The image presents a math exam with several problems related to complex numbers, limits, integrals, differential equations, and 3D geometry. We are given two complex numbers $Z_1 = -1 + i\sqrt{3}$ and $Z_2 = -2\sqrt{3} - 2i$. We are asked to: * I.a) Calculate $Z_1 \times Z_2$ and $\frac{Z_1}{Z_2}$ in algebraic form. * I.b) Write $Z_1 \times Z_2$ and $\frac{Z_1}{Z_2}$ in trigonometric form. * II. Evaluate the limits: $A = \lim_{x \to 0} \frac{x\sin x}{\tan^2 x}$, $B = \lim_{x \to 1} \frac{3x^3 - 2x^2 + x - 2}{x^2 - 3x + 2}$, and $C = \lim_{x \to \frac{\pi}{4}} \frac{\sin x - \cos x}{\sqrt{2}(4x - \pi)}$. * III. Evaluate the integrals: $I = \int_1^2 (x^2 - 2x - 1) dx$, $J = \int_0^{\frac{\pi}{2}} (\sin 3x + 4\sin 4x) dx$, and $K = \int \frac{\sin x}{x} dx$. * IV. Given the differential equation $y'' + 4y = 0$: * a) Solve the differential equation. * b) Find the particular solution if the graph passes through $A(0, 2)$ and the tangent line at $A$ is parallel to the line $y = 2x - 1$. * V. Given points $A(1, -1, 3)$, $B(0, 3, 1)$, $C(6, -7, -1)$, $D(2, 1, 3)$, and $E(4, -6, 2)$: * a) Find the equation of the plane $(ABD)$ and the parametric equation of the line $(EC)$. * b) Show that $(EC)$ is orthogonal to the plane $ABD$.

1. Problem Description

The image presents a math exam with several problems related to complex numbers, limits, integrals, differential equations, and 3D geometry. We are given two complex numbers $Z_1 = -1 + i\sqrt{3}$ and $Z_2 = -2\sqrt{3} - 2i$. We are asked to:

* I.a) Calculate $Z_1 \times Z_2$ and $\frac{Z_1}{Z_2}$ in algebraic form.

* I.b) Write $Z_1 \times Z_2$ and $\frac{Z_1}{Z_2}$ in trigonometric form.

* II. Evaluate the limits: $A = \lim_{x \to 0} \frac{x\sin x}{\tan^2 x}$, $B = \lim_{x \to 1} \frac{3x^3 - 2x^2 + x - 2}{x^2 - 3x + 2}$, and $C = \lim_{x \to \frac{\pi}{4}} \frac{\sin x - \cos x}{\sqrt{2}(4x - \pi)}$.

* III. Evaluate the integrals: $I = \int_1^2 (x^2 - 2x - 1) dx$, $J = \int_0^{\frac{\pi}{2}} (\sin 3x + 4\sin 4x) dx$, and $K = \int \frac{\sin x}{x} dx$.

* IV. Given the differential equation $y'' + 4y = 0$:

* a) Solve the differential equation.

* b) Find the particular solution if the graph passes through $A(0, 2)$ and the tangent line at $A$ is parallel to the line $y = 2x - 1$.

* V. Given points $A(1, -1, 3)$, $B(0, 3, 1)$, $C(6, -7, -1)$, $D(2, 1, 3)$, and $E(4, -6, 2)$:

* a) Find the equation of the plane $(ABD)$ and the parametric equation of the line $(EC)$.

* b) Show that $(EC)$ is orthogonal to the plane $ABD$.

2. Solution Steps

I. Complex Numbers

a) $Z_1 = -1 + i\sqrt{3}$ and $Z_2 = -2\sqrt{3} - 2i$

$Z_1 \times Z_2 = (-1 + i\sqrt{3})(-2\sqrt{3} - 2i) = 2\sqrt{3} + 2i - 6i - 2i^2\sqrt{3} = 2\sqrt{3} + 2\sqrt{3} + 2i - 6i = 4\sqrt{3} - 4i$

$\frac{Z_1}{Z_2} = \frac{-1 + i\sqrt{3}}{-2\sqrt{3} - 2i} = \frac{-1 + i\sqrt{3}}{-2\sqrt{3} - 2i} \times \frac{-2\sqrt{3} + 2i}{-2\sqrt{3} + 2i} = \frac{2\sqrt{3} - 2i - 6i - 2\sqrt{3}}{12 + 4} = \frac{-8i}{16} = -\frac{1}{2}i$

b) Trigonometric form

$Z_1 \times Z_2 = 4\sqrt{3} - 4i = r(\cos\theta + i\sin\theta)$.

