhaustive in solving.
6 / (i)
(ii)P(either Alice is first or Mark is second (or both) in the queue)
(iii)P(there is only one girl chosen if five persons are chosen at random from this group)
(iv)P(all five girls are next to each other | at least fourboys stand next to each other)
Comments on Question 6:
The question was badly attempted by students. The average score for the question is 2 out of 8 only. A number of students gave answers such as 8!. Is it a post Permutation and Combination effect from Question 5? Probability is a number that is between 0 to 1 (both inclusive). Candidates should check their answers if they are logical before submitting their scripts. How can probability be > 1?? An important note to point out that answers should be left in 3 significant figures and not 2 significant figures only. This error is committed by a handful number of students.
(i)This is the simplest part of the question, yet a number of students got it wrong. A number of students attempted this part by . They forgotten that after upon placing Matthew in the position required, one is left with 9 more people left and as such the probability should be . Easier method to tackle the problem would be to see that upon fixing the position of Matthew and Mary, there are 8 more spaces that we can have different arrangements and as such the probability should be .
(ii)A number of students fail to see that the probability that either Alice first of Mark is second (or both) is actually asking for P(Alice first Mark second) i.e. P(A B). This is a typical A’ Level style of questioning and indicates poor preparation on the topic of probability. The question is testing on the concept of P(A B) = P(A) + P(B) – P(A B). A number of students gave their answers as P(A B) = P(A) + P(B) + P(A B). The poor attempt by many students reflect poor concept and grasp of the topic that even a typical Cambridge question they could not handle well.
(iii)This is the only part that is done fairly better. However, it is noticed that some students listed out all the possible combinations for the condition required for the probability. This is unnecessary and wasting time. Candidates should use which is easier as it takes care of the arrangement aspect!
(iv)This part is the toughest part of the question. No students managed to get full marks from this part.
7 / a) X: no. of muffins sold out of 200 muffins
P(=0.542
b) To make a profit, he must earn more than $270, which means he must sell more than $270/$1.80 =150 muffins.
P(he can make a profit) =
= 0.951
For a period of 60 days, since n=60 is large, by CLT
P(
Comments on Question 7:
For (a), quite a significant number of students wrote the distribution as a Poisson distribution with parameter 160, without realizing that in fact, for Poisson distribution, the integer values can range from 0 to infinity. However, the number of muffins sold can at most be 200, as a result it should be a binomial distribution with n = 200 and p = 4/5. Even for those candidates who identified the correct distribution did not understand the meaning of “between 160 to 195 (inclusive)”. Also, quite a handful of candidates made the mistakes of either or .
For (b), most candidates could solve this part of the question. However, there were still quite a significant number of candidates who made the error of claiming that the amount of revenue obtained follows a binomial distribution with n = 360 and p= 4/5. Here, the misconception is that the amount of revenue does not occur in discrete integer values, and neither do they occur in all sorts of denominations. The ideal way is to work along the line of how many muffins should be sold in order to make a profit.
Some candidates also left this part of the question blank.
For last part of the question, many misunderstood the question entirely. Many thought that the question is asking for where Y is defined as the number of days where at least 159 muffins are sold, and thus gave the distribution as . For some of those who managed to understand the question and knew that they have to use Central Limit Theorem to solve the question, didn’t know that continuity correction is not required when the use of CLT is involved. For the rest of the candidates who solve the question by approximation X to a normal distribution, and subsequently had to do continuity correction. Thus a variety of answers are accepted, 0.915; 0.916 and 0.980 (Appropriate mathematical working have to support the suggested answers.)
8 / Let the random variable X be the number of cars arriving at the petrol station during a 10-minute period.
XPo() i.e. XPo()
(i)P(X 5) = 0.858
(ii)Let the random variable Y be the number of cars arriving at the petrol on a particular day.
Y Po() i.e. Y Po(504)
Since = 504 > 10, Y N(504, 504) approximately
P(Y 500) P(Y > 499.5) 0.579
Let the random variable A be the number of 10-minute intervals where more than 5 cars arrive at the petrol station in a day.
Number of 10-minute intervals in a day =
A B(144, P(X > 5) ) i.e. A B(144, 0.1423864467)
P(Ak) 0.05 that is, P(Ak) 0.95
Using GC,
P(A 27) = 0.94803 < 0.95
P(A 28) = 0.96746 > 0.95 Hence least value of k = 28
Comments on Question 8:
Most students were able to score rather well for the question, especially in the first 2 parts. A minority of students were penalized for attempting to scale the variable defined inappropriately as follows:
X – No. of cars arriving in 20 mins, X ~ Po(7)
X – No. of cars arriving in 10 mins, X ~ Po()
or 72X – No. of cars arriving in 1 day, 72X ~ Po(504)
Common mistakes for the respective parts include the following:
(i)A small number of students gave the P(X 5) = 1 – P(X < 4) and used the GC command poissoncdf(3.5,4) for P(X < 4).
(ii)Quite a handful of students forgot to conduct continuity correction on the value 500 to 499.5. Other mistakes include wrong correction of the value to 500.5.
Majority were able to identify the last part of the question to be following a binomial distribution. However, a handful of these students used the wrong probability value for the last part of the question, hence resulting in the wrong value of k found. Some students approximated the variable to Normal though the question did not ask for estimation of answer. Seldom were these students successful in solving for this part as very often continuity correction was not performed.
9 / (a) Stratified Sampling. 3 strata: General Workers, Administrative Staff and Mechanics.
