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Date May 2012 Marks available 6 Reference code 12M.1.hl.TZ2.12
Level HL only Paper 1 Time zone TZ2
Command term Express Question number 12 Adapted from N/A

Question

The graph of a polynomial function f of degree 4 is shown below.

 

Given that \({(x + {\text{i}}y)^2} = - 5 + 12{\text{i}},{\text{ }}x,{\text{ }}y \in \mathbb{R}\) . Show that

(i)     \({x^2} - {y^2} = - 5\) ;

(ii)     \(xy = 6\) .

[2]
A.a.

Hence find the two square roots of \( - 5 + 12{\text{i}}\) .

[5]
A.b.

For any complex number z , show that \({(z^*)^2} = ({z^2})^*\) .

[3]
A.c.

Hence write down the two square roots of \( - 5 - 12{\text{i}}\) .

[2]
A.d.

Explain why, of the four roots of the equation \(f(x) = 0\) , two are real and two are complex.

[2]
B.a.

The curve passes through the point \(( - 1,\, - 18)\) . Find \(f(x)\) in the form

\(f(x) = (x - a)(x - b)({x^2} + cx + d),{\text{ where }}a,{\text{ }}b,{\text{ }}c,{\text{ }}d \in \mathbb{Z}\) .

[5]
B.b.

Find the two complex roots of the equation \(f(x) = 0\) in Cartesian form.

[2]
B.c.

Draw the four roots on the complex plane (the Argand diagram).

[2]
B.d.

Express each of the four roots of the equation in the form \(r{{\text{e}}^{{\text{i}}\theta }}\) .

[6]
B.e.

Markscheme

(i)     \({(x + {\text{i}}y)^2} = - 5 + 12{\text{i}}\)

\({x^2} + 2{\text{i}}xy + {{\text{i}}^2}{y^2} = - 5 + 12{\text{i}}\)     A1

(ii)     equating real and imaginary parts     M1

\({x^2} - {y^2} = - 5\)     AG

\(xy = 6\)     AG

[2 marks]

A.a.

substituting     M1

EITHER

\({x^2} - \frac{{36}}{{{x^2}}} = - 5\)

\({x^4} + 5{x^2} - 36 = 0\)     A1

\({x^2} = 4,\, - 9\)     A1

\(x = \pm 2\) and \(y = \pm 3\)     (A1)

OR

\(\frac{{36}}{{{y^2}}} - {y^2} = - 5\)

\({y^4} - 5{y^2} - 36 = 0\)     A1

\({y^2} = 9,\, - 4\)     A1

\({y^2} = \pm 3\) and \(x = \pm 2\)     (A1) 

Note: Accept solution by inspection if completely correct.

 

THEN

the square roots are \((2 + 3{\text{i}})\) and \(( - 2 - 3{\text{i}})\)     A1

[5 marks]

A.b.

EITHER

consider \(z = x + {\text{i}}y\)

\(z^* = x - {\text{i}}y\)

\({(z^*)^2} = {x^2} - {y^2} - 2{\text{i}}xy\)     A1

\(({z^2}) = {x^2} - {y^2} + 2{\text{i}}xy\)     A1

\(({z^2})^* = {x^2} - {y^2} - 2{\text{i}}xy\)     A1

\({(z^*)^2} = ({z^2})^*\)     AG

OR

\(z^* = r{{\text{e}}^{ - {\text{i}}\theta }}\)

\({(z^*)^2} = {r^2}{{\text{e}}^{ - 2{\text{i}}\theta }}\)     A1

\({z^2} = {r^2}{{\text{e}}^{2{\text{i}}\theta }}\)     A1

\(({z^2})^* = {r^2}{{\text{e}}^{ - 2{\text{i}}\theta }}\)     A1

\({(z^*)^2} = ({z^2})^*\)     AG

[3 marks]

A.c.

