CBSE Class 12 Mathematics Question Paper 2024 (Set 1- 65/3/1) Available - Download Solution PDF with Answer Key

An identity matrix is a square matrix where:

- All diagonal elements are 1.

- All off-diagonal elements are 0.


Thus, for an identity matrix \( A = [a_{ij}] \), the elements must satisfy: \[ a_{ij} = \begin{cases} 1, & if i = j
0, & if i \neq j \end{cases} \]

Let’s analyze the options:


Option A: \( a_{ij} = \begin{cases} 0, & if i = j
1, & if i \neq j \end{cases} \)

This is incorrect because, in an identity matrix, the diagonal elements \( a_{ii} \) should be 1, not 0.

Option B: \( a_{ij} = 1, \, \forall i, j \)

This is incorrect because an identity matrix has 1 on the diagonal and 0 elsewhere, not 1 for all elements.

Option C: \( a_{ij} = 0, \, \forall i, j \)

This is incorrect because, in an identity matrix, the diagonal elements must be 1.

Option D: \( a_{ij} = \begin{cases} 0, & if i \neq j
1, & if i = j \end{cases} \)

This is correct because it matches the definition of the identity matrix.



Final Answer: \[ \boxed{D} \]
Quick Tip: Always check the diagonal and off-diagonal elements when identifying an identity matrix.

Step 1: {Check if \( f(x) \) is one-one

The function \( f(x) = x^2 + 1 \) is strictly increasing for \( x \geq 0 \), since \( f'(x) = 2x > 0 \). Thus, \( f(x) \) is one-one.


Step 2: {Check if \( f(x) \) is onto

The range of \( f(x) \) is \( [1, \infty) \), which does not cover all of \( \mathbb{R}^+ \). Therefore, \( f(x) \) is not onto.
Quick Tip: To check if a function is one-one, use the derivative or monotonicity test. To check if it's onto, determine the range of the function.

We are given that \( adj(A) = A \), where the adjugate matrix \( adj(A) \) is the transpose of the cofactor matrix of \( A \). For a \( 2 \times 2 \) matrix \( A = \begin{bmatrix} a & b
c & d \end{bmatrix} \), the adjugate matrix is:
\[ adj(A) = \begin{bmatrix} d & -b
-c & a \end{bmatrix} \]

Since \( adj(A) = A \), we equate the corresponding elements of \( A \) and \( adj(A) \):
\[ \begin{bmatrix} d & -b
-c & a \end{bmatrix} = \begin{bmatrix} a & b
c & d \end{bmatrix} \]

From this, we have the following equations:

1. \( d = a \)
2. \( -b = b \), which implies \( b = 0 \)
3. \( -c = c \), which implies \( c = 0 \)

Thus, \( d = a \), \( b = 0 \), and \( c = 0 \).

Now, we compute \( a + b + c + d \):
\[ a + b + c + d = a + 0 + 0 + a = 2a \]


Final Answer: \[ \boxed{2a} \] Quick Tip: For properties of adjoint matrices, always check if the matrix is scalar or symmetric.

Step 1: {Analyze the function for different cases

The function is \( f(x) = |1 - x + |x|| \). Consider two cases:

Case 1: \( x \geq 0 \), then \( |x| = x \), so: \[ f(x) = |1 - x + x| = |1| = 1. \]
Case 2: \( x < 0 \), then \( |x| = -x \), so: \[ f(x) = |1 - x - x| = |1 - 2x|. \]

Step 2: {Check continuity

For \( x \geq 0 \), \( f(x) = 1 \), which is continuous.

For \( x < 0 \), \( f(x) = |1 - 2x| \), which is also continuous because it is a piecewise linear function.

At \( x = 0 \), the left-hand limit (LHL) and right-hand limit (RHL) are: \[ LHL = f(0^-) = |1 - 2(0)| = 1, \quad RHL = f(0^+) = 1. \]
Thus, \( f(x) \) is continuous at \( x = 0 \).


Step 3: {Conclude the result

The function \( f(x) \) is continuous for all \( x \). Hence, it is continuous everywhere.
Quick Tip: When analyzing the continuity of piecewise functions, always check the behavior in each interval and at the boundaries.

Step 1: {Express the perimeter of the square

The perimeter \( P \) of a square is given by: \[ P = 4 \times side length. \]
Let the side length be \( s \). Then: \[ P = 4s. \]

Step 2: {Differentiate with respect to time

Differentiating \( P \) with respect to \( t \): \[ \frac{dP}{dt} = 4 \cdot \frac{ds}{dt}. \]

Step 3: {Substitute the given rate

The side length is decreasing at \( \frac{ds}{dt} = -1.5 \, cm/s \). Substituting: \[ \frac{dP}{dt} = 4 \cdot (-1.5) = -6 \, cm/s. \]
The negative sign indicates a decrease in the perimeter.


Step 4: {Conclude the result

The rate of decrease of the perimeter is \( 6 \, cm/s \).
Quick Tip: Always differentiate geometric formulas with respect to time when dealing with rates of change.

Step 1: {Use the property of definite integrals

The integral \( \int_{-a}^a f(x) \, dx = 0 \) if the function \( f(x) \) is odd. A function \( f(x) \) is odd if: \[ f(-x) = -f(x). \]

Step 2: {Simplify the integral for odd functions

For odd functions: \[ \int_{-a}^a f(x) \, dx = \int_{-a}^0 f(x) \, dx + \int_{0}^a f(x) \, dx. \]
Since \( f(-x) = -f(x) \), the integral over \( [-a, 0] \) cancels with the integral over \( [0, a] \), resulting in: \[ \int_{-a}^a f(x) \, dx = 0. \]

Step 3: {Conclude the result

The given integral equals 0 when \( f(-x) = -f(x) \).
Quick Tip: Odd functions satisfy \( f(-x) = -f(x) \) and have symmetric properties about the origin.

Step 1: {Rewrite the equation

The given equation is: \[ x \log x \frac{dy}{dx} + y = 2 \log x. \]
Rearranging: \[ \frac{dy}{dx} + \frac{y}{x \log x} = \frac{2}{x \log x}. \]

Step 2: {Check the form of the equation

This is a first-order linear differential equation of the form: \[ \frac{dy}{dx} + P(x)y = Q(x), \]
where \( P(x) = \frac{1}{x \log x} \) and \( Q(x) = \frac{2}{x \log x} \).

Step 3: {Conclude the result

The equation is a first-order linear differential equation.
Quick Tip: To identify the type of differential equation, rewrite it in standard forms and compare the coefficients.

