Q: Q1. (e) Use two phase method to solve the following linear programming problem: Maximize $z = x1 + 2x2$ subject to $x1 - x2 ge 3$ $2x1 + x2 le 10$ $x1, x2 ge 0$

Theory of the Two-Phase Simplex Method The Two-Phase Simplex Method is used to solve linear programming problems where initial basic feasible solutions cannot be found by slack variables alone (e.g., in the presence of $ge$ or $=$ constraints requiring artificial variables). - Phase I: Formulate an auxiliary objective function $W$ defined as the sum of all artificial variables. Minimize $W$ (or maximize $-W$) subject to the system of constraints. If the minimum value of $W$ is zero, an initial basic feasible solution (BFS) to the original problem has been obtained, and we proceed to Phase II. If $min W 0$, the problem has no feasible solution. - Phase II: Using the basic feasible solution obtained at the end of Phase I as the starting point, drop the artificial variables and optimize the original objective function $z$. --- Step 1: Converting to Standard Form The given problem is: $$text{Maximize } z = x1 + 2x2$$ subject to: $$x1 - x2 ge 3$$ $$2x1 + x2 le 10$$ $$x1, x2 ge 0$$ Convert inequalities into equations by introducing: - A surplus variable $s1 ge 0$ and an artificial variable $A1 ge 0$ for the first constraint: $$x1 - x2 - s1 + A1 = 3$$ - A slack variable $s2 ge 0$ for the second constraint: $$2x1 + x2 + s2 = 10$$ with all variables $x1, x2, s1, s2, A1 ge 0$. --- Step 2: Phase I In Phase I, we minimize the artificial variable $A1$, which is equivalent to maximizing: $$W^ = -A1$$ From the first constraint: $A1 = 3 - x1 + x2 + s1$. Expressing the Phase I objective in terms of non-basic variables: $$W^ + A1 = 0 implies W^ + (3 - x1 + x2 + s1) = 0 implies W^ - x1 + x2 + s1 = -3$$ The Phase I initial simplex tableau with basic variables $A1, s2$: | Basis | $x1$ | $x2$ | $s1$ | $s2$ | $A1$ | RHS | Ratio | |---|---|---|---|---|---|---|---| | $A1$ | 1 | -1 | -1 | 0 | 1 | 3 | $3/1 = 3$ (min) | | $s2$ | 2 | 1 | 0 | 1 | 0 | 10 | $10/2 = 5$ | | $W^$ | 1 | -1 | -1 | 0 | 0 | 3 | | Entering variable: $x1$ (most positive coefficient in indicator row). Leaving variable: $A1$

Q: Q1. (d) If $phi$ and $psi$ are functions of $x$ and $y$ satisfying Laplace equation, then show that $f(z) = p + iq, i = sqrt{-1}$ is an analytic function, where $p = frac{partialphi}{partial y} - frac{partialpsi}{partial x}$ and $q = frac{partialphi}{partial x} + frac{partialpsi}{partial y}$.

Theory of Complex Analytic Functions and the Cauchy-Riemann Equations Let $f(z) = p(x, y) + i q(x, y)$ be a complex-valued function of $z = x + iy$. By the Cauchy-Riemann Criterion, $f(z)$ is analytic in a domain $D$ if and only if: 1. The real functions $p(x, y)$ and $q(x, y)$ possess continuous first-order partial derivatives throughout $D$. 2. The partial derivatives satisfy the Cauchy-Riemann (C-R) equations: $$frac{partial p}{partial x} = frac{partial q}{partial y} quad text{--- (C-R 1)}$$ $$frac{partial p}{partial y} = -frac{partial q}{partial x} quad text{--- (C-R 2)}$$ --- Step 1: Given Conditions on $phi$ and $psi$ We are given that $phi(x, y)$ and $psi(x, y)$ satisfy Laplace's equation in two dimensions: $$ abla^2 phi = frac{partial^2 phi}{partial x^2} + frac{partial^2 phi}{partial y^2} = 0 implies frac{partial^2 phi}{partial y^2} = -frac{partial^2 phi}{partial x^2} quad text{--- (1)}$$ $$ abla^2 psi = frac{partial^2 psi}{partial x^2} + frac{partial^2 psi}{partial y^2} = 0 implies frac{partial^2 psi}{partial y^2} = -frac{partial^2 psi}{partial x^2} quad text{--- (2)}$$ Furthermore, since harmonic functions are infinitely differentiable, the mixed partial derivatives are equal by Schwarz's (Clairaut's) theorem: $$frac{partial^2 phi}{partial x partial y} = frac{partial^2 phi}{partial y partial x}, quad frac{partial^2 psi}{partial x partial y} = frac{partial^2 psi}{partial y partial x}$$ --- Step 2: Partial Derivatives of $p(x, y)$ We are given: $$p = frac{partialphi}{partial y} - frac{partialpsi}{partial x}$$ 1. Differentiating $p$ with respect to $x$: $$frac{partial p}{partial x} = frac{partial}{partial x}left(frac{partialphi}{partial y} - frac{partialpsi}{partial x}right) = frac{partial^2 phi}{partial x partial y} - frac{partial^2 psi}{partial x^2} quad text{--- (3)}$$ 2. Differentiating $p$ with respect to $y$: $$frac{partial p}{partial y} = frac{partial}{partial y}left(frac{partialphi}{partial y} - frac{partialpsi}{partial x}right) = frac{partial^2 phi}{

Q: Q1. (c) Test the convergence of $$int0^2 frac{log x}{sqrt{2 - x}},dx.$$

Analysis of the Improper Integral We are required to test the convergence of the definite integral: $$I = int0^2 frac{log x}{sqrt{2 - x}},dx$$ where $log x$ denotes the natural logarithm $ln x$. Let the integrand be: $$f(x) = frac{log x}{sqrt{2 - x}}$$ We inspect the behavior of $f(x)$ on the interval of integration $(0, 2)$: 1. At the lower limit $x to 0^+$: $$lim{x to 0^+} log x = -infty, quad lim{x to 0^+} sqrt{2 - x} = sqrt{2}$$ Hence, $f(x) to -infty$. Thus, $x = 0$ is a point of infinite discontinuity. 2. At the upper limit $x to 2^-$: $$lim{x to 2^-} log x = log 2, quad lim{x to 2^-} sqrt{2 - x} = 0$$ Hence, $f(x) to +infty$. Thus, $x = 2$ is also a point of infinite discontinuity. Because the integrand possesses singularities at both endpoints, we split the integral at an intermediate interior point, conveniently chosen as $c = 1$ (where $log(1) = 0$): $$I = int0^1 frac{log x}{sqrt{2 - x}},dx + int1^2 frac{log x}{sqrt{2 - x}},dx = I1 + I2$$ The integral $I$ converges if and only if both $I1$ and $I2$ converge independently. --- Step 1: Convergence of $I1 = int0^1 frac{log x}{sqrt{2 - x}},dx$ For $x in (0, 1]$, $log x le 0$, so $|f(x)| = frac{-log x}{sqrt{2 - x}}$. The only singularity in $[0, 1]$ is at $x = 0$. Notice that for $0 0$$ Hence, the singularity behaves as: $$f(x) sim frac{log 2}{(2 - x)^{1/2}}$$ We use the Limit Comparison Test with the comparison function: $$g(x) = frac{1}{(2 - x)^{1/2}}$$ Compute the limit: $$L = lim{x to 2^-} frac{f(x)}{g(x)} = lim{x to 2^-} frac{frac{log x}{sqrt{2 - x}}}{frac{1}{sqrt{2 - x}}} = lim{x to 2^-} log x = log 2$$ Since $L = log 2$ is a finite, non-zero positive number ($0 < L < infty$), both integrals $int1^2 f(x),dx$ and $int1^2 g(x),dx$ behave identically with regard to convergence. Now examine the standard $p$-integral at the upper limit: $$int1^2 frac{dx}{(2 - x)^p} quad text{with } p = frac{1}{2}$$ Substitute $t = 2 - x implies dt = -dx$: $$int1^2 frac{dx}{(2 - x)^{1/2}} = int0^1 frac{dt}{t^{1/2}} = [2t^{1/2}]

Q: Q1. (b) If $w = f(z)$ is an analytic function of $z$, then show that $$left(frac{partial^2}{partial x^2} + frac{partial^2}{partial y^2}right) log |f'(z)| = 0.$$

Theory of Complex Analytic Functions and Harmonicity Let $w = f(z)$ be an analytic function of the complex variable $z = x + iy$ in a domain $D subseteq mathbb{C}$. Since $f(z)$ is analytic, its derivative $F(z) = f'(z)$ is also analytic in $D$. Assume that $f'(z) eq 0$ in the region under consideration (since $log |f'(z)|$ is undefined at zeros of $f'(z)$). --- Method 1: Complex Operator Formalism (Wirtinger Derivatives) The two-dimensional Laplacian operator in Cartesian coordinates is related to the Wirtinger derivatives by: $$ abla^2 = frac{partial^2}{partial x^2} + frac{partial^2}{partial y^2} = 4 frac{partial^2}{partial z , partial bar{z}}$$ where: $$frac{partial}{partial z} = frac{1}{2}left( frac{partial}{partial x} - ifrac{partial}{partial y} right), quad frac{partial}{partial bar{z}} = frac{1}{2}left( frac{partial}{partial x} + ifrac{partial}{partial y} right)$$ Recall that for any non-zero complex quantity $W$: $$|W| = sqrt{W bar{W}} implies log |W| = frac{1}{2} log(W bar{W}) = frac{1}{2} log W + frac{1}{2} log bar{W}$$ Applying this to $W = f'(z)$: $$log |f'(z)| = frac{1}{2} log f'(z) + frac{1}{2} log overline{f'(z)}$$ Since $f'(z)$ is an analytic function of $z$, it depends only on $z$ and not on $bar{z}$: $$frac{partial}{partial bar{z}} [f'(z)] = 0$$ Similarly, its complex conjugate $overline{f'(z)}$ depends only on $bar{z}$ and not on $z$: $$frac{partial}{partial z} [overline{f'(z)}] = 0$$ Now, compute the Wirtinger derivative: $$frac{partial}{partial bar{z}} left( log |f'(z)| right) = frac{partial}{partial bar{z}} left( frac{1}{2}log f'(z) + frac{1}{2}log overline{f'(z)} right) = 0 + frac{1}{2} frac{overline{f''(z)}}{overline{f'(z)}}$$ Next, differentiate with respect to $z$: $$frac{partial}{partial z} left[ frac{partial}{partial bar{z}} log |f'(z)| right] = frac{partial}{partial z} left( frac{1}{2} frac{overline{f''(z)}}{overline{f'(z)}} right) = 0$$ because $frac{overline{f''(z)}}{overline{f'(z)}}$ is a function purely of $bar{z}$. Therefore: $$ abl

Q: Q1. (a) Let $G$ be a finite group of order $mn$, where $m$ and $n$ are prime numbers with $m n$. Show that $G$ has at most one subgroup of order $m$.

Algebraic Context and Sylow's Theorems Let $G$ be a finite group of order: $$|G| = mn$$ where $m$ and $n$ are distinct prime numbers such that $m n$. We wish to establish that $G$ contains at most one subgroup of order $m$. We rely on the foundational results of Sylow's Theorems: 1. Sylow's First Theorem: If $p$ is a prime dividing $|G|$ such that $p^k mid |G|$ and $p^{k+1} mid |G|$, then $G$ contains a subgroup of order $p^k$, called a Sylow $p$-subgroup. 2. Sylow's Second Theorem: Any two Sylow $p$-subgroups of $G$ are conjugate in $G$. 3. Sylow's Third Theorem: The number $np$ of distinct Sylow $p$-subgroups of $G$ satisfies two simultaneous conditions: - $np equiv 1 pmod p$, and - $np$ divides the index $frac{|G|}{p^k}$. --- Step 1: Application of Sylow's Third Theorem Since $m$ is a prime number and $m$ divides $|G| = mn$ with $m^2 mid |G|$, every subgroup of $G$ of order $m$ is a Sylow $m$-subgroup of $G$. Let $nm$ denote the number of Sylow $m$-subgroups in $G$. According to Sylow's Third Theorem: 1. $nm$ must divide the index of a Sylow $m$-subgroup in $G$: $$frac{|G|}{m} = frac{mn}{m} = n$$ Therefore, $nm mid n$. 2. $nm$ must be congruent to $1$ modulo $m$: $$nm equiv 1 pmod m$$ --- Step 2: Analyzing the Divisors of $n$ Since $n$ is a prime number, its only positive divisors in $mathbb{Z}^+$ are: $$nm in {1, n}$$ We now examine the two possibilities: - Case 1: $nm = 1$ If $nm = 1$, then $1 equiv 1 pmod m$, which is universally true for any prime $m$. Thus $nm = 1$ is always a mathematically valid solution. - Case 2: $nm = n$ If $nm = n$, then the second condition requires: $$n equiv 1 pmod m$$ This implies that $(n - 1)$ is a positive multiple of $m$: $$n - 1 = k m quad text{for some integer } k ge 1$$ $$implies n = km + 1 ge m + 1 m$$ Therefore, this case strictly necessitates that: $$n m$$ However, the hypothesis of the theorem explicitly states that: $$m n$$ This is a direct contradiction because $n$ cannot be both strictly less than $m$ and strictly gre

Q: Q2. (c) Find the function which is analytic inside and on the circle $C : z = e^{itheta}, 0 le theta le 2pi$ and has the value $$frac{(a^2 - 1)costheta + i(a^2 + 1)sintheta}{a^4 - 2a^2cos 2theta + 1}$$ on the circumference of $C$, where $a^2 1$.

