P9.1 Change of sign

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Many candidates were successful across all parts of this question demonstrating a good understanding of the Newton Raphson method, but this question did start to discriminate between the least confident and those who were more careful in their processing. Part (a) was attempted by almost all candidates, with the vast majority making good progress, able to find the values of f(3.6) and f(3.7) correctly and observe a sign change. A conclusion was required for the accuracy mark and had to include a reference to a sign change, continuity of the function and mention or indication of a root lying in the interval, as well as both values being correct to 1 significant figure. It appears that reference to continuity is included more frequently by candidates but there was still a good proportion of candidates who did not mention it or who made incorrect statements involving it. Common incorrect statements included “change in sign therefore continuous”, “the interval is continuous” and “ x is continuous”. Radians were required and most candidates recognised this, with very few candidates working in degrees.

Common mistakes

• radians
• f(3.6) = −0.686, f(3.7) = 0.211
• need change of sign, f continuous and a conclusion ("change in sign therefore continuous" or "the interval is continuous" score A0)

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This question brought together a variety of topics and was appropriately placed at the end of the paper. This was very good at discriminating between candidates at the higher end, but it was also pleasing to see a good number of weaker candidates who were still able to attempt later parts such as part (c) and part (d) to a pleasing degree of success. Part (e) was the final part on the paper and for many this resulted in no marks. Most of the candidates who attempted it found a suitable interval but substituted the values in the iterative formula instead of a valid equation set equal to zero. Many candidates wrote the required conclusion with no supporting evidence in an attempt to gain a mark. Some candidates, who used a valid function correctly, failed to score the final mark for not commenting that their function was continuous. A significant proportion of candidates attempted an iterative process with a different iterative formula, including attempting the Newton-Raphson method. The question required used of a suitable interval and so these methods scored no marks. Overall, whilst this question was demanding, there were some very impressive solutions demonstrating an excellent grasp of calculus, strong algebraic skills, and the ability to provide a rigorous argument to prove given answers and results.

Common mistakes

• use h(x) = (2e³ˣ − 4)/(e³ˣ + 4) − x on [0.4315, 0.4325]
• state change of sign and that h is continuous
• conclude α = 0.432

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In part (b), most candidates only referred to the change of sign and didn’t mention continuity. Occasionally a candidate wanted to allude to continuity but was unable to articulate this correctly. For example incorrect wording including referring to the graph as “constant”. Some candidates who referred to continuity used incorrect statements such as “because x is continuous” or “because the interval (or data) is continuous”.

Common mistakes

• f(4) > 0, f(5) < 0 (change of sign) AND f is continuous
• "x is continuous" or "the interval is continuous" scores 0

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

Aspects of this question proved to be more demanding than expected, but overall good marks were scored by candidates with most achieving over half marks in this question. Part (c) Whilst many candidates realised that a sign change of a function between 0.3405 and 0.3415 was required, it was also common to see an attempt at further iteration. This method was not acceptable as a proof of the location of the root. A few chose to substitute into an inappropriate function such as f (x), or the iterative formula, and some failed to specify a function at all. For those who did use (typically) the expression from (a), most selected correct values to substitute in and identify the change of sign, finishing with a clear conclusion. It was very common, however, for the final mark to be lost in this part because there was no mention that the chosen function was continuous.

Common mistakes

• state the function, e.g. h(x) = 2x³ − 4x² + 7x − 2, and evaluate at 0.3405 and 0.3415
• further iteration is not a proof
• must say change of sign, function continuous, and conclude α = 0.341
• continuity statement often missing

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

In general, Q11 discriminated well between candidates of all abilities. Q11(a) was demanding with most candidates scoring no more than one mark. Q11(b), Q11(c) and Q11(d) proved accessible to candidates of all abilities. In Q11(a), most candidates took logs, as advised by the question, and many obtained either ln ln y x x or log log . y x x At this point a significant number of candidates made no further creditable progress. Some candidates used a complete method of applying the product rule on x and implicit differentiation to obtain a correct ln x Most of these candidates then applied d y 1 d y x 1 y x e . x 0 to obtain Some 1 ln . x d d x log ’ method to obtain candidates, who received partial credit, used a ‘ after 0.1, 10 1 d y x finding A few candidates used logarithms without reference to a 1 log . x 10 d y x base. Those who proceeded to obtain received partial credit and those who 0.1 1 e x obtained were allowed full marks. There were some alternative methods employed by a few candidates in Q11(a). A few d y y x ln e x x x ln x x y x candidates rewrote as and applied to give a (1 ln )e 0 x d 1 e . x Others differentiated both sides of the equation ln ln y x x correct with 1 e . x respect to y and usually proceeded to find A very few candidates applied the ln y , x quotient rule to the equation but were rarely successful with this method. ln x xx x x xx or 2 1.6. In Q11(b), most candidates evaluated either at both and 1.5 Many candidates compared their values with either 2 or 0 (depending on whether they xx xx or 2 ). Although many of these concluded that 1.5 1.6, evaluated only a few made any reference to the curve being continuous in this interval. A few candidates, x who made no creditable progress, used values of x outside the given range, e.g. 1.4 x x 1.6. and Very few candidates used values within the range e.g. and 1.51 x 1.59, which meant that they were able to earn both marks. x 1.673, 4x In Q11(c), most candidates found a correct although a few did not state 4 , x x and 3x as part of their correct to 3 decimal places. Some wrote down the iterates 1 2 4x with no intermediate work. solution, while others stated In Q11(d), some candidates gave correct descriptions such as ‘sequence oscillates between 1 and 2’ or ‘the sequence is periodic with period 2’. Other descriptions such as ‘alternates between 1 and 2’, ‘keep getting 1 and 2’ or ‘fluctuates between 1 and 2’ were condoned. Some credit was given for partial explanations such as ‘sequence is non-convergent’, ‘sequence oscillates’ or ‘sequence is divergent’.

Common mistakes

• 1.5^1.5 = 1.84 < 2 and 1.6^1.6 = 2.12 > 2 (or xˣ − 2 change of sign)
• must state C is continuous and conclude

Mark scheme (whole question; this part is marked with the orange bar)

Examiner's comment

This was perhaps the most poorly attempted question on the whole of the paper. Part (a) should have been familiar to centres but one which required the students to set up their own function, usuallyf( ) before substituting in the x values of 3 and 4. This was 2ln(8 ) , x x x well done, but also quite commonly, the only mark gained in the question. To show that a student needed two correct calculations, give a valid reason, which required both 3 4, a change in sign and a mention of continuity, as well as giving a brief deduction. Explanations rarely mentioned the fact that the function needed to be continuous in this interval. Fully correct solutions to part (b) were very rare. Many students assumed that this

Common mistakes

• set up f(x) = 2ln(8 − x) − x
• f(3) = 0.22 and f(4) = −1.23
• must state change of sign AND f continuous on [3, 4] and conclude – continuity rarely mentioned