Increases in average raw grade for both the 2/3 Unit (Common) paper and the 4 Unit (Extra) paper were the most notable features. Instead, they should make a clear note on the cover of both notebooks that part of the answer to question 7 is included in the booklet for question 5.
Following on from the previous question, they then had to calculate how much would be repaid in total over 10 years. This part involved the calculation of the interest paid, thus requiring the subtraction of $85,000 from the previous answer.
Candidates had to find the cancellation fee for each person if the trip was canceled on 6 January. Given the time in Los Angeles, candidates had to work out the time in New York, again using the information on the map indicating time zones.
Candidates had to find the cost of fitting out the bathrooms as a percentage of the total cost of the house. Many did not seem to realize that they had found the total cost of the house in part (i).
Candidates were asked to show two different ways of packing the boxes in the carton. A diagram with concentric squares was provided, and candidates were asked to calculate the fraction shaded.
Candidates were asked to draw the missing column on the bar graph using the picture from the table. Candidates were asked to estimate the number of households watching a particular show based on the fact that 30,000 surveyed households watched it.
Many candidates knew that they should use the numbers 0.3 and 4 but did not know what to do. Regardless of the correctness of their answers to part (c) (i), a large number of candidates knew what was required in this part.
Common errors included confusion over whether to multiply or add the probabilities along the branches, and then whether to add or multiply the results of each branch.
Candidates did not know where the cumulative frequency polygon should start and often drew it using the centers or upper left corners of the columns. Some candidates who correctly drew the polygon in the upper right corners of the columns had problems with two columns of the same height corresponding to 21◦ and 22◦ .
Many mistakenly believed that a circle has e=1, leading them to the wrong conclusion from a correct value for e. Many had no idea what a light-year meant, while others made mistakes such as using the wrong value for the number of days in a year.
This was reasonably well done, with many candidates realizing that the total value of the prizes must be $500. This was reasonably well done, although a lack of understanding of the difference between winning and collecting led many candidates to give the answer $20.
This question was well done, in part because 34◦ was emphasized in the wording of the question. Errors arose from not being able to determine the relevant angle of 132◦, using 1◦ = 60 nautical miles (which only applies to great circles), and omitting the 2 of theπ in the formula for the.
Other candidates incorrectly used 6400 km as the radius of the small circle. b) This part included the use of a premium table where rates were given in $ per $1000. This was poorly done, with only a small majority of candidates answering this part correctly. This part involved using an income tax table, with the added complication of the Medicare fee.
Although the mark was awarded even if candidates omitted the Medicare levy from their calculations, only 60% of candidatures were awarded this mark.
Some candidates were careless when they read the width of the bedroom from the plan and instead used the length of the bathroom (2150 mm) in their calculations. Very poor answers were typical in this part, with most of the candidates misinterpreting the question. Candidates who did not interpret vertical section correctly usually gave a front, side or top view of the house.
The extremely poor response to this part was largely due to misinterpretation of the question.
A large part of the candidacy did not include the constant of integration in their answers. But many tried to treat 1/x in the same way, failing to recognize this as one of the standard integrals and apparently unaware of the problem with their answer, x0/0. On the other hand, some candidates correctly found lnx as the integral of the 1/x term, but tried to treat the expression 1/x2 in the same way and obtained a number of incorrect answers.
A common mistake was for candidates to use the x-coordinate of B as the height of the triangle, rather than their coordinates.
Others solved the equation to determine where the derivative equaled zero, obtained x = −2, and concluded that the gradient of the normal was (always) 1/2. Unsupported statements claiming that the slope of the tangent was 1 or that the slope of the normal was −1 were common. Some candidates misread the question and found the equation of the tangent or just the gradient of the normal.
This part was not done well as a significant number of candidates were unable to correctly state and use some form of the cosine rule.
Other errors included using the incorrect formula Tn = a(n − 1)d or the inappropriate formula Tn =arn−1 or Sn= n2(a+l). Again, most candidates used the correct formula for Tn to estimate the number of rows required. Many candidates wrote both coordinates of point B and not just the x coordinate as asked in the question.
A number of candidates seemed to think that the question required them to find the x-coordinate of the turning point that occurs at (2,4).
Many candidates who scored zero in this part did not seem to know the meaning of the word inequality and gave ordered pairs as their answer. Testing the nature of the stationary points was well done by candidates using the second derivative test, but was often handled poorly when approached by testing the sign of the first derivative on either side of the stationary point. Some candidates seemed to think that the instruction asking them to determine the nature of the stationary points meant that they should find turning points.
Sketches extending to the left of the y-axis were accepted, with markers believing that this restriction had been placed on the sketch to prevent candidates from thinking they had to find the coordinates of the point where the curve crosses the x-axis.
A smaller percentage of candidates did not realize that mass should decrease with time. Most candidates who came up with a negative numerical value for the fortress gave |t| as their answer. Most applications tried to prove that opposite angles are equal or opposite sides are parallel.
Many of the first group failed to achieve full marks when they named angles incorrectly, or only proved.
Too many candidates used properties of the parallelogram OABC to prove that OABC was indeed a parallelogram or that the triangles OAB and OABC were congruent. Among those who recognized the GP, there were many who made mistakes in writing an appropriate expression for the sum of the GP. Unfortunately, most did not understand that their answers to the previous parts meant that the particle moved in the positive direction for the first π/3 seconds and in the negative direction for the rest of the first π seconds.
Most of the incorrect answers were the result of a calculation of particle displacement rather than the total distance traveled.
The rating scheme assigned one grade if two of the three values in the table were correct. A single mark was awarded for correctly obtaining the product values of the three required terms, P(3.5), P(5.3) and P(5.5), or for correctly representing two of these three terms as a sum. One point was awarded for correctly obtaining the values of the three required terms, P(0.3), P(3.3) and P(5.3), in product form, or for correctly expressing two of these three terms as a sum.
The marking scheme awarded one mark for correctly identifying x = 5 as the x-coordinate of the local minimum and one mark for correctly justifying this claim.
Some candidates omitted the parentheses and were treated with doubt as to whether the following line conformed to the implicit parentheses. Arguing the nature of the stationary point by considering the sign of the second derivative was well done. Unfortunately, some candidates were unable to use any notation for the first or second derivatives other than dy.
Values such as 0, 4, or negative values of y were outside the domain of the function, which the physics problem restricted to 0< y < 3.
This part required the candidates to find the coordinates of the point that internally divides an interval in the ratio 2 : 3. The second mark was awarded for correct evaluation to give the coordinates of the point. Even fewer candidates recognized that the second term was of the form f(x)/f(x) and therefore contributed ln(2 sinx+ cosx) to the integral.
Candidates were each given one mark for the module and the argument for the complex numberα= 1 +i√. The most successful was showing that the product of the gradients of PF and FL was -1. A common mistake in the second method was the use of ln(3/2) as the height of the triangle.