B.Tech. DEGREE EXAMINATION, MAY 2022

Detailed Solution:
Step 1: Use properties of eigenvalues.
Let the eigenvalues of the \( 3 \times 3 \) matrix be \( \lambda_1, \lambda_2, \lambda_3 \).
Given: Trace = 2 \( \implies \lambda_1 + \lambda_2 + \lambda_3 = 2 \).
Determinant = 8 \( \implies \lambda_1 \lambda_2 \lambda_3 = 8 \).
One eigenvalue is 2, let's say \( \lambda_1 = 2 \).

Step 2: Find the other two eigenvalues.
Substitute \( \lambda_1 = 2 \) into the trace equation:
\( 2 + \lambda_2 + \lambda_3 = 2 \implies \lambda_2 + \lambda_3 = 0 \implies \lambda_3 = -\lambda_2 \).
Substitute into the determinant equation:
\( 2(\lambda_2)(-\lambda_2) = 8 \implies -2\lambda_2^2 = 8 \implies \lambda_2^2 = -4 \).
So, \( \lambda_2 = 2i \) and \( \lambda_3 = -2i \).

Step 3: Form the characteristic equation.
The characteristic equation is \( (\lambda - \lambda_1)(\lambda - \lambda_2)(\lambda - \lambda_3) = 0 \).
\( (\lambda - 2)(\lambda - 2i)(\lambda + 2i) = 0 \)
\( (\lambda - 2)(\lambda^2 - (2i)^2) = 0 \)
\( (\lambda - 2)(\lambda^2 + 4) = 0 \)
\( \lambda^3 + 4\lambda - 2\lambda^2 - 8 = 0 \)
Rearranging terms: \( \lambda^3 - 2\lambda^2 + 4\lambda - 8 = 0 \).

Detailed Solution:
Step 1: Write the characteristic equation.
The characteristic equation of a \( 2 \times 2 \) matrix \( A \) is given by \( \lambda^2 - (\text{Trace of } A)\lambda + \det(A) = 0 \).

Step 2: Calculate Trace and Determinant.
Trace of \( A = 1 + 4 = 5 \).
Determinant of \( A = (1)(4) - (2)(5) = 4 - 10 = -6 \).

Step 3: Solve for eigenvalues.
Substitute these values into the characteristic equation:
\( \lambda^2 - 5\lambda - 6 = 0 \).
Factoring the quadratic equation:
\( \lambda^2 - 6\lambda + \lambda - 6 = 0 \)
\( \lambda(\lambda - 6) + 1(\lambda - 6) = 0 \)
\( (\lambda - 6)(\lambda + 1) = 0 \)
So, \( \lambda = 6 \) or \( \lambda = -1 \).
The eigenvalues are \( (-1, 6) \).

Detailed Solution:
Step 1: Identify the standard matrix representation.
For a quadratic form \( Q(x_1, x_2, x_3) = \sum_{i=1}^3 \sum_{j=1}^3 a_{ij} x_i x_j \), the symmetric matrix \( A \) has elements:
\( a_{ii} = \text{coefficient of } x_i^2 \)
\( a_{ij} = a_{ji} = \frac{1}{2} \times (\text{coefficient of } x_i x_j) \) for \( i \neq j \).

Step 2: Find the diagonal elements.
\( a_{11} = \text{coefficient of } x_1^2 = 2 \).
\( a_{22} = \text{coefficient of } x_2^2 = 1 \).
\( a_{33} = \text{coefficient of } x_3^2 = 0 \).

Step 3: Find the off-diagonal elements.
\( a_{12} = a_{21} = \frac{1}{2}(0) = 0 \) (since there is no \( x_1 x_2 \) term).
\( a_{13} = a_{31} = \frac{1}{2}(-2) = -1 \).
\( a_{23} = a_{32} = \frac{1}{2}(4) = 2 \).

Step 4: Construct the matrix.
\( A = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix} = \begin{bmatrix} 2 & 0 & -1 \\ 0 & 1 & 2 \\ -1 & 2 & 0 \end{bmatrix} \).

Detailed Solution:
Step 1: Find the eigenvalues of \( A \).
Matrix \( A = \begin{bmatrix} 3 & 4 \\ 0 & -3 \end{bmatrix} \) is an upper triangular matrix.
For an upper or lower triangular matrix, the eigenvalues are simply the elements on its main diagonal.
So, the eigenvalues of \( A \) are \( \lambda_1 = 3 \) and \( \lambda_2 = -3 \).

Step 2: Find the eigenvalues of \( A^2 \).
A fundamental property of eigenvalues states that if \( \lambda \) is an eigenvalue of \( A \), then \( \lambda^k \) is an eigenvalue of \( A^k \).
Therefore, the eigenvalues of \( A^2 \) are \( (\lambda_1)^2 \) and \( (\lambda_2)^2 \).
\( (\lambda_1)^2 = 3^2 = 9 \).
\( (\lambda_2)^2 = (-3)^2 = 9 \).
Thus, the eigenvalues of \( A^2 \) are \( (9, 9) \).

Detailed Solution:
By definition in linear algebra, when a quadratic form is reduced to its canonical (sum of squares) form via an orthogonal or non-singular transformation, the number of non-zero square terms present is exactly equal to the rank of the associated symmetric matrix. Therefore, the rank of the quadratic form is the total number of non-zero square terms in its canonical form.

Detailed Solution:
Step 1: Write \( x \) and \( y \) as functions of \( u \) and \( v \).
Given \( u = x \) and \( v = y \), we can inversely state that \( x = u \) and \( y = v \).

