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Date May 2019 Marks available 5 Reference code 19M.2.AHL.TZ1.H_11
Level Additional Higher Level Paper Paper 2 Time zone Time zone 1
Command term Show that Question number H_11 Adapted from N/A

Question

Consider the equation x53x4+mx3+nx2+px+q=0, where m, n, p, qR.

The equation has three distinct real roots which can be written as log2a, log2b and log2c.

The equation also has two imaginary roots, one of which is di where dR.

The values a, b, and c are consecutive terms in a geometric sequence.

Show that abc=8.

[5]
a.

Show that one of the real roots is equal to 1.

[3]
b.

Given that q=8d2, find the other two real roots.

[9]
c.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

recognition of the other root =di       (A1)

log2a+log2b+log2c+didi=3        M1A1

Note: Award M1 for sum of the roots, A1 for 3. Award A0M1A0 for just log2a+log2b+log2c=3.

log2abc=3       (M1)

abc=23       A1

abc=8       AG

[5 marks]

a.

METHOD 1

let the geometric series be u1u1ru1r2

(u1r)3=8      M1

u1r=2       A1

hence one of the roots is log22=1      R1

 

METHOD 2

ba=cb

b2=acb3=abc=8      M1

b=2       A1

hence one of the roots is log22=1      R1

 

[3 marks]

b.

METHOD 1

product of the roots is r1×r2×1×di×di=8d2       (M1)(A1)

r1×r2=8       A1

sum of the roots is r1+r2+1+di+di=3       (M1)(A1)

r1+r2=2       A1

solving simultaneously       (M1)

r1=2r2=4       A1A1

 

METHOD 2

product of the roots log2a×log2b×log2c×di×di=8d2       M1A1

log2a×log2b×log2c=8       A1

EITHER

abc can be written as 2r22r       M1

(log22r)(log22)(log22r)=8

attempt to solve       M1

(1log2r)(1+log2r)=8

log2r=±3

r=18,8       A1A1

OR

abc can be written as a2, 4a      M1

(log2a)(log22)(log24a)=8

attempt to solve       M1

a=14,16       A1A1

THEN

a and c are 14,16       (A1)

roots are −2, 4       A1

 

[9 marks]

c.

Examiners report

[N/A]
a.
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b.
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c.

Syllabus sections

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