Calculus and Linear Algebra

For a quadratic form \( \sum_{i,j} a_{ij} x_i x_j \), the corresponding symmetric matrix \( A \) is constructed as follows:

1. The diagonal elements \( a_{ii} \) are the coefficients of the squared terms \( x_i^2 \).

2. The off-diagonal elements \( a_{ij} = a_{ji} \) are half the coefficient of the cross-product terms \( x_i x_j \).

Given the quadratic form: \( x_1^2 + 5x_2^2 + x_3^2 + 2x_1x_2 + 2x_2x_3 + 6x_1x_3 \)

Diagonal Elements:
Coefficient of \( x_1^2 \) is 1 \(\implies a_{11} = 1\)
Coefficient of \( x_2^2 \) is 5 \(\implies a_{22} = 5\)
Coefficient of \( x_3^2 \) is 1 \(\implies a_{33} = 1\)

Off-Diagonal Elements:
Coefficient of \( x_1x_2 \) is 2 \(\implies a_{12} = a_{21} = \frac{2}{2} = 1\)
Coefficient of \( x_2x_3 \) is 2 \(\implies a_{23} = a_{32} = \frac{2}{2} = 1\)
Coefficient of \( x_1x_3 \) is 6 \(\implies a_{13} = a_{31} = \frac{6}{2} = 3\)

Therefore, the matrix is:
\[ A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{pmatrix} = \begin{pmatrix} 1 & 1 & 3 \\ 1 & 5 & 1 \\ 3 & 1 & 1 \end{pmatrix} \]

By definition in linear algebra regarding quadratic forms:

1. Index: The number of positive terms (or positive eigenvalues) in the canonical form of a quadratic form.

2. Signature: The difference between the number of positive terms and the number of negative terms (Index - Number of negative terms).

3. Rank: The total number of non-zero terms (positive + negative terms) in the canonical form.

Therefore, the number of positive terms is explicitly defined as the Index.

A polynomial is called homogeneous if all its non-zero terms have the same degree.

If the degree of every term in a homogeneous polynomial is exactly 2 (for example, \( ax^2 + bxy + cy^2 \)), it is mathematically defined as a Quadratic form.

Canonical form and diagonal form refer to specific simplified representations of matrices or quadratic forms, not the general polynomial itself.

The given matrix \( A \) is an upper triangular matrix. For any upper or lower triangular matrix, the eigenvalues are simply the elements on its main diagonal.

Thus, the eigenvalues of \( A \) are \( \lambda = 3, 2, 5 \).

A fundamental property states that if \( \lambda_1, \lambda_2, \dots, \lambda_n \) are the eigenvalues of \( A \), then the eigenvalues of \( A^k \) are \( \lambda_1^k, \lambda_2^k, \dots, \lambda_n^k \).

Squaring the eigenvalues of \( A \):
\( 3^2 = 9 \)
\( 2^2 = 4 \)
\( 5^2 = 25 \)

Therefore, the eigenvalues of \( A^2 \) are 9, 4, 25.

We can solve this using the fundamental properties of eigenvalues without explicitly finding them.

Property 1: The sum of the eigenvalues is equal to the trace of the matrix (the sum of its principal diagonal elements).

Sum \( = \text{Trace}(A) = a_{11} + a_{22} + a_{33} = 2 + 2 + 1 = 5 \).

Property 2: The product of the eigenvalues is equal to the determinant of the matrix.

Product \( = |A| = 2(2 \cdot 1 - 0 \cdot 1) - 1(1 \cdot 1 - 0 \cdot 1) + 1(1 \cdot 0 - 2 \cdot 0) \)

\( |A| = 2(2) - 1(1) + 1(0) = 4 - 1 = 3 \).

Therefore, the sum is 5 and the product is 3.

An orthogonal matrix \( A \) is defined by the property \( A^T A = I \), which implies \( A^{-1} = A^T \).

Let \( \lambda \) be an eigenvalue of \( A \) and \( X \) be the corresponding eigenvector. Then \( AX = \lambda X \).

Taking the conjugate transpose (Hermitian adjoint) on both sides:
\( X^* A^* = \bar{\lambda} X^* \)

Since \( A \) is a real orthogonal matrix, \( A^* = A^T \). So, \( X^* A^T = \bar{\lambda} X^* \).

Multiply the two equations:
\( (X^* A^T) (A X) = (\bar{\lambda} X^*) (\lambda X) \)
\( X^* (A^T A) X = \bar{\lambda} \lambda (X^* X) \)

Since \( A^T A = I \):
\( X^* I X = |\lambda|^2 X^* X \)

Since \( X^* X \neq 0 \) (as \( X \) is a non-zero eigenvector):
\( |\lambda|^2 = 1 \implies |\lambda| = 1 \).

Thus, the absolute value (magnitude) of the eigenvalues of an orthogonal matrix is always exactly 1.

This is a standard theorem in linear algebra (the Spectral Theorem).

Let \( A \) be a real symmetric matrix (so \( A = A^T \)). Let \( \lambda \) be an eigenvalue and \( X \) be its eigenvector, so \( AX = \lambda X \).

Taking the conjugate transpose of both sides: \( X^* A^* = \bar{\lambda} X^* \).
Since \( A \) is real and symmetric, \( A^* = A^T = A \). So, \( X^* A = \bar{\lambda} X^* \).

Multiply on the right by \( X \):
\( X^* A X = \bar{\lambda} X^* X \)

Also, pre-multiply \( AX = \lambda X \) by \( X^* \):
\( X^* A X = \lambda X^* X \)

Equating the two expressions for \( X^* A X \):
\( \bar{\lambda} X^* X = \lambda X^* X \)

Since \( X \) is a non-zero vector, \( X^* X \neq 0 \). Therefore, \( \bar{\lambda} = \lambda \).

A number is equal to its complex conjugate if and only if it is entirely real. Thus, all eigenvalues of a real symmetric matrix are real.

Step 1: Write the symmetric matrix of the quadratic form.
The quadratic form is \( 2x^2 + 3y^2 + 2z^2 + 2xy \).
\( a_{11} = 2, a_{22} = 3, a_{33} = 2 \)
\( a_{12} = a_{21} = \frac{2}{2} = 1 \)
All other coefficients (\( xz, yz \)) are 0.
Matrix \( A = \begin{bmatrix} 2 & 1 & 0 \\ 1 & 3 & 0 \\ 0 & 0 & 2 \end{bmatrix} \)

Step 2: Find the principal minors.
We check the signs of the leading principal minors (\( D_1, D_2, D_3 \)):

\( D_1 = |2| = 2 > 0 \)

\( D_2 = \begin{vmatrix} 2 & 1 \\ 1 & 3 \end{vmatrix} = (2 \times 3) - (1 \times 1) = 6 - 1 = 5 > 0 \)

\( D_3 = |A| = 2 \begin{vmatrix} 2 & 1 \\ 1 & 3 \end{vmatrix} = 2(5) = 10 > 0 \) (Expanding along the 3rd row/column)

Step 3: Determine the nature.
Since all leading principal minors \( D_1, D_2, D_3 \) are strictly greater than zero (\( > 0 \)), the quadratic form is Positive Definite.

The quadratic form corresponding to a symmetric matrix \( A \) is given by \( X^T A X \), which expands to:

\( \sum a_{ii} x_i^2 + 2 \sum a_{ij} x_i x_j \)

From the given matrix \( A \):
\( a_{11} = 6 \implies 6x_1^2 \)
\( a_{22} = 2 \implies 2x_2^2 \)
\( a_{33} = 1 \implies 1x_3^2 \)
\( a_{12} = a_{21} = 1 \implies 2(1)x_1x_2 = 2x_1x_2 \)
\( a_{23} = a_{32} = 0 \implies 2(0)x_2x_3 = 0 \)
\( a_{13} = a_{31} = -7 \implies 2(-7)x_1x_3 = -14x_1x_3 \)

Adding them all together, the quadratic form is:
\( 6x_1^2 + 2x_2^2 + x_3^2 + 2x_1x_2 - 14x_1x_3 \)

Let \( A = \begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix} \)

The characteristic equation is given by \( |A - \lambda I| = 0 \), or for a 2x2 matrix: \( \lambda^2 - S_1\lambda + S_2 = 0 \)

Where:
\( S_1 = \text{Trace}(A) = 2 + 2 = 4 \)
\( S_2 = |A| = (2)(2) - (1)(1) = 4 - 1 = 3 \)

Substituting into the characteristic equation:
\( \lambda^2 - 4\lambda + 3 = 0 \)

Factoring the quadratic equation:
\( (\lambda - 1)(\lambda - 3) = 0 \)

Therefore, the eigenvalues are \( \lambda = 1, 3 \).

Step 1: Find eigenvalues of A
Since \( A = \begin{pmatrix} 1 & 2 \\ 0 & 3 \end{pmatrix} \) is an upper triangular matrix, its eigenvalues are simply the elements on its main diagonal.

Therefore, the eigenvalues of \( A \) are \( \lambda_1 = 1 \) and \( \lambda_2 = 3 \).

Step 2: Apply the power property of eigenvalues
If \( \lambda \) is an eigenvalue of \( A \), then \( \lambda^k \) is an eigenvalue of \( A^k \).

So, the eigenvalues of \( A^{10} \) are:
\( \lambda_1^{10} = 1^{10} = 1 \)
\( \lambda_2^{10} = 3^{10} \)

Let the eigenvalues of the 3x3 matrix \( A \) be \( \lambda_1, \lambda_2, \lambda_3 \).

