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Date May 2022 Marks available 2 Reference code 22M.1.AHL.TZ1.10
Level Additional Higher Level Paper Paper 1 (without calculator) Time zone Time zone 1
Command term Show that Question number 10 Adapted from N/A

Question

Consider the series lnx+plnx+13lnx+, where x, x>1 and p, p0.

Consider the case where the series is geometric.

Now consider the case where the series is arithmetic with common difference d.

Show that p=±13.

[2]
a.i.

Hence or otherwise, show that the series is convergent.

[1]
a.ii.

Given that p>0 and S=3+3, find the value of x.

[3]
a.iii.

Show that p=23.

[3]
b.i.

Write down d in the form klnx, where k.

[1]
b.ii.

The sum of the first n terms of the series is ln1x3.

Find the value of n.

[8]
b.iii.

Markscheme

EITHER

attempt to use a ratio from consecutive terms        M1

plnxlnx=13lnxplnx  OR  13lnx=lnxr2  OR  plnx=lnx13p

 

Note: Candidates may use lnx1+lnxp+lnx13+ and consider the powers of x in geometric sequence

Award M1 for p1=13p.


OR

r=p  and  r2=13        M1


THEN

p2=13  OR  r=±13          A1

p=±13          AG

 

Note: Award M0A0 for r2=13 or p2=13 with no other working seen.

 

[2 marks]

a.i.

EITHER

since, p=13 and 13<1          R1


OR

since, p=13 and -1<p<1          R1


THEN

 the geometric series converges.          AG


Note: Accept r instead of p.
Award R0 if both values of p not considered.

 

[1 mark]

a.ii.

lnx1-13  =3+3           (A1)

lnx=3-33+3-33  OR  lnx=3-3+3-1  lnx=2          A1

x=e2          A1

 

[3 marks]

a.iii.

METHOD 1

attempt to find a difference from consecutive terms or from u2          M1

correct equation          A1

plnx-lnx=13lnx-plnx  OR  13lnx=lnx+2plnx-lnx


Note:
Candidates may use lnx1+lnxp+lnx13+ and consider the powers of x in arithmetic sequence.

Award M1A1 for p-1=13-p

 

2plnx=43lnx  2p=43          A1

p=23          AG

 

METHOD 2

attempt to use arithmetic mean u2=u1+u32          M1

plnx=lnx+13lnx2          A1

2plnx=43lnx  2p=43          A1

p=23          AG

 

METHOD 3

attempt to find difference using u3          M1

13lnx=lnx+2d  d=-13lnx

 

u2=lnx+1213lnx-lnx  OR  plnx-lnx=-13lnx          A1

plnx=23lnx          A1

p=23          AG

 

[3 marks]

b.i.

d=-13lnx       A1

 

[1 mark]

b.ii.

METHOD 1

Sn=n22lnx+n-1×-13lnx

attempt to substitute into Sn and equate to ln1x3           (M1)

n22lnx+n-1×-13lnx=ln1x3

ln1x3=-lnx3=lnx-3           (A1)

=-3lnx           (A1)

correct working with Sn (seen anywhere)           (A1)

n22lnx-n3lnx+13lnx  OR  nlnx-nn-16lnx  OR  n2lnx+4-n3lnx

correct equation without lnx          A1

n273-n3=-3  OR  n-nn-16=-3 or equivalent


Note:
Award as above if the series 1+p+13+ is considered leading to n273-n3=-3.


attempt to form a quadratic =0           (M1)

n2-7n-18=0

attempt to solve their quadratic           (M1)

n-9n+2=0

n=9          A1

 

METHOD 2

ln1x3=-lnx3=lnx-3           (A1)

=-3lnx           (A1)

listing the first 7 terms of the sequence           (A1)

lnx+23lnx+13lnx+0-13lnx-23lnx-lnx+

recognizing first 7 terms sum to 0           M1

8th term is -43lnx           (A1)

9th term is -53lnx           (A1)

sum of 8th and 9th term =-3lnx           (A1)

n=9          A1

 

[8 marks]

b.iii.

Examiners report

Part (a)(i) was well done with few candidates incorrectly using the value of p to verify rather than to 'show' the given result. In part (a)(ii) most did not consider both values of r and some did know the condition for convergence of a geometric series. Part (a)(iii) was generally well done but some had difficulty in simplifying the surd. Part (b) (i) and (ii) was generally well done. Although many completely correct answers to part b (iii) were noted, weaker candidates often made errors in properties of logarithms or algebraic manipulation leading to an incorrect quadratic equation.

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

Syllabus sections

Topic 1—Number and algebra » SL 1.3—Geometric sequences and series
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