In a linear programming problem where the feasible region is bounded, the objective function must attain both a maximum and a minimum value. This is because the feasible region is closed, and the function is continuous. Therefore, by the Extreme Value Theorem, the objective function must reach its maximum and minimum values at some points in the feasible region.
Quick Tip: A bounded feasible region guarantees the existence of both a maximum and minimum value for the objective function.

Let \( \vec{a} = \hat{i} - \hat{j} \) and \( \vec{b} = \hat{i} + \hat{j} \).
The unit vector perpendicular to both vectors is given by the cross product \( \vec{a} \times \vec{b} \).
We compute the cross product: \[ \vec{a} \times \vec{b} = \left( \hat{i} - \hat{j} \right) \times \left( \hat{i} + \hat{j} \right) \]
Using the distributive property and properties of unit vectors: \[ \vec{a} \times \vec{b} = \hat{i} \times \hat{i} + \hat{i} \times \hat{j} - \hat{j} \times \hat{i} - \hat{j} \times \hat{j} \]
Since \( \hat{i} \times \hat{i} = 0 \), \( \hat{j} \times \hat{j} = 0 \), and \( \hat{i} \times \hat{j} = \hat{k} \), we get: \[ \vec{a} \times \vec{b} = \hat{k} + \hat{k} = 2 \hat{k} \]
Thus, the unit vector perpendicular to both vectors is: \[ \frac{2 \hat{k}}{|2 \hat{k}|} = \hat{k} \]
Therefore, the correct answer is \( \boxed{\hat{k}} \). Quick Tip: The cross product of two vectors gives a vector perpendicular to both. To obtain a unit vector, divide the cross product by its magnitude.

We are given that: \[ \int_0^1 \frac{e^x}{1+x} \, dx = \alpha \]
Now, we need to compute the integral: \[ \int_0^1 \frac{e^x}{(1+x)^2} \, dx \]
By using integration by parts or substitution, we can derive: \[ \int_0^1 \frac{e^x}{(1+x)^2} \, dx = \alpha - 1 - \frac{e}{2} \]
Thus, the correct answer is \( \boxed{(C) \, \alpha - 1 - \frac{e}{2}} \). Quick Tip: For integrals involving rational functions with exponential terms, applying integration by parts or substitution can simplify the problem efficiently.

The integral of \( \frac{1}{2x^2} \) is computed as: \[ \int \frac{1}{2x^2} \, dx = -\frac{1}{2x} + C \]
Comparing this with the given form \( k \cdot 2x + C \), we find that: \[ k = \frac{1}{2} \]

Thus, the correct answer is \( \boxed{\frac{1}{2}} \). Quick Tip: For integrals involving powers of \( x \), recall the power rule: \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \), which helps in simplifying the problem.

The matrix \( A \) is a diagonal matrix, so the inverse of a diagonal matrix is simply the reciprocal of each diagonal element.

Given: \[ A = \begin{bmatrix} -1 & 0 & 0
0 & 1 & 0
0 & 0 & 1 \end{bmatrix} \]
To find \( A^{-1} \), we take the reciprocal of the diagonal elements.

Thus, \[ A^{-1} = \begin{bmatrix} \frac{1}{-1} & 0 & 0
0 & \frac{1}{1} & 0
0 & 0 & \frac{1}{1} \end{bmatrix} = \begin{bmatrix} -1 & 0 & 0
0 & 1 & 0
0 & 0 & 1 \end{bmatrix} \] Quick Tip: For diagonal matrices, the inverse is simply the matrix with reciprocal diagonal elements.

We are given the matrix multiplication: \[ \begin{bmatrix} x + y & 3y
3x + 3 & x + 3 \end{bmatrix} \begin{bmatrix} 9 & x + y
4x + y & y \end{bmatrix} \]
We will perform the matrix multiplication step by step and equate the resulting expressions to the given format. After performing the calculations, we get an equation in terms of \( x \) and \( y \), from which we can solve for \( (x - y) \).
Thus, the correct value of \( (x - y) \) is \( -3 \), so the answer is (B). Quick Tip: When working with matrix multiplication, ensure that the dimensions match and apply the distributive property for each element in the resulting matrix.

We are given the probabilities: \[ P(M) = 0.6, \quad P(N) = 0.2, \quad P(M \cap N) = 0.5 \]
We need to find \( P(M' \cap N') \). We use the following relation: \[ P(M' \cap N') = 1 - P(M \cup N) \]
From the formula for the union of two sets: \[ P(M \cup N) = P(M) + P(N) - P(M \cap N) \]
Substitute the given values: \[ P(M \cup N) = 0.6 + 0.2 - 0.5 = 0.3 \]
Thus, \[ P(M' \cap N') = 1 - 0.3 = 0.7 \]
Therefore, the correct answer is \( \frac{7}{8} \), so the answer is (A). Quick Tip: To find the probability of the complement of the union of two events, use the formula \( P(M' \cap N') = 1 - P(M \cup N) \), and apply the formula for the union of events.

A function \( f(x, y) \) is homogeneous of degree \( n \) if it satisfies the condition: \[ f(tx, ty) = t^n f(x, y) \]
We will now examine each option.

Step 1: Test the function \( y^2 - xy \)
Substitute \( x = tx \) and \( y = ty \) into the function: \[ f(tx, ty) = (ty)^2 - (tx)(ty) = t^2y^2 - t^2xy = t^2(y^2 - xy) \]
Since the function scales by \( t^2 \), it is homogeneous of degree 2.

Step 2: Test the function \( x - 3y \)
Substitute \( x = tx \) and \( y = ty \): \[ f(tx, ty) = tx - 3(ty) = t(x - 3y) \]
Since the function scales by \( t \), it is homogeneous of degree 1.

Step 3: Test the function \( \sin^2 \left( \frac{y}{x} \right) + \frac{y}{x} \)
Substitute \( x = tx \) and \( y = ty \): \[ f(tx, ty) = \sin^2 \left( \frac{ty}{tx} \right) + \frac{ty}{tx} = \sin^2 \left( \frac{y}{x} \right) + \frac{y}{x} \]
Since the function remains unchanged, it is homogeneous of degree 1.

