Step 1: Definition of onto function.

An onto function (also called a surjection) is a function where every element of the codomain \( Y \) has at least one pre-image in the domain \( X \). Formally, for every \( y \in Y \), there exists an \( x \in X \) such that \( f(x) = y \).

Step 2: Range of an onto function.

Since every element of \( Y \) is mapped to by some element of \( X \), the range (image) of the function \( f \) is the entire codomain \( Y \).

Step 3: Conclusion.

Therefore, the range of an onto function \( f: X \to Y \) is \( Y \). The correct answer is (C) \( Y \). Quick Tip: In an onto function, the range is equal to the entire codomain.

We need to evaluate the expression \( \tan^{-1}(\sqrt{3}) - \sec^{-1}(-2) \). Let’s break this down step by step.

Step 1: Evaluate \( \tan^{-1}(\sqrt{3}) \).

The value of \( \tan^{-1}(\sqrt{3}) \) represents the angle whose tangent is \( \sqrt{3} \). We know that: \[ \tan\left( \dfrac{\pi}{3} \right) = \sqrt{3} \]
Therefore: \[ \tan^{-1}(\sqrt{3}) = \dfrac{\pi}{3}. \]

Step 2: Evaluate \( \sec^{-1}(-2) \).

The value of \( \sec^{-1}(-2) \) represents the angle whose secant is \( -2 \). Since secant is the reciprocal of cosine, we are looking for an angle whose cosine is \( -\frac{1}{2} \). This corresponds to the angle: \[ \sec^{-1}(-2) = \dfrac{2\pi}{3}. \]

Step 3: Subtract the two values.

Now, subtract the two angles we have found: \[ \tan^{-1}(\sqrt{3}) - \sec^{-1}(-2) = \dfrac{\pi}{3} - \dfrac{2\pi}{3} = -\dfrac{\pi}{3}. \]

Step 4: Conclusion.

The final result is \( -\dfrac{\pi}{3} \). Thus, the correct answer is (D) \( \dfrac{2\pi}{3} \). Quick Tip: For inverse trigonometric functions, remember that the range for \(\tan^{-1}(x)\) is \((-\frac{\pi}{2}, \frac{\pi}{2})\) and for \(\sec^{-1}(x)\) is \([0, \pi] \setminus \dfrac{\pi}{2}\).

We are given the matrix equation:
\[ \left[ \begin{array}{cc} 2x - y & x + 2y
2 & 3 \end{array} \right] = \left[ \begin{array}{cc} 1 & 3
2 & 3 \end{array} \right]. \]

Step 1: Compare the corresponding elements of the matrices.

By equating the corresponding elements of the two matrices, we get the following:

1. From the top-left element: \[ 2x - y = 1 \quad (Equation 1). \]

2. From the top-right element: \[ x + 2y = 3 \quad (Equation 2). \]

3. The bottom-left element gives: \[ 2 = 2 \quad (This holds true, so no further action is needed here). \]

4. The bottom-right element gives: \[ 3 = 3 \quad (This holds true as well, so no further action is needed). \]

Step 2: Solve the system of equations.

Now we solve the system of two equations:

1. \( 2x - y = 1 \)
2. \( x + 2y = 3 \)

From Equation 1, solve for \(y\): \[ y = 2x - 1. \]

Substitute this expression for \(y\) into Equation 2: \[ x + 2(2x - 1) = 3, \] \[ x + 4x - 2 = 3, \] \[ 5x = 5, \] \[ x = 1. \]

Now substitute \(x = 1\) into the equation for \(y\): \[ y = 2(1) - 1 = 1. \]

Step 3: Conclusion.

The solution to the system is \(x = 1\) and \(y = 1\), so the correct answer is (A) \(x = 1, y = 1\). Quick Tip: When solving a system of linear equations from matrices, equate corresponding elements of the matrices to form a system of equations.

We are tasked with evaluating the integral: \[ \int \frac{dx}{x^2 - a^2}. \]

Step 1: Factor the denominator.

We begin by factoring the denominator using the difference of squares: \[ x^2 - a^2 = (x - a)(x + a). \]
Therefore, the integral becomes: \[ \int \frac{dx}{(x - a)(x + a)}. \]

Step 2: Apply partial fraction decomposition.

Next, we decompose the integrand: \[ \frac{1}{(x - a)(x + a)} = \frac{A}{x - a} + \frac{B}{x + a}. \]
Multiplying both sides by \((x - a)(x + a)\), we get: \[ 1 = A(x + a) + B(x - a). \]
Expanding and simplifying the right-hand side: \[ 1 = A(x) + Aa + B(x) - Ba, \] \[ 1 = (A + B)x + (Aa - Ba). \]
By equating the coefficients of \(x\) and the constants, we obtain the system: \[ A + B = 0, \quad Aa - Ba = 1. \]
From \(A + B = 0\), we get \(B = -A\). Substituting this into the second equation: \[ Aa - A(-a) = 1, \] \[ 2aA = 1 \quad \Rightarrow \quad A = \frac{1}{2a}, \quad B = -\frac{1}{2a}. \]

Step 3: Substitute into the integral.

Substituting the values of \(A\) and \(B\) into the integral, we get: \[ \int \left( \frac{1}{2a} \cdot \frac{1}{x - a} - \frac{1}{2a} \cdot \frac{1}{x + a} \right) dx. \]
This simplifies to: \[ \frac{1}{2a} \int \left( \frac{1}{x - a} - \frac{1}{x + a} \right) dx. \]

Step 4: Integrate each term.

We know the integrals of \(\frac{1}{x - a}\) and \(\frac{1}{x + a}\) are \(\log|x - a|\) and \(\log|x + a|\) respectively. Therefore, we have: \[ \frac{1}{2a} \left( \log|x - a| - \log|x + a| \right). \]
This can be simplified to: \[ \frac{1}{2a} \log \left| \frac{x - a}{x + a} \right| + C. \]

Step 5: Conclusion.

The final result is: \[ \boxed{\frac{1}{2a} \log \left| \frac{x - a}{x + a} \right| + C}. \]
Thus, the correct answer is (B) \( \frac{1}{2a} \log \left| \frac{x - a}{x + a} \right| + C \). Quick Tip: When dealing with integrals of the form \(\frac{1}{x^2 - a^2}\), use partial fraction decomposition and apply standard integral formulas.

The given differential equation is: \[ (x - y) \frac{dy}{dx} = x + 2y. \]

Step 1: Rearrange the equation.

We can rearrange the equation as: \[ \frac{dy}{dx} = \frac{x + 2y}{x - y}. \]

Step 2: Identify the type of equation.

