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Date May 2019 Marks available 9 Reference code 19M.2.AHL.TZ1.H_11
Level Additional Higher Level Paper Paper 2 Time zone Time zone 1
Command term Find Question number H_11 Adapted from N/A

Question

Consider the equation x 5 3 x 4 + m x 3 + n x 2 + p x + q = 0 , where m , n , p , q R .

The equation has three distinct real roots which can be written as lo g 2 a , lo g 2 b and lo g 2 c .

The equation also has two imaginary roots, one of which is d i where d R .

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

Show that a b c = 8 .

[5]
a.

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

[3]
b.

Given that q = 8 d 2 , 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  = d i        (A1)

lo g 2 a + lo g 2 b + lo g 2 c + d i d i = 3         M1A1

Note: Award M1 for sum of the roots, A1 for 3. Award A0M1A0 for just  lo g 2 a + lo g 2 b + lo g 2 c = 3 .

lo g 2 a b c = 3        (M1)

a b c = 2 3        A1

a b c = 8        AG

[5 marks]

a.

METHOD 1

let the geometric series be  u 1 u 1 r u 1 r 2

( u 1 r ) 3 = 8       M1

u 1 r = 2        A1

hence one of the roots is  lo g 2 2 = 1       R1

 

METHOD 2

b a = c b

b 2 = a c b 3 = a b c = 8       M1

b = 2        A1

hence one of the roots is  lo g 2 2 = 1       R1

 

[3 marks]

b.

METHOD 1

product of the roots is  r 1 × r 2 × 1 × d i × d i = 8 d 2        (M1)(A1)

r 1 × r 2 = 8        A1

sum of the roots is r 1 + r 2 + 1 + d i + d i = 3        (M1)(A1)

r 1 + r 2 = 2        A1

solving simultaneously       (M1)

r 1 = 2 r 2 = 4        A1A1

 

METHOD 2

product of the roots  lo g 2 a × lo g 2 b × lo g 2 c × d i × d i = 8 d 2        M1A1

lo g 2 a × lo g 2 b × lo g 2 c = 8        A1

EITHER

a b c  can be written as  2 r 2 2 r        M1

( lo g 2 2 r ) ( lo g 2 2 ) ( lo g 2 2 r ) = 8

attempt to solve       M1

( 1 lo g 2 r ) ( 1 + lo g 2 r ) = 8

lo g 2 r = ± 3

r = 1 8 , 8        A1A1

OR

a b c  can be written as  a 2 , 4 a       M1

( lo g 2 a ) ( lo g 2 2 ) ( lo g 2 4 a ) = 8

attempt to solve       M1

a = 1 4 , 16        A1A1

THEN

a and  c are  1 4 , 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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Topic 1—Number and algebra

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