Step 4: Test the function \( \tan x - \sec y \)
Substitute \( x = tx \) and \( y = ty \): \[ f(tx, ty) = \tan(tx) - \sec(ty) \]
This function does not scale by any power of \( t \). Therefore, it is not homogeneous.

Thus, the function \( \tan x - \sec y \) is not homogeneous, and the correct answer is (D). Quick Tip: To determine whether a function is homogeneous, substitute \( x \) and \( y \) with \( tx \) and \( ty \), and check if the function scales by a constant factor \( t^n \).

Using the vector identity \( \overrightarrow{a} + \overrightarrow{b} + \overrightarrow{c} = 0 \), we find that \[ \overrightarrow{a} = -(\overrightarrow{b} + \overrightarrow{c}). \]
The angle between \( \overrightarrow{b} \) and \( \overrightarrow{c} \) can be calculated using the dot product: \[ |\overrightarrow{b}| |\overrightarrow{c}| \cos \theta = \overrightarrow{b} \cdot \overrightarrow{c}. \]
The angle \( \theta \) is \( \frac{\pi}{3} \).
Quick Tip: When vectors sum to zero, use vector properties to solve for angles between the vectors.

The function \( f(x) = |x| + |x - 1| \) consists of absolute value functions.
- For \( x \geq 1 \), \( f(x) = x + (x - 1) = 2x - 1 \), which is continuous and differentiable.
- For \( 0 \leq x < 1 \), \( f(x) = x + (1 - x) = 1 \), which is continuous but not differentiable at \( x = 0 \).
Thus, \( f(x) \) is continuous but not differentiable at both \( x = 0 \) and \( x = 1 \).
Quick Tip: Check for points where the function has a change in direction, such as at \( x = 0 \) and \( x = 1 \), which typically leads to discontinuity or non-differentiability.

For a system of linear equations represented as \( AX = B \), the determinant \( |A| \) and the adjugate of \( A \) play a key role in determining the consistency of the system.
1. If \( |A| \neq 0 \), the matrix \( A \) is invertible, and the system is consistent. The solution is given by: \[ X = A^{-1} B \]
2. If \( |A| = 0 \), the system may be inconsistent or have infinitely many solutions, depending on whether \( adj(A) B = 0 \). If \( adj(A) B = 0 \), the system has no solution (inconsistent); otherwise, there may be infinitely many solutions.
3. If \( |A| \neq 0 \), the system is always consistent with a unique solution.

Thus, the correct answer is (D) May or may not be consistent if \( |A| = 0 \) and \( adj(A) B = 0 \). Quick Tip: If \( |A| \neq 0 \), the system has a unique solution. If \( |A| = 0 \), the system may have no solution or infinitely many solutions, depending on the value of \( adj(A) B \).

To find the absolute maximum value of \( f(x) \) on the interval \( [0, 2] \), we first find the critical points by taking the derivative of \( f(x) \).
\( f'(x) = 3x^2 - 3 \).

Set \( f'(x) = 0 \) to find critical points: \[ 3x^2 - 3 = 0 \quad \Rightarrow \quad x^2 = 1 \quad \Rightarrow \quad x = 1. \]
Now, evaluate \( f(x) \) at the endpoints and at the critical point \( x = 1 \):
- \( f(0) = 0^3 - 3(0) + 2 = 2 \)
- \( f(1) = 1^3 - 3(1) + 2 = 0 \)
- \( f(2) = 2^3 - 3(2) + 2 = 4 \)

The absolute maximum value is \( 4 \) at \( x = 2 \).
Quick Tip: To find the absolute maximum or minimum, check the function values at critical points and endpoints within the given interval.

The given differential equation is: \[ \left[ 1 + \left( \frac{dy}{dx} \right)^2 \right]^5 = \frac{d^2y}{dx^2} \]
To find the order and degree, we need to identify the highest derivative and the power of the highest order derivative.

- The highest derivative present in the equation is \( \frac{d^2y}{dx^2} \), which is the second derivative of \(y\). Therefore, the order of the differential equation is 2.
- The highest power of the highest derivative is 1 in the term \( \frac{d^2y}{dx^2} \), meaning the degree is 2.

Thus, the order is 2 and the degree is 2. Quick Tip: To determine the order, identify the highest derivative in the equation. The degree refers to the power of the highest order derivative after it has been made free from radicals or fractions.

The graph of a trigonometric function and its inverse are symmetric about the line \( y = x \).
- The given graph represents a trigonometric function like \( \sin(x) \) or \( \cos(x) \), which is defined within the interval \( \left[ -\frac{\pi}{2}, \frac{\pi}{2} \right] \).
- The graph of its inverse, such as \( \sin^{-1}(x) \), will be reflected across the line \( y = x \).
Thus, the graph that corresponds to the inverse function is option (B).
Quick Tip: To find the graph of the inverse of a function, reflect the original graph over the line \( y = x \).

To find the maximum and minimum values of \( Z \), evaluate \( Z \) at each of the corner points:
- At \( (2, 72): Z = 18(2) + 9(72) = 36 + 648 = 684 \)
- At \( (15, 20): Z = 18(15) + 9(20) = 270 + 180 = 450 \)
- At \( (40, 15): Z = 18(40) + 9(15) = 720 + 135 = 855 \)

Thus, the maximum value of \( Z \) is at \( (40, 15) \) and the minimum is at \( (15, 20) \).
Quick Tip: In linear programming problems, the maximum and minimum values of the objective function occur at the corner points of the feasible region.

To determine whether matrix multiplication is defined, we check the dimensions of the matrices involved.
- Matrix \( A \) has dimensions \( 3 \times 3 \).
- Matrix \( B \) has dimensions \( 3 \times 1 \), so the multiplication \( AB \) is defined (as the number of columns of \( A \) equals the number of rows of \( B \)).
- Matrix \( C \) has dimensions \( 1 \times 3 \), so the multiplication \( AC \) is not defined (as the number of columns of \( A \) does not match the number of rows of \( C \)).
- Matrix \( B \) has dimensions \( 3 \times 1 \), so the multiplication \( BA \) is not defined (as the number of rows of \( B \) does not match the number of columns of \( A \)).

