Calculus and Linear Algebra - PYQ May 2019

Detailed Solution:
Step 1: Use the properties of Eigenvalues.
1. Sum of eigenvalues = Trace of the matrix (sum of principal diagonal elements).
2. Product of eigenvalues = Determinant of the matrix.

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

Step 3: Verify the options.
We need two numbers whose sum is 5 and product is -6.
Option (B) has roots -1 and 6.
Sum: \( -1 + 6 = 5 \).
Product: \( (-1) \times 6 = -6 \).
This satisfies both conditions. Therefore, the eigenvalues are -1 and 6.

Detailed Solution:
Step 1: Find the eigenvalues of the given matrix.
The matrix \( A = \begin{pmatrix} 1 & 2 \ 0 & 3 \end{pmatrix} \) is an upper triangular matrix.
The eigenvalues of an upper triangular matrix are simply the elements on its main diagonal.
Therefore, the eigenvalues of \( A \) are \( \lambda_1 = 1 \) and \( \lambda_2 = 3 \).

Step 2: Find the inverse of these eigenvalues.
The inverse of \( \lambda_1 \) is \( 1/\lambda_1 = 1/1 = 1 \).
The inverse of \( \lambda_2 \) is \( 1/\lambda_2 = 1/3 \).
The inverse of the eigenvalues are 1 and 1/3.

Detailed Solution:
Step 1: Write the symmetric matrix \( A \) of the quadratic form.
The diagonal elements are the coefficients of \( x_1^2, x_2^2, x_3^2 \) respectively: 1, 2, 3.
The off-diagonal elements \( a_{ij} = a_{ji} \) are half the coefficients of \( x_i x_j \).
\( a_{12} = a_{21} = 2/2 = 1 \)
\( a_{23} = a_{32} = 2/2 = 1 \)
\( a_{13} = a_{31} = -2/2 = -1 \)
\( A = \begin{pmatrix} 1 & 1 & -1 \ 1 & 2 & 1 \ -1 & 1 & 3 \end{pmatrix} \).

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

Step 3: Determine the nature.
Since the principal minors are \( D_1 > 0 \), \( D_2 > 0 \), and \( D_3 < 0 \), the signs are not all positive, nor do they alternate starting with negative. Therefore, the nature of the quadratic form is Indefinite.

Detailed Solution:
Step 1: Identify the number of positive and negative terms in the canonical form.
The canonical form is \( 2y_1^2 - y_2^2 - y_3^2 \).
Number of positive squared terms (\( p \)) = 1 (which is \( 2y_1^2 \)).
Number of negative squared terms (\( q \)) = 2 (which are \( -y_2^2 \) and \( -y_3^2 \)).

Step 2: Calculate the signature.
The signature of a quadratic form is defined as the difference between the number of positive terms and the number of negative terms.
Signature \( s = p - q = 1 - 2 = -1 \).

Detailed Solution:
Two functions \( u(x,y) \) and \( v(x,y) \) are said to be functionally dependent if there exists a relation \( F(u,v) = 0 \) between them, independent of \( x \) and \( y \).
A necessary and sufficient condition for \( u \) and \( v \) to be functionally dependent is that their Jacobian with respect to \( x \) and \( y \) must be identically zero.
That is, \( J = \frac{\partial(u,v)}{\partial(x,y)} = 0 \).

Detailed Solution:
According to the second derivative test for functions of two variables \( f(x,y) \):
Let \( r = f_{xx} \), \( s = f_{xy} \), and \( t = f_{yy} \) evaluated at a critical point \( (a,b) \).
- If \( rt - s^2 > 0 \) and \( r < 0 \), the point is a local maximum.
- If \( rt - s^2 > 0 \) and \( r > 0 \), the point is a local minimum.
- If \( rt - s^2 < 0 \), the function has neither a maximum nor a minimum at the point. This point is called a saddle point.
Therefore, if \( rt - s^2 < 0 \), the point is a saddle point.

Detailed Solution:
Given \( u(x,y) = x^2 + y^2 + 3xy \).
To find the partial derivative \( \frac{\partial u}{\partial x} \), we differentiate \( u \) with respect to \( x \), treating \( y \) as a constant.
\( \frac{\partial u}{\partial x} = \frac{\partial}{\partial x}(x^2) + \frac{\partial}{\partial x}(y^2) + \frac{\partial}{\partial x}(3xy) \)
\( \frac{\partial u}{\partial x} = 2x + 0 + 3y(1) = 2x + 3y \).

Detailed Solution:
Step 1: Define the Jacobian matrix.
\( J = \frac{\partial(u,v)}{\partial(x,y)} = \begin{vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{vmatrix} \).

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

Step 3: Evaluate the determinant.
\( J = \begin{vmatrix} 2x & 0 \ 0 & 2y \end{vmatrix} = (2x)(2y) - (0)(0) = 4xy \).

Detailed Solution:
Step 1: Write the Auxiliary Equation.
The Auxiliary Equation is \( m^2 - 2m + 1 = 0 \).

Step 2: Find the roots of the AE.
The equation can be factored as \( (m - 1)^2 = 0 \).
So the roots are \( m = 1, 1 \) (real and repeated).

Step 3: Write the Complementary Function (C.F.).
For real and repeated roots \( m_1 = m_2 = m \), the C.F. is given by \( (C_1 + C_2 x)e^{mx} \).
Here, \( m = 1 \), so C.F. = \( (C_1 + C_2 x)e^{1x} = (C_1 + C_2 x)e^x \).

Detailed Solution:
Step 1: Set up the Particular Integral expression.
\( P.I. = \frac{1}{D^2 + 9} e^{-2x} \).

Step 2: Apply the rule for \( e^{ax} \).
Substitute \( D = a = -2 \).
\( P.I. = \frac{1}{(-2)^2 + 9} e^{-2x} \)
\( P.I. = \frac{1}{4 + 9} e^{-2x} = \frac{e^{-2x}}{13} \).

Detailed Solution:
Step 1: Identify the type of equation and substitute.
This is a Cauchy-Euler equation. Let \( x = e^z \implies z = \log x \).
Let \( \theta = \frac{d}{dz} \).
Then \( xD = \theta \) and \( x^2D^2 = \theta(\theta - 1) = \theta^2 - \theta \).

Step 2: Transform the differential equation.
Substitute into the equation:
\( (\theta^2 - \theta + \theta + 1)y = 0 \)
\( (\theta^2 + 1)y = 0 \).

Step 3: Solve the constant-coefficient ODE in \( z \).
The Auxiliary Equation is \( m^2 + 1 = 0 \implies m = \pm i \).
The roots are complex conjugates \( \alpha \pm i\beta \) with \( \alpha = 0, \beta = 1 \).
The solution is \( y = e^{0z}(A \cos(1z) + B \sin(1z)) = A \cos z + B \sin z \).

Detailed Solution:
For complex conjugate roots \( m = \alpha \pm i\beta \) of the auxiliary equation, the standard form of the complementary function is given by:
\( y = e^{\alpha x} (c_1 \cos \beta x + c_2 \sin \beta x) \), where \( c_1 \) and \( c_2 \) are arbitrary constants.
Option (D) exactly matches this standard formula.