\((2 - 3{\text{i}})\) and \(( - 2 + 3{\text{i}})\)     A1A1

[2 marks]

A.d.

the graph crosses the x-axis twice, indicating two real roots     R1

since the quartic equation has four roots and only two are real, the other two roots must be complex     R1

[2 marks]

B.a.

\(f(x) = (x + 4)(x - 2)({x^2} + cx + d)\)     A1A1

\(f(0) = - 32 \Rightarrow d = 4\)     A1

Since the curve passes through \(( - 1,\, - 18)\),

\( - 18 = 3 \times ( - 3)(5 - c)\)     M1

\(c = 3\)     A1

Hence \(f(x) = (x + 4)(x - 2)({x^2} + 3x + 4)\)

[5 marks]

B.b.

\(x = \frac{{ - 3 \pm \sqrt {9 - 16} }}{2}\)     (M1)

\( \Rightarrow x = - \frac{3}{2} \pm {\text{i}}\frac{{\sqrt 7 }}{2}\)     A1

[2 marks]

B.c.

     A1A1

 

Note: Accept points or vectors on complex plane.

Award A1 for two real roots and A1 for two complex roots.

 

[2 marks]

B.d.

real roots are \(4{{\text{e}}^{{\text{i}}\pi }}\) and \(2{{\text{e}}^{{\text{i}}0}}\)     A1A1

considering \( - \frac{3}{2} \pm {\text{i}}\frac{{\sqrt 7 }}{2}\)

\(r = \sqrt {\frac{9}{4} + \frac{7}{4}}  = 2\)     A1

finding \(\theta \) using \(\arctan \left( {\frac{{\sqrt 7 }}{3}} \right)\)     M1

\(\theta  = \arctan \left( {\frac{{\sqrt 7 }}{3}} \right) + \pi {\text{ or }}\theta  = \arctan \left( { - \frac{{\sqrt 7 }}{3}} \right) + \pi \)     A1

\( \Rightarrow z = 2{{\text{e}}^{{\text{i}}\left( {\arctan \left( {\frac{{\sqrt 7 }}{3}} \right) + \pi } \right)}}{\text{ or}} \Rightarrow z = 2{{\text{e}}^{{\text{i}}\left( {\arctan \left( {\frac{{ - \sqrt 7 }}{3}} \right) + \pi } \right)}}\)     A1 

Note: Accept arguments in the range \( - \pi {\text{ to }}\pi {\text{ or }}0{\text{ to }}2\pi \) .

Accept answers in degrees.

 

[6 marks]

B.e.

Examiners report

Since (a) was a ‘show that’ question, it was essential for candidates to give a convincing explanation of how the quoted results were obtained. Many candidates just wrote

\[{(x + {\text{i}}y)^2} = {x^2} - {y^2} + 2{\text{i}}xy = - 5 + 12{\text{i}}\]

\[{\text{Therefore }}{x^2} - {y^2} = - 5{\text{ and }}xy = 6\]

This was not given full credit since it simply repeated what was given in the question. Candidates were expected to make it clear that they were equating real and imaginary parts. In (b), candidates who attempted to use de Moivre’s Theorem to find the square roots were given no credit since the question stated ‘hence’.

A.a.

Since (a) was a ‘show that’ question, it was essential for candidates to give a convincing explanation of how the quoted results were obtained. Many candidates just wrote

\[{(x + {\text{i}}y)^2} = {x^2} - {y^2} + 2{\text{i}}xy = - 5 + 12{\text{i}}\]

\[{\text{Therefore }}{x^2} - {y^2} = - 5{\text{ and }}xy = 6\]

This was not given full credit since it simply repeated what was given in the question. Candidates were expected to make it clear that they were equating real and imaginary parts. In (b), candidates who attempted to use de Moivre’s Theorem to find the square roots were given no credit since the question stated ‘hence’.

A.b.

