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Date	May 2022	Marks available	1	Reference code	22M.3.AHL.TZ1.2
Level	Additional Higher Level	Paper	Paper 3	Time zone	Time zone 1
Command term	State	Question number	2	Adapted from	N/A

Question

This question asks you to explore cubic polynomials of the form $(x - r) (x^{2} - 2 a x + a^{2} + b^{2})$ for $x \in ℝ$ and corresponding cubic equations with one real root and two complex roots of the form $(z - r) (z^{2} - 2 a z + a^{2} + b^{2}) = 0$ for $z \in ℂ$ .

In parts (a), (b) and (c), let $r = 1, a = 4$ and $b = 1$ .

Consider the equation $(z - 1) (z^{2} - 8 z + 17) = 0$ for $z \in ℂ$ .

Consider the function $f (x) = (x - 1) (x^{2} - 8 x + 17)$ for $x \in ℝ$ .

Consider the function $g (x) = (x - r) (x^{2} - 2 a x + a^{2} + b^{2})$ for $x \in ℝ$ where $r, a \in ℝ$ and $b \in ℝ, b > 0$ .

The equation $(z - r) (z^{2} - 2 a z + a^{2} + b^{2}) = 0$ for $z \in ℂ$ has roots $r$ and $a \pm b i$ where $r, a \in ℝ$ and $b \in ℝ, b > 0$ .

On the Cartesian plane, the points $C_{1} (a, \sqrt{g' (a)})$ and $C_{2} (a, - \sqrt{g' (a)})$ represent the real and imaginary parts of the complex roots of the equation $(z - r) (z^{2} - 2 a z + a^{2} + b^{2}) = 0$ .

The following diagram shows a particular curve of the form $y = (x - r) (x^{2} - 2 a x + a^{2} + 16)$ and the tangent to the curve at the point $A (a, 80)$ . The curve and the tangent both intersect the $x$ -axis at the point $R (- 2, 0)$ . The points $C_{1}$ and $C_{2}$ are also shown.

Consider the curve $y = (x - r) (x^{2} - 2 a x + a^{2} + b^{2})$ for $a \neq r, b > 0$ . The points $A (a, g (a))$ and $R (r, 0)$ are as defined in part (d)(ii). The curve has a point of inflexion at point $P$ .

Consider the special case where $a = r$ and $b > 0$ .

Given that $1$ and $4 + i$ are roots of the equation, write down the third root.

[1]

a.i.

Verify that the mean of the two complex roots is $4$ .

[1]

a.ii.

Show that the line $y = x - 1$ is tangent to the curve $y = f (x)$ at the point $A (4, 3)$ .

[4]

Sketch the curve $y = f (x)$ and the tangent to the curve at point $A$ , clearly showing where the tangent crosses the $x$ -axis.

[2]

Show that $g' (x) = 2 (x - r) (x - a) + x^{2} - 2 a x + a^{2} + b^{2}$ .

[2]

d.i.

Hence, or otherwise, prove that the tangent to the curve $y = g (x)$ at the point $A (a, g (a))$ intersects the $x$ -axis at the point $R (r, 0)$ .

[6]

d.ii.

Deduce from part (d)(i) that the complex roots of the equation $(z - r) (z^{2} - 2 a z + a^{2} + b^{2}) = 0$ can be expressed as $a \pm i \sqrt{g' (a)}$ .

[1]

Use this diagram to determine the roots of the corresponding equation of the form $(z - r) (z^{2} - 2 a z + a^{2} + 16) = 0$ for $z \in ℂ$ .

[4]

f.i.

State the coordinates of $C_{2}$ .

[1]

f.ii.

Show that the $x$ -coordinate of $P$ is $\frac{1}{3} (2 a + r)$ .

You are not required to demonstrate a change in concavity.

[2]

g.i.

Hence describe numerically the horizontal position of point $P$ relative to the horizontal positions of the points $R$ and $A$ .

[1]

g.ii.

Sketch the curve $y = (x - r) (x^{2} - 2 a x + a^{2} + b^{2})$ for $a = r = 1$ and $b = 2$ .

