Date | May 2018 | Marks available | 7 | Reference code | 18M.1.AHL.TZ1.H_7 |
Level | Additional Higher Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 1 |
Command term | Find | Question number | H_7 | Adapted from | N/A |
Question
Let y=arccos(x2)
Find dydx.
Find ∫10arccos(x2)dx.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
y=arccos(x2)⇒dydx=−12√1−(x2)2(=−1√4−x2) M1A1
Note: M1 is for use of the chain rule.
[2 marks]
attempt at integration by parts M1
u=arccos(x2)⇒dudx=−1√4−x2
dvdx=1⇒v=x (A1)
∫10arccos(x2)dx=[xarccos(x2)]10+∫101√4−x2dx A1
using integration by substitution or inspection (M1)
[xarccos(x2)]10+[−(4−x2)12]10 A1
Note: Award A1 for −(4−x2)12 or equivalent.
Note: Condone lack of limits to this point.
attempt to substitute limits into their integral M1
=π3−√3+2 A1
[7 marks]