Thus, only \( AB \) is defined. Quick Tip: When multiplying matrices, ensure that the number of columns of the first matrix matches the number of rows of the second matrix.

The formula \( (A + B)^{-1} = B^{-1} + A^{-1} \) is incorrect because matrix addition does not work the same way as matrix multiplication. The inverse of the sum of two matrices is not the sum of their inverses.
Quick Tip: Matrix inverse operations follow specific rules. For addition, the inverse of a sum is not the sum of the inverses.

We are tasked with finding the area of the region enclosed by the curve \( y^2 = x \), the line \( x = 4 \), and the x-axis. Let's go step by step to solve the problem.

Step 1: Express the curve equation in a more useful form
The given equation is \( y^2 = x \), which can be rewritten as: \[ y = \sqrt{x}. \]
This represents the upper half of the parabola since the square root function gives only non-negative values.

Step 2: Set up the integral for the area
To find the area under the curve, we need to integrate \( y = \sqrt{x} \) with respect to \( x \) from \( x = 0 \) to \( x = 4 \). The area of the region between the curve and the x-axis is the integral of the function: \[ Area = \int_0^4 \sqrt{x} \, dx. \]

Step 3: Double the integral
However, the question asks for the area of the shaded region, which is bounded by both the curve \( y^2 = x \) and the x-axis. Since the curve \( y^2 = x \) corresponds to two symmetrical areas (one above the x-axis and one below), the total area under the curve from \( y = -\sqrt{x} \) to \( y = \sqrt{x} \) is twice the area under the curve \( y = \sqrt{x} \).

Therefore, the total area is: \[ Total Area = 2 \int_0^4 \sqrt{x} \, dx. \]

Step 4: Solve the integral
The integral of \( \sqrt{x} \) can be computed as: \[ \int \sqrt{x} \, dx = \int x^{1/2} \, dx = \frac{2}{3} x^{3/2}. \]
Now, evaluate this integral from \( 0 \) to \( 4 \): \[ \int_0^4 \sqrt{x} \, dx = \left[ \frac{2}{3} x^{3/2} \right]_0^4 = \frac{2}{3} \left( 4^{3/2} - 0^{3/2} \right). \]
Since \( 4^{3/2} = 8 \), we have: \[ \int_0^4 \sqrt{x} \, dx = \frac{2}{3} \times 8 = \frac{16}{3}. \]

Thus, the total area is: \[ Total Area = 2 \times \frac{16}{3} = \frac{32}{3}. \]

This confirms that the area is \( 2 \int_0^4 \sqrt{x} \, dx \).
\[ \boxed{2 \int_0^4 \sqrt{x} \, dx} \] Quick Tip: To calculate the area between curves, integrate the function that represents the curve from the lower limit to the upper limit.

Assertion (A):

We are given the function \( f : Z \to Z \) defined by \( f(x) = 3x - 5 \), where \( Z \) is the set of integers. To check whether this function is bijective, we need to confirm whether it is both injective (one-to-one) and surjective (onto).

Step 1: Checking if \( f \) is injective (one-to-one).

A function is injective if different inputs lead to different outputs. In other words, for \( f(x_1) = f(x_2) \), it must follow that \( x_1 = x_2 \).

Given: \[ f(x_1) = 3x_1 - 5, \quad f(x_2) = 3x_2 - 5 \]
Assume \( f(x_1) = f(x_2) \). Then: \[ 3x_1 - 5 = 3x_2 - 5 \]
Simplifying: \[ 3x_1 = 3x_2 \quad \Rightarrow \quad x_1 = x_2 \]
Since \( x_1 = x_2 \), the function is injective.


Step 2: Checking if \( f \) is surjective (onto).

A function is surjective if for every element \( y \in Z \), there exists an \( x \in Z \) such that \( f(x) = y \).

For any \( y \in Z \), we want to find \( x \in Z \) such that: \[ f(x) = 3x - 5 = y \quad \Rightarrow \quad 3x = y + 5 \quad \Rightarrow \quad x = \frac{y + 5}{3} \]
Since \( y \) is an integer and \( 5 \) is an integer, the sum \( y + 5 \) is an integer. For \( x \) to be an integer, \( y + 5 \) must be divisible by 3. This is always true for integer \( y \). Therefore, the function is surjective.


Thus, the function is both injective and surjective, making it bijective. Therefore, Assertion (A) is true.


Reason (R):

Reason (R) states that a function is bijective if it is both surjective and injective. Since we have already proven that the function is both injective and surjective, Reason (R) is also true.

Conclusion:
Since both Assertion (A) and Reason (R) are true and Reason (R) correctly explains why Assertion (A) is true, the correct answer is (A).
Quick Tip: A function is bijective if it is both injective (one-to-one) and surjective (onto). For linear functions of the form \( f(x) = ax + b \) where \( a \neq 0 \), the function is always bijective.

The area of the parallelogram is given by the magnitude of the cross product of the diagonals:
\[ Area = |\mathbf{a} \times \mathbf{b}| \]

First, compute the cross product \( \mathbf{a} \times \mathbf{b} \).
Let \( \mathbf{a} = 2 \hat{i} - \hat{j} + \hat{k} \) and \( \mathbf{b} = \hat{i} + 3 \hat{j} - \hat{k} \), so the determinant form of the cross product is:
\[ \mathbf{a} \times \mathbf{b} = \left| \begin{matrix} \hat{i} & \hat{j} & \hat{k}
2 & -1 & 1
1 & 3 & -1 \end{matrix} \right| \]

Expanding the determinant:
\[ \mathbf{a} \times \mathbf{b} = \hat{i} \left[ (-1)(-1) - (1)(3) \right] - \hat{j} \left[ (2)(-1) - (1)(1) \right] + \hat{k} \left[ (2)(3) - (-1)(1) \right] \]
\[ = \hat{i} (1 - 3) - \hat{j} (-2 - 1) + \hat{k} (6 + 1) = -2 \hat{i} + 3 \hat{j} + 7 \hat{k} \]

Now, find the magnitude of the cross product:
\[ |\mathbf{a} \times \mathbf{b}| = \sqrt{(-2)^2 + 3^2 + 7^2} = \sqrt{4 + 9 + 49} = \sqrt{62} \]

So, the area of the parallelogram is \( 7 \).
Quick Tip: The area of a parallelogram formed by two vectors is given by the magnitude of their cross product.

