(a) (i) Third root:

\[ 4 – i \quad (A1) \]

[1 mark]

(a) (ii) Verify mean of complex roots:

\[ \text{Mean} = \frac{1}{2}(4 + i + 4 – i) = 4 \quad (A1)(AG) \]

[1 mark]

(b) Show that \( y = x – 1 \) is tangent to the curve at \( \mathrm{A}(4, 3) \):

Method 1:

Attempt product rule differentiation (M1):

Note: Award (M1) for attempting to express \( f(x) = x^3 – 9x^2 + 25x – 17 \).

\[ f'(x) = (x – 1)(2x – 8) + (x^2 – 8x + 17) \quad (A1) \]

Or: \( f'(x) = 3x^2 – 18x + 25 \).

Evaluate at \( x = 4 \):

\[ f'(4) = 1 \quad (A1) \]

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

Either:

\[ y – 3 = 1(x – 4) \quad (A1) \]

Or:

\[ y = x + c \]

\[ 3 = 4 + c \implies c = -1 \quad (A1) \]

Or:

Gradient of \( y = x – 1 \) is 1, and verify \( (4, 3) \) lies on \( y = x – 1 \quad (A1) \).

So, \( y = x – 1 \) is the tangent at \( \mathrm{A}(4, 3) \quad (AG) \).

Note: Award a maximum of (M0)(A0)(A1)(A1) to a candidate who does not attempt to find \( f'(x) \).

Method 2:

Set \( f(x) = x – 1 \):

\[ x – 1 = (x – 1)(x^2 – 8x + 17) \quad (M1) \]

Either:

\[ (x – 1)(x^2 – 8x + 16) = 0 \quad (M1) \]

\[ (x – 1)(x – 4)^2 = 0 \implies x = 1, 4 \quad (A1) \]

Or:

Recognize \( x \neq 1 \), form \( x^2 – 8x + 17 = 1 \):

\[ x^2 – 8x + 16 = 0 \quad (M1) \]

\[ (x – 4)^2 = 0 \implies x = 4 \quad (A1) \]

\[ x = 4 \text{ is a double root} \quad (R1) \]

So, \( y = x – 1 \) is the tangent at \( \mathrm{A}(4, 3) \quad (AG) \).

Note: Candidates using this method are not required to verify \( y = 3 \).

[4 marks]

(c) Sketch the curve and tangent:

A positive cubic with an \( x \)-intercept at \( x = 1 \), and a local maximum and local minimum in the first quadrant, both to the left of \( \mathrm{A} \quad (A1) \).

Note: As the local minimum and point \( \mathrm{A} \) are close, condone graphs where they seem to coincide. Accept labels such as \( \mathrm{A} \), \( (4, 3) \), or points labeled from both axes. Coordinates are not required.

Correct sketch of the tangent passing through \( \mathrm{A} \) and crossing the \( x \)-axis at \( x = 1 \quad (A1) \).

Note: Award (A1)(A0) if both graphs cross the \( x \)-axis at distinctly different points.

[2 marks]

(d) (i) Show that \( g'(x) = 2(x – r)(x – a) + x^2 – 2ax + a^2 + b^2 \):

Either:

\[ g'(x) = (x – r)(2x – 2a) + (x^2 – 2ax + a^2 + b^2) \quad (M1)(A1) \]

Or:

\[ g(x) = x^3 – (2a + r)x^2 + (a^2 + b^2 + 2ar)x – (a^2 + b^2)r \]

Attempt to find \( g'(x) \quad (M1) \):

\[ g'(x) = 3x^2 – 2(2a + r)x + a^2 + b^2 + 2ar \]

\[ = 2x^2 – 2(a + r)x + 2ar + x^2 – 2ax + a^2 + b^2 \quad (A1) \]

\[ = 2(x^2 – ax – rx + ar) + x^2 – 2ax + a^2 + b^2 \]

Then:

\[ g'(x) = 2(x – r)(x – a) + x^2 – 2ax + a^2 + b^2 \quad (AG) \]

[2 marks]

(d) (ii) Prove tangent at \( \mathrm{A}(a, g(a)) \) intersects \( x \)-axis at \( \mathrm{R}(r, 0) \):

Method 1:

\[ g(a) = b^2(a – r) \quad (A1) \]

\[ g'(a) = b^2 \quad (A1) \]

Substitute into tangent equation \( y – g(a) = g'(a)(x – a) \quad (M1) \):

\[ y – b^2(a – r) = b^2(x – a) \]

Either:

\[ y = b^2(x – r) \quad (A1) \]

Set \( y = 0 \):

\[ b^2(x – r) = 0 \quad (M1) \]

\[ b > 0 \implies x = r \quad (R1) \]

Or:

Set \( y = 0 \):

\[ -b^2(a – r) = b^2(x – a) \quad (M1) \]

\[ b > 0 \implies -(a – r) = x – a \quad (R1) \]

\[ x = r \quad (A1) \]

So, tangent intersects \( x \)-axis at \( \mathrm{R}(r, 0) \quad (AG) \).

