Date | November 2020 | Marks available | 6 | Reference code | 20N.2.AHL.TZ0.H_4 |
Level | Additional Higher Level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Find | Question number | H_4 | Adapted from | N/A |
Question
Find the term independent of x in the expansion of 1x3(13x2-x2)9.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
use of Binomial expansion to find a term in either (13x2-x2)9, (13x73-x232)9, (13-x32)9, (13x3-12)9 or (2-3x3)9 (M1)(A1)
Note: Award M1 for a product of three terms including a binomial coefficient and powers of the two terms, and A1 for a correct expression of a term in the expansion.
finding the powers required to be 2 and 7 (M1)(A1)
constant term is C29×(13)2×(-12)7 (M1)
Note: Ignore all x’s in student’s expression.
therefore term independent of x is -132 (=-0.03125) A1
[6 marks]