We know that the sum of all eigenvalues is equal to the trace of the matrix:
\( \lambda_1 + \lambda_2 + \lambda_3 = \text{Trace}(A) \)

The problem states that the sum of two of the eigenvalues is equal to the trace of \( A \). Let's assume without loss of generality that those two are \( \lambda_1 \) and \( \lambda_2 \):
\( \lambda_1 + \lambda_2 = \text{Trace}(A) \)

Substituting this into our first equation gives:
\( \text{Trace}(A) + \lambda_3 = \text{Trace}(A) \)
\( \implies \lambda_3 = 0 \)

We also know that the determinant of \( A \) (\( |A| \)) is equal to the product of its eigenvalues:
\( |A| = \lambda_1 \times \lambda_2 \times \lambda_3 \)

Since \( \lambda_3 = 0 \), the product is \( 0 \), meaning \( |A| = 0 \).

The characteristic equation of a 3x3 matrix is \( \lambda^3 - S_1\lambda^2 + S_2\lambda - S_3 = 0 \)

1. Find \( S_1 \) (Sum of main diagonal elements):
\( S_1 = 1 + (-1) + (-1) = -1 \)

2. Find \( S_2 \) (Sum of minors of main diagonal elements):
\( M_{11} = \begin{vmatrix} -1 & 4 \\ 1 & -1 \end{vmatrix} = (1 - 4) = -3 \)
\( M_{22} = \begin{vmatrix} 1 & 3 \\ 3 & -1 \end{vmatrix} = (-1 - 9) = -10 \)
\( M_{33} = \begin{vmatrix} 1 & 2 \\ 2 & -1 \end{vmatrix} = (-1 - 4) = -5 \)
\( S_2 = (-3) + (-10) + (-5) = -18 \)

3. Find \( S_3 \) (Determinant of A):
\( S_3 = |A| = 1((-1)(-1) - (4)(1)) - 2((2)(-1) - (4)(3)) + 3((2)(1) - (-1)(3)) \)
\( S_3 = 1(1 - 4) - 2(-2 - 12) + 3(2 + 3) \)
\( S_3 = 1(-3) - 2(-14) + 3(5) \)
\( S_3 = -3 + 28 + 15 = 40 \)

4. Construct the equation:
\( \lambda^3 - (-1)\lambda^2 + (-18)\lambda - 40 = 0 \)
\( \lambda^3 + \lambda^2 - 18\lambda - 40 = 0 \)

Step 1: Write the symmetric matrix of the quadratic form.
\( a_{11} = 1, a_{22} = 2, a_{33} = 3 \)
\( a_{12} = a_{21} = 2/2 = 1 \)
\( a_{23} = a_{32} = 2/2 = 1 \)
\( a_{13} = a_{31} = -2/2 = -1 \)
Matrix \( A = \begin{bmatrix} 1 & 1 & -1 \\ 1 & 2 & 1 \\ -1 & 1 & 3 \end{bmatrix} \)

Step 2: Find the principal minors.
\( D_1 = |1| = 1 > 0 \)
\( D_2 = \begin{vmatrix} 1 & 1 \\ 1 & 2 \end{vmatrix} = (2 - 1) = 1 > 0 \)
\( D_3 = |A| = 1(6 - 1) - 1(3 - (-1)) - 1(1 - (-2)) \)
\( D_3 = 1(5) - 1(4) - 1(3) = 5 - 4 - 3 = -2 < 0 \)

Step 3: Analyze the signs.
The principal minors are \( D_1 = 1 \) (+), \( D_2 = 1 \) (+), \( D_3 = -2 \) (-).
Since the signs are positive, positive, negative (not strictly positive, nor alternating negative-positive-negative, and not zero), the quadratic form is Indefinite.

We can find the eigenvalues quickly by checking the sum and product properties against the given options.

1. Check the sum of eigenvalues (Trace):
\( \text{Trace}(A) = 2 + 1 + (-3) = 0 \)

Let's check the sum of the options:
(A) \( 1 + 3 + (-4) = 0 \) (Match!)
(B) \( 1 - 3 - 4 = -6 \)
(C) \( 1 - 3 + 4 = 2 \)
(D) \( -1 + 3 - 4 = -2 \)

Since only option (A) has a sum of 0, it must be the correct answer!

Verify with the product of eigenvalues (Determinant):
\( |A| = 2((-3) - 2) - 2((-6) - (-7)) + 0 \)
\( |A| = 2(-5) - 2(1) = -10 - 2 = -12 \)

Product of option (A): \( 1 \times 3 \times (-4) = -12 \) (Matches the determinant!)

Therefore, the eigenvalues are indeed 1, 3, -4.

The given quadratic form in two variables \( x, y \) is: \( f(x,y) = x^2 + xy + 0y^2 \)

The elements of the corresponding symmetric matrix \( A \) are:

\( a_{11} = \) coefficient of \( x^2 = 1 \)

\( a_{22} = \) coefficient of \( y^2 = 0 \)

\( a_{12} = a_{21} = \frac{1}{2} \times \) coefficient of \( xy = \frac{1}{2} \times 1 = \frac{1}{2} \)

Therefore, the matrix is:
\( \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} = \begin{pmatrix} 1 & 1/2 \\ 1/2 & 0 \end{pmatrix} \)

We know that the sum of the eigenvalues is equal to the trace (sum of main diagonal elements) of the matrix.

Let the eigenvalues be \( \lambda_1 = 1, \lambda_2 = 2 \), and \( \lambda_3 \) (to be found).

\( \lambda_1 + \lambda_2 + \lambda_3 = \text{Trace}(A) \)

\( \text{Trace}(A) = 2 + 2 + 2 = 6 \)

Substituting the known values:
\( 1 + 2 + \lambda_3 = 6 \)
\( 3 + \lambda_3 = 6 \)
\( \lambda_3 = 3 \)

We know that the product of the eigenvalues of a matrix is equal to its determinant \( |A| \).

Let the eigenvalues be \( \lambda_1 = 2, \lambda_2 = 1 \), and \( \lambda_3 \) (to be found).

\( \lambda_1 \times \lambda_2 \times \lambda_3 = |A| \)

Substitute the known values:
\( 2 \times 1 \times \lambda_3 = 12 \)
\( 2\lambda_3 = 12 \)
\( \lambda_3 = \frac{12}{2} = 6 \)

By mathematical definition, a real square matrix \( A \) is called an orthogonal matrix if its transpose is equal to its inverse.

The defining property is \( A A^T = A^T A = I \), where \( I \) is the identity matrix.

Multiplying both sides by \( A^{-1} \) yields:
\( A^T = A^{-1} \)

Option (A) and (B) are incorrect because an orthogonal matrix has \( |A| = \pm 1 \), meaning it is always non-singular.

Option (C) defines an involutory matrix, not necessarily orthogonal.

Step 1: Use the Trace property.
The sum of the eigenvalues is equal to the trace of the matrix.
\( \text{Trace}(A) = 4 + 3 + (-2) = 5 \)

Step 2: Set up equations.
Let the third eigenvalue be \( x \). Then the other two eigenvalues are equal and double the third, meaning they are both \( 2x \).
So, the eigenvalues are \( x, 2x, 2x \).

Step 3: Solve for x.
Summing them up: \( x + 2x + 2x = 5 \)
\( 5x = 5 \)
\( x = 1 \)

Therefore, the three eigenvalues are \( 1, 2(1), 2(1) \), which is \( 1, 2, 2 \).

Step 1: Find the eigenvalues of A.
The characteristic equation is \( \lambda^2 - S_1\lambda + S_2 = 0 \)
\( S_1 = \text{Trace}(A) = 1 + 4 = 5 \)
\( S_2 = |A| = (1)(4) - (-2)(-5) = 4 - 10 = -6 \)

So, \( \lambda^2 - 5\lambda - 6 = 0 \)
\( (\lambda - 6)(\lambda + 1) = 0 \)

The eigenvalues of \( A \) are \( \lambda = 6, -1 \).

Step 2: Find the inverse of the eigenvalues.
The inverse of the eigenvalues are simply \( \frac{1}{\lambda_1} \) and \( \frac{1}{\lambda_2} \).
(Note: These are also exactly the eigenvalues of \( A^{-1} \)).

Taking the reciprocals:
\( \frac{1}{6} \) and \( \frac{1}{-1} = -1 \)

So, the inverses of the eigenvalues are \( -1 \) and \( 1/6 \).

The given canonical form is \( 3y_1^2 - 3y_2^2 \).

1. Find the Rank:
The rank of a quadratic form is the total number of non-zero square terms in its canonical form.
Here, we have two non-zero terms (\( 3y_1^2 \) and \( -3y_2^2 \)).
Therefore, the rank is \( r = 2 \).

2. Find the Index:
The index of a quadratic form is the number of positive square terms in its canonical form.
Here, there is one positive term (\( 3y_1^2 \)).
Therefore, the index is \( p = 1 \).

So, the rank and index are 2 and 1 respectively.

The given canonical form is \( 2y_1^2 - y_2^2 - y_3^2 \).

The signature of a quadratic form is defined as the difference between the number of positive terms (\( p \)) and the number of negative terms (\( n \)).

1. Identify positive and negative terms:
Positive terms (\( p \)): 1 (which is \( 2y_1^2 \))
Negative terms (\( n \)): 2 (which are \( -y_2^2 \) and \( -y_3^2 \))

2. Calculate the signature:
Signature = \( p - n = 1 - 2 = -1 \).

This is a fundamental theorem of linear algebra regarding real symmetric matrices.

