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Date May 2022 Marks available 3 Reference code 22M.3.AHL.TZ2.2
Level Additional Higher Level Paper Paper 3 Time zone Time zone 2
Command term Show that Question number 2 Adapted from N/A

Question

This question asks you to investigate conditions for the existence of complex roots of polynomial equations of degree 3 and 4.

 
The cubic equation x3+px2+qx+r=0, where p, q, r  , has roots α, β and γ.

Consider the equation x3-7x2+qx+1=0, where q.

Noah believes that if p23q then α, β and γ are all real.

Now consider polynomial equations of degree 4.

The equation x4+px3+qx2+rx+s=0, where p, q, r, s, has roots α, β, γ and δ.

In a similar way to the cubic equation, it can be shown that:

p=-(α+β+γ+δ)

q=αβ+αγ+αδ+βγ+βδ+γδ

r=-(αβγ+αβδ+αγδ+βγδ)

s=αβγδ.

The equation x4-9x3+24x2+22x-12=0, has one integer root.

By expanding x-αx-βx-γ show that:

p=-α+β+γ

q=αβ+βγ+γα

r=-αβγ.

[3]
a.

Show that p2-2q=α2+β2+γ2.

[3]
b.i.

Hence show that α-β2+β-γ2+γ-α2=2p2-6q.

[3]
b.ii.

Given that p2<3q, deduce that α, β and γ cannot all be real.

[2]
c.

Using the result from part (c), show that when q=17, this equation has at least one complex root.

[2]
d.

By varying the value of q in the equation x3-7x2+qx+1=0, determine the smallest positive integer value of q required to show that Noah is incorrect.

[2]
e.i.

Explain why the equation will have at least one real root for all values of q.

[1]
e.ii.

Find an expression for α2+β2+γ2+δ2 in terms of p and q.

[3]
f.i.

Hence state a condition in terms of p and q that would imply x4+px3+qx2+rx+s=0 has at least one complex root.

[1]
f.ii.

Use your result from part (f)(ii) to show that the equation x4-2x3+3x2-4x+5=0 has at least one complex root.

[1]
g.

State what the result in part (f)(ii) tells us when considering this equation x4-9x3+24x2+22x-12=0.

[1]
h.i.

Write down the integer root of this equation.

[1]
h.ii.

By writing x4-9x3+24x2+22x-12 as a product of one linear and one cubic factor, prove that the equation has at least one complex root.

[4]
h.iii.

Markscheme

attempt to expand x-αx-βx-γ            M1

=x2-α+βx+αβx-γ  OR  =x-αx2-β+γx+βγ         A1

x3+px2+qx+r=x3-α+β+γx2+αβ+βγ+γαx-αβγ         A1

comparing coefficients:

p=-α+β+γ         AG 

q=αβ+βγ+γα         AG 

r=-αβγ         AG 

 

Note: For candidates who do not include the AG lines award full marks.

 

[3 marks]

a.

p2-2q=α+β+γ2-2αβ+βγ+γα            (A1)

attempt to expand α+β+γ2            (M1)

=α2+β2+γ2+2αβ+βγ+γα-2αβ+βγ+γα or equivalent         A1

=α2+β2+γ2         AG 

 

Note: Accept equivalent working from RHS to LHS.

 

[3 marks]

b.i.

EITHER

attempt to expand α-β2+β-γ2+γ-α2            (M1)

=α2+β2-2αβ+β2+γ2-2βγ+γ2+α2-2γα         A1

=2α2+β2+γ2-2αβ+βγ+γα

=2p2-2q-2q or equivalent         A1

=2p2-6q         AG 


OR

attempt to write 2p2-6q in terms of α, β, γ            (M1)

=2p2-2q-2q

=2α2+β2+γ2-2αβ+βγ+γα         A1

=α2+β2-2αβ+β2+γ2-2βγ+γ2+α2-2γα         A1

=α-β2+β-γ2+γ-α2         AG 

 

Note: Accept equivalent working where LHS and RHS are expanded to identical expressions.