Step 2: Find the partial derivatives.
\( \frac{\partial x}{\partial u} = 1, \quad \frac{\partial x}{\partial v} = 0 \)
\( \frac{\partial y}{\partial u} = 0, \quad \frac{\partial y}{\partial v} = 1 \)

Step 3: Calculate the Jacobian determinant.
\( \frac{\partial(x,y)}{\partial(u,v)} = \det \begin{bmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{bmatrix} \)
\( = \det \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = (1)(1) - (0)(0) = 1 \).

Detailed Solution:
Step 1: Find the individual partial derivatives.
Given \( u(x, y, z) = x + y + z \).
Partial derivative with respect to \( x \): \( u_x = \frac{\partial u}{\partial x} = 1 + 0 + 0 = 1 \).
Partial derivative with respect to \( y \): \( u_y = \frac{\partial u}{\partial y} = 0 + 1 + 0 = 1 \).
Partial derivative with respect to \( z \): \( u_z = \frac{\partial u}{\partial z} = 0 + 0 + 1 = 1 \).

Step 2: Sum the partial derivatives.
\( u_x + u_y + u_z = 1 + 1 + 1 = 3 \).

Detailed Solution:
By Fermat's theorem for functions of multiple variables, if a differentiable function has a local extremum (maximum or minimum) at a point, then all its first-order partial derivatives must evaluate to zero at that point. A point where the gradient is zero is called a stationary point. Therefore, every extreme point of a differentiable function is necessarily a stationary point.

Detailed Solution:
The given series represents the expansion of a function of two variables \( f(x,y) \) about the origin \( (0,0) \). The Taylor series expansion of a function about the origin is specifically known as Maclaurin's series.

Detailed Solution:
By the Chain Rule of Jacobians, if \( u, v \) are functions of \( x, y \), and \( x, y \) are functions of \( r, \theta \), then the product of their Jacobians simplifies directly as:
\( J(u,v; x,y) \times J(x,y; r,\theta) = J(u,v; r,\theta) \).
In partial derivative notation, this is written as \( \frac{\partial(u,v)}{\partial(x,y)} \times \frac{\partial(x,y)}{\partial(r,\theta)} = \frac{\partial(u,v)}{\partial(r,\theta)} \).

Detailed Solution:
Step 1: Write the auxiliary equation.
The auxiliary equation for the given differential equation is \( (m+3)^2 = 0 \).

Step 2: Solve for the roots.
\( (m+3)(m+3) = 0 \implies m = -3, -3 \).
The roots are real and repeated.

Step 3: Write the general solution.
For a repeated real root \( m \), the complementary function is given by \( y = (c_1 + c_2 x)e^{mx} \).
Substituting \( m = -3 \), we get \( y = (c_1 + c_2 x)e^{-3x} \).

Detailed Solution:
Step 1: Recognize the equation type.
This is a Cauchy-Euler linear differential equation of the form \( (ax^2 D^2 + bx D + c)y = 0 \).

Step 2: Substitute to convert to constant coefficients.
Let \( x = e^z \), which implies \( z = \log x \).
Then, let \( D' \equiv \frac{d}{dz} \). We have \( xD = D' \) and \( x^2 D^2 = D'(D'-1) \).

Step 3: Transform the equation.
Substitute these into the given equation:
\( [D'(D'-1) + D' + 1]y = 0 \)
\( [D'^2 - D' + D' + 1]y = 0 \)
\( (D'^2 + 1)y = 0 \).

Step 4: Solve the new equation.
The auxiliary equation is \( m^2 + 1 = 0 \), so \( m = \pm i \).
The complementary function in terms of \( z \) is \( y = c_1 \cos z + c_2 \sin z \).

Step 5: Back-substitute \( z = \log x \).
\( y = c_1 \cos(\log x) + c_2 \sin(\log x) \).

Detailed Solution:
The Wronskian \( W \) of two functions \( f_1 \) and \( f_2 \) is defined as the determinant of the matrix constructed by placing the functions in the first row and their first derivatives in the second row.
\( W(f_1, f_2) = \det \begin{bmatrix} f_1 & f_2 \\ f_1' & f_2' \end{bmatrix} = f_1 f_2' - f_2 f_1' \).

Detailed Solution:
Step 1: Write the equation in operator form.
\( (D^2 + 2D + 1)y = e^x \).

Step 2: Find the Particular Integral (P.I.).
\( \text{P.I.} = \frac{1}{D^2 + 2D + 1} e^{ax} \), where \( a = 1 \).
Substitute \( D = a = 1 \) into the polynomial \( f(D) \).
\( f(1) = (1)^2 + 2(1) + 1 = 1 + 2 + 1 = 4 \).
Since \( f(1) \neq 0 \), we directly substitute:
\( \text{P.I.} = \frac{1}{4} e^x = \frac{e^x}{4} \).

Detailed Solution:
Step 1: Write the auxiliary equation.
The auxiliary equation is \( m^2 + 2m + 5 = 0 \).

Step 2: Find the roots.
Using the quadratic formula \( m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\( m = \frac{-2 \pm \sqrt{2^2 - 4(1)(5)}}{2} \)
\( m = \frac{-2 \pm \sqrt{4 - 20}}{2} = \frac{-2 \pm \sqrt{-16}}{2} \)
\( m = \frac{-2 \pm 4i}{2} = -1 \pm 2i \).
The roots are complex conjugates of the form \( \alpha \pm i\beta \) where \( \alpha = -1 \) and \( \beta = 2 \).