Detailed Solution:
A straight line has a constant slope, meaning its direction never changes.
Curvature \( \kappa \) is defined as the rate of change of the tangent angle with respect to arc length, \( \kappa = \left| \frac{d\phi}{ds} \right| \).
Since the angle \( \phi \) is constant for a straight line, \( \frac{d\phi}{ds} = 0 \).
Thus, the curvature of any straight line is 0.

Detailed Solution:
Step 1: Write the formula for radius of curvature.
\( \rho = \frac{(1 + (y')^2)^{3/2}}{|y''|} \)

Step 2: Find the first and second derivatives.
Given \( y = 4\sin x \).
\( y' = \frac{dy}{dx} = 4\cos x \).
\( y'' = \frac{d^2y}{dx^2} = -4\sin x \).

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 \( \rho \).
\( \rho = \frac{(1 + (0)^2)^{3/2}}{|-4|} = \frac{1^{3/2}}{4} = \frac{1}{4} \).
Note: Radius of curvature represents a physical distance and is thus typically taken as positive.

Detailed Solution:
By definition in differential geometry, the evolute of a curve is the locus of all its centers of curvature.
Conversely, the original curve is the involute of its evolute.

Detailed Solution:
This is a standard known value of the Gamma function.
\( \Gamma(1/2) = \int_0^\infty t^{-1/2} e^{-t} dt \).
Substituting \( t = x^2 \implies dt = 2x dx \), the integral becomes \( 2\int_0^\infty e^{-x^2} dx \).
Using the Gaussian integral, \( \int_0^\infty e^{-x^2} dx = \frac{\sqrt{\pi}}{2} \).
So \( \Gamma(1/2) = 2 \left( \frac{\sqrt{\pi}}{2} \right) = \sqrt{\pi} \).

Detailed Solution:
This is the standard p-series test. The series \( \sum \frac{1}{n^p} \) converges if and only if \( p > 1 \), and diverges if \( p \le 1 \).

Detailed Solution:
We know the Maclaurin series expansion of \( e^x \) is \( e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \)
Putting \( x = 1 \): \( e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \dots = 1 + \sum_{n=1}^\infty \frac{1}{n!} \).
Thus, \( \sum_{n=1}^\infty \frac{1}{n!} = e - 1 \).
Since the sum is a finite real number, the series is convergent.

Detailed Solution:
By definition, a series \( \sum u_n \) is absolutely convergent if the series formed by the absolute values of its terms, \( \sum |u_n| \), is convergent.

Detailed Solution:
Step 1: Let \( u_n = \frac{n^3}{3^n} \). Use D'Alembert's Ratio Test.
\( u_{n+1} = \frac{(n+1)^3}{3^{n+1}} \).

Step 2: Evaluate the limit.
\( \lim_{n \to \infty} \left| \frac{u_{n+1}}{u_n} \right| = \lim_{n \to \infty} \frac{(n+1)^3}{3^{n+1}} \cdot \frac{3^n}{n^3} \)
\( = \lim_{n \to \infty} \frac{(n+1)^3}{n^3} \cdot \frac{3^n}{3 \cdot 3^n} \)
\( = \frac{1}{3} \lim_{n \to \infty} \left( 1 + \frac{1}{n} \right)^3 \)
\( = \frac{1}{3} (1)^3 = \frac{1}{3} \).

Step 3: Conclusion.
Since \( \frac{1}{3} < 1 \), by the Ratio Test, the series converges. (Note: Since it's a positive term series, convergence implies absolute convergence, but "Convergent" is the primary property being tested).

Detailed Solution:
Step 1: Find the Characteristic Equation.
Let \( A = \begin{pmatrix} 1 & 1 \ 3 & -1 \end{pmatrix} \).
The characteristic equation is \( |A - \lambda I| = 0 \).
\( \begin{vmatrix} 1 - \lambda & 1 \ 3 & -1 - \lambda \end{vmatrix} = 0 \)
\( (1 - \lambda)(-1 - \lambda) - (3)(1) = 0 \)
\( -1 - \lambda + \lambda + \lambda^2 - 3 = 0 \)
\( \lambda^2 - 4 = 0 \).

Step 2: Find the Eigenvalues.
Solving the characteristic equation:
\( \lambda^2 = 4 \implies \lambda = 2, -2 \).
So, the eigenvalues are \( \lambda_1 = 2 \) and \( \lambda_2 = -2 \).

Step 3: Find the Eigenvector for \( \lambda_1 = 2 \).
The eigenvector \( X = \begin{pmatrix} x_1 \ x_2 \end{pmatrix} \) satisfies \( (A - \lambda I)X = 0 \).
Substitute \( \lambda = 2 \):
\( \begin{pmatrix} 1 - 2 & 1 \ 3 & -1 - 2 \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \)
\( \begin{pmatrix} -1 & 1 \ 3 & -3 \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \)
This gives the equation: \( -x_1 + x_2 = 0 \implies x_1 = x_2 \).
Let \( x_1 = 1 \), then \( x_2 = 1 \).
The eigenvector for \( \lambda = 2 \) is \( X_1 = \begin{pmatrix} 1 \ 1 \end{pmatrix} \).

Step 4: Find the Eigenvector for \( \lambda_2 = -2 \).
Substitute \( \lambda = -2 \):
\( \begin{pmatrix} 1 - (-2) & 1 \ 3 & -1 - (-2) \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \)
\( \begin{pmatrix} 3 & 1 \ 3 & 1 \end{pmatrix} \begin{pmatrix} x_1 \ x_2 \end{pmatrix} = \begin{pmatrix} 0 \ 0 \end{pmatrix} \)
This gives the equation: \( 3x_1 + x_2 = 0 \implies x_2 = -3x_1 \).
Let \( x_1 = 1 \), then \( x_2 = -3 \).
The eigenvector for \( \lambda = -2 \) is \( X_2 = \begin{pmatrix} 1 \ -3 \end{pmatrix} \).

Detailed Solution:
Step 1: Substitute the expressions for x, y, and z into u.
Given \( u = x^2 + y^2 + z^2 \).
Substitute \( x = e^t \), \( y = e^t \sin t \), and \( z = e^t \cos t \):
\( u = (e^t)^2 + (e^t \sin t)^2 + (e^t \cos t)^2 \)
\( u = e^{2t} + e^{2t} \sin^2 t + e^{2t} \cos^2 t \).

Step 2: Simplify the expression using trigonometric identities.
Factor out \( e^{2t} \) from the last two terms:
\( u = e^{2t} + e^{2t}(\sin^2 t + \cos^2 t) \)
We know that \( \sin^2 t + \cos^2 t = 1 \), so:
\( u = e^{2t} + e^{2t}(1) = 2e^{2t} \).

Step 3: Differentiate u with respect to t.
\( \frac{du}{dt} = \frac{d}{dt}(2e^{2t}) = 2 \cdot (2e^{2t}) = 4e^{2t} \).
Therefore, \( \frac{du}{dt} = 4e^{2t} \).