Since (a) was a ‘show that’ question, it was essential for candidates to give a convincing explanation of how the quoted results were obtained. Many candidates just wrote

\[{(x + {\text{i}}y)^2} = {x^2} - {y^2} + 2{\text{i}}xy = - 5 + 12{\text{i}}\]

\[{\text{Therefore }}{x^2} - {y^2} = - 5{\text{ and }}xy = 6\]

This was not given full credit since it simply repeated what was given in the question. Candidates were expected to make it clear that they were equating real and imaginary parts. In (b), candidates who attempted to use de Moivre’s Theorem to find the square roots were given no credit since the question stated ‘hence’.

A.c.

Since (a) was a ‘show that’ question, it was essential for candidates to give a convincing explanation of how the quoted results were obtained. Many candidates just wrote

\[{(x + {\text{i}}y)^2} = {x^2} - {y^2} + 2{\text{i}}xy = - 5 + 12{\text{i}}\]

\[{\text{Therefore }}{x^2} - {y^2} = - 5{\text{ and }}xy = 6\]

This was not given full credit since it simply repeated what was given in the question. Candidates were expected to make it clear that they were equating real and imaginary parts. In (b), candidates who attempted to use de Moivre’s Theorem to find the square roots were given no credit since the question stated ‘hence’.

A.d.

In (a), the explanations were often unconvincing. Candidates were expected to make it clear that the two intersections with the x-axis gave two real roots and, since the polynomial was a quartic and therefore had four zeros, the other two roots must be complex. Candidates who made vague statements such as ‘the graph shows two real roots’ were not given full credit. In (b), most candidates stated the values of a and b correctly but algebraic errors often led to incorrect values for the other parameters. Candidates who failed to solve (b) correctly were unable to solve (c), (d) and (e) correctly although follow through was used where possible.

B.a.

In (a), the explanations were often unconvincing. Candidates were expected to make it clear that the two intersections with the x-axis gave two real roots and, since the polynomial was a quartic and therefore had four zeros, the other two roots must be complex. Candidates who made vague statements such as ‘the graph shows two real roots’ were not given full credit. In (b), most candidates stated the values of a and b correctly but algebraic errors often led to incorrect values for the other parameters. Candidates who failed to solve (b) correctly were unable to solve (c), (d) and (e) correctly although follow through was used where possible.

B.b.

In (a), the explanations were often unconvincing. Candidates were expected to make it clear that the two intersections with the x-axis gave two real roots and, since the polynomial was a quartic and therefore had four zeros, the other two roots must be complex. Candidates who made vague statements such as ‘the graph shows two real roots’ were not given full credit. In (b), most candidates stated the values of a and b correctly but algebraic errors often led to incorrect values for the other parameters. Candidates who failed to solve (b) correctly were unable to solve (c), (d) and (e) correctly although follow through was used where possible.

B.c.

In (a), the explanations were often unconvincing. Candidates were expected to make it clear that the two intersections with the x-axis gave two real roots and, since the polynomial was a quartic and therefore had four zeros, the other two roots must be complex. Candidates who made vague statements such as ‘the graph shows two real roots’ were not given full credit. In (b), most candidates stated the values of a and b correctly but algebraic errors often led to incorrect values for the other parameters. Candidates who failed to solve (b) correctly were unable to solve (c), (d) and (e) correctly although follow through was used where possible.

B.d.

In (a), the explanations were often unconvincing. Candidates were expected to make it clear that the two intersections with the x-axis gave two real roots and, since the polynomial was a quartic and therefore had four zeros, the other two roots must be complex. Candidates who made vague statements such as ‘the graph shows two real roots’ were not given full credit. In (b), most candidates stated the values of a and b correctly but algebraic errors often led to incorrect values for the other parameters. Candidates who failed to solve (b) correctly were unable to solve (c), (d) and (e) correctly although follow through was used where possible.

B.e.

Syllabus sections

Topic 1 - Core: Algebra » 1.6 » Modulus–argument (polar) form \(z = r\left( {\cos \theta + {\text{i}}\sin \theta } \right) = r{\text{cis}}\theta = r{e^{{\text{i}}\theta }}\)
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