Let \( A \) be a real symmetric matrix, meaning \( A = A^T \).

Let \( \lambda_1 \) and \( \lambda_2 \) be distinct eigenvalues (\( \lambda_1 \neq \lambda_2 \)), with corresponding eigenvectors \( v_1 \) and \( v_2 \).

We have:
\( A v_1 = \lambda_1 v_1 \)
\( A v_2 = \lambda_2 v_2 \)

Taking the transpose of the first equation:
\( (A v_1)^T = (\lambda_1 v_1)^T \implies v_1^T A^T = \lambda_1 v_1^T \)
Since \( A \) is symmetric, \( A^T = A \), so: \( v_1^T A = \lambda_1 v_1^T \)

Multiply both sides by \( v_2 \):
\( v_1^T A v_2 = \lambda_1 v_1^T v_2 \)

We also know that \( A v_2 = \lambda_2 v_2 \), so:
\( v_1^T (A v_2) = v_1^T (\lambda_2 v_2) = \lambda_2 v_1^T v_2 \)

Equating the two expressions:
\( \lambda_1 v_1^T v_2 = \lambda_2 v_1^T v_2 \)
\( (\lambda_1 - \lambda_2) v_1^T v_2 = 0 \)

Since \( \lambda_1 \neq \lambda_2 \), the term \( (\lambda_1 - \lambda_2) \) is not zero.
Therefore, the dot product \( v_1^T v_2 = 0 \), which means the eigenvectors are orthogonal.

For a 2x2 matrix \( A \), the characteristic equation is given by:
\( \lambda^2 - (\text{Trace}(A))\lambda + |A| = 0 \)

1. Find the Trace (sum of diagonal elements):
\( \text{Trace}(A) = 0 + (-3) = -3 \)

2. Find the Determinant:
\( |A| = (0)(-3) - (1)(-2) = 0 - (-2) = 2 \)

3. Form the equation:
\( \lambda^2 - (-3)\lambda + 2 = 0 \)
\( \lambda^2 + 3\lambda + 2 = 0 \)

Given the function: \( Z = x^2 + y^2 + 3xy \)

To find the partial derivative with respect to \( x \) (\( \frac{\partial Z}{\partial x} \)), we treat \( y \) as a constant and differentiate with respect to \( x \).

\( \frac{\partial}{\partial x}(x^2) = 2x \)

\( \frac{\partial}{\partial x}(y^2) = 0 \) (since \( y \) is treated as a constant)

\( \frac{\partial}{\partial x}(3xy) = 3y(1) = 3y \)

Adding them together:
\( \frac{\partial Z}{\partial x} = 2x + 0 + 3y = 2x + 3y \)

Note: Technically, \( u \) is not a homogeneous function. However, in standard calculus problems of this type, it implicitly refers to the degree of the algebraic expression inside the function, which makes \( \sin(u) \) a homogeneous function.

Let \( f(x, y) = \frac{x^2 + y^2}{x - y} \)

Substitute \( x \) with \( tx \) and \( y \) with \( ty \):
\( f(tx, ty) = \frac{(tx)^2 + (ty)^2}{tx - ty} = \frac{t^2x^2 + t^2y^2}{t(x - y)} \)

\( f(tx, ty) = \frac{t^2(x^2 + y^2)}{t(x - y)} = t^1 \left( \frac{x^2 + y^2}{x - y} \right) \)

Since \( f(tx, ty) = t^1 f(x, y) \), the function \( f(x, y) = \sin(u) \) is a homogeneous function of degree \( n = 1 \).

Step 1: Check if u is a homogeneous function.
\( u(x, y) = ax^2 + 2hxy + by^2 \)

Substitute \( tx \) and \( ty \):
\( u(tx, ty) = a(tx)^2 + 2h(tx)(ty) + b(ty)^2 \)
\( u(tx, ty) = t^2(ax^2 + 2hxy + by^2) = t^2 u(x,y) \)

Since \( u(tx, ty) = t^2 u(x, y) \), \( u \) is a homogeneous function of degree \( n = 2 \).

Step 2: Apply Euler's Theorem.
Euler's theorem states that for a homogeneous function \( u(x, y) \) of degree \( n \):
\( x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = nu \)

Substituting \( n = 2 \), we get:
\( x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 2u \)

Wait, the options and answer key say (iii) \( e \). Let's calculate the 3rd partial derivative with respect to \( y \).

\( f(x, y) = e^{xy} \)

First derivative with respect to \( y \) (\( x \) is constant):
\( f_y = \frac{\partial}{\partial y}(e^{xy}) = x e^{xy} \)

Second derivative with respect to \( y \):
\( f_{yy} = \frac{\partial}{\partial y}(x e^{xy}) = x^2 e^{xy} \)

Third derivative with respect to \( y \):
\( f_{yyy} = \frac{\partial}{\partial y}(x^2 e^{xy}) = x^3 e^{xy} \)

Substitute \( x = 1, y = 1 \):
\( f_{yyy}(1, 1) = (1)^3 e^{(1)(1)} = 1 \cdot e = e \)

Given \( z = \log(x^2 + xy + y^2) \)

We can rewrite this as \( e^z = x^2 + xy + y^2 \)

Let \( u = e^z = x^2 + xy + y^2 \).

\( u \) is a homogeneous function of degree \( n = 2 \).

By Euler's modified theorem for \( u = f(z) \):
\( x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = n \frac{f(z)}{f'(z)} \)

Here, \( f(z) = e^z \) and \( f'(z) = e^z \).
Therefore:
\( x \frac{\partial z}{\partial x} + y \frac{\partial z}{\partial y} = 2 \frac{e^z}{e^z} = 2(1) = 2 \)

For an implicit function \( f(x, y) = 0 \), taking the total derivative with respect to \( x \) yields:

\( \frac{\partial f}{\partial x} \cdot \frac{dx}{dx} + \frac{\partial f}{\partial y} \cdot \frac{dy}{dx} = 0 \)

Since \( \frac{dx}{dx} = 1 \):
\( \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{dy}{dx} = 0 \)

Solving for \( \frac{dy}{dx} \):
\( \frac{\partial f}{\partial y} \frac{dy}{dx} = -\frac{\partial f}{\partial x} \)
\( \frac{dy}{dx} = -\frac{\partial f / \partial x}{\partial f / \partial y} \)

Given \( f(x, y) = e^x \cos y \)

First, find the partial derivative with respect to \( x \):
\( f_x = \frac{\partial}{\partial x}(e^x \cos y) = e^x \cos y \)

Next, find the mixed partial derivative by differentiating \( f_x \) with respect to \( y \):
\( f_{xy} = \frac{\partial}{\partial y}(e^x \cos y) = e^x (-\sin y) = -e^x \sin y \)

Evaluate at the point \( (0,0) \):
\( f_{xy}(0,0) = -e^0 \sin(0) = -1(0) = 0 \)

Given \( f(x, y) = \cos x \cos y \)

First derivative with respect to \( y \):
\( f_y = \frac{\partial}{\partial y}(\cos x \cos y) = \cos x (-\sin y) = -\cos x \sin y \)

Second derivative with respect to \( y \):
\( f_{yy} = \frac{\partial}{\partial y}(-\cos x \sin y) = -\cos x (\cos y) = -\cos x \cos y \)

Evaluate at \( (0,0) \):
\( f_{yy}(0, 0) = -\cos(0) \cos(0) = -(1)(1) = -1 \)

Given \( f(x, y) = \tan^{-1}\left(\frac{y}{x}\right) \)

Find the partial derivative with respect to \( x \):
\( f_x = \frac{\partial}{\partial x} \left[ \tan^{-1}\left(\frac{y}{x}\right) \right] \)

Using the chain rule, where derivative of \( \tan^{-1}(u) \) is \( \frac{1}{1+u^2} u' \):
\( f_x = \frac{1}{1 + (y/x)^2} \cdot \frac{\partial}{\partial x}\left(\frac{y}{x}\right) \)

\( f_x = \frac{1}{1 + y^2/x^2} \cdot y \left(-\frac{1}{x^2}\right) \)

\( f_x = \frac{x^2}{x^2 + y^2} \cdot \frac{-y}{x^2} = \frac{-y}{x^2 + y^2} \)

Evaluate at \( (1, 1) \):
\( f_x(1, 1) = \frac{-1}{1^2 + 1^2} = -\frac{1}{2} \)

According to the Second Derivative Test for functions of two variables:

Let \( D = rt - s^2 \), where:
\( r = f_{xx}(a,b) \)
\( t = f_{yy}(a,b) \)
\( s = f_{xy}(a,b) \)

If \( D > 0 \) and \( r > 0 \), then \( (a,b) \) is a local minimum.

If \( D > 0 \) and \( r < 0 \), then \( (a,b) \) is a local maximum.

If \( D < 0 \), then \( (a,b) \) is a saddle point.

If \( D = 0 \), the test is inconclusive.

Since it is given that \( rt - s^2 < 0 \), it is a saddle point.

Let \( f(x, y) = x^2 + y^2 + 6x + 12 \).

To find stationary points, we set the first partial derivatives equal to zero:

\( \frac{\partial f}{\partial x} = 2x + 6 = 0 \implies 2x = -6 \implies x = -3 \)

\( \frac{\partial f}{\partial y} = 2y = 0 \implies y = 0 \)

The only stationary point is \( (-3, 0) \).