Detailed Solution:
Step 1: Rewrite the equation to form a quadratic in terms of a single parameter.
The given family of lines is \( x \cos \alpha + y \sin \alpha = a \sec \alpha \).
Divide the entire equation by \( \cos \alpha \):
\( x + y \frac{\sin \alpha}{\cos \alpha} = a \frac{\sec \alpha}{\cos \alpha} \)
\( x + y \tan \alpha = a \sec^2 \alpha \).

Step 2: Use the identity \( \sec^2 \alpha = 1 + \tan^2 \alpha \).
\( x + y \tan \alpha = a(1 + \tan^2 \alpha) \)
\( x + y \tan \alpha = a + a \tan^2 \alpha \).

Step 3: Form a quadratic equation in \( t = \tan \alpha \).
Let \( t = \tan \alpha \). The equation becomes:
\( a t^2 - y t + (a - x) = 0 \).
This is a quadratic equation in parameter \( t \) of the form \( At^2 + Bt + C = 0 \), where \( A = a \), \( B = -y \), and \( C = (a - x) \).

Step 4: Find the envelope by setting the discriminant to zero.
The envelope of a family of curves given by a quadratic in its parameter is obtained by setting its discriminant \( B^2 - 4AC = 0 \).
\( (-y)^2 - 4(a)(a - x) = 0 \)
\( y^2 - 4a(a - x) = 0 \)
\( y^2 = 4a(a - x) \) or \( y^2 = -4a(x - a) \).
This is the equation of a parabola, which is the required envelope.

Detailed Solution:
Step 1: Write the definition of the Gamma function.
By definition, \( \Gamma(n) = \int_{0}^{\infty} e^{-x} x^{n-1} dx \) for \( n > 0 \).
Therefore, substituting \( n+1 \) for \( n \), we get:
\( \Gamma(n+1) = \int_{0}^{\infty} e^{-x} x^{(n+1)-1} dx = \int_{0}^{\infty} e^{-x} x^n dx \).

Step 2: Apply integration by parts.
Use the integration by parts formula: \( \int u \, dv = uv - \int v \, du \).
Let \( u = x^n \) and \( dv = e^{-x} dx \).
Then, \( du = n x^{n-1} dx \) and \( v = -e^{-x} \).

Step 3: Evaluate the expression.
\( \Gamma(n+1) = \left[ -x^n e^{-x} \right]_0^\infty - \int_0^\infty (-e^{-x})(n x^{n-1}) dx \)
\( \Gamma(n+1) = \left[ -x^n e^{-x} \right]_0^\infty + n \int_0^\infty e^{-x} x^{n-1} dx \).

Step 4: Evaluate the limits.
For the first term, evaluate at the upper limit (\( x \to \infty \)):
\( \lim_{x \to \infty} \frac{-x^n}{e^x} = 0 \) (since exponential growth is faster than polynomial growth).
At the lower limit (\( x = 0 \)):
\( -0^n e^0 = 0 \).
So, the first term vanishes: \( 0 - 0 = 0 \).

Step 5: Conclude the proof.
\( \Gamma(n+1) = 0 + n \int_0^\infty e^{-x} x^{n-1} dx \)
We recognize the integral as the definition of \( \Gamma(n) \).
Therefore, \( \Gamma(n+1) = n\Gamma(n) \). Hence proved.

Detailed Solution:
Step 1: Find the Complementary Function (C.F.).
The Auxiliary Equation is \( m^2 + 3m + 2 = 0 \).
Factoring the quadratic equation: \( (m + 1)(m + 2) = 0 \).
The roots are \( m_1 = -1 \) and \( m_2 = -2 \) (real and distinct).
So, \( C.F. = C_1 e^{-x} + C_2 e^{-2x} \).

Step 2: Find the Particular Integral (P.I.).
\( P.I. = \frac{1}{D^2 + 3D + 2} \sin x \).
For a term like \( \sin(ax) \), substitute \( D^2 = -a^2 \). Here, \( a = 1 \), so \( D^2 = -(1)^2 = -1 \).
Substitute \( D^2 = -1 \) in the denominator:
\( P.I. = \frac{1}{-1 + 3D + 2} \sin x = \frac{1}{3D + 1} \sin x \).

Step 3: Rationalize the operator.
Multiply numerator and denominator by the conjugate \( (3D - 1) \):
\( P.I. = \frac{3D - 1}{(3D + 1)(3D - 1)} \sin x \)
\( P.I. = \frac{3D - 1}{9D^2 - 1} \sin x \).
Substitute \( D^2 = -1 \) again:
\( P.I. = \frac{3D - 1}{9(-1) - 1} \sin x = \frac{3D - 1}{-10} \sin x \).

Step 4: Apply the differential operator to \( \sin x \).
\( P.I. = -\frac{1}{10} [3D(\sin x) - 1(\sin x)] \)
Since \( D(\sin x) = \cos x \), we have:
\( P.I. = -\frac{1}{10} [3\cos x - \sin x] = \frac{\sin x - 3\cos x}{10} \).

Step 5: Write the complete solution.
The general solution is \( y = C.F. + P.I. \)
\( y = C_1 e^{-x} + C_2 e^{-2x} + \frac{\sin x - 3\cos x}{10} \).

Detailed Solution:
Step 1: Define intermediate variables.
Let \( p = y - z \), \( q = z - x \), and \( r = x - y \).
Then the given function becomes \( u = f(p, q, r) \).

Step 2: Find partial derivatives of \( p, q, r \) with respect to \( x, y, z \).
\( \frac{\partial p}{\partial x} = 0, \quad \frac{\partial p}{\partial y} = 1, \quad \frac{\partial p}{\partial z} = -1 \)
\( \frac{\partial q}{\partial x} = -1, \quad \frac{\partial q}{\partial y} = 0, \quad \frac{\partial q}{\partial z} = 1 \)
\( \frac{\partial r}{\partial x} = 1, \quad \frac{\partial r}{\partial y} = -1, \quad \frac{\partial r}{\partial z} = 0 \)

Step 3: Apply the chain rule to find partial derivatives of \( u \).
\( \frac{\partial u}{\partial x} = \frac{\partial f}{\partial p} \frac{\partial p}{\partial x} + \frac{\partial f}{\partial q} \frac{\partial q}{\partial x} + \frac{\partial f}{\partial r} \frac{\partial r}{\partial x} = \frac{\partial f}{\partial p}(0) + \frac{\partial f}{\partial q}(-1) + \frac{\partial f}{\partial r}(1) = -\frac{\partial f}{\partial q} + \frac{\partial f}{\partial r} \).

\( \frac{\partial u}{\partial y} = \frac{\partial f}{\partial p} \frac{\partial p}{\partial y} + \frac{\partial f}{\partial q} \frac{\partial q}{\partial y} + \frac{\partial f}{\partial r} \frac{\partial r}{\partial y} = \frac{\partial f}{\partial p}(1) + \frac{\partial f}{\partial q}(0) + \frac{\partial f}{\partial r}(-1) = \frac{\partial f}{\partial p} - \frac{\partial f}{\partial r} \).