Problem Statement and Formulation We seek a complex function $f(z)$ which is analytic inside and on the unit circle: $$C = { z in mathbb{C} : |z| le 1 }$$ given its boundary values on the circumference $|z| = 1$, where $z = e^{itheta}$ for $0 le theta le 2pi$: $$f(e^{itheta}) = frac{(a^2 - 1)costheta + i(a^2 + 1)sintheta}{a^4 - 2a^2cos 2theta + 1} quad (text{where } a^2 1)$$ --- Step 1: Converting Boundary Values in Terms of $z = e^{itheta}$ On the boundary circle $C$, we have $z = e^{itheta}$, which gives: $$bar{z} = z^{-1} = frac{1}{z}$$ Using Euler's formulas for trigonometric functions: $$costheta = frac{e^{itheta} + e^{-itheta}}{2} = frac{z + z^{-1}}{2} = frac{z^2 + 1}{2z}$$ $$sintheta = frac{e^{itheta} - e^{-itheta}}{2i} = frac{z - z^{-1}}{2i} = frac{z^2 - 1}{2iz}$$ $$cos 2theta = frac{e^{i2theta} + e^{-i2theta}}{2} = frac{z^2 + z^{-2}}{2} = frac{z^4 + 1}{2z^2}$$ --- Step 2: Simplifying the Numerator Let $N(theta)$ denote the numerator: $$N(theta) = (a^2 - 1)costheta + i(a^2 + 1)sintheta$$ Substitute the expressions for $costheta$ and $sintheta$: $$N(theta) = (a^2 - 1)left( frac{z + z^{-1}}{2} right) + i(a^2 + 1)left( frac{z - z^{-1}}{2i} right)$$ Notice that $frac{i}{2i} = frac{1}{2}$: $$N(theta) = frac{1}{2} left[ (a^2 - 1)(z + z^{-1}) + (a^2 + 1)(z - z^{-1}) right]$$ Expanding the brackets: $$(a^2 - 1)(z + z^{-1}) = a^2 z + a^2 z^{-1} - z - z^{-1}$$ $$(a^2 + 1)(z - z^{-1}) = a^2 z - a^2 z^{-1} + z - z^{-1}$$ Adding these two expressions: $$N(theta) = frac{1}{2} left[ (a^2 z + a^2 z) + (a^2 z^{-1} - a^2 z^{-1}) + (-z + z) + (-z^{-1} - z^{-1}) right]$$ $$N(theta) = frac{1}{2} left[ 2a^2 z - 2z^{-1} right] = a^2 z - z^{-1} = frac{a^2 z^2 - 1}{z}$$ --- Step 3: Simplifying the Denominator Let $D(theta)$ denote the denominator: $$D(theta) = a^4 - 2a^2cos 2theta + 1$$ Substitute $cos 2theta = frac{z^2 + z^{-2}}{2}$: $$D(theta) = a^4 - 2a^2 left( frac{z^2 + z^{-2}}{2} right) + 1 = a^4 - a^2 z^2 - a^2 z^{-2} + 1$$ Rearranging terms: $$D(theta) = (a^4 - a^2 z^2) - (a

Q: Q2. (b) Show that every homomorphic image of an abelian group is abelian, but the converse is not necessarily true.

Abstract Algebraic Setting Let $(G, cdot)$ and $(G', )$ be two groups, and let $phi : G to G'$ be a group homomorphism. Recall the defining property of a group homomorphism: $$phi(a cdot b) = phi(a) phi(b) quad forall a, b in G$$ The homomorphic image of $G$ under $phi$ is the sub-group of $G'$ given by: $$phi(G) = operatorname{Im}(phi) = { phi(g) in G' : g in G }$$ We are required to establish two results: 1. If $G$ is an abelian group, then $phi(G)$ is also abelian. 2. The converse statement is false: the fact that $phi(G)$ is abelian does not imply that $G$ is abelian. --- Part 1: Proving that the Homomorphic Image of an Abelian Group is Abelian Assume that the group $G$ is abelian, meaning: $$a cdot b = b cdot a quad forall a, b in G$$ Let $x, y in phi(G)$ be any two arbitrary elements in the homomorphic image. By definition of the image, there exist elements $a, b in G$ such that: $$x = phi(a) quad text{and} quad y = phi(b)$$ Now, compute their product in $G'$: $$x y = phi(a) phi(b)$$ Since $phi$ is a homomorphism: $$phi(a) phi(b) = phi(a cdot b)$$ Since $G$ is abelian, $a cdot b = b cdot a$: $$phi(a cdot b) = phi(b cdot a)$$ Using the homomorphism property again: $$phi(b cdot a) = phi(b) phi(a) = y x$$ Therefore: $$x y = y x quad forall x, y in phi(G)$$ This proves that the homomorphic image $phi(G)$ is indeed an abelian group. --- Part 2: Proving the Converse is Not Necessarily True (Counterexamples) The converse proposition states: If $phi(G)$ is abelian, then $G$ must be abelian. To disprove this assertion, it suffices to provide a single concrete counterexample. Counterexample 1: The Trivial Homomorphism Let $G$ be any non-abelian group (for instance, the symmetric group $S3$ of degree 3, or the quaternion group $Q8$). Define the trivial homomorphism $phi : G to G'$ by: $$phi(g) = e' quad forall g in G$$ where $e'$ is the identity element of $G'$. The image is the trivial group $phi(G) = {e'}$, which is trivially abelian. However, the domain group $G = S3$

Q: Q3. (a) Locate the poles and their order for the function $$f(z) = frac{1}{z (sin pi z)left(z + frac{1}{2}right)}.$$ Also, find the residue of $f(z)$ at these poles.

Mathematical Formulation of Singularities We consider the complex meromorphic function given by: $$f(z) = frac{1}{z (sin pi z)left(z + frac{1}{2}right)}$$ The isolated singularities of $f(z)$ in the finite complex plane $mathbb{C}$ occur at the points where the denominator $D(z) = z (sin pi z)left(z + frac{1}{2}right)$ vanishes: $$D(z) = 0 iff z = 0, quad sin(pi z) = 0, quad text{or} quad z + frac{1}{2} = 0$$ Solving these individual factors gives: 1. From $sin(pi z) = 0$, we have $pi z = npi implies z = n$ for all integers $n in mathbb{Z}$. 2. From $z = 0$, we have $z = 0$, which coincides with the root $n = 0$ of $sin(pi z)$. 3. From $z + frac{1}{2} = 0$, we have $z = -frac{1}{2}$. --- Step 1: Classification and Order of the Poles Let us determine the multiplicity of the zeros of $D(z)$ at each point: 1. At the Point $z = 0$: The factor $z$ has a simple zero (order 1) at $z = 0$. The factor $sin(pi z)$ expands around $z = 0$ as: $$sin(pi z) = pi z - frac{(pi z)^3}{3!} + frac{(pi z)^5}{5!} - dots = pi z left(1 - frac{pi^2 z^2}{6} + dotsright)$$ which also possesses a zero of order 1 at $z = 0$. The remaining factor evaluated at $z = 0$ is $0 + frac{1}{2} = frac{1}{2} eq 0$. Therefore, the denominator $D(z)$ behaves as: $$D(z) = z cdot (pi z + mathcal{O}(z^3)) cdot left(frac{1}{2} + zright) = frac{pi}{2} z^2 + mathcal{O}(z^3)$$ Since $D(z)$ has a zero of multiplicity $2$ and the numerator is non-zero, $z = 0$ is a pole of order 2 (double pole). 2. At Points $z = n$ where $n in mathbb{Z} setminus {0}$: For any non-zero integer $n$, the factor $z = n eq 0$. The factor $z + frac{1}{2} = n + frac{1}{2} eq 0$ since $n$ is an integer. The factor $sin(pi z)$ has a simple zero at $z = n$ because: $$left[frac{d}{dz} sin(pi z)right]{z=n} = pi cos(npi) = pi (-1)^n eq 0$$ Thus, $D(z)$ has a simple zero at each $z = n$ ($n eq 0$). Therefore, $z = n$ ($n in mathbb{Z}, n eq 0$) are poles of order 1 (simple poles). 3. At the Point $z = -frac{1}{2}$: The factor $z + frac{1}{2}$ has

Q: Q3. (c) Using duality principle, solve the following linear programming problem: Minimize $$z = 4x1 + 3x2 + x3$$ subject to $$begin{aligned} x1 + 2x2 + 4x3 &ge 12 3x1 + 2x2 + x3 &ge 8 x1, x2, x3 &ge 0 end{aligned}$$

Step 1: Formulation of the Dual Problem The given primal linear programming problem is in standard minimization form: $$begin{aligned} text{Minimize } & z = 4x1 + 3x2 + x3 text{subject to } & x1 + 2x2 + 4x3 ge 12 & 3x1 + 2x2 + x3 ge 8 & x1, x2, x3 ge 0 end{aligned}$$ Let $w1$ and $w2$ be the dual variables corresponding to the first and second constraints respectively. According to the principles of duality: 1. The primal is a minimization problem with $ge$ constraints; therefore, the dual is a maximization problem with $le$ constraints and non-negative dual variables ($w1, w2 ge 0$). 2. The primal right-hand side vector $b = begin{pmatrix} 12 8 end{pmatrix}$ becomes the objective coefficient vector of the dual. 3. The primal objective coefficients $c^T = (4, 3, 1)$ become the right-hand sides of the dual constraints. 4. The transpose of the constraint matrix $A = begin{pmatrix} 1 & 2 & 4 3 & 2 & 1 end{pmatrix}$ gives the dual constraint coefficients: $$A^T = begin{pmatrix} 1 & 3 2 & 2 4 & 1 end{pmatrix}$$ Thus, the Dual Problem is formulated as: $$begin{aligned} text{Maximize } & ZD = 12w1 + 8w2 text{subject to } & w1 + 3w2 le 4 quad text{--- (Constraint 1)} & 2w1 + 2w2 le 3 quad text{--- (Constraint 2)} & 4w1 + w2 le 1 quad text{--- (Constraint 3)} & w1, w2 ge 0 end{aligned}$$ --- Step 2: Graphical Solution of the Dual Problem Since the dual involves only two decision variables $w1$ and $w2$, it can be solved accurately using the graphical method in the first quadrant of the $(w1, w2)$ plane: 1. Boundary Line $L1$: $w1 + 3w2 = 4$ - At $w1 = 0 implies w2 = frac{4}{3} approx 1.33$ - At $w2 = 0 implies w1 = 4$ 2. Boundary Line $L2$: $2w1 + 2w2 = 3 iff w1 + w2 = 1.5$ - At $w1 = 0 implies w2 = 1.5$ - At $w2 = 0 implies w1 = 1.5$ 3. Boundary Line $L3$: $4w1 + w2 = 1$ - At $w1 = 0 implies w2 = 1$ - At $w2 = 0 implies w1 = frac{1}{4} = 0.25$ Identifying the Feasible Region: Let us test the boundary points of Constraint 3 against Constraints 1 and 2: - At point $(0, 1)$: -

Question 1

Q2. (a) Using Cauchy's general principle of convergence, examine the convergence of the sequence $langle fn rangle$, where $fn = 1 + frac{1}{1!} + frac{1}{2!} + dots + frac{1}{n!}$.

Accepted Answer

Statement of Cauchy's General Principle of Convergence Cauchy's General Principle of Convergence for Sequences: A sequence of real numbers $langle fn rangle$ converges if and only if it is a Cauchy sequence. That is, for every given $epsilon 0$, there exists a positive integer $N = N(epsilon) in mathbb{N}$ such that for all $n ge N$ and for all positive integers $p ge 1$: $$|f{n+p} - fn| (n + 1)! cdot 2 quad (text{since } n ge 1 implies n + 2 ge 3 2)$$ $$(n + 3)! = (n + 1)!(n + 2)(n + 3) (n + 1)! cdot 2^2$$ In general, for any $k ge 1$: $$(n + k)! = (n + 1)!(n + 2)(n + 3)dots(n + k) ge (n + 1)! cdot 2^{k-1}$$ Taking reciprocals: $$frac{1}{(n + k)!} le frac{1}{(n + 1)!} frac{1}{2^{k-1}}$$ Summing from $k = 1$ to $p$: $$|f{n+p} - fn| = sum{k=1}^p frac{1}{(n+k)!} le frac{1}{(n + 1)!} sum{k=1}^p frac{1}{2^{k-1}}$$ The sum is a finite geometric series with first term $1$ and common ratio $1/2$: $$sum{k=1}^p frac{1}{2^{k-1}} = 1 + frac{1}{2} + frac{1}{2^2} + dots + frac{1}{2^{p-1}} = frac{1 - (1/2)^p}{1 - 1/2} = 2 left(1 - frac{1}{2^p}right) 0$, we can choose a sufficiently large natural number $N in mathbb{N}$ such that: $$frac{2}{(N + 1)!} < epsilon$$ Then, for all $n ge N$ and for every $p ge 1$: $$|f{n+p} - fn| < frac{2}{(n + 1)!} le frac{2}{(N + 1)!} < epsilon$$ This proves that $langle fn rangle$ satisfies Cauchy's convergence criterion on $mathbb{R}$. --- Conclusion Since the real field $mathbb{R}$ is complete (every Cauchy sequence converges), the sequence $langle fn rangle$ is convergent. (Its limit is the mathematical constant $e = sum{k=0}^infty frac{1}{k!} approx 2.71828$).

