Date | November 2021 | Marks available | 1 | Reference code | 21N.1.SL.TZ0.8 |
Level | Standard Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 0 |
Command term | Write down | Question number | 8 | Adapted from | N/A |
Question
Consider the function f(x)=axf(x)=ax where x, a∈ℝ and x>0, a>1.
The graph of f contains the point (23, 4).
Consider the arithmetic sequence log8 27 , log8 p , log8 q , log8 125 , where p>1 and q>1.
Show that a=8.
Write down an expression for f-1(x).
Find the value of f-1(√32).
Show that 27, p, q and 125 are four consecutive terms in a geometric sequence.
Find the value of p and the value of q.
Markscheme
f(23)=4 OR a23=4 (M1)
a=432 OR a=(22)32 OR a2=64 OR 3√a=2 A1
a=8 AG
[2 marks]
f-1(x)=log8 x A1
Note: Accept f-1(x)=loga x.
Accept any equivalent expression for f-1 e.g. f-1(x)=ln xln 8.
[1 mark]
correct substitution (A1)
log8 √32 OR 8x=3212
correct working involving log/index law (A1)
12log8 32 OR 52log8 2 OR log8 2=13 OR log2 252 OR log2 8=3 OR ln 252ln 23 OR 23x=252
f-1(√32)=56 A1
[3 marks]
METHOD 1
equating a pair of differences (M1)
u2-u1=u4-u3(=u3-u2)
log8 p-log8 27=log8 125-log8 q
log8 125-log8 q=log8 q-log8 p
log8(p27)=log8(125q) , log8(125q)=log8(qp) 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]
METHOD 1 (geometric, finding r)
u4=u1r3 OR 125=27(r)3 (M1)
r=53 (seen anywhere) A1
p=27r OR 125q=53 (M1)
p=45, q=75 A1A1
METHOD 2 (arithmetic)
u4=u1+3d OR log8 125=log8 27+3d (M1)
d=log8(53) (seen anywhere) A1
log8 p=log8 27+log8(53) OR log8 q=log8 27+2 log8(53) (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]