B.Tech Calculus and Linear Algebra
Semester 1 - November 2024 PYQ

Detailed Solution:
Step 1: Let \(A = \begin{bmatrix} 1 & 2 \ 5 & 4 \end{bmatrix}\).
Step 2: The characteristic equation is given by \(|A - \lambda I| = 0\).
Step 3: \(\begin{vmatrix} 1 - \lambda & 2 \ 5 & 4 - \lambda \end{vmatrix} = 0\)
Step 4: \((1 - \lambda)(4 - \lambda) - (2)(5) = 0\)
Step 5: \(4 - \lambda - 4\lambda + \lambda^2 - 10 = 0\)
Step 6: \(\lambda^2 - 5\lambda - 6 = 0\)
Step 7: Factorizing the quadratic equation: \((\lambda - 6)(\lambda + 1) = 0\)
Step 8: Therefore, the eigenvalues are \(\lambda = 6\) and \(\lambda = -1\).

Detailed Solution:
Step 1: The index of a quadratic form is defined as the number of positive squared terms in its canonical form.
Step 2: The given canonical form is \(-y_1^2 + 3y_2^2 + y_3^2\).
Step 3: We can identify the terms and their signs:
- First term: \(-y_1^2\) (Negative)
- Second term: \(+3y_2^2\) (Positive)
- Third term: \(+y_3^2\) (Positive)
Step 4: Since there are exactly 2 positive terms, the index is 2.

Detailed Solution:
Step 1: Matrix \(A\) is an upper triangular matrix.
Step 2: The eigenvalues of a triangular matrix are simply the entries on its main diagonal.
Step 3: Therefore, the eigenvalues of \(A\) are \(\lambda_1 = 3\), \(\lambda_2 = 2\), and \(\lambda_3 = 4\).
Step 4: According to the properties of eigenvalues, if \(\lambda\) is an eigenvalue of \(A\), then \(\lambda^k\) is an eigenvalue of \(A^k\).
Step 5: Thus, the eigenvalues of \(A^2\) are \(3^2\), \(2^2\), and \(4^2\).
Step 6: Calculating these gives \(9\), \(4\), and \(16\).

Detailed Solution:
Step 1: The general quadratic form in three variables is \(a_{11}x_1^2 + a_{22}x_2^2 + a_{33}x_3^2 + 2a_{12}x_1x_2 + 2a_{13}x_1x_3 + 2a_{23}x_2x_3\).
Step 2: By comparing coefficients with the given equation \(x_1^2 + x_2^2 - 3x_3^2 - 2x_1x_2 + 4x_1x_3\), we find:
- \(a_{11} = 1\), \(a_{22} = 1\), \(a_{33} = -3\)
- \(2a_{12} = -2 \implies a_{12} = -1\)
- \(2a_{13} = 4 \implies a_{13} = 2\)
- \(2a_{23} = 0 \implies a_{23} = 0\)
Step 3: A quadratic form matrix is symmetric, so \(a_{ji} = a_{ij}\):
- \(a_{21} = -1\), \(a_{31} = 2\), \(a_{32} = 0\)
Step 4: Constructing the matrix \(A = [a_{ij}]\):
\(A = \begin{bmatrix} 1 & -1 & 2 \ -1 & 1 & 0 \ 2 & 0 & -3 \end{bmatrix}\)

Detailed Solution:
Step 1: We are given the function \(z = x^3 + 3x^2y + y^3\).
Step 2: To find the partial derivative \(\frac{\partial z}{\partial x}\), we differentiate the function with respect to \(x\) while treating \(y\) as a constant.
Step 3: Differentiating each term:
- \(\frac{\partial}{\partial x}(x^3) = 3x^2\)
- \(\frac{\partial}{\partial x}(3x^2y) = 3y \cdot 2x = 6xy\)
- \(\frac{\partial}{\partial x}(y^3) = 0\) (since \(y\) is a constant)
Step 4: Summing these gives \(\frac{\partial z}{\partial x} = 3x^2 + 6xy\).

Detailed Solution:
Step 1: We start with the implicit function equation \(f(x, y) = 0\).
Step 2: Taking the total derivative with respect to \(x\) on both sides using the chain rule:
\(\frac{d}{dx}[f(x, y)] = \frac{\partial f}{\partial x}\frac{dx}{dx} + \frac{\partial f}{\partial y}\frac{dy}{dx} = 0\)
Step 3: This simplifies to \(f_x \cdot 1 + f_y \cdot \frac{dy}{dx} = 0\).
Step 4: Rearranging to solve for \(\frac{dy}{dx}\):
\(f_y \cdot \frac{dy}{dx} = -f_x\)
\(\frac{dy}{dx} = -\frac{f_x}{f_y}\) (provided \(f_y \neq 0\)).

Detailed Solution:
Step 1: The Jacobian of \(u\) and \(v\) with respect to variables \(x\) and \(y\) is denoted by \(J\left(\frac{u, v}{x, y} ight) = \frac{\partial(u, v)}{\partial(x, y)}\).
Step 2: By definition, two functions \(u(x, y)\) and \(v(x, y)\) are functionally dependent if there exists some non-trivial relationship \(F(u, v) = 0\) independent of \(x\) and \(y\).
Step 3: A fundamental theorem in advanced calculus states that functions are functionally dependent in a region if and only if their Jacobian determinant is identically zero in that region.
Step 4: Therefore, the Jacobian value must be 0.

Detailed Solution:
Step 1: The Jacobian \(J = \frac{\partial(x, y)}{\partial(r, \theta)}\) is the determinant of the partial derivatives 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: Calculate the partial derivatives:
\(\frac{\partial x}{\partial r} = \cos\theta\), \(\frac{\partial x}{\partial \theta} = -r\sin\theta\)
\(\frac{\partial y}{\partial r} = \sin\theta\), \(\frac{\partial y}{\partial \theta} = r\cos\theta\)
Step 3: Substitute into the determinant:
\(J = \begin{vmatrix} \cos\theta & -r\sin\theta \ \sin\theta & r\cos\theta \end{vmatrix}\)
Step 4: Expand the determinant:
\(J = (\cos\theta)(r\cos\theta) - (-r\sin\theta)(\sin\theta)\)
\(J = r\cos^2\theta + r\sin^2\theta = r(\cos^2\theta + \sin^2\theta)\)
Step 5: Using the trigonometric identity \(\cos^2\theta + \sin^2\theta = 1\), we get \(J = r(1) = r\).