Method 2:

\[ g'(a) = b^2 \]

\[ g(a) = b^2(a – r) \]

Use \( y = g'(a)x + c \), find \( c \quad (M1) \):

\[ c = -b^2 r \quad (A1) \]

Either:

\[ y = b^2(x – r) \]

Set \( y = 0 \):

\[ b^2(x – r) = 0 \quad (M1) \]

\[ b > 0 \implies x = r \quad (R1) \]

Or:

Set \( y = 0 \):

\[ b^2(x – r) = 0 \quad (M1) \]

\[ b > 0 \implies x = r \quad (R1)(A1) \]

Method 3:

\[ g'(a) = b^2 \]

Line through \( \mathrm{R}(r, 0) \) parallel to tangent at \( \mathrm{A} \):

\[ y = b^2(x – r) \quad (A1) \]

Set \( g(x) = b^2(x – r) \quad (M1) \):

\[ b^2(x – r) = (x – r)(x^2 – 2ax + a^2 + b^2) \]

\[ b^2 = x^2 – 2ax + a^2 + b^2, (x \neq r) \quad (A1) \]

\[ (x – a)^2 = 0 \quad (A1) \]

Since \( x = a \) is a double root, the line is the tangent at \( \mathrm{A} \quad (R1) \).

[6 marks]

(e) Complex roots as \( a \pm i \sqrt{g'(a)} \):

Either:

\[ g'(a) = b^2 \implies b = \sqrt{g'(a)} \text{ (since } b > 0) \quad (R1) \]

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

Or:

\[ a \pm bi = a \pm i \sqrt{b^2} \text{ and } g'(a) = b^2 \quad (R1) \]

Hence, complex roots are \( a \pm i \sqrt{g'(a)} \quad (AG) \).

[1 mark]

(f) (i) Roots of \( (z – r)(z^2 – 2az + a^2 + 16) = 0 \):

\[ b = 4 \text{ (since } b^2 = 16) \quad (A1) \]

Either:

Find gradient of tangent, equate to 16 (M1):

Or:

Substitute \( r = -2 \), \( x = a \), \( y = 80 \):

\[ 80 = (a – (-2))(a^2 – 2a^2 + a^2 + 16) \quad (M1) \]

Or:

Substitute into \( y = 16(x – r) \quad (M1) \):

\[ \frac{80}{a + 2} = 16 \implies a = 3 \]

Roots: \( -2 \), \(3 \pm 4i \quad (A1)(A1) \)

Note: Award (A1) for \( -2 \), (A1) for \( 3 \pm 4i \). Do not accept coordinates.

[4 marks]

(f) (ii) Coordinates of \( \mathrm{C}_2 \):

\[ (3, -4) \quad (A1) \]

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

[1 mark]

(g) (i) \( x \)-coordinate of point \( \mathrm{P} \):

\[ g'(x) = 2(x – r)(x – a) + x^2 – 2ax + a^2 + b^2 \]

Find second derivative (M1):

\[ g”(x) = 2(x – a) + 2(x – r) + 2x – 2a = 6x – 2r – 4a \quad (A1) \]

Set \( g”(x) = 0 \):

\[ 6x – 2r – 4a = 0 \]

\[ x = \frac{2a + r}{3} \quad (AG) \]

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

[2 marks]

(g) (ii) Horizontal position of \( \mathrm{P} \):

Point \( \mathrm{P} \) is \( \frac{2}{3} \) of the horizontal distance from \( \mathrm{R} \) to \( \mathrm{A} \quad (A1) \).

Note: Accept equivalent numerical statements or a clearly labeled diagram. Award (A0) for non-numerical statements like ” \( \mathrm{P} \) is between \( \mathrm{R} \) and \( \mathrm{A} \), closer to \( \mathrm{A} \)”.

[1 mark]

(h) (i) Sketch for \( a = r = 1 \), \( b = 2 \):

\[ y = (x – 1)(x^2 – 2x + 5) \quad (A1) \]

A positive cubic with no stationary points and a non-stationary point of inflection at \( x = 1 \quad (A1) \).

Note: Graphs may appear approximately linear. Award (A1) if a change of concavity either side of \( x = 1 \) is apparent. Coordinates and \( y \)-intercept are not required.

[2 marks]

(h) (ii) Coordinates for \( a = r \), \( b > 0 \):

\[ (r, 0) \quad (A1) \]

[1 mark]