\( \frac{\partial u}{\partial z} = \frac{\partial f}{\partial p} \frac{\partial p}{\partial z} + \frac{\partial f}{\partial q} \frac{\partial q}{\partial z} + \frac{\partial f}{\partial r} \frac{\partial r}{\partial z} = \frac{\partial f}{\partial p}(-1) + \frac{\partial f}{\partial q}(1) + \frac{\partial f}{\partial r}(0) = -\frac{\partial f}{\partial p} + \frac{\partial f}{\partial q} \).

Step 4: Sum the partial derivatives.
Add the three partial derivatives obtained in Step 3:
\( \frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} + \frac{\partial u}{\partial z} = \left( -\frac{\partial f}{\partial q} + \frac{\partial f}{\partial r} \right) + \left( \frac{\partial f}{\partial p} - \frac{\partial f}{\partial r} \right) + \left( -\frac{\partial f}{\partial p} + \frac{\partial f}{\partial q} \right) \)
\( = \frac{\partial f}{\partial p} - \frac{\partial f}{\partial p} - \frac{\partial f}{\partial q} + \frac{\partial f}{\partial q} + \frac{\partial f}{\partial r} - \frac{\partial f}{\partial r} = 0 \).
Hence proved.

Detailed Solution:
Step 1: Define the \( n \)-th term.
Let \( \sum u_n \) be the given series, where the \( n \)-th term is \( u_n = \frac{3^n - 1}{2^n + 1} \).

Step 2: Apply the \( n \)-th term test for divergence.
The \( n \)-th term test (or divergence test) states that if \( \lim_{n \to \infty} u_n \neq 0 \), then the series \( \sum u_n \) is divergent.
Let's evaluate the limit of \( u_n \) as \( n \to \infty \).
\( \lim_{n \to \infty} u_n = \lim_{n \to \infty} \frac{3^n - 1}{2^n + 1} \).

Step 3: Simplify the expression to find the limit.
Factor out the dominant terms (\( 3^n \) from the numerator and \( 2^n \) from the denominator):
\( u_n = \frac{3^n(1 - 1/3^n)}{2^n(1 + 1/2^n)} = \left( \frac{3}{2} \right)^n \frac{1 - (1/3)^n}{1 + (1/2)^n} \).
As \( n \to \infty \):
\( (1/3)^n \to 0 \)
\( (1/2)^n \to 0 \)
So, the fraction \( \frac{1 - (1/3)^n}{1 + (1/2)^n} \to \frac{1 - 0}{1 + 0} = 1 \).

However, the term \( \left( \frac{3}{2} \right)^n \to \infty \) since \( \frac{3}{2} > 1 \).
Therefore, \( \lim_{n \to \infty} u_n = \infty \cdot 1 = \infty \).

Step 4: Conclusion.
Since \( \lim_{n \to \infty} u_n \neq 0 \) (in fact, it diverges to infinity), the series \( \sum u_n \) is divergent by the \( n \)-th term divergence test.

Detailed Solution:
Step 1: Write the matrix of the quadratic form.
The matrix \( A \) is formed by taking coefficients of \( x_i^2 \) on the diagonal, and half the coefficients of \( x_i x_j \) off-diagonal.
\( A = \begin{pmatrix} 6 & -2 & 2 \\ -2 & 3 & -1 \\ 2 & -1 & 3 \end{pmatrix} \).

Step 2: Find the Characteristic Equation.
The characteristic equation is \( \lambda^3 - S_1 \lambda^2 + S_2 \lambda - S_3 = 0 \).
\( S_1 = \text{Sum of diagonal elements} = 6 + 3 + 3 = 12 \).
\( S_2 = \text{Sum of principal minors of order 2} = \begin{vmatrix} 3 & -1 \\ -1 & 3 \end{vmatrix} + \begin{vmatrix} 6 & 2 \\ 2 & 3 \end{vmatrix} + \begin{vmatrix} 6 & -2 \\ -2 & 3 \end{vmatrix} \)
\( S_2 = (9 - 1) + (18 - 4) + (18 - 4) = 8 + 14 + 14 = 36 \).
\( S_3 = |A| = 6(9 - 1) - (-2)(-6 + 2) + 2(2 - 6) = 6(8) + 2(-4) + 2(-4) = 48 - 8 - 8 = 32 \).
Equation: \( \lambda^3 - 12\lambda^2 + 36\lambda - 32 = 0 \).

Step 3: Find the Eigenvalues.
By inspection, \( \lambda = 2 \) is a root since \( (2)^3 - 12(2)^2 + 36(2) - 32 = 8 - 48 + 72 - 32 = 0 \).
Factoring out \( (\lambda - 2) \): \( (\lambda - 2)(\lambda^2 - 10\lambda + 16) = 0 \implies (\lambda - 2)(\lambda - 2)(\lambda - 8) = 0 \).
Eigenvalues are \( \lambda = 8, 2, 2 \).

Step 4: Find Eigenvector for \( \lambda = 8 \).
\( (A - 8I)X = 0 \implies \begin{pmatrix} -2 & -2 & 2 \\ -2 & -5 & -1 \\ 2 & -1 & -5 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \).
Using cross-multiplication on the first two equations:
\( -2x_1 - 2x_2 + 2x_3 = 0 \implies x_1 + x_2 - x_3 = 0 \)
\( -2x_1 - 5x_2 - x_3 = 0 \implies 2x_1 + 5x_2 + x_3 = 0 \)
\( \frac{x_1}{1 - (-5)} = \frac{x_2}{-2 - 1} = \frac{x_3}{5 - 2} \implies \frac{x_1}{6} = \frac{x_2}{-3} = \frac{x_3}{3} \implies \frac{x_1}{2} = \frac{x_2}{-1} = \frac{x_3}{1} \).
\( X_1 = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix} \).

Step 5: Find Eigenvectors for \( \lambda = 2, 2 \).
\( (A - 2I)X = 0 \implies \begin{pmatrix} 4 & -2 & 2 \\ -2 & 1 & -1 \\ 2 & -1 & 1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \).
All equations reduce to \( 2x_1 - x_2 + x_3 = 0 \).
Let \( x_1 = 0 \), then \( -x_2 + x_3 = 0 \implies x_2 = 1, x_3 = 1 \). So \( X_2 = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} \).
Let the third eigenvector be \( X_3 \), which must be orthogonal to both \( X_1 \) and \( X_2 \).
\( X_3 = X_1 \times X_2 = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix} \times \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} (-1)(1) - (1)(1) \\ (1)(0) - (2)(1) \\ (2)(1) - (-1)(0) \end{pmatrix} = \begin{pmatrix} -2 \\ -2 \\ 2 \end{pmatrix} \sim \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} \).

Step 6: Normalized Modal Matrix (\( N \)).
Length of \( X_1 = \sqrt{4+1+1} = \sqrt{6} \).
Length of \( X_2 = \sqrt{0+1+1} = \sqrt{2} \).
Length of \( X_3 = \sqrt{1+1+1} = \sqrt{3} \).
\( N = \begin{pmatrix} \frac{2}{\sqrt{6}} & 0 & \frac{1}{\sqrt{3}} \\ \frac{-1}{\sqrt{6}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{3}} \end{pmatrix} \).