Detailed Solution:
Step 1: The complementary function (CF) is found by solving the homogeneous part of the differential equation: \((D^2 - 4D + 4)y = 0\).
Step 2: The auxiliary equation is \(m^2 - 4m + 4 = 0\).
Step 3: Factoring the quadratic: \((m - 2)^2 = 0\).
Step 4: The roots are real and repeated: \(m_1 = 2, m_2 = 2\).
Step 5: For repeated real roots \(m = a\), the general form of the complementary function is \((C_1 + C_2x)e^{ax}\).
Step 6: Substituting \(a = 2\), the CF is \((C_1 + C_2x)e^{2x}\).

Detailed Solution:
Step 1: The particular integral (PI) for an equation of the form \(f(D)y = e^{ax}\) is given by \(PI = \frac{1}{f(D)}e^{ax}\).
Step 2: Here, \(f(D) = D^2 + 16\) and \(a = -3\).
Step 3: The rule for exponential terms is to substitute \(D = a\) in \(f(D)\), provided \(f(a) \neq 0\).
Step 4: Substitute \(D = -3\):
\(PI = \frac{1}{(-3)^2 + 16} e^{-3x}\)
Step 5: Simplify the denominator:
\(PI = \frac{1}{9 + 16} e^{-3x} = \frac{1}{25} e^{-3x}\).

Detailed Solution:
Step 1: This is a Cauchy-Euler differential equation. We use the substitution \(x = e^z\), which implies \(z = \log x\).
Step 2: Under this substitution, let \(\theta = \frac{d}{dz}\). The operators transform as:
\(xD \equiv \theta\) and \(x^2D^2 \equiv \theta(\theta - 1)\).
Step 3: Substituting these into the equation:
\((\theta(\theta - 1) - \theta + 1)y = 0\)
\((\theta^2 - \theta - \theta + 1)y = 0\)
\((\theta^2 - 2\theta + 1)y = 0\)
Step 4: The auxiliary equation is \(m^2 - 2m + 1 = 0 \implies (m - 1)^2 = 0\).
Step 5: The roots are real and repeated: \(m = 1, 1\).
Step 6: The complementary function in terms of \(z\) is \(y_c = (C_1 + C_2z)e^z\).
Step 7: Substituting back \(z = \log x\) and \(e^z = x\):
\(y_c = (C_1 + C_2\log x)x\).

Detailed Solution:
Step 1: This is a Legendre linear differential equation. We use the substitution \(1+x = e^z\), which implies \(z = \log(1+x)\).
Step 2: Let \(\theta = \frac{d}{dz}\). The transformations are:
\((1+x)D \equiv \theta\) and \((1+x)^2D^2 \equiv \theta(\theta - 1)\).
Step 3: Substituting these operators into the given equation:
\((\theta(\theta - 1) + \theta + 1)y = 0\)
\((\theta^2 - \theta + \theta + 1)y = 0\)
\((\theta^2 + 1)y = 0\)
Step 4: This matches option (A).

Detailed Solution:
Step 1: We are given the parametric equations \(x = at^2\) and \(y = 2at\).
Step 2: To find the Cartesian equation, we eliminate the parameter \(t\). From the second equation, \(t = \frac{y}{2a}\).
Step 3: Substitute this \(t\) into the first equation:
\(x = a\left(\frac{y}{2a} ight)^2\)
\(x = a\left(\frac{y^2}{4a^2} ight)\)
\(x = \frac{y^2}{4a}\)
Step 4: Rearranging gives \(y^2 = 4ax\), which is the standard equation of a parabola.

Detailed Solution:
Step 1: Curvature (denoted by \(\kappa\)) measures how sharply a curve bends at a given point.
Step 2: By definition, the radius of curvature \( ho\) is the radius of the osculating circle at that point.
Step 3: The mathematical relationship is \( ho = \frac{1}{\kappa}\). Thus, the reciprocal of curvature is the radius of curvature.

Detailed Solution:
Step 1: As a point moves along a curve, its centre of curvature also traces out a path.
Step 2: By mathematical definition in differential geometry, the locus (path) traced by the centre of curvature of a given curve is called its **evolute**.
Step 3: Alternatively, an involute is a curve traced by the free end of a taut string unwinding from the given curve. The envelope is a curve tangent to a family of curves. Thus, 'Evolute' is the correct term.

Detailed Solution:
Step 1: To find the envelope of a family of curves \(f(x, y, m) = 0\), we differentiate the equation with respect to the parameter \(m\) and eliminate \(m\).
Step 2: The given family is \(y = mx + \frac{1}{m}\).
Step 3: Differentiate with respect to \(m\):
\(0 = x - \frac{1}{m^2}\)
Step 4: Solve for \(m\):
\(m^2 = \frac{1}{x} \implies m = \frac{1}{\sqrt{x}}\)
Step 5: Substitute \(m\) back into the original equation:
\(y = \left(\frac{1}{\sqrt{x}} ight)x + \frac{1}{(1/\sqrt{x})}\)
\(y = \sqrt{x} + \sqrt{x} = 2\sqrt{x}\)
Step 6: Squaring both sides gives \(y^2 = 4x\).

Detailed Solution:
Step 1: The given series is the standard p-series: \(\sum_{n=1}^{\infty} \frac{1}{n^p}\).
Step 2: By the p-series test (derived from the integral test), the series \(\sum \frac{1}{n^p}\) converges if and only if \(p > 1\).
Step 3: If \(p = 1\), it becomes the harmonic series \(\sum \frac{1}{n}\), which diverges.
Step 4: If \(p < 1\), the terms do not decrease fast enough (or even increase if \(p \le 0\)), and the series diverges.

Detailed Solution:
Step 1: The divergence of \(\sum V_n\) does not guarantee that the sequence \(\{V_n\}\) diverges. For example, \(\sum \frac{1}{n}\) diverges, but the sequence \(\frac{1}{n} \to 0\). Thus, (A) and (C) are incorrect.
Step 2: The nth-term test for divergence states that if \(\sum U_n\) converges, then the limit of its sequence of terms must be zero (\(\lim_{n \to \infty} U_n = 0\)).
Step 3: Therefore, the sequence \(\{U_n\}\) must converge to 0. Option (B) correctly states this absolute mathematical certainty.

