Step 1: Understanding the Concept:

This problem is solved using the Principle of Inclusion-Exclusion for three sets.

Let \( n(A), n(B), \) and \( n(C) \) be the number of medals awarded in events A, B, and C respectively.

Let \( n(A \cup B \cup C) \) be the total number of unique men who received at least one medal.

We are asked to find the number of people who received medals in exactly two of the three events.


Step 2: Key Formula or Approach:

1. General Inclusion-Exclusion Formula:
\[ n(A \cup B \cup C) = n(A) + n(B) + n(C) - \left[ n(A \cap B) + n(B \cap C) + n(C \cap A) \right] + n(A \cap B \cap C) \]

2. Formula for "Exactly two events":
\[ Exactly 2 = \left[ n(A \cap B) + n(B \cap C) + n(C \cap A) \right] - 3n(A \cap B \cap C) \]


Step 3: Detailed Explanation:

Given:
\( n(A) = 48, n(B) = 25, n(C) = 18 \)
\( n(A \cup B \cup C) = 60 \)
\( n(A \cap B \cap C) = 5 \)



Let \( S_2 = n(A \cap B) + n(B \cap C) + n(C \cap A) \).

Using the general formula:
\[ 60 = 48 + 25 + 18 - S_2 + 5 \]
\[ 60 = 91 - S_2 + 5 \]
\[ 60 = 96 - S_2 \]
\[ S_2 = 36 \]



Now, find the number of people in exactly two events:
\[ Exactly 2 = S_2 - 3 \times n(A \cap B \cap C) \]
\[ Exactly 2 = 36 - 3(5) \]
\[ Exactly 2 = 36 - 15 = 21 \]


Step 4: Final Answer:

The number of men who received medals in exactly two of the three events is 21.
Quick Tip: In Venn diagrams, the sum of intersections \( \sum n(A \cap B) \) counts the "exactly 2" region once and the "exactly 3" region three times. To isolate "exactly 2", subtract \( 3 \times n(A \cap B \cap C) \).

Step 1: Understanding the Concept:

The image of a point in a plane is found by extending a perpendicular line from the point through the plane to an equal distance on the other side.


Step 2: Key Formula or Approach:

The coordinates of the image \( (\alpha, \beta, \gamma) \) of point \( (x_1, y_1, z_1) \) in plane \( ax + by + cz + d = 0 \) are:
\[ \frac{\alpha - x_1}{a} = \frac{\beta - y_1}{b} = \frac{\gamma - z_1}{c} = -2 \left( \frac{ax_1 + by_1 + cz_1 + d}{a^2 + b^2 + c^2} \right) \]


Step 3: Detailed Explanation:

Given point \( (x_1, y_1, z_1) = (2, 3, 5) \) and plane \( 2x + y - 3z - 6 = 0 \).

Coefficients: \( a = 2, b = 1, c = -3, d = -6 \).

Calculate the multiplier \( k \):
\[ k = -2 \left( \frac{2(2) + 1(3) - 3(5) - 6}{2^2 + 1^2 + (-3)^2} \right) \]
\[ k = -2 \left( \frac{4 + 3 - 15 - 6}{4 + 1 + 9} \right) \]
\[ k = -2 \left( \frac{-14}{14} \right) = -2(-1) = 2 \]

Now, calculate \( \alpha, \beta, \gamma \):
\[ \alpha - 2 = 2(2) \implies \alpha = 4 + 2 = 6 \]
\[ \beta - 3 = 1(2) \implies \beta = 2 + 3 = 5 \]
\[ \gamma - 5 = -3(2) \implies \gamma = -6 + 5 = -1 \]

Sum: \( \alpha + \beta + \gamma = 6 + 5 - 1 = 10 \).


Step 4: Final Answer:

The sum \( \alpha + \beta + \gamma = 10 \).
Quick Tip: The multiplier for the foot of the perpendicular is \( -1 \), while for the image it is \( -2 \). Ensure the plane equation is in the form \( ax + by + cz + d = 0 \).

Step 1: Understanding the Concept:

Rewrite ellipse \( E_k \): \( \frac{x^2}{1/k} + \frac{y^2}{1/k^2} = 1 \).

The semi-major axis is \( a = \frac{1}{\sqrt{k}} \) and semi-minor axis is \( b = \frac{1}{k} \).

The circle touches the line joining \( (a, 0) \) and \( (0, b) \). This line is \( \frac{x}{a} + \frac{y}{b} = 1 \).


Step 2: Key Formula or Approach:

The perpendicular distance from origin to \( \frac{x}{a} + \frac{y}{b} = 1 \) is the radius \( r_k \):
\[ \frac{1}{r_k^2} = \frac{1}{a^2} + \frac{1}{b^2} \]


Step 3: Detailed Explanation:

For \( E_k \), \( a^2 = \frac{1}{k} \) and \( b^2 = \frac{1}{k^2} \).

Substitute:
\[ \frac{1}{r_k^2} = \frac{1}{1/k} + \frac{1}{1/k^2} = k + k^2 \]

We need:
\[ S = \sum_{k=1}^{20} (k^2 + k) = \sum_{k=1}^{20} k^2 + \sum_{k=1}^{20} k \]

Using standard sum formulas:
\[ \sum k^2 = \frac{n(n+1)(2n+1)}{6} = \frac{20 \times 21 \times 41}{6} = 2870 \]
\[ \sum k = \frac{n(n+1)}{2} = \frac{20 \times 21}{2} = 210 \]

Total sum \( S = 2870 + 210 = 3080 \).


Step 4: Final Answer:

The required value is 3080.
Quick Tip: The radius of a circle inscribed in a rhombus formed by axis-endpoints of an ellipse satisfies \( \frac{1}{r^2} = \frac{1}{a^2} + \frac{1}{b^2} \). Memorizing this saves time in JEE-level problems.

Step 1: Understanding the Concept:

We need to find the number of non-negative integer solutions to \( x+y+z=15 \) such that no two variables are equal.


Step 2: Key Formula or Approach:

1. Total non-negative solutions for \( x+y+z=n \) is \( \binom{n+r-1}{r-1} \).

2. Subtract cases where at least two variables are equal.


Step 3: Detailed Explanation:

Total solutions = \( \binom{15+3-1}{3-1} = \binom{17}{2} = 136 \).

Cases where variables are equal:

1. All three equal: \( x = y = z \implies 3x = 15 \implies x=5 \). Triplet: \( (5,5,5) \). (1 case)

2. Exactly two equal:

Assume \( x = y \). Then \( 2x + z = 15 \).