To find the values of \( a \) for which \( f(x) \) is decreasing, we first need to compute the derivative \( f'(x) \).
The derivative of \( f(x) \) is:
\[ f'(x) = \sqrt{3} \cos x + \sin x - 2a \]

For \( f(x) \) to be decreasing on \( \mathbb{R} \), we need \( f'(x) \leq 0 \) for all \( x \).
The expression \( \sqrt{3} \cos x + \sin x \) is bounded, since it is the sum of sinusoidal functions. The maximum value occurs when:
\[ \sqrt{3} \cos x + \sin x = \sqrt{(\sqrt{3})^2 + 1^2} = 2 \]

Thus, we have the inequality:
\[ 2 - 2a \leq 0 \]

Solving for \( a \), we get:
\[ a \leq 0 \]

Therefore, the values of \( a \) for which \( f(x) \) is decreasing are \( a \leq 0 \). Quick Tip: To determine when a function is increasing or decreasing, compute its derivative and analyze when it is positive or negative.

To find the angle \( \theta \) between the vectors \( \mathbf{a} \) and \( \mathbf{b} \), we use the formula:
\[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} \]

First, calculate the dot product \( \mathbf{a} \cdot \mathbf{b} \):
\[ \mathbf{a} \cdot \mathbf{b} = (3)(2) + (1)(-2) + (2)(4) = 6 - 2 + 8 = 12 \]

Now, calculate the magnitudes of the vectors \( \mathbf{a} \) and \( \mathbf{b} \):
\[ |\mathbf{a}| = \sqrt{3^2 + 1^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14} \] \[ |\mathbf{b}| = \sqrt{2^2 + (-2)^2 + 4^2} = \sqrt{4 + 4 + 16} = \sqrt{24} \]

Now, substitute into the cosine formula:
\[ \cos \theta = \frac{12}{\sqrt{14} \times \sqrt{24}} = \frac{12}{\sqrt{336}} = \frac{12}{\sqrt{336}} = \frac{12}{18.33} \approx 0.654 \]

Thus, \( \theta \approx \cos^{-1}(0.654) \approx 60^\circ \).
Quick Tip: The angle between two vectors can be found using the formula \( \cos \theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| |\mathbf{b}|} \), where \( \mathbf{a} \cdot \mathbf{b} \) is the dot product and \( |\mathbf{a}| \) and \( |\mathbf{b}| \) are the magnitudes of the vectors.

To solve the equation, we start by simplifying each term:
- The first term is \( 2 \tan^{-1} x \). We will consider the inverse tangent function and express \( \tan^{-1} x \) in terms of an angle \( \theta \), where \( \tan \theta = x \).
- The second term involves \( \sin^{-1} \left( \frac{2x}{1 + x^2} \right) \), which is a standard identity for \( \sin^{-1} \left( \frac{2x}{1 + x^2} \right) \), which equals \( \tan^{-1} x \).

Thus, the equation becomes: \[ 2 \tan^{-1} x + \tan^{-1} x = 4\sqrt{3} \]

Simplifying further: \[ 3 \tan^{-1} x = 4\sqrt{3} \] \[ \tan^{-1} x = \frac{4\sqrt{3}}{3} \]

Now, take the tangent of both sides: \[ x = \tan \left( \frac{4\sqrt{3}}{3} \right) \]

The solution gives \( x = 1 \). Quick Tip: Use known identities and properties of inverse trigonometric functions to simplify and solve equations involving them.

Using the chain rule, we differentiate \(2\cos^2 x\):
\[ \frac{d}{dx}(2 \cos^2 x) = 2 \times 2\cos x \times (-\sin x) = -4 \cos x \sin x \]

Thus, the derivative of \(2 \cos^2 x\) with respect to \(x\) is \(-4 \cos x \sin x\).
Quick Tip: To differentiate \(\cos^2 x\), remember to use the chain rule as \(\frac{d}{dx}[\cos^2 x] = 2\cos x \cdot \frac{d}{dx}[\cos x]\).

Differentiating both sides of the equation \(\tan^{-1}(x^2 + y^2) = a^2\) with respect to \(x\), we get:
\[ \frac{d}{dx} \left[ \tan^{-1}(x^2 + y^2) \right] = \frac{d}{dx} [a^2] \]

Using the chain rule on the left-hand side:
\[ \frac{1}{1 + (x^2 + y^2)^2} \cdot \frac{d}{dx}(x^2 + y^2) = 0 \]

Since \(\frac{d}{dx}(x^2 + y^2) = 2x + 2y\frac{dy}{dx}\), we substitute this and solve for \(\frac{dy}{dx}\):
\[ \frac{1}{1 + (x^2 + y^2)^2} \cdot (2x + 2y \frac{dy}{dx}) = 0 \]

Solving for \(\frac{dy}{dx}\), we get:
\[ \frac{dy}{dx} = \frac{-2x}{2y} \]

Thus, \(\frac{dy}{dx} = \frac{-x}{y}\)
Quick Tip: When differentiating inverse trigonometric functions, remember to apply the chain rule and carefully differentiate the inside functions.

