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

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

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

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

Show that $g\text{'}\left(x\right)=2\left(x-r\right)\left(x-a\right)+x^2-2ax+a^2+b^2$.

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

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

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

State the coordinates of ${\text{C}}_2$.

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

You are not required to demonstrate a change in concavity.

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

Sketch the curve $y=\left(x-r\right)\left(x^2-2ax+a^2+b^2\right)$ for $a=r=1$ and $b=2$.

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

$4-\text{i}$        A1

[1 mark]

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

$=4$          AG

[1 mark]

METHOD 1

attempts product rule differentiation        (M1)

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

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

$f\text{'}\left(4\right)=1$        A1

Note: Where $f\text{'}\left(x\right)$ is correct, award A1 for solving $f\text{'}\left(x\right)=1$ and obtaining $x=4$.

EITHER

$y-3=1\left(x-4\right)$        A1

OR

$y=x+c$

$3=4+c\Rightarrow c=-1$        A1

OR

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

THEN

so $y=x-1$ is the tangent to the curve at $\text{A}\left(4, 3\right)$        AG

Note: Award a maximum of (M0)A0A1A1 to a candidate who does not attempt to find $f\text{'}\left(x\right)$.

METHOD 2

sets $f\left(x\right)=x-1$ to form $x-1=\left(x-1\right)\left(x^2-8x+17\right)$        (M1)

EITHER

$\left(x-1\right)\left(x^2-8x+16\right)=0 \left(x^3-9x^2+24x-16=0\right)$        A1

attempts to solve a correct cubic equation        (M1)

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

OR

recognises that $x\ne 1$ and forms $x^2-8x+17=1 \left(x^2-8x+16=0\right)$        A1

attempts to solve a correct quadratic equation        (M1)

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

THEN

$x=4$ is a double root        R1

so $y=x-1$ is the tangent to the curve at $\text{A}\left(4, 3\right)$        AG

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

[4 marks]

a positive cubic with an  $x$-intercept $\left(x=1\right)$, and a local maximum and local minimum in the first quadrant both positioned to the left of $\text{A}$        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 $\text{A}, \left(4,3\right)$ or the point labelled from both axes. Coordinates are not required.

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

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

[2 marks]

EITHER

$g\text{'}\left(x\right)=\left(x-r\right)\left(2x-2a\right)+x^2-2ax+a^2+b^2$         (M1)A1

OR

$g\left(x\right)=x^3-\left(2a+r\right)x^2+\left(a^2+b^2+2ar\right)x-\left(a^2+b^2\right)r$

attempts to find $g\text{'}\left(x\right)$        M1

$g\text{'}\left(x\right)=3x^2-2\left(2a+r\right)x+a^2+b^2+2ar$

$=2x^2-2\left(a+r\right)x+2ar+x^2-2ax+a^2+b^2$        A1

$\left(=2\left(x^2-ax-rx+ar\right)+x^2-2ax+a^2+b^2\right)$

THEN

$g\text{'}\left(x\right)=2\left(x-r\right)\left(x-a\right)+x^2-2ax+a^2+b^2$        AG

[2 marks]

METHOD 1

$g\left(a\right)=b^2\left(a-r\right)$         (A1)

$g\text{'}\left(a\right)=b^2$         (A1)

attempts to substitute their $g\left(a\right)$ and $g\text{'}\left(a\right)$ into $y-g\left(a\right)=g\text{'}\left(a\right)\left(x-a\right)$        M1

$y-b^2\left(a-r\right)=b^2\left(x-a\right)$

EITHER

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

sets $y=0$ so $b^2\left(x-r\right)=0$        M1

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

OR 

sets $y=0$ so $-b^2\left(a-r\right)=b^2\left(x-a\right)$        M1

$b>0$ OR $b\ne 0\Rightarrow -\left(a-r\right)=x-a$        R1

$x=r$        A1

THEN

so the tangent intersects the $x$-axis at the point $\text{R}\left(r, 0\right)$        AG

METHOD 2

$g\text{'}\left(a\right)=b^2$         (A1)

$g\left(a\right)=b^2\left(a-r\right)$         (A1)

attempts to substitute their $g\left(a\right)$ and $g\text{'}\left(a\right)$ into $y=g\text{'}\left(a\right)x+c$ and attempts to find $c$        M1

$c=-b^{2}r$

EITHER

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

sets $y=0$ so $b^2\left(x-r\right)=0$        M1

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

OR

sets $y=0$ so $b^2\left(x-r\right)=0$        M1

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

$x=r$        A1

METHOD 3

$g\text{'}\left(a\right)=b^2$         (A1)

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

sets $g\left(x\right)=b^2\left(x-r\right)$ to form $b^2\left(x-r\right)=\left(x-r\right)\left(x^2-2ax+a^2+b^2\right)$        M1

$b^2=x^2-2ax+a^2+b^2, \left(x\ne r\right)$        A1

${\left(x-a\right)}^2=0$        A1

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

[6 marks]

EITHER

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


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

OR

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

THEN

hence the complex roots can be expressed as $a\pm \text{i}\sqrt{g\text{'}\left(a\right)}$        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 $80=\left(a-\left(-2\right)\right)\left(a^2-2a^2+a^2+16\right)$       (M1)

OR

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

THEN

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

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

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

[4 marks]

$\left(3, -4\right)$        A1

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

[1 mark]

$g\text{'}\left(x\right)=2\left(x-r\right)\left(x-a\right)+x^2-2ax+a^2+b^2$

attempts to find $g\text{'}\text{'}\left(x\right)$        M1

$g\text{'}\text{'}\left(x\right)=2\left(x-a\right)+2\left(x-r\right)+2x-2a \left(=6x-2r-4a\right)$

sets $g\text{'}\text{'}\left(x\right)=0$ and correctly solves for $x$        A1

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

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

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

[2 marks]

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

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

[1 mark]

$y=\left(x-1\right)\left(x^2-2x+5\right)$       (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]

$\left(r, 0\right)$         A1

[1 mark]

Part (a) (i) was generally well done with a significant majority of candidates using the conjugate root theorem to state $4-\text{i}$ 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 $4+\text{i}$ and $4-\text{i}$.

Part (b) was reasonably well done by a significant number of candidates. Most were able to find a correct expression for $f\text{'}\left(x\right)$ and a good number of those candidates were able to determine that $f\text{'}\left(4\right)=1$. 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 $f\text{'}\left(x\right)$.

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 $g\left(x\right)$, then differentiating to find $g\text{'}\left(x\right)$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 $g\text{'}\left(a\right)=b^2$ and/or $g\left(a\right)=b^2(a-r)$. Only the very best candidates obtained full marks by concluding that as $b>0$ or $b\ne 0$, then $x=r$ when $y=0$.

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

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 $3-4\text{i}$.

Of the candidates who attempted part (g) (i), most were able to find an expression for $g\text{'}\text{'}\left(x\right)$ and a reasonable number of these were then able to convincingly show that $x=\frac{1}{3}(2a+r)$. 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 $x=\frac{1}{3}(2a+r)$ in words.

Of the candidates who attempted part (h) (i), most were able to determine that $y=(x-1)\left(x^2-2x+5\right)$. 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).