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Date May 2013 Marks available 2 Reference code 13M.2.sl.TZ1.3
Level SL only Paper 2 Time zone TZ1
Command term Find Question number 3 Adapted from N/A

Question

In the expansion of (3x2)12(3x2)12 , the term in x5x5 can be expressed as (12r)×(3x)p×(2)q(12r)×(3x)p×(2)q .

(a)     Write down the value of pp , of qq and of rr .

(b)     Find the coefficient of the term in x5x5 .

[5]
.

Write down the value of pp , of qq and of rr .

[3]
a.

Find the coefficient of the term in x5x5 .

[2]
b.

Markscheme

(a)     p=5p=5 , q=7q=7 , r=7r=7    (accept r=5r=5)     A1A1A1     N3

[3 marks]

 

(b)     correct working     (A1)

eg   (127)×(3x)5×(2)7(127)×(3x)5×(2)7 , 792792 , 243243 , 2727 , 2463436824634368

coefficient of term in x5x5 is 2463436824634368     A1     N2

Note: Do not award the final A1 for an answer that contains xx.

[2 marks]

 

Total [5 marks]

.

p=5p=5 , q=7q=7 , r=7r=7    (accept r=5r=5)     A1A1A1     N3

[3 marks]

a.

correct working     (A1)

eg   (127)×(3x)5×(2)7(127)×(3x)5×(2)7 , 792792 , 243243 , 2727 , 2463436824634368

coefficient of term in x5x5 is 2463436824634368     A1     N2

Note: Do not award the final A1 for an answer that contains xx.

[2 marks]

 

Total [5 marks]

b.

Examiners report

Candidates frequently made reasonable attempts at both parts of the question. Those who correctly stated the values in (a) were generally successful in part (b). Many candidates offered the whole term rather than the coefficient in part (b) and lost the final mark. Some candidates appeared to have misread the order of the variables, stating that p=7p=7 (instead of r=7r=7 ), q=5q=5 (instead of p=5p=5) and r=5r=5 or 77 (instead of q=5q=5 or 77). A large number of candidates did not make the connection between parts (a) and (b).

.

Candidates frequently made reasonable attempts at both parts of the question. Those who correctly stated the values in (a) were generally successful in part (b). Many candidates offered the whole term rather than the coefficient in part (b) and lost the final mark. Some candidates appeared to have misread the order of the variables, stating that p=7p=7 (instead of r=7r=7 ), q=5q=5 (instead of p=5p=5) and r=5r=5 or 77 (instead of q=5q=5 or 77). A large number of candidates did not make the connection between parts (a) and (b).

a.

Candidates frequently made reasonable attempts at both parts of the question. Those who correctly stated the values in (a) were generally successful in part (b). Many candidates offered the whole term rather than the coefficient in part (b) and lost the final mark. Some candidates appeared to have misread the order of the variables, stating that p=7p=7 (instead of r=7r=7 ), q=5q=5 (instead of p=5p=5) and r=5r=5 or 77 (instead of q=5q=5 or 77). A large number of candidates did not make the connection between parts (a) and (b).

b.

Syllabus sections

Topic 1 - Algebra » 1.3 » The binomial theorem: expansion of (a+b)n(a+b)n, nN .
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