Detailed Solution:
Step 1: The given matrix $A$ is a diagonal matrix. For any diagonal or triangular matrix, the eigenvalues are simply the elements present on its principal diagonal.
Step 2: Therefore, the eigenvalues of $A$ ($\lambda_1, \lambda_2, \lambda_3$) are $1, 2,$ and $5$.
Step 3: By the property of eigenvalues, if $\lambda$ is an eigenvalue of a matrix $A$, then $\lambda^n$ is an eigenvalue of the matrix $A^n$.
Step 4: The eigenvalues of $A^2$ will be $1^2, 2^2, 5^2$, which are $1, 4,$ and $25$.

Detailed Solution:
Step 1: The characteristic equation of a $2 \times 2$ matrix $A$ is given by $\lambda^2 - (\text{Trace of } A)\lambda + \text{Det}(A) = 0$.
Step 2: Calculate the Trace (sum of principal diagonal elements): $\text{Trace}(A) = 1 + 3 = 4$.
Step 3: Calculate the Determinant ($\text{Det}(A)$): $\text{Det}(A) = (1)(3) - (4)(2) = 3 - 8 = -5$.
Step 4: Substitute the trace and determinant into the formula: $\lambda^2 - 4\lambda + (-5) = 0$, which yields $\lambda^2 - 4\lambda - 5 = 0$.

Detailed Solution:
Step 1: The nature of a quadratic form is determined by the signs of the principal minors of its symmetric matrix $A$.
Step 2: Find the first principal minor $D_1$. $D_1 = |1| = 1$, which is positive ($D_1 > 0$).
Step 3: Find the second principal minor $D_2$ (the determinant of the whole matrix). $D_2 = |A| = (1)(1) - (2)(2) = 1 - 4 = -3$, which is negative ($D_2 < 0$).
Step 4: Since $D_1 > 0$ but $D_2 < 0$, the signs of the principal minors alternate incorrectly for both positive and negative definiteness. Hence, the quadratic form is indefinite. Alternatively, calculating eigenvalues yields $-1, 3$ (mixed signs), which confirms the indefinite nature.

Detailed Solution:
Step 1: The signature of a quadratic form is defined as the difference between the number of positive square terms ($p$) and the number of negative square terms ($n$) in its canonical form.
Step 2: Observe the canonical form $y_1^2 + y_2^2 + 4y_3^2$. All three terms have positive coefficients ($+1, +1, +4$).
Step 3: Number of positive terms $p = 3$. Number of negative terms $n = 0$.
Step 4: Signature = $p - n = 3 - 0 = 3$.

Detailed Solution:
Step 1: We can use the total derivative formula or direct substitution. Direct substitution is simpler here.
Step 2: Substitute $x = at$ and $y = 2at$ into $u$.
$u = (at)^2 + (2at)^2 + (at)^2(2at)$
$u = a^2t^2 + 4a^2t^2 + a^2t^2 \cdot 2at$
$u = 5a^2t^2 + 2a^3t^3$
Step 3: Differentiate $u$ with respect to $t$.
$\frac{du}{dt} = \frac{d}{dt}(5a^2t^2 + 2a^3t^3)$
$\frac{du}{dt} = 5a^2(2t) + 2a^3(3t^2) = 10a^2t + 6a^3t^2$.

Detailed Solution:
Step 1: To find the stationary points, we must set the first-order partial derivatives to zero. $\frac{\partial f}{\partial x} = 0$ and $\frac{\partial f}{\partial y} = 0$.
Step 2: Calculate $\frac{\partial f}{\partial x} = \frac{\partial}{\partial x}(3x^2 - y^2 + x^3) = 6x + 3x^2$.
Step 3: Calculate $\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(3x^2 - y^2 + x^3) = -2y$.
Step 4: Solve the equations:
$6x + 3x^2 = 0 \Rightarrow 3x(2 + x) = 0 \Rightarrow x = 0 \text{ or } x = -2$.
$-2y = 0 \Rightarrow y = 0$.
Step 5: The stationary points $(x,y)$ are $(0, 0)$ and $(-2, 0)$.

Detailed Solution:
Step 1: The Jacobian $J$ is defined as the determinant of the Jacobian matrix: $\begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix}$.
Step 2: Compute the partial derivatives:
$\frac{\partial x}{\partial r} = \cos \theta$, $\frac{\partial x}{\partial \theta} = -r \sin \theta$
$\frac{\partial y}{\partial r} = \sin \theta$, $\frac{\partial y}{\partial \theta} = r \cos \theta$
Step 3: Evaluate the determinant:
$J = (\cos \theta)(r \cos \theta) - (-r \sin \theta)(\sin \theta)$
$J = r \cos^2 \theta + r \sin^2 \theta = r(\cos^2 \theta + \sin^2 \theta) = r(1) = r$.

