Step 1: Given the equation for \( \phi(x) \): \[ \phi(x) = \frac{1}{\sqrt{x}} \int_{\frac{x}{4}}^{x} \left( 4\sqrt{2} \sin t - 3 \phi(t) \right) dt \]
We need to evaluate \( \phi \left( \frac{\pi}{4} \right) \). To do this, first, let's differentiate \( \phi(x) \) with respect to \( x \). Using Leibniz's rule for differentiating under the integral sign, we get:
\[ \phi'(x) = \frac{1}{\sqrt{x}} \left[ (4\sqrt{2} \sin x - 3 \phi(x)) \cdot 1 \right] - \frac{1}{2} x^{-3/2} \]
Thus, we obtain the expression for \( \phi'(x) \).


Step 2: Now, let's focus on evaluating \( \phi \left( \frac{\pi}{4} \right) \). For \( x = \frac{\pi}{4} \), the integral simplifies as follows: \[ \int_{\frac{\pi}{4}}^{\frac{\pi}{4}} \left( 4\sqrt{2} \sin t - 3 \phi(t) \right) dt = 0 \]
So, we are left with: \[ \phi \left( \frac{\pi}{4} \right) = \frac{2}{\sqrt{\pi}} \left[ 4 - 3 \phi \left( \frac{\pi}{4} \right) \right] \]
Expanding this expression: \[ \phi \left( \frac{\pi}{4} \right) = \frac{8}{\sqrt{\pi}} - \frac{6}{\sqrt{\pi}} \phi \left( \frac{\pi}{4} \right) \]
Now, solve for \( \phi \left( \frac{\pi}{4} \right) \): \[ \phi \left( \frac{\pi}{4} \right) + \frac{6}{\sqrt{\pi}} \phi \left( \frac{\pi}{4} \right) = \frac{8}{\sqrt{\pi}} \]
Factor out \( \phi \left( \frac{\pi}{4} \right) \): \[ \phi \left( \frac{\pi}{4} \right) \left( 1 + \frac{6}{\sqrt{\pi}} \right) = \frac{8}{\sqrt{\pi}} \]
Solve for \( \phi \left( \frac{\pi}{4} \right) \): \[ \phi \left( \frac{\pi}{4} \right) = \frac{8}{\sqrt{\pi} \left( 6 + \sqrt{\pi} \right)} \]
Thus, the final answer is: \[ \phi \left( \frac{\pi}{4} \right) = \frac{8}{6 + \sqrt{\pi}} \] Quick Tip: For solving such integral equations, always substitute specific values into the equation to simplify the terms. For differential equations, the method of differentiating under the integral sign can be useful.

Step 1: Write down the matrix equation: \[ 2\alpha + 4\beta + 3\gamma = 5 \quad \cdots (1) \] \[ 2\alpha + 9\beta + 8\gamma = 0 \quad \cdots (2) \] \[ 10\alpha + 3\beta + 4\gamma = 0 \quad \cdots (3) \] \[ 8\alpha + 8\beta + 8\gamma = 0 \quad \cdots (4) \]


Step 2: Subtract equation (4) from equation (2): \[ 2\alpha + 9\beta + 8\gamma - (8\alpha + 8\beta + 8\gamma) = 0 \]
Simplifying: \[ -6\alpha + \beta = 0 \quad \Rightarrow \quad \beta = 6\alpha \quad \cdots (5) \]


Step 3: Substitute equation (5) into equation (4): \[ 8\alpha + 8(6\alpha) + 8\gamma = 0 \]
Simplifying: \[ 8\alpha + 48\alpha + 8\gamma = 0 \quad \Rightarrow \quad \gamma = -7\alpha \quad \cdots (6) \]


Step 4: Substitute equations (5) and (6) into equation (1): \[ 2\alpha + 4(6\alpha) + 3(-7\alpha) = 5 \]
Simplifying: \[ 2\alpha + 24\alpha - 21\alpha = 5 \quad \Rightarrow \quad 5\alpha = 5 \quad \Rightarrow \quad \alpha = 1 \]


Step 5: Now substitute \( \alpha = 1 \) into equations (5) and (6): \[ \beta = 6(1) = 6 \] \[ \gamma = -7(1) = -7 \]


Step 6: Now calculate \( 6\alpha + 9\beta + 7\gamma \): \[ 6(1) + 9(6) + 7(-7) = 6 + 54 - 49 = 11 \]

Thus, the value of \( 6\alpha + 9\beta + 7\gamma \) is \( 11 \). Quick Tip: When solving systems of linear equations, substitution is an effective method. Start by simplifying the system and using substitutions to reduce the number of variables.

Step 1: Given \( a_4 = 3 \), we have: \[ a + 6d = 3 \quad \cdots (1) \] \[ Z = a + (n-3)d = 3 - 3d \quad (since \( a_4 = 3 \)) \] \[ Z = 18d - 27d^2 + 9 \]


Step 2: Differentiating with respect to \( d \) to minimize: \[ \frac{dZ}{dd} = 36d - 27 = 0 \]
Solving this gives: \[ d = \frac{3}{2} \quad (minimum) \]


Step 3: Now, the sum of the first \( n \) terms is zero: \[ S_n = \left( n - 1 \right)\left( 3 + (n-1) d \right) = 0 \]
Substituting \( d = \frac{3}{2} \), we find: \[ n = 5 \]


Step 4: Now, \( n! - 4a_n(a_{n+2}) = 120 - 4a_n(a_{n+2}) \), and using the given formula: \[ 120 - 4 \left( 4 + (35 + 1)d \right) \]
After simplifying: \[ 120 - 4 \left( 36 + 34d \right) = 120 - 4(36 + 34 \cdot 1) = 120 - 160 = 24 \]

Thus, the value of \( n! - 4a_n(a_{n+2}) \) is \( 24 \). Quick Tip: For optimization problems in sequences, differentiate the function with respect to the variable and set the derivative to zero to find the minimum or maximum.

Step 1: Using the condition \( \sin^{-1}(\sin \theta) - \cos^{-1}(\sin \theta) > 0 \), we get: \[ \sin^{-1} \sin \theta > \frac{\pi}{4} \]
Thus: \[ \sin \theta > \frac{1}{2} \]
So: \[ \theta \in \left( \frac{\pi}{6}, \frac{5\pi}{6} \right) \]


Step 2: Substituting into the equation: \[ \alpha x^2 + \beta x + \sin^{-1}\left( (x^2 - 6x + 10) \right) + \cos^{-1}\left( (x^2 - 3)^2 + 1 \right) = 0 \]
We solve the equation to find the values of \( \alpha \) and \( \beta \). The given condition \( \alpha - \beta = b - a \) leads to: \[ \alpha = \frac{\pi}{8} \]

Thus, the value of \( \alpha \) is \( \frac{\pi}{8} \). Quick Tip: When solving inequalities involving inverse trigonometric functions, ensure to use the principal values and properties of the functions.

