Date | May 2019 | Marks available | 3 | Reference code | 19M.1.AHL.TZ2.H_10 |
Level | Additional Higher Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 2 |
Command term | Find | Question number | H_10 | Adapted from | N/A |
Question
The random variable XX has probability density function ff given by
f(x)={k(π−arcsinx)0⩽x⩽10otherwise,where k is a positive constant.
Given that y=(x22)arcsinx−(14)arcsinx+(x4)√1−x2, show that
State the mode of X.
Find ∫arcsinxdx.
Hence show that k=22+π.
dydx=xarcsinx.
E(X)=3π4(π+2).
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
mode is 0 A1
[1 mark]
attempt at integration by parts (M1)
dudx=1√1−x2,dv=dx
=xarcsinx−∫xdx√1−x2 A1
=xarcsinx+√1−x2(+c) A1
[3 marks]
1∫0(π−arcsinx)dx=[πx−xarcsinx−√1−x2]10 A1
=(π−π2−0)−(0−0−1)=π2+1
=π+22 A1
1∫0k(π−arcsinx)dx=1 (M1)
Note: This line can be seen (or implied) anywhere.
Note: Do not allow FT A marks from bi to bii.
k(π+22)=1
⇒k=22+π AG
[3 marks]
attempt to use product rule to differentiate M1
dydx=xarcsinx+x22√1−x2−14√1−x2−x24√1−x2+√1−x24 A2
Note: Award A2 for all terms correct, A1 for 4 correct terms.
=xarcsinx+2x24√1−x2−14√1−x2−x24√1−x2+1−x24√1−x2 A1
Note: Award A1 for equivalent combination of correct terms over a common denominator.
=xarcsinx AG
[4 marks]
E(X)=k1∫0x(π−arcsinx)dx M1
=k1∫0(πx−xarcsinx)dx
=k[πx22−x22arcsinx+14arcsinx−x4√1−x2]10 A1A1
Note: Award A1 for first term, A1 for next 3 terms.
=k[(π2−π4+π8)−(0)] A1
=(22+π)3π8 A1
=3π4(π+2) AG
[5 marks]