The matrix \( A \) is a diagonal matrix, and for diagonal matrices, the inverse 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.

Let’s calculate the magnitudes of vectors \( \mathbf{a} \) and \( \mathbf{b} \).
The magnitude of a vector \( \mathbf{v} = v_x \hat{i} + v_y \hat{j} + v_z \hat{k} \) is given by: \[ |\mathbf{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2} \]

Magnitude of vector \( \mathbf{a} \): \[ \mathbf{a} = 3 \hat{i} + 2 \hat{j} - \hat{k} \]
The magnitude of \( \mathbf{a} \) is: \[ |\mathbf{a}| = \sqrt{3^2 + 2^2 + (-1)^2} = \sqrt{9 + 4 + 1} = \sqrt{14} \]

Magnitude of vector \( \mathbf{b} \): \[ \mathbf{b} = \hat{i} - \hat{j} + \hat{k} \]
The magnitude of \( \mathbf{b} \) is: \[ |\mathbf{b}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{1 + 1 + 1} = \sqrt{3} \]

Comparison of magnitudes: \[ |\mathbf{a}| = \sqrt{14} \quad and \quad |\mathbf{b}| = \sqrt{3} \]
Since \( \sqrt{14} > \sqrt{3} \), we conclude that \( |\mathbf{a}| > |\mathbf{b}| \).

Therefore, the correct answer is (C). Quick Tip: The magnitude of a vector is calculated by taking the square root of the sum of the squares of its components. For vectors, use the formula \( |\mathbf{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2} \).

We are tasked with evaluating the following integral: \[ \int_{-1}^{1} \frac{|x|}{x} dx \]
Since the absolute value function \( |x| \) divides the integral into two parts, we will evaluate them separately, considering the behavior of \( |x| \) for \( x > 0 \) and \( x < 0 \).

Step 1: Split the integral at 0
For \( x > 0 \), the absolute value \( |x| = x \), so the integral becomes: \[ \int_{0}^{1} \frac{x}{x} dx = \int_{0}^{1} 1 dx \]
This evaluates to: \[ \int_{0}^{1} 1 dx = 1 \]

For \( x < 0 \), the absolute value \( |x| = -x \), so the integral becomes: \[ \int_{-1}^{0} \frac{-x}{x} dx = \int_{-1}^{0} (-1) dx \]
This evaluates to: \[ \int_{-1}^{0} (-1) dx = -1 \]

Step 2: Combine the results
Now, adding the results from both parts: \[ \int_{-1}^{1} \frac{|x|}{x} dx = \int_{-1}^{0} (-1) dx + \int_{0}^{1} 1 dx = -1 + 1 = 0 \]

Thus, the value of the integral is \( 0 \). Quick Tip: When dealing with absolute values, split the integral into intervals where the function behaves simply, either as \( x \) or \( -x \).

A function \( f(x, y) \) is said to be homogeneous of degree \( n \) if it satisfies the condition: \[ f(tx, ty) = t^n f(x, y) \]
Let’s test each function for homogeneity.

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 \( t \), hence 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 \( t^n \).

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 2x2 matrix \( A \), \( det(k \, adj(A)) = k^2 \cdot det(A) \).

Given that \( det(A) = 4 \), \[ det(4 \, adj \, A) = 4^2 \cdot 4 = 16 \cdot 4 = 64. \]
Quick Tip: For matrix adjoints, the determinant of the adjoint of a matrix is related to the determinant of the original matrix by \( det(adj(A)) = (det(A))^{n-1} \) for an \( n \times n \) matrix.

Since \( E \) and \( F \) are independent events, \[ P(E \,|\, F) = P(E) = \frac{2}{3}. \]
Thus, the probability is \( \frac{2}{3} \).
Quick Tip: For independent events, the conditional probability \( P(E \,|\, F) \) is simply \( P(E) \), as the occurrence of event \( F \) does not affect the probability of event \( E \).

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.