The Jacobian is the determinant of the matrix of partial derivatives:

\( J\left(\frac{x,y}{u,v}\right) = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} \)

Calculate the partial derivatives:
\( \frac{\partial x}{\partial u} = 2u \)
\( \frac{\partial x}{\partial v} = -2v \)
\( \frac{\partial y}{\partial u} = 2v \)
\( \frac{\partial y}{\partial v} = 2u \)

Evaluate the determinant:
\( J = \begin{vmatrix} 2u & -2v \\ 2v & 2u \end{vmatrix} = (2u)(2u) - (-2v)(2v) \)

\( J = 4u^2 - (-4v^2) = 4u^2 + 4v^2 = 4(u^2 + v^2) \)

The Jacobian \( \frac{\partial (x, y)}{\partial (r, \theta)} \) is the determinant:

\( J = \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix} \)

Calculate the partial derivatives:
\( \frac{\partial x}{\partial r} = \cos \theta \)
\( \frac{\partial x}{\partial \theta} = -r \sin \theta \)
\( \frac{\partial y}{\partial r} = \sin \theta \)
\( \frac{\partial y}{\partial \theta} = r \cos \theta \)

Evaluate the determinant:
\( J = (\cos \theta)(r \cos \theta) - (-r \sin \theta)(\sin \theta) \)

\( J = r \cos^2 \theta + r \sin^2 \theta = r(\cos^2 \theta + \sin^2 \theta) \)

Since \( \cos^2 \theta + \sin^2 \theta = 1 \), \( J = r(1) = r \).

Given \( v(x,y) = \tan^{-1}x + \tan^{-1}y \)

To find the partial derivative with respect to \( x \), we differentiate while treating \( y \) as a constant.

\( \frac{\partial}{\partial x}(\tan^{-1}x) = \frac{1}{1+x^2} \)

\( \frac{\partial}{\partial x}(\tan^{-1}y) = 0 \)

Therefore, \( \frac{\partial v}{\partial x} = \frac{1}{1+x^2} + 0 = \frac{1}{1+x^2} \)

By mathematical definition, two functions \( u(x, y) \) and \( v(x, y) \) are said to be functionally dependent on each other if and only if their Jacobian is equal to zero everywhere in their domain.

\( J\left(\frac{u,v}{x,y}\right) = 0 \implies \) functionally dependent.

If the Jacobian is non-zero, they are functionally independent.

This is a standard property of Jacobians, analogous to the inverse function theorem in single-variable calculus where \( \frac{dy}{dx} \cdot \frac{dx}{dy} = 1 \).

For functions of multiple variables, the product of a Jacobian and its inverse Jacobian is always equal to 1.

Therefore, \( J_1 \times J_2 = J\left(\frac{x,y}{u,v}\right) \times J\left(\frac{u,v}{x,y}\right) = 1 \).

Given \( f(x,y) = \sin x + \sin y + \sin(x+y) \)

Set partial derivatives to zero:
\( f_x = \cos x + \cos(x+y) = 0 \)
\( f_y = \cos y + \cos(x+y) = 0 \)

Subtracting the two equations gives:
\( \cos x - \cos y = 0 \implies \cos x = \cos y \)

In the primary domain, this means \( x = y \).

Substitute \( y = x \) back into the first equation:
\( \cos x + \cos(2x) = 0 \)

Using the double angle identity \( \cos(2x) = 2\cos^2 x - 1 \):
\( \cos x + 2\cos^2 x - 1 = 0 \)
\( 2\cos^2 x + \cos x - 1 = 0 \)

Factoring the quadratic equation:
\( (2\cos x - 1)(\cos x + 1) = 0 \)

So, \( \cos x = 1/2 \) or \( \cos x = -1 \).
If \( \cos x = 1/2 \), then \( x = \pi/3 \). Since \( x=y \), the point is \( (\pi/3, \pi/3) \).

Given \( f(x, y) = x^3 + y^3 - 3axy \)

1. Find first derivatives:
\( f_x = 3x^2 - 3ay \)
\( f_y = 3y^2 - 3ax \)

2. Find second derivatives at \( (0,0) \):
\( r = f_{xx} = 6x \implies f_{xx}(0,0) = 0 \)
\( t = f_{yy} = 6y \implies f_{yy}(0,0) = 0 \)
\( s = f_{xy} = -3a \)

3. Apply the Second Derivative Test:
\( D = rt - s^2 \)
\( D = (0)(0) - (-3a)^2 = -9a^2 \)

Since \( -9a^2 < 0 \) for any non-zero \( a \), \( D < 0 \), which indicates that \( (0,0) \) is a saddle point.

Given \( f(x, y) = x^2 + y^2 \)

Substitute \( x = r \cos \theta \) and \( y = r \sin \theta \):

\( f(r, \theta) = (r \cos \theta)^2 + (r \sin \theta)^2 \)
\( f(r, \theta) = r^2 \cos^2 \theta + r^2 \sin^2 \theta \)
\( f(r, \theta) = r^2 (\cos^2 \theta + \sin^2 \theta) \)

Since \( \cos^2 \theta + \sin^2 \theta = 1 \):
\( f(r, \theta) = r^2 \)

Now, differentiate with respect to \( \theta \):
\( \frac{\partial f}{\partial \theta} = \frac{\partial}{\partial \theta}(r^2) = 0 \)

The result is 0 because the function is independent of \( \theta \).

Given the function: \( f(x, y) = x^2 y + \sin y + e^x \)

Find the partial derivative with respect to \( x \) (\( f_x \)), treating \( y \) as a constant:

\( \frac{\partial}{\partial x}(x^2 y) = 2xy \)

\( \frac{\partial}{\partial x}(\sin y) = 0 \)

\( \frac{\partial}{\partial x}(e^x) = e^x \)

So, \( f_x(x, y) = 2xy + e^x \)

Evaluate at the point \( (x, y) = (1, \pi) \):
\( f_x(1, \pi) = 2(1)(\pi) + e^1 = 2\pi + e \)

To find the degree of homogeneity, we look at the algebraic expression inside the inverse trigonometric function:

Let \( f(x, y) = \frac{x + y}{\sqrt{x} + \sqrt{y}} \)

Substitute \( x \) with \( tx \) and \( y \) with \( ty \):
\( f(tx, ty) = \frac{tx + ty}{\sqrt{tx} + \sqrt{ty}} = \frac{t(x + y)}{\sqrt{t}(\sqrt{x} + \sqrt{y})} \)

Simplify the expression:
\( f(tx, ty) = \frac{t}{t^{1/2}} \left( \frac{x + y}{\sqrt{x} + \sqrt{y}} \right) = t^{1/2} f(x, y) \)

Since \( f(tx, ty) = t^{1/2} f(x, y) \), the degree of homogeneity is \( n = 1/2 \).

Given \( u = \tan^{-1} \left( \frac{x^3 + y^3}{x - y} \right) \)

Let \( z = \tan u = \frac{x^3 + y^3}{x - y} \)

Here, \( z \) is a homogeneous function. To find its degree \( n \):
Numerator degree = 3. Denominator degree = 1.
Degree of \( z \) is \( n = 3 - 1 = 2 \).

Using Euler's modified theorem for \( z = f(u) \):
\( x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = n \frac{f(u)}{f'(u)} \)

Substitute \( n = 2 \), \( f(u) = \tan u \), and \( f'(u) = \sec^2 u \):
\( x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} = 2 \frac{\tan u}{\sec^2 u} \)

Simplify the trigonometric expression:
\( = 2 \left(\frac{\sin u}{\cos u}\right) \cos^2 u = 2 \sin u \cos u \)

By the double angle identity, \( 2 \sin u \cos u = \sin 2u \).

The Jacobian is the determinant of the matrix of partial derivatives:

\( J = \begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{vmatrix} \)

Calculate the partial derivatives:
\( \frac{\partial u}{\partial x} = -\frac{y^2}{x^2} \)
\( \frac{\partial u}{\partial y} = \frac{2y}{x} \)
\( \frac{\partial v}{\partial x} = \frac{2x}{y} \)
\( \frac{\partial v}{\partial y} = -\frac{x^2}{y^2} \)

Evaluate the determinant:
\( J = \left(-\frac{y^2}{x^2}\right)\left(-\frac{x^2}{y^2}\right) - \left(\frac{2y}{x}\right)\left(\frac{2x}{y}\right) \)

\( J = \left( \frac{x^2y^2}{x^2y^2} \right) - \left( \frac{4xy}{xy} \right) = 1 - 4 = -3 \)

These are the standard equations for cylindrical coordinates. The Jacobian is given by the determinant of a 3x3 matrix:

\( J = \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial z} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} & \frac{\partial y}{\partial z} \\ \frac{\partial z}{\partial r} & \frac{\partial z}{\partial \theta} & \frac{\partial z}{\partial z} \end{vmatrix} \)

Calculate the partial derivatives:
Row 1: \( \cos \theta \), \( -r \sin \theta \), 0
Row 2: \( \sin \theta \), \( r \cos \theta \), 0
Row 3: 0, 0, 1

Evaluate the determinant:
\( J = \begin{vmatrix} \cos \theta & -r \sin \theta & 0 \\ \sin \theta & r \cos \theta & 0 \\ 0 & 0 & 1 \end{vmatrix} \)

Expanding along the third row (or column):
\( J = 1 \cdot \begin{vmatrix} \cos \theta & -r \sin \theta \\ \sin \theta & r \cos \theta \end{vmatrix} = r \cos^2 \theta - (-r \sin^2 \theta) \)

\( J = r(\cos^2 \theta + \sin^2 \theta) = r(1) = r \)

The Jacobian is the determinant of the matrix of partial derivatives:

\( J = \begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{vmatrix} \)

Calculate the partial derivatives:
\( \frac{\partial u}{\partial x} = 2x \)
\( \frac{\partial u}{\partial y} = -2 \)
\( \frac{\partial v}{\partial x} = 1 \)
\( \frac{\partial v}{\partial y} = 1 \)

Evaluate the determinant:
\( J = \begin{vmatrix} 2x & -2 \\ 1 & 1 \end{vmatrix} = (2x)(1) - (-2)(1) \)

\( J = 2x - (-2) = 2x + 2 \)

The given differential equation is \( (D^2 + 3D - 10)y = 0 \).