[2]

h.i.

For $a = r$ and $b > 0$ , state in terms of $r$ , the coordinates of points $P$ and $A$ .

[1]

h.ii.

Markscheme

$4 - i$ A1

[1 mark]

a.i.

mean $= \frac{1}{2} (4 + i + 4 - i)$ A1

$= 4$ AG

[1 mark]

a.ii.

METHOD 1

attempts product rule differentiation (M1)

Note: Award (M1) for attempting to express $f (x)$ as $f (x) = x^{3} - 9 x^{2} + 25 x - 17$

$f' (x) = (x - 1) (2 x - 8) + x^{2} - 8 x + 17 (f' (x) = 3 x^{2} - 18 x + 25)$ A1

$f' (4) = 1$ A1

Note: Where $f' (x)$ is correct, award A1 for solving $f' (x) = 1$ and obtaining $x = 4$ .

EITHER

$y - 3 = 1 (x - 4)$ A1

$y = x + c$

$3 = 4 + c \Rightarrow c = - 1$ A1

states the gradient of $y = x - 1$ is also $1$ and verifies that $(4, 3)$ lies on the line $y = x - 1$ A1

THEN

so $y = x - 1$ is the tangent to the curve at $A (4, 3)$ AG

Note: Award a maximum of (M0)A0A1A1 to a candidate who does not attempt to find $f' (x)$ .

METHOD 2

sets $f (x) = x - 1$ to form $x - 1 = (x - 1) (x^{2} - 8 x + 17)$ (M1)

EITHER

$(x - 1) (x^{2} - 8 x + 16) = 0 (x^{3} - 9 x^{2} + 24 x - 16 = 0)$ A1

attempts to solve a correct cubic equation (M1)

$(x - 1) {(x - 4)}^{2} = 0 \Rightarrow x = 1, 4$

recognises that $x \neq 1$ and forms $x^{2} - 8 x + 17 = 1 (x^{2} - 8 x + 16 = 0)$ A1

attempts to solve a correct quadratic equation (M1)

${(x - 4)}^{2} = 0 \Rightarrow x = 4$

THEN

$x = 4$ is a double root R1

so $y = x - 1$ is the tangent to the curve at $A (4, 3)$ AG

Note: Candidates using this method are not required to verify that $y = 3$ .

[4 marks]

a positive cubic with an $x$ -intercept $(x = 1)$ , and a local maximum and local minimum in the first quadrant both positioned to the left of $A$ A1

Note: As the local minimum and point A are very close to each other, condone graphs that seem to show these points coinciding.
For the point of tangency, accept labels such as $A, (4, 3)$ or the point labelled from both axes. Coordinates are not required.

a correct sketch of the tangent passing through $A$ and crossing the $x$ -axis at the same point $(x = 1)$ as the curve A1

Note: Award A1A0 if both graphs cross the $x$ -axis at distinctly different points.

[2 marks]

EITHER

$g' (x) = (x - r) (2 x - 2 a) + x^{2} - 2 a x + a^{2} + b^{2}$ (M1)A1

$g (x) = x^{3} - (2 a + r) x^{2} + (a^{2} + b^{2} + 2 a r) x - (a^{2} + b^{2}) r$

attempts to find $g' (x)$ M1

$g' (x) = 3 x^{2} - 2 (2 a + r) x + a^{2} + b^{2} + 2 a r$

$= 2 x^{2} - 2 (a + r) x + 2 a r + x^{2} - 2 a x + a^{2} + b^{2}$ A1

$(= 2 (x^{2} - a x - r x + a r) + x^{2} - 2 a x + a^{2} + b^{2})$

THEN

$g' (x) = 2 (x - r) (x - a) + x^{2} - 2 a x + a^{2} + b^{2}$ AG

[2 marks]

d.i.