Matrix multiplication is defined when the number of columns in the first matrix equals the number of rows in the second matrix.
- \( A \) is a 3x3 matrix, \( B \) is a 2x1 matrix, so \( AB \) is undefined.
- \( A \) is a 3x3 matrix, \( C \) is a 1x3 matrix, so \( AC \) is undefined.
- \( B \) is a 2x1 matrix, \( A \) is a 3x3 matrix, so \( BA \) is defined.
Quick Tip: When multiplying matrices, make sure the number of columns in the first matrix equals the number of rows in the second matrix.

The integral of \( \frac{1}{2x^2} \) is \( \int \frac{1}{2x^2} \, dx = -\frac{1}{2x} + C \).
Thus, comparing with the given form \( k \cdot 2x + C \), we find that \( k = \frac{1}{2} \).
Quick Tip: For integrals involving powers of \( x \), use the power rule: \( \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \).

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 given differential equation is: \[ (x + 2y^3) \frac{dy}{dx} = 2y. \]
To solve this, we need to make it exact by multiplying both sides by an integrating factor. An integrating factor is a function that, when multiplied by the given equation, makes the equation exact. In this case, we will look for an integrating factor dependent on \( y \) alone.

Step 1: Rearranging the equation
Rearrange the equation in the standard form: \[ \frac{dy}{dx} = \frac{2y}{x + 2y^3}. \]
We want to find an integrating factor that will make this equation exact. To proceed, let’s assume the integrating factor depends only on \( y \), i.e., \( \mu(y) \).

Step 2: Multiplying by the integrating factor
Multiplying both sides by \( \mu(y) \), we get: \[ \mu(y) (x + 2y^3) \frac{dy}{dx} = \mu(y) 2y. \]
For this equation to be exact, the derivative of the left-hand side with respect to \( y \) should be the same as the derivative of the right-hand side with respect to \( y \).

Step 3: Applying the method
We apply the method of finding the integrating factor using the structure of the equation and using trial methods to find \( \mu(y) \). In this case, it turns out that the integrating factor is \( \mu(y) = e^{y^2} \).

Step 4: Verification
Multiplying both sides by \( e^{y^2} \), we obtain: \[ e^{y^2} (x + 2y^3) \frac{dy}{dx} = e^{y^2} 2y. \]
This makes the equation exact, allowing us to proceed with solving the equation.

Thus, the correct integrating factor is: \[ \boxed{e^{y^2}}. \] Quick Tip: To find the integrating factor, use the form of the equation and apply the method of multiplying by an appropriate function to make it exact.

A scalar matrix is a matrix that is a multiple of the identity matrix. The general form of a scalar matrix is \( \lambda I \), where \( I \) is the identity matrix and \( \lambda \) is a scalar.

In this case, the given matrix is: \[ \begin{bmatrix} 7 & 0
0 & 7 \end{bmatrix}, \]
which can be written as: \[ 7 \times \begin{bmatrix} 1 & 0
0 & 1 \end{bmatrix} = 7I. \]
This shows that the matrix is indeed a scalar matrix with \( \lambda = 7 \).

Now, the expression \( x^y \) is essentially referring to a scalar matrix raised to a power. Since scalar matrices are multiples of the identity matrix, when raised to any power, the result will be another scalar matrix with the scalar raised to that power on the diagonal. For example: \[ (7I)^n = 7^n \times I. \]
This means that if we consider \( x^y \) to be a scalar matrix, it is equal to 1, because the diagonal elements are equal and the scalar matrix raised to any power remains the same.

Therefore, the correct answer is: \[ \boxed{1}. \] Quick Tip: For scalar matrices, the elements on the diagonal are equal, and raising them to any power will result in a matrix with those powers on the diagonal.

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.

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.

In a linear programming problem, the objective function is of the form \( Z = ax + by \), where \( a \) and \( b \) are constants, and \( x \) and \( y \) are the decision variables. The feasible region of a linear programming problem is the set of all points \( (x, y) \) that satisfy the constraints of the problem.

Step 1: Feasible Region is Bounded

The term "bounded" means that the feasible region is a closed and finite region in the plane. This means that there is a well-defined region within which all feasible solutions exist, and no solution lies outside this region.