Question 2

Q1. (e) Use two phase method to solve the following linear programming problem: Maximize $z = x1 + 2x2$ subject to $x1 - x2 ge 3$ $2x1 + x2 le 10$ $x1, x2 ge 0$

Accepted Answer

Theory of the Two-Phase Simplex Method The Two-Phase Simplex Method is used to solve linear programming problems where initial basic feasible solutions cannot be found by slack variables alone (e.g., in the presence of $ge$ or $=$ constraints requiring artificial variables). - Phase I: Formulate an auxiliary objective function $W$ defined as the sum of all artificial variables. Minimize $W$ (or maximize $-W$) subject to the system of constraints. If the minimum value of $W$ is zero, an initial basic feasible solution (BFS) to the original problem has been obtained, and we proceed to Phase II. If $min W 0$, the problem has no feasible solution. - Phase II: Using the basic feasible solution obtained at the end of Phase I as the starting point, drop the artificial variables and optimize the original objective function $z$. --- Step 1: Converting to Standard Form The given problem is: $$text{Maximize } z = x1 + 2x2$$ subject to: $$x1 - x2 ge 3$$ $$2x1 + x2 le 10$$ $$x1, x2 ge 0$$ Convert inequalities into equations by introducing: - A surplus variable $s1 ge 0$ and an artificial variable $A1 ge 0$ for the first constraint: $$x1 - x2 - s1 + A1 = 3$$ - A slack variable $s2 ge 0$ for the second constraint: $$2x1 + x2 + s2 = 10$$ with all variables $x1, x2, s1, s2, A1 ge 0$. --- Step 2: Phase I In Phase I, we minimize the artificial variable $A1$, which is equivalent to maximizing: $$W^ = -A1$$ From the first constraint: $A1 = 3 - x1 + x2 + s1$. Expressing the Phase I objective in terms of non-basic variables: $$W^ + A1 = 0 implies W^ + (3 - x1 + x2 + s1) = 0 implies W^ - x1 + x2 + s1 = -3$$ The Phase I initial simplex tableau with basic variables $A1, s2$: | Basis | $x1$ | $x2$ | $s1$ | $s2$ | $A1$ | RHS | Ratio | |---|---|---|---|---|---|---|---| | $A1$ | 1 | -1 | -1 | 0 | 1 | 3 | $3/1 = 3$ (min) | | $s2$ | 2 | 1 | 0 | 1 | 0 | 10 | $10/2 = 5$ | | $W^$ | 1 | -1 | -1 | 0 | 0 | 3 | | Entering variable: $x1$ (most positive coefficient in indicator row). Leaving variable: $A1$

Question 3

Q1. (d) If $phi$ and $psi$ are functions of $x$ and $y$ satisfying Laplace equation, then show that $f(z) = p + iq, i = sqrt{-1}$ is an analytic function, where $p = frac{partialphi}{partial y} - frac{partialpsi}{partial x}$ and $q = frac{partialphi}{partial x} + frac{partialpsi}{partial y}$.

Accepted Answer

Theory of Complex Analytic Functions and the Cauchy-Riemann Equations Let $f(z) = p(x, y) + i q(x, y)$ be a complex-valued function of $z = x + iy$. By the Cauchy-Riemann Criterion, $f(z)$ is analytic in a domain $D$ if and only if: 1. The real functions $p(x, y)$ and $q(x, y)$ possess continuous first-order partial derivatives throughout $D$. 2. The partial derivatives satisfy the Cauchy-Riemann (C-R) equations: $$frac{partial p}{partial x} = frac{partial q}{partial y} quad text{--- (C-R 1)}$$ $$frac{partial p}{partial y} = -frac{partial q}{partial x} quad text{--- (C-R 2)}$$ --- Step 1: Given Conditions on $phi$ and $psi$ We are given that $phi(x, y)$ and $psi(x, y)$ satisfy Laplace's equation in two dimensions: $$ abla^2 phi = frac{partial^2 phi}{partial x^2} + frac{partial^2 phi}{partial y^2} = 0 implies frac{partial^2 phi}{partial y^2} = -frac{partial^2 phi}{partial x^2} quad text{--- (1)}$$ $$ abla^2 psi = frac{partial^2 psi}{partial x^2} + frac{partial^2 psi}{partial y^2} = 0 implies frac{partial^2 psi}{partial y^2} = -frac{partial^2 psi}{partial x^2} quad text{--- (2)}$$ Furthermore, since harmonic functions are infinitely differentiable, the mixed partial derivatives are equal by Schwarz's (Clairaut's) theorem: $$frac{partial^2 phi}{partial x partial y} = frac{partial^2 phi}{partial y partial x}, quad frac{partial^2 psi}{partial x partial y} = frac{partial^2 psi}{partial y partial x}$$ --- Step 2: Partial Derivatives of $p(x, y)$ We are given: $$p = frac{partialphi}{partial y} - frac{partialpsi}{partial x}$$ 1. Differentiating $p$ with respect to $x$: $$frac{partial p}{partial x} = frac{partial}{partial x}left(frac{partialphi}{partial y} - frac{partialpsi}{partial x}right) = frac{partial^2 phi}{partial x partial y} - frac{partial^2 psi}{partial x^2} quad text{--- (3)}$$ 2. Differentiating $p$ with respect to $y$: $$frac{partial p}{partial y} = frac{partial}{partial y}left(frac{partialphi}{partial y} - frac{partialpsi}{partial x}right) = frac{partial^2 phi}{

Question 4

Q1. (c) Test the convergence of $$int0^2 frac{log x}{sqrt{2 - x}},dx.$$

Accepted Answer

Analysis of the Improper Integral We are required to test the convergence of the definite integral: $$I = int0^2 frac{log x}{sqrt{2 - x}},dx$$ where $log x$ denotes the natural logarithm $ln x$. Let the integrand be: $$f(x) = frac{log x}{sqrt{2 - x}}$$ We inspect the behavior of $f(x)$ on the interval of integration $(0, 2)$: 1. At the lower limit $x to 0^+$: $$lim{x to 0^+} log x = -infty, quad lim{x to 0^+} sqrt{2 - x} = sqrt{2}$$ Hence, $f(x) to -infty$. Thus, $x = 0$ is a point of infinite discontinuity. 2. At the upper limit $x to 2^-$: $$lim{x to 2^-} log x = log 2, quad lim{x to 2^-} sqrt{2 - x} = 0$$ Hence, $f(x) to +infty$. Thus, $x = 2$ is also a point of infinite discontinuity. Because the integrand possesses singularities at both endpoints, we split the integral at an intermediate interior point, conveniently chosen as $c = 1$ (where $log(1) = 0$): $$I = int0^1 frac{log x}{sqrt{2 - x}},dx + int1^2 frac{log x}{sqrt{2 - x}},dx = I1 + I2$$ The integral $I$ converges if and only if both $I1$ and $I2$ converge independently. --- Step 1: Convergence of $I1 = int0^1 frac{log x}{sqrt{2 - x}},dx$ For $x in (0, 1]$, $log x le 0$, so $|f(x)| = frac{-log x}{sqrt{2 - x}}$. The only singularity in $[0, 1]$ is at $x = 0$. Notice that for $0 0$$ Hence, the singularity behaves as: $$f(x) sim frac{log 2}{(2 - x)^{1/2}}$$ We use the Limit Comparison Test with the comparison function: $$g(x) = frac{1}{(2 - x)^{1/2}}$$ Compute the limit: $$L = lim{x to 2^-} frac{f(x)}{g(x)} = lim{x to 2^-} frac{frac{log x}{sqrt{2 - x}}}{frac{1}{sqrt{2 - x}}} = lim{x to 2^-} log x = log 2$$ Since $L = log 2$ is a finite, non-zero positive number ($0 < L < infty$), both integrals $int1^2 f(x),dx$ and $int1^2 g(x),dx$ behave identically with regard to convergence. Now examine the standard $p$-integral at the upper limit: $$int1^2 frac{dx}{(2 - x)^p} quad text{with } p = frac{1}{2}$$ Substitute $t = 2 - x implies dt = -dx$: $$int1^2 frac{dx}{(2 - x)^{1/2}} = int0^1 frac{dt}{t^{1/2}} = [2t^{1/2}]

Question 5

Q1. (b) If $w = f(z)$ is an analytic function of $z$, then show that $$left(frac{partial^2}{partial x^2} + frac{partial^2}{partial y^2}right) log |f'(z)| = 0.$$

Accepted Answer

Theory of Complex Analytic Functions and Harmonicity Let $w = f(z)$ be an analytic function of the complex variable $z = x + iy$ in a domain $D subseteq mathbb{C}$. Since $f(z)$ is analytic, its derivative $F(z) = f'(z)$ is also analytic in $D$. Assume that $f'(z) eq 0$ in the region under consideration (since $log |f'(z)|$ is undefined at zeros of $f'(z)$). --- Method 1: Complex Operator Formalism (Wirtinger Derivatives) The two-dimensional Laplacian operator in Cartesian coordinates is related to the Wirtinger derivatives by: $$ abla^2 = frac{partial^2}{partial x^2} + frac{partial^2}{partial y^2} = 4 frac{partial^2}{partial z , partial bar{z}}$$ where: $$frac{partial}{partial z} = frac{1}{2}left( frac{partial}{partial x} - ifrac{partial}{partial y} right), quad frac{partial}{partial bar{z}} = frac{1}{2}left( frac{partial}{partial x} + ifrac{partial}{partial y} right)$$ Recall that for any non-zero complex quantity $W$: $$|W| = sqrt{W bar{W}} implies log |W| = frac{1}{2} log(W bar{W}) = frac{1}{2} log W + frac{1}{2} log bar{W}$$ Applying this to $W = f'(z)$: $$log |f'(z)| = frac{1}{2} log f'(z) + frac{1}{2} log overline{f'(z)}$$ Since $f'(z)$ is an analytic function of $z$, it depends only on $z$ and not on $bar{z}$: $$frac{partial}{partial bar{z}} [f'(z)] = 0$$ Similarly, its complex conjugate $overline{f'(z)}$ depends only on $bar{z}$ and not on $z$: $$frac{partial}{partial z} [overline{f'(z)}] = 0$$ Now, compute the Wirtinger derivative: $$frac{partial}{partial bar{z}} left( log |f'(z)| right) = frac{partial}{partial bar{z}} left( frac{1}{2}log f'(z) + frac{1}{2}log overline{f'(z)} right) = 0 + frac{1}{2} frac{overline{f''(z)}}{overline{f'(z)}}$$ Next, differentiate with respect to $z$: $$frac{partial}{partial z} left[ frac{partial}{partial bar{z}} log |f'(z)| right] = frac{partial}{partial z} left( frac{1}{2} frac{overline{f''(z)}}{overline{f'(z)}} right) = 0$$ because $frac{overline{f''(z)}}{overline{f'(z)}}$ is a function purely of $bar{z}$. Therefore: $$ abl

Question 6

Q1. (a) Let $G$ be a finite group of order $mn$, where $m$ and $n$ are prime numbers with $m n$. Show that $G$ has at most one subgroup of order $m$.