Detailed Solution:
Step 1: The standard Maclaurin series expansion for the exponential function $e^z$ is $\sum_{n=0}^{\infty} \frac{z^n}{n!} = 1 + z + \frac{z^2}{2!} + \frac{z^3}{3!} + ...$
Step 2: Let $z = (x - y)$. Substitute this expression into the standard expansion.
Step 3: $e^{x-y} = 1 + (x-y) + \frac{(x-y)^2}{2!} + \frac{(x-y)^3}{3!} + ...$. This matches option (A).

Detailed Solution:
Step 1: This is a second-order linear homogeneous differential equation with constant coefficients. We write its auxiliary equation by substituting $D = m$.
Step 2: The auxiliary equation is $m^2 + 4 = 0$.
Step 3: Solving for $m$, we get $m^2 = -4 \Rightarrow m = \pm \sqrt{-4} = \pm 2i$. These are complex conjugate roots of the form $\alpha \pm i\beta$ where $\alpha = 0, \beta = 2$.
Step 4: The general solution for complex roots is $y = e^{\alpha x}(A \cos \beta x + B \sin \beta x)$. Substituting $\alpha$ and $\beta$, we get $y = e^{0}(A \cos 2x + B \sin 2x) = A \cos 2x + B \sin 2x$.

Detailed Solution:
Step 1: The particular integral (PI) for $f(D)y = \cos ax$ is found by applying $\frac{1}{f(D)} \cos ax$ and replacing $D^2$ with $-a^2$. Here, $a = 1$, so replace $D^2$ with $-1^2 = -1$.
Step 2: PI = $\frac{1}{D^2 - 3D + 2} \cos x = \frac{1}{-1 - 3D + 2} \cos x = \frac{1}{1 - 3D} \cos x$.
Step 3: Rationalize the operator by multiplying numerator and denominator by the conjugate $(1 + 3D)$.
PI = $\frac{1 + 3D}{(1 - 3D)(1 + 3D)} \cos x = \frac{1 + 3D}{1 - 9D^2} \cos x$.
Step 4: Replace $D^2$ with $-1$ in the denominator again.
PI = $\frac{1 + 3D}{1 - 9(-1)} \cos x = \frac{1 + 3D}{10} \cos x$.
Step 5: Apply the operator $(1 + 3D)$ to $\cos x$. Note that $D(\cos x) = -\sin x$.
PI = $\frac{1}{10} (\cos x + 3(-\sin x)) = \frac{1}{10} (\cos x - 3\sin x)$.

Detailed Solution:
Step 1: The complementary function (CF) is the general solution of the associated homogeneous equation $(D^2 + 4D + 4)y = 0$.
Step 2: Write the auxiliary equation: $m^2 + 4m + 4 = 0$.
Step 3: Factor the quadratic equation: $(m + 2)^2 = 0 \Rightarrow m = -2, -2$.
Step 4: Since the roots are real and repeated (identical), the form of the complementary function is $CF = (c_1x + c_2)e^{mx}$. Using constants $A$ and $B$, we write $CF = (Ax + B)e^{-2x}$.

Detailed Solution:
Step 1: For a Cauchy-Euler equation, we make the substitution $x = e^z$, which implies $z = \ln x$. Let $\theta = \frac{d}{dz}$. We substitute $xD = \theta$ and $x^2D^2 = \theta(\theta - 1)$.
Step 2: Substitute these into the homogeneous part of the equation: $(\theta(\theta - 1) + \theta + 2)y = 0$.
Step 3: Simplify the operator: $(\theta^2 - \theta + \theta + 2)y = 0 \Rightarrow (\theta^2 + 2)y = 0$.
Step 4: The auxiliary equation is $m^2 + 2 = 0$, giving complex roots $m = \pm \sqrt{2}i$.
Step 5: The CF in terms of $z$ is $y_c = A \cos(\sqrt{2} z) + B \sin(\sqrt{2} z)$.
Step 6: Substitute back $z = \ln x$ to get the final CF: $CF = A \cos(\sqrt{2} \ln x) + B \sin(\sqrt{2} \ln x)$.

Detailed Solution:
Step 1: The standard formula for the radius of curvature at the vertex of a parabola $y^2 = 4ax$ is $\rho = 2a$. Alternatively, we can use the Cartesian formula $\rho = \frac{(1 + y_1^2)^{3/2}}{|y_2|}$.
Step 2: Compare $y^2 = 4x$ with $y^2 = 4ax$. Here, $4a = 4 \Rightarrow a = 1$.
Step 3: Using the shortcut formula for the vertex $(0,0)$, $\rho = 2a = 2(1) = 2$.
Step 4: (Using derivation) Differentiating implicitly: $2y y_1 = 4 \Rightarrow y_1 = 2/y$. Since $y=0$ at origin, $y_1$ is infinite. We must invert: $x = y^2 / 4 \Rightarrow x_1 = dx/dy = y/2$. At $(0,0), x_1 = 0$. $x_2 = d^2x/dy^2 = 1/2$. $\rho = \frac{(1 + x_1^2)^{3/2}}{|x_2|} = \frac{(1 + 0)^{3/2}}{1/2} = 2$.