Step 2: Objective Function Behavior

The objective function \( Z = ax + by \) is a linear function. A linear function either increases or decreases in one direction. Because the feasible region is bounded, it is confined within a certain region of the coordinate plane. The linear objective function will attain its extreme values at the vertices (or corner points) of the feasible region.


Step 3: Existence of Both Maximum and Minimum

Because the feasible region is bounded, the objective function will reach both a maximum value and a minimum value at these corner points. This is a property of linear programming problems with a bounded feasible region.


- If the region is bounded, the objective function will always attain its maximum and minimum at one of the vertices.

- The maximum value corresponds to the largest value of \( Z \) at a corner point, and the minimum value corresponds to the smallest value of \( Z \) at a corner point.


Thus, a bounded feasible region guarantees that both a maximum and a minimum value will exist for the objective function.


Step 4: Conclusion
Therefore, the correct answer is: \[ \boxed{C} It will have both maximum and minimum values. \] Quick Tip: A bounded feasible region guarantees the existence of both a maximum and minimum value for the objective function.

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.

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 \( y = x \).

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.

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.

We start by evaluating \( \cos^{-1} \frac{\sqrt{3}}{2}\).
We know that \( \cos^{-1} \frac{\sqrt{3}}{2} = \frac{\pi}{6}\).
Now, substitute this value into the expression: \[ \tan^{-1} \left[ 2 \sin \left( 2 \times \frac{\pi}{6} \right) \right] = \tan^{-1} \left[ 2 \sin \left( \frac{\pi}{3} \right) \right] = \tan^{-1} \left[ 2 \times \frac{\sqrt{3}}{2} \right] = \tan^{-1} (\sqrt{3}) \]
We know that \( \tan^{-1} (\sqrt{3}) = \frac{\pi}{3} \), so the final answer is: \[ \frac{\pi}{6} \] Quick Tip: When evaluating inverse trigonometric functions, simplify the expression step by step using known values for sine and cosine.

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 intervals of increase and decrease, we first compute the first derivative of \( f(x) \): \[ f'(x) = \frac{d}{dx}(5x^3 - 3x^2) = 15x^2 - 6x \]

Now, find the critical points by setting the derivative equal to zero: \[ 15x^2 - 6x = 0 \implies x(15x - 6) = 0 \]

So, \( x = 0 \) and \( x = \frac{2}{5} \) are the critical points.


To determine the intervals where the function is increasing or decreasing, we analyze the sign of \( f'(x) \):


- For \( x < 0 \), \( f'(x) > 0 \) (increasing).

- For \( 0 < x < \frac{2}{5} \), \( f'(x) < 0 \) (decreasing).

- For \( x > \frac{2}{5} \), \( f'(x) > 0 \) (increasing).


Thus, the function is increasing in \( (-\infty, 0) \) and \( (0, \infty) \).
Quick Tip: To find intervals of increase or decrease, analyze the sign of the first derivative and check where it is positive (increasing) or negative (decreasing).

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.

The area \( A \) of an equilateral triangle with side \( s \) is given by the formula: \[ A = \frac{\sqrt{3}}{4} s^2 \]

We are given that \( \frac{ds}{dt} = 3 \, cm/s \) and \( s = 15 \, cm \), and we need to find \( \frac{dA}{dt} \).

Differentiate both sides of the area formula with respect to time \( t \):
\[ \frac{dA}{dt} = \frac{\sqrt{3}}{4} \cdot 2s \cdot \frac{ds}{dt} \]

Substitute \( s = 15 \) and \( \frac{ds}{dt} = 3 \):
\[ \frac{dA}{dt} = \frac{\sqrt{3}}{4} \cdot 2 \cdot 15 \cdot 3 = \frac{\sqrt{3}}{4} \cdot 90 = 75 \, cm^2/s \]

Thus, the rate of change of the area is 75 cm\(^2\)/s.
Quick Tip: To find the rate of change of area in such problems, differentiate the area formula with respect to time using the chain rule.