Step 7: Canonical Form.
By orthogonal transformation \( X = NY \), the quadratic form reduces to \( \lambda_1 y_1^2 + \lambda_2 y_2^2 + \lambda_3 y_3^2 \).
Canonical Form = \( 8y_1^2 + 2y_2^2 + 2y_3^2 \).

Step 8: Find Rank, Index, and Signature.
Rank (\( r \)) = number of non-zero eigenvalues = 3.
Index (\( p \)) = number of positive eigenvalues = 3.
Signature (\( s \)) = (number of positive eigenvalues) - (number of negative eigenvalues) = 3 - 0 = 3.

Detailed Solution:
Step 1: Characteristic Equation of A.
The characteristic equation is \( \lambda^3 - S_1 \lambda^2 + S_2 \lambda - S_3 = 0 \).
\( S_1 = 1 + 2 + 1 = 4 \).
\( S_2 = \begin{vmatrix} 2 & 3 \\ 2 & 1 \end{vmatrix} + \begin{vmatrix} 1 & 7 \\ 1 & 1 \end{vmatrix} + \begin{vmatrix} 1 & 3 \\ 4 & 2 \end{vmatrix} = (2-6) + (1-7) + (2-12) = -4 - 6 - 10 = -20 \).
\( S_3 = |A| = 1(2 - 6) - 3(4 - 3) + 7(8 - 2) = -4 - 3 + 42 = 35 \).
So, characteristic equation is \( \lambda^3 - 4\lambda^2 - 20\lambda - 35 = 0 \).

Step 2: Cayley-Hamilton Theorem Statement.
Every square matrix satisfies its own characteristic equation.
We must show that \( A^3 - 4A^2 - 20A - 35I = 0 \).

Step 3: Calculate \( A^2 \) and \( A^3 \).
\( A^2 = \begin{pmatrix} 1 & 3 & 7 \\ 4 & 2 & 3 \\ 1 & 2 & 1 \end{pmatrix} \begin{pmatrix} 1 & 3 & 7 \\ 4 & 2 & 3 \\ 1 & 2 & 1 \end{pmatrix} = \begin{pmatrix} 20 & 23 & 23 \\ 15 & 22 & 37 \\ 10 & 9 & 14 \end{pmatrix} \).

\( A^3 = A \cdot A^2 = \begin{pmatrix} 1 & 3 & 7 \\ 4 & 2 & 3 \\ 1 & 2 & 1 \end{pmatrix} \begin{pmatrix} 20 & 23 & 23 \\ 15 & 22 & 37 \\ 10 & 9 & 14 \end{pmatrix} = \begin{pmatrix} 135 & 152 & 232 \\ 140 & 163 & 208 \\ 60 & 76 & 111 \end{pmatrix} \).

Step 4: Verify the equation.
\( A^3 - 4A^2 - 20A - 35I = \begin{pmatrix} 135 & 152 & 232 \\ 140 & 163 & 208 \\ 60 & 76 & 111 \end{pmatrix} - \begin{pmatrix} 80 & 92 & 92 \\ 60 & 88 & 148 \\ 40 & 36 & 56 \end{pmatrix} - \begin{pmatrix} 20 & 60 & 140 \\ 80 & 40 & 60 \\ 20 & 40 & 20 \end{pmatrix} - \begin{pmatrix} 35 & 0 & 0 \\ 0 & 35 & 0 \\ 0 & 0 & 35 \end{pmatrix} \)
Let's evaluate element (1,1): \( 135 - 80 - 20 - 35 = 0 \).
Element (1,2): \( 152 - 92 - 60 - 0 = 0 \).
It can be seen that all elements sum to zero. Thus, \( A^3 - 4A^2 - 20A - 35I = 0 \). The theorem is verified.

Step 5: Find \( A^{-1} \).
Multiply the characteristic equation by \( A^{-1} \):
\( A^2 - 4A - 20I - 35A^{-1} = 0 \implies 35A^{-1} = A^2 - 4A - 20I \).
\( 35A^{-1} = \begin{pmatrix} 20 & 23 & 23 \\ 15 & 22 & 37 \\ 10 & 9 & 14 \end{pmatrix} - \begin{pmatrix} 4 & 12 & 28 \\ 16 & 8 & 12 \\ 4 & 8 & 4 \end{pmatrix} - \begin{pmatrix} 20 & 0 & 0 \\ 0 & 20 & 0 \\ 0 & 0 & 20 \end{pmatrix} \).
\( 35A^{-1} = \begin{pmatrix} 20 - 4 - 20 & 23 - 12 - 0 & 23 - 28 - 0 \\ 15 - 16 - 0 & 22 - 8 - 20 & 37 - 12 - 0 \\ 10 - 4 - 0 & 9 - 8 - 0 & 14 - 4 - 20 \end{pmatrix} = \begin{pmatrix} -4 & 11 & -5 \\ -1 & -6 & 25 \\ 6 & 1 & -10 \end{pmatrix} \).
\( A^{-1} = \frac{1}{35} \begin{pmatrix} -4 & 11 & -5 \\ -1 & -6 & 25 \\ 6 & 1 & -10 \end{pmatrix} \).

Detailed Solution:
Step 1: Define variables and the function to minimize.
Let the dimensions of the rectangular box be \( x \) (length), \( y \) (width), and \( z \) (height).
The volume of the box is \( V = xyz = 32 \).
The surface area of a box open at the top is \( S = xy + 2yz + 2zx \).
We want to minimize \( S(x, y, z) \) subject to the constraint \( g(x, y, z) = xyz - 32 = 0 \).

Step 2: Use Lagrange Multipliers.
Define the Lagrangian function: \( F(x, y, z, \lambda) = xy + 2yz + 2zx + \lambda(xyz - 32) \).
Find the partial derivatives and set them to zero:
\( \frac{\partial F}{\partial x} = y + 2z + \lambda yz = 0 \implies x(y + 2z) = -\lambda xyz \) ... (Eq. 1)
\( \frac{\partial F}{\partial y} = x + 2z + \lambda xz = 0 \implies y(x + 2z) = -\lambda xyz \) ... (Eq. 2)
\( \frac{\partial F}{\partial z} = 2y + 2x + \lambda xy = 0 \implies z(2y + 2x) = -\lambda xyz \) ... (Eq. 3)

Step 3: Solve the system of equations.
From Eq. 1 and Eq. 2:
\( xy + 2xz = xy + 2yz \)
\( 2xz = 2yz \implies x = y \) (since \( z \neq 0 \)).

From Eq. 2 and Eq. 3:
\( xy + 2yz = 2yz + 2xz \)
\( xy = 2xz \implies y = 2z \) (since \( x \neq 0 \)).

Since \( x = y \), we have \( x = 2z \).

Step 4: Use the volume constraint to find the dimensions.
Substitute \( x = 2z \) and \( y = 2z \) into the volume equation \( xyz = 32 \):
\( (2z)(2z)(z) = 32 \)
\( 4z^3 = 32 \implies z^3 = 8 \implies z = 2 \).

Substitute \( z = 2 \) back to find \( x \) and \( y \):
\( x = 2(2) = 4 \)
\( y = 2(2) = 4 \).

Step 5: Conclusion.
The dimensions for minimum surface area are length = 4 feet, width = 4 feet, and height = 2 feet.