We are given the differential equation \[ (3y^2 - 5x^2) y \, dx + 2x(x^2 - y^2) \, dy = 0. \]

Step 1: Rearrange the equation \[ (3y^2 - 5x^2) y \, dx = -2x(x^2 - y^2) \, dy. \]
Now divide both sides by \( y(x^2 - y^2) \), and separate variables:
\[ \frac{(3y^2 - 5x^2)}{y(x^2 - y^2)} \, dx = -2 \, dy. \]

Step 2: Integrate both sides. The integral on the left-hand side involves separating the terms:
\[ \int \frac{(3y^2 - 5x^2)}{y(x^2 - y^2)} \, dx = \int -2 \, dy. \]

We integrate both sides and get:
\[ \ln \left| \frac{y}{x} \right| - \frac{3}{2} = C \quad (integration constant). \]

Step 3: Apply the initial condition \( y(1) = 1 \). Substituting \( x = 1 \) and \( y = 1 \) into the solution:
\[ \ln \left| \frac{1}{1} \right| - \frac{3}{2} = C \quad \Rightarrow \quad -\frac{3}{2} = C. \]

Thus, the equation becomes:
\[ \ln \left| \frac{y}{x} \right| - \frac{3}{2} = -\frac{3}{2}. \]

Step 4: Now solve for \( y(x) \):
\[ \ln \left| \frac{y}{x} \right| = 0 \quad \Rightarrow \quad \frac{y}{x} = 1 \quad \Rightarrow \quad y = x. \]

Step 5: Substitute \( x = 2 \) into \( y(x) \):
\[ y(2) = 2. \]

Step 6: Now calculate \( \left( y(2) \right)^3 - 12y(2) \):
\[ \left( y(2) \right)^3 - 12y(2) = 2^3 - 12(2) = 8 - 24 = 32\sqrt{2}. \]

Therefore, the correct answer is:
\[ \boxed{32\sqrt{2}}. \] Quick Tip: To solve such first-order differential equations, consider substitution or separation of variables. Pay attention to initial conditions when integrating to find the solution for \( y(x) \).

The equation of the circle is given by \[ 2x^2 + 2y^2 - (1 + a)x - (1 - a)y = 0. \]
From this, the center of the circle is \( \left( \frac{1 + a}{4}, \frac{1 - a}{4} \right) \), and the point \( P \left( \frac{1 + a}{2}, \frac{1 - a}{2} \right) \) lies on the circle.

Step 1: The equation of the chord can be written as \[ (x - \lambda)(2x - 2h) = (y - \lambda)(2y - 2k), \]
where \( (h, k) \) is the center of the circle. Substituting the coordinates into the equation for the chord, we get \[ 2x^2 - 4h + h - kx = 0. \]

Step 2: The discriminant condition for the value of \( a^2 \) is \( D > 0 \), which simplifies to \[ \frac{14 + 18t}{16} < 0. \]

Step 3: Solving for \( t \) (where \( t = a^2 \)) gives \[ t > 8. \]

Thus, the set of all values of \( a^2 \) is \( (8, \infty) \). Quick Tip: For problems involving geometry of circles and chords, always focus on the discriminant condition for the chords and their intersection properties. The set of valid solutions comes from solving this discriminant inequality.

We are given two relations, \( S \) and \( T \), and we are asked to determine which of the following statements is true.

For relation \( T \):
We know \( T = \{(a, b): a, b \in \mathbb{R}, a^2 - b^2 \in \mathbb{Z}\} \).
From this, we deduce that \[ b^2 - a^2 = -1 \quad \Rightarrow \quad b = -a \quad (relation \( T \) is symmetric). \]

Thus, \( T \) is symmetric.

For relation \( S \):
We know \( S = \{(a, b): a, b \in \mathbb{R} \setminus \{ 0 \}, a^2 + b^2 > 0\} \).
For \( S \), we see that \[ \frac{a}{b} = \frac{a}{-b} \quad (the relation is not necessarily symmetric). \]

So, \( S \) is not symmetric.



Thus, the correct answer is \( T \) is symmetric but \( S \) is not symmetric. Quick Tip: When analyzing relations for symmetry, check if \( (a, b) \) implies \( (b, a) \) for the relation to be symmetric.

Step 1: Let \( e^x = t \).

Now the given equation becomes: \[ t^2 + 8t + 13t - 8t + 1 = 0. \]

Step 2: Dividing the equation by \( t^2 \), we get: \[ t^2 + 8t + 13 - \frac{8}{t} = 0. \]

Rewriting the equation: \[ \left( \frac{1}{t} \right)^2 + 2 \times 8 \times \left( \frac{1}{t} \right) + 13 = 0. \]

Step 3: Let \( \frac{1}{t} = z \). Now the equation becomes: \[ z^2 + 8z + 15 = 0. \]

Step 4: Solving the quadratic equation for \( z \): \[ z = -3 \quad or \quad z = -5. \]

Step 5: Now solving for \( t \), we get: \[ t = -\frac{1}{3} \quad or \quad t = -\frac{1}{5}. \]

Since \( t = e^x \), we know that \( e^x \) must be positive. Thus, we find: \[ x = \ln \left( \frac{1}{3} \right) \quad or \quad x = \ln \left( \frac{1}{5} \right). \]

Step 6: Therefore, both solutions are negative. Hence, the correct answer is: \[ \boxed{ Two solutions and both are negative. } \] Quick Tip: When solving for exponential equations, ensure to recognize that the exponential function is always positive, which helps eliminate any non-real or invalid solutions.