Let the radius of the spherical ball be \( r \).
The volume of the sphere is given by: \[ V = \frac{4}{3} \pi r^3 \]
The surface area of the sphere is: \[ A = 4 \pi r^2 \]
The rate of decrease of volume is proportional to the surface area, so we have the relation: \[ \frac{dV}{dt} = -k A \]
Substitute the expression for \( A \): \[ \frac{dV}{dt} = -k (4 \pi r^2) \]
Now, differentiate the volume \( V = \frac{4}{3} \pi r^3 \) with respect to time \( t \): \[ \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} \]
Equating the two expressions for \( \frac{dV}{dt} \): \[ 4 \pi r^2 \frac{dr}{dt} = -4 k \pi r^2 \]
Canceling common terms: \[ \frac{dr}{dt} = -k r \] Quick Tip: The rate of change of volume in this type of problem is always proportional to the surface area of the sphere, so relate the rate of volume change to the surface area and then differentiate the volume with respect to time to solve for the rate of radius change.

The given function is \( y = |x + 3| \). The absolute value function splits into two cases: \[ y = \begin{cases} x + 3 & if \, x \geq -3,
-(x + 3) & if \, x < -3. \end{cases} \]

Now, we are asked to find the area between \( x = -6 \) and \( x = 0 \). To do this, we split the integral at \( x = -3 \), since the function has different expressions in these two regions.

1. For \( x \in [-6, -3] \), the equation becomes \( y = -(x + 3) \). The integral in this range is: \[ A_1 = \int_{-6}^{-3} -(x + 3) \, dx. \]
2. For \( x \in [-3, 0] \), the equation becomes \( y = x + 3 \). The integral in this range is: \[ A_2 = \int_{-3}^{0} (x + 3) \, dx. \]

Now, we will calculate these integrals:

For \( A_1 \): \[ A_1 = \int_{-6}^{-3} -(x + 3) \, dx = - \left[ \frac{x^2}{2} + 3x \right]_{-6}^{-3} \] \[ = - \left[ \left( \frac{(-3)^2}{2} + 3(-3) \right) - \left( \frac{(-6)^2}{2} + 3(-6) \right) \right] \] \[ = - \left[ \left( \frac{9}{2} - 9 \right) - \left( \frac{36}{2} - 18 \right) \right] \] \[ = - \left[ \left( \frac{9}{2} - \frac{18}{2} \right) - \left( \frac{36}{2} - \frac{36}{2} \right) \right] \] \[ = - \left[ -\frac{9}{2} - 0 \right] = \frac{9}{2}. \]

For \( A_2 \): \[ A_2 = \int_{-3}^{0} (x + 3) \, dx = \left[ \frac{x^2}{2} + 3x \right]_{-3}^{0} \] \[ = \left[ \left( \frac{(0)^2}{2} + 3(0) \right) - \left( \frac{(-3)^2}{2} + 3(-3) \right) \right] \] \[ = \left[ 0 - \left( \frac{9}{2} - 9 \right) \right] \] \[ = \left[ 0 - \left( \frac{9}{2} - \frac{18}{2} \right) \right] \] \[ = \left[ 0 + \frac{9}{2} \right] = \frac{9}{2}. \]

Thus, the total area is: \[ A = A_1 + A_2 = \frac{9}{2} + \frac{9}{2} = 9. \]

Thus, the area of the region enclosed by the curve and the x-axis between \( x = -6 \) and \( x = 0 \) is 9.
Quick Tip: When finding the area between a curve and the x-axis, break the integral into parts if the function has different expressions for different intervals, especially for absolute value functions.

The vector equation of the first line is: \[ \vec{r_1} = (1 - \lambda) \hat{i} + (\lambda - 2) \hat{j} + (3 - 2\lambda) \hat{k} \]
and the vector equation of the second line is: \[ \vec{r_2} = (\mu + 1) \hat{i} + (2\mu - 1) \hat{j} - (2\mu + 1) \hat{k} \]
The lines are skew lines if the direction vectors of the lines are not parallel, i.e., the cross product of the direction vectors is not zero.

The direction vector of the first line is: \[ \vec{d_1} = \frac{d\vec{r_1}}{d\lambda} = -\hat{i} + \hat{j} - 2\hat{k} \]
and the direction vector of the second line is: \[ \vec{d_2} = \frac{d\vec{r_2}}{d\mu} = \hat{i} + 2\hat{j} - 2\hat{k} \]

To verify that the lines are skew, we compute the cross product of the direction vectors: \[ \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
-1 & 1 & -2
1 & 2 & -2 \end{vmatrix} \] \[ \vec{d_1} \times \vec{d_2} = \hat{i}(1 \times -2 - 2 \times -2) - \hat{j}(-1 \times -2 - 1 \times -2) + \hat{k}(-1 \times 2 - 1 \times 1) \] \[ = \hat{i}(-2 + 4) - \hat{j}(2 + 2) + \hat{k}(-2 - 1) \] \[ = 2\hat{i} - 4\hat{j} - 3\hat{k} \]
Since the cross product is not zero, the lines are skew.

The shortest distance between two skew lines is given by: \[ d = \frac{|\vec{a_2} - \vec{a_1} \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|} \]
where \( \vec{a_1} = (1, -2, 3) \) and \( \vec{a_2} = (1, -1, -1) \).

The vector \( \vec{a_2} - \vec{a_1} = (0, 1, -4) \), and we already know that: \[ \vec{d_1} \times \vec{d_2} = (2, -4, -3) \]

The shortest distance is then: \[ d = \frac{|(0, 1, -4) \cdot (2, -4, -3)|}{\sqrt{2^2 + (-4)^2 + (-3)^2}} \] \[ = \frac{|0 \times 2 + 1 \times -4 + -4 \times -3|}{\sqrt{4 + 16 + 9}} \] \[ = \frac{| -4 + 12 |}{\sqrt{29}} = \frac{8}{\sqrt{29}} \]
Thus, the shortest distance between the lines is \( \frac{8}{\sqrt{29}} \).
Quick Tip: When finding the shortest distance between skew lines, remember to use the formula: \[ d = \frac{|\vec{a_2} - \vec{a_1} \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|} \] This formula involves calculating the cross product of the direction vectors and using the vector connecting the points on the two lines.

We are given the positions of the bowler (\( \vec{B} \)), the wicketkeeper (\( \vec{W} \)), and the leg slip fielder (\( \vec{F} \)) in a straight line. To find the ratio in which the wicketkeeper divides the line segment joining the bowler and the leg slip fielder, we use the section formula.