Detailed Solution for (i):
Step 1: Evaluate the function and its derivatives at \( (a,b) = (-1, \pi/4) \).
\( f(x,y) = e^x \sin y \implies f(-1, \pi/4) = e^{-1} \sin(\pi/4) = \frac{1}{e\sqrt{2}} \).

First-order derivatives:
\( f_x = e^x \sin y \implies f_x(-1, \pi/4) = \frac{1}{e\sqrt{2}} \).
\( f_y = e^x \cos y \implies f_y(-1, \pi/4) = \frac{1}{e\sqrt{2}} \).

Second-order derivatives:
\( f_{xx} = e^x \sin y \implies f_{xx}(-1, \pi/4) = \frac{1}{e\sqrt{2}} \).
\( f_{xy} = e^x \cos y \implies f_{xy}(-1, \pi/4) = \frac{1}{e\sqrt{2}} \).
\( f_{yy} = -e^x \sin y \implies f_{yy}(-1, \pi/4) = -\frac{1}{e\sqrt{2}} \).

Third-order derivatives:
\( f_{xxx} = e^x \sin y \implies f_{xxx}(-1, \pi/4) = \frac{1}{e\sqrt{2}} \).
\( f_{xxy} = e^x \cos y \implies f_{xxy}(-1, \pi/4) = \frac{1}{e\sqrt{2}} \).
\( f_{xyy} = -e^x \sin y \implies f_{xyy}(-1, \pi/4) = -\frac{1}{e\sqrt{2}} \).
\( f_{yyy} = -e^x \cos y \implies f_{yyy}(-1, \pi/4) = -\frac{1}{e\sqrt{2}} \).

Step 2: Write the Taylor series expansion.
The expansion up to 3rd degree terms is given by:
\( f(x,y) \approx f(a,b) + [ (x-a)f_x + (y-b)f_y ] + \frac{1}{2!} [ (x-a)^2f_{xx} + 2(x-a)(y-b)f_{xy} + (y-b)^2f_{yy} ] + \frac{1}{3!} [ (x-a)^3f_{xxx} + 3(x-a)^2(y-b)f_{xxy} + 3(x-a)(y-b)^2f_{xyy} + (y-b)^3f_{yyy} ] \).

Substituting the values:
\( f(x,y) = \frac{1}{e\sqrt{2}} \left[ 1 + (x+1) + (y-\pi/4) + \frac{1}{2} \left( (x+1)^2 + 2(x+1)(y-\pi/4) - (y-\pi/4)^2 \right) + \frac{1}{6} \left( (x+1)^3 + 3(x+1)^2(y-\pi/4) - 3(x+1)(y-\pi/4)^2 - (y-\pi/4)^3 \right) \right] \).

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Detailed Solution for (ii):
Step 1: Calculate the first Jacobian, \( J = \frac{\partial(x,y)}{\partial(r,\theta)} \).
Given \( x = r \cos \theta \) and \( y = r \sin \theta \).
\( \frac{\partial x}{\partial r} = \cos \theta \), \( \frac{\partial x}{\partial \theta} = -r \sin \theta \).
\( \frac{\partial y}{\partial r} = \sin \theta \), \( \frac{\partial y}{\partial \theta} = r \cos \theta \).
\( J = \begin{vmatrix} \cos \theta & -r \sin \theta \\ \sin \theta & r \cos \theta \end{vmatrix} = (\cos \theta)(r \cos \theta) - (-r \sin \theta)(\sin \theta) = r(\cos^2 \theta + \sin^2 \theta) = r \).

Step 2: Express \( r \) and \( \theta \) in terms of \( x \) and \( y \).
\( r = \sqrt{x^2 + y^2} = (x^2 + y^2)^{1/2} \).
\( \theta = \tan^{-1}\left(\frac{y}{x}\right) \).

Step 3: Calculate the second Jacobian, \( J' = \frac{\partial(r,\theta)}{\partial(x,y)} \).
\( \frac{\partial r}{\partial x} = \frac{1}{2}(x^2 + y^2)^{-1/2}(2x) = \frac{x}{\sqrt{x^2 + y^2}} \).
\( \frac{\partial r}{\partial y} = \frac{1}{2}(x^2 + y^2)^{-1/2}(2y) = \frac{y}{\sqrt{x^2 + y^2}} \).
\( \frac{\partial \theta}{\partial x} = \frac{1}{1 + (y/x)^2} \left(-\frac{y}{x^2}\right) = \frac{x^2}{x^2 + y^2} \left(-\frac{y}{x^2}\right) = -\frac{y}{x^2 + y^2} \).
\( \frac{\partial \theta}{\partial y} = \frac{1}{1 + (y/x)^2} \left(\frac{1}{x}\right) = \frac{x^2}{x^2 + y^2} \left(\frac{1}{x}\right) = \frac{x}{x^2 + y^2} \).
\( J' = \begin{vmatrix} \frac{x}{\sqrt{x^2 + y^2}} & \frac{y}{\sqrt{x^2 + y^2}} \\ -\frac{y}{x^2 + y^2} & \frac{x}{x^2 + y^2} \end{vmatrix} = \frac{x^2}{(x^2 + y^2)^{3/2}} - \left( -\frac{y^2}{(x^2 + y^2)^{3/2}} \right) = \frac{x^2 + y^2}{(x^2 + y^2)^{3/2}} = \frac{1}{\sqrt{x^2 + y^2}} = \frac{1}{r} \).

Step 4: Verify the property.
\( J \times J' = r \times \frac{1}{r} = 1 \).
The given property is verified.

Detailed Solution:
Step 1: Write the equations in differential operator form.
Let \( D = \frac{d}{dt} \).
\( Dx + 2y = -\sin t \) ... (Eq. 1)
\( -2x + Dy = \cos t \) ... (Eq. 2)

Step 2: Eliminate \( x \).
From Eq. 2, we can write \( 2x = Dy - \cos t \implies x = \frac{1}{2}Dy - \frac{1}{2}\cos t \).
Differentiate \( x \) with respect to \( t \):
\( Dx = \frac{1}{2}D^2y + \frac{1}{2}\sin t \).

Step 3: Substitute \( Dx \) in Eq. 1.
\( \left(\frac{1}{2}D^2y + \frac{1}{2}\sin t\right) + 2y = -\sin t \).
Multiply the entire equation by 2:
\( D^2y + \sin t + 4y = -2\sin t \)
\( (D^2 + 4)y = -3\sin t \).

Step 4: Solve the second-order linear ODE for \( y \).
The Auxiliary Equation is \( m^2 + 4 = 0 \implies m = \pm 2i \).
So, Complementary Function \( C.F. = C_1 \cos 2t + C_2 \sin 2t \).

Find the Particular Integral \( P.I. \):
\( P.I. = \frac{1}{D^2 + 4} (-3\sin t) \).
Substitute \( D^2 = -1^2 = -1 \):
\( P.I. = \frac{-3}{-1 + 4} \sin t = \frac{-3}{3} \sin t = -\sin t \).
Therefore, \( y = C_1 \cos 2t + C_2 \sin 2t - \sin t \).