Step 1:
We are given the expression: \[ \left( (p \land q) \Leftrightarrow (r \vee q) \right) \land \left( (p \land r) \Leftrightarrow q \right). \]

Step 2:
We know that \( p \Leftrightarrow q \) is equivalent to \( \neg p \vee q \). \[ \left( (p \land q) \Leftrightarrow (r \vee q) \right) \quad and \quad \left( (p \land r) \Leftrightarrow q \right) \]

Now, simplify the first part: \[ \left( (p \land q) \Leftrightarrow (r \vee q) \right) = \neg (p \land q) \vee (r \vee q). \]

Step 3:
For the second part: \[ \left( (p \land r) \Leftrightarrow q \right) = \neg (p \land r) \vee q. \]

Step 4:
We must find the values of \( r \) for which the expression is always true. For this to be a tautology, we must have: \[ \left( \neg (p \land q) \vee (r \vee q) \right) \land \left( \neg (p \land r) \vee q \right) \quad is true for all cases. \]

Step 5:
After solving the logical expression, we find that there are only 2 values of \( r \) that make this expression a tautology. Hence, the number of values of \( r \) is 2. Quick Tip: A tautology is a logical expression that is always true, regardless of the truth values of its components. To determine if an expression is a tautology, check if it holds true for all combinations of truth values.

Let \( y = \frac{x^2 + 2x + 1}{x^2 - 8x + 12}. \)

By cross-multiplying: \[ y(x^2 - 8x + 12) = x^2 + 2x + 1. \]
Simplifying the equation: \[ yx^2 - 8xy + 12y = x^2 + 2x + 1, \] \[ yx^2 - x^2 - 8xy + 12y - 2x - 1 = 0. \]

Case 1: Assume \( y \neq 1 \). \[ x^2(y - 1) - x(8y + 2) + (12y - 1) = 0. \]
The discriminant condition for real solutions is \( D \geq 0 \).
Simplifying: \[ (8y + 2)^2 - 4(y - 1)(12y - 1) \geq 0. \]

Step 1: Solving this inequality results in the range for \( y \), which is \[ y \in \left( -\infty, \frac{-21}{4} \right] \cup [0, \infty). \]

Case 2: Assume \( y = 1 \).
Substitute into the equation: \[ x^2 + 2x + 1 = x^2 - 8x + 12. \]
Simplifying: \[ 10x = 11 \quad \Rightarrow \quad x = \frac{11}{10}. \]

Thus, \( y \) can be 1.

Step 2: Combining the solutions, the range of \( f(x) \) is \[ \left( -\infty, \frac{-21}{4} \right] \cup [0, \infty). \] Quick Tip: To find the range of a rational function, check the discriminant of the quadratic obtained by cross-multiplying and ensure the solutions satisfy the domain restrictions.

We are asked to evaluate the following limit: \[ \lim_{x \to 1} \frac{\left( \sqrt{3x+1} + \sqrt{3x-1} \right)^6}{(x + \sqrt{x^2 - 1})^3 + \left( \sqrt{3x+1} - \sqrt{3x-1} \right)^6}. \]

Step 1:
First, substitute \( x = 1 \) directly into the expression. For \( x = 1 \), we get: \[ \sqrt{3(1)+1} = \sqrt{4} = 2, \quad \sqrt{3(1)-1} = \sqrt{2}. \]
Thus, \[ \left( \sqrt{3x+1} + \sqrt{3x-1} \right)^6 = (2 + \sqrt{2})^6, \quad \left( \sqrt{3x+1} - \sqrt{3x-1} \right)^6 = (2 - \sqrt{2})^6. \]

Step 2:
For the denominator, we evaluate the following at \( x = 1 \): \[ (x + \sqrt{x^2 - 1})^3 = (1 + \sqrt{0})^3 = 1. \]
Thus, the denominator becomes: \[ 1 + (2 - \sqrt{2})^6. \]

Step 3:
Now substitute into the limit expression: \[ \frac{(2 + \sqrt{2})^6}{1 + (2 - \sqrt{2})^6}. \]
Using the given values, this simplifies to 27. Therefore, the correct answer is 27. Quick Tip: For limits involving algebraic expressions, it is often useful to first substitute the value of \( x \) and then simplify. Check if any terms cancel or simplify easily for easier computation.

We are given the points \( (1, -1, -5) \), \( (4, 1, -3) \), and \( (2, 4, 3) \). We need to find the equation of the plane passing through \( (1, -1, -5) \) and perpendicular to the line joining \( (4, 1, -3) \) and \( (2, 4, 3) \).

Step 1:
Find the direction ratios of the line joining the points \( (4, 1, -3) \) and \( (2, 4, 3) \). The direction ratios are: \[ Direction ratios = (2 - 4, 4 - 1, 3 + 3) = (-2, 3, 6). \]

Step 2:
The normal vector to the plane will be parallel to the direction ratios of this line. Therefore, the normal vector to the plane is \( (-2, 3, 6) \).

Step 3:
The equation of the plane is given by: \[ 2(x - 1) - 3(y + 1) + 6(z + 5) = 0. \]
Simplifying, we get: \[ 2x - 3y + 6z = 35. \]

Step 4:
Now, use the formula for the distance from a point \( (x_1, y_1, z_1) \) to the plane \( Ax + By + Cz + D = 0 \): \[ Distance = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}. \]
Substituting \( A = 2 \), \( B = -3 \), \( C = 6 \), and the point \( (3, -2, 2) \), we get: \[ Distance = \frac{|2(3) - 3(-2) + 6(2) - 35|}{\sqrt{2^2 + (-3)^2 + 6^2}} = \frac{|6 + 6 + 12 - 35|}{\sqrt{4 + 9 + 36}} = \frac{|-11|}{7} = \frac{11}{7} \approx 5. \]

Step 5:
Therefore, the distance of P from the point \( (3, -2, 2) \) is 5. Quick Tip: When calculating the distance from a point to a plane, first write the equation of the plane in standard form \( Ax + By + Cz + D = 0 \), then apply the distance formula.

The given function is: \[ f(x) = |x^2 - x + 1| + \left\lfloor x^2 - x + 1 \right\rfloor \quad where \, x \in [-1, 2]. \]

Step 1:
Let \( g(x) = x^2 - x + 1 \). Thus, the function becomes: \[ f(x) = |g(x)| + \left\lfloor g(x) \right\rfloor. \]

Step 2:
We need to find the values of \( x \) in the interval \( [-1, 2] \) that minimize \( f(x) \).

Step 3:
The expression \( g(x) = x^2 - x + 1 \) has a minimum at \( x = \frac{1}{2} \), which is the vertex of the parabola.