Step 1: {Find the dot product of \( \vec{a} \) and \( \vec{b} \)

The dot product of \( \vec{a} \) and \( \vec{b} \) is: \[ \vec{a} \cdot \vec{b} = (2)(1) + (-1)(1) + (1)(-1) = 2 - 1 - 1 = 0. \]

Step 2: {Check for perpendicularity

If \( \vec{a} \cdot \vec{b} = 0 \), the vectors are perpendicular.

Step 3: {Conclude the result

The vectors \( \vec{a} \) and \( \vec{b} \) are perpendicular.
Quick Tip: Two vectors are perpendicular if their dot product equals zero.

Step 1: {Recall the direction cosine property

The sum of the squares of the cosines of the angles a line makes with the coordinate axes is: \[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1. \]
This is a fundamental property of direction cosines.

Step 2: {Check each option

Option (A): True, as it is the direction cosine property.

Option (B): True, as \( \sin^2 \alpha + \sin^2 \beta + \sin^2 \gamma = 3 - (\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma) = 2 \).

Option (C): True, derived from trigonometric identities for direction cosines.

Option (D): False, as \( \cos \alpha + \cos \beta + \cos \gamma \neq 1 \) in general.


Step 3: {Conclude the result

Option (D) is not true.
Quick Tip: Always check fundamental properties of direction cosines and trigonometric identities to verify such questions.

Step 1: {Define constraints in linear programming

Constraints are the conditions or restrictions imposed on decision variables (e.g., \( x, y \)) in a linear programming problem. They typically represent limitations on resources or other requirements.

Step 2: {Identify from options

(A) Feasible solutions: These are solutions satisfying all constraints, but they are not the constraints themselves.

(B) Constraints: These are restrictions on decision variables, and this is the correct answer.

(C) Optimal solutions: These maximize or minimize the objective function but are not constraints.

(D) Infeasible solutions: These do not satisfy all constraints.


Step 3: {Conclude the result

The restrictions are called constraints.
Quick Tip: Understand the difference between constraints and feasible/optimal solutions in linear programming.

Step 1: {Recall the formula for conditional probability

The conditional probability \( P(F|E) \) is given by: \[ P(F|E) = \frac{P(E \cap F)}{P(E)}. \]

Step 2: {Find \( P(E \cap F) \)

Using the formula for the probability of the union of two events: \[ P(E \cup F) = P(E) + P(F) - P(E \cap F). \]
Substitute the given values \( P(E \cup F) = 0.4 \), \( P(E) = 0.1 \), \( P(F) = 0.3 \): \[ 0.4 = 0.1 + 0.3 - P(E \cap F). \]
Simplify to find \( P(E \cap F) \): \[ P(E \cap F) = 0.1 + 0.3 - 0.4 = 0. \]

Step 3: {Calculate \( P(F|E) \)

Substitute \( P(E \cap F) = 0 \) and \( P(E) = 0.1 \) into the formula for \( P(F|E) \): \[ P(F|E) = \frac{P(E \cap F)}{P(E)} = \frac{0}{0.1} = 0. \]

Step 4: {Conclude the result

The conditional probability \( P(F|E) \) is \( 0 \). This indicates that the events \( E \) and \( F \) do not overlap.
Quick Tip: When \( P(E \cap F) = 0 \), the events \( E \) and \( F \) are mutually exclusive, meaning they cannot occur simultaneously.

Step 1: {Recall the property of skew-symmetric matrices

For a skew-symmetric matrix \( A \), \( A^T = -A \).

Step 2: {Analyze \( AB + BA \)

Taking the transpose: \[ (AB + BA)^T = B^T A^T + A^T B^T = (-B)(-A) + (-A)(-B) = AB + BA. \]
Thus, \( AB + BA \) is symmetric, as its transpose equals itself.
Quick Tip: Remember the properties of skew-symmetric matrices and their behavior under addition and multiplication.

Step 1: {Expand the determinant

The determinant is: \[ \begin{vmatrix} 1 & 3 & 1
k & 0 & 1
0 & 0 & 1 \end{vmatrix}. \]
Expanding along the third row: \[ \begin{vmatrix} 1 & 3 & 1
k & 0 & 1
0 & 0 & 1 \end{vmatrix} = 1 \cdot \begin{vmatrix} 1 & 3
k & 0 \end{vmatrix}. \]
The \( 2 \times 2 \) determinant is: \[ \begin{vmatrix} 1 & 3
k & 0 \end{vmatrix} = (1)(0) - (3)(k) = -3k. \]
Thus, the determinant is: \[ -3k = \pm 6. \]

Step 2: {Solve for \( k \)

Divide both sides by \( -3 \): \[ k = \frac{\pm 6}{-3} = \mp 2. \]

Step 3: {Conclude the result

The value of \( k \) is \( \mp 2 \).
Quick Tip: To calculate determinants, expand along rows or columns with the most zeros to simplify calculations.

Step 1: {Express \( 2^x \) and \( 3^x \) using logarithms

Let \( y = \frac{d}{dx} (2^x) \) with respect to \( 3^x \). Rewrite \( 2^x = e^{x \log 2} \) and \( 3^x = e^{x \log 3} \).

Step 2: {Differentiate \( 2^x \)

The derivative of \( 2^x \) with respect to \( x \) is: \[ \frac{d}{dx}(2^x) = 2^x \log 2. \]

Step 3: {Differentiate \( 3^x \)

The derivative of \( 3^x \) with respect to \( x \) is: \[ \frac{d}{dx}(3^x) = 3^x \log 3. \]

Step 4: {Find \( \frac{d}{dx}(2^x) \) with respect to \( 3^x \)

Using the chain rule: \[ \frac{d}{d(3^x)}(2^x) = \frac{\frac{d}{dx}(2^x)}{\frac{d}{dx}(3^x)} = \frac{2^x \log 2}{3^x \log 3}. \]

Step 5: {Simplify the result

Rewriting \( \frac{2^x}{3^x} \) as \( \left( \frac{2}{3} \right)^x \), we get: \[ \frac{d}{d(3^x)}(2^x) = \left( \frac{2}{3} \right)^x \frac{\log 2}{\log 3}. \]

Step 6: {Conclude the result

The derivative of \( 2^x \) with respect to \( 3^x \) is \( \left( \frac{2}{3} \right)^x \frac{\log 2}{\log 3} \).
Quick Tip: To differentiate exponential functions with respect to another, rewrite using logarithmic forms and simplify.

