Date | May 2017 | Marks available | 7 | Reference code | 17M.2.sl.TZ1.6 |
Level | SL only | Paper | 2 | Time zone | TZ1 |
Command term | Find | Question number | 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
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)6−r(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)7−r(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]