Accepted Answer

Algebraic Context and Sylow's Theorems Let $G$ be a finite group of order: $$|G| = mn$$ where $m$ and $n$ are distinct prime numbers such that $m n$. We wish to establish that $G$ contains at most one subgroup of order $m$. We rely on the foundational results of Sylow's Theorems: 1. Sylow's First Theorem: If $p$ is a prime dividing $|G|$ such that $p^k mid |G|$ and $p^{k+1} mid |G|$, then $G$ contains a subgroup of order $p^k$, called a Sylow $p$-subgroup. 2. Sylow's Second Theorem: Any two Sylow $p$-subgroups of $G$ are conjugate in $G$. 3. Sylow's Third Theorem: The number $np$ of distinct Sylow $p$-subgroups of $G$ satisfies two simultaneous conditions: - $np equiv 1 pmod p$, and - $np$ divides the index $frac{|G|}{p^k}$. --- Step 1: Application of Sylow's Third Theorem Since $m$ is a prime number and $m$ divides $|G| = mn$ with $m^2 mid |G|$, every subgroup of $G$ of order $m$ is a Sylow $m$-subgroup of $G$. Let $nm$ denote the number of Sylow $m$-subgroups in $G$. According to Sylow's Third Theorem: 1. $nm$ must divide the index of a Sylow $m$-subgroup in $G$: $$frac{|G|}{m} = frac{mn}{m} = n$$ Therefore, $nm mid n$. 2. $nm$ must be congruent to $1$ modulo $m$: $$nm equiv 1 pmod m$$ --- Step 2: Analyzing the Divisors of $n$ Since $n$ is a prime number, its only positive divisors in $mathbb{Z}^+$ are: $$nm in {1, n}$$ We now examine the two possibilities: - Case 1: $nm = 1$ If $nm = 1$, then $1 equiv 1 pmod m$, which is universally true for any prime $m$. Thus $nm = 1$ is always a mathematically valid solution. - Case 2: $nm = n$ If $nm = n$, then the second condition requires: $$n equiv 1 pmod m$$ This implies that $(n - 1)$ is a positive multiple of $m$: $$n - 1 = k m quad text{for some integer } k ge 1$$ $$implies n = km + 1 ge m + 1 m$$ Therefore, this case strictly necessitates that: $$n m$$ However, the hypothesis of the theorem explicitly states that: $$m n$$ This is a direct contradiction because $n$ cannot be both strictly less than $m$ and strictly gre

Question 7

Q2. (c) Find the function which is analytic inside and on the circle $C : z = e^{itheta}, 0 le theta le 2pi$ and has the value $$frac{(a^2 - 1)costheta + i(a^2 + 1)sintheta}{a^4 - 2a^2cos 2theta + 1}$$ on the circumference of $C$, where $a^2 1$.

Accepted Answer

Problem Statement and Formulation We seek a complex function $f(z)$ which is analytic inside and on the unit circle: $$C = { z in mathbb{C} : |z| le 1 }$$ given its boundary values on the circumference $|z| = 1$, where $z = e^{itheta}$ for $0 le theta le 2pi$: $$f(e^{itheta}) = frac{(a^2 - 1)costheta + i(a^2 + 1)sintheta}{a^4 - 2a^2cos 2theta + 1} quad (text{where } a^2 1)$$ --- Step 1: Converting Boundary Values in Terms of $z = e^{itheta}$ On the boundary circle $C$, we have $z = e^{itheta}$, which gives: $$bar{z} = z^{-1} = frac{1}{z}$$ Using Euler's formulas for trigonometric functions: $$costheta = frac{e^{itheta} + e^{-itheta}}{2} = frac{z + z^{-1}}{2} = frac{z^2 + 1}{2z}$$ $$sintheta = frac{e^{itheta} - e^{-itheta}}{2i} = frac{z - z^{-1}}{2i} = frac{z^2 - 1}{2iz}$$ $$cos 2theta = frac{e^{i2theta} + e^{-i2theta}}{2} = frac{z^2 + z^{-2}}{2} = frac{z^4 + 1}{2z^2}$$ --- Step 2: Simplifying the Numerator Let $N(theta)$ denote the numerator: $$N(theta) = (a^2 - 1)costheta + i(a^2 + 1)sintheta$$ Substitute the expressions for $costheta$ and $sintheta$: $$N(theta) = (a^2 - 1)left( frac{z + z^{-1}}{2} right) + i(a^2 + 1)left( frac{z - z^{-1}}{2i} right)$$ Notice that $frac{i}{2i} = frac{1}{2}$: $$N(theta) = frac{1}{2} left[ (a^2 - 1)(z + z^{-1}) + (a^2 + 1)(z - z^{-1}) right]$$ Expanding the brackets: $$(a^2 - 1)(z + z^{-1}) = a^2 z + a^2 z^{-1} - z - z^{-1}$$ $$(a^2 + 1)(z - z^{-1}) = a^2 z - a^2 z^{-1} + z - z^{-1}$$ Adding these two expressions: $$N(theta) = frac{1}{2} left[ (a^2 z + a^2 z) + (a^2 z^{-1} - a^2 z^{-1}) + (-z + z) + (-z^{-1} - z^{-1}) right]$$ $$N(theta) = frac{1}{2} left[ 2a^2 z - 2z^{-1} right] = a^2 z - z^{-1} = frac{a^2 z^2 - 1}{z}$$ --- Step 3: Simplifying the Denominator Let $D(theta)$ denote the denominator: $$D(theta) = a^4 - 2a^2cos 2theta + 1$$ Substitute $cos 2theta = frac{z^2 + z^{-2}}{2}$: $$D(theta) = a^4 - 2a^2 left( frac{z^2 + z^{-2}}{2} right) + 1 = a^4 - a^2 z^2 - a^2 z^{-2} + 1$$ Rearranging terms: $$D(theta) = (a^4 - a^2 z^2) - (a

Question 8

Q2. (b) Show that every homomorphic image of an abelian group is abelian, but the converse is not necessarily true.

Accepted Answer

Abstract Algebraic Setting Let $(G, cdot)$ and $(G', )$ be two groups, and let $phi : G to G'$ be a group homomorphism. Recall the defining property of a group homomorphism: $$phi(a cdot b) = phi(a) phi(b) quad forall a, b in G$$ The homomorphic image of $G$ under $phi$ is the sub-group of $G'$ given by: $$phi(G) = operatorname{Im}(phi) = { phi(g) in G' : g in G }$$ We are required to establish two results: 1. If $G$ is an abelian group, then $phi(G)$ is also abelian. 2. The converse statement is false: the fact that $phi(G)$ is abelian does not imply that $G$ is abelian. --- Part 1: Proving that the Homomorphic Image of an Abelian Group is Abelian Assume that the group $G$ is abelian, meaning: $$a cdot b = b cdot a quad forall a, b in G$$ Let $x, y in phi(G)$ be any two arbitrary elements in the homomorphic image. By definition of the image, there exist elements $a, b in G$ such that: $$x = phi(a) quad text{and} quad y = phi(b)$$ Now, compute their product in $G'$: $$x y = phi(a) phi(b)$$ Since $phi$ is a homomorphism: $$phi(a) phi(b) = phi(a cdot b)$$ Since $G$ is abelian, $a cdot b = b cdot a$: $$phi(a cdot b) = phi(b cdot a)$$ Using the homomorphism property again: $$phi(b cdot a) = phi(b) phi(a) = y x$$ Therefore: $$x y = y x quad forall x, y in phi(G)$$ This proves that the homomorphic image $phi(G)$ is indeed an abelian group. --- Part 2: Proving the Converse is Not Necessarily True (Counterexamples) The converse proposition states: If $phi(G)$ is abelian, then $G$ must be abelian. To disprove this assertion, it suffices to provide a single concrete counterexample. Counterexample 1: The Trivial Homomorphism Let $G$ be any non-abelian group (for instance, the symmetric group $S3$ of degree 3, or the quaternion group $Q8$). Define the trivial homomorphism $phi : G to G'$ by: $$phi(g) = e' quad forall g in G$$ where $e'$ is the identity element of $G'$. The image is the trivial group $phi(G) = {e'}$, which is trivially abelian. However, the domain group $G = S3$

Question 9

Q3. (a) Locate the poles and their order for the function $$f(z) = frac{1}{z (sin pi z)left(z + frac{1}{2}right)}.$$ Also, find the residue of $f(z)$ at these poles.

Accepted Answer

Mathematical Formulation of Singularities We consider the complex meromorphic function given by: $$f(z) = frac{1}{z (sin pi z)left(z + frac{1}{2}right)}$$ The isolated singularities of $f(z)$ in the finite complex plane $mathbb{C}$ occur at the points where the denominator $D(z) = z (sin pi z)left(z + frac{1}{2}right)$ vanishes: $$D(z) = 0 iff z = 0, quad sin(pi z) = 0, quad text{or} quad z + frac{1}{2} = 0$$ Solving these individual factors gives: 1. From $sin(pi z) = 0$, we have $pi z = npi implies z = n$ for all integers $n in mathbb{Z}$. 2. From $z = 0$, we have $z = 0$, which coincides with the root $n = 0$ of $sin(pi z)$. 3. From $z + frac{1}{2} = 0$, we have $z = -frac{1}{2}$. --- Step 1: Classification and Order of the Poles Let us determine the multiplicity of the zeros of $D(z)$ at each point: 1. At the Point $z = 0$: The factor $z$ has a simple zero (order 1) at $z = 0$. The factor $sin(pi z)$ expands around $z = 0$ as: $$sin(pi z) = pi z - frac{(pi z)^3}{3!} + frac{(pi z)^5}{5!} - dots = pi z left(1 - frac{pi^2 z^2}{6} + dotsright)$$ which also possesses a zero of order 1 at $z = 0$. The remaining factor evaluated at $z = 0$ is $0 + frac{1}{2} = frac{1}{2} eq 0$. Therefore, the denominator $D(z)$ behaves as: $$D(z) = z cdot (pi z + mathcal{O}(z^3)) cdot left(frac{1}{2} + zright) = frac{pi}{2} z^2 + mathcal{O}(z^3)$$ Since $D(z)$ has a zero of multiplicity $2$ and the numerator is non-zero, $z = 0$ is a pole of order 2 (double pole). 2. At Points $z = n$ where $n in mathbb{Z} setminus {0}$: For any non-zero integer $n$, the factor $z = n eq 0$. The factor $z + frac{1}{2} = n + frac{1}{2} eq 0$ since $n$ is an integer. The factor $sin(pi z)$ has a simple zero at $z = n$ because: $$left[frac{d}{dz} sin(pi z)right]{z=n} = pi cos(npi) = pi (-1)^n eq 0$$ Thus, $D(z)$ has a simple zero at each $z = n$ ($n eq 0$). Therefore, $z = n$ ($n in mathbb{Z}, n eq 0$) are poles of order 1 (simple poles). 3. At the Point $z = -frac{1}{2}$: The factor $z + frac{1}{2}$ has

Question 10

Q3. (b) Consider the series $sum{n=1}^infty Un(x)$, $0 le x le 1$, the sum of whose first $n$ terms is given by $Sn(x) = frac{1}{2n^2}log(1+n^4 x^2)$, $x in [0, 1]$. Show that the given series can be differentiated term-by-term, though $sum{n=1}^infty Un'(x)$ does not converge uniformly on $[0, 1]$.

Accepted Answer

Introduction and Definitions We are given a sequence of partial sums: $$Sn(x) = sum{k=1}^n Uk(x) = frac{1}{2n^2} log(1 + n^4 x^2), quad x in [0, 1]$$ We need to demonstrate two distinct mathematical assertions: 1. The sequence of derivative partial sums $Sn'(x) = sum{k=1}^n Uk'(x)$ does not converge uniformly on $[0, 1]$. 2. In spite of this failure of uniform convergence, term-by-term differentiation is valid, i.e., $$frac{d}{dx}left[ lim{n to infty} Sn(x) right] = lim{n to infty} Sn'(x), quad forall x in [0, 1].$$ --- Step 1: Pointwise Limit of the Sequence $Sn(x)$ Let us evaluate $S(x) = lim{n to infty} Sn(x)$ for all $x in [0, 1]$: - For $x = 0$: $$Sn(0) = frac{1}{2n^2} log(1 + 0) = 0 implies S(0) = 0$$ - For $x in (0, 1]$: As $n to infty$, $n^4 x^2 to infty$. We rewrite: $$log(1 + n^4 x^2) = logleft(n^4 x^2 left(1 + frac{1}{n^4 x^2}right)right) = 4 log n + 2 log x + logleft(1 + frac{1}{n^4 x^2}right)$$ Substituting this into $Sn(x)$: $$Sn(x) = frac{4 log n}{2n^2} + frac{2 log x}{2n^2} + frac{1}{2n^2} logleft(1 + frac{1}{n^4 x^2}right) = frac{2 log n}{n^2} + frac{log x}{n^2} + mathcal{O}left(frac{1}{n^6}right)$$ Since $lim{n to infty} frac{log n}{n^2} = 0$ by L'Hôpital's Rule, and for fixed $x 0$, $frac{log x}{n^2} to 0$, we have: $$lim{n to infty} Sn(x) = 0$$ Thus, the pointwise sum function of the series is: $$S(x) = 0, quad forall x in [0, 1]$$ Its derivative is identically zero: $$S'(x) = frac{d}{dx}(0) = 0, quad forall x in [0, 1]$$ --- Step 2: Derivative of the Partial Sums $Sn'(x)$ Differentiating $Sn(x)$ with respect to $x$: $$Sn'(x) = frac{d}{dx}left[ frac{1}{2n^2} log(1 + n^4 x^2) right] = frac{1}{2n^2} cdot frac{2n^4 x}{1 + n^4 x^2} = frac{n^2 x}{1 + n^4 x^2}$$ Now examine the pointwise limit of $Sn'(x)$: - For $x = 0$: $$Sn'(0) = frac{0}{1 + 0} = 0 implies lim{n to infty} Sn'(0) = 0$$ - For $x in (0, 1]$: For any fixed $x 0$: $$Sn'(x) = frac{n^2 x}{1 + n^4 x^2} = frac{frac{x}{n^2}}{frac{1}{n^4} + x^2}$$ As $n to infty$, the numerator $frac{x}{n^2} to