Step 3: Form the complementary function.
For complex roots, the complementary function is \( y_c = e^{\alpha x} (c_1 \cos \beta x + c_2 \sin \beta x) \).
Substitute \( \alpha = -1 \) and \( \beta = 2 \):
\( y_c = e^{-x} (c_1 \cos 2x + c_2 \sin 2x) \).
Expanding this, we get \( y_c = c_1 e^{-x} \cos 2x + c_2 e^{-x} \sin 2x \).

Detailed Solution:
By definition in differential geometry, the curvature \( \kappa \) of a curve at a point measures how fast the curve is changing direction. The reciprocal of the curvature is defined as the radius of curvature \( \rho \).
Formula: \( \rho = \frac{1}{\kappa} \).
Therefore, the reciprocal of the curvature is the radius of curvature.

Detailed Solution:
The envelope of a family of curves is defined as a curve that is tangent to each member of the family at some point. Alternatively, it can be defined as the locus of the points of intersection of consecutive (or infinitesimally close) members of the family of curves. Thus, the correct term is "Envelope".

Detailed Solution:
The evolute of a curve is the locus of its centers of curvature. A unique and well-known geometric property of a cycloid is that its evolute is another identical cycloid, but it is translated in space. Therefore, the evolute of a cycloid is a cycloid.

Detailed Solution:
Step 1: Find the first derivative \( y' \).
Given \( y = 4 \sin x \).
\( y' = \frac{dy}{dx} = 4 \cos x \).

Step 2: Find the second derivative \( y'' \).
\( y'' = \frac{d^2y}{dx^2} = -4 \sin x \).

Step 3: Evaluate derivatives at \( x = \pi/2 \).
At \( x = \pi/2 \):
\( y' = 4 \cos(\pi/2) = 4(0) = 0 \).
\( y'' = -4 \sin(\pi/2) = -4(1) = -4 \).

Step 4: Calculate the algebraic radius of curvature \( \rho \).
The formula for algebraic radius of curvature in Cartesian coordinates is:
\( \rho = \frac{(1 + (y')^2)^{3/2}}{y''} \).
\( \rho = \frac{(1 + 0^2)^{3/2}}{-4} = \frac{1}{-4} = -\frac{1}{4} \).
Note: While radius is often taken as positive magnitude, multiple choice options offering both positive and negative signs imply the algebraic radius of curvature is required.

Detailed Solution:
Step 1: Convert the mixed fraction.
\( 3 \frac{1}{2} = \frac{7}{2} \). We need to find \( \Gamma(7/2) \).

Step 2: Use the Gamma function property.
The property is \( \Gamma(n+1) = n \Gamma(n) \).
Applying this recursively:
\( \Gamma(7/2) = \Gamma(5/2 + 1) = \frac{5}{2} \Gamma(5/2) \)
\( \Gamma(5/2) = \frac{3}{2} \Gamma(3/2) \)
\( \Gamma(3/2) = \frac{1}{2} \Gamma(1/2) \)

Step 3: Calculate the final value.
Substitute the known value \( \Gamma(1/2) = \sqrt{\pi} \):
\( \Gamma(7/2) = \frac{5}{2} \times \frac{3}{2} \times \frac{1}{2} \times \sqrt{\pi} \)
\( \Gamma(7/2) = \frac{5 \times 3 \times 1}{2 \times 2 \times 2} \sqrt{\pi} = \frac{15}{8} \sqrt{\pi} \).

Detailed Solution:
The behavior of an infinite series \( \sum u_n \) depends on the limit of its sequence of partial sums, \( S_n \):
1. If \( \lim_{n \to \infty} S_n = L \) (finite), the series is convergent.
2. If \( \lim_{n \to \infty} S_n = \pm \infty \), the series is divergent.
3. If the limit neither exists finitely nor diverges to infinity (i.e., it fluctuates), the series is called an oscillatory series.

Detailed Solution:
By the standard p-series test (or Hyperharmonic series test), an infinite series of the form \( \sum \frac{1}{n^p} \):
- Converges if \( p > 1 \).
- Diverges if \( p \leq 1 \).
Thus, the condition for divergence is \( p \leq 1 \).

Detailed Solution:
According to Leibnitz's Test for alternating series, a series of the form \( \sum_{n=1}^{\infty} (-1)^{n-1} u_n \) converges if it satisfies two conditions:
1. The terms must decrease numerically in magnitude, i.e., \( u_{n+1} \leq u_n \) for all \( n \). This means \( \{u_n\} \) is a monotonic decreasing sequence.
2. The limit of the nth term must be zero as n approaches infinity: \( \lim_{n \to \infty} u_n = 0 \).
Only option (B) lists both necessary conditions.

Detailed Solution:
By the mathematical definition of absolute convergence, an infinite series \( \sum u_n \) is said to be absolutely convergent if the series of absolute values of its terms, \( \sum |u_n| \), is convergent.

Detailed Solution:
Step 1: Define the function and find its derivatives.
\( f(x) = \log(1+x) \implies f(0) = \log(1) = 0 \).
\( f'(x) = \frac{1}{1+x} = (1+x)^{-1} \implies f'(0) = 1 \).
\( f''(x) = -1(1+x)^{-2} \implies f''(0) = -1 \).
\( f'''(x) = 2(1+x)^{-3} \implies f'''(0) = 2 \).