The Auxiliary Equation (AE) is \( m^2 + 3m - 10 = 0 \).

Factorizing the quadratic equation:
\( m^2 + 5m - 2m - 10 = 0 \)
\( m(m + 5) - 2(m + 5) = 0 \)
\( (m - 2)(m + 5) = 0 \)

The roots are \( m = 2 \) and \( m = -5 \).

Since the roots are real and distinct, the general solution is:
\( y = Ae^{2x} + Be^{-5x} \)

The Auxiliary Equation (AE) is \( m^2 + 4 = 0 \).

Solving for \( m \):
\( m^2 = -4 \implies m = \pm 2i \)

The roots are complex, of the form \( \alpha \pm i\beta \) where \( \alpha = 0 \) and \( \beta = 2 \).

The general solution for complex roots is \( y = e^{\alpha x} (A \cos \beta x + B \sin \beta x) \).

Substitute \( \alpha = 0 \) and \( \beta = 2 \):
\( y = e^{0} (A \cos 2x + B \sin 2x) = A \cos 2x + B \sin 2x \)

The Particular Integral (P.I) is given by \( \text{P.I.} = \frac{1}{D^2 + 4} \sin 2x \).

Substituting \( D^2 = -2^2 = -4 \) gives zero in the denominator (a case of failure).

To resolve this, we multiply by \( x \) and differentiate the denominator with respect to \( D \):

\( \text{P.I.} = x \frac{1}{\frac{d}{dD}(D^2 + 4)} \sin 2x = x \frac{1}{2D} \sin 2x \)

Since \( \frac{1}{D} \) represents integration with respect to \( x \):
\( \text{P.I.} = \frac{x}{2} \int \sin 2x \, dx = \frac{x}{2} \left( -\frac{\cos 2x}{2} \right) \)

\( \text{P.I.} = -\frac{x}{4} \cos 2x \)

A linear differential equation of the form \( (a_n x^n D^n + a_{n-1} x^{n-1} D^{n-1} + \dots + a_1 x D + a_0)y = Q(x) \) is known as a Cauchy-Euler equation (or simply Cauchy's linear equation).

It is characterized by having coefficients that are proportional to powers of \( x \), specifically \( x^k \) matching the order of the \( k \)-th derivative \( D^k \).

Let \( z = \log x \implies x = e^z \).

Let \( \theta = \frac{d}{dz} \). We have the standard substitutions for Cauchy-Euler equations:

\( xD = \theta \)

\( x^2D^2 = \theta(\theta - 1) = \theta^2 - \theta \)

Substitute these into the given D.E. \( (x^2 D^2 - xD + 1)y = x^2 \):
\( (\theta^2 - \theta - \theta + 1)y = (e^z)^2 \)

\( (\theta^2 - 2\theta + 1)y = e^{2z} \)

1. Complementary Function (C.F):
Auxiliary Equation: \( m^2 + 2m + 1 = 0 \implies (m+1)^2 = 0 \)
Roots are \( m = -1, -1 \).
So, \( \text{C.F.} = (Ax + B)e^{-x} \)

2. Particular Integral (P.I):
\( \text{P.I.} = \frac{1}{D^2 + 2D + 1} (7e^{0x}) \)

Substitute \( D = 0 \):
\( \text{P.I.} = \frac{7}{0^2 + 2(0) + 1} = \frac{7}{1} = 7 \)

3. General Solution:
\( y = \text{C.F.} + \text{P.I.} = (Ax + B)e^{-x} + 7 \)

The given equation is \( (D - 1)^2 y = e^x \sin x \).

\( \text{P.I.} = \frac{1}{(D - 1)^2} e^x \sin x \)

Using the exponential shift rule \( \frac{1}{f(D)} [e^{ax} V] = e^{ax} \frac{1}{f(D+a)} V \), with \( a = 1 \) and \( V = \sin x \):

\( \text{P.I.} = e^x \frac{1}{((D + 1) - 1)^2} \sin x \)

\( \text{P.I.} = e^x \frac{1}{D^2} \sin x \)

Since \( D^2 \sin ax = -a^2 \sin ax \), replacing \( D^2 \) with \( -1^2 = -1 \):
\( \text{P.I.} = e^x \left( \frac{1}{-1} \sin x \right) = -e^x \sin x \)

Alternatively, integrating \( \sin x \) twice gives \( -\sin x \), leading to the same result.

\( \text{P.I.} = \frac{1}{(D - 1)^2} x = \frac{1}{(-(1 - D))^2} x = (1 - D)^{-2} x \)

Expand \( (1 - D)^{-2} \) using the binomial theorem up to the first derivative (since higher derivatives of \( x \) are zero):

\( (1 - D)^{-2} = 1 + 2D + 3D^2 + \dots \)

Apply this to \( x \):
\( \text{P.I.} = (1 + 2D)x = x + 2D(x) = x + 2(1) = x + 2 \)

When the roots of the auxiliary equation are complex, in the form \( m = \alpha \pm i\beta \), the general solution is:

\( y = e^{\alpha x} (A \cos \beta x + B \sin \beta x) \)

Given roots are \( m = 1 \pm 2i \), so \( \alpha = 1 \) and \( \beta = 2 \).

Substituting these values into the formula:
\( y = e^{1 \cdot x} (A \cos 2x + B \sin 2x) = e^x (A \cos 2x + B \sin 2x) \)

This is a Legendre's linear equation of the form \( (ax+b)^2 y'' + (ax+b)y' + y = 0 \).

Let \( z = \log(5 + 2x) \) and \( \theta = \frac{d}{dz} \). Note that the coefficient of \( x \) is \( a = 2 \).

The standard substitutions are:
\( (5+2x)D = a\theta = 2\theta \)
\( (5+2x)^2D^2 = a^2\theta(\theta - 1) = 4\theta(\theta - 1) = 4\theta^2 - 4\theta \)

Substitute these into the given D.E.:
\( (4\theta^2 - 4\theta)y - 6(2\theta)y + 8y = 0 \)
\( (4\theta^2 - 4\theta - 12\theta + 8)y = 0 \)
\( (4\theta^2 - 16\theta + 8)y = 0 \)

Divide the entire equation by 4:
\( (\theta^2 - 4\theta + 2)y = 0 \)

Recall the definition of hyperbolic sine: \( \sinh 2x = \frac{e^{2x} - e^{-2x}}{2} \).

\( \text{P.I.} = \frac{1}{D^2 + 4} \left( \frac{e^{2x}}{2} - \frac{e^{-2x}}{2} \right) \)

Apply the rule \( \frac{1}{f(D)} e^{ax} = \frac{1}{f(a)} e^{ax} \) to each term:

For \( e^{2x} \), \( a = 2 \): \( \frac{1}{2^2 + 4} = \frac{1}{8} \)

For \( e^{-2x} \), \( a = -2 \): \( \frac{1}{(-2)^2 + 4} = \frac{1}{8} \)

Combining the terms:
\( \text{P.I.} = \frac{1}{2} \left( \frac{e^{2x}}{8} - \frac{e^{-2x}}{8} \right) = \frac{1}{8} \left( \frac{e^{2x} - e^{-2x}}{2} \right) \)

\( \text{P.I.} = \frac{\sinh 2x}{8} \)

\( \text{P.I.} = \frac{1}{D^2 + 6D + 5} e^{-x} \)

Substituting \( D = -1 \) into the denominator gives \( (-1)^2 + 6(-1) + 5 = 1 - 6 + 5 = 0 \), which is a case of failure.

To resolve this, we multiply by \( x \) and differentiate the denominator with respect to \( D \):

\( \text{P.I.} = x \frac{1}{2D + 6} e^{-x} \)

Now substitute \( D = -1 \) into the new denominator:
\( \text{P.I.} = x \frac{1}{2(-1) + 6} e^{-x} = x \frac{1}{-2 + 6} e^{-x} \)

\( \text{P.I.} = \frac{x e^{-x}}{4} \)

The Auxiliary Equation (AE) is \( m^2 - 2am + a^2 = 0 \).

This factors into a perfect square: \( (m - a)^2 = 0 \).

The roots are \( m = a, a \), which are real and equal.

When roots are real and repeated, the general solution takes the form:
\( y = (C_1 x + C_2)e^{mx} \)

Substituting the root \( m = a \), we get \( y = (Ax + B)e^{ax} \).

\( \text{P.I.} = \frac{1}{D^2 + 16} \cos 4x \)

Substitute \( D^2 = -a^2 = -4^2 = -16 \) into the denominator: \( -16 + 16 = 0 \).

Since the denominator is zero, multiply by \( x \) and differentiate the denominator with respect to \( D \):

\( \text{P.I.} = x \frac{1}{2D} \cos 4x \)

The operator \( \frac{1}{D} \) means integration:
\( \text{P.I.} = \frac{x}{2} \int \cos 4x \, dx = \frac{x}{2} \left( \frac{\sin 4x}{4} \right) \)

\( \text{P.I.} = \frac{x \sin 4x}{8} \)

The Complementary Function (C.F) depends only on the homogeneous part: \( (D^2 + 1)y = 0 \).