Step 5: Find \( x \).
We have \( x = \frac{1}{2}Dy - \frac{1}{2}\cos t \).
First, find \( Dy \) (derivative of \( y \)):
\( Dy = -2C_1 \sin 2t + 2C_2 \cos 2t - \cos t \).

Now substitute \( Dy \) into the equation for \( x \):
\( x = \frac{1}{2}(-2C_1 \sin 2t + 2C_2 \cos 2t - \cos t) - \frac{1}{2}\cos t \)
\( x = -C_1 \sin 2t + C_2 \cos 2t - \frac{1}{2}\cos t - \frac{1}{2}\cos t \)
\( x = -C_1 \sin 2t + C_2 \cos 2t - \cos t \).

Detailed Solution:
Step 1: Find the Complementary Function (C.F.).
Auxiliary Equation: \( m^2 + 4 = 0 \implies m = \pm 2i \).
\( C.F. = c_1 \cos 2x + c_2 \sin 2x \).
Let \( y_1 = \cos 2x \) and \( y_2 = \sin 2x \).

Step 2: Calculate the Wronskian (\( W \)).
\( W = y_1 y_2' - y_2 y_1' \)
\( y_1' = -2\sin 2x \)
\( y_2' = 2\cos 2x \)
\( W = (\cos 2x)(2\cos 2x) - (\sin 2x)(-2\sin 2x) = 2\cos^2 2x + 2\sin^2 2x = 2(\cos^2 2x + \sin^2 2x) = 2 \).

Step 3: Apply the method of variation of parameters.
The particular integral is given by \( P.I. = u y_1 + v y_2 \), where:
\( X = 4\tan 2x \)
\( u = -\int \frac{y_2 X}{W} dx \) and \( v = \int \frac{y_1 X}{W} dx \).

Step 4: Evaluate the integral for \( u \).
\( u = -\int \frac{\sin 2x (4\tan 2x)}{2} dx = -2 \int \sin 2x \left(\frac{\sin 2x}{\cos 2x}\right) dx = -2 \int \frac{\sin^2 2x}{\cos 2x} dx \).
Using \( \sin^2 2x = 1 - \cos^2 2x \):
\( u = -2 \int \frac{1 - \cos^2 2x}{\cos 2x} dx = -2 \int (\sec 2x - \cos 2x) dx \).
\( u = -2 \left[ \frac{1}{2} \log(\sec 2x + \tan 2x) - \frac{1}{2} \sin 2x \right] = -\log(\sec 2x + \tan 2x) + \sin 2x \).

Step 5: Evaluate the integral for \( v \).
\( v = \int \frac{\cos 2x (4\tan 2x)}{2} dx = 2 \int \cos 2x \left(\frac{\sin 2x}{\cos 2x}\right) dx = 2 \int \sin 2x dx \).
\( v = 2 \left( \frac{-\cos 2x}{2} \right) = -\cos 2x \).

Step 6: Write the Particular Integral and the Complete Solution.
\( P.I. = [-\log(\sec 2x + \tan 2x) + \sin 2x](\cos 2x) + (-\cos 2x)(\sin 2x) \)
\( P.I. = -\cos 2x \log(\sec 2x + \tan 2x) + \sin 2x \cos 2x - \sin 2x \cos 2x \)
\( P.I. = -\cos 2x \log(\sec 2x + \tan 2x) \).

The general solution is \( y = C.F. + P.I. \):
\( y = c_1 \cos 2x + c_2 \sin 2x - \cos 2x \log(\sec 2x + \tan 2x) \).

Detailed Solution:
Step 1: Find the first and second derivatives at the given point.
Given equation of the parabola is \( y^2 = 12x \).
Differentiate with respect to \( x \):
\( 2y \frac{dy}{dx} = 12 \implies \frac{dy}{dx} = y' = \frac{6}{y} \).
At the point \( (3, 6) \), \( y' = \frac{6}{6} = 1 \).

Differentiate again to find the second derivative:
\( y'' = \frac{d}{dx}\left(\frac{6}{y}\right) = -\frac{6}{y^2} y' \).
At \( (3, 6) \), \( y'' = -\frac{6}{(6)^2} (1) = -\frac{6}{36} = -\frac{1}{6} \).

Step 2: Find the center of curvature \( (\bar{x}, \bar{y}) \).
The formulas for the center of curvature are:
\( \bar{x} = x - \frac{y'(1 + (y')^2)}{y''} \)
\( \bar{y} = y + \frac{1 + (y')^2}{y''} \)

Substitute the values \( x=3 \), \( y=6 \), \( y'=1 \), and \( y''=-\frac{1}{6} \):
\( \bar{x} = 3 - \frac{1(1 + 1^2)}{-1/6} = 3 - \frac{2}{-1/6} = 3 - (-12) = 15 \).
\( \bar{y} = 6 + \frac{1 + 1^2}{-1/6} = 6 + \frac{2}{-1/6} = 6 - 12 = -6 \).
So, the center of curvature is \( C(15, -6) \).

Step 3: Find the radius of curvature \( \rho \).
\( \rho = \frac{(1 + (y')^2)^{3/2}}{|y''|} = \frac{(1 + 1^2)^{3/2}}{|-1/6|} = \frac{2^{3/2}}{1/6} = 6(2\sqrt{2}) = 12\sqrt{2} \).
\( \rho^2 = (12\sqrt{2})^2 = 144 \times 2 = 288 \).

Step 4: Write the equation of the circle of curvature.
The equation is \( (x - \bar{x})^2 + (y - \bar{y})^2 = \rho^2 \).
\( (x - 15)^2 + (y - (-6))^2 = 288 \)
\( (x - 15)^2 + (y + 6)^2 = 288 \)
Expanding the terms:
\( x^2 - 30x + 225 + y^2 + 12y + 36 = 288 \)
\( x^2 + y^2 - 30x + 12y + 261 - 288 = 0 \)
\( x^2 + y^2 - 30x + 12y - 27 = 0 \).

Detailed Solution:
Step 1: Write the parametric equations of the ellipse.
\( x = a \cos t \), \( y = b \sin t \).

Step 2: Find the first and second derivatives \( y' \) and \( y'' \).
\( \frac{dx}{dt} = -a \sin t \), \( \frac{dy}{dt} = b \cos t \).
\( y' = \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{b \cos t}{-a \sin t} = -\frac{b}{a} \cot t \).

\( y'' = \frac{d}{dt}(y') \cdot \frac{dt}{dx} = \frac{d}{dt}\left(-\frac{b}{a} \cot t\right) \cdot \left(\frac{1}{-a \sin t}\right) \).
\( y'' = \left(\frac{b}{a} \csc^2 t\right) \left(-\frac{1}{a \sin t}\right) = -\frac{b}{a^2} \csc^3 t \).

Step 3: Find the coordinates of the center of curvature \( (\bar{x}, \bar{y}) \).
First, simplify the term \( \frac{1 + (y')^2}{y''} \):
\( 1 + (y')^2 = 1 + \frac{b^2}{a^2} \cot^2 t = \frac{a^2 \sin^2 t + b^2 \cos^2 t}{a^2 \sin^2 t} \).
\( \frac{1 + (y')^2}{y''} = \frac{\frac{a^2 \sin^2 t + b^2 \cos^2 t}{a^2 \sin^2 t}}{-\frac{b}{a^2 \sin^3 t}} = -\frac{(a^2 \sin^2 t + b^2 \cos^2 t) \sin t}{b} \).