Step 2: Apply the Maclaurin series formula.
The Maclaurin series is given by:
\( f(x) = f(0) + \frac{x}{1!} f'(0) + \frac{x^2}{2!} f''(0) + \frac{x^3}{3!} f'''(0) + \dots \)
Substitute the values evaluated at \( x = 0 \):
\( f(x) = 0 + \frac{x}{1}(1) + \frac{x^2}{2}(-1) + \frac{x^3}{6}(2) + \dots \)
\( f(x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \dots \)

Solution for (a):
Step 1: Define an orthogonal matrix.
Let \( A \) be a real orthogonal matrix of order \( n \).
By definition of an orthogonal matrix, \( A A^T = A^T A = I \), where \( A^T \) is the transpose of \( A \) and \( I \) is the identity matrix.

Step 2: Set up the eigenvalue equation.
Let \( \lambda \) be an eigenvalue of \( A \), and let \( X \) be the corresponding non-zero eigenvector.
Then, \( A X = \lambda X \).      ... (Equation 1)

Step 3: Take the conjugate transpose (theta) of both sides.
Let \( X^\theta \) denote the conjugate transpose of \( X \), and \( A^\theta \) denote the conjugate transpose of \( A \).
Taking the conjugate transpose of Equation 1:
\( (A X)^\theta = (\lambda X)^\theta \)
Using the property \( (AB)^\theta = B^\theta A^\theta \):
\( X^\theta A^\theta = \bar{\lambda} X^\theta \)
Since \( A \) is a real matrix, \( A^\theta = A^T \). Therefore:
\( X^\theta A^T = \bar{\lambda} X^\theta \).      ... (Equation 2)

Step 4: Multiply Equation 2 and Equation 1.
Multiply the left hand side of Equation 2 with the left hand side of Equation 1, and similarly for the right hand side:
\( (X^\theta A^T) (A X) = (\bar{\lambda} X^\theta) (\lambda X) \)
By associativity of matrix multiplication:
\( X^\theta (A^T A) X = (\bar{\lambda} \lambda) (X^\theta X) \)

Step 5: Simplify using the orthogonal property.
We know that \( A^T A = I \), and \( \bar{\lambda} \lambda = |\lambda|^2 \). Substituting these:
\( X^\theta (I) X = |\lambda|^2 X^\theta X \)
\( X^\theta X = |\lambda|^2 X^\theta X \)

Step 6: Conclude the proof.
Since \( X \) is a non-zero eigenvector, its inner product with itself is strictly positive, meaning \( X^\theta X \neq 0 \).
Dividing both sides by the scalar \( X^\theta X \):
\( 1 = |\lambda|^2 \)
Taking the positive square root:
\( |\lambda| = 1 \).
Hence, it is proved that the eigenvalues of an orthogonal matrix are of unit modulus.

Solution for (b):
Step 1: Identify the quadratic form of the conic.
The given equation of the conic is \( x^2 + y^2 - 2xy - 2x - 2y + 1 = 0 \).
The quadratic part is \( Q(x,y) = x^2 - 2xy + y^2 \).
The symmetric matrix \( A \) associated with this quadratic form is:
\( A = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \).

Step 2: Find the eigenvalues of \( A \).
The characteristic equation is \( |A - \lambda I| = 0 \).
\( \begin{vmatrix} 1 - \lambda & -1 \\ -1 & 1 - \lambda \end{vmatrix} = 0 \)
\( (1 - \lambda)^2 - (-1)(-1) = 0 \)
\( 1 - 2\lambda + \lambda^2 - 1 = 0 \implies \lambda^2 - 2\lambda = 0 \implies \lambda(\lambda - 2) = 0 \).
The eigenvalues are \( \lambda_1 = 2 \) and \( \lambda_2 = 0 \).

Step 3: Find the corresponding normalized eigenvectors.
For \( \lambda_1 = 2 \):
\( (A - 2I)X = 0 \implies \begin{bmatrix} -1 & -1 \\ -1 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \).
\( -x - y = 0 \implies x = -y \).
Let \( y = -1 \), then \( x = 1 \). Eigenvector \( X_1 = \begin{bmatrix} 1 \\ -1 \end{bmatrix} \).
Normalized \( X_1 = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ -1 \end{bmatrix} \).

For \( \lambda_2 = 0 \):
\( (A - 0I)X = 0 \implies \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} \).
\( x - y = 0 \implies x = y \).
Let \( y = 1 \), then \( x = 1 \). Eigenvector \( X_2 = \begin{bmatrix} 1 \\ 1 \end{bmatrix} \).
Normalized \( X_2 = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix} \).

Step 4: Form the orthogonal transformation matrix \( P \).
\( P = [X_1 \ X_2] = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 & 1 \\ -1 & 1 \end{bmatrix} \).
This transformation \( \begin{bmatrix} x \\ y \end{bmatrix} = P \begin{bmatrix} X \\ Y \end{bmatrix} \) rotates the axes to align with the principal axes.
\( x = \frac{X + Y}{\sqrt{2}} \) and \( y = \frac{-X + Y}{\sqrt{2}} \).

Step 5: Apply the transformation to the original equation.
The quadratic part \( x^2 - 2xy + y^2 \) becomes \( \lambda_1 X^2 + \lambda_2 Y^2 = 2X^2 + 0Y^2 = 2X^2 \).
The linear part \( -2x - 2y \) becomes:
\( -2\left( \frac{X + Y}{\sqrt{2}} \right) - 2\left( \frac{-X + Y}{\sqrt{2}} \right) = \frac{-2X - 2Y + 2X - 2Y}{\sqrt{2}} = \frac{-4Y}{\sqrt{2}} = -2\sqrt{2}Y \).

