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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)=(x2+3)7. Find the term in x5 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(x2+3)6(x2)

recognizing need to find x4 term in (x2+3)6 (seen anywhere)     R1

eg14x (term in x4)

valid approach to find the terms in (x2+3)6     (M1)

eg(6r)(x2)6r(3)r, (x2)6(3)0+(x2)5(3)1+, Pascal’s triangle to 6th row

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

eg5th term, r=2, (64), (x2)2(3)4

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

eg(64)(x2)2(3)4, 15×34, 14x×15×81(x2)2

17010x5     A1     N3

METHOD 2

recognition of need to find x6 in (x2+3)7 (seen anywhere) R1 

valid approach to find the terms in (x2+3)7     (M1)

eg(7r)(x2)7r(3)r, (x2)7(3)0+(x2)6(3)1+, Pascal’s triangle to 7th row

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

eg6th term, r=3, (73), (x2)3(3)4

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

eg(74)(x2)3(3)4, 35×34

correct term     (A1)

2835x6

differentiating their term in x6     (M1)

eg(2835x6), (6)(2835x5)

17010x5     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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