We are asked to solve several math problems, including finding limits, probabilities, integrals, complex numbers, differential equations, and analytic geometry.

1. Problem Description

We are asked to solve several math problems, including finding limits, probabilities, integrals, complex numbers, differential equations, and analytic geometry.

2. Solution Steps

a. Limits

1. lim(x->0) $(2e^{3x} - xe^x + 2e^2 + 4)$

Since the expression is continuous at $x=0$, we can evaluate the limit by direct substitution:

$2e^{3(0)} - (0)e^0 + 2e^2 + 4 = 2(1) - 0 + 2e^2 + 4 = 6 + 2e^2$

2. lim(x->-infinity) $(x + \sqrt{x^2+9})$

We can multiply and divide by the conjugate:

lim(x->-infinity) $(x + \sqrt{x^2+9}) * \frac{x - \sqrt{x^2+9}}{x - \sqrt{x^2+9}} = lim(x->-infinity) \frac{x^2 - (x^2+9)}{x - \sqrt{x^2+9}} = lim(x->-infinity) \frac{-9}{x - \sqrt{x^2+9}}$

Since x approaches negative infinity, $\sqrt{x^2} = |x| = -x$. Then,

$lim(x->-infinity) \frac{-9}{x - \sqrt{x^2+9}} = lim(x->-infinity) \frac{-9}{x - \sqrt{x^2(1 + 9/x^2)}} = lim(x->-infinity) \frac{-9}{x - |x|\sqrt{1 + 9/x^2}} = lim(x->-infinity) \frac{-9}{x - (-x)\sqrt{1 + 9/x^2}} = lim(x->-infinity) \frac{-9}{x + x\sqrt{1 + 9/x^2}} = lim(x->-infinity) \frac{-9}{x(1 + \sqrt{1 + 9/x^2})}$

As $x$ approaches negative infinity, $9/x^2$ approaches

0. Then, the limit becomes

$lim(x->-infinity) \frac{-9}{x(1 + \sqrt{1})} = lim(x->-infinity) \frac{-9}{2x} = 0$

3. lim(x->0) $\frac{8x^2 + x + 1}{ax^2 + 1} = 1$

Since the limit exists and is equal to 1, we can directly substitute x=0:

$\frac{8(0)^2 + 0 + 1}{a(0)^2 + 1} = \frac{1}{1} = 1$

The expression is equal to 1 for all values of a.

4. lim(x->0) $\frac{sinx - sinxcosx}{x^3}$

$\frac{sinx - sinxcosx}{x^3} = \frac{sinx(1-cosx)}{x^3}$.

We know that $lim(x->0) \frac{sinx}{x} = 1$ and $1 - cosx = 2sin^2(x/2)$. Therefore,

$lim(x->0) \frac{sinx(1-cosx)}{x^3} = lim(x->0) \frac{sinx}{x} * \frac{2sin^2(x/2)}{x^2} = 1 * 2 * lim(x->0) \frac{sin^2(x/2)}{x^2} = 2 * lim(x->0) \frac{sin(x/2)}{x} * \frac{sin(x/2)}{x}$

$2 * lim(x->0) \frac{sin(x/2)}{x} * \frac{sin(x/2)}{x} = 2 * lim(x->0) \frac{sin(x/2)}{x/2} * \frac{1}{2} * \frac{sin(x/2)}{x/2} * \frac{1}{2} = 2 * 1 * \frac{1}{2} * 1 * \frac{1}{2} = \frac{1}{2}$

b. Probability

Total number of students =

2

4. 3 English, 2 French, 5 Chinese, 4 Korean, 10 Khmer.

We choose 3 students. The number of ways to do this is $C(24, 3) = \frac{24*23*22}{3*2*1} = 4*23*22 = 2024$.

A: All 3 are from Europe.

Number of European students = 3 English + 2 French =

5. Number of ways to choose 3 = $C(5, 3) = \frac{5*4*3}{3*2*1} = 10$.

$P(A) = \frac{10}{2024} = \frac{5}{1012}$.

B: All 3 from Asia, not Khmer.

Number of Asian students (excluding Khmer) = 5 Chinese + 4 Korean =

9. Number of ways to choose 3 = $C(9, 3) = \frac{9*8*7}{3*2*1} = 3*4*7 = 84$.

$P(B) = \frac{84}{2024} = \frac{21}{506}$.

C: One Khmer, one European, one Asian (not Khmer)

Number of ways to choose 1 Khmer = $C(10, 1) = 10$.

Number of ways to choose 1 European = $C(5, 1) = 5$.

Number of ways to choose 1 Asian (not Khmer) = $C(9, 1) = 9$.

Number of ways to have this event = $10 * 5 * 9 = 450$.

$P(C) = \frac{450}{2024} = \frac{225}{1012}$

c. Integrals

1. $I = \int_1^e (2x + 1 + \frac{lnx}{x})dx = \int_1^e 2x dx + \int_1^e dx + \int_1^e \frac{lnx}{x}dx = [x^2]_1^e + [x]_1^e + [\frac{(lnx)^2}{2}]_1^e$

$I = (e^2 - 1) + (e-1) + (\frac{(lne)^2}{2} - \frac{(ln1)^2}{2}) = e^2 - 1 + e - 1 + \frac{1}{2} - 0 = e^2 + e - \frac{3}{2}$

2. $J = \int \frac{6x^2 + 4x}{x^3 + x^2 - x - 1} dx$

Note that $x^3+x^2-x-1 = x^2(x+1) - (x+1) = (x^2-1)(x+1) = (x-1)(x+1)(x+1) = (x-1)(x+1)^2$

Let $\frac{6x^2+4x}{(x-1)(x+1)^2} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{C}{(x+1)^2}$

$6x^2+4x = A(x+1)^2 + B(x-1)(x+1) + C(x-1)$

$6x^2+4x = A(x^2+2x+1) + B(x^2-1) + C(x-1) = (A+B)x^2 + (2A+C)x + (A-B-C)$

Then:

$A+B = 6$

$2A+C = 4$

$A-B-C = 0$

From the first two equations $B = 6-A$ and $C=4-2A$. Substituting into the last equation:

$A - (6-A) - (4-2A) = 0$

$A - 6 + A - 4 + 2A = 0$

$4A = 10$, so $A = \frac{5}{2}$

$B = 6 - \frac{5}{2} = \frac{7}{2}$

$C = 4 - 2(\frac{5}{2}) = -1$

$J = \int (\frac{5/2}{x-1} + \frac{7/2}{x+1} - \frac{1}{(x+1)^2})dx = \frac{5}{2}ln|x-1| + \frac{7}{2}ln|x+1| + \frac{1}{x+1} + C$

3. $K = \int_0^1 (x^2+1)xdx = \int_0^1 (x^3+x) dx = [\frac{x^4}{4} + \frac{x^2}{2}]_0^1 = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}$

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

3. Final Answer

Limits: $6 + 2e^2$, 0, for any a, 1/2

Probabilities: 5/1012, 21/506, 225/1012

Integrals: $e^2 + e - \frac{3}{2}$, $\frac{5}{2}ln|x-1| + \frac{7}{2}ln|x+1| + \frac{1}{x+1} + C$, 3/4, 1