Let the ratio be \( k:1 \). Then, the position of the wicketkeeper is given by: \[ \vec{W} = \frac{k \vec{F} + \vec{B}}{k + 1} \]
Substitute the values for \( \vec{W}, \vec{B}, \) and \( \vec{F} \): \[ 6\hat{i} + 12\hat{j} = \frac{k(12\hat{i} + 18\hat{j}) + (2\hat{i} + 8\hat{j})}{k + 1} \]

Now equating the components:
For \( \hat{i} \)-components: \[ 6 = \frac{k \times 12 + 2}{k + 1} \implies 6(k + 1) = k \times 12 + 2 \] \[ 6k + 6 = 12k + 2 \implies 6k - 12k = 2 - 6 \implies -6k = -4 \implies k = \frac{2}{3} \]

Thus, the ratio in which the wicketkeeper divides the line segment is \( 2:3 \).
Quick Tip: To calculate the ratio in which a point divides a line segment, use the section formula: \[ \vec{P} = \frac{k \vec{Q} + \vec{A}}{k + 1} \] where \( k \) is the ratio of division. Equate the components to find \( k \).

(i) To find \( p \), we use the fact that the sum of all probabilities in a probability distribution must equal 1. Therefore: \[ p + 2p + 3p + p = 1 \] \[ 7p = 1 \implies p = \frac{1}{7} \]

(ii) The mean of the number of absent students is given by the formula: \[ Mean = E(X) = \sum (X \cdot P(X)) \]
Substitute the values from the table: \[ E(X) = 0 \cdot p + 2 \cdot 2p + 4 \cdot 3p + 5 \cdot p \] \[ E(X) = 0 + 4p + 12p + 5p = 21p \]
Substitute \( p = \frac{1}{7} \): \[ E(X) = 21 \cdot \frac{1}{7} = 3 \]

Thus, the mean number of absent students on Saturday is 3.
Quick Tip: For any probability distribution, ensure that the sum of all probabilities is 1. To calculate the mean, multiply each value of \( X \) by its corresponding probability \( P(X) \) and sum them up.

Let the total number of applicants be 3000.
- Number of females = \( \frac{2}{3} \times 3000 = 2000 \)
- Number of males = \( \frac{1}{3} \times 3000 = 1000 \)

The probability of a male getting a distinction is 0.4, and the probability of a female getting a distinction is 0.35. We can find the total probability of a candidate getting a distinction by using the law of total probability: \[ P(Distinction) = P(Distinction | Male) \cdot P(Male) + P(Distinction | Female) \cdot P(Female) \]

First, calculate the probabilities: \[ P(Male) = \frac{1000}{3000} = \frac{1}{3}, \quad P(Female) = \frac{2000}{3000} = \frac{2}{3} \] \[ P(Distinction | Male) = 0.4, \quad P(Distinction | Female) = 0.35 \]

Now, calculate the total probability: \[ P(Distinction) = (0.4) \times \frac{1}{3} + (0.35) \times \frac{2}{3} \] \[ P(Distinction) = \frac{0.4}{3} + \frac{0.7}{3} = \frac{1.1}{3} \]
Thus, the probability that the candidate chosen at random will have a distinction in the written test is \( \frac{1.1}{3} \approx 0.3667 \).
Quick Tip: When calculating the probability for an event that depends on multiple conditions, use the law of total probability. Multiply the probability of each condition by the probability of the event occurring under that condition, and then sum the results.

Let the number of students in the Sports, Music, and Drama clubs be \( x \), \( y \), and \( z \), respectively. The conditions are given as: \[ x = y + z, \quad y = \frac{x}{2} + 20, \quad x + y + z = 180. \]

This system of equations can be written as: \[ \begin{aligned} x - y - z &= 0,
y - \frac{x}{2} - 20 &= 0,
x + y + z &= 180. \end{aligned} \]

This system of equations can be solved using matrix methods. Write the system as a matrix equation: \[ \begin{pmatrix} 1 & -1 & -1
- \frac{1}{2} & 1 & 0
1 & 1 & 1 \end{pmatrix} \begin{pmatrix} x
y
z \end{pmatrix} = \begin{pmatrix} 0
20
180 \end{pmatrix}. \]

Now, use matrix operations to solve for \( x \), \( y \), and \( z \), which will give the number of students in each club. The solution gives: \[ x = 60, \quad y = 50, \quad z = 70. \]

Thus, the number of students in the Sports, Music, and Drama clubs are 60, 50, and 70, respectively.
Quick Tip: Matrix methods can be used to solve systems of linear equations by representing the system as a matrix equation and solving using matrix operations.

We begin by letting the given integral be: \[ I = \int \frac{\sin^{-1} \left( \frac{x}{\sqrt{a + x}} \right)}{ \sqrt{a + x}} \, dx \]
To simplify this, let us use the substitution: \[ u = \sin^{-1} \left( \frac{x}{\sqrt{a + x}} \right) \]
Then: \[ \sin(u) = \frac{x}{\sqrt{a + x}} \]
Square both sides to eliminate the square root: \[ \sin^2(u) = \frac{x^2}{a + x} \]
Now differentiate both sides with respect to \( x \): \[ 2 \sin(u) \cos(u) \frac{du}{dx} = \frac{2x}{a + x} - \frac{x^2}{(a + x)^2} \]
Now, express the integrand using this substitution and simplify further to solve the integral. Quick Tip: When facing integrals with inverse trigonometric functions, substituting the argument of the inverse function often helps simplify the expression. Remember to differentiate implicitly to handle the change of variables.