METHOD 1

$g (a) = b^{2} (a - r)$ (A1)

$g' (a) = b^{2}$ (A1)

attempts to substitute their $g (a)$ and $g' (a)$ into $y - g (a) = g' (a) (x - a)$ M1

$y - b^{2} (a - r) = b^{2} (x - a)$

EITHER

$y = b^{2} (x - r) (y = b^{2} x - b^{2} r)$ A1

sets $y = 0$ so $b^{2} (x - r) = 0$ M1

$b > 0 \Rightarrow x = r$ OR $b \neq 0 \Rightarrow x = r$ R1

sets $y = 0$ so $- b^{2} (a - r) = b^{2} (x - a)$ M1

$b > 0$ OR $b \neq 0 \Rightarrow - (a - r) = x - a$ R1

$x = r$ A1

THEN

so the tangent intersects the $x$ -axis at the point $R (r, 0)$ AG

METHOD 2

$g' (a) = b^{2}$ (A1)

$g (a) = b^{2} (a - r)$ (A1)

attempts to substitute their $g (a)$ and $g' (a)$ into $y = g' (a) x + c$ and attempts to find $c$ M1

$c = - b^{2} r$

EITHER

$y = b^{2} (x - r) (y = b^{2} x - b^{2} r)$ A1

sets $y = 0$ so $b^{2} (x - r) = 0$ M1

$b > 0 \Rightarrow x = r$ OR $b \neq 0 \Rightarrow x = r$ R1

sets $y = 0$ so $b^{2} (x - r) = 0$ M1

$b > 0$ OR $b \neq 0 \Rightarrow x - r = 0$ R1

$x = r$ A1

METHOD 3

$g' (a) = b^{2}$ (A1)

the line through $R (r, 0)$ parallel to the tangent at $A$ has equation
$y = b^{2} (x - r)$ A1

sets $g (x) = b^{2} (x - r)$ to form $b^{2} (x - r) = (x - r) (x^{2} - 2 a x + a^{2} + b^{2})$ M1

$b^{2} = x^{2} - 2 a x + a^{2} + b^{2}, (x \neq r)$ A1

${(x - a)}^{2} = 0$ A1

since there is a double root $(x = a)$ , this parallel line through $R (r, 0)$ is the required tangent at $A$ R1

[6 marks]

d.ii.

EITHER

$g' (a) = b^{2} \Rightarrow b = \sqrt{g' (a)}$ (since $b > 0$ ) R1

Note: Accept $b = \pm \sqrt{g' (a)}$ .

$(a \pm b i=) a \pm i \sqrt{b^{2}}$ and $g' (a) = b^{2}$ R1

THEN

hence the complex roots can be expressed as $a \pm i \sqrt{g' (a)}$ AG

[1 mark]

$b = 4$ (seen anywhere) A1

EITHER

attempts to find the gradient of the tangent in terms of $a$ and equates to $16$ (M1)

OR

substitutes $r = - 2, x = a$ and $y = 80$ to form $80 = (a - (- 2)) (a^{2} - 2 a^{2} + a^{2} + 16)$ (M1)

substitutes $r = - 2, x = a$ and $y = 80$ into $y = 16 (x - r)$ (M1)

THEN

$\frac{80}{a + 2} = 16 \Rightarrow a = 3$

roots are $- 2$ (seen anywhere) and $3 \pm 4 i$ A1A1

Note: Award A1 for $- 2$ and A1 for $3 \pm 4 i$ . Do not accept coordinates.

[4 marks]

f.i.

$(3, - 4)$ A1

Note: Accept “ $x = 3$ and $y = - 4$ ”.
Do not award A1FT for $(a, - 4)$ .

[1 mark]

f.ii.

$g' (x) = 2 (x - r) (x - a) + x^{2} - 2 a x + a^{2} + b^{2}$

attempts to find $g'' (x)$ M1

$g'' (x) = 2 (x - a) + 2 (x - r) + 2 x - 2 a (= 6 x - 2 r - 4 a)$

sets $g'' (x) = 0$ and correctly solves for $x$ A1

for example, obtaining $x - r + 2 (x - a) = 0$ leading to $3 x = 2 a + r$

so $x = \frac{1}{3} (2 a + r)$ AG

Note: Do not award A1 if the answer does not lead to the AG.