Detailed Solution:
Step 1: Curvature ($\kappa$) defines how much a curve bends at a specific point. For a circle, the bending is uniform everywhere.
Step 2: The curvature of a circle is mathematically defined as $\kappa = \frac{1}{R}$, where $R$ is the constant radius of the circle.
Step 3: The radius of curvature ($\rho$) is the reciprocal of the curvature ($\rho = \frac{1}{\kappa}$).
Step 4: Therefore, for a circle, $\rho = \frac{1}{1/R} = R$. The radius of curvature is identical to the radius of the circle at every point.

Detailed Solution:
Step 1: The Gamma function is defined as $\Gamma(n) = \int_0^{\infty} x^{n-1} e^{-x} dx$.
Step 2: To find $\Gamma(1/2)$, substitute $n = 1/2$: $\Gamma(1/2) = \int_0^{\infty} x^{-1/2} e^{-x} dx$.
Step 3: Make the substitution $x = t^2 \Rightarrow dx = 2t dt$. Limits remain $0$ to $\infty$.
Step 4: $\Gamma(1/2) = \int_0^{\infty} (t^2)^{-1/2} e^{-t^2} (2t dt) = 2 \int_0^{\infty} e^{-t^2} dt$.
Step 5: The Gaussian integral $\int_0^{\infty} e^{-t^2} dt = \frac{\sqrt{\pi}}{2}$.
Step 6: Therefore, $\Gamma(1/2) = 2 \left( \frac{\sqrt{\pi}}{2} \right) = \sqrt{\pi}$. This is a standard identity.

Detailed Solution:
Step 1: The Gamma function acts as an extension of the factorial function to complex and real numbers.
Step 2: A fundamental property of the Gamma function is the recurrence relation: $\Gamma(z + 1) = z \Gamma(z)$.
Step 3: For a positive integer $n$, applying the relation recursively gives: $\Gamma(n+1) = n \Gamma(n) = n(n-1)\Gamma(n-1) = ... = n(n-1)...(2)(1)\Gamma(1)$.
Step 4: Since $\Gamma(1) = 1$, the product evaluates directly to $n!$. Therefore, $\Gamma(n+1) = n!$.

Detailed Solution:
Step 1: By the integral test, the convergence of the series $\sum_{n=1}^{\infty} \frac{1}{n^p}$ behaves identically to the improper integral $\int_{1}^{\infty} \frac{1}{x^p} dx$.
Step 2: Evaluating the integral: $\int_{1}^{\infty} x^{-p} dx = \left[ \frac{x^{-p+1}}{-p+1} \right]_{1}^{\infty}$ (for $p \neq 1$).
Step 3: If $p > 1$, then $-p+1$ is negative. As $x \to \infty$, $x^{-p+1} = \frac{1}{x^{p-1}} \to 0$. The integral evaluates to a finite number ($\frac{1}{p-1}$), so the series converges.
Step 4: If $p < 1$, the term $x^{-p+1} \to \infty$, and the integral diverges.
Step 5: If $p = 1$, the integral is $\ln|x|$, which diverges to $\infty$. Thus, it only converges strictly when $p > 1$.

Detailed Solution:
Step 1: The given series is the alternating harmonic series: $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n}$.
Step 2: First, test for absolute convergence by taking the absolute value of the terms: $\sum_{n=1}^{\infty} \left| \frac{(-1)^{n-1}}{n} \right| = \sum_{n=1}^{\infty} \frac{1}{n}$. This is the standard harmonic series (a p-series with $p=1$), which is known to diverge.
Step 3: Since it does not converge absolutely, test for conditional convergence using the Alternating Series Test (Leibniz Test). The sequence $a_n = \frac{1}{n}$ is positive, strictly decreasing, and $\lim_{n \to \infty} \frac{1}{n} = 0$.
Step 4: Because it satisfies the Alternating Series Test but its absolute series diverges, the series is said to be conditionally convergent. (It happens to converge exactly to $\ln 2$).

Detailed Solution:
Step 1: D'Alembert's ratio test evaluates the limit of the absolute ratio of consecutive terms: $l = \lim_{n \to \infty} \left| \frac{u_{n+1}}{u_n} \right|$.
Step 2: The test concludes the following:
- If $l < 1$, the series converges absolutely.
- If $l > 1$ (or $l = \infty$), the series diverges.
- If $l = 1$, the test is inconclusive (the series could converge or diverge).
Step 3: Therefore, the explicit condition required for the series to be convergent is $l < 1$.