Possible values for \( x \): \( 0, 1, 2, 3, 4, 5, 6, 7 \). (8 values)

Out of these, \( x=5 \) leads to \( z=5 \) (all equal), which we handle separately.

So, there are 7 cases where \( x=y \neq z \).

Similarly, 7 cases for \( y=z \neq x \) and 7 cases for \( z=x \neq y \).

Total cases with exactly two equal = \( 7 \times 3 = 21 \).



Cases with at least two equal = \( 21 + 1 = 22 \).

Number of distinct triplets = \( 136 - 22 = 114 \).


Step 4: Final Answer:

The number of distinct non-negative integer triplets is 114.
Quick Tip: To find distinct solutions, calculate Total Solutions - (Cases with 2 equal) - (Cases with 3 equal). Ensure you don't double-count the "all equal" case when analyzing "two equal" variables.

Step 1: Understanding the Concept:

For \( \log_B A \ge 0 \), we require:

1. \( B > 0 \) and \( B \neq 1 \).

2. \( A > 0 \).

3. If \( B > 1 \), then \( A \ge 1 \).

4. If \( 0 < B < 1 \), then \( 0 < A \le 1 \).


Step 2: Detailed Explanation:

Base \( B = x + 3.5 \). Condition: \( x > -3.5 \) and \( x \neq -2.5 \).

Argument \( A = \left( \frac{x-7}{2x-3} \right)^2 \). Condition: \( x \neq 7, x \neq 1.5 \).



Case 1: \( B > 1 \implies x > -2.5 \).

Then \( \left( \frac{x-7}{2x-3} \right)^2 \ge 1 \implies \frac{(x-7)^2}{(2x-3)^2} - 1 \ge 0 \).
\[ \frac{(x-7-2x+3)(x-7+2x-3)}{(2x-3)^2} \ge 0 \]
\[ \frac{(-x-4)(3x-10)}{(2x-3)^2} \ge 0 \implies \frac{(x+4)(3x-10)}{(2x-3)^2} \le 0 \]

Interval: \( [-4, 10/3] \).

Intersect with \( x > -2.5, x \neq 1.5 \): \( x \in (-2.5, 1.5) \cup (1.5, 10/3] \).

Integers: \( -2, -1, 0, 1, 2, 3 \). (6 solutions).



Case 2: \( 0 < B < 1 \implies -3.5 < x < -2.5 \).

Then \( \left( \frac{x-7}{2x-3} \right)^2 \le 1 \).

Interval: \( x \le -4 \) or \( x \ge 10/3 \). No intersection with \( (-3.5, -2.5) \).


Step 3: Final Answer:

Total integral solutions = 6.
Quick Tip: Always check the validity of the logarithm base and argument first. When solving inequalities, the direction of the inequality sign flips if the base is between 0 and 1.

Step 1: Understanding the Concept:

We use the property of transposes and basic matrix algebra to determine the form of matrix \( A \).


Step 2: Detailed Explanation:

Given: \( A^T = \alpha A + I \).

Take transpose on both sides: \( (A^T)^T = (\alpha A + I)^T \implies A = \alpha A^T + I \).

Substitute \( A^T = \alpha A + I \) into the second equation:
\[ A = \alpha(\alpha A + I) + I = \alpha^2 A + (\alpha + 1)I \]
\[ A(1 - \alpha^2) = (\alpha + 1)I \]

Since \( \alpha \neq \pm 1 \), divide by \( 1 - \alpha^2 \):
\[ A = \frac{\alpha+1}{(1-\alpha)(1+\alpha)} I = \frac{1}{1-\alpha} I \]

Let \( k = \frac{1}{1-\alpha} \), so \( A = kI \).

Then \( A^2 - A = (k^2 - k)I \).
\( \det(A^2 - A) = (k^2 - k)^2 \det(I) = (k^2 - k)^2 = 4 \).

So, \( k^2 - k = 2 \) or \( k^2 - k = -2 \).

1) \( k^2 - k - 2 = 0 \implies (k-2)(k+1) = 0 \implies k=2, k=-1 \).

If \( k=2 \implies \frac{1}{1-\alpha} = 2 \implies \alpha = 1/2 \).

If \( k=-1 \implies \frac{1}{1-\alpha} = -1 \implies \alpha = 2 \).

2) \( k^2 - k + 2 = 0 \). No real roots.

Sum of \( \alpha = 1/2 + 2 = 5/2 \).


Step 3: Final Answer:

The sum of all possible values of \( \alpha \) is \( \frac{5}{2} \).
Quick Tip: If a matrix equation involves \( A \) and \( A^T \), taking the transpose again is the standard way to solve for \( A \). For \( n \times n \) matrices, \( \det(cA) = c^n \det(A) \).

Step 1: Understanding the Concept:

We need to check LHL, RHL, and the function value at \( x=0 \) and \( x=1 \). Note \( |-x + [x]| = |-(x - [x])| = \{x\} \).


Step 2: Detailed Explanation:

At \( x=0 \):
\( f(0) = [0] + \{0\} = 0 \).

RHL: \( \lim_{x \to 0^+} [x^2-x] + \{x\} \). As \( x \to 0^+ \), \( x^2-x \to 0^- \), so \( [x^2-x] = -1 \). \( \{x\} \to 0 \). \( RHL = -1 \).

LHL: \( \lim_{x \to 0^-} [x^2-x] + \{x\} \). As \( x \to 0^- \), \( x^2-x \to 0^+ \), so \( [x^2-x] = 0 \). \( \{x\} \to 1 \). \( LHL = 1 \).

Not continuous at \( x=0 \).



At \( x=1 \):
\( f(1) = [0] + \{1\} = 0 \).

RHL: \( \lim_{x \to 1^+} [x^2-x] + \{x\} \). \( x^2-x \to 0^+ \), so \( [x^2-x] = 0 \). \( \{x\} \to 0 \). \( RHL = 0 \).

LHL: \( \lim_{x \to 1^-} [x^2-x] + \{x\} \). \( x^2-x \to 0^- \), so \( [x^2-x] = -1 \). \( \{x\} \to 1 \). \( LHL = -1 + 1 = 0 \).

Continuous at \( x=1 \).


Step 3: Final Answer:

The function is continuous at \( x=1 \) but not at \( x=0 \).
Quick Tip: The fractional part function \( \{x\} \) has a limit of 1 as \( x \to n^- \) and 0 as \( x \to n^+ \). This property helps in evaluating limits involving GIF and absolute values.