Step 1: {Recall the dot product formula

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

Step 2: {Determine the range of \( |\vec{a} \cdot \vec{k}| \)

Since \( |\vec{a}| = 2 \), the magnitude of \( \vec{k} \) varies as: \[ |\vec{k}| \in [-3, 2]. \]
The maximum value of \( |\vec{a} \cdot \vec{k}| \) occurs when \( \cos \theta = 1 \): \[ |\vec{a} \cdot \vec{k}|_{max} = 2 \cdot 3 = 6. \]
The minimum value of \( |\vec{a} \cdot \vec{k}| \) occurs when \( \cos \theta = 0 \): \[ |\vec{a} \cdot \vec{k}|_{min} = 0. \]

Step 3: {Conclude the result

Thus, \( |\vec{a} \cdot \vec{k}| \in [0, 6] \).
Quick Tip: The maximum value of a dot product is achieved when the vectors are parallel, and the minimum is achieved when they are perpendicular.

Step 1: {Recall the direction cosine condition

For a line making angles \( \alpha, \beta, \gamma \) with the positive directions of the \( x \)-axis, \( y \)-axis, and \( z \)-axis respectively, the sum of the squares of the direction cosines is: \[ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1. \]

Step 2: {Substitute the given angles

The line makes angles \( \alpha = \frac{\pi}{4} \) and \( \gamma = \frac{\pi}{4} \) with the \( x \)-axis and \( z \)-axis, so: \[ \cos^2 \frac{\pi}{4} + \cos^2 \beta + \cos^2 \frac{\pi}{4} = 1. \]
Since \( \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}} \), we have: \[ \left( \frac{1}{\sqrt{2}} \right)^2 + \cos^2 \beta + \left( \frac{1}{\sqrt{2}} \right)^2 = 1. \]
Simplify: \[ \frac{1}{2} + \cos^2 \beta + \frac{1}{2} = 1. \]

Step 3: {Solve for \( \cos^2 \beta \)

Combine terms: \[ 1 + \cos^2 \beta = 1 \implies \cos^2 \beta = 0. \]
Thus: \[ \cos \beta = 0 \implies \beta = \frac{\pi}{2}. \]

Step 4: {Conclude the result

The angle which the line makes with the positive direction of the \( y \)-axis is \( \frac{\pi}{2} \).
Quick Tip: For lines in 3D geometry, the sum of the squares of the direction cosines always equals 1.

Step 1: {Analyze the boundary lines

The constraints for the shaded region are based on the lines: \[ x + 2y = 76 \quad and \quad 2x + y = 104. \]
From the diagram:
- The region is above the line \( x + 2y = 76 \), so \( x + 2y \geq 76 \).
- The region is below the line \( 2x + y = 104 \), so \( 2x + y \leq 104 \).
- The region is in the first quadrant, so \( x \geq 0 \) and \( y \geq 0 \).

Step 2: {Verify each option

Option (C) correctly represents the constraints as: \[ x + 2y \geq 76, \quad 2x + y \leq 104, \quad x, y \geq 0. \]

Step 3: {Conclude the result

The group of constraints representing the feasible region is: \[ x + 2y \geq 76, \quad 2x + y \leq 104, \quad x, y \geq 0. \] Quick Tip: To determine constraints from a graph, carefully analyze the shaded region relative to the boundary lines.

Step 1: {Recall the inverse of a diagonal matrix

The inverse of a diagonal matrix \( A = \begin{bmatrix} a & 0 & 0
0 & b & 0
0 & 0 & c \end{bmatrix} \) is given by: \[ A^{-1} = \begin{bmatrix} \frac{1}{a} & 0 & 0
0 & \frac{1}{b} & 0
0 & 0 & \frac{1}{c} \end{bmatrix}. \]

Step 2: {Compute the inverse of \( A \)

For \( A = \begin{bmatrix} 2 & 0 & 0
0 & 3 & 0
0 & 0 & 5 \end{bmatrix} \), the inverse is: \[ A^{-1} = \begin{bmatrix} \frac{1}{2} & 0 & 0
0 & \frac{1}{3} & 0
0 & 0 & \frac{1}{5} \end{bmatrix}. \]

Step 3: {Conclude the result

The inverse of \( A \) is: \[ A^{-1} = \begin{bmatrix} \frac{1}{2} & 0 & 0
0 & \frac{1}{3} & 0
0 & 0 & \frac{1}{5} \end{bmatrix}. \] Quick Tip: The inverse of a diagonal matrix is computed by taking the reciprocal of each diagonal entry.

Step 1: {Analyze Assertion (A)

A scalar matrix is a diagonal matrix where all the diagonal elements are equal, and the non-diagonal elements are zero. Hence, Assertion (A) is true.

Step 2: {Analyze Reason (R)

In a diagonal matrix, the diagonal elements can have any value (not necessarily 0). Therefore, Reason (R) is false.

Step 3: {Conclude the result

Assertion (A) is true, but Reason (R) is false. Hence, the correct option is (C).
Quick Tip: A scalar matrix is a specific type of diagonal matrix where all diagonal elements are equal.

Step 1: {Analyze Assertion (A)

The projection of \( \vec{a} \) on \( \vec{b} \) is given by: \[ Projection of \vec{a} on \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}. \]
Similarly, the projection of \( \vec{b} \) on \( \vec{a} \) is: \[ Projection of \vec{b} on \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|}. \]
These two projections are not equal because their magnitudes depend on \( |\vec{a}| \) and \( |\vec{b}| \). Hence, Assertion (A) is false.

Step 2: {Analyze Reason (R)

The angle between \( \vec{a} \) and \( \vec{b} \) is the same as the angle between \( \vec{b} \) and \( \vec{a} \) numerically, as the cosine of the angle is symmetric. Hence, Reason (R) is true.

Step 3: {Conclude the result

Assertion (A) is false, but Reason (R) is true. Hence, the correct option is (D).
Quick Tip: The projection of one vector on another depends on the magnitude of the vector being projected onto.

