Consider the polynomial $P(z) = z^4 - 6z^3 + 18z^2 - 30z + 25$, where $z \in \mathbb{C}$.
Given that $z_1 = 1 + 2\mathrm{i}$ is a root of $P(z) = 0$.
(a) Write down the root $z_2$, the complex conjugate of $z_1$. [1]
(b) Show that $z^2 - 2z + 5$ is a factor of $P(z)$. [2]
(c) Hence, determine the other two roots of the equation $P(z) = 0$. [3]
(d) Let the four roots of $P(z) = 0$ be represented by the points $A, B, C,$ and $D$ on an Argand diagram. Calculate the area of the quadrilateral formed by these points. [3]
Let $P(x) = x^3 + ax^2 + bx + c$, where $a, b, c \in \mathbb{R}$.
When $P(x)$ is divided by $(x-1)$, the remainder is $8$.
When $P(x)$ is divided by $(x+1)$, the remainder is $-2$.
When $P(x)$ is divided by $(x-2)$, the remainder is $22$.
(a) Use the Remainder Theorem to form a system of three linear equations in $a, b,$ and $c$. [3]
(b) Solve the system to find the values of $a, b,$ and $c$. [3]
(c) Let $\alpha, \beta,$ and $\gamma$ be the roots of $P(x) = 0$. Determine the value of $\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}$. [2]
Consider the cubic equation $x^3 - 7x^2 + kx - 8 = 0$, where $k \in \mathbb{R}$. The roots of this equation are in a Geometric Progression (G.P.).
(a) Determine the real value of the common ratio, $r$, and the value of the parameter $k$. [5]
(b) Consider a general cubic equation $x^3 + px^2 + qx + r = 0$ where the roots are in an Arithmetic Progression (A.P.).
Prove that $2p^3 - 9pq + 27r = 0$. [6]
Consider the function $f(x) = \frac{x^2 - 4x + 5}{x - 3}$, for $x \in \mathbb{R}, x \neq 3$.
(a) Find constants $A, B,$ and $C$ such that $f(x) = Ax + B + \frac{C}{x-3}$. [2]
(b) Hence, write down the equation of the oblique asymptote of the graph of $y = f(x)$. [1]
(c) Determine the range of $f$. [5]
(d) Let $g(x) = 2x^2 - 8x + 5$ for $x \in D$, where $D = \{x \in \mathbb{R} \mid x \ge k\}$.
(i) The inverse function $g^{-1}$ exists. State the smallest possible value of $k$. [1]
(ii) For this value of $k$, find an expression for $g^{-1}(x)$ and state its domain. [4]
A ceramic component is removed from a kiln at an initial temperature of $1200^\circ\text{C}$ and placed in a room where the ambient temperature is maintained at $25^\circ\text{C}$. The temperature of the component, $T(t)$, in degrees Celsius, after $t$ minutes is modelled by Newton's Law of Cooling:
$$T(t) = 25 + A e^{-kt}, \quad t \ge 0$$
After 15 minutes, the temperature of the component is $800^\circ\text{C}$.
(a) Show that $A = 1175$. [1]
(b) Show that the constant $k \approx 0.0277$. [3]
(c) Calculate the time required for the component to cool to $100^\circ\text{C}$. [3]
(d) The rate of cooling is defined as $\left| \frac{\mathrm{d}T}{\mathrm{d}t} \right|$. Determine the temperature of the component at the instant when the rate of cooling has dropped to $10^\circ\text{C min}^{-1}$. [5]
Mark Scheme
(Confidential: For Examiner Use Only)
Question 1
(a) $z_2 = 1 - 2\mathrm{i}$
A1
(b) Attempt to multiply factors: $(z - (1+2\mathrm{i}))(z - (1-2\mathrm{i})) = ((z-1) - 2\mathrm{i})((z-1) + 2\mathrm{i})$
M1
$= (z-1)^2 - (2\mathrm{i})^2 = z^2 - 2z + 1 + 4 = z^2 - 2z + 5$
AG
(c) Long division $(z^4 - 6z^3 + 18z^2 - 30z + 25) \div (z^2 - 2z + 5)$ gives $z^2 - 4z + 5$
M1 A1
Solve $z^2 - 4z + 5 = 0 \implies z = \frac{4 \pm \sqrt{16-20}}{2}$
M1
Roots are $2 + \mathrm{i}$ and $2 - \mathrm{i}$
A1
(d) Vertices: $(1,2), (1,-2), (2,1), (2,-1)$. Trapezoid/Shoelace method.