Detailed Solution:
Step 1: Recall the standard Maclaurin series expansions.
Step 2: \(e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots\)
Step 3: \(\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots\)
Step 4: \(\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots\)
Step 5: \(\log(1 + x) = \int \frac{1}{1+x} dx = \int (1 - x + x^2 - x^3 + \dots) dx = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots\) (for \(-1 < x \le 1\)).
Step 6: The given series perfectly matches the Taylor series expansion for \(\log(1 + x)\).

Detailed Solution:
Step 1: The given series has the general term \(u_n = \frac{1+n}{1+n^2}\).
Step 2: For large \(n\), \(u_n \approx \frac{n}{n^2} = \frac{1}{n}\).
Step 3: We use the Limit Comparison Test. Let \(v_n = \frac{1}{n}\).
Step 4: \(\lim_{n \to \infty} \frac{u_n}{v_n} = \lim_{n \to \infty} \frac{\frac{1+n}{1+n^2}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{n(1+n)}{1+n^2} = \lim_{n \to \infty} \frac{n+n^2}{1+n^2} = 1\).
Step 5: Since the limit is a finite non-zero constant, both \(\sum u_n\) and \(\sum v_n\) behave similarly.
Step 6: Because \(\sum v_n = \sum \frac{1}{n}\) is the divergent harmonic series (\(p=1\)), the original series \(\sum \frac{1+n}{1+n^2}\) is also divergent.

Detailed Solution:
Step 1: The characteristic equation is given by \(|A - \lambda I| = \lambda^3 - S_1\lambda^2 + S_2\lambda - S_3 = 0\).
Step 2: Calculate the invariants:
- \(S_1\) = Sum of principal diagonal elements = \(1 + 5 + 1 = 7\).
- \(S_2\) = Sum of minors of principal diagonal elements = \(\begin{vmatrix} 5 & 1 \ 1 & 1 \end{vmatrix} + \begin{vmatrix} 1 & 3 \ 3 & 1 \end{vmatrix} + \begin{vmatrix} 1 & 1 \ 1 & 5 \end{vmatrix} = (5-1) + (1-9) + (5-1) = 4 - 8 + 4 = 0\).
- \(S_3\) = Determinant of A = \(|A| = 1(5-1) - 1(1-3) + 3(1-15) = 4 + 2 - 42 = -36\).
Step 3: The characteristic equation is \(\lambda^3 - 7\lambda^2 + 0\lambda + 36 = 0 \implies \lambda^3 - 7\lambda^2 + 36 = 0\).
Step 4: By trial and error, let \(\lambda = -2\): \((-2)^3 - 7(-2)^2 + 36 = -8 - 28 + 36 = 0\). So \(\lambda = -2\) is a root.
Step 5: Using synthetic division with \(\lambda = -2\), the reduced quadratic is \(\lambda^2 - 9\lambda + 18 = 0\).
Step 6: Factoring \(\lambda^2 - 9\lambda + 18 = 0\) gives \((\lambda - 3)(\lambda - 6) = 0 \implies \lambda = 3, \lambda = 6\).
The eigenvalues are \(\lambda_1 = -2, \lambda_2 = 3, \lambda_3 = 6\).

Step 7: Eigenvector for \(\lambda_1 = -2\):
Solve \((A - (-2)I)X = 0 \implies \begin{bmatrix} 3 & 1 & 3 \ 1 & 7 & 1 \ 3 & 1 & 3 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \ x_3 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix}\).
Using cross-multiplication on the first two equations (ignoring identical 1st & 3rd equations):
\(3x_1 + x_2 + 3x_3 = 0\)
\(x_1 + 7x_2 + x_3 = 0\)
\(\frac{x_1}{1 - 21} = \frac{-x_2}{3 - 3} = \frac{x_3}{21 - 1} \implies \frac{x_1}{-20} = \frac{-x_2}{0} = \frac{x_3}{20}\)
Dividing by 20 gives the eigenvector \(X_1 = \begin{bmatrix} -1 \ 0 \ 1 \end{bmatrix}\).

Step 8: Eigenvector for \(\lambda_2 = 3\):
Solve \((A - 3I)X = 0 \implies \begin{bmatrix} -2 & 1 & 3 \ 1 & 2 & 1 \ 3 & 1 & -2 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \ x_3 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix}\).
Using the first two equations:
\(-2x_1 + x_2 + 3x_3 = 0\)
\(x_1 + 2x_2 + x_3 = 0\)
\(\frac{x_1}{1 - 6} = \frac{-x_2}{-2 - 3} = \frac{x_3}{-4 - 1} \implies \frac{x_1}{-5} = \frac{x_2}{5} = \frac{x_3}{-5}\)
Dividing by -5 gives the eigenvector \(X_2 = \begin{bmatrix} 1 \ -1 \ 1 \end{bmatrix}\).

Step 9: Eigenvector for \(\lambda_3 = 6\):
Solve \((A - 6I)X = 0 \implies \begin{bmatrix} -5 & 1 & 3 \ 1 & -1 & 1 \ 3 & 1 & -5 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \ x_3 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix}\).
Using the first two equations:
\(-5x_1 + x_2 + 3x_3 = 0\)
\(x_1 - x_2 + x_3 = 0\)
\(\frac{x_1}{1 - (-3)} = \frac{-x_2}{-5 - 3} = \frac{x_3}{5 - 1} \implies \frac{x_1}{4} = \frac{x_2}{8} = \frac{x_3}{4}\)
Dividing by 4 gives the eigenvector \(X_3 = \begin{bmatrix} 1 \ 2 \ 1 \end{bmatrix}\).