Substitute these into the original conic equation:
\( 2X^2 - 2\sqrt{2}Y + 1 = 0 \).

Step 6: Identify the nature of the conic.
Rearrange the equation into standard form:
\( 2X^2 = 2\sqrt{2}Y - 1 \)
\( X^2 = \sqrt{2}Y - \frac{1}{2} \)
\( X^2 = \sqrt{2} \left( Y - \frac{1}{2\sqrt{2}} \right) \).

This is an equation of the form \( X^2 = 4a(Y - k) \), which represents a parabola.
The principal axis is the line \( X = 0 \), which corresponds to the eigenvector for \( \lambda_1 = 2 \), i.e., \( x + y = 0 \).

Solution for (a):
Step 1: Write down Maclaurin's series for a function of two variables.
Let \( f(x,y) = e^x \cos y \). The expansion around \( (0,0) \) up to the third degree is given by:
\( f(x,y) = f(0,0) + (x f_x + y f_y) + \frac{1}{2!} (x^2 f_{xx} + 2xy f_{xy} + y^2 f_{yy}) + \frac{1}{3!} (x^3 f_{xxx} + 3x^2y f_{xxy} + 3xy^2 f_{xyy} + y^3 f_{yyy}) + \dots \)
where all partial derivatives are evaluated at \( (0,0) \).

Step 2: Evaluate the function at \( (0,0) \).
\( f(0,0) = e^0 \cos 0 = 1 \cdot 1 = 1 \).

Step 3: Evaluate first-order partial derivatives at \( (0,0) \).
\( f_x = \frac{\partial f}{\partial x} = e^x \cos y \implies f_x(0,0) = e^0 \cos 0 = 1 \).
\( f_y = \frac{\partial f}{\partial y} = -e^x \sin y \implies f_y(0,0) = -e^0 \sin 0 = 0 \).

Step 4: Evaluate second-order partial derivatives at \( (0,0) \).
\( f_{xx} = \frac{\partial^2 f}{\partial x^2} = e^x \cos y \implies f_{xx}(0,0) = 1 \).
\( f_{xy} = \frac{\partial^2 f}{\partial x \partial y} = -e^x \sin y \implies f_{xy}(0,0) = 0 \).
\( f_{yy} = \frac{\partial^2 f}{\partial y^2} = -e^x \cos y \implies f_{yy}(0,0) = -1 \).

Step 5: Evaluate third-order partial derivatives at \( (0,0) \).
\( f_{xxx} = \frac{\partial^3 f}{\partial x^3} = e^x \cos y \implies f_{xxx}(0,0) = 1 \).
\( f_{xxy} = \frac{\partial^3 f}{\partial x^2 \partial y} = -e^x \sin y \implies f_{xxy}(0,0) = 0 \).
\( f_{xyy} = \frac{\partial^3 f}{\partial x \partial y^2} = -e^x \cos y \implies f_{xyy}(0,0) = -1 \).
\( f_{yyy} = \frac{\partial^3 f}{\partial y^3} = e^x \sin y \implies f_{yyy}(0,0) = 0 \).

Step 6: Substitute values into the Maclaurin series formula.
\( f(x,y) = 1 + (x \cdot 1 + y \cdot 0) + \frac{1}{2!} (x^2 \cdot 1 + 2xy \cdot 0 + y^2 \cdot (-1)) + \frac{1}{3!} (x^3 \cdot 1 + 3x^2y \cdot 0 + 3xy^2 \cdot (-1) + y^3 \cdot 0) \)
\( f(x,y) = 1 + x + \frac{1}{2}(x^2 - y^2) + \frac{1}{6}(x^3 - 3xy^2) \)
\( e^x \cos y = 1 + x + \frac{x^2}{2} - \frac{y^2}{2} + \frac{x^3}{6} - \frac{xy^2}{2} \).
This is the required expansion up to the third degree.

Solution for (b):
Step 1: Check for functional dependence using the Jacobian.
Three functions \( u, v, w \) of variables \( x, y, z \) are functionally dependent if their Jacobian \( J(u,v,w) = \frac{\partial(u,v,w)}{\partial(x,y,z)} = 0 \).
Given:
\( u = x + y - z \)
\( v = x - y + z \)
\( w = x^2 + y^2 + z^2 - 2yz \)

Step 2: Calculate the partial derivatives.
For \( u \): \( u_x = 1, u_y = 1, u_z = -1 \).
For \( v \): \( v_x = 1, v_y = -1, v_z = 1 \).
For \( w \): \( w_x = 2x, w_y = 2y - 2z, w_z = 2z - 2y \).

Step 3: Evaluate the Jacobian determinant.
\( J = \det \begin{bmatrix} u_x & u_y & u_z \\ v_x & v_y & v_z \\ w_x & w_y & w_z \end{bmatrix} = \det \begin{bmatrix} 1 & 1 & -1 \\ 1 & -1 & 1 \\ 2x & 2(y-z) & 2(z-y) \end{bmatrix} \)
Factor out 2 from the third row:
\( J = 2 \det \begin{bmatrix} 1 & 1 & -1 \\ 1 & -1 & 1 \\ x & y-z & z-y \end{bmatrix} \)
Apply column operation: \( C_2 \to C_2 + C_3 \)
\( J = 2 \det \begin{bmatrix} 1 & 0 & -1 \\ 1 & 0 & 1 \\ x & 0 & z-y \end{bmatrix} \)
Since the entire second column is zero, the determinant \( J = 0 \).
Therefore, the functions \( u, v, w \) are functionally dependent.