 

[3 marks]

b.ii.

p2<3q2p2-6q<0

α-β2+β-γ2+γ-α2<0         A1

if all roots were real α-β2+β-γ2+γ-α20         R1


Note:
Condone strict inequality in the R1 line.
Note: Do not award A0R1.


roots cannot all be real         AG 

 

[2 marks]

c.

p2=-72=49 and 3q=51         A1

so p2<3q the equation has at least one complex root         R1

 

Note: Allow equivalent comparisons; e.g. checking p2<6q

 

[2 marks]

d.

use of GDC (eg graphs or tables)         (M1)

q=12         A1

 

[2 marks]

e.i.

complex roots appear in conjugate pairs (so if complex roots occur the other root will be real OR all 3 roots will be real).

OR

a cubic curve always crosses the x-axis at at least one point.       R1

 

[1 mark]

e.ii.

attempt to expand α+β+γ+δ2           (M1)

α+β+γ+δ2=α2+β2+γ2+δ2+2αβ+αγ+αδ+βγ+βδ+γδ           (A1)

α2+β2+γ2+δ2=α+β+γ+δ2-2αβ+αγ+αδ+βγ+βδ+γδ

α2+β2+γ2+δ2=p2-2q          A1

 

[3 marks]

f.i.

p2<2q  OR  p2-2q<0          A1

 

Note: Allow FT on their result from part (f)(i).

 

[1 mark]

f.ii.

4<6  OR  22-2×3<0          R1

hence there is at least one complex root.         AG

 

Note: Allow FT from part (f)(ii) for the R mark provided numerical reasoning is seen.

 

[1 mark]

g.

p2>2q 81>2×24  (so) nothing can be deduced         R1

 

Note: Do not allow FT for the R mark.

 

[1 mark]

h.i.

-1          A1

 

[1 mark]

h.ii.

attempt to express as a product of a linear and cubic factor           M1

x+1x3-10x2+34x-12          A1A1

 

Note: Award A1 for each factor. Award at most A1A0 if not written as a product.

 

since for the cubic, p2<3q 100<102          R1

there is at least one complex root          AG

 

[4 marks]

h.iii.

Examiners report

The first part of this question proved to be very accessible, with the majority of candidates expanding their brackets as required, to find the coefficients p, q and r.

a.

The first part of this question was usually answered well, though presentation in the second part sometimes left a lot to be desired. The expression 2p2-2q-2q was expected to be seen more often, as a 'pivot' to reaching the required result. Algebra was often lengthy, but untidily so, sometimes leaving examiners to do some mental tidying up on behalf of the candidate.

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

A good number of candidates recognised the reasoning required in this part of the question and were able to score both marks.

c.

Most candidates found applying this specific case to be very straightforward.

d.

Most candidates offered incorrect answers in the first part; despite their working suggested utilisation of the GDC, it was clear that many did not appreciate what the question was asking. The second part was usually answered well, with the idea of complex roots occurring in conjugate pairs being put to good use.

e.i.
[N/A]
e.ii.

Some very dubious algebra was seen here, and often no algebra at all. Despite this, a good number of candidates seemed to make the 'leap' to the correct expression p2-2q, perhaps fortuitously so in a number of cases.

f.i.
[N/A]
f.ii.

Of those finding p2-2q in part f, a surprising number of answers seen employed the test of checking whether p2<3q.

g.

Part i was usually not answered successfully, which may have been due to shortage of time. However, it was pleasing to see a number of candidates reach the end of the paper and successfully factorise the given quartic using a variety of methods. The final part required the p2<3q test. Though correct reasoning was sometimes seen, it was rare for this final mark to be gained.

h.i.
[N/A]
h.ii.
[N/A]
h.iii.

Syllabus sections

Topic 1—Number and algebra » SL 1.6—Simple proof
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Topic 1—Number and algebra

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