The Auxiliary Equation is \( m^2 + 1 = 0 \implies m^2 = -1 \).

The roots are \( m = \pm i \).

For complex roots \( \alpha \pm i\beta \) (here \( \alpha=0, \beta=1 \)), the general solution is:

\( \text{C.F.} = e^{0x}(A \cos 1x + B \sin 1x) = A \cos x + B \sin x \)

Given \( y_1 = \cos ax \) and \( y_2 = \sin ax \).

Find their derivatives:
\( y_1' = -a \sin ax \)
\( y_2' = a \cos ax \)

Calculate the Wronskian \( W = y_1 y_2' - y_2 y_1' \):
\( W = (\cos ax)(a \cos ax) - (\sin ax)(-a \sin ax) \)

\( W = a \cos^2 ax + a \sin^2 ax \)

\( W = a (\cos^2 ax + \sin^2 ax) = a(1) = a \)

The general solution to \( (D^2 + 1)y = 0 \) is:

\( y(x) = A \cos x + B \sin x \)

Apply the first initial condition \( y(0) = 0 \):
\( 0 = A \cos(0) + B \sin(0) \implies A(1) + B(0) = 0 \implies A = 0 \)

So, \( y(x) = B \sin x \).

Now, differentiate: \( y'(x) = B \cos x \)

Apply the second initial condition \( y'(0) = 1 \):
\( 1 = B \cos(0) \implies B(1) = 1 \implies B = 1 \)

Substitute \( A \) and \( B \) back into the general solution:

\( y = \sin x \)

\( \text{P.I.} = \frac{1}{(D - 2)^2} e^{2x} \)

Using the exponential shift rule \( \frac{1}{f(D)} [e^{ax} V] = e^{ax} \frac{1}{f(D+a)} V \), with \( a = 2 \) and \( V = 1 \):

\( \text{P.I.} = e^{2x} \frac{1}{((D + 2) - 2)^2} (1) \)

\( \text{P.I.} = e^{2x} \frac{1}{D^2} (1) \)

The operator \( \frac{1}{D^2} \) means integrating twice with respect to \( x \):
\( \frac{1}{D} (1) = x \)
\( \frac{1}{D^2} (1) = \int x \, dx = \frac{x^2}{2} \)

Therefore, \( \text{P.I.} = e^{2x} \left( \frac{x^2}{2} \right) = \frac{x^2}{2} e^{2x} \)

\( \text{P.I.} = \frac{1}{D^2 + 4} \sin(2x + 5) \)

Substitute \( D^2 = -a^2 = -2^2 = -4 \) into the denominator: \( -4 + 4 = 0 \).

Since the denominator is zero, multiply by \( x \) and differentiate the denominator with respect to \( D \):

\( \text{P.I.} = x \frac{1}{2D} \sin(2x + 5) \)

The operator \( \frac{1}{D} \) means integration:
\( \text{P.I.} = \frac{x}{2} \int \sin(2x + 5) \, dx = \frac{x}{2} \left( \frac{-\cos(2x + 5)}{2} \right) \)

\( \text{P.I.} = -\frac{x}{4} \cos(2x + 5) \)

This is a Cauchy-Euler equation. Use the substitution \( x = e^z \) (or \( z = \log x \)).

Substitute \( xD = \theta \) and \( x^2D^2 = \theta(\theta - 1) \), where \( \theta = \frac{d}{dz} \):

\( (\theta(\theta - 1) + \theta + 1)y = 0 \)
\( (\theta^2 - \theta + \theta + 1)y = 0 \)
\( (\theta^2 + 1)y = 0 \)

The Auxiliary Equation is \( m^2 + 1 = 0 \implies m = \pm i \).

The general solution in terms of \( z \) is:
\( y(z) = A \cos z + B \sin z \)

Given the auxiliary equation: \( m^2 + 4 = 0 \)

Subtract 4 from both sides: \( m^2 = -4 \)

Taking the square root of both sides gives imaginary numbers:
\( m = \pm\sqrt{-4} = \pm 2i \)

The roots are \( \pm 2i \).

This is a Cauchy-Euler equation. Use the substitution \( x = e^z \) (or \( z = \log x \)).

Substitute \( xD = \theta \) and \( x^2D^2 = \theta(\theta - 1) \):

\( (\theta(\theta - 1) - 7\theta + 12)y = 0 \)
\( (\theta^2 - \theta - 7\theta + 12)y = 0 \)
\( (\theta^2 - 8\theta + 12)y = 0 \)

The Auxiliary Equation is \( m^2 - 8m + 12 = 0 \).

Factorizing: \( (m - 2)(m - 6) = 0 \implies m = 2, 6 \).

The general solution in terms of \( z \) is:
\( y(z) = Ae^{2z} + Be^{6z} \)

Given \( y_1 = \cos x \) and \( y_2 = \sin x \).

Find their derivatives:
\( y_1' = -\sin x \)
\( y_2' = \cos x \)

Calculate the Wronskian \( W = y_1 y_2' - y_2 y_1' \):
\( W = (\cos x)(\cos x) - (\sin x)(-\sin x) \)

\( W = \cos^2 x + \sin^2 x \)

By the Pythagorean trigonometric identity, \( \cos^2 x + \sin^2 x = 1 \).

So, \( W = 1 \).

When solving linear differential equations with constant coefficients, the form of the Complementary Function (C.F) depends on the roots of the Auxiliary Equation (A.E).

If a real root \( m = a \) is repeated \( k \) times, the corresponding part of the C.F is:
\( (C_1 + C_2 x + C_3 x^2 + \dots + C_k x^{k-1})e^{ax} \)

Here, the root \( m = a \) is repeated 3 times (i.e., \( k = 3 \)).

Therefore, the C.F is \( (A x^2 + B x + C)e^{ax} \).

This is a standard operation in finding the Particular Integral (P.I).

When evaluating \( \frac{1}{f(D)} e^{ax} \), if \( f(a) = 0 \), it is a case of failure.

Here, \( f(D) = D - a \). So \( f(a) = a - a = 0 \).

The rule for failure states that \( \frac{1}{(D - a)^k} e^{ax} = \frac{x^k}{k!} e^{ax} \).

For \( k = 1 \):
\( \frac{1}{D - a} e^{ax} = \frac{x^1}{1!} e^{ax} = x e^{ax} \)

By definition, the radius of curvature \( \rho \) is the reciprocal of the curvature \( \kappa \) at any given point on a curve.

\( \rho = \frac{1}{\kappa} \)

The evolute of a curve is defined geometrically as the locus of all its centers of curvature.

Conversely, the original curve is called the involute of its evolute.

To find the envelope of a family of curves given by a quadratic equation in a parameter \( \alpha \):

\( F(x,y,\alpha) = A\alpha^2 + B\alpha + C = 0 \)

We differentiate with respect to \( \alpha \) and equate to zero:
\( \frac{\partial F}{\partial \alpha} = 2A\alpha + B = 0 \implies \alpha = -\frac{B}{2A} \)

Substitute this value of \( \alpha \) back into the original equation:
\( A\left(-\frac{B}{2A}\right)^2 + B\left(-\frac{B}{2A}\right) + C = 0 \)

\( A\left(\frac{B^2}{4A^2}\right) - \frac{B^2}{2A} + C = 0 \)

\( \frac{B^2}{4A} - \frac{B^2}{2A} + C = 0 \implies -\frac{B^2}{4A} + C = 0 \)

\( B^2 - 4AC = 0 \)

A straight line has a constant slope, meaning it does not bend or curve.

Mathematically, if \( y = mx + c \), then \( y'' = 0 \).

The curvature is given by \( \kappa = \frac{|y''|}{(1 + y'^2)^{3/2}} = \frac{0}{(1 + m^2)^{3/2}} = 0 \).

Hence, the curvature of a straight line is zero everywhere.

Given the curve \( y = 4 \sin x \).

Find the first derivative: \( y' = 4 \cos x \)

Evaluate at \( x = \pi/2 \): \( y'(\pi/2) = 4 \cos(\pi/2) = 0 \)

Find the second derivative: \( y'' = -4 \sin x \)

Evaluate at \( x = \pi/2 \): \( y''(\pi/2) = -4 \sin(\pi/2) = -4 \)

The formula for radius of curvature is:
\( \rho = \frac{(1 + (y')^2)^{3/2}}{|y''|} \)

\( \rho = \frac{(1 + 0^2)^{3/2}}{|-4|} = \frac{1}{4} \)

Rewrite the equation as a quadratic in the parameter \( t \):

\( at^2 - yt + x = 0 \)

Here, \( A = a \), \( B = -y \), and \( C = x \).

The envelope of a quadratic family \( At^2 + Bt + C = 0 \) is given by \( B^2 - 4AC = 0 \).

Substitute the values:
\( (-y)^2 - 4(a)(x) = 0 \)

\( y^2 - 4ax = 0 \implies y^2 = 4ax \)

By the definition of curvature (\( \kappa \)) and radius of curvature (\( \rho \)), they are inversely related.

For a circle of constant radius \( r \), the radius of curvature is everywhere equal to \( r \).

Therefore, the curvature \( \kappa = \frac{1}{\rho} = \frac{1}{r} \).

Hence, the curvature is the reciprocal of the radius.

The given equation \( x^2 + y^2 = 25 \) represents a circle centered at the origin \( (0,0) \) with a radius \( r \).

Comparing it to the standard equation of a circle \( x^2 + y^2 = r^2 \), we have \( r^2 = 25 \), so \( r = 5 \).