Step 4: Find the relation between them.
We need to express \( w \) in terms of \( u \) and \( v \).
Notice that \( w = x^2 + y^2 + z^2 - 2yz \) can be rewritten as a perfect square:
\( w = x^2 + (y - z)^2 \).

Let's find expressions for \( x \) and \( (y - z) \) using \( u \) and \( v \):
Adding \( u \) and \( v \):
\( u + v = (x + y - z) + (x - y + z) = 2x \implies x = \frac{u+v}{2} \).

Subtracting \( v \) from \( u \):
\( u - v = (x + y - z) - (x - y + z) = 2y - 2z = 2(y - z) \implies (y - z) = \frac{u-v}{2} \).

Substitute these into the expression for \( w \):
\( w = \left( \frac{u+v}{2} \right)^2 + \left( \frac{u-v}{2} \right)^2 \)
\( w = \frac{u^2 + 2uv + v^2}{4} + \frac{u^2 - 2uv + v^2}{4} \)
\( w = \frac{2u^2 + 2v^2}{4} \)
\( w = \frac{u^2 + v^2}{2} \)
\( 2w = u^2 + v^2 \).

The required relation is \( u^2 + v^2 = 2w \).

Solution for (a):
Step 1: Find the Complementary Function (C.F.).
The auxiliary equation is \( m^2 - 4m + 4 = 0 \).
\( (m - 2)^2 = 0 \)
Roots are \( m = 2, 2 \) (real and repeated).
Therefore, C.F. = \( (c_1 + c_2 x)e^{2x} \).

Step 2: Find the Particular Integral (P.I.).
\( \text{P.I.} = \frac{1}{D^2 - 4D + 4} (x^2 e^{2x}) \)
\( \text{P.I.} = \frac{1}{(D - 2)^2} (x^2 e^{2x}) \)

Use the exponential shift rule: \( \frac{1}{f(D)} [e^{ax} V] = e^{ax} \frac{1}{f(D+a)} V \). Here \( a = 2 \), \( V = x^2 \).
\( \text{P.I.} = e^{2x} \frac{1}{(D + 2 - 2)^2} (x^2) \)
\( \text{P.I.} = e^{2x} \frac{1}{D^2} (x^2) \)

The operator \( \frac{1}{D} \) means integration with respect to \( x \). So \( \frac{1}{D^2} \) means integrating twice.
First integration: \( \frac{1}{D} (x^2) = \int x^2 \, dx = \frac{x^3}{3} \).
Second integration: \( \frac{1}{D} \left( \frac{x^3}{3} \right) = \int \frac{x^3}{3} \, dx = \frac{x^4}{12} \).

So, \( \text{P.I.} = e^{2x} \left( \frac{x^4}{12} \right) = \frac{x^4 e^{2x}}{12} \).

Step 3: Write the complete solution.
The general solution is \( y = \text{C.F.} + \text{P.I.} \)
\( y = (c_1 + c_2 x)e^{2x} + \frac{x^4 e^{2x}}{12} \).

Solution for (b):
Step 1: Find the Complementary Function (C.F.).
The auxiliary equation is \( m^2 - 2m + 2 = 0 \).
Using the quadratic formula, \( m = \frac{2 \pm \sqrt{4 - 8}}{2} = \frac{2 \pm \sqrt{-4}}{2} = 1 \pm i \).
Here \( \alpha = 1 \) and \( \beta = 1 \).
Therefore, C.F. = \( e^x (c_1 \cos x + c_2 \sin x) \).

Step 2: Find the Particular Integral (P.I.).
\( \text{P.I.} = \frac{1}{D^2 - 2D + 2} (e^x \sin x) \)

Use the exponential shift rule: \( \frac{1}{f(D)} [e^{ax} V] = e^{ax} \frac{1}{f(D+a)} V \). Here \( a = 1 \), \( V = \sin x \).
\( \text{P.I.} = e^x \frac{1}{(D + 1)^2 - 2(D + 1) + 2} (\sin x) \)
\( \text{P.I.} = e^x \frac{1}{D^2 + 2D + 1 - 2D - 2 + 2} (\sin x) \)
\( \text{P.I.} = e^x \frac{1}{D^2 + 1} (\sin x) \)

Now, for \( \sin(bx) \), we substitute \( D^2 = -b^2 \). Here \( b = 1 \), so \( D^2 = -1 \).
Substituting \( D^2 = -1 \) in the denominator gives \( -1 + 1 = 0 \) (case of failure).
When failure occurs, multiply the numerator by \( x \) and differentiate the denominator with respect to \( D \):
\( \text{P.I.} = e^x \cdot x \cdot \frac{1}{\frac{d}{dD}(D^2 + 1)} (\sin x) \)
\( \text{P.I.} = x e^x \frac{1}{2D} (\sin x) \)

The operator \( \frac{1}{D} \) stands for integration.
\( \text{P.I.} = \frac{x e^x}{2} \int \sin x \, dx \)
\( \text{P.I.} = \frac{x e^x}{2} (-\cos x) = -\frac{x e^x \cos x}{2} \).

Step 3: Write the complete solution.
The general solution is \( y = \text{C.F.} + \text{P.I.} \)
\( y = e^x (c_1 \cos x + c_2 \sin x) - \frac{x e^x \cos x}{2} \).

