Detailed Solution:
The function \( f(x) \) is undefined at \( x = 1 \) because the denominator becomes zero. While \( \frac{x^2-1}{x-1} \) simplifies to \( x+1 \), this is only valid where the original expression is defined. Therefore, the functions are identical everywhere except at \( x = 1 \).

Detailed Solution:
Powers of \( i \) repeat in a cycle of four: \( i, -1, -i, 1 \). Since 2024 is a multiple of 4, \( i^{2024} = 1 \). The sequence is \( 1 + i + (-1) + (-i) \), which sums exactly to \( 0 \).

Detailed Solution:
Complete the square for both \( x \) and \( y \):
\( (x^2 – 6x + 9) + (y^2 + 8y + 16) = 0 + 9 + 16 \)
\( (x-3)^2 + (y+4)^2 = 25 \)
The radius squared \( r^2 \) is 25. Area = \( \pi r^2 = 25\pi \).

Detailed Solution:
Use the product rule: \( \log_2(x(x-2)) = 3 \).
Convert to exponential form: \( x^2 – 2x = 2^3 \), so \( x^2 – 2x – 8 = 0 \).
Factoring gives \( (x-4)(x+2) = 0 \). We must have \( x > 2 \) for the logs to be defined, so \( x = 4 \).

Detailed Solution:
No solution occurs when the lines are parallel. The ratio of coefficients of \( x \) and \( y \) must be the same: \( \frac{k}{4} = \frac{-3}{-6} \).
\( \frac{k}{4} = \frac{1}{2} \implies k = 2 \).

Detailed Solution:
Let \( R(x) = ax + b \).
\( P(2) = 2a + b = 5 \)
\( P(-1) = -a + b = 3 \)
Subtracting the equations: \( 3a = 2 \implies a = 2/3 \).
Then \( b = 3 + 2/3 = 11/3 \). So \( R(x) = \frac{2}{3}x + \frac{11}{3} \).

Detailed Solution:
Set 1: \( 2x – 5 = x + 4 \implies x = 9 \).
Set 2: \( 2x – 5 = -(x + 4) \implies 3x = 1 \implies x = 1/3 \).
Sum: \( 9 + 1/3 = 27/3 + 1/3 = 28/3 \).

Detailed Solution:
In Quadrant II, sine is positive but cosine is negative. Using \( \sin^2 + \cos^2 = 1 \):
\( \cos^2(\theta) = 1 – (3/5)^2 = 16/25 \implies \cos(\theta) = -4/5 \).
\( \tan(\theta) = \frac{\sin}{\cos} = \frac{3/5}{-4/5} = -3/4 \).

Detailed Solution:
Surface Area of Cube = \( 6s^2 = 216 \implies s^2 = 36 \implies s = 6 \).
The diameter of the inscribed sphere equals the side of the cube, so \( d = 6 \) and \( r = 3 \).
Volume \( V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi(27) = 36\pi \).

Detailed Solution:
First term \( a = 3(1/4)^1 = 3/4 \). Common ratio \( r = 1/4 \).
Sum \( S = \frac{a}{1-r} = \frac{3/4}{1 – 1/4} = \frac{3/4}{3/4} = 1 \).

Detailed Solution:
Square both sides: \( (x + \frac{1}{x})^2 = 3^2 \implies x^2 + 2 + \frac{1}{x^2} = 9 \implies x^2 + \frac{1}{x^2} = 7 \).
Square again: \( (x^2 + \frac{1}{x^2})^2 = 7^2 \implies x^4 + 2 + \frac{1}{x^4} = 49 \implies x^4 + \frac{1}{x^4} = 47 \).

Detailed Solution:
Let \( x = 2 \): \( f(2) + 2f(-1) = 4 \).
Let \( x = -1 \): \( f(-1) + 2f(2) = 1 \).
This is a system: \( A + 2B = 4 \) and \( B + 2A = 1 \).
Multiply the second by 2: \( 2B + 4A = 2 \).
Subtract the first: \( 3A = -2 \implies A = -2/3 \). Wait, let’s re-solve carefully.
\( f(-1) = 1 – 2f(2) \). Substitute: \( f(2) + 2(1 – 2f(2)) = 4 \implies f(2) + 2 – 4f(2) = 4 \implies -3f(2) = 2 \implies f(2) = -2/3 \). (Note: If choice isn’t there, recalculated: \( f(2) + 2f(-1) = 4 \), \( f(-1) + 2f(2) = 1 \). \( f(-1) = 1-2f(2) \). \( f(2) + 2 – 4f(2) = 4 \rightarrow -3f(2) = 2 \). Let’s assume A was meant as -2/3).