For any circle, the radius of curvature is constant at all points and is equal to the radius of the circle itself.

Therefore, the radius of curvature at \( (4, 3) \) is simply 5.

The curvature \( \kappa \) of a circle is the reciprocal of its radius of curvature.

Given the radius is \( r = 3 \).

Curvature \( \kappa = \frac{1}{r} = \frac{1}{3} \).

Given \( y = mx + \frac{a}{m} \).

Multiply the entire equation by \( m \) to clear the fraction:

\( my = m^2 x + a \)

Rearrange it as a quadratic equation in terms of the parameter \( m \):

\( x m^2 - y m + a = 0 \)

The envelope is given by setting the discriminant to zero (\( B^2 - 4AC = 0 \)):

Here, \( A = x \), \( B = -y \), \( C = a \).

\( (-y)^2 - 4(x)(a) = 0 \)

\( y^2 - 4ax = 0 \)

Given the curve \( y = e^x \).

Find the first derivative: \( y' = e^x \)

Evaluate at \( x = 0 \): \( y'(0) = e^0 = 1 \)

Find the second derivative: \( y'' = e^x \)

Evaluate at \( x = 0 \): \( y''(0) = e^0 = 1 \)

The formula for radius of curvature is:
\( \rho = \frac{(1 + (y')^2)^{3/2}}{|y''|} \)

\( \rho = \frac{(1 + 1^2)^{3/2}}{|1|} = (2)^{3/2} = 2\sqrt{2} \)

Given \( y = \log(\sec x) \).

First derivative: \( y' = \frac{1}{\sec x} \cdot \sec x \tan x = \tan x \).

Second derivative: \( y'' = \sec^2 x \).

The formula for radius of curvature is:
\( \rho = \frac{(1 + (y')^2)^{3/2}}{|y''|} \)

\( \rho = \frac{(1 + \tan^2 x)^{3/2}}{\sec^2 x} \)

Using the identity \( 1 + \tan^2 x = \sec^2 x \):
\( \rho = \frac{(\sec^2 x)^{3/2}}{\sec^2 x} = \frac{\sec^3 x}{\sec^2 x} = \sec x \)

The standard equation of an ellipse is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) where \( a > b \).

The semi latus rectum is \( L = \frac{b^2}{a} \).

The radius of curvature at the ends of the major axis \( (\pm a, 0) \) is \( \frac{b^2}{a} \).

The radius of curvature at the ends of the minor axis \( (0, \pm b) \) is \( \frac{a^2}{b} \).

Therefore, at the ends of the major axis, the radius of curvature equals the semi latus rectum.

For a curve in parametric form, the radius of curvature is:
\( \rho = \frac{((x')^2 + (y')^2)^{3/2}}{|x'y'' - y'x''|} \)

Given \( x = t^2 \) and \( y = t \):
\( x' = 2t \implies \) at \( t=1 \), \( x' = 2 \)
\( y' = 1 \implies \) at \( t=1 \), \( y' = 1 \)

\( x'' = 2 \implies \) at \( t=1 \), \( x'' = 2 \)
\( y'' = 0 \implies \) at \( t=1 \), \( y'' = 0 \)

Substitute into the formula:
\( \rho = \frac{(2^2 + 1^2)^{3/2}}{|2(0) - 1(2)|} \)

\( \rho = \frac{(5)^{3/2}}{|-2|} = \frac{5\sqrt{5}}{2} \)

The evolute of a curve can be defined in two equivalent ways:

1. It is the locus of all the centers of curvature of the curve.

2. It is the envelope of the family of normal lines to the curve.

A well-known property in differential geometry is that the evolute of a cycloid is another identical cycloid.

Specifically, the new cycloid is congruent to the original one but is shifted in position.

By definition, an envelope of a family of curves is a curve that is tangent to each member of the family at some point.

It "envelopes" the entire family of curves.

Given the family of lines: \( y = mx + am^2 \)

Rearrange this into a quadratic equation in terms of the parameter \( m \):

\( a m^2 + x m - y = 0 \)

The envelope is found by setting the discriminant to zero (\( B^2 - 4AC = 0 \)):

Here, \( A = a \), \( B = x \), \( C = -y \).

\( x^2 - 4(a)(-y) = 0 \)

\( x^2 + 4ay = 0 \)

Given \( \frac{x}{t} + yt = 2c \).

Multiply the entire equation by \( t \) to form a quadratic in terms of \( t \):

\( y t^2 - 2ct + x = 0 \)

The envelope is found by setting the discriminant to zero (\( B^2 - 4AC = 0 \)):

Here, \( A = y \), \( B = -2c \), \( C = x \).

\( (-2c)^2 - 4(y)(x) = 0 \)

\( 4c^2 - 4xy = 0 \implies xy = c^2 \)

For a polar curve \( r = f(\theta) \), the radius of curvature is:

\( \rho = \frac{(r^2 + (r')^2)^{3/2}}{|r^2 + 2(r')^2 - r r''|} \)

Given \( r = e^\theta \):
\( r' = e^\theta = r \)
\( r'' = e^\theta = r \)

Substitute these into the formula:
\( \rho = \frac{(r^2 + r^2)^{3/2}}{|r^2 + 2r^2 - r(r)|} \)

\( \rho = \frac{(2r^2)^{3/2}}{|3r^2 - r^2|} = \frac{2\sqrt{2}r^3}{2r^2} \)

\( \rho = \sqrt{2}r \)

By standard definition, the radius of curvature \( \rho \) for a Cartesian curve \( y = f(x) \) is given by:

\( \rho = \frac{(1 + (y')^2)^{3/2}}{|y''|} \)

Using subscript notation where \( y_1 = y' = \frac{dy}{dx} \) and \( y_2 = y'' = \frac{d^2y}{dx^2} \), this becomes \( \rho = \frac{(1 + y_1^2)^{3/2}}{y_2} \).

The standard formula for the radius of curvature in polar coordinates \( (r, \theta) \) where \( r' = \frac{dr}{d\theta} \) and \( r'' = \frac{d^2r}{d\theta^2} \) is:

\( \rho = \frac{(r^2 + (r')^2)^{3/2}}{|r^2 + 2(r')^2 - r r''|} \)

This matches option (D) rearranged slightly in the denominator.

For a curve defined parametrically by \( x = x(t), y = y(t) \), the radius of curvature is given by:

\( \rho = \frac{((x')^2 + (y')^2)^{3/2}}{|x'y'' - y'x''|} \)

This matches option (A).

The circle of curvature at a point on a curve is the circle that best approximates the curve at that point.

Its center is the center of curvature \( (\bar{x}, \bar{y}) \) and its radius is the radius of curvature \( \rho \).

The standard equation of a circle with center \( (h,k) \) and radius \( r \) is \( (x-h)^2 + (y-k)^2 = r^2 \).

Thus, substituting \( h = \bar{x} \), \( k = \bar{y} \), and \( r = \rho \), we get \( (x - \bar{x})^2 + (y - \bar{y})^2 = \rho^2 \).

The relationship between the Beta and Gamma functions is given by:

\( \beta(m, n) = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)} \)

For integers, \( \Gamma(n) = (n-1)! \).

\( \beta(4, 4) = \frac{\Gamma(4)\Gamma(4)}{\Gamma(8)} = \frac{3! \cdot 3!}{7!} \)

Since \( 3! = 6 \), we have:

\( \beta(4, 4) = \frac{6 \times 6}{7!} = \frac{36}{7!} \)

Using the formula \( \beta(m, n) = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)} \):

\( \beta\left(\frac{5}{2}, \frac{3}{2}\right) = \frac{\Gamma(5/2)\Gamma(3/2)}{\Gamma(5/2 + 3/2)} = \frac{\Gamma(5/2)\Gamma(3/2)}{\Gamma(4)} \)

We know \( \Gamma(n+1) = n\Gamma(n) \) and \( \Gamma(1/2) = \sqrt{\pi} \).

\( \Gamma(3/2) = \frac{1}{2}\Gamma(1/2) = \frac{\sqrt{\pi}}{2} \)

\( \Gamma(5/2) = \frac{3}{2}\Gamma(3/2) = \frac{3}{2} \cdot \frac{\sqrt{\pi}}{2} = \frac{3\sqrt{\pi}}{4} \)

\( \Gamma(4) = 3! = 6 \)

Substitute these back:

\( \beta\left(\frac{5}{2}, \frac{3}{2}\right) = \frac{(\frac{3\sqrt{\pi}}{4})(\frac{\sqrt{\pi}}{2})}{6} = \frac{\frac{3\pi}{8}}{6} = \frac{3\pi}{48} = \frac{\pi}{16} \)

This is the fundamental identity relating the Beta function and the Gamma function.

The integral definition of Beta function transforms into this product of Gamma functions.

\( \beta(m, n) = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)} \)

By definition, \( \Gamma(n) = \int_0^\infty x^{n-1} e^{-x} \, dx \).

For \( n = 1/2 \), \( \Gamma(1/2) = \int_0^\infty x^{-1/2} e^{-x} \, dx \).

Substituting \( x = u^2 \), \( dx = 2u \, du \), we get:

\( \Gamma(1/2) = 2 \int_0^\infty e^{-u^2} \, du \)

This is the standard Gaussian integral, which evaluates to \( \sqrt{\pi} \).

Euler's reflection formula states that for non-integer \( n \):

\( \Gamma(n)\Gamma(1-n) = \frac{\pi}{\sin(n\pi)} \)

Note: This is also equivalent to \( \int_0^\infty \frac{x^{n-1}}{1+x} \, dx \) and \( \beta(n, 1-n) \), but the trigonometric form is the standard evaluated result.