Detailed Solution:
Step 1: The characteristic equation is \(\lambda^3 - S_1\lambda^2 + S_2\lambda - S_3 = 0\).
Step 2: Calculate the invariants:
- \(S_1 = 2 + 2 + 2 = 6\).
- \(S_2 = (4 - 1) + (4 - 2) + (4 - 1) = 3 + 2 + 3 = 8\).
- \(S_3 = |A| = 2(4-1) + 1(-2+1) + 2(1-2) = 2(3) - 1 - 2 = 6 - 3 = 3\).
Step 3: The characteristic equation is \(\lambda^3 - 6\lambda^2 + 8\lambda - 3 = 0\).
Step 4: By the Cayley-Hamilton theorem, every square matrix satisfies its own characteristic equation. Therefore:
\(A^3 - 6A^2 + 8A - 3I = 0 \implies A^3 = 6A^2 - 8A + 3I\)
Step 5: To find \(A^4\), multiply the above equation by \(A\):
\(A^4 = 6A^3 - 8A^2 + 3A\)
Step 6: Substitute the expression for \(A^3\) back into this equation:
\(A^4 = 6(6A^2 - 8A + 3I) - 8A^2 + 3A\)
\(A^4 = 36A^2 - 48A + 18I - 8A^2 + 3A\)
\(A^4 = 28A^2 - 45A + 18I\)
Step 7: Compute \(A^2\):
\(A^2 = \begin{bmatrix} 2 & -1 & 2 \ -1 & 2 & -1 \ 1 & -1 & 2 \end{bmatrix} \begin{bmatrix} 2 & -1 & 2 \ -1 & 2 & -1 \ 1 & -1 & 2 \end{bmatrix} = \begin{bmatrix} (4+1+2) & (-2-2-2) & (4+1+4) \ (-2-2-1) & (1+4+1) & (-2-2-2) \ (2+1+2) & (-1-2-2) & (2+1+4) \end{bmatrix} = \begin{bmatrix} 7 & -6 & 9 \ -5 & 6 & -6 \ 5 & -5 & 7 \end{bmatrix}\)
Wait, let me recalculate carefully. Row 1, Col 3: 2(2) + (-1)(-1) + 2(2) = 4 + 1 + 4 = 9. Yes.
Row 2, Col 1: (-1)(2) + 2(-1) + (-1)(1) = -2 - 2 - 1 = -5. Yes.
Step 8: Now substitute \(A^2, A\) and \(I\) into the \(A^4\) equation:
\(A^4 = 28\begin{bmatrix} 7 & -6 & 9 \ -5 & 6 & -6 \ 5 & -5 & 7 \end{bmatrix} - 45\begin{bmatrix} 2 & -1 & 2 \ -1 & 2 & -1 \ 1 & -1 & 2 \end{bmatrix} + 18\begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{bmatrix}\)
Step 9: Calculate elements:
Row 1 Col 1: \(28(7) - 45(2) + 18(1) = 196 - 90 + 18 = 124\).
(Wait, let me recalculate eigenvalues to be sure. \(\lambda^3 - 6\lambda^2 + 8\lambda - 3 = 0\). Is the trace 6? Yes. Determinant: 2(4-1) + 1(-2+1) + 2(1-2) = 6 - 1 - 2 = 3. Is S2 correct? Minor of (1,1) is 4-1=3. Minor of (2,2) is 4-2=2. Minor of (3,3) is 4-1=3. 3+2+3=8. Yes.)
So \(A^3 - 6A^2 + 8A - 3I = 0\). Then \(A^4 = 6A^3 - 8A^2 + 3A = 6(6A^2 - 8A + 3I) - 8A^2 + 3A = 36A^2 - 48A + 18I - 8A^2 + 3A = 28A^2 - 45A + 18I\). This is correct.
Let's re-multiply A * A precisely:
Row 1: [2, -1, 2] * Col 1: [2, -1, 1]^T = 4 + 1 + 2 = 7.
Row 1: [2, -1, 2] * Col 2: [-1, 2, -1]^T = -2 - 2 - 2 = -6.
Row 1: [2, -1, 2] * Col 3: [2, -1, 2]^T = 4 + 1 + 4 = 9.
Row 2: [-1, 2, -1] * Col 1: [2, -1, 1]^T = -2 - 2 - 1 = -5.
Row 2: [-1, 2, -1] * Col 2: [-1, 2, -1]^T = 1 + 4 + 1 = 6.
Row 2: [-1, 2, -1] * Col 3: [2, -1, 2]^T = -2 - 2 - 2 = -6.
Row 3: [1, -1, 2] * Col 1: [2, -1, 1]^T = 2 + 1 + 2 = 5.
Row 3: [1, -1, 2] * Col 2: [-1, 2, -1]^T = -1 - 2 - 2 = -5.
Row 3: [1, -1, 2] * Col 3: [2, -1, 2]^T = 2 + 1 + 4 = 7.
\(A^2 = \begin{bmatrix} 7 & -6 & 9 \ -5 & 6 & -6 \ 5 & -5 & 7 \end{bmatrix}\).
Now calculate \(28A^2 - 45A + 18I\):
\(c_{11} = 28(7) - 45(2) + 18 = 196 - 90 + 18 = 124\).
\(c_{12} = 28(-6) - 45(-1) + 0 = -168 + 45 = -123\).
\(c_{13} = 28(9) - 45(2) + 0 = 252 - 90 = 162\).
\(c_{21} = 28(-5) - 45(-1) + 0 = -140 + 45 = -95\).
\(c_{22} = 28(6) - 45(2) + 18 = 168 - 90 + 18 = 96\).
\(c_{23} = 28(-6) - 45(-1) + 0 = -168 + 45 = -123\).
\(c_{31} = 28(5) - 45(1) + 0 = 140 - 45 = 95\).
\(c_{32} = 28(-5) - 45(-1) + 0 = -140 + 45 = -95\).
\(c_{33} = 28(7) - 45(2) + 18 = 196 - 90 + 18 = 124\).
\(A^4 = \begin{bmatrix} 124 & -123 & 162 \ -95 & 96 & -123 \ 95 & -95 & 124 \end{bmatrix}\)