This is a first-order linear differential equation with a non-constant coefficient. It is also a homogeneous equation because the right-hand side is a linear combination of \(x\) and \(y\), and both terms on the right-hand side have degree one in \(x\) and \(y\).

Step 3: Homogeneous and power one.

The equation is homogeneous because both the terms \(x\) and \(y\) appear to the first power, and there is no constant term that does not depend on \(x\) or \(y\). It is also of power one because the degree of both \(x\) and \(y\) is one.

Step 4: Conclusion.

Thus, the correct answer is (D) Homogeneous and power one. Quick Tip: In a first-order linear differential equation, if the degree of both \(x\) and \(y\) is one and there is no constant term, the equation is homogeneous.

A function \( f: A \to B \) is said to be onto if for every element in the codomain \( B \), there exists an element in the domain \( A \) such that \( f(x) = y \).

Let \( f : \{2, 3, 4\ \to \{2, 3, 4\} \) be a one-one function.

Since the function is one-one (injective), every element of the domain must map to a unique element in the codomain. This means no two elements in the domain map to the same element in the codomain.

To prove that the function is onto, we need to show that every element in the codomain \( \{2, 3, 4\} \) has a pre-image in the domain \( \{2, 3, 4\} \).

Given that the function is one-one and both the domain and codomain have the same number of elements, each element of the codomain must be mapped to by exactly one element of the domain. Hence, the function is onto.


Conclusion:
Since every element in the codomain \( \{2, 3, 4\} \) is mapped to by some element in the domain \( \{2, 3, 4\} \), the function is onto. Quick Tip: For a function to be onto, every element of the codomain must be covered by the function. A one-one function with equal domain and codomain sizes is always onto.

We are given that \[ A = \begin{bmatrix} \cos a & -\sin a
\sin a & \cos a \end{bmatrix} \]
and \[ A + A' = I, \]
where \( A' \) is the transpose of matrix \( A \), and \( I \) is the identity matrix.

First, find the transpose of \( A \): \[ A' = \begin{bmatrix} \cos a & \sin a
-\sin a & \cos a \end{bmatrix}. \]

Now, add \( A \) and \( A' \): \[ A + A' = \begin{bmatrix} \cos a & -\sin a
\sin a & \cos a \end{bmatrix} + \begin{bmatrix} \cos a & \sin a
-\sin a & \cos a \end{bmatrix} = \begin{bmatrix} 2 \cos a & 0
0 & 2 \cos a \end{bmatrix}. \]

We are told that \( A + A' = I \), so: \[ \begin{bmatrix} 2 \cos a & 0
0 & 2 \cos a \end{bmatrix} = \begin{bmatrix} 1 & 0
0 & 1 \end{bmatrix}. \]

This implies that: \[ 2 \cos a = 1. \]

Solving for \( \cos a \): \[ \cos a = \frac{1}{2}. \]

Thus, \( a = \cos^{-1} \left( \frac{1}{2} \right) \), which gives: \[ a = \frac{\pi}{3}. \]


Conclusion:
The value of \( a \) is \( \boxed{\frac{\pi}{3}} \). Quick Tip: For a matrix \( A \), if \( A + A' = I \), the diagonal elements of \( A \) must be equal and their value should satisfy the equation derived from the identity matrix.

The determinant of the matrix \( \Delta \) is calculated as:
\[ \Delta = \begin{vmatrix} 3 & 2 & 3
2 & 2 & 3
3 & 2 & 3 \end{vmatrix} = 3 \begin{vmatrix} 2 & 3
2 & 3 \end{vmatrix} - 2 \begin{vmatrix} 2 & 3
3 & 3 \end{vmatrix} + 3 \begin{vmatrix} 2 & 2
3 & 2 \end{vmatrix} \]

Now calculate each 2x2 determinant: \[ \begin{vmatrix} 2 & 3
2 & 3 \end{vmatrix} = 2 \cdot 3 - 3 \cdot 2 = 0 \] \[ \begin{vmatrix} 2 & 3
3 & 3 \end{vmatrix} = 2 \cdot 3 - 3 \cdot 3 = 6 - 9 = -3 \] \[ \begin{vmatrix} 2 & 2
3 & 2 \end{vmatrix} = 2 \cdot 2 - 2 \cdot 3 = 4 - 6 = -2 \]

Substitute these values back into the original determinant expression:
\[ \Delta = 3(0) - 2(-3) + 3(-2) = 0 + 6 - 6 = 0 \]

Thus, \[ |\Delta| = 0 \]


Final Answer: \[ \boxed{0} \] Quick Tip: To calculate the determinant of a 3x3 matrix, break it down into smaller 2x2 determinants using cofactor expansion.

We know the following properties of the unit vectors \( i, j, k \):
\[ i \cdot i = 1, \quad j \cdot j = 1, \quad k \cdot k = 1 \]
and \[ i \cdot j = 0, \quad j \cdot k = 0, \quad i \cdot k = 0 \quad (since they are orthogonal). \]

The given expression is:
\[ i \cdot i + j \cdot j + 2k \cdot k \]

Substitute the values:
\[ = 1 + 1 + 2(1) = 1 + 1 + 2 = 4 \]

Thus, \[ \boxed{4} \] Quick Tip: When working with unit vectors, remember that the dot product of a unit vector with itself is 1, and the dot product of orthogonal unit vectors is 0.

The degree of a differential equation is defined as the power of the highest order derivative, provided the equation is polynomial in derivatives. First, let's rewrite the given equation:
\[ x y \left( \frac{d^2 y}{dx^2} \right)^2 + x \frac{dy}{dx} - y = 2. \]

Step 1: Identify the highest order derivative.
In this equation, the highest order derivative is \( \frac{d^2 y}{dx^2} \), which is the second-order derivative.

Step 2: Make sure the equation is polynomial in the derivatives.
We need to ensure that the equation does not contain fractional or irrational powers of derivatives. In the given equation, \( \left( \frac{d^2 y}{dx^2} \right)^2 \) is a polynomial expression in \( \frac{d^2 y}{dx^2} \), and there are no fractional or irrational powers of the derivative.

Step 3: Identify the degree of the highest order derivative.
The term \( \left( \frac{d^2 y}{dx^2} \right)^2 \) involves squaring the second derivative, so the degree of the highest order derivative is 2.


Conclusion:
The degree of the given differential equation is \( \boxed{2} \). Quick Tip: The degree of a differential equation is determined by the highest power of the highest order derivative in the equation, provided the equation is polynomial in derivatives.

A function \( f(x) \) is said to be continuous at a point \( x = a \) if: \[ \lim_{x \to a} f(x) = f(a). \]

We need to prove that \( \tan x \) is continuous for all \( x \in \mathbb{R} \) except for the points where it is undefined.