Step 1: Understanding the Concept:

The normal vector of the required plane is perpendicular to the normals of the two given planes. Therefore, it is parallel to the cross product of their normal vectors.


Step 2: Detailed Explanation:

Normals: \( \vec{n_1} = (2, 4, 5) \), \( \vec{n_2} = (3, -2, 3) \).

Direction of normal \( \vec{n} = \vec{n_1} \times \vec{n_2} \):
\[ \vec{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}
2 & 4 & 5
3 & -2 & 3 \end{vmatrix} = (12+10)\mathbf{i} - (6-15)\mathbf{j} + (-4-12)\mathbf{k} = (22, 9, -16) \]

Equation of plane through \( (-2, 3, 5) \):
\[ 22(x+2) + 9(y-3) - 16(z-5) = 0 \]
\[ 22x + 44 + 9y - 27 - 16z + 80 = 0 \]
\[ 22x + 9y - 16z + 97 = 0 \]

Comparing, \( \alpha = 22, \beta = 9, \gamma = -16 \).

Sum \( = 22 + 9 - 16 = 15 \).


Step 3: Final Answer:

The sum \( \alpha + \beta + \gamma \) is 15.
Quick Tip: When a plane is perpendicular to two others, use the cross product of their normal vectors to find the new normal. Always verify that the constant term matches the given form before finding variables.

Step 1: Understanding the Concept:

The region is inside a circle of radius 2 centered at \( (0, 2) \), and below the parabola \( y = x^2/2 \).


Step 2: Detailed Explanation:

Intersection: \( 2y + (y-2)^2 = 4 \implies y^2 - 2y = 0 \implies y=0, 2 \).

Points are \( (0,0), (2,2), (-2,2) \).

Total area = Area of lower semi-circle - Area between parabola and line \( y=2 \) (inside circle).

Area of semi-circle \( = 2\pi \).

Area below parabola from \( x=-2 \) to \( 2 \): \( \int_{-2}^2 (2 - \frac{x^2}{2}) dx = [2x - \frac{x^3}{6}]_{-2}^2 = (4 - 8/6) - (-4 + 8/6) = 16/3 \).

Required area \( = 2\pi - 16/3 \).


Step 3: Final Answer:

The area is \( 2\pi - 16/3 \).
Quick Tip: Identify the symmetric regions to simplify integration. Area below a standard parabola \( y=cx^2 \) in a rectangular region is often \( \frac{2}{3} \times Base \times Height \).

Step 1: Understanding the Concept:

The given differential inequality can be written as the derivative of a product.


Step 2: Detailed Explanation:

LHS is \( \frac{d}{dx} [x \ln x \cdot f(x)] \).

So, \( \frac{d}{dx} [x \ln x \cdot f(x)] \ge 1 \).

Let \( g(x) = x \ln x \cdot f(x) \). Integrating from 2 to 4:
\[ g(4) - g(2) = \int_2^4 g'(x) dx \ge \int_2^4 1 dx = 2 \]

Check given values:
\( g(4) = 4 \ln 4 (1/4) = 2 \ln 2 \approx 1.386 \).
\( g(2) = 2 \ln 2 (1/2) = \ln 2 \approx 0.693 \).
\( g(4) - g(2) = \ln 2 \approx 0.693 \).

Since \( 0.693 < 2 \), the condition \( g'(x) \ge 1 \) is not satisfied by the boundary values. Therefore, no such function exists.


Step 3: Final Answer:

Neither statement is true.
Quick Tip: When dealing with differential inequalities, look for the derivative of a compound function. Integrating the inequality over the domain helps test the validity of the conditions against the boundary values.

Step 1: Understanding the Concept:

A plane described by two vectors \( \vec{u} \) and \( \vec{v} \) has a normal vector \( \vec{n} = \vec{u} \times \vec{v} \).

The line of intersection of two planes is parallel to the cross product of their respective normal vectors.


Step 2: Key Formula or Approach:

1. Normal vector \( \vec{n} = \vec{u} \times \vec{v} \).

2. Direction of intersection \( \vec{d} = \vec{n_1} \times \vec{n_2} \).

3. Use dot product for angle and cross product magnitude for the second part.


Step 3: Detailed Explanation:

For Plane 1, vectors are \( \vec{u_1} = (1, 1, 0) \) and \( \vec{v_1} = (1, 0, 1) \).
\[ \vec{n_1} = \vec{u_1} \times \vec{v_1} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & 1 & 0
1 & 0 & 1 \end{vmatrix} = \hat{i}(1) - \hat{j}(1) + \hat{k}(-1) = (1, -1, -1) \]

For Plane 2, vectors are \( \vec{u_2} = (1, -1, 0) \) and \( \vec{v_2} = (0, 1, -1) \).
\[ \vec{n_2} = \vec{u_2} \times \vec{v_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & -1 & 0
0 & 1 & -1 \end{vmatrix} = \hat{i}(1) - \hat{j}(-1) + \hat{k}(1) = (1, 1, 1) \]

Direction of intersection \( \vec{d} = \vec{n_1} \times \vec{n_2} \):
\[ \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & -1 & -1
1 & 1 & 1 \end{vmatrix} = \hat{i}(0) - \hat{j}(2) + \hat{k}(2) = (0, -2, 2) \]

Let \( \vec{a} = k(0, -2, 2) \). Given \( \vec{a} \cdot \vec{b} = 6 \) with \( \vec{b} = (2, -2, 1) \):
\[ k(0 + 4 + 2) = 6 \implies 6k = 6 \implies k = 1 \]

So, \( \vec{a} = -2\hat{j} + 2\hat{k} \).

Now, \( |\vec{a}| = \sqrt{0^2 + (-2)^2 + 2^2} = 2\sqrt{2} \) and \( |\vec{b}| = \sqrt{2^2 + (-2)^2 + 1^2} = 3 \).

Angle \( \theta \):
\[ \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{6}{2\sqrt{2} \cdot 3} = \frac{1}{\sqrt{2}} \implies \theta = \frac{\pi}{4} \]

Magnitude of cross product:
\[ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta = 2\sqrt{2} \cdot 3 \cdot \frac{1}{\sqrt{2}} = 6 \]


Step 4: Final Answer:

The ordered pair \( (\theta, |\vec{a} \times \vec{b}|) \) is \( (\frac{\pi}{4}, 6) \).
Quick Tip: If a plane is defined by two vectors starting from the origin, its normal is their cross product. The line of intersection of two such planes is perpendicular to both normals, hence it follows the direction of the cross product of the normals.