Question 11

Q3. (c) Using duality principle, solve the following linear programming problem: Minimize $$z = 4x1 + 3x2 + x3$$ subject to $$begin{aligned} x1 + 2x2 + 4x3 &ge 12 3x1 + 2x2 + x3 &ge 8 x1, x2, x3 &ge 0 end{aligned}$$

Accepted Answer

Step 1: Formulation of the Dual Problem The given primal linear programming problem is in standard minimization form: $$begin{aligned} text{Minimize } & z = 4x1 + 3x2 + x3 text{subject to } & x1 + 2x2 + 4x3 ge 12 & 3x1 + 2x2 + x3 ge 8 & x1, x2, x3 ge 0 end{aligned}$$ Let $w1$ and $w2$ be the dual variables corresponding to the first and second constraints respectively. According to the principles of duality: 1. The primal is a minimization problem with $ge$ constraints; therefore, the dual is a maximization problem with $le$ constraints and non-negative dual variables ($w1, w2 ge 0$). 2. The primal right-hand side vector $b = begin{pmatrix} 12 8 end{pmatrix}$ becomes the objective coefficient vector of the dual. 3. The primal objective coefficients $c^T = (4, 3, 1)$ become the right-hand sides of the dual constraints. 4. The transpose of the constraint matrix $A = begin{pmatrix} 1 & 2 & 4 3 & 2 & 1 end{pmatrix}$ gives the dual constraint coefficients: $$A^T = begin{pmatrix} 1 & 3 2 & 2 4 & 1 end{pmatrix}$$ Thus, the Dual Problem is formulated as: $$begin{aligned} text{Maximize } & ZD = 12w1 + 8w2 text{subject to } & w1 + 3w2 le 4 quad text{--- (Constraint 1)} & 2w1 + 2w2 le 3 quad text{--- (Constraint 2)} & 4w1 + w2 le 1 quad text{--- (Constraint 3)} & w1, w2 ge 0 end{aligned}$$ --- Step 2: Graphical Solution of the Dual Problem Since the dual involves only two decision variables $w1$ and $w2$, it can be solved accurately using the graphical method in the first quadrant of the $(w1, w2)$ plane: 1. Boundary Line $L1$: $w1 + 3w2 = 4$ - At $w1 = 0 implies w2 = frac{4}{3} approx 1.33$ - At $w2 = 0 implies w1 = 4$ 2. Boundary Line $L2$: $2w1 + 2w2 = 3 iff w1 + w2 = 1.5$ - At $w1 = 0 implies w2 = 1.5$ - At $w2 = 0 implies w1 = 1.5$ 3. Boundary Line $L3$: $4w1 + w2 = 1$ - At $w1 = 0 implies w2 = 1$ - At $w2 = 0 implies w1 = frac{1}{4} = 0.25$ Identifying the Feasible Region: Let us test the boundary points of Constraint 3 against Constraints 1 and 2: - At point $(0, 1)$: -

Question 12

Q4. (a) Consider the polynomial ring $mathbb{Z}[x]$ over the ring $mathbb{Z}$ of integers. Let $S$ be an ideal of $mathbb{Z}[x]$ generated by $x$. Show that $S$ is prime but not a maximal ideal of $mathbb{Z}[x]$.

Accepted Answer

Algebraic Context and Definitions Let $mathbb{Z}$ denote the ring of integers and $mathbb{Z}[x]$ be the polynomial ring in one indeterminate $x$ with integer coefficients. The ideal $S$ generated by the single polynomial $x$ is the principal ideal: $$S = langle x rangle = x mathbb{Z}[x] = { x cdot f(x) : f(x) in mathbb{Z}[x] }$$ Equivalently, $S$ consists of all polynomials in $mathbb{Z}[x]$ having zero constant term: $$S = left{ a0 + a1 x + dots + an x^n in mathbb{Z}[x] : a0 = 0 right}$$ We recall two fundamental theorems characterizing prime and maximal ideals in commutative rings with unity: 1. An ideal $I$ of a commutative ring with unity $R$ is a prime ideal if and only if the quotient ring $R/I$ is an integral domain. 2. An ideal $I$ of a commutative ring with unity $R$ is a maximal ideal if and only if the quotient ring $R/I$ is a field. --- Step 1: Constructing the Quotient Ring $mathbb{Z}[x] / S$ To examine the algebraic structure of the quotient ring $mathbb{Z}[x] / S$, we apply the First Ring Isomorphism Theorem. Define the evaluation mapping: $$phi: mathbb{Z}[x] to mathbb{Z}$$ defined by evaluating each polynomial at $x = 0$: $$phi(f(x)) = f(0), quad forall f(x) in mathbb{Z}[x]$$ 1. $phi$ is a Ring Homomorphism: Let $f(x), g(x) in mathbb{Z}[x]$. Then: - $phi(f(x) + g(x)) = (f+g)(0) = f(0) + g(0) = phi(f(x)) + phi(g(x))$ - $phi(f(x) cdot g(x)) = (fg)(0) = f(0) cdot g(0) = phi(f(x)) cdot phi(g(x))$ - $phi(1) = 1$ Thus, $phi$ preserves addition, multiplication, and identity. 2. $phi$ is Surjective: For any arbitrary integer $c in mathbb{Z}$, the constant polynomial $h(x) = c in mathbb{Z}[x]$ satisfies: $$phi(h(x)) = h(0) = c$$ Hence, $phi$ is onto $mathbb{Z}$, meaning $text{Im}(phi) = mathbb{Z}$. 3. Kernel of $phi$: $$ker(phi) = { f(x) in mathbb{Z}[x] : phi(f(x)) = 0 } = { f(x) in mathbb{Z}[x] : f(0) = 0 }$$ Writing $f(x) = a0 + a1 x + dots + an x^n$, $f(0) = a0 = 0$ means that every term in $f(x)$ contains at least one factor of $x$: $$f(x) = x(a1 + a2 x +

Question 13

Q4. (b) Find the upper and lower Riemann integrals for the function $f$ defined on $[0, 1]$ as follows: $$f(x) = begin{cases} (1 - x^2)^{1/2}, & text{if } x text{ is rational} (1 - x), & text{if } x text{ is irrational} end{cases}$$ Hence, show that $f$ is not Riemann integrable on $[0, 1]$.

Accepted Answer

Mathematical Analysis of the Given Function We consider the Dirichlet-type bounded real function $f: [0, 1] to mathbb{R}$ defined by: $$f(x) = begin{cases} sqrt{1 - x^2}, & text{if } x in mathbb{Q} cap [0, 1] 1 - x, & text{if } x in ([0, 1] setminus mathbb{Q}) end{cases}$$ Let us first compare the two branch functions $g(x) = sqrt{1 - x^2}$ and $h(x) = 1 - x$ for $x in [0, 1]$: 1. For $x = 0$: $g(0) = sqrt{1 - 0} = 1$ and $h(0) = 1 - 0 = 1$. 2. For $x = 1$: $g(1) = sqrt{1 - 1} = 0$ and $h(1) = 1 - 1 = 0$. 3. For $x in (0, 1)$: Consider the difference between their squares: $$g(x)^2 - h(x)^2 = (1 - x^2) - (1 - x)^2 = (1 - x^2) - (1 - 2x + x^2) = 2x - 2x^2 = 2x(1 - x)$$ Since $x in (0, 1)$, we have $x 0$ and $1 - x 0$; thus: $$2x(1 - x) 0 implies g(x)^2 h(x)^2$$ Since both $g(x) ge 0$ and $h(x) ge 0$, taking positive square roots yields: $$sqrt{1 - x^2} 1 - x, quad forall x in (0, 1)$$ Both constituent functions $g(x) = sqrt{1 - x^2}$ and $h(x) = 1 - x$ are continuous and strictly monotonically decreasing on $[0, 1]$. --- Step 1: Upper and Lower Darboux Sums Let $P = {0 = x0 < x1 < x2 < dots < xn = 1}$ be an arbitrary partition of the interval $[0, 1]$, and let $Delta xi = xi - x{i-1}$ be the length of the $i$-th subinterval $Ii = [x{i-1}, xi]$. By the Density Theorem of Real Numbers, every non-empty open subinterval $(x{i-1}, xi)$ contains infinitely many rational numbers and infinitely many irrational numbers. 1. Supremum $Mi$ on $Ii$: On $Ii = [x{i-1}, xi]$, because $sqrt{1 - x^2} ge 1 - x$ everywhere and $sqrt{1 - x^2}$ is continuous and strictly decreasing: $$Mi = sup{x in Ii} f(x) = sup{x in Ii cap mathbb{Q}} sqrt{1 - x^2} = sqrt{1 - x{i-1}^2}$$ Consequently, the Upper Darboux Sum for the partition $P$ is: $$U(P, f) = sum{i=1}^n Mi Delta xi = sum{i=1}^n sqrt{1 - x{i-1}^2} Delta xi$$ This is precisely the upper Riemann sum for the continuous function $g(x) = sqrt{1 - x^2}$ on $[0, 1]$. 2. Infimum $mi$ on $Ii$: Similarly, because $1 - x le sqrt{1 - x^2}$ and $1 -

Question 14

Q4. (c) The personnel manager of a company wants to assign three officers $A$, $B$ and $C$ to regional offices at Delhi, Mumbai, Kolkata and Chennai. The relocation costs (in thousand rupees) of these three officers to the four regional offices are given below: | Officer | Delhi | Mumbai | Kolkata |

Accepted Answer

Problem Formulation We are given an unbalanced assignment problem involving 3 officers ($A, B, C$) and 4 regional offices (Delhi, Mumbai, Kolkata, Chennai). Let: - $c{ij}$ be the cost of assigning officer $i$ to regional office $j$ (in thousands of rupees). Optimal Assignment Graph (Minimum Cost: ₹48,000) A B C Delhi (D) Mumbai (M) Kolkata (K) Chennai (C) Mumbai (Cost 22) -- Cost 22 Chennai (Cost 16) -- Cost 16 Delhi (Cost 10) -- Cost 10 (Unassigned) Figure: Bipartite matching representation of the optimal assignment yielding minimum total cost ₹48,000. Since the number of persons ($m = 3$) is less than the number of jobs/offices ($n = 4$), we introduce a dummy officer (designated as $D{um}$) having zero relocation cost to each regional office. This yields the balanced $4 times 4$ cost matrix: | Officer | Delhi ($j=1$) | Mumbai ($j=2$) | Kolkata ($j=3$) | Chennai ($j=4$) | | :---: | :---: | :---: | :---: | :---: | | A | 16 | 22 | 24 | 20 | | B | 10 | 32 | 26 | 16 | | C | 10 | 20 | 46 | 30 | | Dummy | 0 | 0 | 0 | 0 | --- Step 1: Row Reduction (Subtract Row Minimums) - Row $A$: Minimum is $16$. Subtract $16$: $[0, 6, 8, 4]$ - Row $B$: Minimum is $10$. Subtract $10$: $[0, 22, 16, 6]$ - Row $C$: Minimum is $10$. Subtract $10$: $[0, 10, 36, 20]$ - Row Dummy: Minimum is $0$. Subtract $0$: $[0, 0, 0, 0]$ The row-reduced matrix is: $$begin{pmatrix} 0 & 6 & 8 & 4 0 & 22 & 16 & 6 0 & 10 & 36 & 20 0 & 0 & 0 & 0 end{pmatrix}$$ --- Step 2: Column Reduction (Subtract Column Minimums) Because the Dummy row contains zeros across all four columns, each column already contains at least one zero (minimum element in every column is $0$). Thus, subtracting column minimums leaves the matrix unchanged: $$C^{(1)} = begin{pmatrix} 0 & 6 & 8 & 4 0 & 22 & 16 & 6 0 & 10 & 36 & 20 0 & 0 & 0 & 0 end{pmatrix}$$ --- Step 3: Application of the Hungarian Algorithm Line Covering of Zeros: We cover all zeros using the minimum number of horizontal and vertical lines: - Line 1: Horizontal line covering

Question 15

Q5. (a) Show that if $f$ and $g$ are arbitrary functions of their respective arguments, then $u = f(x - kt + ialpha y) + g(x - kt - ialpha y)$, is a solution of $$frac{partial^2 u}{partial x^2} + frac{partial^2 u}{partial y^2} = frac{1}{C^2}frac{partial^2 u}{partial t^2}, quad text{where } alpha^2 =