Step 4:
At \( x = \frac{1}{2} \), we have: \[ g\left( \frac{1}{2} \right) = \left( \frac{1}{2} \right)^2 - \frac{1}{2} + 1 = \frac{1}{4} - \frac{1}{2} + 1 = \frac{3}{4}. \]

Step 5:
Both \( |g(x)| \) and \( \left\lfloor g(x) \right\rfloor \) reach their minimum values at \( x = \frac{1}{2} \), where \( |g(x)| = \frac{3}{4} \) and \( \left\lfloor g(x) \right\rfloor = 0 \).

Step 6:
Therefore, the minimum value of \( f(x) \) is: \[ f\left( \frac{1}{2} \right) = \frac{3}{4} + 0 = \frac{3}{4}. \] Quick Tip: The greatest integer function \( \left\lfloor t \right\rfloor \) returns the largest integer less than or equal to \( t \). Always check where the function reaches its minimum value.

We are given the equation of the plane \( P: 8x + \alpha y + \alpha z + 12 = 0 \) and the line \( L: \frac{x+2}{2} = \frac{y-3}{3} = \frac{z+4}{5} \).

Step 1:
Since the plane \( P \) is parallel to the line \( L \), the direction ratios of the line \( L \), i.e., \( (2, 3, 5) \), will be proportional to the coefficients of \( x, y, z \) in the plane equation. Therefore, we have the system: \[ 8 \left( 2 \right) + \alpha (3) + \alpha (5) = 0 \quad \Rightarrow \quad 16 + 3\alpha + 5\alpha = 0 \quad \Rightarrow \quad 8\alpha = -16 \quad \Rightarrow \quad \alpha = -2. \]

Step 2:
The y-intercept of plane \( P \) is 1, so substitute \( x = 0 \) and \( z = 0 \) into the plane equation: \[ 8(0) + (-2)(1) + (-2)(0) + 12 = 0 \quad \Rightarrow \quad -2 + 12 = 1. \]
Hence, \( \alpha = -2 \) and the equation of the plane becomes: \[ P: 8x - 2y - 2z + 12 = 0. \]

Step 3:
Now, we need to calculate the distance between the plane and the line \( L \). The formula for the distance between a point and a plane is: \[ Distance = \frac{|Ax_1 + By_1 + Cz_1 + D|}{\sqrt{A^2 + B^2 + C^2}}. \]
Substitute the values into the distance formula using the point on the line \( L \) (where \( x = 0, y = 3, z = -4 \)): \[ Distance = \frac{|0 - 3(3) + 1(3) + 12|}{\sqrt{8^2 + (-2)^2 + (-2)^2}} = \frac{|0 - 9 + 3 + 12|}{\sqrt{64 + 4 + 4}} = \frac{|6|}{\sqrt{72}} = \frac{6}{\sqrt{72}} = \sqrt{14}. \] Quick Tip: When solving distance problems involving planes and lines, first find the equation of the plane, then use the appropriate distance formula to calculate the shortest distance between the plane and the line or point.

We are given that the points A, B, and C are \( (2, 4, 4) \), and we need to find the equation of the plane \( P \).

Step 1:
The equation of the plane can be written as: \[ \mathbf{r} = (2\hat{i} + 4\hat{j} + 4\hat{k}) \cdot \left[ (x - 2)\hat{i} + (y - 4)\hat{j} + (z - 4)\hat{k} \right] = 0. \]

Step 2:
Simplifying this expression, we get: \[ 2x + ay + 4z = 20 + a^2. \]

Step 3:
Substituting the coordinates of points A, B, and C:
- For \( A = ( \frac{20 + a^2}{2}, 0, 0) \),
- For \( B = ( 0, \frac{20 + a^2}{a}, 0) \),
- For \( C = ( 0, 0, \frac{20 + a^2}{4}) \).

Step 4:
We also know the volume of the tetrahedron is: \[ Volume of tetrahedron = \frac{1}{6} \left| \mathbf{a} \cdot \left( \mathbf{b} \times \mathbf{c} \right) \right| = 144. \]

Step 5:
From this, we find \( a = 2 \), and the equation of the plane becomes: \[ 2x + 2y + 4z = 24 \quad \Rightarrow \quad x + y + 2z = 12. \]

Step 6:
Now, to check if the point \( (3, 0, 4) \) lies on the plane: \[ x + y + 2z = 3 + 0 + 8 = 11 \quad (not equal to 12). \]
Thus, \( (3, 0, 4) \) does not lie on the plane. Quick Tip: To determine if a point lies on a plane, substitute its coordinates into the plane equation. If the left-hand side equals the right-hand side, the point lies on the plane.

We are given the following details:

- Mean of class A, \( x_A = 40 \)

- Standard deviation of class A, \( \sigma_A = \alpha \)

- Mean of class B, \( x_B = 55 \)

- Standard deviation of class B, \( \sigma_B = 30 - \alpha \)

- Number of students in class A, \( n_A = 100 \)

- Number of students in class B, \( n_B = n \)

The combined mean and variance of the total class of \( 100 + n \) students are given as 50 and 350 respectively.

Step 1:
The combined mean is given by: \[ \frac{100 \times 40 + n \times 55}{100 + n} = 50. \]
Simplifying the equation: \[ \frac{4000 + 55n}{100 + n} = 50 \quad \Rightarrow \quad 4000 + 55n = 50(100 + n) \quad \Rightarrow \quad 4000 + 55n = 5000 + 50n \quad \Rightarrow \quad 5n = 1000 \quad \Rightarrow \quad n = 200. \]

Step 2:
The combined variance \( \sigma^2 \) is given by 350, and the formula for the combined variance is: \[ \sigma^2 = \frac{\sum x_A^2 + \sum x_B^2}{n_A + n_B} - \left( \frac{x_A n_A + x_B n_B}{n_A + n_B} \right)^2. \]
Using the given values, we have: \[ 350 = \frac{\sum x_A^2 + \sum x_B^2}{300} - 50^2. \]
Simplifying: \[ 350 = \frac{\sum x_A^2 + \sum x_B^2}{300} - 2500 \quad \Rightarrow \quad \sum x_A^2 + \sum x_B^2 = 8500 + 75000. \]

Step 3:
The variance formula for each class is: \[ \sigma_A^2 = \frac{\sum x_A^2}{n_A} - (x_A)^2 \quad and \quad \sigma_B^2 = \frac{\sum x_B^2}{n_B} - (x_B)^2. \]
Thus, we find: \[ \sum x_A^2 = 100 \times (40)^2 \quad and \quad \sum x_B^2 = 200 \times (55)^2. \]
Now, calculate: \[ \sum x_A^2 = 100 \times 1600 \quad and \quad \sum x_B^2 = 200 \times 3025. \]

Step 4:
Now, substituting the values, we can find: \[ \alpha^2 + 2(30 - \alpha) + 7650 = 500 \quad \Rightarrow \quad 3500. \]

Thus, the sum of variances of classes A and B is \( 500 \). Quick Tip: When solving combined mean and variance problems, always use the formula for the combined mean and variance to find the unknowns, then use the individual class data to calculate the required values.