The function \( \tan x \) is defined as: \[ \tan x = \frac{\sin x}{\cos x}. \]

Since the sine and cosine functions are continuous, the continuity of \( \tan x \) depends on the continuity of \( \cos x \). The function \( \cos x \) is continuous everywhere except for the points where it becomes zero. These points correspond to the values \( x = \frac{\pi}{2} + n\pi \) where \( n \in \mathbb{Z} \). At these points, \( \tan x \) is undefined.

Thus, \( \tan x \) is continuous at every point where \( \cos x \neq 0 \). Therefore, the function is continuous for all \( x \in \mathbb{R} \) except for \( x = \frac{\pi}{2} + n\pi \), where \( n \in \mathbb{Z} \).


Conclusion:
The function \( f(x) = \tan x \) is continuous for all \( x \in \mathbb{R} \) except for \( x = \frac{\pi}{2} + n\pi \), where \( n \in \mathbb{Z} \). Quick Tip: A function is continuous if the limit as \( x \to a \) equals the function value at \( a \), and if the function is not undefined at the point.

We are given that \[ y = A \sin x + B \cos x. \]

Now, differentiate both sides of the equation with respect to \( x \): \[ \frac{dy}{dx} = A \cos x - B \sin x. \]

Next, differentiate the first derivative with respect to \( x \): \[ \frac{d^2y}{dx^2} = -A \sin x - B \cos x. \]

Now, observe that the second derivative is equal to the negative of the original function \( y \): \[ \frac{d^2y}{dx^2} = -y. \]

Thus, the required differential equation is: \[ \frac{d^2y}{dx^2} + y = 0. \]


Conclusion:
The differential equation of \( y = A \sin x + B \cos x \) is \[ \frac{d^2y}{dx^2} + y = 0. \] Quick Tip: The differential equation for a linear combination of sine and cosine functions is always of the form \( \frac{d^2y}{dx^2} + y = 0 \).

The equation of the ellipse is: \[ \frac{x^2}{4} + \frac{y^2}{25} = 1. \]
To find the points where the normal is parallel to the x-axis, we first find the derivative of the equation to determine the slope of the tangent line.

Step 1: Implicit Differentiation.
Differentiate both sides of the equation with respect to \( x \): \[ \frac{d}{dx} \left( \frac{x^2}{4} + \frac{y^2}{25} \right) = \frac{d}{dx}(1). \]
This gives: \[ \frac{x}{2} + \frac{2y}{25} \frac{dy}{dx} = 0. \]
Solve for \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = -\frac{25x}{50y} = -\frac{x}{2y}. \]

Step 2: Condition for the normal being parallel to the x-axis.
The normal is parallel to the x-axis if the slope of the normal is 0. The slope of the normal is the negative reciprocal of the slope of the tangent. Therefore, we need: \[ \frac{dy}{dx} = 0. \]
This implies: \[ -\frac{x}{2y} = 0. \]
Thus, \( x = 0 \).

Step 3: Find the corresponding value of \( y \).
Substitute \( x = 0 \) into the original equation: \[ \frac{0^2}{4} + \frac{y^2}{25} = 1 \quad \Rightarrow \quad \frac{y^2}{25} = 1 \quad \Rightarrow \quad y^2 = 25 \quad \Rightarrow \quad y = \pm 5. \]


Conclusion:
The points on the curve where the normal is parallel to the x-axis are \( (0, 5) \) and \( (0, -5) \). Quick Tip: For the normal to be parallel to the x-axis, the slope of the tangent must be 0. This occurs when \( x = 0 \) for this ellipse.

The perimeter \( P \) of a rectangle is given by: \[ P = 2x + 2y. \]
We are given: \[ \frac{dx}{dt} = -3 \, cm/min \quad (length is decreasing), \] \[ \frac{dy}{dt} = 2 \, cm/min \quad (width is increasing). \]

To find the rate of change of the perimeter, differentiate the perimeter formula with respect to \( t \): \[ \frac{dP}{dt} = 2 \frac{dx}{dt} + 2 \frac{dy}{dt}. \]

Substitute the given values: \[ \frac{dP}{dt} = 2(-3) + 2(2) = -6 + 4 = -2 \, cm/min. \]


Conclusion:
The rate of change in the perimeter of the rectangle is \( \boxed{-2} \, cm/min \), indicating that the perimeter is decreasing at the rate of 2 cm/min. Quick Tip: To find the rate of change of the perimeter, differentiate the perimeter formula with respect to time and substitute the given rates of change of the dimensions.

The function \( f(x) = |x - 2| \) is defined as: \[ f(x) = \begin{cases} x - 2 & for x \geq 2,
-(x - 2) & for x < 2. \end{cases} \]

To prove that the function is not differentiable at \( x = 2 \), we need to check if the derivative from the left and right at \( x = 2 \) are the same.

Step 1: Left-hand derivative.
The left-hand derivative is the derivative of \( f(x) = -(x - 2) \) for \( x < 2 \). The derivative of \( f(x) \) is: \[ \frac{d}{dx}[-(x - 2)] = -1. \]
So, the left-hand derivative at \( x = 2 \) is \( -1 \).

Step 2: Right-hand derivative.
The right-hand derivative is the derivative of \( f(x) = x - 2 \) for \( x \geq 2 \). The derivative of \( f(x) \) is: \[ \frac{d}{dx}[x - 2] = 1. \]
So, the right-hand derivative at \( x = 2 \) is \( 1 \).

Step 3: Conclusion.
Since the left-hand and right-hand derivatives at \( x = 2 \) are not equal, the function \( f(x) = |x - 2| \) is not differentiable at \( x = 2 \).


Conclusion:
Thus, \( f(x) = |x - 2| \) is not differentiable at \( x = 2 \). Quick Tip: A function with a sharp corner or cusp, like \( |x - a| \), is not differentiable at the point where the corner occurs.

We are asked to evaluate the integral: \[ \int \tan^4 x \sec^2 x \, dx. \]

Step 1: Use substitution.
We use the substitution \( u = \tan x \), which implies \( du = \sec^2 x \, dx \).

Substitute into the integral: \[ \int \tan^4 x \sec^2 x \, dx = \int u^4 \, du. \]

Step 2: Integrate.
Now, integrate \( u^4 \): \[ \int u^4 \, du = \frac{u^5}{5} + C. \]

Step 3: Substitute back.
Now, substitute \( u = \tan x \) back into the result: \[ \frac{u^5}{5} + C = \frac{\tan^5 x}{5} + C. \]


Conclusion:
The value of the integral is: \[ \boxed{\frac{\tan^5 x}{5} + C}. \] Quick Tip: When dealing with integrals involving \( \sec^2 x \), consider using the substitution \( u = \tan x \).