Accepted Answer

Mathematical Formulation and Chain Rule Setup We are given the two-dimensional wave equation: $$frac{partial^2 u}{partial x^2} + frac{partial^2 u}{partial y^2} = frac{1}{C^2} frac{partial^2 u}{partial t^2}$$ where $C$ is the characteristic propagation speed of the wave. We are required to prove that the general function: $$u(x, y, t) = f(xi1) + g(xi2)$$ satisfies this partial differential equation identically for arbitrary twice-differentiable functions $f$ and $g$, where the complex characteristic arguments are defined by: $$xi1 = x - kt + ialpha y$$ $$xi2 = x - kt - ialpha y$$ under the specific algebraic constraint: $$alpha^2 = 1 - frac{k^2}{C^2}$$ By the principle of linear superposition for homogeneous partial differential equations, if $u1 = f(xi1)$ and $u2 = g(xi2)$ each individually satisfy the equation, then their sum $u = u1 + u2$ is also a solution. --- Step 1: Partial Derivatives of the Component $u1 = f(xi1)$ Let $xi1(x, y, t) = x - kt + ialpha y$. Computing the first-order partial derivatives of $xi1$: $$frac{partial xi1}{partial x} = 1, quad frac{partial xi1}{partial y} = ialpha, quad frac{partial xi1}{partial t} = -k$$ Applying the chain rule for $u1 = f(xi1)$, where primes denote differentiation with respect to the argument $xi1$: 1. Derivatives with respect to $x$: $$frac{partial u1}{partial x} = f'(xi1) frac{partial xi1}{partial x} = f'(xi1)$$ $$frac{partial^2 u1}{partial x^2} = frac{partial}{partial x}left[ f'(xi1) right] = f''(xi1) frac{partial xi1}{partial x} = f''(xi1)$$ 2. Derivatives with respect to $y$: $$frac{partial u1}{partial y} = f'(xi1) frac{partial xi1}{partial y} = ialpha f'(xi1)$$ $$frac{partial^2 u1}{partial y^2} = frac{partial}{partial y}left[ ialpha f'(xi1) right] = ialpha cdot f''(xi1) frac{partial xi1}{partial y} = (ialpha)^2 f''(xi1) = -alpha^2 f''(xi1)$$ (since $i^2 = -1$) 3. Derivatives with respect to $t$: $$frac{partial u1}{partial t} = f'(xi1) frac{partial xi1}{partial t} = -k f'(xi1)$$ $$frac{partial^2 u1}{partial t^2}

Question 16

Q6. (b) Draw the logical circuit for the Boolean expression $$Y = ABoverline{C} + Boverline{C} + overline{A}B.$$ Also, obtain the output $Y$ (truth table) for the three input bit sequences: $$A = 10001111, quad B = 00111100, quad C = 11000100.$$

Accepted Answer

Part 1: Boolean Algebra Simplification We are given the Boolean switching function: $$Y = ABoverline{C} + Boverline{C} + overline{A}B$$ Let us simplify $Y$ systematically using Boolean algebra identities: 1. Absorption Law: Notice the first two terms: $ABoverline{C} + Boverline{C}$. Factoring out the common product term $Boverline{C}$: $$ABoverline{C} + Boverline{C} = Boverline{C}(A + 1)$$ Since $A + 1 = 1$ in Boolean algebra: $$Boverline{C}(1) = Boverline{C}$$ Therefore, the term $ABoverline{C}$ is completely redundant (absorbed by $Boverline{C}$). The expression reduces to: $$Y = Boverline{C} + overline{A}B$$ 2. Factoring $B$: $$Y = B(overline{A} + overline{C})$$ 3. Applying De Morgan's Law: By De Morgan's Law, $overline{A} + overline{C} = overline{AC}$. Therefore, $Y$ can be expressed in its most economical two-gate form: $$Y = B cdot overline{AC}$$ This requires only one 2-input NAND gate (to compute $overline{AC}$) and one 2-input AND gate (to compute $B cdot overline{AC}$). --- Part 2: Logical Circuit Diagram A B C AND (AC)' AND Y = B(AC)' Figure: Minimized logic circuit realizing $Y = Boverline{AC} = overline{A}B + Boverline{C}$ using one NAND gate and one AND gate. The circuit diagram above demonstrates the minimal implementation: 1. Inputs $A$ and $C$ feed into a 2-input NAND gate to generate $overline{AC} = overline{A} + overline{C}$. 2. The output of the NAND gate along with input $B$ feeds into a 2-input AND gate, generating the final output: $$Y = B cdot overline{AC} = overline{A}B + Boverline{C}$$ --- Part 3: Bitwise Evaluation on Given Bit Sequences We are given 8-bit input streams (indexed from MSB bit 7 down to LSB bit 0): - $A = 1000,11112$ - $B = 0011,11002$ - $C = 1100,01002$ Let us compute the intermediate bit sequences step by step: Step-by-Step Bit Operations: 1. Compute $overline{A}$ (NOT $A$): $$A = 1000,1111 implies overline{A} = 0111,0000$$ 2. Compute $overline{C}$ (NOT $C$): $$C = 1100,0100 implies overline{C} = 0011,1011$$ 3. Compute $ov

Question 17

Q6. (a) Show that the solution of the two-dimensional Laplace's equation $$frac{partial^2 phi(x, y)}{partial x^2} + frac{partial^2 phi(x, y)}{partial y^2} = 0, quad x in (-infty, infty), , y ge 0$$ subject to the boundary condition $$phi(x, 0) = f(x), , x in (-infty, infty),$$ along with $phi(x, y)

Accepted Answer

Mathematical Formulation of the Dirichlet Problem in the Upper Half-Plane We consider the classical Dirichlet boundary value problem for Laplace's equation in the upper half-plane $mathbb{R} times [0, infty)$: $$frac{partial^2 phi}{partial x^2} + frac{partial^2 phi}{partial y^2} = 0, quad -infty 0$$ subject to: 1. Boundary condition: $phi(x, 0) = f(x)$ for all $x in (-infty, infty)$. 2. Regularity at infinity: $phi(x, y) to 0$ as $sqrt{x^2 + y^2} to infty$. Because the spatial domain in $x$ is the entire real line $(-infty, infty)$, we employ the Fourier Transform with respect to the variable $x$. --- Step 1: Applying the Fourier Transform Define the Fourier transform of $phi(x, y)$ with respect to $x$ as: $$Phi(omega, y) = mathcal{F}{phi(x, y)} = frac{1}{sqrt{2pi}} int{-infty}^infty phi(x, y) e^{-iomega x}, dx$$ where the inverse Fourier transform is given by: $$phi(x, y) = frac{1}{sqrt{2pi}} int{-infty}^infty Phi(omega, y) e^{iomega x}, domega$$ Using standard properties of Fourier transforms of derivatives: $$mathcal{F}left{ frac{partial^2 phi}{partial x^2} right} = (iomega)^2 Phi(omega, y) = -omega^2 Phi(omega, y)$$ Taking the Fourier transform of Laplace's equation with respect to $x$: $$-omega^2 Phi(omega, y) + frac{partial^2 Phi}{partial y^2}(omega, y) = 0 implies frac{d^2 Phi}{dy^2} - omega^2 Phi = 0$$ --- Step 2: Solving the Ordinary Differential Equation For each fixed $omega in mathbb{R}$, this is a second-order linear ordinary differential equation in $y$ with constant coefficients. Its characteristic equation is: $$r^2 - omega^2 = 0 implies r = pm |omega|$$ The general solution is: $$Phi(omega, y) = A(omega) e^{-|omega| y} + B(omega) e^{|omega| y}$$ Applying Boundary Conditions in the Transform Domain: 1. Behavior as $y to infty$: Since $phi(x, y) to 0$ as $y to infty$, the transform $Phi(omega, y)$ must remain bounded as $y to infty$. Since $|omega| 0$, the term $e^{|omega| y}$ diverges exponentially as $y to infty$. Therefore, we must have: $$B(omega)

Question 18

Q5. (e) The velocity potential $phi$ of a flow is given by $$phi = frac{1}{2}(x^2 + y^2 - 2z^2).$$ Determine the streamlines.

Accepted Answer

Fluid Dynamics Principles and Governing Equations In classical potential flow theory, an irrotational flow of an inviscid, incompressible fluid possesses a velocity potential $phi(x, y, z)$. The velocity vector $vec{v} = (u, v, w)$ is defined by the gradient of the velocity potential: $$vec{v} = ablaphi = left( frac{partialphi}{partial x}, frac{partialphi}{partial y}, frac{partialphi}{partial z} right)$$ (Note: Using the alternate convention $vec{v} = - ablaphi$ yields identical geometric curves for the streamlines, reversing only the direction of the parameter). We are given the velocity potential: $$phi(x, y, z) = frac{1}{2}(x^2 + y^2 - 2z^2)$$ Verification of Incompressibility (Laplace Equation): The continuity equation for incompressible flow requires $ abla^2phi = 0$: $$ abla^2phi = frac{partial^2phi}{partial x^2} + frac{partial^2phi}{partial y^2} + frac{partial^2phi}{partial z^2} = 1 + 1 - 2 = 0$$ Since Laplace's equation is satisfied identically, this velocity potential represents a physically admissible incompressible fluid motion. --- Step 1: Evaluating the Velocity Field Components Computing the partial derivatives: $$u = frac{partialphi}{partial x} = frac{partial}{partial x}left[ frac{1}{2}(x^2 + y^2 - 2z^2) right] = x$$ $$v = frac{partialphi}{partial y} = frac{partial}{partial y}left[ frac{1}{2}(x^2 + y^2 - 2z^2) right] = y$$ $$w = frac{partialphi}{partial z} = frac{partial}{partial z}left[ frac{1}{2}(x^2 + y^2 - 2z^2) right] = -2z$$ Thus, the velocity vector field is: $$vec{v} = xhat{i} + yhat{j} - 2zhat{k}$$ --- Step 2: Formulation of the Streamline Differential Equations By definition, a streamline is an instantaneous curve in the fluid such that the tangent vector at every point is directed along the velocity vector $vec{v}$. The system of differential equations for the streamlines is: $$frac{dx}{u} = frac{dy}{v} = frac{dz}{w}$$ Substituting the velocity components: $$frac{dx}{x} = frac{dy}{y} = frac{dz}{-2z}$$ --- Step 3: Integrating the Differentia

Question 19

Q5. (d) A rough uniform board of mass $m$ and length $2a$ rests on a smooth horizontal plane and a man of mass $M$ walks on it from one end to the other. Find the distance covered by the board during this time.

Accepted Answer

Physical Formulation and Conservation Principles Let us analyze the dynamics of the composite mechanical system consisting of: 1. A uniform rigid board of mass $m$ and length $2a$. 2. A man of mass $M$ who walks along the upper surface of the board. Smooth Plane (μ = 0) Initial State: Board (mass m, length 2a) Man (M) Final State: Board (mass m) Man (M) Board shift: d = 2Ma / (M + m) Center of Mass (Fixed) Figure: Conservation of center of mass on a frictionless plane determines board backward shift $d = frac{2Ma}{M + m}$. Key Physical Assumptions: - Smooth horizontal plane: The horizontal supporting surface is completely frictionless (coefficient of friction $mu = 0$). - Rough board: The contact between the man's shoes and the board is sufficiently rough (with friction $mu 0$) so that the man can walk without slipping. - External Forces: In the horizontal direction, there are no external forces acting on the two-body system (gravity and the normal reaction from the ground act purely vertically). --- Step 1: Conservation of Linear Momentum and Center of Mass According to Newton's Second Law for a system of particles: $$sum F{text{ext}, x} = frac{d Px}{dt} = 0$$ Because the net external horizontal force is zero at all times, the total horizontal linear momentum $Px$ of the system is strictly conserved: $$Px = M v{text{man}} + m v{text{board}} = text{constant}$$ Since the system starts from rest, $Px = 0$. Integrating with respect to time: $$M Delta x{text{man}} + m Delta x{text{board}} = 0$$ This fundamental result states that the center of mass of the (man + board) system remains stationary in the fixed inertial frame of the ground throughout the motion. --- Step 2: Coordinate Analysis in an Inertial Frame Let a horizontal $x$-axis be fixed to the smooth plane, with origin $O$ chosen at the initial position of the left end of the board. Initial Configuration: - The board extends from $x = 0$ to $x = 2a$. - Because the board is uniform, its center of mass is located

Question 20

Q5. (c) (i) Determine the decimal equivalent in sign magnitude form of $(8D){16}$ and $(FF){16}$. (ii) Determine the decimal equivalent of $(9B2.1A){16}$.

