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Date November 2021 Marks available 4 Reference code 21N.1.SL.TZ0.8
Level Standard Level Paper Paper 1 (without calculator) Time zone Time zone 0
Command term Show that Question number 8 Adapted from N/A

Question

Consider the function fx=ax where x, a and x>0, a>1.

The graph of f contains the point 23,4.

Consider the arithmetic sequence log827 , log8p , log8q , log8125 , where p>1 and q>1.

Show that a=8.

[2]
a.

Write down an expression for f-1x.

[1]
b.

Find the value of f-132.

[3]
c.

Show that 27, p, q and 125 are four consecutive terms in a geometric sequence.

[4]
d.i.

Find the value of p and the value of q.

[5]
d.ii.

Markscheme

f23=4   OR   a23=4             (M1)

a=432   OR   a=2232   OR   a2=64   OR   a3=2                 A1

a=8                 AG

 

[2 marks]

a.

f-1x=log8x                 A1


Note:
Accept f-1x=logax.
         Accept any equivalent expression for f-1 e.g. f-1x=lnxln8.

 

[1 mark]

b.

correct substitution                 (A1)

log832   OR   8x=3212

correct working involving log/index law                 (A1)

12log832   OR   52log82   OR   log82=13   OR   log2252   OR   log28=3   OR   ln252ln23   OR   23x=252

f-132=56                 A1

 

[3 marks]

c.

METHOD 1

equating a pair of differences               (M1)

u2-u1=u4-u3=u3-u2

log8p-log827=log8125-log8q

log8125-log8q=log8q-log8p

log8p27=log8125q, log8125q=log8qp           A1A1

p27=125q  and  125q=qp           A1

27, p, q and 125 are in geometric sequence           AG


Note:
If candidate assumes the sequence is geometric, award no marks for part (i). If r=53 has been found, this will be awarded marks in part (ii).

 

METHOD 2

expressing a pair of consecutive terms, in terms of d               (M1)

p=8d×27 and q=82d×27   OR   q=82d×27 and 125=83d×27

two correct pairs of consecutive terms, in terms of d                 A1

8d×2727=82d×278d×27=83d×2782d×27  (must include 3 ratios)                 A1

all simplify to 8d                 A1

27, p, q and 125 are in geometric sequence           AG

 

[4 marks]

d.i.

METHOD 1 (geometric, finding r)

u4=u1r3   OR   125=27r3                 (M1)

r=53  (seen anywhere)                 A1

p=27r   OR   125q=53                 (M1)

p=45, q=75       A1A1

 

METHOD 2 (arithmetic)

u4=u1+3d   OR   log8125=log827+3d                 (M1)

d=log853  (seen anywhere)                 A1

log8p=log827+log853   OR   log8q=log827+2log853                 (M1)

p=45, q=75       A1A1

 

METHOD 3 (geometric using proportion)

recognizing proportion                 (M1)

pq=125×27   OR   q2=125p   OR   p2=27q

two correct proportion equations                 A1

attempt to eliminate either p or q                 (M1)

q2=125×125×27q   OR   p2=27×125×27p

p=45, q=75       A1A1

 

[5 marks]

d.ii.

Examiners report

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

Syllabus sections

Topic 1—Number and algebra » SL 1.2—Arithmetic sequences and series
Show 82 related questions
Topic 1—Number and algebra » SL 1.3—Geometric sequences and series
Topic 2—Functions » SL 2.5—Composite functions, identity, finding inverse
Topic 1—Number and algebra » SL 1.7—Laws of exponents and logs
Topic 2—Functions » SL 2.9—Exponential and logarithmic functions
Topic 1—Number and algebra
Topic 2—Functions

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