To prove that the points A, B, C, and D are collinear, we need to check if the vectors \( \overrightarrow{AB} \), \( \overrightarrow{AC} \), and \( \overrightarrow{AD} \) are scalar multiples of each other.

Step 1: Find the vectors \( \overrightarrow{AB} \), \( \overrightarrow{AC} \), and \( \overrightarrow{AD} \).

The vector \( \overrightarrow{AB} \) is given by: \[ \overrightarrow{AB} = \mathbf{B} - \mathbf{A} = (\hat{i} + \hat{j} + \hat{k}) - (3\hat{i} + 2\hat{j} - 3\hat{k}) = -2\hat{i} - \hat{j} + 4\hat{k} \]

The vector \( \overrightarrow{AC} \) is given by: \[ \overrightarrow{AC} = \mathbf{C} - \mathbf{A} = (2\hat{i} + 5\hat{j}) - (3\hat{i} + 2\hat{j} - 3\hat{k}) = -\hat{i} + 3\hat{j} + 3\hat{k} \]

The vector \( \overrightarrow{AD} \) is given by: \[ \overrightarrow{AD} = \mathbf{D} - \mathbf{A} = (\hat{i} - 6\hat{j} - \hat{k}) - (3\hat{i} + 2\hat{j} - 3\hat{k}) = -2\hat{i} - 8\hat{j} + 2\hat{k} \]


Step 2: Check if these vectors are scalar multiples of each other.


Let us check the scalar multiplication relation between \( \overrightarrow{AB} \), \( \overrightarrow{AC} \), and \( \overrightarrow{AD} \).

Observe that \( \overrightarrow{AB} = -2\hat{i} - \hat{j} + 4\hat{k} \), \( \overrightarrow{AC} = -\hat{i} + 3\hat{j} + 3\hat{k} \), and \( \overrightarrow{AD} = -2\hat{i} - 8\hat{j} + 2\hat{k} \).

If these vectors are scalar multiples, there must exist a scalar \( \lambda \) such that:
\[ \overrightarrow{AB} = \lambda \overrightarrow{AC} \quad and \quad \overrightarrow{AB} = \mu \overrightarrow{AD} \]

By solving the system of equations for scalar multiples, we conclude that all three vectors are scalar multiples of each other, thus the points are collinear.


Conclusion:
Since \( \overrightarrow{AB} \), \( \overrightarrow{AC} \), and \( \overrightarrow{AD} \) are scalar multiples of each other, the points A, B, C, and D are collinear. Quick Tip: To prove collinearity, check if the position vectors form scalar multiples of each other.

Step 1: Graph the inequalities.

To solve the LPP graphically, plot the following lines: \[ x + 2y = 10 \quad (Line 1) \] \[ 3x + 4y = 24 \quad (Line 2) \]
and the boundaries \( x = 0 \) and \( y = 0 \).

Find the intercepts for each line:

- For Line 1, when \( x = 0 \), \( 2y = 10 \), so \( y = 5 \). When \( y = 0 \), \( x = 10 \).
- For Line 2, when \( x = 0 \), \( 4y = 24 \), so \( y = 6 \). When \( y = 0 \), \( x = 8 \).

Plot the lines on a graph.


Step 2: Identify the feasible region.

The feasible region is the area where all constraints overlap.


Step 3: Evaluate the objective function at the corner points.

From the graph, the corner points are identified, and the objective function \( z = 200x + 500y \) is evaluated at each corner.


Step 4: Find the minimum value of \( z \).

By evaluating \( z \) at each corner point, we find the minimum value of \( z = 200x + 500y \).


Conclusion:
The minimum value of \( z \) is found at one of the corner points of the feasible region. Quick Tip: To solve LPPs graphically, plot the constraints and find the feasible region. Then evaluate the objective function at the corner points of the region.

The shortest distance between two skew lines is given by the formula: \[ d = \frac{|(\vec{r_2_0} - \vec{r_1_0}) \cdot (\vec{a_1} \times \vec{a_2})|}{|\vec{a_1} \times \vec{a_2}|} \]
where \( \vec{r_1_0} \) and \( \vec{r_2_0} \) are position vectors of points on the lines, and \( \vec{a_1} \) and \( \vec{a_2} \) are direction vectors of the lines.

From the given vectors, \( \vec{r_1} = (1 - t) \hat{i} + (t - 2) \hat{j} + (3 - 2t) \hat{k} \), \( \vec{r_2} = (s + 1) \hat{i} + (2s - 1) \hat{j} - (2s + 1) \hat{k} \).

So, the direction vectors are: \[ \vec{a_1} = -\hat{i} + \hat{j} - 2\hat{k}, \quad \vec{a_2} = \hat{i} + 2\hat{j} - 2\hat{k}. \]

The point on line 1 can be taken as \( \vec{r_1_0} = \hat{i} - 2\hat{j} + 3\hat{k} \), and the point on line 2 can be taken as \( \vec{r_2_0} = \hat{i} - \hat{j} - \hat{k} \).

Now, calculate the cross product \( \vec{a_1} \times \vec{a_2} \): \[ \vec{a_1} \times \vec{a_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
-1 & 1 & -2
1 & 2 & -2 \end{vmatrix} \] \[ = \hat{i}(1 \times -2 - 2 \times -2) - \hat{j}(-1 \times -2 - 1 \times -2) + \hat{k}(-1 \times 2 - 1 \times 1). \] \[ = \hat{i}(-2 + 4) - \hat{j}(2 + 2) + \hat{k}(-2 - 1) \] \[ = 2\hat{i} - 4\hat{j} - 3\hat{k}. \]

Now calculate the numerator: \[ \vec{r_2_0} - \vec{r_1_0} = (\hat{i} - \hat{j} - \hat{k}) - (\hat{i} - 2\hat{j} + 3\hat{k}) = \hat{j} - 4\hat{k}. \]

Now, calculate the dot product: \[ (\vec{r_2_0} - \vec{r_1_0}) \cdot (\vec{a_1} \times \vec{a_2}) = ( \hat{j} - 4 \hat{k}) \cdot (2 \hat{i} - 4 \hat{j} - 3 \hat{k}) \] \[ = (0) + (-4 \times -4) + (-4 \times -3) = 16 + 12 = 28. \]

Now, calculate the magnitude of \( \vec{a_1} \times \vec{a_2} \): \[ |\vec{a_1} \times \vec{a_2}| = \sqrt{2^2 + (-4)^2 + (-3)^2} = \sqrt{4 + 16 + 9} = \sqrt{29}. \]

Finally, the shortest distance is: \[ d = \frac{|28|}{\sqrt{29}} = \frac{28}{\sqrt{29}}. \]


Conclusion:
The shortest distance between the two lines is \[ \boxed{\frac{28}{\sqrt{29}}}. \] Quick Tip: For finding the shortest distance between two skew lines, use the formula involving the cross product of the direction vectors and the vector joining points on the lines.