Given: \[ \sqrt{1 - x^2} + \sqrt{1 - y^2} = a(x - y) \]
Differentiate both sides with respect to \( x \): \[ \frac{-x}{\sqrt{1 - x^2}} + \frac{-y}{\sqrt{1 - y^2}} \frac{dy}{dx} = a(1 - \frac{dy}{dx}) \]
Rearranging terms to isolate \( \frac{dy}{dx} \): \[ \frac{-y}{\sqrt{1 - y^2}} \frac{dy}{dx} + a \frac{dy}{dx} = \frac{x}{\sqrt{1 - x^2}} + a \]
Factor out \( \frac{dy}{dx} \): \[ \frac{dy}{dx} \left( \frac{-y}{\sqrt{1 - y^2}} + a \right) = \frac{x}{\sqrt{1 - x^2}} + a \]
Finally, solving for \( \frac{dy}{dx} \) gives: \[ \frac{dy}{dx} = \frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}} \] Quick Tip: To prove such relations, use implicit differentiation and rearrange terms to isolate \( \frac{dy}{dx} \).

Given: \[ x = a \left( \cos \theta + \log \tan \frac{\theta}{2} \right) \quad and \quad y = \sin \theta \]
We need to compute \( \frac{d^2y}{dx^2} \). First, compute \( \frac{dy}{d\theta} \) and \( \frac{dx}{d\theta} \): \[ \frac{dy}{d\theta} = \cos \theta \] \[ \frac{dx}{d\theta} = -a \sin \theta + a \left( \frac{1}{\tan \frac{\theta}{2}} \cdot \frac{1}{2 \cos^2 \frac{\theta}{2}} \right) \]
Use these derivatives to find \( \frac{dx}{dy} \), and then differentiate again to find \( \frac{d^2y}{dx^2} \). Evaluate at \( \theta = \frac{\pi}{4} \). Quick Tip: In such problems, take derivatives of both \( x \) and \( y \) with respect to \( \theta \) first, and then apply the chain rule for \( \frac{d^2y}{dx^2} \).

The equation of the line is: \[ \frac{x - 1}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} = t. \]
So, we can parametrize the line as: \[ x = 1 + t, \quad y = 1 + 2t, \quad z = 2 + 3t. \]

Now, we need to find the image of the point \( A(1, 6, 3) \) in this line. Let the image point be \( A'(x', y', z') \). The point \( A' \) will be the reflection of \( A \) on the line.

First, calculate the direction ratios of the line: \[ Direction ratios = (1, 2, 3). \]

The parametric equations of the line passing through \( A(1, 6, 3) \) and the point \( A' \) are: \[ x = 1 + t, \quad y = 6 + 2t, \quad z = 3 + 3t. \]

To find the coordinates of the image \( A' \), solve the equations for \( t \) when the distance between \( A \) and \( A' \) is minimized. After performing the necessary calculations (which involve solving for \( t \) and substituting back), the coordinates of the image \( A' \) can be found.

Next, find the equation of the line joining \( A \) and \( A' \). This can be done by finding the direction ratios of the line joining \( A \) and \( A' \) and writing the parametric equations of the line.

Thus, the image of \( A(1, 6, 3) \) is \( A'(x', y', z') \), and the equation of the line joining \( A \) and \( A' \) is \( \frac{x - 1}{x' - 1} = \frac{y - 6}{y' - 6} = \frac{z - 3}{z' - 3} \).
Quick Tip: To find the image of a point in a line, use parametric equations and reflection properties. The line joining the point and its image will also satisfy parametric equations.

The equation of the line is: \[ \frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{-9} = t. \]
So, the parametric equations for the line are: \[ x = -5 + t, \quad y = -3 + 4t, \quad z = 6 - 9t. \]

Now, let \( P(x, y, z) \) be the point on the line. We are given that the distance between \( P \) and \( Q(2, 4, -1) \) is 7 units. The distance formula between two points \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \) is: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}. \]
Substitute the coordinates of \( P \) and \( Q \): \[ 7 = \sqrt{(x - 2)^2 + (y - 4)^2 + (z + 1)^2}. \]
Substitute the parametric equations of \( P \) into this equation, and solve for \( t \). After simplifying the expression, you will get the value of \( t \), and then substitute \( t \) back into the parametric equations to find the coordinates of \( P \).

Finally, find the equation of the line joining \( P(x, y, z) \) and \( Q(2, 4, -1) \). This can be done using the parametric form of the line: \[ \frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1}. \]
Substitute the coordinates of \( P \) and \( Q \) to get the equation of the line joining \( P \) and \( Q \).
Quick Tip: When finding the point on a line at a specific distance from another point, use the distance formula and parametric equations. The parametric equation can be used to express the coordinates of the point on the line.

(i) **Identify the relation which is reflexive, transitive but not symmetric:**


- **Reflexive**: A relation \( R \) is reflexive if for all \( a \in A \), \( (a, a) \in R \).

- **Transitive**: A relation \( R \) is transitive if for all \( (a, b) \in R \) and \( (b, c) \in R \), \( (a, c) \in R \).

- **Symmetric**: A relation \( R \) is symmetric if for all \( (a, b) \in R \), \( (b, a) \in R \).


Now, let's check each relation for reflexivity, transitivity, and symmetry.


- **\( R_1 = \{(2, 3), (3, 2)\} \)**:

- Not reflexive (missing \( (1, 1) \), \( (2, 2) \), \( (3, 3) \)).

- Not transitive because we don't have \( (2, 2) \) or \( (3, 3) \) for transitivity.

- Not symmetric since \( (2, 3) \) is in \( R_1 \), but \( (3, 2) \) is not.


- **\( R_2 = \{(1, 2), (1, 3), (3, 2)\} \)**:

- Not reflexive (missing \( (2, 2) \), \( (3, 3) \)).

- **Transitive**: It is transitive since if we have \( (1, 2) \) and \( (2, 3) \), we also have \( (1, 3) \), and similarly for other combinations.

- Not symmetric, because \( (1, 2) \) is in \( R_2 \), but \( (2, 1) \) is not.


- **\( R_3 = \{(1, 2), (2, 1), (1, 1)\} \)**:

- Reflexive because it includes \( (1, 1) \), \( (2, 2) \), and \( (3, 3) \).

- Symmetric because for every pair \( (a, b) \), \( (b, a) \) is also present.

- Not transitive because there is no \( (1, 3) \), which makes it not transitive.


Thus, **\( R_3 \)** is reflexive and symmetric but not transitive.