Solution for (a):
Step 1: Write the parametric equations of the parabola.
For the parabola \( y^2 = 4ax \), any point on the curve can be represented parametrically as:
\( x = at^2 \)
\( y = 2at \)

Step 2: Find the first and second derivatives of \( y \) with respect to \( x \).
Differentiate \( x \) and \( y \) with respect to \( t \):
\( \frac{dx}{dt} = 2at \)
\( \frac{dy}{dt} = 2a \)

First derivative \( y_1 \):
\( y_1 = \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2a}{2at} = \frac{1}{t} \).

Second derivative \( y_2 \):
\( y_2 = \frac{d^2y}{dx^2} = \frac{d}{dx} \left( \frac{1}{t} \right) = \frac{d}{dt} \left( \frac{1}{t} \right) \cdot \frac{dt}{dx} \)
\( y_2 = \left( -\frac{1}{t^2} \right) \cdot \frac{1}{2at} = -\frac{1}{2at^3} \).

Step 3: Find the coordinates of the center of curvature \( (\bar{x}, \bar{y}) \).
Let's calculate \( 1 + y_1^2 \):
\( 1 + y_1^2 = 1 + \frac{1}{t^2} = \frac{t^2 + 1}{t^2} \).

The formula for the x-coordinate of the center of curvature is:
\( \bar{x} = x - \frac{y_1(1 + y_1^2)}{y_2} \)
\( \bar{x} = at^2 - \frac{\frac{1}{t} \left( \frac{t^2 + 1}{t^2} \right)}{-\frac{1}{2at^3}} \)
\( \bar{x} = at^2 + \frac{t^2 + 1}{t^3} \cdot 2at^3 \)
\( \bar{x} = at^2 + 2a(t^2 + 1) = at^2 + 2at^2 + 2a = 3at^2 + 2a \).

The formula for the y-coordinate of the center of curvature is:
\( \bar{y} = y + \frac{1 + y_1^2}{y_2} \)
\( \bar{y} = 2at + \frac{\frac{t^2 + 1}{t^2}}{-\frac{1}{2at^3}} \)
\( \bar{y} = 2at - \left( \frac{t^2 + 1}{t^2} \right)(2at^3) \)
\( \bar{y} = 2at - 2at(t^2 + 1) = 2at - 2at^3 - 2at = -2at^3 \).

Step 4: Eliminate the parameter 't' to find the locus (the evolute).
From the \( \bar{x} \) equation:
\( 3at^2 = \bar{x} - 2a \implies t^2 = \frac{\bar{x} - 2a}{3a} \).

Square the \( \bar{y} \) equation to eliminate \( t \):
\( \bar{y}^2 = (-2at^3)^2 = 4a^2 t^6 = 4a^2 (t^2)^3 \).

Substitute the expression for \( t^2 \):
\( \bar{y}^2 = 4a^2 \left( \frac{\bar{x} - 2a}{3a} \right)^3 = 4a^2 \frac{(\bar{x} - 2a)^3}{27a^3} = \frac{4}{27a}(\bar{x} - 2a)^3 \).

Cross-multiplying, we get:
\( 27a \bar{y}^2 = 4(\bar{x} - 2a)^3 \).

Replacing \( (\bar{x}, \bar{y}) \) with \( (x, y) \), the equation of the evolute is:
\( 27ay^2 = 4(x - 2a)^3 \). (This represents a semi-cubical parabola).

Solution for (b):
Step 1: Write the parametric equations of the curve.
\( x = a(\cos t + t \sin t) \)
\( y = a(\sin t - t \cos t) \)

Step 2: Differentiate \( x \) and \( y \) with respect to \( t \).
Using the product rule on the \( t \sin t \) and \( t \cos t \) terms:
\( \frac{dx}{dt} = a \left[ -\sin t + (\sin t + t \cos t) \right] = a(t \cos t) \).
\( \frac{dy}{dt} = a \left[ \cos t - (\cos t - t \sin t) \right] = a(t \sin t) \).

Step 3: Find the first derivative \( y_1 \).
\( y_1 = \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \)
\( y_1 = \frac{a t \sin t}{a t \cos t} = \tan t \).

Step 4: Find the second derivative \( y_2 \).
\( y_2 = \frac{d^2y}{dx^2} = \frac{d}{dx}(\tan t) = \frac{d}{dt}(\tan t) \cdot \frac{dt}{dx} \)
\( y_2 = \sec^2 t \cdot \frac{1}{at \cos t} \)
\( y_2 = \frac{\sec^2 t \cdot \sec t}{at} = \frac{\sec^3 t}{at} \).

Step 5: Calculate the radius of curvature \( \rho \).
The formula for radius of curvature is \( \rho = \frac{(1 + y_1^2)^{3/2}}{|y_2|} \).
Let's evaluate \( 1 + y_1^2 \):
\( 1 + y_1^2 = 1 + \tan^2 t = \sec^2 t \).

Now substitute into the formula:
\( \rho = \frac{(\sec^2 t)^{3/2}}{\frac{\sec^3 t}{at}} \)
\( \rho = \frac{\sec^3 t}{\frac{\sec^3 t}{at}} \)
\( \rho = \sec^3 t \cdot \frac{at}{\sec^3 t} \)
\( \rho = at \).

Therefore, the radius of curvature at any point 't' is \( at \).