Now, calculate \( \bar{x} \):
\( \bar{x} = x - y' \frac{1 + (y')^2}{y''} \)
\( \bar{x} = a \cos t - \left(-\frac{b \cos t}{a \sin t}\right) \left( -\frac{(a^2 \sin^2 t + b^2 \cos^2 t) \sin t}{b} \right) \)
\( \bar{x} = a \cos t - \frac{\cos t (a^2 \sin^2 t + b^2 \cos^2 t)}{a} \)
\( \bar{x} = \frac{a^2 \cos t - a^2 \cos t \sin^2 t - b^2 \cos^3 t}{a} \)
Using \( 1 - \sin^2 t = \cos^2 t \):
\( \bar{x} = \frac{a^2 \cos^3 t - b^2 \cos^3 t}{a} = \frac{a^2 - b^2}{a} \cos^3 t \).

Calculate \( \bar{y} \):
\( \bar{y} = y + \frac{1 + (y')^2}{y''} \)
\( \bar{y} = b \sin t - \frac{(a^2 \sin^2 t + b^2 \cos^2 t) \sin t}{b} \)
\( \bar{y} = \frac{b^2 \sin t - a^2 \sin^3 t - b^2 \cos^2 t \sin t}{b} \)
Using \( 1 - \cos^2 t = \sin^2 t \):
\( \bar{y} = \frac{b^2 \sin^3 t - a^2 \sin^3 t}{b} = \frac{b^2 - a^2}{b} \sin^3 t = -\frac{a^2 - b^2}{b} \sin^3 t \).

Step 4: Eliminate parameter \( t \) to find the locus (evolute).
From \( \bar{x} \): \( \cos t = \left( \frac{a \bar{x}}{a^2 - b^2} \right)^{1/3} \).
From \( \bar{y} \): \( \sin t = \left( \frac{-b \bar{y}}{a^2 - b^2} \right)^{1/3} \).
Using the identity \( \cos^2 t + \sin^2 t = 1 \):
\( \left( \frac{a \bar{x}}{a^2 - b^2} \right)^{2/3} + \left( \frac{-b \bar{y}}{a^2 - b^2} \right)^{2/3} = 1 \).
\( (a \bar{x})^{2/3} + (b \bar{y})^{2/3} = (a^2 - b^2)^{2/3} \).
Replacing \( (\bar{x}, \bar{y}) \) with \( (x, y) \), the evolute is \( (ax)^{2/3} + (by)^{2/3} = (a^2 - b^2)^{2/3} \).

Detailed Solution for (i):
Step 1: Find the \( n \)-th term of the series, \( u_n \).
Numerator sequence: \( 1, 3, 5, \dots \) (Arithmetic Progression with \( a=1, d=2 \)) \( \implies 1 + (n-1)2 = 2n - 1 \).
Denominator: The factors are \( n, (n+1), (n+2) \).
Thus, \( u_n = \frac{2n - 1}{n(n+1)(n+2)} \).

Step 2: Choose an auxiliary series \( v_n \).
For large \( n \), the highest power of \( n \) in numerator is \( n^1 \) and in denominator is \( n^3 \).
So, let \( v_n = \frac{n}{n^3} = \frac{1}{n^2} \).

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

Step 4: Conclusion.
Since the limit is a finite, non-zero constant, both \( \sum u_n \) and \( \sum v_n \) behave alike.
The series \( \sum v_n = \sum \frac{1}{n^2} \) is a p-series with \( p = 2 > 1 \), which is convergent.
Therefore, the given series \( \sum u_n \) is also convergent.

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Detailed Solution for (ii):
Step 1: Identify the \( n \)-th term.
Let \( u_n = (\log n)^{-2n} = \frac{1}{(\log n)^{2n}} \).

Step 2: Apply Cauchy's Root Test.
Cauchy's Root Test states that if \( \lim_{n \to \infty} (u_n)^{1/n} = L \):
- If \( L < 1 \), the series converges.
- If \( L > 1 \), the series diverges.

Step 3: Evaluate the limit.
\( \lim_{n \to \infty} (u_n)^{1/n} = \lim_{n \to \infty} \left( (\log n)^{-2n} \right)^{1/n} \)
\( = \lim_{n \to \infty} (\log n)^{-2} = \lim_{n \to \infty} \frac{1}{(\log n)^2} \).

As \( n \to \infty \), \( \log n \to \infty \), so \( (\log n)^2 \to \infty \).
Therefore, \( \lim_{n \to \infty} \frac{1}{(\log n)^2} = 0 \).

Step 4: Conclusion.
Since \( L = 0 < 1 \), the series \( \sum (\log n)^{-2n} \) is convergent by Cauchy's Root Test.

Detailed Solution:
Step 1: Simplify the \( n \)-th term \( u_n \).
\( u_n = \frac{1}{\sqrt{n+1} - 1} \).
Multiply the numerator and denominator by the conjugate \( \sqrt{n+1} + 1 \):
\( u_n = \frac{1 \times (\sqrt{n+1} + 1)}{(\sqrt{n+1} - 1)(\sqrt{n+1} + 1)} \)
\( u_n = \frac{\sqrt{n+1} + 1}{(n+1) - 1} = \frac{\sqrt{n+1} + 1}{n} \).

Step 2: Choose an auxiliary series \( v_n \).
For large \( n \), the numerator is dominated by \( \sqrt{n} = n^{1/2} \) and the denominator is \( n^1 \).
Let \( v_n = \frac{n^{1/2}}{n} = \frac{1}{n^{1/2}} = \frac{1}{\sqrt{n}} \).

Step 3: Apply the Limit Comparison Test.
Evaluate \( \lim_{n \to \infty} \frac{u_n}{v_n} \):
\( \lim_{n \to \infty} \frac{u_n}{v_n} = \lim_{n \to \infty} \left[ \frac{\sqrt{n+1} + 1}{n} \times \frac{\sqrt{n}}{1} \right] \)
\( = \lim_{n \to \infty} \frac{\sqrt{n(n+1)} + \sqrt{n}}{n} \)
\( = \lim_{n \to \infty} \frac{n\sqrt{1 + 1/n} + \sqrt{n}}{n} \)
Divide numerator and denominator by \( n \):
\( = \lim_{n \to \infty} \left( \sqrt{1 + \frac{1}{n}} + \frac{1}{\sqrt{n}} \right) \).
As \( n \to \infty \), \( 1/n \to 0 \) and \( 1/\sqrt{n} \to 0 \).
Limit \( = \sqrt{1 + 0} + 0 = 1 \).

Step 4: Conclusion.
Since the limit is a finite, non-zero constant (1), both series behave alike.
The series \( \sum v_n = \sum \frac{1}{n^{1/2}} \) is a p-series with \( p = 1/2 \leq 1 \), which is divergent.
Therefore, by the Limit Comparison Test, the given series \( \sum u_n \) is also divergent.