Question 2

(a) Topic-AHL 1.14 Complex conjugate roots of quadratic and polynomial equations with real coefficients.

(b) Topic-SL 5.6 The product and quotient rules.

(c) Topic-SL 5.8  Local maximum and minimum points

(d) Topic-SL 5.6 The product and quotient rules.

(e) Topic-SL 5.6 The product and quotient rules.

(f) Topic-SL 5.1 Derivative interpreted as gradient function and as rate of change.

(g) Topic-SL 5.6 The product and quotient rules.

(h) Topic-AHL 5.12 Higher derivatives.

This question asks you to explore cubic polynomials of the form
\(
(x – r)(x^2 – 2ax + a^2 + b^2)
\)
for $x \in \mathbb{R}$ and corresponding cubic equations with one real root and two complex roots of the form
\(
(z – r)(z^2 – 2az + a^2 + b^2) = 0
\)
for $z \in \mathbb{C}$.
In parts (a), (b), and (c), let $r = 1$, $a = 4$, and $b = 1$.
Consider the equation
\(
(z – 1)(z^2 – 8z + 17) = 0 \quad \text{for } z \in \mathbb{C}.
\)
(a)
(i) Given that $1$ and $4 + i$ are roots of the equation, write down the third root.
(ii) Verify that the mean of the two complex roots is $4$.
Consider the function
\(
f(x) = (x – 1)(x^2 – 8x + 17) \quad \text{for } x \in \mathbb{R}.
\)
(b) Show that the line $y = x – 1$ is tangent to the curve $y = f(x)$ at the point $A(4, 3)$.
(c) Sketch the curve $y = f(x)$ and the tangent to the curve at point $A$, clearly showing where the tangent crosses the $x$-axis.
Consider the function
\(
g(x) = (x – r)(x^2 – 2ax + a^2 + b^2) \quad \text{for } x \in \mathbb{R}, \, a \in \mathbb{R}, \, b \in \mathbb{R}, \, b > 0.
\)
(d)
(i) Show that
\(
g'(x) = 2(x – r)(x – a) + x^2 – 2ax + a^2 + b^2.
\)
(ii) 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 $R(r, 0)$.
The equation
\(
(z – r)(z^2 – 2az + a^2 + b^2) = 0
\)
for $z \in \mathbb{C}$ has roots $r$ and $a \pm bi$ where $r, a \in \mathbb{R}$ and $b \in \mathbb{R}, b > 0$.
(e) Deduce from part (d)(i) that the complex roots of the equation
\(
(z – r)(z^2 – 2az + a^2 + b^2) = 0
\)
can be expressed as
\(
a \pm i \sqrt{g'(a)}.
\)
On the Cartesian plane, the points
\(
C_1\left(a, \sqrt{g'(a)}\right) \quad \text{and} \quad C_2\left(a, -\sqrt{g'(a)}\right)
\)
represent the real and imaginary parts of the complex roots of the equation
\(
(z – r)(z^2 – 2az + a^2 + b^2) = 0.
\)
The following diagram shows a particular curve of the form
\(
y = (x – r)(x^2 – 2ax + a^2 + 16)
\)
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.

(f)
i) Use this diagram to determine the roots of the corresponding equation of the form
\(
(z – r)(z^2 – 2az + a^2 + 16) = 0 \quad \text{for } z \in \mathbb{C}.
\)
(ii) State the coordinates of $C_2$.
Consider the curve
\(
y = (x – r)(x^2 – 2ax + a^2 + b^2) \quad \text{for } a \neq r, \, b > 0.
\)
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 $P$.
(g)
(i) Show that the $x$-coordinate of $P$ is
\(
\frac{1}{3}(2a + r).
\)
You are \textbf{not} required to demonstrate a change in concavity.
(ii) Hence describe numerically the horizontal position of point $P$ relative to the horizontal positions of the points $R$ and $A$.
Consider the special case where $a = r$ and $b > 0$.
(h)
(i) Sketch the curve
\(
y = (x – r)(x^2 – 2ax + a^2 + b^2)
\)
for $a = r = 1$ and $b = 2$.
(ii) For $a = r$ and $b > 0$, state in terms of $r$, the coordinates of points $P$ and $A$.