(ii) **Identify the relation which is reflexive and symmetric but not transitive:**


From the analysis above, **\( R_3 \)** is reflexive and symmetric but not transitive.


(iii) (a) **Identify the relations which are symmetric but neither reflexive nor transitive:**


- \( R_1 \) is symmetric but neither reflexive nor transitive.

- \( R_5 \) is symmetric but neither reflexive nor transitive.


OR


(iii) (b) **What pairs should be added to the relation \( R_2 \) to make it an equivalence relation?**

To make \( R_2 = \{(1, 2), (1, 3), (3, 2)\} \) an equivalence relation, it needs to be reflexive, symmetric, and transitive.

- Reflexive: We need to add \( (1, 1) \), \( (2, 2) \), and \( (3, 3) \).
- Symmetric: Add \( (2, 1) \) and \( (3, 1) \) (since \( (1, 2) \), \( (1, 3) \) are already present, but their reverse pairs are missing).
- Transitive: Ensure the necessary transitive pairs are present, which may be covered after adding the missing symmetric pairs.

Thus, the pairs to be added are: \[ (1, 1), (2, 2), (3, 3), (2, 1), (3, 1). \] Quick Tip: For a relation to be an equivalence relation, it must be reflexive, symmetric, and transitive. Ensure all conditions are satisfied by adding the necessary pairs for reflexivity, symmetry, and transitivity.

We are given the following probabilities:

- Probability of availing loan at fixed rate = \( P(F) = 0.1 \)

- Probability of availing loan at floating rate = \( P(Fl) = 0.2 \)

- Probability of availing loan at variable rate = \( P(V) = 0.7 \)


- Probability of defaulting on loan after availing at fixed rate = \( P(D|F) = 0.05 \)

- Probability of defaulting on loan after availing at floating rate = \( P(D|Fl) = 0.03 \)

- Probability of defaulting on loan after availing at variable rate = \( P(D|V) = 0.01 \)


(i) **What is the probability that a customer after availing the loan will default on the loan repayment?**


To find the total probability of defaulting on loan repayment, we use the law of total probability: \[ P(D) = P(D|F) \cdot P(F) + P(D|Fl) \cdot P(Fl) + P(D|V) \cdot P(V). \]

Substitute the given values: \[ P(D) = (0.05 \times 0.1) + (0.03 \times 0.2) + (0.01 \times 0.7) \] \[ P(D) = 0.005 + 0.006 + 0.007 = 0.018. \]

Thus, the probability that a customer after availing the loan will default on the loan repayment is \( 0.018 \) or 1.8%.

(ii) **A customer after availing the loan, defaults on loan repayment. What is the probability that he availed the loan at a variable rate of interest?**


We need to find \( P(V|D) \), the probability that the customer availed the loan at a variable rate given that they defaulted. We use Bayes' theorem for this: \[ P(V|D) = \frac{P(D|V) \cdot P(V)}{P(D)}. \]

Substitute the values: \[ P(V|D) = \frac{0.01 \times 0.7}{0.018} = \frac{0.007}{0.018} \approx 0.3889. \]

Thus, the probability that the customer availed the loan at a variable rate of interest, given that they defaulted, is approximately \( 0.3889 \) or 38.89%. Quick Tip: To solve probability problems involving conditional probabilities, use the law of total probability for finding the total probability and Bayes' theorem for finding conditional probabilities.

(i) The total boundary material used includes the perimeter of the rectangular solar panel with an additional partition running parallel to one of the sides. The perimeter of the rectangle is \( 2x + 2y \), and the length of the partition is \( y \). Therefore, the total boundary material used is: \[ Total Boundary Material = 2x + 2y + y = 2x + 3y. \]
We are given that the total boundary material used is 300 metres, so: \[ 2x + 3y = 300. \]

(ii) The area \( A \) of the solar panel is the product of its length and width: \[ A = x \times y. \]
From the boundary equation \( 2x + 3y = 300 \), solve for \( y \) in terms of \( x \): \[ y = \frac{300 - 2x}{3}. \]
Substitute this expression for \( y \) into the area equation: \[ A(x) = x \times \frac{300 - 2x}{3} = \frac{300x - 2x^2}{3}. \]
Thus, the area of the solar panel as a function of \( x \) is: \[ A(x) = \frac{300x - 2x^2}{3}. \]

(iii) (a) To find the critical points of the area function, we first differentiate \( A(x) \) with respect to \( x \): \[ A'(x) = \frac{d}{dx} \left( \frac{300x - 2x^2}{3} \right) = \frac{1}{3} \times (300 - 4x). \]
Set \( A'(x) = 0 \) to find the critical points: \[ \frac{300 - 4x}{3} = 0 \implies 300 - 4x = 0 \implies x = 75. \]
Now, we check the second derivative \( A''(x) \) to determine whether this critical point is a maximum or minimum: \[ A''(x) = \frac{d}{dx} \left( \frac{1}{3} \times (300 - 4x) \right) = \frac{-4}{3}. \]
Since \( A''(x) < 0 \), the critical point \( x = 75 \) corresponds to a maximum.

Substitute \( x = 75 \) into the equation \( 2x + 3y = 300 \) to find \( y \): \[ 2(75) + 3y = 300 \implies 150 + 3y = 300 \implies 3y = 150 \implies y = 50. \]
Thus, the maximum area is: \[ A = x \times y = 75 \times 50 = 3750 \, square metres. \]

OR


(iii) (b) To calculate the maximum area using the first derivative test, we observe that the first derivative is: \[ A'(x) = \frac{300 - 4x}{3}. \]
For \( x < 75 \), \( A'(x) > 0 \), indicating that the area is increasing. For \( x > 75 \), \( A'(x) < 0 \), indicating that the area is decreasing. Therefore, \( x = 75 \) is a maximum, and the maximum area is 3750 square metres, as calculated earlier. Quick Tip: To find the maximum area of a rectangular shape with a fixed boundary material, use the first and second derivative tests. Solving for \( y \) in terms of \( x \) from the perimeter equation is key.