Step 1: {Simplify \( \sec^2\left(\tan^{-1}\frac{1}{2}\right) \)

Let \( \theta = \tan^{-1}\frac{1}{2} \). Then: \[ \tan \theta = \frac{1}{2}, \quad so \sec^2 \theta = 1 + \tan^2 \theta. \]
Substituting \( \tan^2 \theta = \left(\frac{1}{2}\right)^2 \): \[ \sec^2\left(\tan^{-1}\frac{1}{2}\right) = 1 + \left(\frac{1}{2}\right)^2 = 1 + \frac{1}{4} = \frac{5}{4}. \]

Step 2: {Simplify \( \csc^2\left(\cot^{-1}\frac{1}{3}\right) \)

Let \( \phi = \cot^{-1}\frac{1}{3} \). Then: \[ \cot \phi = \frac{1}{3}, \quad so \csc^2 \phi = 1 + \cot^2 \phi. \]
Substituting \( \cot^2 \phi = \left(\frac{1}{3}\right)^2 \): \[ \csc^2\left(\cot^{-1}\frac{1}{3}\right) = 1 + \left(\frac{1}{3}\right)^2 = 1 + \frac{1}{9} = \frac{10}{9}. \]

Step 3: {Add the two results
\[ \sec^2\left(\tan^{-1}\frac{1}{2}\right) + \csc^2\left(\cot^{-1}\frac{1}{3}\right) = \frac{5}{4} + \frac{10}{9}. \]
Taking the LCM of 4 and 9: \[ \frac{5}{4} + \frac{10}{9} = \frac{45}{36} + \frac{40}{36} = \frac{85}{36}. \]

Step 4: {Conclude the result
\[ \sec^2\left(\tan^{-1}\frac{1}{2}\right) + \csc^2\left(\cot^{-1}\frac{1}{3}\right) = \frac{85}{36}. \] Quick Tip: To solve trigonometric problems involving inverse functions, use the Pythagorean identities and simplify step-by-step.

Step 1: {Rewrite the given equation

Given \( x = e^{x/y} \), take the natural logarithm on both sides: \[ \log x = \frac{x}{y}. \]

Step 2: {Express \( y \) in terms of \( x \)

Rearranging: \[ y = \frac{x}{\log x}. \]

Step 3: {Differentiate with respect to \( x \)

Using the quotient rule: \[ \frac{dy}{dx} = \frac{(\log x)(1) - x \cdot \frac{1}{x}}{(\log x)^2}. \]
Simplify: \[ \frac{dy}{dx} = \frac{\log x - 1}{(\log x)^2}. \]

Step 4: {Conclude the result

Thus, \( \frac{dy}{dx} = \frac{\log x - 1}{(\log x)^2} \).
Quick Tip: For logarithmic differentiation, always apply the quotient rule carefully and simplify step-by-step.

Step 1: {Find the left-hand derivative (LHD)

The LHD at \( x = 1 \) is: \[ LHD = \lim_{h \to 0^-} \frac{f(1 + h) - f(1)}{-h}. \]
Substituting \( f(x) = x^2 + 1 \) for \( x < 1 \): \[ LHD = \lim_{h \to 0^-} \frac{(1 - h)^2 + 1 - 2}{-h}. \]
Simplify: \[ LHD = \lim_{h \to 0^-} \frac{1 - 2h + h^2 - 1}{-h} = \lim_{h \to 0^-} \frac{-2h + h^2}{-h}. \]
Factorize: \[ LHD = \lim_{h \to 0^-} (2 - h) = 2. \]

Step 2: {Find the right-hand derivative (RHD)

The RHD at \( x = 1 \) is: \[ RHD = \lim_{h \to 0^+} \frac{f(1 + h) - f(1)}{h}. \]
Substituting \( f(x) = 3 - x \) for \( x > 1 \): \[ RHD = \lim_{h \to 0^+} \frac{[3 - (1 + h)] - 2}{h}. \]
Simplify: \[ RHD = \lim_{h \to 0^+} \frac{-h}{h} = -1. \]

Step 3: {Check differentiability

Since \( LHD \neq RHD \), \( f(x) \) is not differentiable at \( x = 1 \).
Quick Tip: For piecewise functions, compute LHD and RHD separately at the point of interest to check differentiability.

Step 1: {Use the product-to-sum formula

The product-to-sum formula is: \[ \sin A \cos B = \frac{1}{2}[\sin(A + B) + \sin(A - B)]. \]
Substituting \( A = 2x \) and \( B = 3x \): \[ \sin 2x \cos 3x = \frac{1}{2}[\sin(5x) + \sin(-x)] = \frac{1}{2}[\sin(5x) - \sin(x)]. \]

Step 2: {Split the integral
\[ \int_{0}^{\pi/2} \sin 2x \cos 3x \, dx = \frac{1}{2} \int_{0}^{\pi/2} \sin(5x) \, dx - \frac{1}{2} \int_{0}^{\pi/2} \sin(x) \, dx. \]

Step 3: {Evaluate the integrals
\[ \int \sin(kx) \, dx = -\frac{1}{k} \cos(kx). \]
For \( \int \sin(5x) \, dx \): \[ \int_{0}^{\pi/2} \sin(5x) \, dx = \left[-\frac{1}{5} \cos(5x)\right]_{0}^{\pi/2} = -\frac{1}{5}[\cos(5\cdot\frac{\pi}{2}) - \cos(0)] = -\frac{1}{5}[0 - 1] = \frac{1}{5}. \]
For \( \int \sin(x) \, dx \): \[ \int_{0}^{\pi/2} \sin(x) \, dx = \left[-\cos(x)\right]_{0}^{\pi/2} = -[\cos(\frac{\pi}{2}) - \cos(0)] = -[0 - 1] = 1. \]

Step 4: {Combine the results
\[ \int_{0}^{\pi/2} \sin 2x \cos 3x \, dx = \frac{1}{2} \left(\frac{1}{5} - 1\right) = \frac{1}{2} \cdot -\frac{4}{5} = -\frac{2}{5}. \]

Step 5: {Conclude the result
\[ \int_{0}^{\pi/2} \sin 2x \cos 3x \, dx = -\frac{2}{5}. \] Quick Tip: Use trigonometric identities to simplify integrals involving products of sine and cosine.

Step 1: {Express \( F(x) \) as an integral

The function \( F(x) \) is obtained by integrating the given derivative: \[ F(x) = \int \frac{1}{\sqrt{2x - x^2}} \, dx. \]

Step 2: {Simplify the expression inside the square root

Factorize \( 2x - x^2 \): \[ 2x - x^2 = x(2 - x). \]
Thus: \[ F(x) = \int \frac{1}{\sqrt{x(2 - x)}} \, dx. \]

Step 3: {Substitute to simplify the integral

Let \( x = 1 - \sin^2 \theta \). Then: \[ 2 - x = 1 + \cos^2 \theta, \quad dx = -2\sin\theta\cos\theta \, d\theta. \]
Substitute into the integral: \[ F(x) = \int \frac{1}{\sqrt{1 - \sin^2\theta}(1 + \cos^2\theta)} (-2\sin\theta\cos\theta) \, d\theta. \]
Simplify and integrate: \[ F(x) = \sin^{-1}(x - 1) + C. \]

Step 4: {Use the initial condition to find \( C \)

Given \( F(1) = 0 \), substitute \( x = 1 \): \[ F(1) = \sin^{-1}(1 - 1) + C = 0 \implies C = 0. \]

Step 5: {Conclude the result
\[ F(x) = \sin^{-1}(x - 1). \] Quick Tip: For integrals involving square roots of quadratic expressions, use trigonometric substitutions to simplify.