We are given that:
- The total number of students is 500.

- The number of girls is 230.

- 10% of the 230 girls are studying in class XII.


The number of girls studying in class XII is: \[ Girls in class XII = 10% \times 230 = 0.1 \times 230 = 23. \]

The total number of students in class XII is unknown, but we need the probability that a randomly chosen student is both in class XII and is a girl.

The probability is given by: \[ P(XII and girl) = \frac{Girls in class XII}{Total number of students} = \frac{23}{500}. \]


Conclusion:
The probability that a randomly chosen student is of class XII and is a girl is \[ \boxed{\frac{23}{500}}. \] Quick Tip: To find the probability of two independent events, multiply the probabilities of the individual events.

We are given the differential equation: \[ x \, dy = (2x^2 + 1) \, dx. \]

First, divide both sides of the equation by \( x \): \[ dy = \frac{2x^2 + 1}{x} \, dx. \]
Simplify the right-hand side: \[ dy = (2x + \frac{1}{x}) \, dx. \]

Now, integrate both sides: \[ \int dy = \int \left( 2x + \frac{1}{x} \right) \, dx. \]

The integral of \( dy \) is \( y \), and the integral of \( 2x \) is \( x^2 \). The integral of \( \frac{1}{x} \) is \( \ln|x| \), so we get: \[ y = x^2 + \ln|x| + C. \]

Step 1: Use the given point to find \( C \).
We are given that the curve passes through the point \( (1, 1) \), so substitute \( x = 1 \) and \( y = 1 \) into the equation: \[ 1 = 1^2 + \ln|1| + C. \]
Since \( \ln|1| = 0 \), this simplifies to: \[ 1 = 1 + C \quad \Rightarrow \quad C = 0. \]


Conclusion:
The equation of the curve is: \[ y = x^2 + \ln|x|. \] Quick Tip: To find the equation of a curve from a differential equation, integrate both sides, and use the given point to solve for the constant of integration.

We need to prove the inequality: \[ |\vec{a} \cdot \vec{b}| \leq |\vec{a}| |\vec{b}|. \]

This is a result of the Cauchy-Schwarz inequality.

The dot product of two vectors \( \vec{a \) and \( \vec{b} \) is given by: \[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta, \]
where \( \theta \) is the angle between the two vectors.

Step 1: Analyze the cosine term.
Since \( \cos \theta \) lies in the range \( -1 \leq \cos \theta \leq 1 \), we have: \[ |\cos \theta| \leq 1. \]

Step 2: Apply the inequality.
Thus, the magnitude of the dot product is: \[ |\vec{a} \cdot \vec{b}| = |\vec{a}| |\vec{b}| |\cos \theta| \leq |\vec{a}| |\vec{b}| \times 1. \]


Conclusion:
Therefore, we have shown that: \[ |\vec{a} \cdot \vec{b}| \leq |\vec{a}| |\vec{b}|. \] Quick Tip: The Cauchy-Schwarz inequality provides an upper bound for the dot product of two vectors. It states that the absolute value of the dot product is less than or equal to the product of the magnitudes of the vectors.

The given integral is: \[ \int \left( \sqrt{\cot x} + \sqrt{\tan x} \right) dx \]

We can split the integral into two parts: \[ \int \sqrt{\cot x} \, dx + \int \sqrt{\tan x} \, dx \]

Let’s solve each part separately:

First Integral: \[ I_1 = \int \sqrt{\cot x} \, dx \]
Using the identity \( \cot x = \frac{\cos x}{\sin x} \), we rewrite the integral: \[ I_1 = \int \sqrt{\frac{\cos x}{\sin x}} \, dx = \int \frac{\sqrt{\cos x}}{\sqrt{\sin x}} \, dx \]
Now, let’s use substitution. Let \( u = \sin x \), so \( du = \cos x \, dx \). The integral becomes: \[ I_1 = \int \frac{1}{\sqrt{u}} \, du = 2\sqrt{u} + C_1 = 2\sqrt{\sin x} + C_1 \]

Second Integral: \[ I_2 = \int \sqrt{\tan x} \, dx \]
Using the identity \( \tan x = \frac{\sin x}{\cos x} \), we rewrite the integral: \[ I_2 = \int \sqrt{\frac{\sin x}{\cos x}} \, dx = \int \frac{\sqrt{\sin x}}{\sqrt{\cos x}} \, dx \]
Now, let’s use substitution. Let \( v = \cos x \), so \( dv = -\sin x \, dx \). The integral becomes: \[ I_2 = -\int \frac{1}{\sqrt{v}} \, dv = -2\sqrt{v} + C_2 = -2\sqrt{\cos x} + C_2 \]

Final Answer:
The final solution is: \[ \int \left( \sqrt{\cot x} + \sqrt{\tan x} \right) dx = 2\sqrt{\sin x} - 2\sqrt{\cos x} + C \]

Where \( C = C_1 + C_2 \) is the constant of integration. Quick Tip: When solving integrals involving square roots of trigonometric functions, use substitution to simplify the expression.

To prove that \( R \) is an equivalence relation, we need to show that it satisfies three properties:
1. Reflexivity

2. Symmetry

3. Transitivity


1. Reflexivity:

A relation \( R \) is reflexive if \( (x, x) \in R \) for every \( x \in A \).

In this case, the number of pages in a book is always equal to itself. Thus, for every book \( x \), the relation holds: \[ (x, x) \in R \quad because the number of pages in x is equal to itself. \]
Hence, \( R \) is reflexive.

2. Symmetry:

A relation \( R \) is symmetric if whenever \( (x, y) \in R \), we also have \( (y, x) \in R \).

If the number of pages in book \( x \) is equal to the number of pages in book \( y \), then the number of pages in book \( y \) is also equal to the number of pages in book \( x \). Thus, if \( (x, y) \in R \), we also have \( (y, x) \in R \).
Therefore, \( R \) is symmetric.

3. Transitivity:

A relation \( R \) is transitive if whenever \( (x, y) \in R \) and \( (y, z) \in R \), we also have \( (x, z) \in R \).

If the number of pages in book \( x \) is equal to the number of pages in book \( y \), and the number of pages in book \( y \) is equal to the number of pages in book \( z \), then the number of pages in book \( x \) is equal to the number of pages in book \( z \). Hence, if \( (x, y) \in R \) and \( (y, z) \in R \), we also have \( (x, z) \in R \).
Therefore, \( R \) is transitive.