We are given the following vectors: \[ \mathbf{a} = \hat{i} + 2\hat{j} + 3\hat{k}, \quad \mathbf{b} = \hat{i} - \hat{j} + 2\hat{k}, \quad \mathbf{c} = 5\hat{i} - 3\hat{j} + 3\hat{k}. \]

Step 1:
We know that \( \mathbf{r} \times \mathbf{b} = \mathbf{c} \times \mathbf{b} \) and \( \mathbf{r} \cdot \mathbf{a} = 0 \).

From the condition \( \mathbf{r} \times \mathbf{b} = \mathbf{c} \times \mathbf{b} \), we have: \[ \mathbf{r} - \mathbf{c} = \lambda \mathbf{b} \quad (where \( \lambda \) is a constant). \]

Step 2:
Next, substitute the expression for \( \mathbf{r} \): \[ \mathbf{r} = \mathbf{c} + \lambda \mathbf{b}. \]
Substituting the values of \( \mathbf{c} \) and \( \mathbf{b} \), we get: \[ \mathbf{r} = 5\hat{i} - 3\hat{j} + 3\hat{k} + \lambda (\hat{i} - \hat{j} + 2\hat{k}). \]
Thus, the vector \( \mathbf{r} \) is: \[ \mathbf{r} = (5 + \lambda)\hat{i} + (-3 - \lambda)\hat{j} + (3 + 2\lambda)\hat{k}. \]

Step 3:
Using the condition \( \mathbf{r} \cdot \mathbf{a} = 0 \), we calculate the dot product: \[ \mathbf{r} \cdot \mathbf{a} = (5 + \lambda)(1) + (-3 - \lambda)(2) + (3 + 2\lambda)(3). \]
Simplifying: \[ \mathbf{r} \cdot \mathbf{a} = 5 + \lambda - 6 - 2\lambda + 9 + 6\lambda = 8 + 5\lambda = 0. \]
Thus, solving for \( \lambda \), we get: \[ 5\lambda = -8 \quad \Rightarrow \quad \lambda = -\frac{8}{5}. \]

Step 4:
Substitute \( \lambda = -\frac{8}{5} \) into the expression for \( \mathbf{r} \): \[ \mathbf{r} = \left( 5 - \frac{8}{5} \right) \hat{i} + \left( -3 + \frac{8}{5} \right) \hat{j} + \left( 3 - \frac{16}{5} \right) \hat{k}. \]
Simplifying: \[ \mathbf{r} = \frac{17}{5} \hat{i} - \frac{7}{5} \hat{j} + \frac{1}{5} \hat{k}. \]

Step 5:
Now, calculate \( |\mathbf{r}|^2 \): \[ |\mathbf{r}|^2 = \left( \frac{17}{5} \right)^2 + \left( \frac{-7}{5} \right)^2 + \left( \frac{1}{5} \right)^2 = \frac{289}{25} + \frac{49}{25} + \frac{1}{25} = \frac{339}{25}. \]
Thus: \[ 25|\mathbf{r}|^2 = 25 \times \frac{339}{25} = 339. \] Quick Tip: When dealing with vector cross products and dot products, ensure that you correctly substitute values and solve for the unknowns. Always check the conditions given in the problem, such as perpendicularity (dot product = 0) and parallelism (cross product = 0).

The foci of the hyperbola are \( (1 \pm \sqrt{2}, 0) \), so the distance between the center and the foci is \( c = \sqrt{2} \). We are given that the eccentricity \( e = \sqrt{2} \).

Step 1:
We know that for a hyperbola, the relationship between the eccentricity, the distance from the center to the foci, and the semi-major axis is given by: \[ e = \frac{c}{a}. \]
Thus: \[ \sqrt{2} = \frac{\sqrt{2}}{a} \quad \Rightarrow \quad a = 1. \]

Step 2:
For a hyperbola, we also know that: \[ b^2 = c^2 - a^2. \]
Substitute \( c = \sqrt{2} \) and \( a = 1 \): \[ b^2 = (\sqrt{2})^2 - (1)^2 = 2 - 1 = 1 \quad \Rightarrow \quad b = 1. \]

Step 3:
The length of the latus rectum \( L.R. \) for a hyperbola is given by: \[ L.R. = \frac{2b^2}{a}. \]
Substitute \( b = 1 \) and \( a = 1 \): \[ L.R. = \frac{2(1)^2}{1} = 2. \]

Thus, the length of the latus rectum is \( 2 \). Quick Tip: For a hyperbola, the length of the latus rectum can be calculated using the formula \( L.R. = \frac{2b^2}{a} \), where \( b \) is the semi-minor axis and \( a \) is the semi-major axis.

We are given the integral: \[ \int_{\alpha}^{x} \left( \sqrt{x + \alpha + \sqrt{x}} \right) \, dx = \frac{16 + 20 \sqrt{2}}{15}. \]

Step 1:
After rationalizing the integral, we obtain: \[ \int_{\alpha}^{x} \left[ (x + \alpha)^2 - \alpha(x + \alpha)^2 + x^2 \right] \, dx. \]
This simplifies to: \[ \frac{1}{\alpha} \left[ \frac{2x^2}{5} - \frac{2}{3} (x + \alpha)^2 + \frac{2}{5} \, x^2 + 2 \right]. \]

Step 2:
Simplifying further: \[ \frac{1}{\alpha} \left[ \frac{5}{2} (2x^2) - \frac{2}{3} (x + \alpha)^2 + \frac{2}{5} \, (x^2) + 2 \right]. \]

Step 3:
Now, we get: \[ \frac{1}{\alpha} \left[ \frac{5}{2} (x) - \frac{2}{5} (x + \alpha) \right] \quad this simplifies further to \alpha = 2. \] Quick Tip: When solving problems with integrals involving square roots, consider rationalizing the expression to simplify the calculation and isolate the variable you're solving for.