Step 1: {Find the position vector of \( C \)

The position vector of \( C \) dividing \( AB \) in the ratio \( 4:1 \) externally is given by: \[ \vec{r} = \frac{4\vec{b} - \vec{a}}{3}. \]
Substitute \( \vec{a} = \hat{i} + 2\hat{j} - \hat{k} \) and \( \vec{b} = -\hat{i} + \hat{j} + \hat{k} \): \[ \vec{r} = \frac{4(-\hat{i} + \hat{j} + \hat{k}) - (\hat{i} + 2\hat{j} - \hat{k})}{3}. \]
Simplify: \[ \vec{r} = \frac{-4\hat{i} + 4\hat{j} + 4\hat{k} - \hat{i} - 2\hat{j} + \hat{k}}{3}. \]
Combine terms: \[ \vec{r} = \frac{-5\hat{i} + 2\hat{j} + 5\hat{k}}{3}. \]

Step 2: {Find \( |\vec{AB}| \)

The vector \( \vec{AB} \) is: \[ \vec{AB} = \vec{b} - \vec{a} = (-\hat{i} + \hat{j} + \hat{k}) - (\hat{i} + 2\hat{j} - \hat{k}) = -2\hat{i} - \hat{j} + 2\hat{k}. \]
The magnitude is: \[ |\vec{AB}| = \sqrt{(-2)^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3. \]

Step 3: {Find \( |\vec{BC}| \)

The vector \( \vec{BC} \) is: \[ \vec{BC} = \vec{b} - \vec{r} = (-\hat{i} + \hat{j} + \hat{k}) - \frac{-5\hat{i} + 2\hat{j} + 5\hat{k}}{3}. \]
Simplify: \[ \vec{BC} = \frac{2\hat{i} - \hat{j} - 2\hat{k}}{3}. \]
The magnitude is: \[ |\vec{BC}| = \sqrt{\left(\frac{2}{3}\right)^2 + \left(\frac{-1}{3}\right)^2 + \left(\frac{-2}{3}\right)^2} = \sqrt{\frac{4}{9} + \frac{1}{9} + \frac{4}{9}} = \sqrt{\frac{9}{9}} = 1. \]

Step 4: {Find the ratio \( |\vec{AB}| : |\vec{BC}| \)
\[ |\vec{AB}| : |\vec{BC}| = 3 : 1. \]

Step 5: {Conclude the result

The position vector of \( C \) is: \[ \vec{r} = \frac{-5\hat{i} + 2\hat{j} + 5\hat{k}}{3}. \]
The ratio \( |\vec{AB}| : |\vec{BC}| \) is \( 3:1 \).
Quick Tip: To find the ratio of line segments, calculate the magnitudes of the respective vectors and simplify.

Step 1: {Recall the magnitude of the cross product

The magnitude of the cross product of \( \vec{a} \) and \( \vec{b} \) is given by: \[ |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}||\sin \theta|, \]
where \( \theta \) is the angle between \( \vec{a} \) and \( \vec{b} \).

Step 2: {Analyze the sine function

Since \( |\sin \theta| \leq 1 \), it follows that: \[ |\vec{a} \times \vec{b}| \leq |\vec{a}||\vec{b}|. \]

Step 3: {Equality condition

Equality holds when \( |\sin \theta| = 1 \), which occurs when \( \theta = \frac{\pi}{2} \). In this case, \( \vec{a} \) is perpendicular to \( \vec{b} \).

Step 4: {Conclude the result
\[ |\vec{a} \times \vec{b}| \leq |\vec{a}||\vec{b}|, \quad with equality when \( \vec{a \perp \vec{b} \)}. \] Quick Tip: The cross product achieves its maximum magnitude when the vectors are perpendicular.

Step 1: {Rearrange the given equation

The given equation is: \[ x \cos(p + y) + \cos p \sin(p + y) = 0. \]
Divide throughout by \( \cos(p + y) \): \[ x + \cos p \tan(p + y) = 0. \]
This implies: \[ \tan(p + y) = -\frac{x}{\cos p}. \]

Step 2: {Differentiate with respect to \( x \)

Differentiate both sides with respect to \( x \): \[ \sec^2(p + y) \cdot \frac{d}{dx}(p + y) = -\frac{d}{dx}\left(\frac{x}{\cos p}\right). \]
Simplify: \[ \sec^2(p + y) \frac{dy}{dx} = -\frac{1}{\cos p}. \]

Step 3: {Express \( \frac{dy}{dx} \) in terms of \( \cos^2(p + y) \)

Using \( \sec^2(p + y) = \frac{1}{\cos^2(p + y)} \), we get: \[ \frac{1}{\cos^2(p + y)} \cdot \frac{dy}{dx} = -\frac{1}{\cos p}. \]
Multiply through by \( \cos^2(p + y) \): \[ \frac{dy}{dx} = -\cos^2(p + y) \cdot \cos p. \]

Step 4: {Conclude the result
\[ \cos p \frac{dy}{dx} = -\cos^2(p + y). \] Quick Tip: For trigonometric identities involving derivatives, simplify using tangent and secant relationships before differentiating.

Step 1: {Find the limits at \( x = 2 \)

For \( x < 2 \): \[ f(x) = \frac{x - 2}{|x - 2|} + a = -1 + a. \]
For \( x > 2 \): \[ f(x) = \frac{x - 2}{|x - 2|} + b = 1 + b. \]
At \( x = 2 \): \[ f(x) = a + b. \]

Step 2: {Set up the continuity condition

For \( f(x) \) to be continuous at \( x = 2 \): \[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2). \]
Substitute the values: \[ -1 + a = 1 + b = a + b. \]

Step 3: {Solve for \( a \) and \( b \)

From \( -1 + a = 1 + b \): \[ a - b = 2. \]
From \( 1 + b = a + b \): \[ a = 1. \]
Substitute \( a = 1 \) into \( a - b = 2 \): \[ 1 - b = 2 \implies b = -1. \]

Step 4: {Conclude the result

The values of \( a \) and \( b \) are \( a = 1 \) and \( b = -1 \).
Quick Tip: For continuity of piecewise functions, equate the left-hand limit, right-hand limit, and the value of the function at the given point.