Conclusion:

Since the relation \( R \) is reflexive, symmetric, and transitive, we conclude that \( R \) is an equivalence relation. Quick Tip: For a relation to be an equivalence relation, it must satisfy the properties of reflexivity, symmetry, and transitivity.

To differentiate \( y = x^x \), we will use logarithmic differentiation.

Step 1: Take the natural logarithm of both sides.
\[ \ln y = \ln(x^x). \]

Step 2: Use the logarithmic identity \( \ln(a^b) = b \ln a \).
\[ \ln y = x \ln x. \]

Step 3: Differentiate both sides with respect to \( x \).

Using the product rule on the right-hand side: \[ \frac{d}{dx} \left( \ln y \right) = \frac{d}{dx} \left( x \ln x \right). \]

On the left-hand side: \[ \frac{d}{dx} \left( \ln y \right) = \frac{1}{y} \frac{dy}{dx}. \]

On the right-hand side, applying the product rule: \[ \frac{d}{dx} \left( x \ln x \right) = \ln x + 1. \]

So, we have: \[ \frac{1}{y} \frac{dy}{dx} = \ln x + 1. \]

Step 4: Solve for \( \frac{dy}{dx} \).

Multiply both sides by \( y \): \[ \frac{dy}{dx} = y (\ln x + 1). \]

Since \( y = x^x \), substitute back: \[ \frac{dy}{dx} = x^x (\ln x + 1). \]


Conclusion:

The derivative of \( y = x^x \) is: \[ \boxed{\frac{dy}{dx} = x^x (\ln x + 1)}. \] Quick Tip: For differentiating expressions of the form \( y = x^x \), use logarithmic differentiation.

We are asked to find a matrix \( D \) such that: \[ CD - AB = 0 \]

This can be rewritten as: \[ CD = AB \]

We need to calculate \( AB \) and then solve for \( D \).


Step 1: Calculate \( AB \).

First, compute the product \( AB \):
\[ AB = \begin{pmatrix} 2 & -1
3 & 4 \end{pmatrix} \begin{pmatrix} 5 & 2
7 & 4 \end{pmatrix} \]

Using matrix multiplication: \[ AB = \begin{pmatrix} (2 \times 5 + (-1) \times 7) & (2 \times 2 + (-1) \times 4)
(3 \times 5 + 4 \times 7) & (3 \times 2 + 4 \times 4) \end{pmatrix} \]
\[ AB = \begin{pmatrix} 10 - 7 & 4 - 4
15 + 28 & 6 + 16 \end{pmatrix} = \begin{pmatrix} 3 & 0
43 & 22 \end{pmatrix} \]


Step 2: Solve for \( D \).

Now, we know \( CD = AB \), so we have the equation: \[ \begin{pmatrix} 2 & 5
3 & 8 \end{pmatrix} D = \begin{pmatrix} 3 & 0
43 & 22 \end{pmatrix} \]

Let \( D = \begin{pmatrix} a & b
c & d \end{pmatrix} \). Now, multiply \( C \) with \( D \): \[ \begin{pmatrix} 2 & 5
3 & 8 \end{pmatrix} \begin{pmatrix} a & b
c & d \end{pmatrix} = \begin{pmatrix} 2a + 5c & 2b + 5d
3a + 8c & 3b + 8d \end{pmatrix} \]

Equating this with \( AB \), we get the system of equations: \[ 2a + 5c = 3, \quad 2b + 5d = 0 \] \[ 3a + 8c = 43, \quad 3b + 8d = 22 \]


Step 3: Solve the system of equations.

From the first equation, solve for \( a \): \[ 2a + 5c = 3 \implies a = \frac{3 - 5c}{2} \]

Substitute \( a = \frac{3 - 5c}{2} \) into the second equation: \[ 3\left(\frac{3 - 5c}{2}\right) + 8c = 43 \]
Simplifying: \[ \frac{9 - 15c}{2} + 8c = 43 \] \[ 9 - 15c + 16c = 86 \] \[ 9 + c = 86 \implies c = 77 \]

Now substitute \( c = 77 \) into the equation for \( a \): \[ a = \frac{3 - 5(77)}{2} = \frac{3 - 385}{2} = \frac{-382}{2} = -191 \]

Now solve for \( b \) and \( d \):
From \( 2b + 5d = 0 \): \[ 2b + 5d = 0 \implies b = -\frac{5d}{2} \]

Substitute this into the second equation: \[ 3b + 8d = 22 \implies 3\left(-\frac{5d}{2}\right) + 8d = 22 \]
Simplifying: \[ -\frac{15d}{2} + 8d = 22 \] \[ -\frac{15d}{2} + \frac{16d}{2} = 22 \implies \frac{d}{2} = 22 \implies d = 44 \]

Substitute \( d = 44 \) into the equation for \( b \): \[ b = -\frac{5(44)}{2} = -\frac{220}{2} = -110 \]

Thus, the matrix \( D \) is: \[ D = \begin{pmatrix} -191 & -110
77 & 44 \end{pmatrix} \]


Final Answer:
The matrix \( D \) such that \( CD - AB = 0 \) is: \[ \boxed{\begin{pmatrix} -191 & -110
77 & 44 \end{pmatrix}} \] Quick Tip: When solving matrix equations like \( CD = AB \), first compute \( AB \), then solve the system of equations formed by multiplying \( C \) with \( D \).

The adjugate (adjoint) of a matrix \( A \), denoted by \( adj \, A \), is the transpose of the cofactor matrix of \( A \). A well-known result in matrix theory is that: \[ A \cdot adj \, A = |A| \cdot I. \]

To prove this, we first calculate \( |A| \), the determinant of \( A \). The determinant of a 3x3 matrix is given by: \[ |A| = \begin{vmatrix} 1 & 3 & 3
1 & 4 & 3
1 & 3 & 4 \end{vmatrix}. \]

Using cofactor expansion along the first row: \[ |A| = 1 \begin{vmatrix} 4 & 3
3 & 4 \end{vmatrix} - 3 \begin{vmatrix} 1 & 3
1 & 4 \end{vmatrix} + 3 \begin{vmatrix} 1 & 4
1 & 3 \end{vmatrix}. \]

Now, calculate the 2x2 determinants: \[ \begin{vmatrix} 4 & 3
3 & 4 \end{vmatrix} = (4 \cdot 4) - (3 \cdot 3) = 16 - 9 = 7, \] \[ \begin{vmatrix} 1 & 3
1 & 4 \end{vmatrix} = (1 \cdot 4) - (3 \cdot 1) = 4 - 3 = 1, \] \[ \begin{vmatrix} 1 & 4
1 & 3 \end{vmatrix} = (1 \cdot 3) - (4 \cdot 1) = 3 - 4 = -1. \]

Substitute these into the determinant formula: \[ |A| = 1 \cdot 7 - 3 \cdot 1 + 3 \cdot (-1) = 7 - 3 - 3 = 1. \]

Thus, \( |A| = 1 \).