M1
Area $= \frac{1}{2}(4+2)(1) = 3$
A1 A1
Question 2
(a) $P(1)=8 \Rightarrow a+b+c=7$; $P(-1)=-2 \Rightarrow a-b+c=-1$; $P(2)=22 \Rightarrow 4a+2b+c=14$
M1 A1 A1
(b) Solve system: Subtracting (1)-(2) gives $2b=8 \Rightarrow b=4$
A1
Sub $b=4$ implies $a+c=3$ and $4a+c=6$. Solving gives $a=1, c=2$.
M1 A1
(c) $\sum \frac{1}{\alpha} = \frac{\alpha\beta + \beta\gamma + \gamma\alpha}{\alpha\beta\gamma}$. Vieta: $\sum \alpha\beta = 4$, $\alpha\beta\gamma = -2$.
M1
Value $= 4 / (-2) = -2$
A1
Question 3
(a) Roots $u/r, u, ur$. Product $u^3 = -(-8) \Rightarrow u=2$.
M1 A1
$P(2)=0 \Rightarrow 8 - 28 + 2k - 8 = 0 \Rightarrow 2k=28 \Rightarrow k=14$.
M1 A1
Poly is $(x-2)(x^2 - 5x + 4)$. Roots $1, 2, 4 \Rightarrow r=2$.
A1
(b) AP Roots $u-d, u, u+d$. Sum $3u = -p \Rightarrow u = -p/3$.
M1
Sub root $u$ into eqn: $(-p/3)^3 + p(-p/3)^2 + q(-p/3) + r = 0$
M1 A1
Multiply by 27: $-p^3 + 3p^3 - 9pq + 27r = 0 \Rightarrow 2p^3 - 9pq + 27r = 0$
M1 A1 AG
Question 4
(a) Division: $x-1$ rem $2$. $f(x) = x-1 + \frac{2}{x-3}$. ($A=1, B=-1, C=2$)
M1 A1
(c) Set $y = \dots$ rearrange to quadratic in $x$: $x^2 - (4+y)x + (5+3y) = 0$. $\Delta \ge 0$.
M1 M1
$(4+y)^2 - 4(5+3y) \ge 0 \Rightarrow y^2 - 4y - 4 \ge 0$. Range: $y \le 2 - 2\sqrt{2}, y \ge 2 + 2\sqrt{2}$
A1 A1 A1
(d)(i) Vertex at $x=2$, so $k=2$.
A1
(d)(ii) Swap $x, y$: $x = 2(y-2)^2 - 3 \Rightarrow y = 2 + \sqrt{\frac{x+3}{2}}$.
M1 A1
Question 5
(a) $t=0, T=1200 \Rightarrow 1200 = 25 + A \Rightarrow A = 1175$.
A1
(b) $800 = 25 + 1175e^{-15k} \Rightarrow e^{-15k} = 775/1175 = 31/47$. $k = \frac{\ln(31/47)}{-15} \approx 0.0277$.
M1 M1 A1
(c) $100 = 25 + 1175e^{-0.0277t} \Rightarrow t = \frac{\ln(75/1175)}{-0.0277} \approx 99.3$ min.
M1 A1 A1
(d) Rate $\left|\frac{dT}{dt}\right| = |-k(T-25)| = 10$.
M1
$T-25 = \frac{10}{0.0277} \approx 361 \Rightarrow T \approx 386^\circ\text{C}$.
M1 A1 A1