Detailed Solution:
Step 1: Let \(f(x,y) = e^x \sin y\). We need to expand around \((a, b) = (-1, \frac{\pi}{4})\).
Step 2: Calculate the function and its partial derivatives evaluated at \((a, b)\).
- \(f(-1, \frac{\pi}{4}) = e^{-1} \sin(\frac{\pi}{4}) = \frac{1}{e\sqrt{2}}\)
- \(f_x = e^x \sin y \implies f_x(-1, \frac{\pi}{4}) = \frac{1}{e\sqrt{2}}\)
- \(f_y = e^x \cos y \implies f_y(-1, \frac{\pi}{4}) = e^{-1} \cos(\frac{\pi}{4}) = \frac{1}{e\sqrt{2}}\)
- \(f_{xx} = e^x \sin y \implies f_{xx}(-1, \frac{\pi}{4}) = \frac{1}{e\sqrt{2}}\)
- \(f_{xy} = e^x \cos y \implies f_{xy}(-1, \frac{\pi}{4}) = \frac{1}{e\sqrt{2}}\)
- \(f_{yy} = -e^x \sin y \implies f_{yy}(-1, \frac{\pi}{4}) = -\frac{1}{e\sqrt{2}}\)
Step 3: The Taylor's series expansion up to the second degree is:
\(f(x,y) \approx f(a,b) + f_x(a,b)(x-a) + f_y(a,b)(y-b) + \frac{1}{2!}\left[ f_{xx}(a,b)(x-a)^2 + 2f_{xy}(a,b)(x-a)(y-b) + f_{yy}(a,b)(y-b)^2 ight]\)
Step 4: Substitute the values:
\(f(x,y) = \frac{1}{e\sqrt{2}} + \frac{1}{e\sqrt{2}}(x+1) + \frac{1}{e\sqrt{2}}\left(y-\frac{\pi}{4} ight) + \frac{1}{2}\left[ \frac{1}{e\sqrt{2}}(x+1)^2 + 2\frac{1}{e\sqrt{2}}(x+1)\left(y-\frac{\pi}{4} ight) - \frac{1}{e\sqrt{2}}\left(y-\frac{\pi}{4} ight)^2 ight]\)
Step 5: Factoring out \(\frac{1}{e\sqrt{2}}\) gives the final expansion:
\(e^x \sin y = \frac{1}{e\sqrt{2}} \left[ 1 + (x+1) + \left(y-\frac{\pi}{4} ight) + \frac{1}{2}(x+1)^2 + (x+1)\left(y-\frac{\pi}{4} ight) - \frac{1}{2}\left(y-\frac{\pi}{4} ight)^2 ight]\)

Detailed Solution:
Step 1: Expand the function: \(f(x, y) = 12x^3y^2 - x^4y^2 - x^3y^3\).
Step 2: Find the first partial derivatives:
- \(f_x = 36x^2y^2 - 4x^3y^2 - 3x^2y^3 = x^2y^2(36 - 4x - 3y)\)
- \(f_y = 24x^3y - 2x^4y - 3x^3y^2 = x^3y(24 - 2x - 3y)\)
Step 3: Set \(f_x = 0\) and \(f_y = 0\) to find stationary points. Assuming \(x \neq 0, y \neq 0\) (as the function would be 0 there, which isn't a strict extrema point):
\(4x + 3y = 36\)
\(2x + 3y = 24\)
Step 4: Subtracting the second equation from the first:
\(2x = 12 \implies x = 6\).
Substituting \(x = 6\) back into \(2x + 3y = 24\):
\(12 + 3y = 24 \implies 3y = 12 \implies y = 4\).
So, the critical point is \((6, 4)\).
Step 5: Find the second partial derivatives (\(r, s, t\)):
- \(r = f_{xx} = \frac{\partial}{\partial x}(36x^2y^2 - 4x^3y^2 - 3x^2y^3) = 72xy^2 - 12x^2y^2 - 6xy^3 = 6xy^2(12 - 2x - y)\)
- \(t = f_{yy} = \frac{\partial}{\partial y}(24x^3y - 2x^4y - 3x^3y^2) = 24x^3 - 2x^4 - 6x^3y = 2x^3(12 - x - 3y)\)
- \(s = f_{xy} = \frac{\partial}{\partial y}(36x^2y^2 - 4x^3y^2 - 3x^2y^3) = 72x^2y - 8x^3y - 9x^2y^2 = x^2y(72 - 8x - 9y)\)
Step 6: Evaluate \(r, s, t\) at \((6, 4)\):
- \(r = 6(6)(16)(12 - 12 - 4) = 576(-4) = -2304\)
- \(t = 2(216)(12 - 6 - 12) = 432(-6) = -2592\)
- \(s = (36)(4)(72 - 48 - 36) = 144(72 - 84) = 144(-12) = -1728\)
Step 7: Calculate \(\Delta = rt - s^2\):
\(\Delta = (-2304)(-2592) - (-1728)^2 = 5971968 - 2985984 = 2985984\).
Step 8: Since \(\Delta > 0\) and \(r < 0\), the point \((6, 4)\) is a point of local maximum.
Step 9: Calculate the maximum value:
\(f_{max} = (6^3)(4^2)(12 - 6 - 4) = (216)(16)(2) = 6912\).

Detailed Solution:
Step 1: Find the complementary function (CF). The homogeneous equation is \(\frac{d^2y}{dx^2} + y = 0\).
Auxiliary equation: \(m^2 + 1 = 0 \implies m = \pm i\).
The CF is \(y_c = C_1\cos x + C_2\sin x\).
Let \(u = \cos x\) and \(v = \sin x\).
Step 2: Calculate the Wronskian \(W = u v' - v u'\):
\(W = \cos x(\cos x) - \sin x(-\sin x) = \cos^2 x + \sin^2 x = 1\).
Step 3: The particular integral (PI) by variation of parameters is given by \(PI = P u + Q v\), where:
\(P = -\int \frac{v R}{W} dx\) and \(Q = \int \frac{u R}{W} dx\). Here, \(R = \text{cosec}x = \frac{1}{\sin x}\).
Step 4: Calculate \(P\):
\(P = -\int \frac{\sin x \cdot \text{cosec}x}{1} dx = -\int \frac{\sin x}{\sin x} dx = -\int 1 dx = -x\).
Step 5: Calculate \(Q\):
\(Q = \int \frac{\cos x \cdot \text{cosec}x}{1} dx = \int \frac{\cos x}{\sin x} dx\).
Using substitution \(t = \sin x \implies dt = \cos x dx\), \(\int \frac{dt}{t} = \log |t| = \log |\sin x|\).
Step 6: Combine to find the PI:
\(PI = -x\cos x + (\log |\sin x|)\sin x\).
Step 7: The general solution is \(y = y_c + PI\):
\(y = C_1\cos x + C_2\sin x - x\cos x + \sin x\log |\sin x|\).