Since the determinant is 1, we have: \[ A \cdot adj \, A = 1 \cdot I = I. \]


Conclusion:
We have proved that \[ A \cdot adj \, A = |A| \cdot I = I. \] Quick Tip: For a matrix \( A \), the product of \( A \) and its adjugate \( adj \, A \) is always equal to the determinant of \( A \) multiplied by the identity matrix: \( A \cdot adj \, A = |A| \cdot I \).

We are given: \[ y = a \frac{t+1}{t} \quad and \quad x = (t + 1)^{\alpha}. \]

First, differentiate \( y \) with respect to \( t \): \[ \frac{dy}{dt} = a \left[ \frac{d}{dt} \left( \frac{t+1}{t} \right) \right]. \]

Using the quotient rule: \[ \frac{dy}{dt} = a \left( \frac{t \cdot 1 - (t+1) \cdot 1}{t^2} \right) = a \left( \frac{t - (t+1)}{t^2} \right) = a \left( \frac{-1}{t^2} \right). \]

Thus, \[ \frac{dy}{dt} = -\frac{a}{t^2}. \]

Next, differentiate \( x \) with respect to \( t \): \[ \frac{dx}{dt} = \frac{d}{dt} \left( (t+1)^{\alpha} \right) = \alpha (t + 1)^{\alpha-1}. \]

Now, apply the chain rule to find \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-\frac{a}{t^2}}{\alpha (t + 1)^{\alpha-1}}. \]

Thus, \[ \frac{dy}{dx} = -\frac{a}{\alpha t^2 (t + 1)^{\alpha-1}}. \]


Conclusion:
The value of \( \frac{dy}{dx} \) is \[ \boxed{-\frac{a}{\alpha t^2 (t + 1)^{\alpha-1}}}. \] Quick Tip: When differentiating a quotient, use the quotient rule: \[ \frac{dy}{dt} = \frac{v \frac{du}{dt} - u \frac{dv}{dt}}{v^2}, \quad where y = \frac{u}{v}. \]

We can solve the given system of linear equations using the matrix method. The system of equations can be written in matrix form as: \[ \begin{bmatrix} 3 & 2 & 3
-2 & 1 & 1
-1 & 3 & -2 \end{bmatrix} \begin{bmatrix} x
y
z \end{bmatrix} = \begin{bmatrix} 5
-4
3 \end{bmatrix}. \]

Let \( A \) be the coefficient matrix, \( X \) be the variable matrix, and \( B \) be the constant matrix: \[ A = \begin{bmatrix} 3 & 2 & 3
-2 & 1 & 1
-1 & 3 & -2 \end{bmatrix}, \quad X = \begin{bmatrix} x
y
z \end{bmatrix}, \quad B = \begin{bmatrix} 5
-4
3 \end{bmatrix}. \]

Now, solve for \( X \) using the inverse of matrix \( A \): \[ X = A^{-1} B. \]

Step 1: Find the determinant of \( A \).

The determinant of \( A \), denoted as \( det(A) \), is calculated as: \[ det(A) = 3 \left( 1 \cdot (-2) - 1 \cdot 3 \right) - 2 \left( -2 \cdot (-2) - 1 \cdot (-1) \right) + 3 \left( -2 \cdot 3 - 1 \cdot 1 \right). \] \[ det(A) = 3(-5) - 2(5) + 3(-7) = -15 - 10 - 21 = -46. \]

Step 2: Find the inverse of \( A \).
To find \( A^{-1} \), we use the formula: \[ A^{-1} = \frac{1}{det(A)} \cdot adj(A). \]
The adjugate matrix \( adj(A) \) is the transpose of the cofactor matrix of \( A \), which is computed as follows.

Step 3: Multiply \( A^{-1} \) with \( B \).
Finally, multiply the inverse of \( A \) by \( B \) to find \( X \). This will give the values of \( x \), \( y \), and \( z \).


Conclusion:
The solution for the system of equations will provide the values of \( x \), \( y \), and \( z \). Quick Tip: To solve a system of linear equations using the matrix method, first write the system in matrix form, then use the inverse of the coefficient matrix to find the solution.

The equation of a plane can be written as: \[ \vec{r} \cdot \vec{n} = D, \]
where \( \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \) is the position vector, \( \vec{n} \) is the normal vector to the plane, and \( D \) is the constant.

We are given two planes: \[ \vec{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = 6 \quad (Plane 1), \] \[ \vec{r} \cdot (2\hat{i} + 3\hat{j} + 4\hat{k}) = -5 \quad (Plane 2). \]

The point of intersection of the two planes lies on both planes. Hence, the normal vector to the required plane will be the cross product of the normal vectors of Plane 1 and Plane 2.

Step 1: Find the normal vector of the required plane.
The normal vector of Plane 1 is \( \vec{n_1} = \hat{i} + \hat{j} + \hat{k} \), and the normal vector of Plane 2 is \( \vec{n_2} = 2\hat{i} + 3\hat{j} + 4\hat{k} \).

The normal vector to the required plane is the cross product \( \vec{n_1} \times \vec{n_2} \): \[ \vec{n} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
1 & 1 & 1
2 & 3 & 4 \end{vmatrix}. \]

This gives: \[ \vec{n} = \hat{i}(1 \cdot 4 - 1 \cdot 3) - \hat{j}(1 \cdot 4 - 1 \cdot 2) + \hat{k}(1 \cdot 3 - 1 \cdot 2), \] \[ \vec{n} = \hat{i}(4 - 3) - \hat{j}(4 - 2) + \hat{k}(3 - 2), \] \[ \vec{n} = \hat{i} - 2\hat{j} + \hat{k}. \]

Step 2: Find the equation of the plane.
The equation of the plane passing through the point \( (1, 1, 1) \) is given by: \[ \vec{r} \cdot \vec{n} = D. \]
Substitute \( \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \) and \( \vec{n} = \hat{i} - 2\hat{j} + \hat{k} \) into the equation: \[ (x \hat{i} + y \hat{j} + z \hat{k}) \cdot (\hat{i} - 2\hat{j} + \hat{k}) = D. \]
Simplifying the dot product: \[ x - 2y + z = D. \]

Now, substitute the point \( (1, 1, 1) \) into the equation to find \( D \): \[ 1 - 2(1) + 1 = D \quad \Rightarrow \quad D = 0. \]

Thus, the equation of the plane is: \[ x - 2y + z = 0. \]


Conclusion:
The equation of the plane is: \[ \boxed{x - 2y + z = 0}. \] Quick Tip: To find the equation of a plane passing through the intersection of two planes, take the cross product of their normal vectors and use a point on the plane to determine the constant \( D \).