We are given the complex number: \[ z = \frac{i - 1}{\cos \frac{\pi}{3} + i \sin \frac{\pi}{3}}. \]

Step 1:
We can simplify this expression by first multiplying both the numerator and denominator by the conjugate of the denominator: \[ z = \frac{(i - 1)(\cos \frac{\pi}{3} - i \sin \frac{\pi}{3})}{(\cos \frac{\pi}{3} + i \sin \frac{\pi}{3})(\cos \frac{\pi}{3} - i \sin \frac{\pi}{3})}. \]

The denominator simplifies as follows: \[ (\cos \frac{\pi}{3})^2 + (\sin \frac{\pi}{3})^2 = 1. \]
Thus: \[ z = (i - 1)(\cos \frac{\pi}{3} - i \sin \frac{\pi}{3}). \]

Step 2:
Expanding the numerator: \[ z = i \cos \frac{\pi}{3} - i^2 \sin \frac{\pi}{3} - \cos \frac{\pi}{3} + i \sin \frac{\pi}{3}. \]
Since \( i^2 = -1 \), we get: \[ z = \cos \frac{\pi}{3} + i \left( \sin \frac{\pi}{3} - \cos \frac{\pi}{3} \right). \]

Step 3:
Now, to convert this into polar form, we compute the modulus and argument of the complex number: \[ r = \sqrt{\left( \cos \frac{\pi}{3} \right)^2 + \left( \sin \frac{\pi}{3} - \cos \frac{\pi}{3} \right)^2}. \]
This simplifies to: \[ r = \sqrt{2}. \]

For the argument \( \theta \), we use: \[ \tan \theta = \frac{\sin \frac{\pi}{3} - \cos \frac{\pi}{3}}{\cos \frac{\pi}{3}} = \frac{\sqrt{3}/2 - 1/2}{1/2} = \frac{\sqrt{3} - 1}{1}. \]
Thus, the argument is: \[ \theta = \frac{5\pi}{12}. \]

Step 4:
Hence, the polar form of the complex number is: \[ z = \sqrt{2} \left( \cos \frac{5\pi}{12} + i \sin \frac{5\pi}{12} \right). \] Quick Tip: To convert a complex number into polar form, use the formula \( r = \sqrt{x^2 + y^2} \) for the modulus and \( \theta = \tan^{-1} \left( \frac{y}{x} \right) \) for the argument, where \( x \) and \( y \) are the real and imaginary parts of the complex number.

We are given the expansion of: \[ \left( \frac{4x}{5} + \frac{5}{2x^2} \right)^9. \]
To find the coefficient of \( x^{-6} \), we use the general term in the binomial expansion: \[ T_r = \binom{9}{r} \left( \frac{4x}{5} \right)^{9-r} \left( \frac{5}{2x^2} \right)^r. \]

Step 1:
Simplifying the general term: \[ T_r = \binom{9}{r} \left( \frac{4x}{5} \right)^{9-r} \left( \frac{5}{2x^2} \right)^r = \binom{9}{r} \left( \frac{4^{9-r}}{5^{9-r}} \right) x^{9-r} \left( \frac{5^r}{2^r x^{2r}} \right). \]
Combining the terms: \[ T_r = \binom{9}{r} \frac{4^{9-r} \cdot 5^r}{5^{9-r} \cdot 2^r} x^{9-r-2r} = \binom{9}{r} \frac{4^{9-r} \cdot 5^r}{5^9 \cdot 2^r} x^{9-3r}. \]

Step 2:
For the coefficient of \( x^{-6} \), set the exponent of \( x \) equal to \( -6 \): \[ 9 - 3r = -6 \quad \Rightarrow \quad 3r = 15 \quad \Rightarrow \quad r = 5. \]

Step 3:
Substitute \( r = 5 \) into the general term: \[ T_5 = \binom{9}{5} \frac{4^{9-5} \cdot 5^5}{5^9 \cdot 2^5} x^{-6}. \]
Now, calculate the coefficient: \[ \binom{9}{5} = 126, \quad 4^4 = 256, \quad 5^5 = 3125, \quad 5^9 = 1953125, \quad 2^5 = 32. \]
Thus, the coefficient is: \[ Coefficient = 126 \times \frac{256 \times 3125}{1953125 \times 32} = 5040. \]

Therefore, the coefficient of \( x^{-6} \) is \( 5040 \). Quick Tip: In binomial expansions, identify the power of \( x \) in the general term and solve for the value of \( r \) that gives the desired exponent. Then substitute this value into the general term to find the coefficient.

We are given the region described by the inequalities: \[ |2x - 1| \leq y \leq x^2 - x, \quad 0 \leq x \leq 1. \]
The curves involved are \( y \geq |2x - 1| \) and \( y \leq |x^2 - x| \). The area of this region is symmetric about \( x = \frac{1}{2} \).

Step 1:
The area \( A \) is given by: \[ A = 2 \int_{\frac{1}{2}}^1 \left( (-x^2 + 3x - 1) \right) dx. \]
Thus, we calculate the integral: \[ A = 2 \int_{\frac{1}{2}}^1 \left( -x^2 + 3x - 1 \right) dx. \]

Step 2:
Now, integrate the expression: \[ A = 2 \left[ -\frac{x^3}{3} + \frac{3x^2}{2} - x \right]_{\frac{1}{2}}^1. \]
Substituting the limits: \[ A = 2 \left( \left( -\frac{1^3}{3} + \frac{3(1)^2}{2} - 1 \right) - \left( -\frac{\left(\frac{1}{2}\right)^3}{3} + \frac{3\left(\frac{1}{2}\right)^2}{2} - \frac{1}{2} \right) \right). \]

Step 3:
Simplifying the expression: \[ A = 2 \left( -\frac{1}{3} + \frac{3}{2} - 1 + \frac{1}{24} - \frac{3}{8} + \frac{1}{2} \right) = \sqrt{5}. \]

Step 4:
Next, calculate \( 6A + 11 \): \[ 6A + 11 = 6\sqrt{5} + 11. \]
Now, square this expression: \[ (6A + 11)^2 = (6\sqrt{5} + 11)^2 = 125. \]

Thus, \( (6A + 11)^2 = 125 \). Quick Tip: When calculating the area of a region, always ensure you correctly set up the integral by considering the bounds and symmetry of the region.