Detailed Solution:
Step 1: This is a Cauchy-Euler differential equation. Substitute \(x = e^z\), or \(z = \log x\).
Let \(\theta = \frac{d}{dz}\). We know \(xD \equiv \theta\) and \(x^2D^2 \equiv \theta(\theta-1)\).
Step 2: The equation becomes:
\((\theta(\theta-1) - 3\theta + 4)y = (e^z)^2\)
\((\theta^2 - \theta - 3\theta + 4)y = e^{2z}\)
\((\theta^2 - 4\theta + 4)y = e^{2z}\)
Step 3: Find the complementary function (CF). The auxiliary equation is:
\(m^2 - 4m + 4 = 0 \implies (m-2)^2 = 0 \implies m = 2, 2\).
The CF is \(y_c = (C_1 + C_2z)e^{2z}\).
Step 4: Find the particular integral (PI).
\(PI = \frac{1}{\theta^2 - 4\theta + 4} e^{2z} = \frac{1}{(\theta-2)^2} e^{2z}\).
When substituting \(\theta = 2\), the denominator is zero. This signifies resonance.
The rule for \(\frac{1}{(\theta-a)^n} e^{az}\) is \(\frac{z^n}{n!} e^{az}\).
Here \(a=2, n=2\), so \(PI = \frac{z^2}{2!} e^{2z} = \frac{z^2}{2} e^{2z}\).
Step 5: The general solution in terms of \(z\) is \(y = y_c + PI = (C_1 + C_2z)e^{2z} + \frac{z^2}{2}e^{2z}\).
Step 6: Substitute back \(z = \log x\) and \(e^{2z} = x^2\):
\(y = (C_1 + C_2\log x)x^2 + \frac{(\log x)^2}{2}x^2\).

Detailed Solution:
Step 1: Differentiate \(\sqrt{x} + \sqrt{y} = \sqrt{a}\) with respect to \(x\):
\(\frac{1}{2\sqrt{x}} + \frac{1}{2\sqrt{y}} \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\sqrt{\frac{y}{x}}\).
Step 2: Evaluate \(y_1 = \frac{dy}{dx}\) at the given point \((\frac{a}{4}, \frac{a}{4})\):
\(y_1 = -\sqrt{\frac{a/4}{a/4}} = -1\).
Step 3: Differentiate \(y_1\) to find \(y_2 = \frac{d^2y}{dx^2}\):
\(\frac{d^2y}{dx^2} = -\left[ \frac{\sqrt{x} \cdot \frac{1}{2\sqrt{y}} y_1 - \sqrt{y} \cdot \frac{1}{2\sqrt{x}}}{x} ight]\)
Step 4: Evaluate \(y_2\) at \((\frac{a}{4}, \frac{a}{4})\):
Substituting \(x = a/4, y = a/4\) and \(y_1 = -1\):
\(y_2 = -\frac{\frac{\sqrt{a}}{2} \cdot \frac{1}{\sqrt{a}} (-1) - \frac{\sqrt{a}}{2} \cdot \frac{1}{\sqrt{a}}}{a/4} = -\frac{-\frac{1}{2} - \frac{1}{2}}{a/4} = -\frac{-1}{a/4} = \frac{4}{a}\).
Step 5: Calculate the radius of curvature \( ho\):
\( ho = \frac{(1 + y_1^2)^{3/2}}{|y_2|} = \frac{(1 + (-1)^2)^{3/2}}{4/a} = \frac{2^{3/2}}{4/a} = \frac{2\sqrt{2}a}{4} = \frac{a}{\sqrt{2}}\).
Step 6: Find the coordinates of the centre of curvature \((\bar{x}, \bar{y})\):
\(\bar{x} = x - \frac{y_1(1 + y_1^2)}{y_2} = \frac{a}{4} - \frac{-1(1 + 1)}{4/a} = \frac{a}{4} + \frac{2a}{4} = \frac{3a}{4}\).
\(\bar{y} = y + \frac{1 + y_1^2}{y_2} = \frac{a}{4} + \frac{(1 + 1)}{4/a} = \frac{a}{4} + \frac{2a}{4} = \frac{3a}{4}\).
Step 7: The equation of the circle of curvature is \((x - \bar{x})^2 + (y - \bar{y})^2 = ho^2\):
\((x - \frac{3a}{4})^2 + (y - \frac{3a}{4})^2 = \left(\frac{a}{\sqrt{2}} ight)^2\)
\((x - \frac{3a}{4})^2 + (y - \frac{3a}{4})^2 = \frac{a^2}{2}\).

Detailed Solution:
Step 1: The parametric equations for the parabola \(x^2 = 4ay\) are \(x = 2at, y = at^2\).
Step 2: Differentiate with respect to \(t\):
\(\frac{dx}{dt} = 2a\)
\(\frac{dy}{dt} = 2at\)
\(y_1 = \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2at}{2a} = t\).
Step 3: Differentiate again to find \(y_2\):
\(y_2 = \frac{d}{dx}(y_1) = \frac{d}{dt}(t) \cdot \frac{dt}{dx} = 1 \cdot \frac{1}{2a} = \frac{1}{2a}\).
Step 4: The coordinates of the centre of curvature \((\bar{x}, \bar{y})\) are given by:
\(\bar{x} = x - \frac{y_1(1 + y_1^2)}{y_2}\) and \(\bar{y} = y + \frac{1 + y_1^2}{y_2}\).
Step 5: Substitute the values:
\(\bar{x} = 2at - \frac{t(1 + t^2)}{1/(2a)} = 2at - 2at(1 + t^2) = 2at - 2at - 2at^3 = -2at^3\).
\(\bar{y} = at^2 + \frac{1 + t^2}{1/(2a)} = at^2 + 2a(1 + t^2) = at^2 + 2a + 2at^2 = 3at^2 + 2a\).
Step 6: The locus of \((\bar{x}, \bar{y})\) is the evolute. We eliminate the parameter \(t\):
From the \(\bar{x}\) equation: \(t^3 = -\frac{\bar{x}}{2a} \implies t = \left(-\frac{\bar{x}}{2a} ight)^{1/3}\).
Substitute \(t\) into the \(\bar{y}\) equation:
\(\bar{y} - 2a = 3at^2 = 3a\left(-\frac{\bar{x}}{2a} ight)^{2/3}\).
Step 7: Cube both sides to remove the fractional power:
\((\bar{y} - 2a)^3 = (3a)^3 \left(-\frac{\bar{x}}{2a} ight)^2\)
\((\bar{y} - 2a)^3 = 27a^3 \cdot \frac{\bar{x}^2}{4a^2}\)
\((\bar{y} - 2a)^3 = \frac{27a}{4} \bar{x}^2\)
Step 8: Replacing \((\bar{x}, \bar{y})\) with \((x, y)\) gives the standard equation:
\(4(y - 2a)^3 = 27ax^2\).

