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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(πarcsinx)0x10otherwise,where k is a positive constant.

Given that y=(x22)arcsinx(14)arcsinx+(x4)1x2, show that

State the mode of X.

[1]
a.

Find arcsinxdx.

[3]
b.i.

Hence show that k=22+π.

[3]
b.ii.

dydx=xarcsinx.

[4]
c.i.

E(X)=3π4(π+2).

[5]
c.ii.

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]

a.

attempt at integration by parts      (M1)

dudx=11x2,dv=dx

=xarcsinxxdx1x2    A1

=xarcsinx+1x2(+c)    A1

[3 marks]

b.i.

10(πarcsinx)dx=[πxxarcsinx1x2]10   A1

=(ππ20)(001)=π2+1

=π+22    A1

10k(πarcsinx)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]

b.ii.

attempt to use product rule to differentiate    M1

dydx=xarcsinx+x221x2141x2x241x2+1x24  A2

Note: Award A2 for all terms correct, A1 for 4 correct terms.

=xarcsinx+2x241x2141x2x241x2+1x241x2    A1

Note: Award A1 for equivalent combination of correct terms over a common denominator.

=xarcsinx    AG

[4 marks]

c.i.

E(X)=k10x(πarcsinx)dx    M1

=k10(πxxarcsinx)dx

=k[πx22x22arcsinx+14arcsinxx41x2]10      A1A1

Note: Award A1 for first term, A1 for next 3 terms.

=k[(π2π4+π8)(0)]      A1

=(22+π)3π8      A1

=3π4(π+2)    AG

[5 marks]

c.ii.

Examiners report

[N/A]
a.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
c.i.
[N/A]
c.ii.

Syllabus sections

Topic 4—Statistics and probability » AHL 4.14—Properties of discrete and continuous random variables
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Topic 4—Statistics and probability

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