Candidates should be aware of the fact that they are responsible for providing the examiners with the process by which they obtained their answer. This is always to the benefit of the candidate as marks are awarded for correct elements of the solution and not deducted for mistakes made. Many candidates graphed x > −4 incorrectly, marking the segment of the number line to the left of −4 instead of the segment to the right.
Candidates were required to show at least some understanding of the process of solving simultaneous equations.
The grading scheme accepted the answer 20−−04 or an equivalent numerical expression to assign a grade. Most candidates recognized the need to substitute in the equation of the line found in part (c). Alternative approaches included showing that the gradient of the line through M and (0,−3) was 2, showing that the equation of that line was the same as that obtained in part (c), or comparing their equation from (c) with general form y=mx+b.
The marking scheme awarded a mark for correctly substituting into the circle equation from part (f) or for correctly using any of the other approaches described above.
Those who tried to solve both the equation of the circle and the equation of the line P Q had only limited success. The product rule form dxd(uv) =uv[dxd(lnu)+dxd(lnv)] was used by several centers. Most candidates who answered this part used the table of standard integrals to obtain 15tan 5x(+C).
The biggest mistake involved misusing the 2 in the numerator, leading to answers like 12ln(x+ 3) or ln(x+ 3).
A few candidates claimed without support that m = 2 and proceeded with the correct tangent equation, y= 2x−2. Arithmetic errors by candidates who used the correct and appropriate formula were treated more leniently and almost every such candidate scored at least one mark in this section. Some candidates visualized the depot placed somewhere other than at the start of the highway.
There was a problem for the candidates who thought of the problem in terms of round trips.
Most students who completed the tree diagram knew how to multiply the two probabilities along the W W branch to answer this part of the question. Some candidates did not answer this part of the question and went directly to part (iii). Although the correct wording for this part was included in their answer to part (iii), a mark for this part could not be given unless the candidate clearly indicated the link to part (ii).
Candidates had to substitute their value of k and t = 12 into the formula to obtain the first mark.
After discovering that the maximum occurs when t = 3, a statement was required that the maximum value is 256. Partial marks were awarded for finding the correct derivative, solving when it is zero, and deriving the maximum value. Common errors included not recognizing t = 0 as a possible solution, obtaining the wrong derivative (32t − 4t3 and 36t − 3t3 occurred regularly), using solutions for the second derivative instead of the first, and simply stating the time when the maximum occurred as the maximum value calculation .
A worrying number of candidates failed to see the need to use calculus to determine when a maximum occurred and simply tabulated a table of values of N for t = 0, 1, . This required understanding that the point where the rate of change was greatest corresponds to the inflection point on the graph. Only the best candidates were able to score full marks here and roughly half failed to score any marks.
Candidates were expected to show units on each axis and a curve starting with a stationary minimum point on the vertical (N) axis, leading to an inflection at t. 3 followed by a downward concave section with a maximum at t = 3 that sweeps down to the horizontal axis at t= 5. Section grades are given for starting the curve from the exact point with the exact concavity leading to refraction and for a maximum turning point followed by no change in concavity.
Many appeared to be unaware of any relationship between maximum and inflection found here and their answers to parts (iii) and (iv). Overall, attention must be drawn to the need for a clear, smooth curve when sketching a curve.
Many candidates tackled this part of the question in isolation from the previous parts, working through the whole calculation again. Examiners strongly recommend using a fine-point pencil for each curve sketch so that corrections can be made without having to redraw the entire sketch. This part is particularly well done by a number of approaches, including the use of point tables.
Many candidates who scored zero in the rest of question 7 managed to score points here. Unsurprisingly, candidates made the question more difficult by breaking up the area into endless permutations, usually with no supporting diagram. A successful strategy used by candidates uncomfortable with a region below the x-axis was to translate both curves up 2 units before integration.
Very few candidates who received a negative answer indicated that they understood that this was not possible.
Most candidates who attempted this part of the question were able to demonstrate an understanding that the initial velocity refers to t = 0. A consequence of this approach to marking was that candidates who showed their work scored much better than those who didn't. A second mark was awarded to candidates who recognized that the particle came to rest at t = 0.5 sec. and then found its displacement at that time.
The third mark was given to those who correctly combined this information with understanding. Candidates who did this were still able to gain the first two grades provided they showed their replacement line, but were unable to gain the third grade. Candidates who tried to find the correct answer, 4 + 6 ln(1.5)− 3 ln 4 were generally less successful than those who worked with numerical approximations found using a calculator.
It was pleasing to see that many candidates took advantage of the two entry points in this section. Candidates who could not show that the shaded area was A = 117x− 32x2 in part (i) were still able to get full marks in part (ii) using the given result. Those who were unable to acquire this sign usually mistakenly assumed that the small rectangle was a square.
Most candidates who were able to find an expression for area in terms of x and y were then only able to derive the expression in terms of x by substituting y = 117−x. To achieve the 3 marks for this part a candidate had to solve A = 0, demonstrate that a maximum occurred at this stationary point and show that this point corresponded to toy = 2x.
Most candidates did not appreciate the fact that they could use their existing sketch, with the addition of the line=x−2. Many candidates were able to achieve at least one mark and understood what the question required. Candidates were more likely to score in this part than any other parts of Question 9.
Candidates were less successful when they started with the equation 62 =x2+22−2·x·2, as it offered additional opportunities for error in factoring cosα. About a third of candidates who were successful in part (i) scored here, with grades 1 or 2 being equally common. Candidates who gained 1 mark usually marked the similar form of the answer in part (i) and started by writing sinα.
However, examiners were not always sure that candidates understood why it was appropriate to use sinα in the cosine rule here, and the other mark awarded for showing the equality of sinα and cos(90◦ − α) was not awarded unless convincing evidence was provided that the candidate had made this connection. Many were able to eliminate negative solutions, but elimination ofx= 4 was much rarer. Those who earned the grade usually showed that substituting x= 4 into the sinα or cosα formulas produces an angle of either 0 or 180 degrees.
Many candidates wrongly reasoned that the length of BC must be less than the length P C of the triangle P BC. A few candidates realized that if x = 4, then the sides of the 'triangle' would be 2, 4 and 6, violating the triangle inequality.
A minority of candidates attempted the question using the relationship between roots and coefficients. Many candidates assumed that they were required to find the equation of the tangent and wasted considerable time in the process. A small number of candidates used first principles by finding the intercept of the tangent to the curve y =f(r) with the r-axis.
Based on the context of the question, the testers decided that this was acceptable in this case. Most of the candidates who undertook this work did not complete their work, allegedly due to lack of time. One point was awarded for substituting u = sinx, du = cosx dx, or equivalently, an answer of the form (sinx)n; the second mark was given for the correct values of the constant and n.
The second mark was for the correct center and radius, and the third mark to indicate that the inside of the circle was intended. In the main the graph of y = |f(x)| was well drawn, although a number of candidates mishandled the portions af(x) which were to the left of the y-axis or below the x-axis. Many candidates drew only the top half of the chart, thus earning only one mark.
This question proved challenging for the average candidate, but almost 10% of candidates were able to score full marks. A minority of candidates noticed that they could use the sum of the roots and quickly found Q. b). A large number of candidates tried to explain the left side of the inequality by considering lim.
This question was found very difficult by the vast majority of candidates. Many correct methods have been used to show that the nested summation of the left-hand side is equal to the expression on the right-hand side. Many did not understand the correct order of operations when they took the imaginary parts of the terms.
Unit (Additional) and 3/4 Unit (Common)
Unit (Additional)