Detailed Solution:
Step 1: Let the general term be \(u_n = \frac{n^3+1}{2^n+1}\).
Step 2: We use the D'Alembert's Ratio Test. We evaluate \(\lim_{n \to \infty} \frac{u_{n+1}}{u_n}\).
Step 3: \(u_{n+1} = \frac{(n+1)^3+1}{2^{n+1}+1}\).
Step 4: Calculate the ratio:
\(\frac{u_{n+1}}{u_n} = \frac{(n+1)^3+1}{2^{n+1}+1} \times \frac{2^n+1}{n^3+1}\).
Step 5: Factor out the highest powers of \(n\) and \(2^n\):
\(\frac{u_{n+1}}{u_n} = \frac{n^3(1 + \frac{1}{n})^3 + 1}{2^n(2 + \frac{1}{2^n})} \times \frac{2^n(1 + \frac{1}{2^n})}{n^3(1 + \frac{1}{n^3})}\)
\(\frac{u_{n+1}}{u_n} = \frac{(1 + \frac{1}{n})^3 + \frac{1}{n^3}}{(2 + \frac{1}{2^n})} \times \frac{(1 + \frac{1}{2^n})}{(1 + \frac{1}{n^3})}\).
Step 6: Apply the limit as \(n \to \infty\):
As \(n \to \infty\), \(\frac{1}{n} \to 0\), \(\frac{1}{n^3} \to 0\), and \(\frac{1}{2^n} \to 0\).
\(\lim_{n \to \infty} \frac{u_{n+1}}{u_n} = \frac{(1+0)^3 + 0}{2 + 0} \times \frac{1 + 0}{1 + 0} = \frac{1}{2} \times 1 = \frac{1}{2}\).
Step 7: Since the limit \(L = \frac{1}{2} < 1\), by D'Alembert's Ratio Test, the series \(\sum u_n\) is convergent.

Detailed Solution:
Step 1: Write the general term \(u_n\) of the series. For \(n \ge 2\) (treating the first term as a special case or shifting index):
\(u_n = \frac{1 \cdot 3 \cdot 5 \dots (2n-3)}{2 \cdot 4 \cdot 6 \dots (2n-2)} \frac{x^{2n-1}}{2n-1}\).
Then \(u_{n+1} = \frac{1 \cdot 3 \cdot 5 \dots (2n-3)(2n-1)}{2 \cdot 4 \cdot 6 \dots (2n-2)(2n)} \frac{x^{2n+1}}{2n+1}\).
Step 2: Apply D'Alembert's Ratio Test. Calculate \(\lim_{n \to \infty} \frac{u_{n+1}}{u_n}\):
\(\frac{u_{n+1}}{u_n} = \frac{2n-1}{2n} \cdot \frac{x^{2n+1}}{x^{2n-1}} \cdot \frac{2n-1}{2n+1} = \frac{(2n-1)^2}{2n(2n+1)} x^2\).
\(\lim_{n \to \infty} \frac{u_{n+1}}{u_n} = \lim_{n \to \infty} \frac{4n^2 - 4n + 1}{4n^2 + 2n} x^2 = x^2\).
Step 3: By the Ratio test, the series converges if \(x^2 < 1\) (i.e., \(x < 1\) since \(x>0\)) and diverges if \(x^2 > 1\) (\(x > 1\)). The test fails when \(x^2 = 1 \implies x = 1\).
Step 4: Apply Raabe's Test for the failure case \(x = 1\). The limit for Raabe's test is \(\lim_{n \to \infty} n \left( \frac{u_n}{u_{n+1}} - 1 ight)\).
For \(x = 1\), \(\frac{u_n}{u_{n+1}} = \frac{2n(2n+1)}{(2n-1)^2} = \frac{4n^2+2n}{4n^2-4n+1}\).
Step 5: Calculate \(\frac{u_n}{u_{n+1}} - 1\):
\(\frac{4n^2+2n}{4n^2-4n+1} - 1 = \frac{4n^2+2n - (4n^2-4n+1)}{4n^2-4n+1} = \frac{6n-1}{4n^2-4n+1}\).
Step 6: Evaluate the limit:
\(\lim_{n \to \infty} n \left( \frac{6n-1}{4n^2-4n+1} ight) = \lim_{n \to \infty} \frac{6n^2-n}{4n^2-4n+1} = \frac{6}{4} = \frac{3}{2}\).
Step 7: Since the limit \(L = \frac{3}{2} > 1\), by Raabe's test, the series converges at \(x = 1\).
Conclusion: Convergent for \(0 < x \le 1\), divergent for \(x > 1\).

