GCE A Level 数学统计真题

FURTHER MATHEMATICS

Paper 9231/01 Paper 1

General comments

Only the occasional script of outstanding quality and a small number of good quality scripts were received in response to this examination. There were a significant number of very poor scripts. The better candidates presented their work well, while the work of weaker candidates was untidy with many incomplete or deleted attempts at solutions. Many candidates were unable to differentiate or integrate accurately and algebraic technique was often weak. Numerical answers were frequently inaccurate.

There was no evidence to suggest that candidates had any difficulty completing the paper in the time allowed. The vast majority of candidates made substantial attempts at nearly all the questions. Once again there were very few misreads and this year there were few rubric infringements. The occasional candidate, who could not fully attempt either of the two alternatives in the final question, would submit two partial solutions.

It was felt that candidates had been well prepared for all parts of the syllabus and had a sound knowledge of certain areas. Induction, linear spaces and complex numbers, however, still remained problematical for many candidates. Recent improvements in vector work seem to have been maintained.

Comments on specific questions

Question 1

Many candidates were able to gain the first mark by producing the result t t t a S d 1422

022∫+=π. The

number of candidates who could successfully perform the integration, either directly, or using a suitable substitution, was disappointingly small. The fact that the answer was given in the question meant that full working was required.

Question 2

There were many complete and accurate answers to this question. Almost all candidates were able to

establish the result 1

13)1(32+?=++n n n n n . Many were then able to use the method of differences to show that 11131)1(13131)1(32+=+??????+?=??????++∑

N N n n N n n n , from which they were able to deduce the sum to infinity correctly. Answers : ()131311+?N N , 3

1. 9231 Further Mathematics November 2007

1

Question 3

This proved to be a most difficult question for a large number of candidates.

Often only one or two marks were gained for stating inductive hypothesis and/or demonstrating that the result was true for n = 1.

There seemed to be much confusion in the minds of candidates over what constituted a polynomial in x of degree n , so few were able to show that H k is true ? H k +1 is true.

The most concise solutions stated that )(e )(e 2)(e d d 222x P x P x x P x x k x k x ′+=?????

? and explained that the first term was the product of 2

e x and a polynomial in x o

f degree k + 1, while the second term was the product of 2e x and a polynomial in x of degree k – 1, thus producin

g 2e x P k + 1(x ). Occasionally a candidate, who did get to this stage, did not write a full conclusion and so dropped the final mark.

Question 4

The proof of the first result was done well by many candidates. There were two common methods, both of which required forming a pair of simultaneous equations.

Firstly: α is a root of the equation α3 – 8α2 + 5 = 0 hence α2(α – 8) + 5 = 0.

Also α + β + γ = 8 hence α ? 8 = ?(β + γ). The result is then obvious.

Secondly: αβ + βγ + γα = 0 hence α(β + γ) = –βγ.

Also αβγ = ?5 hence βγ = α

5?. The result is again obvious.

Two strategies were evident in the second part of the question, but fewer candidates were able to complete this part of the question. Since the product of the roots is negative, there are one or three negative roots. Call a negative root α