P9.2 Iteration, cobweb & staircase

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

Examiner's comment

This question was generally well-answered, with most learners demonstrating a good grasp of the concepts. However, some common pitfalls were noted, particularly in differentiation and the inclusion of units. Part (c) involved using the given iterative formula and appeared accessible to almost all learners. In part (i), the vast majority gained the two marks in the first part with only occasional rounding inaccuracies such as giving the answer to 3 significant figures instead of 3 decimal places or only calculating the second iteration 2 t . In part (ii) many responses did not include units in their final answer or used incorrect units. A large number also spent time writing down a list of subsequent iterations, not realising that for a one-mark question the final answer is sufficient. 14

Common mistakes

• t₂ = 8.758, t₃ = 8.478 (3 d.p., not 3 s.f.)
• repeated iteration gives 8.55 seconds (units needed)

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

Examiner's comment

In part (c) many candidates scored all three marks often with no working, just by writing down the two numerical answers required. Candidates should however be advised to show their method as they risk losing both marks. In contrast, some candidates wrote down all the iterations they thought necessary, including the working for each one. Amongst those that didn’t get the final mark, the common mistakes were, stopping at too early an iterate such as x2 or x3, stating an answer of 0.6707 instead of 0.6706 and not rounding to 4dp. There were a small number of candidates who didn’t seem to realise they needed to keep applying the iteration to find the value for alpha and instead tried to solve the equation given in part (b). Others thought 0.6 or 0.7109 should be substituted into the equation for part (b), not realising a sequence needed to be generated.

Common mistakes

• x₂ = 0.7109
• keep iterating until stable: α = 0.6706 (not 0.6707, not x₂ or x₃)
• 4 d.p.
• generate the sequence, do not substitute into the equation in (b)

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. In part (c), the majority of candidates scored the mark for a correct diagram, although some started their initial line higher up than the x-axis, which was condoned on this occasion. Several candidates used a cobweb diagram, and for others there were often lines drawn to the left of 1 x = . Only a handful of candidates put arrows on their lines, but their omission was, again, condoned.

Common mistakes

• staircase from x = 1 converging to β (cobweb is wrong)

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 (d), for many candidates who had not scored in or attempted earlier parts, was an opportunity to score full marks relatively easily. This was very well done. Occasionally, rounding 2 x to 1.50 rather than 1.502 was seen. However, those who wrote down an unrounded answer which rounded to 1.502 first were able to get the A mark. Candidates who showed the value embedded in the formula and then did not achieve 1.502 were able to get the M mark, too. A minority of candidates skipped 1.502 to go to the next term, 1.873 but this still was able to score the method mark as this was still evidence that the iterative formula had been used. The correct answer 1.968 = was widely achieved for part (b)(ii), although 1.967 was a common incorrect answer. The final mark was also lost for a small number of candidates who failed to round to 3 decimal places.

Common mistakes

• x₂ = 1.502
• β = 1.968 (3 d.p.)

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

Examiner's comment

This was the second modelling question on the paper. The given equation modelled the speed of and scored marks in the more accessible parts (a) and (c). Most candidates managed to reach the correct answer of 25 in part (a). A common incorrect answer was 10, Part (c) was the most successful part of the question and familiar to most candidates. Many were able to carry out the iteration process correctly, and write down 7.298 without any problems. In part (c) (ii), the omission of the units, “seconds”, was the most common slip.

Common mistakes

• t₂ = 26/(1 + ln 8) − 1 = 7.44, t₃ = 7.298
• repeated iteration to 7.33 seconds (units needed)

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 (b) A significant majority of candidates obtained both required answers to this part. Occasionally only x2 and x3 were found. There were just a few occasions where a candidate had misread the 1/7 in the iterative formula as ½. If the working was shown for this situation, then a method mark could still have been awarded.

Common mistakes

• x₂ = 0.3294, x₄ = 0.3398 to 4 d.p.
• misreading 1/7 as 1/2

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

Examiner's comment

Part (c) was generally well done, with most candidates getting part (i) correct. Some candidates left out part (ii) and a few made an error with their accuracy here, 1.15651 and 1.1565 being seen a few times.

Common mistakes

• x₂ = 1.13894
• x coordinate of P = 1.15650 to 5 d.p. (not 1.15651 or 1.1565)

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

• x₂ = 1.633, x₃ = 1.466, x₄ = 1.673 (3 d.p.)

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

• sequence is periodic with period 2 / oscillates between 1 and 2 (both marks)
• "non-convergent", "oscillates" or "divergent" alone gets one mark

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

Examiner's comment

interval. Fully correct solutions to part (b) were very rare. Many students assumed that this was a question on continued iteration and gave all values of x from 1x to 8x without any consideration of the demand of the question. Students were required to ''use the graph'' to show whether or not the iteration formula could be used to find an approximation for . x Hence, to satisfy this demand, a cobweb diagram starting at was required followed by 4 1 an explanation that ''it can be used'' because ''the cobweb spirals inwards towards .

Common mistakes

• draw a cobweb on Figure 2 starting at x₁ = 4 (listing iterates x₁ to x₈ scores nothing)
• conclude it CAN be used because the cobweb spirals inwards/converges to α