[2 marks]

g.i.

point $P$ is $\frac{2}{3}$ of the horizontal distance (way) from point $R$ to point $A$ A1

Note: Accept equivalent numerical statements or a clearly labelled diagram displaying the numerical relationship.
Award A0 for non-numerical statements such as “ $P$ is between $R$ and $A$ , closer to $A$ ”.

[1 mark]

g.ii.

$y = (x - 1) (x^{2} - 2 x + 5)$ (A1)

a positive cubic with no stationary points and a non-stationary point of inflexion at $x = 1$ A1

Note: Graphs may appear approximately linear. Award this A1 if a change of concavity either side of $x = 1$ is apparent.
Coordinates are not required and the $y$ -intercept need not be indicated.

[2 marks]

h.i.

$(r, 0)$ A1

[1 mark]

h.ii.

Examiners report

Part (a) (i) was generally well done with a significant majority of candidates using the conjugate root theorem to state $4 - i$ as the third root. A number of candidates, however, wasted considerable time attempting an algebraic method to determine the third root. Part (a) (ii) was reasonably well done. A few candidates however attempted to calculate the product of $4 + i$ and $4 - i$ .

Part (b) was reasonably well done by a significant number of candidates. Most were able to find a correct expression for $f' (x)$ and a good number of those candidates were able to determine that $f' (4) = 1$ . Candidates that did not determine the equation of the tangent had to state that the gradient of $y = x - 1$ is also 1 and verify that the point (4,3) lies on the line. A few candidates only met one of those requirements. Weaker candidates tended to only verify that the point (4,3) lies on the curve and the tangent line without attempting to find $f' (x)$ .

Part (c) was not answered as well as anticipated. A number of sketches were inaccurate and carelessly drawn with many showing both graphs crossing the x-axis at distinctly different points.

Part (d) (i) was reasonably well done by a good number of candidates. Most successful responses involved use of the product rule. A few candidates obtained full marks by firstly expanding $g (x)$ , then differentiating to find $g' (x)$ and finally simplifying to obtain the desired result. A number of candidates made elementary mistakes when differentiating. In general, the better candidates offered reasonable attempts at showing the general result in part (d) (ii). A good number gained partial credit by determining that $g' (a) = b^{2}$ and/or $g (a) = b^{2} (a - r)$ . Only the very best candidates obtained full marks by concluding that as $b > 0$ or $b \neq 0$ , then $x = r$ when $y = 0$ .

In general, only the best candidates were able to use the result $g' (a) = b^{2}$ to deduce that the complex roots of the equation can be expressed as $a \pm i \sqrt{g' (a)}$ . Although given the complex roots $a \pm b i$ , a significant number of candidates attempted, with mixed success, to use the quadratic formula to solve the equation $z^{2} - 2 a z + a^{2} + b^{2} = 0$ .

In part (f) (i), only a small number of candidates were able to determine all the roots of the equation. Disappointingly, a large number did not state $- 2$ as a root. Some candidates determined that $b = 4$ but were unable to use the diagram to determine that $a = 3$ . Of the candidates who determined all the roots in part (f) (i), very few gave the correct coordinates for C₂ . The most frequent error was to give the y-coordinate as $3 - 4 i$ .

Of the candidates who attempted part (g) (i), most were able to find an expression for $g'' (x)$ and a reasonable number of these were then able to convincingly show that $x = \frac{1}{3} (2 a + r)$ . It was very rare to see a correct response to part (g) (ii). A few candidates stated that P is between R and A with some stating that P was closer to A. A small number restated $x = \frac{1}{3} (2 a + r)$ in words.

Of the candidates who attempted part (h) (i), most were able to determine that $y = (x - 1) (x^{2} - 2 x + 5)$ . However, most graphs were poorly drawn with many showing a change in concavity at $x = 0$ rather than at $x = 1$ . In part (h) (ii), only a very small number of candidates determined that A and P coincide at (r,0).