$r = \sqrt{(4\sqrt{3})^2 + (-4)^2} = \sqrt{48 + 16} = \sqrt{64} = 8$.

$\cos\theta = \frac{4\sqrt{3}}{8} = \frac{\sqrt{3}}{2}$ and $\sin\theta = \frac{-4}{8} = -\frac{1}{2}$.

Thus, $\theta = -\frac{\pi}{6}$.

$Z_1 \times Z_2 = 8(\cos(-\frac{\pi}{6}) + i\sin(-\frac{\pi}{6})) = 8(\cos(\frac{11\pi}{6}) + i\sin(\frac{11\pi}{6}))$

$\frac{Z_1}{Z_2} = -\frac{1}{2}i = 0 - \frac{1}{2}i$.

$r = \sqrt{0^2 + (-\frac{1}{2})^2} = \frac{1}{2}$.

$\cos\theta = 0$ and $\sin\theta = -1$.

Thus, $\theta = -\frac{\pi}{2}$.

$\frac{Z_1}{Z_2} = \frac{1}{2}(\cos(-\frac{\pi}{2}) + i\sin(-\frac{\pi}{2})) = \frac{1}{2}(\cos(\frac{3\pi}{2}) + i\sin(\frac{3\pi}{2}))$

II. Limits

A = $\lim_{x \to 0} \frac{x\sin x}{\tan^2 x} = \lim_{x \to 0} \frac{x\sin x}{\frac{\sin^2 x}{\cos^2 x}} = \lim_{x \to 0} \frac{x\cos^2 x}{\sin x} = \lim_{x \to 0} \frac{x}{\sin x} \times \lim_{x \to 0} \cos^2 x = 1 \times 1 = 1$.

B = $\lim_{x \to 1} \frac{3x^3 - 2x^2 + x - 2}{x^2 - 3x + 2} = \lim_{x \to 1} \frac{3x^3 - 2x^2 + x - 2}{(x - 1)(x - 2)}$.

Since the limit is indeterminate, we have $3(1)^3 - 2(1)^2 + 1 - 2 = 3 - 2 + 1 - 2 = 0$. This means $x-1$ is a factor.

Polynomial long division gives us $3x^3 - 2x^2 + x - 2 = (x-1)(3x^2 + x + 2)$.

So, $\lim_{x \to 1} \frac{(x - 1)(3x^2 + x + 2)}{(x - 1)(x - 2)} = \lim_{x \to 1} \frac{3x^2 + x + 2}{x - 2} = \frac{3(1)^2 + 1 + 2}{1 - 2} = \frac{6}{-1} = -6$.

C = $\lim_{x \to \frac{\pi}{4}} \frac{\sin x - \cos x}{\sqrt{2}(4x - \pi)}$.

Using L'Hopital's rule:

$\lim_{x \to \frac{\pi}{4}} \frac{\cos x + \sin x}{4\sqrt{2}} = \frac{\cos(\frac{\pi}{4}) + \sin(\frac{\pi}{4})}{4\sqrt{2}} = \frac{\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}}{4\sqrt{2}} = \frac{\sqrt{2}}{4\sqrt{2}} = \frac{1}{4}$.

III. Integrals

$I = \int_1^2 (x^2 - 2x - 1) dx = [\frac{x^3}{3} - x^2 - x]_1^2 = (\frac{8}{3} - 4 - 2) - (\frac{1}{3} - 1 - 1) = \frac{8}{3} - 6 - \frac{1}{3} + 2 = \frac{7}{3} - 4 = \frac{7 - 12}{3} = -\frac{5}{3}$.