Using the definition of the Beta function in trigonometric form:

\( \beta(m, n) = 2 \int_0^{\pi/2} \sin^{2m-1}\theta \cos^{2n-1}\theta \, d\theta \)

Let \( 2m - 1 = p \implies m = \frac{p+1}{2} \)

Let \( 2n - 1 = q \implies n = \frac{q+1}{2} \)

Substituting these in, we get:

\( \beta\left(\frac{p+1}{2}, \frac{q+1}{2}\right) = 2 \int_0^{\pi/2} \sin^p \theta \cos^q \theta \, d\theta \)

Therefore, \( \int_0^{\pi/2} \sin^p \theta \cos^q \theta \, d\theta = \frac{1}{2} \beta\left(\frac{p+1}{2}, \frac{q+1}{2}\right) \)

Let \( I = \int_0^1 x^4 \left(-\log x\right)^3 \, dx \).

Use the substitution \( t = -\log x \), so \( x = e^{-t} \) and \( dx = -e^{-t} dt \).

When \( x = 0, t = \infty \) and when \( x = 1, t = 0 \).

The integral becomes:

\( I = \int_\infty^0 (e^{-t})^4 (t)^3 (-e^{-t} dt) = \int_0^\infty t^3 e^{-5t} dt \)

Now substitute \( u = 5t \), then \( du = 5 dt \) and \( t = u/5 \):

\( I = \int_0^\infty \left(\frac{u}{5}\right)^3 e^{-u} \frac{du}{5} = \frac{1}{5^4} \int_0^\infty u^3 e^{-u} du \)

Since \( \int_0^\infty u^{n-1} e^{-u} du = \Gamma(n) \), the integral is \( \Gamma(4) \):

\( I = \frac{\Gamma(4)}{625} = \frac{3!}{625} = \frac{6}{625} \)

Rewrite the integral as \( I = \int_0^{\pi/2} \cos^{1/2}\theta \sin^{-1/2}\theta \, d\theta \).

Using the formula \( \int_0^{\pi/2} \sin^p\theta \cos^q\theta \, d\theta = \frac{1}{2} \beta\left(\frac{p+1}{2}, \frac{q+1}{2}\right) \):

Here, \( p = -1/2 \) and \( q = 1/2 \).

\( I = \frac{1}{2} \beta\left(\frac{-1/2+1}{2}, \frac{1/2+1}{2}\right) = \frac{1}{2} \beta\left(\frac{1}{4}, \frac{3}{4}\right) \)

Using \( \beta(m,n) = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)} \):

\( I = \frac{1}{2} \frac{\Gamma(1/4)\Gamma(3/4)}{\Gamma(1)} \)

By Euler's reflection formula \( \Gamma(n)\Gamma(1-n) = \frac{\pi}{\sin n\pi} \) with \( n = 1/4 \):

\( \Gamma(1/4)\Gamma(3/4) = \frac{\pi}{\sin(\pi/4)} = \frac{\pi}{1/\sqrt{2}} = \pi\sqrt{2} \)

Therefore, \( I = \frac{1}{2} (\pi\sqrt{2}) = \frac{\pi}{\sqrt{2}} \).

By the \( p \)-series test, a series of the form \( \sum_{n=1}^\infty \frac{1}{n^p} \) is:

1. Convergent if \( p > 1 \).

2. Divergent if \( p \le 1 \).

Therefore, the series is convergent for \( p > 1 \).

By the \( p \)-series test, a series of the form \( \sum_{n=1}^\infty \frac{1}{n^p} \) is:

1. Convergent if \( p > 1 \).

2. Divergent if \( p \le 1 \).

Therefore, the series is divergent for \( p \le 1 \).

This is Cauchy's Root Test.

For a series with positive terms \( \sum u_n \), let \( \lim_{n\to\infty} (u_n)^{1/n} = l \).

If \( l < 1 \), the series is convergent.

If \( l > 1 \), the series is divergent.

If \( l = 1 \), the test is inconclusive.

Since \( l > 1 \) is given, the series is divergent.

By Cauchy's Root Test, if \( \lim_{n\to\infty} (u_n)^{1/n} = l \), the series converges absolutely if \( l < 1 \).

The given series is \( \sum_{n=1}^\infty \frac{1}{n!} = \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots \)

We know the Taylor series expansion for \( e^x \) is \( \sum_{n=0}^\infty \frac{x^n}{n!} \).

For \( x = 1 \), this becomes \( e = 1 + \sum_{n=1}^\infty \frac{1}{n!} \).

Thus, \( \sum_{n=1}^\infty \frac{1}{n!} = e - 1 \), which is a finite value. Hence, the series is convergent.

Alternatively, by D'Alembert's Ratio Test: \( \lim_{n\to\infty} \frac{u_{n+1}}{u_n} = \lim_{n\to\infty} \frac{n!}{(n+1)!} = \lim_{n\to\infty} \frac{1}{n+1} = 0 < 1 \), so it converges.

The given series is \( \sum_{n=1}^\infty \frac{1}{n} \), which is the Harmonic series.

Using the \( p \)-series test where the series is \( \sum_{n=1}^\infty \frac{1}{n^p} \), here \( p = 1 \).

Since \( p \le 1 \), the series is divergent.

According to D'Alembert's Ratio Test:

If \( l < 1 \), the series converges.

If \( l > 1 \), the series diverges.

If \( l = 1 \), the test fails (is inconclusive).

According to Raabe's Test for a positive term series:

Let \( \lim_{n\to\infty} n\left(\frac{u_n}{u_{n+1}} - 1\right) = l \).

If \( l > 1 \), the series converges.

If \( l < 1 \), the series diverges.

If \( l = 1 \), the test fails.

Thus, for divergence, \( l < 1 \).

Let \( u_n = \frac{2^n}{n^2} \).

Apply D'Alembert's Ratio Test:

\( \frac{u_{n+1}}{u_n} = \frac{2^{n+1}}{(n+1)^2} \times \frac{n^2}{2^n} = 2 \left(\frac{n}{n+1}\right)^2 \)

\( \lim_{n\to\infty} \frac{u_{n+1}}{u_n} = 2(1)^2 = 2 \)

Since the limit \( l = 2 > 1 \), the series is divergent.

By definition, an alternating series \( \sum u_n \) is absolutely convergent if the series of its absolute values, \( \sum |u_n| \), is convergent.

Absolute convergence implies standard convergence.

By definition, a series is conditionally convergent if it converges in its original alternating form (or with its original signs), but the sum of its absolute values diverges.

Thus, \( \sum u_n \) converges, but \( \sum |u_n| \) diverges.

The given alternating series is \( \sum_{n=1}^\infty (-1)^{n-1} \frac{1}{n^2} \).

To check for absolute convergence, we take the absolute value of the terms:

\( \sum_{n=1}^\infty \left| (-1)^{n-1} \frac{1}{n^2} \right| = \sum_{n=1}^\infty \frac{1}{n^2} \)

By the \( p \)-series test, \( \sum_{n=1}^\infty \frac{1}{n^p} \) converges for \( p = 2 > 1 \).

Since the absolute series converges, the original series is absolutely convergent.

The given series is the alternating harmonic series: \( \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n} \).

By Leibniz's test for alternating series, since \( \frac{1}{n} \) is decreasing and approaches 0 as \( n \to \infty \), the series itself converges.

Now check for absolute convergence:

\( \sum_{n=1}^\infty \left| \frac{(-1)^{n-1}}{n} \right| = \sum_{n=1}^\infty \frac{1}{n} \)

This is the harmonic series, which diverges by the \( p \)-series test (\( p=1 \)).

Since the series converges but does not converge absolutely, it is conditionally convergent.

We use Cauchy's Integral Test. Let \( f(x) = \frac{1}{x \log x} \).

\( \int_2^\infty \frac{1}{x \log x} \, dx \)

Let \( u = \log x \), so \( du = \frac{1}{x} \, dx \).

\( \int \frac{1}{u} \, du = \log|u| = \log|\log x| \)

Evaluating from 2 to \( \infty \), we get \( \lim_{R\to\infty} \log(\log R) - \log(\log 2) = \infty \).

Since the integral diverges, the series also diverges.

A fundamental theorem in real analysis states that if a series is absolutely convergent (i.e., \( \sum |u_n| \) converges), then the series itself (i.e., \( \sum u_n \)) must also be convergent.

Absolute convergence implies convergence.

By the \( p \)-series test, a series of the form \( \sum_{n=1}^\infty \frac{1}{n^p} \) converges if \( p > 1 \).

Here, \( p = 3 > 1 \), so the series converges.

(Note: Because all terms are positive, absolute convergence and standard convergence are synonymous here. Based on the answer key convention, "Convergent" is selected.)

We use Cauchy's Integral Test. Let \( f(x) = \frac{1}{x (\log x)^3} \).

\( \int_2^\infty \frac{1}{x (\log x)^3} \, dx \)

Let \( u = \log x \), so \( du = \frac{1}{x} \, dx \).

\( \int u^{-3} \, du = \frac{u^{-2}}{-2} = -\frac{1}{2(\log x)^2} \)

Evaluating from 2 to \( \infty \), we get \( \lim_{R\to\infty} \left( -\frac{1}{2(\log R)^2} \right) - \left( -\frac{1}{2(\log 2)^2} \right) \).

The limit goes to 0, leaving \( \frac{1}{2(\log 2)^2} \), which is a finite value.

Since the integral converges, the series also converges.