Detailed Solution:
Substitute: \( 3^b = 4 \implies (2^a)^b = 4 \implies 2^{ab} = 4 \).
Then \( 4^c = 8 \implies (2^{ab})^c = 8 \implies 2^{abc} = 2^3 \).
Therefore, \( abc = 3 \).

Detailed Solution:
Calculate terms: \( a_1 = 2 \), \( a_2 = 1/(1-2) = -1 \), \( a_3 = 1/(1-(-1)) = 1/2 \), \( a_4 = 1/(1-1/2) = 2 \).
The sequence cycles every 3 terms: \( (2, -1, 1/2) \). Since 2025 is a multiple of 3, \( a_{2025} = a_3 = 1/2 \).

Detailed Solution:
The maximum value of the difference \( \sqrt{x+3} – \sqrt{x-2} \) occurs as \( x \) approaches its minimum allowed value (\( x=2 \)). At \( x=2 \), the value is \( \sqrt{5} \approx 2.23 \). As \( x \to \infty \), the difference approaches 0. Since 5 is greater than the maximum possible value, there are 0 solutions.

Detailed Solution:
Let roots be \( r_1, r_2 \). By Vieta’s: \( r_1 + r_2 = 5 \) and \( r_1r_2 = 3 \).
\( r_1^2 + r_2^2 = (r_1 + r_2)^2 – 2r_1r_2 = 5^2 – 2(3) = 25 – 6 = 19 \).

Detailed Solution:
\( y = x^2 \implies x = y^{1/2} \).
\( z = y^3 \).
\( \log_{xz}(y) = \log_{y^{1/2} \cdot y^3}(y) = \log_{y^{3.5}}(y) = 1/3.5 = 2/7 \).

Detailed Solution:
Let dimensions be \( L, W, H \).
\( LW=12, WH=15, LH=20 \).
Multiply them: \( (LWH)^2 = 12 \cdot 15 \cdot 20 = 3600 \).
Volume \( V = LWH = \sqrt{3600} = 60 \).

Detailed Solution:
\( x^2 – 3x + 2 = (x-2)(x-1) \). Let \( R(x) = ax + b \).
\( P(2) = 2^{100} = 2a + b \).
\( P(1) = 1^{100} = a + b \).
Subtract: \( a = 2^{100} – 1 \).
Then \( b = 1 – (2^{100} – 1) = 2 – 2^{100} \).

Detailed Solution:
Use \( \sin(45-30) = \sin 45 \cos 30 – \cos 45 \sin 30 \).
\( = (\frac{\sqrt{2}}{2})(\frac{\sqrt{3}}{2}) – (\frac{\sqrt{2}}{2})(\frac{1}{2}) = \frac{\sqrt{6}-\sqrt{2}}{4} \).

Detailed Solution:
Using Vieta’s formulas, \( a+b = m \) and \( ab = 1 \).
The identity for the sum of cubes is \( a^3 + b^3 = (a+b)^3 – 3ab(a+b) \).
Substituting the values: \( m^3 – 3(1)(m) = m^3 – 3m \).

Detailed Solution:
Total ways to choose 3 marbles: \( \binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 \).
Ways to choose 2 red and 1 blue: \( \binom{4}{2} \times \binom{6}{1} = 6 \times 6 = 36 \).
Probability = \( 36/120 = 3/10 \).

Detailed Solution:
By the AM-GM inequality: \( \frac{x^2 + \frac{16}{x^2}}{2} \geq \sqrt{x^2 \cdot \frac{16}{x^2}} \).
\( \frac{f(x)}{2} \geq \sqrt{16} = 4 \).
So, \( f(x) \geq 8 \). The minimum value is 8.

Detailed Solution:
Since \( 5^2 + 12^2 = 13^2 \), triangle \( ABC \) is a right triangle with the right angle at \( B \).
Area = \( \frac{1}{2} \times 5 \times 12 = 30 \).
Also, Area = \( \frac{1}{2} \times \text{base}(AC) \times \text{altitude}(h) \).
\( 30 = \frac{1}{2} \times 13 \times h \implies h = 60/13 \).

Detailed Solution:
Using the change of base formula: \( \log_5(12) = \frac{\log_{10}(12)}{\log_{10}(5)} \).
\( \log_{10}(12) = \log_{10}(2^2 \cdot 3) = 2a + b \).
\( \log_{10}(5) = \log_{10}(10/2) = \log_{10}(10) – \log_{10}(2) = 1 – a \).
Result: \( \frac{2a+b}{1-a} \).