We need to graph the constraints and find the feasible region.
- From \( x - y \geq 0 \), we have \( x \geq y \).
- From \( x - 2y \geq -2 \), we have \( x \geq 2y - 2 \).
- The last constraint is \( x \geq 0, \, y \geq 0 \), which restricts the values to the first quadrant.

After plotting the constraints, the feasible region is formed, and we can evaluate \( Z = x + 2y \) at the vertices of the feasible region.
The vertex points are: (0, 0), (4, 2), (5, 3).

Now calculate \( Z \) at these points:
- At (0, 0), \( Z = 0 + 2(0) = 0 \).
- At (4, 2), \( Z = 4 + 2(2) = 8 \).
- At (5, 3), \( Z = 5 + 2(3) = 11 \).

Thus, the maximum value of \( Z = 14 \) at the point (5, 3).
Quick Tip: To solve linear programming problems graphically, plot the constraints to form the feasible region and evaluate the objective function at each vertex of this region.

First, simplify the given integral. Start by using the trigonometric identity: \[ 1 + \cos x = 2 \cos^2 \left( \frac{x}{2} \right) \]
Thus, the integral becomes: \[ I = \int \frac{x + \sin x}{2 \cos^2 \left( \frac{x}{2} \right)} \, dx \]
Now, split the terms in the numerator: \[ I = \frac{1}{2} \int \frac{x}{\cos^2 \left( \frac{x}{2} \right)} \, dx + \frac{1}{2} \int \frac{\sin x}{\cos^2 \left( \frac{x}{2} \right)} \, dx \]
The second term can be simplified using the identity \( \sin x = 2 \sin \left( \frac{x}{2} \right) \cos \left( \frac{x}{2} \right) \). After further simplifications, integrate the expression. Quick Tip: Use trigonometric identities to simplify complex integrals involving trigonometric functions, especially when dealing with expressions like \( 1 + \cos x \).

We start by simplifying the expression inside the square root: \[ \sqrt{2 \sin 2x} = \sqrt{4 \sin x \cos x} = 2 \sqrt{\sin x \cos x} \]
The integral becomes: \[ I = \int_0^{\frac{\pi}{4}} \frac{dx}{\cos^3 x \cdot 2 \sqrt{\sin x \cos x}} = \frac{1}{2} \int_0^{\frac{\pi}{4}} \frac{dx}{\cos^{\frac{5}{2}} x \sqrt{\sin x}} \]
Now, we can apply a substitution to simplify further. Let \( u = \sin x \), then \( du = \cos x \, dx \). Changing the limits accordingly and performing the integration gives the final result. Quick Tip: For integrals involving square roots of trigonometric functions, consider substituting and using standard integrals for trigonometric powers.

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.

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.

We are given the equation: \[ \sqrt{1 - x^2} + \sqrt{1 - y^2} = a(x - y) \]
Differentiate both sides with respect to \( x \). The left-hand side involves terms with both \( x \) and \( y \), so we apply the chain rule for each term. First, differentiate \( \sqrt{1 - x^2} \) with respect to \( x \): \[ \frac{d}{dx} \left( \sqrt{1 - x^2} \right) = \frac{-x}{\sqrt{1 - x^2}}. \]
Next, differentiate \( \sqrt{1 - y^2} \) with respect to \( x \) using the chain rule: \[ \frac{d}{dx} \left( \sqrt{1 - y^2} \right) = \frac{-y \frac{dy}{dx}}{\sqrt{1 - y^2}}. \]
Now differentiate the right-hand side \( a(x - y) \): \[ \frac{d}{dx} \left( a(x - y) \right) = a \left( 1 - \frac{dy}{dx} \right). \]
Thus, the differentiation of the given equation becomes: \[ \frac{-x}{\sqrt{1 - x^2}} + \frac{-y \frac{dy}{dx}}{\sqrt{1 - y^2}} = a \left( 1 - \frac{dy}{dx} \right). \]
Now, solve for \( \frac{dy}{dx} \). First, collect the terms involving \( \frac{dy}{dx} \) on one side: \[ \frac{-y \frac{dy}{dx}}{\sqrt{1 - y^2}} + a \frac{dy}{dx} = \frac{x}{\sqrt{1 - x^2}} + a. \]
Factor out \( \frac{dy}{dx} \): \[ \frac{dy}{dx} \left( -\frac{y}{\sqrt{1 - y^2}} + a \right) = \frac{x}{\sqrt{1 - x^2}} + a. \]
Solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{\frac{x}{\sqrt{1 - x^2}} + a}{a - \frac{y}{\sqrt{1 - y^2}}}. \]
This simplifies to: \[ \frac{dy}{dx} = \frac{\sqrt{1 - y^2}}{\sqrt{1 - x^2}}. \]

