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Date May 2017 Marks available 7 Reference code 17M.2.SL.TZ1.S_6
Level Standard Level Paper Paper 2 Time zone Time zone 1
Command term Find Question number S_6 Adapted from N/A

Question

Let f ( x ) = ( x 2 + 3 ) 7 . Find the term in x 5 in the expansion of the derivative, f ( x ) .

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

METHOD 1 

derivative of f ( x )     A2

7 ( x 2 + 3 ) 6 ( x 2 )

recognizing need to find x 4 term in ( x 2 + 3 ) 6 (seen anywhere)     R1

eg 14 x  (term in  x 4 )

valid approach to find the terms in ( x 2 + 3 ) 6     (M1)

eg ( 6 r ) ( x 2 ) 6 r ( 3 ) r ,   ( x 2 ) 6 ( 3 ) 0 + ( x 2 ) 5 ( 3 ) 1 + , Pascal’s triangle to 6th row

identifying correct term (may be indicated in expansion)     (A1)

eg 5th term,  r = 2 ,   ( 6 4 ) ,   ( x 2 ) 2 ( 3 ) 4

correct working (may be seen in expansion)     (A1)

eg ( 6 4 ) ( x 2 ) 2 ( 3 ) 4 ,   15 × 3 4 ,   14 x × 15 × 81 ( x 2 ) 2

17010 x 5     A1     N3

METHOD 2

recognition of need to find x 6 in ( x 2 + 3 ) 7 (seen anywhere) R1 

valid approach to find the terms in ( x 2 + 3 ) 7     (M1)

eg ( 7 r ) ( x 2 ) 7 r ( 3 ) r ,   ( x 2 ) 7 ( 3 ) 0 + ( x 2 ) 6 ( 3 ) 1 + , Pascal’s triangle to 7th row

identifying correct term (may be indicated in expansion)     (A1)

eg 6th term, r = 3 ,   ( 7 3 ) ,  ( x 2 ) 3 ( 3 ) 4

correct working (may be seen in expansion)     (A1)

eg ( 7 4 ) ( x 2 ) 3 ( 3 ) 4 ,   35 × 3 4

correct term     (A1)

2835 x 6

differentiating their term in x 6     (M1)

eg ( 2835 x 6 ) ,  (6)(2835 x 5 )

17010 x 5     A1     N3

[7 marks]

Examiners report

[N/A]

Syllabus sections

Topic 5 —Calculus » SL 5.3—Differentiating polynomials, n E Z
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Topic 1—Number and algebra » SL 1.9—Binomial theorem where n is an integer
Topic 1—Number and algebra
Topic 5 —Calculus

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