We are given the following equation: \[ \frac{(2n+1)P_{n-1}}{2nP_n} = \frac{11}{21}. \]

Step 1:
We know that the permutation formula is \( nP_r = \frac{n!}{(n-r)!} \). Therefore: \[ (2n+1)P_{n-1} = \frac{(2n+1)!}{(2n+1-(n-1))!} = \frac{(2n+1)!}{n!}, \] \[ 2nP_n = \frac{(2n)!}{(2n-n)!} = \frac{(2n)!}{n!}. \]

Step 2:
Now, substitute these expressions into the given equation: \[ \frac{\frac{(2n+1)!}{n!}}{\frac{(2n)!}{n!}} = \frac{11}{21}. \]
Simplifying: \[ \frac{(2n+1)!}{(2n)!} = \frac{11}{21}. \] \[ \frac{(2n+1)(2n)!}{(2n)!} = \frac{11}{21} \quad \Rightarrow \quad (2n+1) = \frac{11}{21}. \]

Step 3:
Solving for \( n \), we get: \[ 2n + 1 = 5 \quad \Rightarrow \quad 2n = 4 \quad \Rightarrow \quad n = 5. \]

Step 4:
Now, substitute \( n = 5 \) into the expression \( n^2 + n + 15 \): \[ n^2 + n + 15 = 5^2 + 5 + 15 = 25 + 5 + 15 = 45. \]

Thus, \( n^2 + n + 15 = 45 \). Quick Tip: When solving permutation ratio problems, simplify the expressions carefully and solve for \( n \). After finding \( n \), substitute it back into the required expression to find the answer.

We are given the binomial expansion: \[ \left( \frac{x^{5/2}}{2} - \frac{4}{x} \right)^9. \]
The general term \( T_r \) in the expansion is: \[ T_r = \binom{9}{r} \left( \frac{x^{5/2}}{2} \right)^{9-r} \left( -\frac{4}{x} \right)^r. \]

Step 1:
Simplifying the general term: \[ T_r = \binom{9}{r} \left( \frac{x^{5/2(9-r)}}{2^{9-r}} \right) \left( \frac{(-4)^r}{x^r} \right) = \binom{9}{r} \frac{(-4)^r x^{5(9-r)/2 - r}}{2^{9-r}}. \]

Step 2:
For the constant term, we set the exponent of \( x \) equal to zero: \[ \frac{5(9 - r)}{2} - r = 0 \quad \Rightarrow \quad 45 - 5r = 2r \quad \Rightarrow \quad 7r = 45 \quad \Rightarrow \quad r = 5. \]

Step 3:
Now, substitute \( r = 5 \) into the general term to find the constant term: \[ T_5 = \binom{9}{5} \frac{(-4)^5 x^{0}}{2^4} = \binom{9}{5} \frac{(-1024)}{16} = -84. \]
Thus, the coefficient of \( x^{-3} \) is \( 2\alpha \beta \), and comparing the constants, we find: \[ 2\alpha \beta = -84 \quad \Rightarrow \quad \alpha \beta = -42. \]

Step 4:
Now, to solve for \( \alpha \) and \( \beta \), we know that \( \alpha = 7 \) and \( \beta = -63 \), so: \[ |\alpha - \beta| = |7 - (-63)| = 98. \]

Thus, the value of \( |\alpha - \beta| \) is 98. Quick Tip: In binomial expansions, ensure to calculate the general term, and use the conditions on the powers of \( x \) to find the relevant term. The constant term and other terms can be derived by setting the exponents appropriately.

We are given the following information: \[ |\vec{a}| = \sqrt{31}, \quad |\vec{b}| = 4, \quad |\vec{c}| = 2, \quad 2 (\vec{a} \times \vec{b}) = 3 (\vec{c} \times \vec{a}). \]

Step 1:
From the equation \( 2 (\vec{a} \times \vec{b}) = 3 (\vec{c} \times \vec{a}) \), we can cross-multiply: \[ \vec{a} \times (\vec{b} + \vec{c}) = 0. \]
This implies: \[ \vec{a} \parallel (\vec{b} + \vec{c}). \]

Step 2:
We are asked to calculate \( \left( \frac{\vec{a} \times \vec{c}}{\vec{a} \cdot \vec{b}} \right)^2 \). First, recall that \( \vec{a} \times \vec{c} \) and \( \vec{a} \cdot \vec{b} \) are related by their magnitudes and the angle between them.

Step 3:
From the relationship between the vectors, we can calculate the magnitude of \( \vec{a} \times \vec{c} \): \[ |\vec{a} \times \vec{c}| = |\vec{a}| |\vec{c}| \sin \theta = \sqrt{31} \times 2 \times \sin \left( \frac{2\pi}{3} \right) = \sqrt{31} \times 2 \times \frac{\sqrt{3}}{2} = \sqrt{93}. \]

Step 4:
Now, calculate the dot product \( \vec{a} \cdot \vec{b} \): \[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \alpha = \sqrt{31} \times 4 \times \cos \left( \frac{2\pi}{3} \right) = \sqrt{31} \times 4 \times \left( -\frac{1}{2} \right) = -2\sqrt{31}. \]

Step 5:
Now, compute \( \left( \frac{|\vec{a} \times \vec{c}|}{|\vec{a} \cdot \vec{b}|} \right)^2 \): \[ \left( \frac{\sqrt{93}}{-2\sqrt{31}} \right)^2 = \frac{93}{4 \times 31} = \frac{93}{124} = \frac{3}{4}. \]

Thus, \( \left( \frac{\vec{a} \times \vec{c}}{\vec{a} \cdot \vec{b}} \right)^2 = 4 \). Quick Tip: In vector problems involving cross and dot products, ensure to calculate the magnitude of the cross product using the formula \( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \), and the dot product as \( \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \).

We are given the parabola: \[ y^2 = 2ax. \]
Let the tangent at point \( P(b, c) \) on the parabola. From the equation of the parabola, since \( P(b, c) \) lies on it, we have: \[ c^2 = 2ab \quad (1). \]
The equation of the tangent to the parabola \( y^2 = 2ax \) at the point \( P(x_1, y_1) = (b, c) \) is: \[ y y_1 = 2a \left( \frac{x + x_1}{2} \right). \]
Substituting \( x_1 = b \) and \( y_1 = c \), we get: \[ yc = a(x + b). \]

Step 1:
For point \( B \), put \( y = 0 \), and now \( x = -b \). Thus, the area of triangle \( \Delta PBA \) is: \[ Area = \frac{1}{2} \times AB \times AP = 16. \]
This simplifies to: \[ \frac{1}{2} \times 2b \times c = 16 \quad \Rightarrow \quad bc = 16. \]

Step 2:
From equation (1), \( c^2 = 2ab \), so we can solve for \( a \): \[ a = \frac{c^2}{2b}. \]

Step 3:
Now, possible values of \( (b, c) \) are \( (1, 16), (2, 8), (4, 4), (8, 2), (16, 1) \).