Step 1: {Find the derivative of \( f(x) \)

The given function is \( f(x) = \frac{\log x}{x} \). Differentiate using the quotient rule: \[ f'(x) = \frac{x \cdot \frac{1}{x} - \log x \cdot 1}{x^2} = \frac{1 - \log x}{x^2}. \]

Step 2: {Find critical points

For \( f'(x) = 0 \): \[ 1 - \log x = 0 \implies \log x = 1 \implies x = e. \]

Step 3: {Determine intervals of increase and decrease

For \( x \in (0, e) \): \[ 1 - \log x > 0 \implies f'(x) > 0 \quad (strictly increasing). \]
For \( x \in (e, \infty) \): \[ 1 - \log x < 0 \implies f'(x) < 0 \quad (strictly decreasing). \]

Step 4: {Conclude the result

The function is strictly increasing on \( (0, e) \) and strictly decreasing on \( (e, \infty) \).
Quick Tip: To determine monotonicity, find \( f'(x) \), solve \( f'(x) = 0 \), and test the sign of \( f'(x) \) in intervals.

Step 1: {Find the derivative of \( f(x) \)

The given function is \( f(x) = \frac{x}{2} + \frac{2}{x} \). Differentiate: \[ f'(x) = \frac{1}{2} - \frac{2}{x^2}. \]

Step 2: {Find critical points

Set \( f'(x) = 0 \): \[ \frac{1}{2} - \frac{2}{x^2} = 0 \implies \frac{2}{x^2} = \frac{1}{2} \implies x^2 = 4 \implies x = 2. \]

Step 3: {Evaluate \( f(x) \) at critical points and endpoints

At \( x = 1 \): \[ f(1) = \frac{1}{2} + \frac{2}{1} = \frac{1}{2} + 2 = \frac{5}{2}. \]
At \( x = 2 \): \[ f(2) = \frac{2}{2} + \frac{2}{2} = 1 + 1 = 2. \]

Step 4: {Conclude the result

The absolute maximum value is \( \frac{5}{2} \) at \( x = 1 \), and the absolute minimum value is \( 2 \) at \( x = 2 \).
Quick Tip: To find absolute extrema on a closed interval, evaluate the function at critical points and endpoints.

Step 1: {Break the integrand into partial fractions

Let: \[ \frac{x^2 + 1}{(x^2 + 2)(x^2 + 4)} = \frac{A}{x^2 + 2} + \frac{B}{x^2 + 4}. \]
Multiply through by \( (x^2 + 2)(x^2 + 4) \): \[ x^2 + 1 = A(x^2 + 4) + B(x^2 + 2). \]
Simplify: \[ x^2 + 1 = A x^2 + 4A + B x^2 + 2B \implies (A + B)x^2 + (4A + 2B) = x^2 + 1. \]
Equating coefficients: \[ A + B = 1, \quad 4A + 2B = 1. \]

Step 2: {Solve for \( A \) and \( B \)

From \( A + B = 1 \), \( B = 1 - A \). Substitute into \( 4A + 2B = 1 \): \[ 4A + 2(1 - A) = 1 \implies 4A + 2 - 2A = 1 \implies 2A = -1 \implies A = -\frac{1}{2}. \]
Then: \[ B = 1 - A = 1 - \left(-\frac{1}{2}\right) = \frac{3}{2}. \]

Step 3: {Write the integral in terms of partial fractions
\[ I = \int \frac{-\frac{1}{2}}{x^2 + 2} \, dx + \int \frac{\frac{3}{2}}{x^2 + 4} \, dx. \]

Step 4: {Integrate each term

The integral of \( \frac{1}{x^2 + a^2} \) is \( \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) \). Thus: \[ \int \frac{-\frac{1}{2}}{x^2 + 2} \, dx = -\frac{1}{2\sqrt{2}} \tan^{-1}\left(\frac{x}{\sqrt{2}}\right), \] \[ \int \frac{\frac{3}{2}}{x^2 + 4} \, dx = \frac{3}{4} \tan^{-1}\left(\frac{x}{2}\right). \]

Step 5: {Combine the results
\[ I = -\frac{1}{2\sqrt{2}} \tan^{-1}\left(\frac{x}{\sqrt{2}}\right) + \frac{3}{4} \tan^{-1}\left(\frac{x}{2}\right) + C. \] Quick Tip: For integrals involving quadratic terms, use partial fractions and standard inverse trigonometric integrals.

Step 1: {Simplify the integrand

Use the identity \( 1 + \cos 2x = 2\cos^2 x \) and \( \sin 2x = 2\sin x \cos x \): \[ \frac{2 + \sin 2x}{1 + \cos 2x} = \frac{2 + 2\sin x \cos x}{2\cos^2 x} = \sec^2 x + \tan x. \]

Step 2: {Rewrite the integral
\[ I = \int (\sec^2 x + \tan x) e^x \, dx. \]

Step 3: {Integrate term by term

For \( \int \sec^2 x e^x \, dx \), use substitution \( u = \tan x \): \[ \int \sec^2 x e^x \, dx = e^x \tan x. \]
For \( \int \tan x e^x \, dx \), combine it with the first term: \[ I = e^x \tan x + C. \] Quick Tip: For trigonometric integrals, simplify using identities before integrating.

Step 1: {Simplify the integrand

The given integral is: \[ I = \int_{0}^{\pi/4} \frac{1}{\sin x + \cos x} \, dx. \]
Use the identity \( \sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) \). Substituting: \[ I = \int_{0}^{\pi/4} \frac{1}{\sqrt{2} \sin\left(x + \frac{\pi}{4}\right)} \, dx = \frac{1}{\sqrt{2}} \int_{0}^{\pi/4} \csc\left(x + \frac{\pi}{4}\right) \, dx. \]

Step 2: {Integrate \( \csc(x + \pi/4) \)

The integral of \( \csc(x) \) is: \[ \int \csc x \, dx = \log|\csc x - \cot x| + C. \]
Substituting this, we have: \[ I = \frac{1}{\sqrt{2}} \left[\log\left|\csc\left(x + \frac{\pi}{4}\right) - \cot\left(x + \frac{\pi}{4}\right)\right|\right]_{0}^{\pi/4}. \]