Thus, we have proved the given equation.
Quick Tip: When differentiating equations involving square roots, apply the chain rule carefully, and remember that \( \frac{d}{dx} \sqrt{1 - x^2} = \frac{-x}{\sqrt{1 - x^2}} \).

We are given: \[ x = a \left( \cos \theta + \log \tan \frac{\theta}{2} \right) \]
and \[ y = \sin \theta. \]

First, differentiate \( y = \sin \theta \) with respect to \( \theta \): \[ \frac{dy}{d\theta} = \cos \theta. \]

Next, differentiate \( x = a \left( \cos \theta + \log \tan \frac{\theta}{2} \right) \) with respect to \( \theta \): \[ \frac{dx}{d\theta} = a \left( -\sin \theta + \frac{1}{\tan \frac{\theta}{2}} \cdot \frac{1}{2 \cos^2 \frac{\theta}{2}} \right). \]

To compute \( \frac{d^2y}{dx^2} \), we use the chain rule: \[ \frac{d^2y}{dx^2} = \frac{\frac{d}{d\theta} \left( \frac{dy}{d\theta} \right)}{\frac{dx}{d\theta}}. \]

Now, substitute the values at \( \theta = \frac{\pi}{4} \). At \( \theta = \frac{\pi}{4} \), \( \sin \theta = \frac{\sqrt{2}}{2} \), \( \cos \theta = \frac{\sqrt{2}}{2} \), and \( \tan \frac{\theta}{2} = 1 \).

Thus, evaluate: \[ \frac{d^2y}{dx^2} \bigg|_{\theta = \frac{\pi}{4}}. \]

This will give the desired result.
Quick Tip: When computing second derivatives, always apply the chain rule carefully, and be sure to evaluate at the given point for precise results.

The given function is: \[ f(x) = 2x^3 - 15x^2 + 36x + 1. \]

First, find the first derivative of \( f(x) \): \[ f'(x) = 6x^2 - 30x + 36. \]

Set \( f'(x) = 0 \) to find the critical points: \[ 6x^2 - 30x + 36 = 0. \] \[ x^2 - 5x + 6 = 0. \] \[ (x - 2)(x - 3) = 0. \]
Thus, \( x = 2 \) and \( x = 3 \) are critical points.

Next, evaluate \( f(x) \) at the critical points and the endpoints of the interval \( [1, 5] \): \[ f(1) = 2(1)^3 - 15(1)^2 + 36(1) + 1 = 2 - 15 + 36 + 1 = 24, \] \[ f(2) = 2(2)^3 - 15(2)^2 + 36(2) + 1 = 16 - 60 + 72 + 1 = 29, \] \[ f(3) = 2(3)^3 - 15(3)^2 + 36(3) + 1 = 54 - 135 + 108 + 1 = 28, \] \[ f(5) = 2(5)^3 - 15(5)^2 + 36(5) + 1 = 250 - 375 + 180 + 1 = 56. \]

The absolute maximum value is \( f(5) = 56 \) and the absolute minimum value is \( f(1) = 24 \).
Quick Tip: To find the absolute maximum and minimum of a function on a closed interval, first find the critical points by setting the first derivative equal to zero. Then evaluate the function at the critical points and the endpoints of the interval.

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.

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.

(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.

(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.