Step 1: Understanding the Concept:

This question explores the relationship between the components of a vector and its Euclidean norm (magnitude).

Magnitude \( |\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2} \).


Step 2: Detailed Explanation:

Statement (A): Let \( M = \max\{|a_1|, |a_2|, |a_3|\} \).

Since \( a_1^2, a_2^2, a_3^2 \ge 0 \), it follows that \( a_i^2 \le a_1^2 + a_2^2 + a_3^2 \) for any \( i \in \{1, 2, 3\} \).

Taking the square root, \( |a_i| \le \sqrt{a_1^2 + a_2^2 + a_3^2} = |\vec{a}| \).

This is true for each \( i \), thus \( \max\{|a_1|, |a_2|, |a_3|\} \le |\vec{a}| \). Statement (A) is true.



Statement (B): Let \( M = \max\{|a_1|, |a_2|, |a_3|\} \).

Then \( a_1^2 \le M^2, a_2^2 \le M^2, a_3^2 \le M^2 \).

Adding these inequalities: \( a_1^2 + a_2^2 + a_3^2 \le 3M^2 \).

Taking the square root: \( |\vec{a}| \le \sqrt{3}M \).

Since \( \sqrt{3} \approx 1.732 < 3 \), it is certainly true that \( |\vec{a}| \le 3M \).

Thus, \( |\vec{a}| \le 3 \max\{|a_1|, |a_2|, |a_3|\} \). Statement (B) is true.


Step 3: Final Answer:

Both statements (A) and (B) are true based on general vector magnitude properties.
Quick Tip: For a vector in \( \mathbb{R}^n \), the relationship between the max norm and the Euclidean norm is \( \|x\|_{\infty} \le \|x\|_2 \le \sqrt{n} \|x\|_{\infty} \). These inequalities hold for all vectors regardless of component constraints.

Step 1: Understanding the Concept:

The mean of an Arithmetic Progression (AP) is given by \( \frac{x_1 + x_n}{2} \). We use this to find the common difference and then compute the sum of the new sequence \( y_i \).


Step 2: Key Formula or Approach:

1. Mean of AP \( = \frac{a + (a + (n-1)d)}{2} \).

2. \( \sum i = \frac{n(n+1)}{2} \) and \( \sum i^2 = \frac{n(n+1)(2n+1)}{6} \).


Step 3: Detailed Explanation:

Given \( x_1 = 2 \) and mean \( \bar{x} = 200 \).
\[ \frac{x_1 + x_{100}}{2} = 200 \implies 2 + x_{100} = 400 \implies x_{100} = 398 \]

Using \( x_{100} = x_1 + 99d \):
\[ 398 = 2 + 99d \implies 396 = 99d \implies d = 4 \]

General term \( x_i = 2 + (i-1)4 = 4i - 2 \).

Now, \( y_i = i(x_i - i) = i(4i - 2 - i) = i(3i - 2) = 3i^2 - 2i \).

Mean of \( y_i = \frac{1}{100} \sum_{i=1}^{100} (3i^2 - 2i) \):
\[ Mean(y) = \frac{3}{100} \sum_{i=1}^{100} i^2 - \frac{2}{100} \sum_{i=1}^{100} i \]
\[ Mean(y) = \frac{3}{100} \left( \frac{100 \cdot 101 \cdot 201}{6} \right) - \frac{2}{100} \left( \frac{100 \cdot 101}{2} \right) \]
\[ Mean(y) = \frac{101 \cdot 201}{2} - 101 = 101 \left( \frac{201}{2} - 1 \right) = 101 \left( 100.5 - 1 \right) = 101 \times 99.5 \]
\[ Mean(y) = 10049.50 \]


Step 4: Final Answer:

The mean of the sequence \( y_i \) is 10049.50.
Quick Tip: The mean of an AP is always its average of first and last terms. When calculating the mean of a sequence derived from indices, use the power sum formulas \( \sum i^k \) to expedite the arithmetic.

Step 1: Understanding the Concept:

Rotating a complex number \( z \) by \( 90^\circ \) (right angle) anticlockwise corresponds to multiplying by \( i \).

Rotating by \( 90^\circ \) clockwise corresponds to multiplying by \( -i \).


Step 2: Key Formula or Approach:

1. \( w = z \cdot e^{i\theta} \).

2. \( Arg(x + iy) = \pi - \tan^{-1}|y/x| \) if the point is in the second quadrant.


Step 3: Detailed Explanation:

Given \( z_1 = 5 + 4i \). Anticlockwise rotation \( 90^\circ \):
\[ w_1 = (5 + 4i) \cdot i = 5i + 4i^2 = -4 + 5i \]

Given \( z_2 = 3 + 5i \). Clockwise rotation \( 90^\circ \):
\[ w_2 = (3 + 5i) \cdot (-i) = -3i - 5i^2 = 5 - 3i \]

Difference \( w_1 - w_2 \):
\[ w_1 - w_2 = (-4 + 5i) - (5 - 3i) = -9 + 8i \]

The point \( (-9, 8) \) lies in the second quadrant.

The principal argument \( \phi \) is:
\[ \phi = \pi - \tan^{-1} \left| \frac{8}{-9} \right| = \pi - \tan^{-1} \frac{8}{9} \]


Step 4: Final Answer:

The principal argument is \( \pi - \tan^{-1} \frac{8}{9} \).
Quick Tip: Multiplication by \( i \) rotates a vector by \( 90^\circ \) counter-clockwise. Always visualize the resulting complex number in the Cartesian plane to correctly determine the quadrant for the principal argument.

Step 1: Understanding the Concept:

Substitution simplifies the exponential terms. The resulting logarithmic integral can be solved using Integration by Parts.


Step 2: Key Formula or Approach:

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

2. \( \int \ln(t + \sqrt{1+t^2}) dt = t \ln(t + \sqrt{1+t^2}) - \sqrt{1+t^2} + C \).


Step 3: Detailed Explanation:

Let \( e^x = t \implies e^x dx = dt \).

Limits: \( x = -\ln 2 \implies t = 1/2 \); \( x = \ln 2 \implies t = 2 \).

The integral becomes \( I = \int_{1/2}^2 \ln(t + \sqrt{1+t^2}) dt \).