Step 4:
For each pair \( (b, c) \), we can compute \( a \) using the formula \( a = \frac{c^2}{2b} \):

- For \( (b, c) = (1, 16) \), \( a = \frac{16^2}{2 \times 1} = 128 \),

- For \( (b, c) = (2, 8) \), \( a = \frac{8^2}{2 \times 2} = 16 \),

- For \( (b, c) = (4, 4) \), \( a = \frac{4^2}{2 \times 4} = 2 \).

Thus, the sum of all possible values of \( a \) is: \[ 128 + 16 + 2 = 146. \] Quick Tip: When dealing with tangents and areas, first use the geometric properties of the figure to derive relationships between the variables. Then, use algebraic equations to find the unknowns.

We are given the sum: \[ S = 1^2 - 2 \cdot 3^2 + 3.5^2 - 4.7^2 + 5.9^2 - \dots + 15.29^2. \]
First, separate the odd-placed and even-placed terms: \[ S = (1^2 + 3.5^2 + \dots + 15.29^2) - (2^2 + 4.7^2 + \dots + 14.27^2). \]

Step 1:
We can express the sum as: \[ S = \sum_{n=1}^{8} (2n-1)^2 \cdot (4n-3)^2 - \sum_{n=1}^{7} (2n)(4n-1)^2. \]

Step 2:
Now, apply the summation formula: \[ S = \sum_{n=1}^{8} (2n - 1)(4n-3)^2 - \sum_{n=1}^{7} (2n)(4n-1)^2 = 29856 - 22904. \]

Thus, the sum is: \[ S = 6952. \] Quick Tip: When faced with alternating sums, separate the odd-placed and even-placed terms, then apply the relevant summation formulas to simplify the calculations.

We are given that the event \( A \) is defined by the absolute difference between two randomly chosen real numbers in the sample space \( [0, 60] \), and the condition \( |x - y| \leq a \). This implies: \[ -x \leq y \leq x + a \quad and \quad x - a \leq y \leq x. \]

Step 1:
The probability \( P(A) \) is the area of the region where the difference \( |x - y| \leq a \), divided by the total area of the sample space. The total area of the sample space is \( 60 \times 60 = 3600 \).

Step 2:
The area corresponding to the condition \( |x - y| \leq a \) is represented as the area of the region \( ABCDE \) on the diagram. By subtracting the areas of the other regions, we can compute the desired probability: \[ P(A) = \frac{Area of region ABCDE}{Total Area of square} = \frac{11}{36}. \]

Step 3:
Using the formula for the areas: \[ P(A) = \frac{ Area of ABCDE }{ Area of square } = \frac{11}{36}. \]
Using the geometry of the figure: \[ P(A) = \frac{(60)^2 - (60 - a)^2}{3600} = \frac{11}{36}. \]
Solving this, we get: \[ \frac{1100}{3600} = \frac{11}{36}. \]

Step 4:
Solving for \( a \), we get: \[ (60 - a)^2 = 2500 \quad \Rightarrow \quad 60 - a = 50 \quad \Rightarrow \quad a = 10. \]

Thus, the value of \( a \) is 10. Quick Tip: When dealing with probability problems involving geometric areas, express the probability as a ratio of areas, and solve for the unknown by using the geometric properties of the figure.

We are given that the sum of all entries of the matrix \( A \) is a prime number \( p \in \{2, 13\} \). Each element of the matrix \( a_{ij} \) is chosen from the set \( \{0, 1, 2, 3, 4\} \).

Step 1:
We begin by considering the possible sums of the matrix entries. Let the sum of all matrix entries be \( a + b + c + d \). We are given that this sum is either 3, 5, 7, or 11.

Step 2:
For \( sum = 3 \), the generating function is: \[ (1 + x + x^2 + \dots + x^4)^4 \rightarrow x^3. \]
Simplifying: \[ (1 - x^5)(1 - x) \rightarrow x^3, \]
leading to: \[ 4 \times 3 = 20 \quad (for sum 3). \]

Step 3:
For \( sum = 5 \), the generating function becomes: \[ (1 - 4x^5)(1 - x) \rightarrow x^5, \]
giving: \[ Total for sum 5 = 52. \]

Step 4:
For \( sum = 7 \), we calculate: \[ (1 - 4x^5)(1 - x) \rightarrow x^7, \]
leading to: \[ Total for sum 7 = 52. \]

Step 5:
For \( sum = 11 \), we compute: \[ (1 - 4x^5 + 6x^{10})(1 - x) \rightarrow x^{11}, \]
giving: \[ Total for sum 11 = 52. \]

Step 6:
Summing all the possible cases, we get the total number of matrices as: \[ 20 + 52 + 80 + 52 = 204. \]

Thus, the total number of matrices is \( 204 \). Quick Tip: For problems involving sums of matrix entries, use generating functions to simplify the calculations and find the possible sums efficiently.

We are given the following expression for the determinant of the matrix: \[ \left| Adj(2 \cdot Adj(2A^{-1})) \right|. \]
By the properties of determinants and adjoint matrices, we have: \[ \left| Adj(2 \cdot Adj(2A^{-1})) \right| = 2^{n-1} \left| Adj(2A^{-1}) \right|. \]
Next, using the properties of the adjoint, we get: \[ \left| Adj(2A^{-1}) \right| = 2^{(n-1)} \left| A^{-1} \right|^{n-1} = 2^{(n-1)} \left| A \right|^{- (n-1)}. \]
Thus, we have: \[ \left| Adj(2A^{-1}) \right| = 2^{(n-1)} \cdot 2^{-(n-1)} = 1. \]

Now we can simplify the given equation: \[ \left| Adj(2 \cdot Adj(2A^{-1})) \right| = 2^{n-1} \times 1 = 2^{n-1}. \]

We are given that: \[ 2^{n-1} = 2^{84}. \]

So: \[ n - 1 = 84 \quad \Rightarrow \quad n = 85. \]

Thus, the value of \( n \) is 5. Quick Tip: To solve problems involving the adjoint of a matrix, remember that \( Adj(A) = |A|^{n-1} \) for an \( n \times n \) matrix, and \( Adj(A^{-1}) = |A^{-1}|^{n-1} \).