Step 3: {Evaluate the limits

At \( x = \pi/4 \): \[ \csc\left(\frac{\pi}{4} + \frac{\pi}{4}\right) = \csc\left(\frac{\pi}{2}\right) = 1, \quad \cot\left(\frac{\pi}{4} + \frac{\pi}{4}\right) = \cot\left(\frac{\pi}{2}\right) = 0. \]
At \( x = 0 \): \[ \csc\left(0 + \frac{\pi}{4}\right) = \csc\left(\frac{\pi}{4}\right) = \sqrt{2}, \quad \cot\left(0 + \frac{\pi}{4}\right) = \cot\left(\frac{\pi}{4}\right) = 1. \]
Substitute these values: \[ I = \frac{1}{\sqrt{2}} \left[\log\left(1 - 0\right) - \log\left(\sqrt{2} - 1\right)\right] = \frac{1}{\sqrt{2}} \left[\log(1) - \log(\sqrt{2} - 1)\right]. \]

Step 4: {Simplify the result
\[ I = \frac{1}{\sqrt{2}} \log(\sqrt{2} + 1) - \frac{1}{\sqrt{2}} \log(\sqrt{2} - 1). \] Quick Tip: To evaluate integrals involving \( \sin x + \cos x \), simplify using trigonometric identities, such as expressing the sum as a single sine or cosine term.

Step 1: {Plot the constraints

The inequalities form the following lines: \[ x + y = 800, \quad 2x + y = 1000, \quad x = 400, \quad y = 0. \]
Plot these lines on a graph, and shade the region satisfying all constraints.

Step 2: {Find corner points

The feasible region is a polygon, and the vertices (corner points) are: \[ (0, 0), \quad (400, 0), \quad (400, 200), \quad (200, 600), \quad (0, 800). \]

Step 3: {Evaluate \( z = 4x + 3y \) at each corner point
\[ z(0, 0) = 4(0) + 3(0) = 0, \] \[ z(400, 0) = 4(400) + 3(0) = 1600, \] \[ z(400, 200) = 4(400) + 3(200) = 2200, \] \[ z(200, 600) = 4(200) + 3(600) = 2600, \] \[ z(0, 800) = 4(0) + 3(800) = 2400. \]

Step 4: {Conclude the result

The maximum value of \( z = 2600 \) occurs at \( (200, 600) \).
Quick Tip: For graphical solutions of linear programming problems, always evaluate the objective function at the corner points of the feasible region.

Step 1: {Assign probabilities

Let: \[ P(E_1) = \frac{4}{7}, \quad P(E_2) = \frac{1}{7}, \quad P(E_3) = \frac{2}{7}. \]
The probabilities of increased profits under each CEO are: \[ P(A | E_1) = 0.3, \quad P(A | E_2) = 0.8, \quad P(A | E_3) = 0.5. \]

Step 2: {Apply Bayes' theorem

The probability that the profits increased due to \( R \) as CEO is: \[ P(E_3 | A) = \frac{P(E_3) P(A | E_3)}{P(E_1) P(A | E_1) + P(E_2) P(A | E_2) + P(E_3) P(A | E_3)}. \]

Step 3: {Substitute the values
\[ P(E_3 | A) = \frac{\frac{2}{7} \cdot 0.5}{\frac{4}{7} \cdot 0.3 + \frac{1}{7} \cdot 0.8 + \frac{2}{7} \cdot 0.5}. \]
Simplify the denominator: \[ P(E_3 | A) = \frac{\frac{2}{7} \cdot 0.5}{\frac{4}{7} \cdot 0.3 + \frac{1}{7} \cdot 0.8 + \frac{2}{7} \cdot 0.5} = \frac{1}{3}. \] Quick Tip: Use Bayes' theorem to compute conditional probabilities by carefully analyzing the given data.

Step 1: {Check reflexivity

For any \( x \in A \): \[ x + x = 2x \quad is divisible by 2. \]
Thus, \( (x, x) \in R \), and \( R \) is reflexive.

Step 2: {Check symmetry

If \( (x, y) \in R \), then \( x + y \) is divisible by 2. Since \( x + y = y + x \), \( (y, x) \in R \). Thus, \( R \) is symmetric.

Step 3: {Check transitivity

If \( (x, y) \in R \) and \( (y, z) \in R \), then: \[ x + y is divisible by 2 and y + z is divisible by 2. \]
Adding: \[ (x + y) + (y + z) = x + z + 2y is divisible by 2. \]
Thus, \( (x, z) \in R \), and \( R \) is transitive.

Step 4: {Find equivalence class \([2]\)

Equivalence class \([2]\) contains all \( y \in A \) such that \( (2, y) \in R \). This means: \[ 2 + y is divisible by 2. \]
The elements are: \[ y = -4, -2, 0, 2, 4. \]
Thus, \([2] = \{-4, -2, 0, 2, 4\}\). Quick Tip: To check equivalence relations, verify reflexivity, symmetry, and transitivity systematically.

Step 1: {Differentiate \( f(x) \) to find critical points

The derivative is: \[ f'(x) = 4x^3 - 124x + a. \]
Since \( f(x) \) attains a local maximum at \( x = 1 \), we have: \[ f'(1) = 4(1)^3 - 124(1) + a = 0. \]
Solve for \( a \): \[ 4 - 124 + a = 0 \implies a = 120. \]

Step 2: {Find critical points

Substitute \( a = 120 \) into \( f'(x) \): \[ f'(x) = 4x^3 - 124x + 120. \]
Factorize: \[ f'(x) = 4(x - 1)(x^2 + x - 30). \]
Further factorize: \[ f'(x) = 4(x - 1)(x - 5)(x + 6). \]
The critical points are \( x = -6, 1, 5 \).

Step 3: {Determine the nature of critical points using \( f''(x) \)

The second derivative is: \[ f''(x) = 12x^2 - 124. \]
Evaluate \( f''(x) \) at each critical point: \[ f''(-6) = 12(-6)^2 - 124 = 432 - 124 = 308 > 0 \quad (local minimum at \( x = -6 \)). \] \[ f''(1) = 12(1)^2 - 124 = 12 - 124 = -112 < 0 \quad (local maximum at \( x = 1 \)). \] \[ f''(5) = 12(5)^2 - 124 = 300 - 124 = 176 > 0 \quad (local minimum at \( x = 5 \)). \]

Step 4: {Conclusion

The function \( f(x) \) attains: \[ Local maximum at x = 1, \quad local minima at x = -6, 5. \] Quick Tip: To identify the nature of critical points, use the second derivative test. If \( f''(x) > 0 \), it's a local minimum; if \( f''(x) < 0 \), it's a local maximum.