$J = \int_0^{\frac{\pi}{2}} (\sin 3x + 4\sin 4x) dx = [-\frac{1}{3}\cos 3x - \cos 4x]_0^{\frac{\pi}{2}} = (-\frac{1}{3}\cos(\frac{3\pi}{2}) - \cos(2\pi)) - (-\frac{1}{3}\cos(0) - \cos(0)) = (0 - 1) - (-\frac{1}{3} - 1) = -1 + \frac{1}{3} + 1 = \frac{1}{3}$.

$K = \int \frac{\sin x}{x} dx$ This integral cannot be expressed in terms of elementary functions. It is related to the sine integral function, denoted as $Si(x) = \int_0^x \frac{\sin t}{t} dt$. Thus, $\int \frac{\sin x}{x} dx = Si(x) + C$.

IV. Differential Equation

a) $y'' + 4y = 0$. Characteristic equation: $r^2 + 4 = 0$.

$r^2 = -4$, so $r = \pm 2i$.

The general solution is $y(x) = c_1\cos(2x) + c_2\sin(2x)$.

b) $y(0) = 2$, so $c_1\cos(0) + c_2\sin(0) = 2$, which means $c_1 = 2$.

$y'(x) = -2c_1\sin(2x) + 2c_2\cos(2x)$.

The tangent line at $A(0, 2)$ has a slope of 2 (since it's parallel to $y = 2x - 1$).

So, $y'(0) = 2$. Therefore, $-2c_1\sin(0) + 2c_2\cos(0) = 2$, which means $2c_2 = 2$, so $c_2 = 1$.

The particular solution is $y(x) = 2\cos(2x) + \sin(2x)$.

V. 3D Geometry

a) Points $A(1, -1, 3)$, $B(0, 3, 1)$, $D(2, 1, 3)$, $E(4, -6, 2)$, $C(6, -7, -1)$.

$\vec{AB} = \langle -1, 4, -2 \rangle$, $\vec{AD} = \langle 1, 2, 0 \rangle$

The normal vector to the plane $ABD$ is $\vec{n} = \vec{AB} \times \vec{AD} = \begin{vmatrix} i & j & k \\ -1 & 4 & -2 \\ 1 & 2 & 0 \end{vmatrix} = i(0 - (-4)) - j(0 - (-2)) + k(-2 - 4) = 4i - 2j - 6k = \langle 4, -2, -6 \rangle$. We can divide this by 2 to simplify, giving us $\langle 2, -1, -3\rangle$.

The equation of the plane $ABD$ is $2(x - 1) - 1(y + 1) - 3(z - 3) = 0$, which simplifies to $2x - 2 - y - 1 - 3z + 9 = 0$, or $2x - y - 3z + 6 = 0$.

$\vec{EC} = \langle 6-4, -7-(-6), -1-2 \rangle = \langle 2, -1, -3 \rangle$.

Parametric equation of line $(EC)$:

$x = 4 + 2t$, $y = -6 - t$, $z = 2 - 3t$.

b) The direction vector of line $(EC)$ is $\vec{EC} = \langle 2, -1, -3 \rangle$. The normal vector to plane $(ABD)$ is $\vec{n} = \langle 2, -1, -3 \rangle$. Since $\vec{EC}$ is parallel to $\vec{n}$, the line $(EC)$ is orthogonal to the plane $ABD$.

3. Final Answer

I. a) $Z_1 \times Z_2 = 4\sqrt{3} - 4i$, $\frac{Z_1}{Z_2} = -\frac{1}{2}i$

I. b) $Z_1 \times Z_2 = 8(\cos(\frac{11\pi}{6}) + i\sin(\frac{11\pi}{6}))$, $\frac{Z_1}{Z_2} = \frac{1}{2}(\cos(\frac{3\pi}{2}) + i\sin(\frac{3\pi}{2}))$

II. $A = 1$, $B = -6$, $C = \frac{1}{4}$

III. $I = -\frac{5}{3}$, $J = \frac{1}{3}$, $K = Si(x) + C$

IV. a) $y(x) = c_1\cos(2x) + c_2\sin(2x)$

IV. b) $y(x) = 2\cos(2x) + \sin(2x)$

V. a) Plane $(ABD): 2x - y - 3z + 6 = 0$, Line $(EC): x = 4 + 2t, y = -6 - t, z = 2 - 3t$

V. b) $(EC)$ is orthogonal to plane $(ABD)$.