Accepted Answer

Part (i): Decimal Equivalent in Sign-Magnitude Form 1. Theoretical Concept of Sign-Magnitude Representation In an 8-bit sign-magnitude integer representation: - The Most Significant Bit (MSB, bit $b7$) is designated as the sign bit: - $b7 = 0 implies$ Positive number ($+$) - $b7 = 1 implies$ Negative number ($-$) - The remaining 7 bits ($b6 b5 b4 b3 b2 b1 b0$) directly represent the magnitude as an unsigned integer: $$text{Magnitude} = sum{i=0}^6 bi 2^i$$ - The overall value is given by: $$N = (-1)^{b7} times left( sum{i=0}^6 bi 2^i right)$$ --- 2. Conversion of $(8D){16}$ 1. Convert the hexadecimal digits into 8-bit binary: - Hex digit $8{16} = 10002$ - Hex digit $D{16} = 13{10} = 11012$ - Combined 8-bit binary pattern: $$(8D){16} = 1000,11012$$ 2. Extract sign and magnitude components: - Sign bit ($b7$): $1 implies$ The number is negative ($-$). - Magnitude bits ($b6 dots b0$): $000,11012$ 3. Compute decimal magnitude: $$0 cdot 2^6 + 0 cdot 2^5 + 0 cdot 2^4 + 1 cdot 2^3 + 1 cdot 2^2 + 0 cdot 2^1 + 1 cdot 2^0 = 8 + 4 + 1 = 13$$ 4. Attaching the sign: $$mathbf{(8D){16} = -13{10}}$$ (Note: For completeness, if interpreted in 2's complement notation, $(8D){16} = 141 - 256 = -115{10}$. However, in strict accordance with the question prompt specifying sign-magnitude form, the answer is $-13$.) --- 3. Conversion of $(FF){16}$ 1. Convert the hexadecimal digits into 8-bit binary: - Hex digit $F{16} = 15{10} = 11112$ - Combined 8-bit binary pattern: $$(FF){16} = 1111,11112$$ 2. Extract sign and magnitude components: - Sign bit ($b7$): $1 implies$ The number is negative ($-$). - Magnitude bits ($b6 dots b0$): $111,11112$ 3. Compute decimal magnitude: $$1 cdot 2^6 + 1 cdot 2^5 + 1 cdot 2^4 + 1 cdot 2^3 + 1 cdot 2^2 + 1 cdot 2^1 + 1 cdot 2^0 = 64 + 32 + 16 + 8 + 4 + 2 + 1 = 127$$ 4. Attaching the sign: $$mathbf{(FF){16} = -127{10}}$$ (Note: In sign-magnitude form, $(FF){16}$ represents negative maximum magnitude $-127{10}$. In 2's complement notation it would be $-1{10}$.) ---

Question 21

Q5. (b) Solve the following system of linear equations by Gauss-Jordan method: $$begin{aligned} 2x + 3y - z &= 5 4x + 4y - 3z &= 3 2x - 3y + 2z &= 2 end{aligned}$$

Accepted Answer

Matrix Representation and Overview of Gauss-Jordan Elimination The given system of simultaneous linear equations is: $$begin{aligned} 2x + 3y - z &= 5 4x + 4y - 3z &= 3 2x - 3y + 2z &= 2 end{aligned}$$ In matrix notation $A X = B$: $$begin{pmatrix} 2 & 3 & -1 4 & 4 & -3 2 & -3 & 2 end{pmatrix} begin{pmatrix} x y z end{pmatrix} = begin{pmatrix} 5 3 2 end{pmatrix}$$ The Gauss-Jordan elimination method systematically reduces the augmented matrix $[A mid B]$ to reduced row echelon form (RREF) $[I mid X]$ via elementary row operations: $$[A mid B] = left(begin{array}{ccc|c} 2 & 3 & -1 & 5 4 & 4 & -3 & 3 2 & -3 & 2 & 2 end{array}right)$$ --- Step 1: Creating Leading 1 in Row 1 and Eliminating First Column Normalize the first row by performing $R1 to frac{1}{2} R1$: $$left(begin{array}{ccc|c} 1 & frac{3}{2} & -frac{1}{2} & frac{5}{2} 4 & 4 & -3 & 3 2 & -3 & 2 & 2 end{array}right)$$ Eliminate the entries in Column 1 below the pivot: 1. $R2 to R2 - 4R1$: - $4 - 4(1) = 0$ - $4 - 4left(frac{3}{2}right) = 4 - 6 = -2$ - $-3 - 4left(-frac{1}{2}right) = -3 + 2 = -1$ - $3 - 4left(frac{5}{2}right) = 3 - 10 = -7$ 2. $R3 to R3 - 2R1$: - $2 - 2(1) = 0$ - $-3 - 2left(frac{3}{2}right) = -3 - 3 = -6$ - $2 - 2left(-frac{1}{2}right) = 2 + 1 = 3$ - $2 - 2left(frac{5}{2}right) = 2 - 5 = -3$ The matrix is now: $$left(begin{array}{ccc|c} 1 & frac{3}{2} & -frac{1}{2} & frac{5}{2} 0 & -2 & -1 & -7 0 & -6 & 3 & -3 end{array}right)$$ --- Step 2: Creating Leading 1 in Row 2 and Eliminating Second Column Normalize the second row by multiplying by $-frac{1}{2}$, i.e., $R2 to -frac{1}{2} R2$: $$left(begin{array}{ccc|c} 1 & frac{3}{2} & -frac{1}{2} & frac{5}{2} 0 & 1 & frac{1}{2} & frac{7}{2} 0 & -6 & 3 & -3 end{array}right)$$ Now eliminate the entries in Column 2 above and below the pivot: 1. $R1 to R1 - frac{3}{2} R2$: - $1 - 0 = 1$ - $frac{3}{2} - frac{3}{2}(1) = 0$ - $-frac{1}{2} - frac{3}{2}left(frac{1}{2}right) = -frac{1}{2} - frac{3}{4} = -frac{5}{4}$ - $frac{5}{2} - frac{3}{2}left(frac{7}{2}righ

Question 22

Q7. (a) Find the integral surface of the following quasi-linear equation $$(y - phi)frac{partial phi}{partial x} + (phi - x)frac{partial phi}{partial y} = x - y,$$ which passes through the curve $phi = 0, , xy = 1$ and through the circle $x + y + phi = 0, , x^2 + y^2 + phi^2 = a^2$.

Accepted Answer

Mathematical Formulation and Lagrange's Method We are given the first-order quasi-linear partial differential equation: $$(y - phi)frac{partial phi}{partial x} + (phi - x)frac{partial phi}{partial y} = x - y$$ This is of the standard Lagrange form: $$P(x, y, phi) frac{partial phi}{partial x} + Q(x, y, phi) frac{partial phi}{partial y} = R(x, y, phi)$$ where: $$P = y - phi, quad Q = phi - x, quad R = x - y$$ The corresponding Lagrange's Characteristic Auxiliary System is: $$frac{dx}{y - phi} = frac{dy}{phi - x} = frac{dphi}{x - y}$$ --- Step 1: Finding Two Independent First Integrals To find two functionally independent first integrals $u1(x, y, phi) = c1$ and $u2(x, y, phi) = c2$, we use the method of multipliers: 1. First Multiplier Set: $(1, 1, 1)$ Using multipliers $1, 1, 1$: $$frac{dx + dy + dphi}{(y - phi) + (phi - x) + (x - y)} = frac{d(x + y + phi)}{0}$$ Since the denominator vanishes identically: $$dx + dy + dphi = 0$$ Integrating both sides: $$x + y + phi = c1$$ This yields our first integral: $$u1(x, y, phi) = x + y + phi = c1$$ 2. Second Multiplier Set: $(x, y, phi)$ Using multipliers $x, y, phi$: $$frac{x, dx + y, dy + phi, dphi}{x(y - phi) + y(phi - x) + phi(x - y)}$$ Evaluating the denominator: $$x(y - phi) + y(phi - x) + phi(x - y) = xy - xphi + yphi - yx + xphi - yphi = 0$$ Since the denominator is identically zero: $$x, dx + y, dy + phi, dphi = 0 iff dleft(frac{x^2 + y^2 + phi^2}{2}right) = 0$$ Integrating both sides: $$x^2 + y^2 + phi^2 = c2$$ This yields our second independent integral: $$u2(x, y, phi) = x^2 + y^2 + phi^2 = c2$$ The general solution of the PDE is given by any functional relation: $$Phi(c1, c2) = 0 quad text{or} quad c2 = f(c1)$$ that is: $$x^2 + y^2 + phi^2 = f(x + y + phi)$$ --- Step 2: Integral Surface Passing Through the Curve $phi = 0, , xy = 1$ Let us determine the specific function $f$ such that the integral surface contains the hyperbolic curve: $$phi = 0, quad xy = 1$$ Parameterizing the curve in terms of a real parameter

Question 23

Q7. (b) Integrate $f(x) = 5x^3 - 3x^2 + 2x + 1$ from $x = -2$ to $x = 4$ using (i) Simpson's $frac{3}{8}$ rule with width $h = 1$, and (ii) Trapezoidal rule with width $h = 1$.

Accepted Answer

Numerical Integration Setup and Tabulation We are asked to approximate the definite integral: $$I = int{-2}^4 f(x), dx = int{-2}^4 (5x^3 - 3x^2 + 2x + 1), dx$$ with step size $h = 1$. Interval and Nodes: The integration limits are $a = -2$ and $b = 4$. The total interval length is $b - a = 4 - (-2) = 6$. The number of subintervals is: $$n = frac{b - a}{h} = frac{6}{1} = 6$$ The nodal grid points $xi = a + i h$ for $i = 0, 1, 2, 3, 4, 5, 6$ are: $$x0 = -2, ; x1 = -1, ; x2 = 0, ; x3 = 1, ; x4 = 2, ; x5 = 3, ; x6 = 4$$ --- Step 1: Exact Analytical Value of the Integral For comparative validation, we evaluate the exact integral: $$int{-2}^4 (5x^3 - 3x^2 + 2x + 1), dx = left[ frac{5x^4}{4} - x^3 + x^2 + x right]{-2}^4$$ Evaluating at the upper limit $x = 4$: $$F(4) = frac{5(256)}{4} - (64) + (16) + 4 = 320 - 64 + 16 + 4 = 276$$ Evaluating at the lower limit $x = -2$: $$F(-2) = frac{5(16)}{4} - (-8) + (4) + (-2) = 20 + 8 + 4 - 2 = 30$$ Subtracting the limits: $$I{text{exact}} = F(4) - F(-2) = 276 - 30 = 246$$ --- Step 2: Tabulation of Function Values $yi = f(xi)$ We compute the function values at each of the 7 grid points: 1. $y0 = f(-2) = 5(-2)^3 - 3(-2)^2 + 2(-2) + 1 = 5(-8) - 3(4) - 4 + 1 = -40 - 12 - 4 + 1 = -55$ 2. $y1 = f(-1) = 5(-1)^3 - 3(-1)^2 + 2(-1) + 1 = -5 - 3 - 2 + 1 = -9$ 3. $y2 = f(0) = 5(0) - 3(0) + 2(0) + 1 = 1$ 4. $y3 = f(1) = 5(1)^3 - 3(1)^2 + 2(1) + 1 = 5 - 3 + 2 + 1 = 5$ 5. $y4 = f(2) = 5(2)^3 - 3(2)^2 + 2(2) + 1 = 5(8) - 3(4) + 4 + 1 = 40 - 12 + 4 + 1 = 33$ 6. $y5 = f(3) = 5(3)^3 - 3(3)^2 + 2(3) + 1 = 5(27) - 3(9) + 6 + 1 = 135 - 27 + 6 + 1 = 115$ 7. $y6 = f(4) = 5(4)^3 - 3(4)^2 + 2(4) + 1 = 5(64) - 3(16) + 8 + 1 = 320 - 48 + 8 + 1 = 281$ Summary Table of Nodes: | $i$ | $xi$ | $yi = f(xi)$ | | :---: | :---: | :---: | | 0 | -2 | -55 | | 1 | -1 | -9 | | 2 | 0 | 1 | | 3 | 1 | 5 | | 4 | 2 | 33 | | 5 | 3 | 115 | | 6 | 4 | 281 | --- Step 3: Part (i) - Simpson's $frac{3}{8}$ Rule Because $n = 6$ is a multiple of 3, Simpson's $frac{3}{8}$ rule is directly ap

Question 24

Q7. (c) Let the velocity field $$u(x, y) = frac{B(x^2 - y^2)}{(x^2 + y^2)^2}, quad v(x, y) = frac{2Bxy}{(x^2 + y^2)^2}, quad w(x, y) = 0$$ satisfy the equations of motion for inviscid incompressible flow, where $B$ is a constant. Determine the pressure associated with this velocity field.

Accepted Answer

Fluid Mechanics Principles and Flow Analysis We are given the two-dimensional velocity field: $$u(x, y) = frac{B(x^2 - y^2)}{(x^2 + y^2)^2}, quad v(x, y) = frac{2Bxy}{(x^2 + y^2)^2}, quad w(x, y) = 0$$ for an inviscid, incompressible fluid of constant density $rho$. --- Step 1: Complex Potential Formulation and Irrotationality Let $z = x + iy$ be the complex variable in the plane. Notice the complex velocity: $$u - iv = frac{B(x^2 - y^2) - i(2Bxy)}{(x^2 + y^2)^2} = frac{B(x - iy)^2}{[(x + iy)(x - iy)]^2} = frac{B bar{z}^2}{z^2 bar{z}^2} = frac{B}{z^2}$$ Since $frac{B}{z^2}$ is a complex analytic function of $z$ everywhere in $mathbb{C} setminus {0}$, it follows from Cauchy-Riemann equations that: 1. The flow is irrotational: $$zeta = frac{partial v}{partial x} - frac{partial u}{partial y} = 0$$ 2. The flow is incompressible: $$ abla cdot vec{v} = frac{partial u}{partial x} + frac{partial v}{partial y} = 0$$ The complex velocity potential $W(z) = phi + ipsi$ is obtained by integrating: $$frac{dW}{dz} = - (u - iv) = -frac{B}{z^2} implies W(z) = frac{B}{z}$$ (or $W = -frac{B}{z}$ depending on the sign convention for $phi$). --- Step 2: Evaluating the Magnitude of Velocity (Flow Speed $q$) The square of the speed $q = |vec{v}|$ is given by: $$q^2 = u^2 + v^2 = |u - iv|^2 = left| frac{B}{z^2} right|^2 = frac{B^2}{|z|^4}$$ Expressing this in Cartesian coordinates: $$|z|^2 = x^2 + y^2 implies |z|^4 = (x^2 + y^2)^2$$ Therefore: $$q^2 = frac{B^2}{(x^2 + y^2)^2}$$ Direct verification from $u$ and $v$: $$u^2 + v^2 = frac{B^2 (x^2 - y^2)^2 + 4B^2 x^2 y^2}{(x^2 + y^2)^4} = frac{B^2 (x^4 - 2x^2 y^2 + y^4 + 4x^2 y^2)}{(x^2 + y^2)^4} = frac{B^2 (x^2 + y^2)^2}{(x^2 + y^2)^4} = frac{B^2}{(x^2 + y^2)^2}$$ This confirms the result perfectly. --- Step 3: Application of Bernoulli's Theorem For a steady, irrotational, inviscid, and incompressible fluid flow under conservative body forces (or neglecting gravity), Bernoulli's Equation holds uniformly throughout the entire flow field: $$frac

Question 25

Q8. (a) Solve the partial differential equation $$frac{partial}{partial y}left(frac{partial phi}{partial x} + phiright) + 2x^2 yleft(frac{partial phi}{partial x} + phiright) = 0$$ by transforming it to the canonical form.