Using integration by parts: \( u = \ln(t + \sqrt{1+t^2}), dv = dt \implies du = \frac{1}{\sqrt{1+t^2}}dt, v = t \).
\[ I = [t \ln(t + \sqrt{1+t^2})]_{1/2}^2 - \int_{1/2}^2 \frac{t}{\sqrt{1+t^2}} dt \]
\[ I = [t \ln(t + \sqrt{1+t^2}) - \sqrt{1+t^2}]_{1/2}^2 \]
\[ I = \left( 2 \ln(2 + \sqrt{5}) - \sqrt{5} \right) - \left( \frac{1}{2} \ln \left( \frac{1}{2} + \frac{\sqrt{5}}{2} \right) - \frac{\sqrt{5}}{2} \right) \]
\[ I = 2 \ln(2 + \sqrt{5}) - \frac{1}{2} \ln \left( \frac{1 + \sqrt{5}}{2} \right) - \frac{\sqrt{5}}{2} \]
\[ I = \ln(2 + \sqrt{5})^2 - \ln \sqrt{\frac{1 + \sqrt{5}}{2}} - \frac{\sqrt{5}}{2} = \ln \left( \frac{(2 + \sqrt{5})^2 \cdot \sqrt{2}}{\sqrt{1 + \sqrt{5}}} \right) - \frac{\sqrt{5}}{2} \]


Step 4: Final Answer:

The integral value is \( \log_e \left( \frac{\sqrt{2}(2+\sqrt{5})^2}{\sqrt{1+\sqrt{5}}} \right) - \frac{\sqrt{5}}{2} \).
Quick Tip: Notice that \( \ln(t + \sqrt{1+t^2}) \) is the standard form for \( \sinh^{-1}(t) \). Remembering its integral \( x \sinh^{-1}x - \sqrt{1+x^2} \) makes such problems trivial.

Step 1: Understanding the Concept:

Expressing the equation in terms of a single trigonometric function (e.g., \( \cos^2 \theta \)) allows us to solve it as a polynomial equation.


Step 2: Detailed Explanation:

Replace \( \sin^6 \theta \) with \( (1 - \cos^2 \theta)^3 \). Let \( x = \cos^2 \theta \). Note \( x \in [0, 1] \).
\[ 3x^2 - 5x - 2(1-x)^3 + 2 = 0 \]
\[ 3x^2 - 5x - 2(1 - 3x + 3x^2 - x^3) + 2 = 0 \]
\[ 3x^2 - 5x - 2 + 6x - 6x^2 + 2x^3 + 2 = 0 \]
\[ 2x^3 - 3x^2 + x = 0 \implies x(2x^2 - 3x + 1) = 0 \]
\[ x(2x - 1)(x - 1) = 0 \implies x = 0, x = 1/2, x = 1 \]

Analyze the roots for \( \theta \in [0, 2\pi] \):

1. \( \cos^2 \theta = 0 \implies \cos \theta = 0 \implies \theta = \frac{\pi}{2}, \frac{3\pi}{2} \) (2 solutions).

2. \( \cos^2 \theta = 1 \implies \cos \theta = \pm 1 \implies \theta = 0, \pi, 2\pi \) (3 solutions).

3. \( \cos^2 \theta = 1/2 \implies \cos \theta = \pm \frac{1}{\sqrt{2}} \implies \theta = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4} \) (4 solutions).

Total number of solutions \( = 2 + 3 + 4 = 9 \).


Step 3: Final Answer:

The total number of elements in the set is 9.
Quick Tip: Substitute \( \cos^2 \theta = x \) to turn complex trigonometric equations into polynomials. Don't forget to include both endpoints of the interval if they satisfy the equation.

Step 1: Understanding the Concept:

A matrix \( \begin{pmatrix} a & b
c & d \end{pmatrix} \) is invertible if its determinant \( ad - bc \neq 0 \). The total number of matrices is \( 3^4 = 81 \).


Step 2: Detailed Explanation:

It is easier to count singular matrices (\( ad = bc \)). Possible products using \( \{0, 1, 2\} \) are \( \{0, 1, 2, 4\} \).

Count pairs \( (x, y) \) such that \( xy = k \):

- \( k=0 \): \( (0,0), (0,1), (0,2), (1,0), (2,0) \) (5 pairs).

- \( k=1 \): \( (1,1) \) (1 pair).

- \( k=2 \): \( (1,2), (2,1) \) (2 pairs).

- \( k=4 \): \( (2,2) \) (1 pair).

Total singular matrices (\( ad = bc \)):

- Both products are 0: \( 5 \times 5 = 25 \).

- Both products are 1: \( 1 \times 1 = 1 \).

- Both products are 2: \( 2 \times 2 = 4 \).

- Both products are 4: \( 1 \times 1 = 1 \).

Total singular \( = 25 + 1 + 4 + 1 = 31 \).

Invertible matrices \( = 81 - 31 = 50 \).
\( P(A) = 50/81 \).


Step 3: Final Answer:

The probability is \( \frac{50}{81} \).
Quick Tip: When counting invertible matrices from a finite set, subtraction from the total is usually faster. Group by the product value of the diagonals to count singular cases systematically.

Step 1: Understanding the Concept:

The mean of data shifts with addition/subtraction, while variance remains unchanged. The combined mean and variance use pooling formulas.


Step 2: Detailed Explanation:

For Set A': Mean \( \mu_1 = 5 - 3 = 2 \), Variance \( \sigma_1^2 = 12 \), Size \( n_1 = 5 \).

For Set B': Mean \( \mu_2 = 8 + 2 = 10 \), Variance \( \sigma_2^2 = 20 \), Size \( n_2 = 5 \).

Combined Mean \( \bar{x}_c \):
\[ \bar{x}_c = \frac{5(2) + 5(10)}{10} = 6 \]

Combined Variance \( \sigma_c^2 \):
\[ \sigma_c^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2} \]

where \( d_1 = 2 - 6 = -4 \) and \( d_2 = 10 - 6 = 4 \).
\[ \sigma_c^2 = \frac{5(12 + 16) + 5(20 + 16)}{10} = \frac{140 + 180}{10} = 32 \]

Sum \( = \bar{x}_c + \sigma_c^2 = 6 + 32 = 38 \).


Step 3: Final Answer:

The sum of the mean and variance of set C is 38.
Quick Tip: Combined variance for two sets of equal size is simply the average of their individual variances plus the average of the squared deviations from the combined mean.

Step 1: Understanding the Concept:

The area of the triangle formed by AB and the axes must be a specific fraction of the total rectangle's area. We then find the locus of the midpoint.


Step 2: Detailed Explanation:

Area of rectangle \( R = 2 \times 5 = 10 \).

Area of triangle \( OAB = \frac{1}{2}\alpha\beta \).

