MaxWong

c + di lies on the 1st Quadrant, which means the polar form of c + di is $re^{i\theta}$, where $r = \sqrt{c^2 + d^2}$, $\theta \in \left(0, \dfrac\pi2\right)$.

e - fi lies on the 4th Quadrant, which means the polar form of e - fi is $Re^{i\phi}$, where $R = \sqrt{e^2 + f^2}$, $\phi \in \left(\dfrac{3\pi}2, 2\pi\right)$

Therefore, (c + di)(e - fi) = $rRe^{i\left(\theta + \phi\right)}$

For $\theta \in \left(0, \dfrac\pi2\right)$ and $\phi \in \left(\dfrac{3\pi}2, 2\pi\right)$, $\theta + \phi \in \left(\dfrac{3\pi}{2} , \dfrac{5\pi}{2}\right)$

Therefore (c + di)(e - fi) can lie inside the 4th Quadrant and the 1st Quadrant.

a)

$f'(x) \text{ is }\left\{\begin{matrix}\text{positive}\\\text{negative}\end{matrix}\right\}\text{ when }f(x)\text{ is }\left\{\begin{matrix}\text{strictly increasing}\\\text{strictly decreasing}\end{matrix}\right\}.$

$\implies f(x) \text{ is strictly increasing on the interval }(-\infty, 1)\cup (3, \infty)\text{ and strictly decreasing on the interval } (1, 3).$

What about when x = 1 or x = 3?

Let ε be an arbitrary constant, infinitesmally small.

$f'(1 -\varepsilon) > 0\text{ and }f'(1 + \varepsilon) < 0$

At x = 1, local maximum of f(x) occurs.

$f'(3 -\varepsilon) < 0\text{ and }f'(3 + \varepsilon) > 0$

At x = 3, local minimum of f(x) occurs.

$f'(x) \text{ is }\left\{\begin{matrix}\text{increasing}\\\text{decreasing}\\\text{constant}\end{matrix}\right\}\text{ when }f(x)\text{ }\left\{\begin{matrix}\text{is convex}\\\text{is concave}\\\text{has a point of inflexion}\end{matrix}\right\}.$

$\implies f(x) \text{ is concave on the interval } (-\infty, 2)\text{ and is convex on the interval }(2, \infty).$

$\text{Point of inflexion of }f(x)\text{ occurs at } x =2.$

b)

$f'(x) \text{ is }\left\{\begin{matrix}\text{increasing}\\\text{decreasing}\\\text{constant}\end{matrix}\right\}\text{ when }f''(x)\text{ is }\left\{\begin{matrix}\text{positive}\\\text{negative}\\\text{0}\end{matrix}\right\}.$

$\implies f''(x) \text{ is negative on the interval } (-\infty, 2)\text{ and is positive on the interval }(2, \infty).$

$f''(2) = 0$

You have two sides and the included angle, so you can find the area by using $\text{Area} = \dfrac12 bc \sin \theta$, where b and c are the side lengths and $\theta$ is the included angle.

This equation cannot be solved with algebra, and you can only approximate its roots by Newton-Raphson iteration.

If $\mathbf A^2 = \mathbf A$, then $\left(\det \mathbf A\right)^2 = \det \mathbf A$,

which means $\det \mathbf A$ = 0 or 1

Let f(x) = x(x - 1)(x - 2) Q(x) + R(x), where Q and R are polynomials which represents quotient and remainder respectively.

Let R(x) = ax^2 + bx + c.

f(0) = 4

R(0) = 4

c = 4

f(1) = 5 

R(1) = 5

a + b + c = 5

a + b = 1 --- (1)

f(2) = 10

R(2) = 10

4a + 2b + c = 10

4a + 2b = 6 --- (2)

(2) - 2 * (1) : 2a = 4

a = 2

2 + b = 1

b = -1

Therefore

R(x) = 2x^2 - x + 4

arc EH = arc FG

$x^2 - 2x = 56 - 3x\\ x^2 + x - 56 = 0\\ (x + 8)(x - 7) = 0\\ x = -8\text{(rej.) or }x = 7\\ $

Both arcs are 35 degrees.

arc EPF = $360^\circ - 70^\circ - 2\cdot 35^\circ = \boxed{220^\circ}$

By power of points,

$4(4 + 18) = 6(6 + x)\\ x = \dfrac{26}3$

By Lagrange multipliers,

$\begin{cases} \nabla_{x,y,z}{\left(4x^2 + y^2 + 16z^2 - 1\right)} = \lambda \nabla_{x,y,z}\left(7x + 2y + 8z\right)\\ 4x^2 + y^2 + 16z^2 = 1 \end{cases}$

$\begin{cases} x = \dfrac78 \lambda\\ y = \lambda\\ z = \dfrac14\lambda\\ 4x^2 + y^2 + 16z^2 = 1 \end{cases}$

Substitute the first three equations to the fourth one,

$\dfrac{49}{16}\lambda^2 + \lambda^2 + \lambda^2 = 1\\ \lambda^2 = \dfrac{16}{81}\\ \lambda = \pm \dfrac 49$

Optimum occurs when $(x,y,z) = \left(\dfrac7{18}, \dfrac49, \dfrac19\right)$ or $(x,y,z) = \left(-\dfrac7{18}, -\dfrac49, -\dfrac19\right)$

Substitute these values to find that the maximum value of 7x + 2y + 8z occurs when $(x,y,z) = \left(\dfrac7{18}, \dfrac49, \dfrac19\right)$, and the value is $\dfrac92$.

By power of points,

$7(7 + x) = 6(6 + 15)\\ 49 + 7x = 126\\ x = 11$