Solution for (a):
Step 1: Simplify the nth term of the series.
Let \( u_n = \sqrt{n^3 + 1} - \sqrt{n^3} \).
To simplify this, we multiply and divide by its conjugate \( (\sqrt{n^3 + 1} + \sqrt{n^3}) \):
\( u_n = \frac{(\sqrt{n^3 + 1} - \sqrt{n^3})(\sqrt{n^3 + 1} + \sqrt{n^3})}{\sqrt{n^3 + 1} + \sqrt{n^3}} \)

Using the algebraic identity \( (a-b)(a+b) = a^2 - b^2 \):
\( u_n = \frac{(\sqrt{n^3 + 1})^2 - (\sqrt{n^3})^2}{\sqrt{n^3 + 1} + \sqrt{n^3}} \)
\( u_n = \frac{n^3 + 1 - n^3}{\sqrt{n^3 + 1} + \sqrt{n^3}} \)
\( u_n = \frac{1}{\sqrt{n^3 + 1} + \sqrt{n^3}} \).

Step 2: Choose an auxiliary series \( v_n \) for limit comparison.
Let's extract the highest power of \( n \) from the denominator to find the asymptotic behavior.
\( u_n = \frac{1}{n^{3/2} \left( \sqrt{1 + \frac{1}{n^3}} + 1 \right)} \).
Let \( v_n = \frac{1}{n^{3/2}} \).

Step 3: Apply the Limit Comparison Test.
Evaluate \( \lim_{n \to \infty} \frac{u_n}{v_n} \):
\( \lim_{n \to \infty} \frac{u_n}{v_n} = \lim_{n \to \infty} \frac{ \frac{1}{n^{3/2} \left( \sqrt{1 + \frac{1}{n^3}} + 1 \right)} }{ \frac{1}{n^{3/2}} } \)
\( \lim_{n \to \infty} \frac{u_n}{v_n} = \lim_{n \to \infty} \frac{1}{\sqrt{1 + \frac{1}{n^3}} + 1} \)
\( = \frac{1}{\sqrt{1 + 0} + 1} = \frac{1}{1 + 1} = \frac{1}{2} \).

Since the limit is a finite, non-zero constant (\( 1/2 \)), by the Limit Comparison Test, both series \( \sum u_n \) and \( \sum v_n \) converge or diverge together.

Step 4: Analyze the convergence of \( \sum v_n \).
The series \( \sum v_n = \sum \frac{1}{n^{3/2}} \) is a p-series of the form \( \sum \frac{1}{n^p} \) with \( p = 3/2 \).
Since \( p = 3/2 > 1 \), the series \( \sum v_n \) converges by the p-series test.

Step 5: Conclusion.
Since \( \sum v_n \) converges, the given series \( \sum_{n=1}^\infty \left( \sqrt{n^3 + 1} - \sqrt{n^3} \right) \) also converges.

Solution for (b):
Step 1: Test for absolute convergence.
A series \( \sum u_n \) is absolutely convergent if the series of its absolute values \( \sum |u_n| \) converges.
Here, \( u_n = \frac{(-1)^n n}{n^2+1} \).
So, the absolute series is \( \sum |u_n| = \sum_{n=1}^\infty \frac{n}{n^2+1} \).

Step 2: Use Limit Comparison Test for \( \sum |u_n| \).
Let \( a_n = \frac{n}{n^2+1} \).
Choose an auxiliary series based on the leading terms: let \( v_n = \frac{n}{n^2} = \frac{1}{n} \).
Evaluate the limit:
\( \lim_{n \to \infty} \frac{a_n}{v_n} = \lim_{n \to \infty} \frac{ \frac{n}{n^2+1} }{ \frac{1}{n} } \)
\( = \lim_{n \to \infty} \frac{n^2}{n^2+1} = \lim_{n \to \infty} \frac{1}{1 + \frac{1}{n^2}} = \frac{1}{1+0} = 1 \).

Since the limit is a finite, non-zero constant, \( \sum a_n \) and \( \sum v_n \) converge or diverge together.
The series \( \sum v_n = \sum \frac{1}{n} \) is the harmonic series (p-series with \( p=1 \)), which is known to be divergent.
Therefore, \( \sum |u_n| \) is divergent. This means the given series is not absolutely convergent.

Step 3: Test for conditional convergence (Leibnitz's Test).
Since it is an alternating series, let's test if it converges conditionally using Leibnitz's Alternating Series Test.
Let \( b_n = \frac{n}{n^2+1} \).
Condition 1: Check the limit as \( n \to \infty \).
\( \lim_{n \to \infty} b_n = \lim_{n \to \infty} \frac{n}{n^2(1+1/n^2)} = \lim_{n \to \infty} \frac{1}{n(1+1/n^2)} = 0 \). (Condition satisfied).

Condition 2: Check for monotonically decreasing terms.
Let \( f(x) = \frac{x}{x^2+1} \).
\( f'(x) = \frac{(x^2+1)(1) - x(2x)}{(x^2+1)^2} = \frac{x^2+1-2x^2}{(x^2+1)^2} = \frac{1-x^2}{(x^2+1)^2} \).
For \( x > 1 \), \( 1-x^2 < 0 \), meaning \( f'(x) < 0 \).
Therefore, \( b_{n+1} < b_n \) for all \( n \ge 1 \), so the sequence is monotonically decreasing. (Condition satisfied).

Conclusion:
Since both conditions of Leibnitz's test are satisfied, the series converges.
Because the absolute series diverges but the alternating series converges, the given series is conditionally convergent, but not absolutely convergent.