Stratum / General Workers / Administrative Staff / Mechanics
Sample Size / / /
Conduct simple random sampling for each stratum to select the required sample.
(b)(i), . Thus s = 7.055
(ii)Let X be the life span of a randomly chosen light bulb. Let be the mean life span of a randomly chosen light bulb.
To test against
Let be the level of significance.
Since Xfollows a normal distribution, and we estimate the population variance, we will conduct a T test.(Approximate by Z test accepted)
.
Thus p – value = 0.058696(Z test p-value = 0.056197)
For to be rejected, p – value <
Thus the smallest value of the level of significance is.(Z-test: 5.7%)
(iii)To test against
Level of significance: 8%
Since X follows a normal distribution, and we estimate the population variance, we will conduct a T test.(Approximate by Z test accepted)
Reject H0 if (Reject H0 if )
.
For H0 to be not rejected, .(or )Thus
or
Comments on Question 9:
(a) The majority of candidates were able to identify the correct sampling method (stratified random sampling) and the 3 strata. The number of people to be selected in each stratum was correctly given. However, many candidates did not state how the people in each stratum was to be selected – by simple random sampling.
(b)(i) The majority of candidates knew how to obtain the unbiased estimates of the population mean and variance with only a few who used the wrong formula. The formula for s2 could be obtained from the formulae booklet and candidates should be familiar with the booklet.
(ii) One striking observation is the obvious lack of fluency in expressing mathematical ideas using the language of Statistics, particularly in defining H0 and H1. Bad expressions like
‘H0 : 0 = 1000; H1 : 1 < 1000’,
‘H0 : 0 = 1000; H1 : 10’,
‘H0 = 1000; H1 < 1000’,
‘H0 : = 1000; H1 : < H0’,
‘H0 = 0; H10’ were often seen.
On a more happy note, most candidates were able to determine the correct direction of the inequality in H1, which is < 1000.
Most candidates who defined H0 and H1 correctly generally could proceed to obtain the correct p-value and claimed that p for H0 to be rejected.
One common error that appeared was giving the final answer to 1 decimal place. Many did not adhere strictly to this requirement and simply wrote, by the T-test, > 5.870 smallest = 5.87 when the correct answer should be 5.9 to 1 decimal place. The same error was seen for candidates who used the Z-test. Many wrote smallest = 5.62 when the correct answer should be 5.7 to 1 decimal place.
(iii) Many candidates were unable to define H0 and H1 correctly and some the definitions skipped completely. It must be emphasised that defining H0 and H1 is very important in Statistical testing. One common error was writing for H0 not to be rejected.
10 / i)Let X be the weight of the boys and Y be the weight of the girls. and
Since and , thus
and . Thus
ii)
P(= P(=0.0850 (3 s.f)
iii)
P(0.264 (3 s.f)
iv) Let W be the no. of boys who weigh heavier than 46kg, out of 50 boys.
P(
W ~ B(50,0.93319)
Since n is large, np = 50*0.93319= 46.7 (>5), nq = 50*0.06681 = 3.34(<5)
~ B(50, 0.06681)
Since n is large, np = 3.34 (<5)
~ Po(3.34) approximately
P(= 0.755 (3 s.f)
Comments on Question 10:
10 i) A handful students did not leave their answers to 3 significant figures as required by the question. Some of them were unable to find the value of and because of one or more of the following errors:
a)
b)
c)
ii) Some students do not know that “thrice” mean 3 X and not 2 X.
iv) A number of students had attempted the question by “Central Limit Theorem” which is incorrect. The probable reason for them to do so is because they have mixed up approximation with “CLT”.
While attempting to check the conditions for approximation, some students tend to leave out the condition “n is large”.
11 / (i)
(ii)(a) For model A, . For model B, . (Answers to 4decimal places)
(b)Since the value of r is closer to 1 for model B, model B is an appropriate model.
(iii)For model B, . Thus .
Using the regression line may not be appropriate as the value of T is not in the data range.
(iv)Need to add the point . Since and , thus and .
Comments on Question 11:
(i)Majority of the candidates made use of graph paper to present their scatter diagram. Some candidates choose to sketch the diagram on the writing paper. However, they fail to give proper scale to the diagram and resulted in no credit. Common mistakes are interchanging of the two variables wrongly, indicating the points wrongly. Candidates are once again reminded that the regression line is NOT required in a scatter diagram.
(ii)It is shocking to see that candidates were unable to make use of the graphic calculator to arrive at the answers required. Given the allocated marks, they should be able to deduce that algebraic method is not required in this question. They were confused by the complexity of the algebraic calculations involving T2 and ln l and gave erroneous workings. Common mistakes were reading the values wrongly and failed to give answers to 4 decimal places. It is alarming to note that a considerable number of candidates decided that model A is better as it has a r value that is closer to the original r value for T and l. Candidates are reminded that a r value that has magnitude closer to 1 is a better model.
(iii)Most candidates who has identified the correct model were unable to get full credit for this part. The main reason was due to the use of the wrong regression line for the estimation. Since l is to be estimated, we should use the regression line of l on T2. Despite identifying the wrong line, they were able to recognize that T = 3 will result in extrapolation, and hence, making this method unsuitable.
(iv)This part proved to be quite a challenge to the candidates. Most of them were able to recognize that the required pointis . However, they failed to notice that the mean of l and the mean of T2 should be used instead. Those who were able to get the mean failed to give the value of T as required by the question. It is interesting to note that quite a large number of candidates make use of the regression line to find the value of , which could be easily obtained from the graphic calculator via the 2-Var Stats command.