Case 1: Area \( OAB = \frac{1}{5} \times 10 = 2 \implies \alpha\beta = 4 \).

Case 2: Area \( OAB = \frac{4}{5} \times 10 = 8 \implies \alpha\beta = 16 \).

Since \( \alpha \le 2 \) and \( \beta \le 5 \), the maximum product \( \alpha\beta \) is 10. Thus, only \( \alpha\beta = 4 \) is possible.

Let midpoint be \( (h, k) \). Then \( h = \alpha/2, k = \beta/2 \implies \alpha = 2h, \beta = 2k \).

Substitute: \( (2h)(2k) = 4 \implies hk = 1 \).

The locus \( xy = 1 \) is a rectangular hyperbola.


Step 3: Final Answer:

The mid-point lies on a hyperbola.
Quick Tip: Always check the domain of your variables. Here, the larger area ratio was physically impossible given the rectangle's dimensions, narrowing down the locus to a single branch of a hyperbola.

Step 1: Understanding the Concept:

This differential equation contains the term \( y dx + x dy \), which is the exact differential \( d(xy) \).


Step 2: Detailed Explanation:

Let \( xy = v \). The equation becomes:
\[ (1 - v^2) dx = dv \implies dx = \frac{dv}{1 - v^2} \]

Integrating both sides:
\[ x + C = \frac{1}{2} \ln \left| \frac{1+v}{1-v} \right| \]

Given \( x = 1, y = 2 \implies v = 2 \).
\[ 1 + C = \frac{1}{2} \ln \left| \frac{3}{-1} \right| = \frac{1}{2} \ln 3 \implies C = \frac{1}{2} \ln 3 - 1 \]

Now for \( x = 2, y = \alpha \implies v = 2\alpha \):
\[ 2 + \frac{1}{2} \ln 3 - 1 = \frac{1}{2} \ln \left| \frac{1+2\alpha}{1-2\alpha} \right| \]
\[ 1 + \frac{1}{2} \ln 3 = \frac{1}{2} \ln \left| \frac{1+2\alpha}{1-2\alpha} \right| \implies 2 + \ln 3 = \ln \left| \frac{1+2\alpha}{1-2\alpha} \right| \]
\[ 3e^2 = \frac{2\alpha+1}{2\alpha-1} \implies 6\alpha e^2 - 3e^2 = 2\alpha + 1 \]
\[ 2\alpha(3e^2 - 1) = 3e^2 + 1 \implies \alpha = \frac{1+3e^2}{2(3e^2-1)} \]


Step 3: Final Answer:

The value of \( \alpha \) is \( \frac{1+3e^2}{2(3e^2-1)} \).
Quick Tip: Substitution \( xy = v \) is very common when terms like \( x dy + y dx \) appear. Always use absolute values when integrating \( 1/(1-v^2) \) to handle cases where the product \( xy \) exceeds 1.

Step 1: Understanding the Concept:

The general term in the binomial expansion of \( (a + x)^n \) is \( T_{r+1} = \binom{n}{r} a^{n-r} x^r \).

The coefficient of \( x^r \) is \( C_r = \binom{n}{r} a^{n-r} \).

We need to find the sum of coefficients from \( x^1 \) to \( x^7 \) and divide by the number of terms (7).


Step 2: Key Formula or Approach:

1. Sum of all coefficients \( \sum_{r=0}^9 C_r = (2+1)^9 \).

2. Mean of required coefficients \( = \frac{1}{7} \sum_{r=1}^7 C_r \).


Step 3: Detailed Explanation:

The sum of all coefficients in \( (2+x)^9 \) is obtained by setting \( x = 1 \):
\[ \sum_{r=0}^9 \binom{9}{r} 2^{9-r} = (2+1)^9 = 3^9 = 19683 \]

The sum we need is \( S = C_1 + C_2 + \dots + C_7 \).
\[ S = \left( \sum_{r=0}^9 C_r \right) - (C_0 + C_8 + C_9) \]

Calculating the excluded terms:
\( C_0 = \binom{9}{0} 2^9 = 1 \times 512 = 512 \)
\( C_8 = \binom{9}{8} 2^1 = 9 \times 2 = 18 \)
\( C_9 = \binom{9}{9} 2^0 = 1 \times 1 = 1 \)

Sum of excluded terms \( = 512 + 18 + 1 = 531 \).
\[ S = 19683 - 531 = 19152 \]

The mean of the 7 coefficients is:
\[ Mean = \frac{19152}{7} = 2736 \]


Step 4: Final Answer:

The mean of the coefficients of \( x, x^2, \dots, x^7 \) is 2736.
Quick Tip: To find the sum of specific binomial coefficients, always evaluate the whole sum \( (a+b)^n \) and subtract the terms that are not needed. This avoids large individual calculations for each coefficient.

Step 1: Understanding the Concept:

The direction of line \( l \) is found using the cross product of the directions of \( l_1 \) and \( l_2 \). We then find the intersection point P and project it onto \( l_2 \) to find the coordinates of Q.


Step 3: Detailed Explanation:

1. Direction of line \( l \) (\( \vec{d} \)):
\[ \vec{d} = (1\hat{i} + 2\hat{j} + 3\hat{k}) \times (2\hat{i} + 2\hat{j} + 1\hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & 2 & 3
2 & 2 & 1 \end{vmatrix} = -4\hat{i} + 5\hat{j} - 2\hat{k} \]

Line \( l \): \( \vec{r} = t(-4, 5, -2) \).

2. Intersection P of \( l \) and \( l_1 \):

Setting coordinates equal: \( -4t = 1+\lambda \), \( 5t = -11+2\lambda \), \( -2t = -7+3\lambda \).

From the first: \( \lambda = -4t - 1 \). Substitute into second:
\( 5t = -11 + 2(-4t-1) \implies 13t = -13 \implies t = -1 \).

So, \( P = (4, -5, 2) \).

3. Foot of perpendicular Q on \( l_2 \):

General point \( Q = (-1+2\mu, 2\mu, 1+\mu) \). Vector \( \vec{PQ} = (-5+2\mu, 5+2\mu, -1+\mu) \).

Since \( \vec{PQ} \perp (2, 2, 1) \):
\[ 2(-5+2\mu) + 2(5+2\mu) + 1(-1+\mu) = 0 \implies 9\mu - 1 = 0 \implies \mu = 1/9 \]

4. Find \( \alpha, \beta, \gamma \):
\( \alpha = -1 + 2/9 = -7/9 \), \( \beta = 2/9 \), \( \gamma = 1 + 1/9 = 10/9 \).

Sum \( \alpha + \beta + \gamma = (-7 + 2 + 10)/9 = 5/9 \).