Step 1: {Define the variables

Let the length of the rectangle be \( x \) cm and the breadth be \( (150 - x) \) cm (since the perimeter is \( 2(x + breadth) = 300 \)).

When the rectangle is rolled along its length, the radius \( r \) and height \( h \) of the cylinder formed are: \[ 2\pi r = x \implies r = \frac{x}{2\pi}, \quad h = (150 - x). \]

Step 2: {Write the volume of the cylinder

The volume \( V \) of the cylinder is given by: \[ V = \pi r^2 h = \pi \left(\frac{x}{2\pi}\right)^2 (150 - x). \]
Simplify: \[ V = \pi \frac{x^2}{4\pi^2} (150 - x) = \frac{x^2}{4\pi} (150 - x). \]

Step 3: {Differentiate \( V \) with respect to \( x \)

The derivative of \( V \) is: \[ \frac{dV}{dx} = \frac{1}{4\pi} \left(2x(150 - x) - x^2\right). \]
Simplify: \[ \frac{dV}{dx} = \frac{1}{4\pi} \left(300x - 3x^2\right). \]

Step 4: {Set \( \frac{dV}{dx} = 0 \) to find critical points
\[ 300x - 3x^2 = 0 \implies x(300 - 3x) = 0 \implies x = 0 or x = 100. \]

Step 5: {Use the second derivative test to confirm maxima

The second derivative is: \[ \frac{d^2V}{dx^2} = \frac{1}{4\pi} (300 - 6x). \]
At \( x = 100 \): \[ \frac{d^2V}{dx^2} = \frac{1}{4\pi} (300 - 600) = \frac{-300}{4\pi} = \frac{-75}{\pi} < 0. \]
Hence, \( V \) is maximum when \( x = 100 \).

Step 6: {Calculate the dimensions of the rectangle

When \( x = 100 \), the length is \( 100 \) cm, and the breadth is: \[ 150 - x = 50 \, cm. \] Quick Tip: For optimization problems, write the expression for the quantity to be optimized, differentiate it, and use the second derivative test to confirm maxima or minima.

Step 1: {Rewrite the equation of the circle

The equation \( x^2 + y^2 = 16 \) can be rewritten as: \[ y = \pm\sqrt{16 - x^2}. \]

Step 2: {Set up the integral

The required area is: \[ A = 2 \int_{-2}^{2} \sqrt{16 - x^2} \, dx. \]

Step 3: {Solve the integral

Use the trigonometric substitution \( x = 4\sin\theta \), \( dx = 4\cos\theta \, d\theta \), and bounds \( x = -2 \) to \( x = 2 \) correspond to \( \theta = -\pi/6 \) to \( \pi/6 \): \[ \int_{-2}^{2} \sqrt{16 - x^2} \, dx = 2 \cdot \frac{1}{2} \int_{-\pi/6}^{\pi/6} 16 \, d\theta = \frac{16}{2} \int_{-\pi/6}^{\pi/6} \cos\theta \, d\theta. \]

Evaluate: \[ A = 8\sqrt{3} + \frac{16\pi}{3}. \] Quick Tip: For areas under curves, set up definite integrals and use symmetry to simplify computations.

The equation of lines \( l_1 \) and \( l_2 \) are:
\[ l_1: \frac{x-1}{1} = \frac{y-1}{2} = \frac{z-2}{3} = \lambda, \quad l_2: \frac{x-1}{0} = \frac{y-3}{-3} = \frac{z-7}{2} = \mu \]

Any point on \( l_1 \) is \( (\lambda, 2\lambda + 1, 3\lambda + 2) \) and any point on \( l_2 \) is \( (1, -3\mu, 2\mu + 7) \).
If \( l_1 \) and \( l_2 \) intersect, then we solve for \( \lambda \) and \( \mu \):
\[ \lambda = 1, \quad 2\lambda + 1 = -3\lambda + 2 \quad and \quad 3\lambda + 2 = 2\mu + 7 \]

Solving the system:
\[ \lambda = 1 \quad and \quad \mu = -1 \]

The point of intersection of \( l_1 \) and \( l_2 \) is \( (1, 3, 5) \).

Let the direction ratios of the required line be \( \). Then,
\[ a + 2b + 3c = 0 \quad and \quad -3b + 2c = 0 \]

Thus, we have:
\[ \frac{a}{13} = \frac{b}{-2} = \frac{c}{-3} \]

The required equation of the line is:
\[ \frac{x-1}{13} = \frac{y-3}{-2} = \frac{z-5}{-3} \] Quick Tip: To find the line perpendicular to two given lines, use the cross product of their direction vectors to obtain the direction ratios.

Step 1: {Find direction ratios of \( AB \) and \( CD \)

For \( AB \): \[ \vec{AB} = \langle 1 - (-1), -2 - 2, 5 - 1 \rangle = \langle 2, -4, 4 \rangle. \]
For \( CD \): \[ Direction ratios of \( CD \) = \langle 1, -2, 2 \rangle. \]

Step 2: {Find the shortest distance between \( AB \) and \( CD \)

The equation of \( AB \) is: \[ \frac{x + 1}{2} = \frac{y - 2}{-4} = \frac{z - 1}{4}. \]
Take points \( A(-1, 2, 1) \) and \( C(4, -7, 8) \). Let: \[ \vec{AC} = \langle 4 - (-1), -7 - 2, 8 - 1 \rangle = \langle 5, -9, 7 \rangle. \]
The shortest distance is: \[ d = \frac{| \vec{AC} \cdot (\vec{AB} \times \vec{CD}) |}{| \vec{AB} \times \vec{CD} |}. \]
Compute \( \vec{AB} \times \vec{CD} \): \[ \vec{AB} \times \vec{CD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}
2 & -4 & 4
1 & -2 & 2 \end{vmatrix} = \hat{i}(8 - (-8)) - \hat{j}(4 - 4) + \hat{k}(-4 - (-4)). \] \[ \vec{AB} \times \vec{CD} = \langle 16, 0, 0 \rangle. \]
Thus, the shortest distance is: \[ d = \frac{\sqrt{26}}{3}. \]

Step 3: {Calculate the area of parallelogram ABCD

The area is: \[ Area = Base \times Height = 6 \times \frac{\sqrt{26}}{3} = 2\sqrt{26}. \] Quick Tip: The shortest distance between two skew lines can be calculated using the cross product of their direction ratios and a vector joining any point on one line to the other.