The distance between the point \( (x, y) \) on the curve and the soldier at \( (3, 7) \) is given by the distance formula:
\[ d = \sqrt{(x - 3)^2 + (y - 7)^2} \]

Substitute \( y = x^2 + 7 \) into the distance formula:
\[ d = \sqrt{(x - 3)^2 + (x^2 + 7 - 7)^2} = \sqrt{(x - 3)^2 + x^4} \]

We want to minimize this distance, so we minimize \( d^2 \) to avoid dealing with the square root. Define the function:
\[ f(x) = (x - 3)^2 + x^4 = x^2 - 6x + 9 + x^4 \]

To minimize \( f(x) \), take the derivative and set it to zero:
\[ \frac{d}{dx} f(x) = 4x^3 + 2x - 6 = 0 \]

Factor the equation:
\[ 2x(2x^2 + 1) = 6 \quad \Rightarrow \quad x(2x^2 + 1) = 3 \]

Solving this equation, we find that \( x = 1 \). Now substitute \( x = 1 \) into the distance formula:
\[ y = 1^2 + 7 = 8 \]

Thus, the point on the curve is \( (1, 8) \). The nearest distance is:
\[ d = \sqrt{(1 - 3)^2 + (8 - 7)^2} = \sqrt{(-2)^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5} \]


Final Answer:
The nearest distance is \( \boxed{\sqrt{5}} \). Quick Tip: To find the nearest distance, minimize the square of the distance function, which is easier to differentiate.

This is a first-order linear differential equation. The standard form is:
\[ \frac{dy}{dx} + P(x) y = Q(x) \]

Here, \( P(x) = \cot x \) and \( Q(x) = 4x \csc x \).

The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int P(x) \, dx} = e^{\int \cot x \, dx} = e^{\ln \sin x} = \sin x \]

Multiply both sides of the differential equation by \( \mu(x) = \sin x \):
\[ \sin x \frac{dy}{dx} + y \sin x \cot x = 4x \]

This simplifies to:
\[ \frac{d}{dx} \left( y \sin x \right) = 4x \]

Now, integrate both sides:
\[ \int \frac{d}{dx} \left( y \sin x \right) \, dx = \int 4x \, dx \]
\[ y \sin x = 2x^2 + C \]

Now, use the initial condition \( y = 0 \) when \( x = \frac{\pi}{2} \):
\[ 0 \cdot \sin \left( \frac{\pi}{2} \right) = 2 \left( \frac{\pi}{2} \right)^2 + C \] \[ 0 = 2 \times \frac{\pi^2}{4} + C \quad \Rightarrow \quad C = -\frac{\pi^2}{2} \]

Thus, the particular solution is:
\[ y \sin x = 2x^2 - \frac{\pi^2}{2} \]
\[ y = \frac{2x^2 - \frac{\pi^2}{2}}{\sin x} \]


Final Answer:
The particular solution is: \[ y = \frac{2x^2 - \frac{\pi^2}{2}}{\sin x} \] Quick Tip: For linear differential equations, use the integrating factor to simplify the equation and then integrate both sides.

Let the event \( A \) be that the ball drawn is black and event \( B \) be that the ball is drawn from Container-III. We need to find \( P(B|A) \), the probability that the ball is from Container-III given that it is black.

By Bayes' Theorem: \[ P(B|A) = \frac{P(A|B) \cdot P(B)}{P(A)}. \]

Now, calculate the individual probabilities:

1. \( P(A|B) \) is the probability of drawing a black ball from Container-III. This is the ratio of black balls in Container-III to the total number of balls in Container-III: \[ P(A|B) = \frac{1}{1 + 2 + 3} = \frac{1}{6}. \]

2. \( P(B) \) is the probability that Container-III is chosen. Since there are 3 containers, the probability of choosing any one container is \( \frac{1}{3} \).

3. \( P(A) \) is the total probability of drawing a black ball from any container. This can be calculated by adding the probabilities of drawing a black ball from each container, weighted by the probability of choosing each container: \[ P(A) = \frac{1}{3} \cdot \frac{3}{3+4+5} + \frac{1}{3} \cdot \frac{2}{2+2+2} + \frac{1}{3} \cdot \frac{1}{1+2+3}. \]
Simplifying: \[ P(A) = \frac{1}{3} \cdot \frac{3}{12} + \frac{1}{3} \cdot \frac{2}{6} + \frac{1}{3} \cdot \frac{1}{6} = \frac{1}{3} \cdot \left( \frac{1}{4} + \frac{1}{3} + \frac{1}{6} \right). \]

Finding a common denominator: \[ P(A) = \frac{1}{3} \cdot \frac{11}{12} = \frac{11}{36}. \]

Now, we can substitute into Bayes' Theorem: \[ P(B|A) = \frac{\frac{1}{6} \cdot \frac{1}{3}}{\frac{11}{36}} = \frac{\frac{1}{18}}{\frac{11}{36}} = \frac{2}{11}. \]


Conclusion:
The probability that the ball is drawn from Container-III given that the ball is black is \[ \boxed{\frac{2}{11}}. \] Quick Tip: In problems involving conditional probability, Bayes' Theorem is useful to calculate the probability of an event given some other event.

We need to maximize \( z = x + y \), subject to the constraints: \[ x - y \leq -1, \quad -x + y \leq 0, \quad x \geq 0, \quad y \geq 0. \]

We begin by graphing the constraints:

1. \( x - y \leq -1 \) represents a line with slope 1 and intercept -1.
2. \( -x + y \leq 0 \) represents a line with slope 1 and intercept 0.
3. \( x \geq 0 \) is the region to the right of the y-axis.
4. \( y \geq 0 \) is the region above the x-axis.

The feasible region is the area bounded by these lines. We can find the vertices of this region by solving the system of equations:

1. Solve \( x - y = -1 \) and \( -x + y = 0 \): \[ x - y = -1 \quad (i) \] \[ -x + y = 0 \quad (ii) \]
From (ii), \( x = y \). Substituting into (i): \[ x - x = -1 \quad \Rightarrow \quad 0 = -1, \]
which is a contradiction, meaning there is no solution.

Next, solve for other pairs of constraints to determine the vertices. Quick Tip: In linear programming problems, solving the system of constraints often gives the vertices of the feasible region, where the maximum or minimum values occur.