Value required: \( 9 \times (5/9) = 5 \).


Step 4: Final Answer:

The value of \( 9(\alpha + \beta + \gamma) \) is 5.
Quick Tip: The direction of a line perpendicular to two given lines is always parallel to the cross product of their direction vectors. Use this direction vector to write the parametric form of the line.

Step 1: Understanding the Concept:

The statement involves logical implication. An implication \( A \Rightarrow B \) is False only when \( A \) is True and \( B \) is False. We can simplify the premise using distributive laws.


Step 3: Detailed Explanation:

1. Simplify the premise:
\[ (p \lor q) \land (p \lor r) \equiv p \lor (q \land r) \]

2. Let \( S = [p \lor (q \land r)] \Rightarrow (q \lor r) \).

Find when the statement \( S \) is False:

For \( S \) to be False, we need \( q \lor r \) to be False AND \( p \lor (q \land r) \) to be True.
\( q \lor r \) is False only when \( q = F \) and \( r = F \).

Substitute \( q=F, r=F \) into the premise:
\( p \lor (F \land F) \equiv p \lor F \equiv p \).

For this to be True, \( p \) must be True.

3. Thus, the only triplet that makes the statement False is \( (T, F, F) \).

4. Total triplets for \( (p, q, r) = 2^3 = 8 \).

5. Number of True triplets \( = 8 - 1 = 7 \).


Step 4: Final Answer:

The number of ordered triplets for which the statement is True is 7.
Quick Tip: Instead of building a truth table for 8 cases, analyze when the implication is False. This is almost always faster for logical deduction problems in competitive exams.

Step 1: Understanding the Concept:

Since \( a \) and \( b \) are roots, they satisfy the quadratic equation. Also, the product of roots \( ab = -1 \) allows us to relate terms like \( a^{n} \) and \( a^{n-2} \) with \( b \).


Step 3: Detailed Explanation:

1. Roots property: \( a+b = 7 \) and \( ab = -1 \).

2. Squaring the sum: \( (a+b)^2 = 49 \implies a^2 + b^2 + 2ab = 49 \).

Since \( ab = -1 \), we have \( a^2 + b^2 - 2 = 49 \implies a^2 + b^2 = 51 \).

3. Simplify the numerator:

Group terms: \( (a^{21} + a^{17}) + (b^{21} + b^{17}) \).
\[ a^{21} + a^{17} = a^{19}(a^2 + \frac{1}{a^2}) \]

Since \( ab = -1 \), \( b = -1/a \), so \( b^2 = 1/a^2 \).

Thus, \( a^{21} + a^{17} = a^{19}(a^2 + b^2) = 51a^{19} \).

Similarly, \( b^{21} + b^{17} = b^{19}(b^2 + \frac{1}{b^2}) = b^{19}(b^2 + a^2) = 51b^{19} \).

4. The numerator becomes \( 51(a^{19} + b^{19}) \).

5. Result \( = \frac{51(a^{19} + b^{19})}{a^{19} + b^{19}} = 51 \).


Step 4: Final Answer:

The value of the expression is 51.
Quick Tip: Whenever you see high powers like 21, 19, 17, look for a pattern involving \( a^2 + 1/a^2 \) or Newton's sums. The property \( ab = -1 \) suggests that \( a^{-n} \) is related to \( b^n \).

Step 1: Understanding the Concept:

This is an Arithmetico-Geometric Progression (AGP). The standard method to find its sum is to multiply the series by the common ratio and subtract it from the original series.


Step 3: Detailed Explanation:

1. Write the series:
\[ S = 109 + \frac{108}{5} + \frac{107}{5^2} + \dots + \frac{1}{5^{108}} \]

2. Multiply by \( 1/5 \):
\[ \frac{S}{5} = \frac{109}{5} + \frac{108}{5^2} + \dots + \frac{2}{5^{108}} + \frac{1}{5^{109}} \]

3. Subtract \( \frac{S}{5} \) from \( S \):
\[ S - \frac{S}{5} = 109 - \left( \frac{1}{5} + \frac{1}{5^2} + \dots + \frac{1}{5^{108}} \right) - \frac{1}{5^{109}} \]
\[ \frac{4S}{5} = 109 - \frac{\frac{1}{5}(1 - 5^{-108})}{1 - \frac{1}{5}} - \frac{1}{5^{109}} = 109 - \frac{1}{4}(1 - 5^{-108}) - \frac{1}{5^{109}} \]
\[ \frac{4S}{5} = 109 - \frac{1}{4} + \frac{1}{4 \cdot 5^{108}} - \frac{1}{5^{109}} = \frac{435}{4} + \frac{1}{4 \cdot 5^{109}} \]

4. Multiply by 20 to get \( 16S \):
\[ 16S = 5 \times 435 + \frac{5}{4 \cdot 5^{109}} \times 4 = 2175 + \frac{1}{5^{108}} \]

Note \( \frac{1}{5^{108}} = \frac{1}{(5^2)^{54}} = (25)^{-54} \).
\[ 16S - (25)^{-54} = 2175 \]


Step 4: Final Answer:

The final value is 2175.
Quick Tip: For any AGP, the difference \( S - rS \) always simplifies into a pure geometric progression (GP). Summing the resulting GP correctly is the key to the final answer.

Step 1: Understanding the Concept:

This problem uses the reduction formula obtained via integration by parts to evaluate the given linear combination of integrals.


Step 3: Detailed Explanation:

1. Observe the function \( h(t) = t^{m+1}(1+3t)^n \).

2. Calculate its derivative:
\[ \frac{d}{dt} [t^{m+1}(1+3t)^n] = (m+1)t^m(1+3t)^n + n(1+3t)^{n-1} \cdot 3 \cdot t^{m+1} \]
\[ = (m+1)t^m(1+3t)^n + 3nt^{m+1}(1+3t)^{n-1} \]

3. Integrating both sides from 0 to 2:
\[ [t^{m+1}(1+3t)^n]_0^2 = (m+1) \alpha(m, n) + 3n \alpha(m+1, n-1) \]

4. Substitute \( m = 10, n = 6 \):
\[ [t^{11}(1+3t)^6]_0^2 = 11 \alpha(10, 6) + 3(6) \alpha(11, 5) \]
\[ 2^{11}(1+6)^6 - 0 = 11 \alpha(10, 6) + 18 \alpha(11, 5) \]
\[ 2^{11} \cdot 7^6 = 11 \alpha(10, 6) + 18 \alpha(11, 5) \]

5. Given: \( 2^{11} \cdot 7^6 = p(14)^6 = p \cdot 2^6 \cdot 7^6 \).
\[ p = 2^{11}/2^6 = 2^5 = 32 \]


Step 4: Final Answer:

The value of \( p \) is 32.
Quick Tip: When you see a sum of two related integrals, check if it forms the derivative of a product \( t^{k}(f(t))^n \). Integrating this product directly is faster than traditional integration by parts.