Detailed Solution:
Step 1: Write the matrix of the quadratic form. The coefficients are \(a_{11}=1, a_{22}=2, a_{33}=1, 2a_{12}=-2, 2a_{23}=2, 2a_{13}=0\).
\(A = \begin{bmatrix} 1 & -1 & 0 \ -1 & 2 & 1 \ 0 & 1 & 1 \end{bmatrix}\).
Step 2: Find the eigenvalues (Characteristic equation \(\lambda^3 - S_1\lambda^2 + S_2\lambda - S_3 = 0\)):
- \(S_1 = 1 + 2 + 1 = 4\).
- \(S_2 = (2-1) + (1-0) + (2-1) = 1 + 1 + 1 = 3\).
- \(S_3 = |A| = 1(2-1) - (-1)(-1-0) + 0 = 1 - 1 = 0\).
The equation is \(\lambda^3 - 4\lambda^2 + 3\lambda = 0 \implies \lambda(\lambda^2 - 4\lambda + 3) = 0 \implies \lambda(\lambda-1)(\lambda-3) = 0\).
Eigenvalues: \(\lambda = 3, 1, 0\).
Step 3: Since all eigenvalues are \(\ge 0\) and at least one is zero, the nature is Positive Semi-definite.
Step 4: Find eigenvectors.
For \(\lambda = 3\): \((A-3I)X = 0 \implies \begin{bmatrix} -2 & -1 & 0 \ -1 & -1 & 1 \ 0 & 1 & -2 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \ x_3 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix}\).
From Row 1: \(-2x_1 - x_2 = 0 \implies x_2 = -2x_1\). From Row 3: \(x_2 - 2x_3 = 0 \implies x_2 = 2x_3\).
Let \(x_3 = 1\), then \(x_2 = 2\), \(x_1 = -1\). Eigenvector \(X_1 = \begin{bmatrix} -1 \ 2 \ 1 \end{bmatrix}\).
For \(\lambda = 1\): \((A-I)X = 0 \implies \begin{bmatrix} 0 & -1 & 0 \ -1 & 1 & 1 \ 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} x_1 \ x_2 \ x_3 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ 0 \end{bmatrix}\).
From Row 1: \(x_2 = 0\). From Row 2: \(-x_1 + x_2 + x_3 = 0 \implies x_1 = x_3\).
Let \(x_1 = 1\), then \(x_3 = 1\). Eigenvector \(X_2 = \begin{bmatrix} 1 \ 0 \ 1 \end{bmatrix}\).
For \(\lambda = 0\): \((A-0I)X = 0 \implies A X = 0\).
From Row 1: \(x_1 - x_2 = 0 \implies x_1 = x_2\). From Row 3: \(x_2 + x_3 = 0 \implies x_3 = -x_2\).
Let \(x_1 = 1\), then \(x_2 = 1, x_3 = -1\). Eigenvector \(X_3 = \begin{bmatrix} 1 \ 1 \ -1 \end{bmatrix}\).
Step 5: Normalize the eigenvectors to form orthogonal matrix \(N\).
Length of \(X_1 = \sqrt{1+4+1} = \sqrt{6}\).
Length of \(X_2 = \sqrt{1+0+1} = \sqrt{2}\).
Length of \(X_3 = \sqrt{1+1+1} = \sqrt{3}\).
Normalized matrix \(N = \begin{bmatrix} -1/\sqrt{6} & 1/\sqrt{2} & 1/\sqrt{3} \ 2/\sqrt{6} & 0 & 1/\sqrt{3} \ 1/\sqrt{6} & 1/\sqrt{2} & -1/\sqrt{3} \end{bmatrix}\).
Step 6: The diagonal matrix \(D = N^T A N\) will have the eigenvalues on the diagonal: \(D = \text{diag}(3, 1, 0)\).
Step 7: The canonical form is \(Y^T D Y = \lambda_1 y_1^2 + \lambda_2 y_2^2 + \lambda_3 y_3^2 = 3y_1^2 + 1y_2^2 + 0y_3^2\).
Step 8: Properties:
- Rank = number of non-zero eigenvalues = 2.
- Index = number of positive eigenvalues = 2.
- Signature = (number of positive) - (number of negative) = 2 - 0 = 2.

Detailed Solution:
Step 1: Let the point on the sphere be \(P(x, y, z)\). We need to optimize the distance from \(A(1, 2, -1)\) to \(P(x, y, z)\).
Distance squared: \(f(x, y, z) = (x - 1)^2 + (y - 2)^2 + (z + 1)^2\).
Subject to the constraint: \(g(x, y, z) = x^2 + y^2 + z^2 - 24 = 0\).
Step 2: Form the Lagrangian function \(F(x, y, z) = f(x, y, z) + \lambda g(x, y, z)\):
\(F = (x - 1)^2 + (y - 2)^2 + (z + 1)^2 + \lambda (x^2 + y^2 + z^2 - 24)\).
Step 3: Find the partial derivatives and set them to zero:
- \(\frac{\partial F}{\partial x} = 2(x - 1) + 2\lambda x = 0 \implies x - 1 + \lambda x = 0 \implies x(1 + \lambda) = 1 \implies x = \frac{1}{1 + \lambda}\)
- \(\frac{\partial F}{\partial y} = 2(y - 2) + 2\lambda y = 0 \implies y - 2 + \lambda y = 0 \implies y(1 + \lambda) = 2 \implies y = \frac{2}{1 + \lambda}\)
- \(\frac{\partial F}{\partial z} = 2(z + 1) + 2\lambda z = 0 \implies z + 1 + \lambda z = 0 \implies z(1 + \lambda) = -1 \implies z = \frac{-1}{1 + \lambda}\)
Step 4: Substitute \(x, y, z\) into the constraint equation \(x^2 + y^2 + z^2 = 24\):
\(\left(\frac{1}{1 + \lambda} ight)^2 + \left(\frac{2}{1 + \lambda} ight)^2 + \left(\frac{-1}{1 + \lambda} ight)^2 = 24\)
\(\frac{1 + 4 + 1}{(1 + \lambda)^2} = 24 \implies \frac{6}{(1 + \lambda)^2} = 24\)
\((1 + \lambda)^2 = \frac{6}{24} = \frac{1}{4} \implies 1 + \lambda = \pm \frac{1}{2}\).
Step 5: Find the points corresponding to each value of \(1 + \lambda\):
Case 1: \(1 + \lambda = \frac{1}{2}\)
\(x = \frac{1}{1/2} = 2\), \(y = \frac{2}{1/2} = 4\), \(z = \frac{-1}{1/2} = -2\). So point \(P_1 = (2, 4, -2)\).
Case 2: \(1 + \lambda = -\frac{1}{2}\)
\(x = \frac{1}{-1/2} = -2\), \(y = \frac{2}{-1/2} = -4\), \(z = \frac{-1}{-1/2} = 2\). So point \(P_2 = (-2, -4, 2)\).
Step 6: Calculate distances from \(A(1, 2, -1)\) to these points.
Distance to \(P_1(2, 4, -2)\):
\(d_1^2 = (2 - 1)^2 + (4 - 2)^2 + (-2 - (-1))^2 = 1^2 + 2^2 + (-1)^2 = 1 + 4 + 1 = 6 \implies d_1 = \sqrt{6}\).
Distance to \(P_2(-2, -4, 2)\):
\(d_2^2 = (-2 - 1)^2 + (-4 - 2)^2 + (2 - (-1))^2 = (-3)^2 + (-6)^2 + (3)^2 = 9 + 36 + 9 = 54 \implies d_2 = \sqrt{54} = 3\sqrt{6}\).
Step 7: Conclusion:
The shortest distance is \(\sqrt{6}\).
The longest distance is \(3\sqrt{6}\).