Step 1: Understanding the Concept:

A term in the binomial expansion is integral if the exponents of both irrational bases are non-negative integers.


Step 2: Key Formula or Approach:

General term \( T_{r+1} = \binom{n}{r} (3^{1/2})^{n-r} (5^{1/4})^r \).


Step 3: Detailed Explanation:

1. For \( (3^{1/2} + 5^{1/4})^{680} \), the general term is:
\[ T_{r+1} = \binom{680}{r} 3^{(680-r)/2} 5^{r/4}, where r \in \{0, 1, \dots, 680\} \]

2. Conditions for integral terms:

- \( \frac{680-r}{2} \) must be an integer \( \implies r \) is even.

- \( \frac{r}{4} \) must be an integer \( \implies r \) is a multiple of 4.

3. Combining these, \( r \) must be a multiple of 4.

4. Possible values for \( r \): \( 0, 4, 8, \dots, 680 \).

5. Number of terms \( = \frac{680 - 0}{4} + 1 = 170 + 1 = 171 \).


Step 4: Final Answer:

The number of integral terms is 171.
Quick Tip: The number of rational terms is given by \( \lfloor \frac{n}{LCM of denominators} \rfloor + 1 \), provided that the starting term (at \( r=0 \)) is also rational.

Step 1: Understanding the Concept:

Eccentricity \( e = \sqrt{1 + \frac{b^2}{a^2}} \). For \( e \) to be rational, \( 1 + \frac{b^2}{a^2} \) must be a perfect square of a rational number.


Step 3: Detailed Explanation:

1. Calculate eccentricity squared:
\[ e^2 = 1 + \frac{3+n}{1+n} = \frac{1+n+3+n}{1+n} = \frac{2n+4}{n+1} = \frac{2(n+2)}{n+1} \]

2. We need \( \frac{2(n+2)}{n+1} = q^2 \) for some rational \( q \).

Since \( n+1 \) and \( n+2 \) are coprime, we need \( n+1 = m^2 \) and \( n+2 = 2p^2 \) (or vice versa).

Checking even values of \( n \):

- \( n=48 \): \( e^2 = \frac{2(50)}{49} = \frac{100}{49} = (10/7)^2 \). This is rational and \( n=48 \) is even.

Checking smaller even values: \( n=2, 4, \dots, 46 \) do not satisfy the condition.

3. For \( n=48 \): \( a^2 = 49 \implies a = 7 \), \( b^2 = 3+48 = 51 \).

4. Length of latus rectum \( l = \frac{2b^2}{a} = \frac{2 \times 51}{7} = \frac{102}{7} \).

5. Required value: \( 21 \times \frac{102}{7} = 3 \times 102 = 306 \).


Step 4: Final Answer:

The value of \( 21l \) is 306.
Quick Tip: Coprime numerator and denominator in a square root \( \sqrt{P/Q} \) must each be perfect squares for the result to be rational. Always test simple values of \( n \) before moving to Pell's equation methods.

Step 1: Understanding the Concept:

If a matrix satisfies \( A^3 = A \), its eigenvalues must satisfy \( \lambda^3 - \lambda = 0 \), meaning \( \lambda \in \{0, 1, -1\} \). We compare the characteristic polynomial of \( A \) with \( \lambda^3 - \lambda = 0 \).


Step 3: Detailed Explanation:

1. Find the characteristic equation \( \det(A - \lambda I) = 0 \):
\[ \begin{vmatrix} -\lambda & 1 & 2
a & -\lambda & 3
1 & c & -\lambda \end{vmatrix} = -\lambda(\lambda^2 - 3c) - 1(-a\lambda - 3) + 2(ac + \lambda) \]
\[ = -\lambda^3 + (3c + a + 2)\lambda + (3 + 2ac) = 0 \]

2. For eigenvalues to be \( 0, 1, -1 \), the equation must be \( \lambda^3 - \lambda = 0 \).

3. Match coefficients:

- Constant term: \( 3 + 2ac = 0 \implies ac = -3/2 \).

- Coefficient of \( \lambda \): \( -(a + 3c + 2) = -1 \implies a + 3c + 2 = 1 \implies a + 3c = -1 \).

4. Solve for \( a \):

Substitute \( c = -3/(2a) \):
\[ a + 3(-3/2a) = -1 \implies 2a^2 + 2a - 9 = 0 \]

5. Positive root \( a = \frac{-2 + \sqrt{4 + 72}}{4} = \frac{-2 + \sqrt{76}}{4} = \frac{-1 + \sqrt{19}}{2} \).

Since \( \sqrt{19} \approx 4.35 \), \( a \approx (4.35 - 1)/2 = 1.675 \).
\( a \in (1, 2] \), so \( n = 2 \).


Step 4: Final Answer:

The value of \( n \) is 2.
Quick Tip: Using properties of eigenvalues (\( \sum \lambda_i = Tr(A) \) and \( \prod \lambda_i = \det(A) \)) is much more efficient than calculating matrix powers for polynomial matrix equations.

Step 1: Understanding the Concept:

This is a problem of Derangements. A derangement is a permutation of elements of a set such that no element appears in its original position.


Step 2: Key Formula or Approach:

The number of derangements of \( n \) items is \( D_n = n! \left[ \frac{1}{2!} - \frac{1}{3!} + \dots + \frac{(-1)^n}{n!} \right] \).


Step 3: Detailed Explanation:

For \( n = 5 \) students:
\[ D_5 = 5! \left( \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \frac{1}{5!} \right) \]
\[ D_5 = 120 \left( \frac{1}{2} - \frac{1}{6} + \frac{1}{24} - \frac{1}{120} \right) \]
\[ D_5 = 60 - 20 + 5 - 1 = 44 \]


Step 4: Final Answer:

The number of ways is 44.
Quick Tip: Memorize basic derangement values for speed: \( D_2=1, D_3=2, D_4=9, D_5=44 \). You can also use the recursion \( D_n = (n-1)(D_{n-1} + D_{n-2}) \).