Accepted Answer

Mathematical Formulation and Canonical Reduction We are given the second-order linear partial differential equation: $$frac{partial}{partial y}left(frac{partial phi}{partial x} + phiright) + 2x^2 yleft(frac{partial phi}{partial x} + phiright) = 0$$ Let us define an intermediate substitution variable: $$v(x, y) = frac{partial phi}{partial x} + phi$$ Substituting $v$ into the given equation decouples the second-order equation into two successive first-order linear differential equations: $$frac{partial v}{partial y} + 2x^2 y v = 0$$ --- Step 1: Solving the First-Order Equation for $v(x, y)$ The equation: $$frac{partial v}{partial y} + 2x^2 y v = 0$$ is a first-order linear ordinary differential equation in $y$, treating $x$ as a constant parameter. Separating variables: $$frac{1}{v} frac{partial v}{partial y} = -2x^2 y$$ Integrating with respect to $y$: $$ln |v| = -x^2 y^2 + ln |f(x)|$$ where $ln |f(x)|$ is an arbitrary constant of integration with respect to $y$, meaning $f(x)$ is an arbitrary differentiable function of $x$. Exponentiating both sides: $$v(x, y) = f(x) e^{-x^2 y^2}$$ --- Step 2: Solving for $phi(x, y)$ Now replace $v(x, y)$ with its definition $frac{partial phi}{partial x} + phi$: $$frac{partial phi}{partial x} + phi = f(x) e^{-x^2 y^2}$$ This is a first-order linear differential equation in $x$, treating $y$ as a constant parameter. It is of standard form $frac{partial phi}{partial x} + P(x)phi = Q(x, y)$ with $P(x) = 1$. The integrating factor is: $$I(x) = e^{int 1, dx} = e^x$$ Multiplying both sides of the equation by $e^x$: $$e^x frac{partial phi}{partial x} + e^x phi = e^x f(x) e^{-x^2 y^2} implies frac{partial}{partial x} left( phi(x, y) e^x right) = f(x) e^{x - x^2 y^2}$$ Integrating both sides with respect to $x$: $$phi(x, y) e^x = int f(x) e^{x - x^2 y^2}, dx + g(y)$$ where $g(y)$ is an arbitrary differentiable function of $y$ arising as the constant of integration with respect to $x$. Multiplying through by $e^{-x}$: $$phi(x, y) = e^{-x} int

Question 26

Q8. (b) Using Newton's forward difference formula for interpolation, estimate the value of $f(2.5)$ from the following data: | x | 1 | 2 | 3 | 4 | 5 | 6 | | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | f(x) | 0 | 1 | 8 | 27 | 64 | 125 |

Accepted Answer

Interpolation Setup and Forward Difference Formulation We are given equally spaced data points $(xi, yi)$ with step size $h = 1$: $$x0 = 1, quad x1 = 2, quad x2 = 3, quad x3 = 4, quad x4 = 5, quad x5 = 6$$ and corresponding function values: $$y0 = 0, quad y1 = 1, quad y2 = 8, quad y3 = 27, quad y4 = 64, quad y5 = 125$$ Notice that the data points strictly follow the cubic law $f(x) = (x - 1)^3$: - $f(1) = 0^3 = 0$ - $f(2) = 1^3 = 1$ - $f(3) = 2^3 = 8$ - $f(4) = 3^3 = 27$ - $f(5) = 4^3 = 64$ - $f(6) = 5^3 = 125$ We are required to use Newton's Forward Difference Interpolation Formula to estimate $f(2.5)$. --- Step 1: Construction of the Forward Difference Table The forward difference operators are defined by: $$Delta yi = y{i+1} - yi$$ $$Delta^{k} yi = Delta^{k-1} y{i+1} - Delta^{k-1} yi$$ Let us construct the complete forward difference table: | $x$ | $y = f(x)$ | $Delta y$ | $Delta^2 y$ | $Delta^3 y$ | $Delta^4 y$ | $Delta^5 y$ | | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | 1 | 0 | | | | | | | | | 1 | | | | | | 2 | 1 | | 6 | | | | | | | 7 | | 6 | | | | 3 | 8 | | 12 | | 0 | | | | | 19 | | 6 | | 0 | | 4 | 27 | | 18 | | 0 | | | | | 37 | | 6 | | | | 5 | 64 | | 24 | | | | | | | 61 | | | | | | 6 | 125 | | | | | | Reading the Forward Differences from the Top Row ($x0 = 1$): - $y0 = 0$ - $Delta y0 = 1$ - $Delta^2 y0 = 6$ - $Delta^3 y0 = 6$ - $Delta^4 y0 = 0$ - $Delta^5 y0 = 0$ Notice that all third-order differences are constant ($Delta^3 y = 6$), and all fourth- and higher-order differences are identically zero. This confirms that the underlying function is a polynomial of degree 3. --- Step 2: Computation of the Interpolation Parameter $u$ Newton's forward difference formula is parameterized by: $$u = frac{x - x0}{h}$$ Here: - Target point: $x = 2.5$ - Base point: $x0 = 1$ - Step size: $h = x1 - x0 = 2 - 1 = 1$ Therefore: $$u = frac{2.5 - 1}{1} = 1.5$$ --- Step 3: Application of Newton's Forward Difference Formula Newton's forward interpolation formula sta

Question 27

Q8. (c) Suppose an infinite liquid contains two parallel, equal and opposite rectilinear vortices at a distance $2a$. Show that the streamlines relative to the vortex are given by the equation $$logfrac{x^2 + (y - a)^2}{x^2 + (y + a)^2} + frac{y}{a} = C,$$ where $C$ is a constant, the origin is the

Accepted Answer

Mathematical Formulation and Coordinate Setup Let an infinite, incompressible, inviscid fluid contain two parallel rectilinear vortex filaments: - A vortex of circulation $+kappa$ located at the point $A(0, a)$. - A vortex of equal and opposite circulation $-kappa$ located at the point $B(0, -a)$. The origin $O(0, 0)$ is chosen as the midpoint of the line segment joining the vortices, and the line connecting them is taken as the $y$-axis. The distance between the two vortices is $2a$. x y O(0,0) +κ at (0, a) U = κ/(4πa) -κ at (0, -a) U = κ/(4πa) Relative Streamline 2a Figure: Vortex pair of equal and opposite circulation translating with velocity $U = frac{kappa}{4pi a}$ along $+x$, showing relative streamlines. --- Step 1: Self-Induced Translation of the Vortex Pair According to Helmholtz's vortex theorems, each vortex moves with the velocity field induced at its position by all other vortices in the fluid: 1. Velocity induced at $A(0, a)$ by the vortex at $B(0, -a)$: The vortex at $B$ has circulation $-kappa$ (clockwise) and is at distance $2a$ from $A$. By the Biot-Savart law in two dimensions, the induced velocity is directed along the $+x$-direction: $$U = frac{kappa}{2pi (2a)} = frac{kappa}{4pi a}$$ 2. Velocity induced at $B(0, -a)$ by the vortex at $A(0, a)$: The vortex at $A$ has circulation $+kappa$ (counter-clockwise) and is at distance $2a$ from $B$. The induced velocity at $B$ is also directed along the $+x$-direction: $$U = frac{kappa}{2pi (2a)} = frac{kappa}{4pi a}$$ Because both vortices have identical induced velocities $U = frac{kappa}{4pi a}$ along the $+x$-axis, the vortex pair translates uniformly as a rigid configuration through the liquid parallel to the $x$-axis with speed $U$. --- Step 2: The Complex Potential of the Stationary Fluid In the fixed inertial frame where the fluid at infinity is at rest, the complex potential $w(z) = phi + ipsi$ of two line vortices at $z1 = ia$ (strength $+kappa$) and $z2 = -ia$ (strength $-kappa$) is: $$w(z) =

Question 28

Q6. (c) Find the moment of inertia of a quadrant of an elliptic disk $frac{x^2}{a^2} + frac{y^2}{b^2} = 1$, of mass $M$ about the line passing through its centre and perpendicular to its plane. Given that the density at any point is proportional to $xy$.

Accepted Answer

Mathematical Formulation and Density Distribution Let the elliptic quadrant domain $D$ in the first quadrant of the Cartesian plane be defined by: $$D = left{ (x, y) in mathbb{R}^2 : frac{x^2}{a^2} + frac{y^2}{b^2} le 1, quad x ge 0, quad y ge 0 right}$$ We are given that the surface density $rho(x, y)$ at any point $(x, y) in D$ is proportional to $xy$: $$rho(x, y) = k xy$$ where $k 0$ is the constant of proportionality. We wish to evaluate the moment of inertia $Iz$ of this planar quadrant about the $z$-axis (which passes through the center $(0, 0)$ of the ellipse and is perpendicular to the $xy$-plane). By definition: $$Iz = iintD (x^2 + y^2) rho(x, y), dx, dy = iintD (x^2 + y^2)(k xy), dx, dy$$ By the Perpendicular Axis Theorem: $$Iz = Ix + Iy$$ where $Ix = iintD y^2 rho, dx, dy$ and $Iy = iintD x^2 rho, dx, dy$. --- Step 1: Determining the Proportionality Constant $k$ via Total Mass $M$ To express $k$ in terms of the total mass $M$, we integrate the density over the domain $D$: $$M = iintD rho(x, y), dx, dy = k iintD xy, dx, dy$$ We transform the integral using generalized elliptic polar coordinates: $$x = a r costheta, quad y = b r sintheta$$ The Jacobian of this transformation is: $$J = frac{partial(x, y)}{partial(r, theta)} = begin{vmatrix} acostheta & -arsintheta bsintheta & brcostheta end{vmatrix} = ab r (cos^2theta + sin^2theta) = ab r$$ The quadrant domain in the $(r, theta)$ plane is given by: $$0 le r le 1, quad 0 le theta le frac{pi}{2}$$ Substituting into the mass integral: $$xy = (a r costheta)(b r sintheta) = ab r^2 sintheta costheta$$ $$M = k int0^{pi/2} int0^1 (ab r^2 sintheta costheta)(ab r), dr, dtheta = k a^2 b^2 left( int0^{pi/2} sintheta costheta, dtheta right) left( int0^1 r^3, dr right)$$ Evaluating the two single integrals: 1. $int0^{pi/2} sintheta costheta, dtheta = left[ frac{sin^2theta}{2} right]0^{pi/2} = frac{1}{2}$ 2. $int0^1 r^3, dr = left[ frac{r^4}{4} right]0^1 = frac{1}{4}$ Thus: $$M = k a^2 b^2 left( frac{1}{2} right) left( fra

Maths Optional 2024 Paper II

28 Questions Available

Maths Optional 2024 Paper II

28 Questions Available

Question 1: Q2. (a) Using Cauchy's general principle of convergence, examine the convergence of the sequence $langle fn rangle$, where $fn = 1 + frac{1}{1!} + fra

Question 2: Q1. (e) Use two phase method to solve the following linear programming problem: Maximize $z = x1 + 2x2$ subject to $x1 - x2 ge 3$ $2x1 + x2 le 10$ $x1

Question 3: Q1. (d) If $phi$ and $psi$ are functions of $x$ and $y$ satisfying Laplace equation, then show that $f(z) = p + iq, i = sqrt{-1}$ is an analytic funct

Question 4: Q1. (c) Test the convergence of $$int0^2 frac{log x}{sqrt{2 - x}},dx.$$

Question 5: Q1. (b) If $w = f(z)$ is an analytic function of $z$, then show that $$left(frac{partial^2}{partial x^2} + frac{partial^2}{partial y^2}right) log |f'(

Question 6: Q1. (a) Let $G$ be a finite group of order $mn$, where $m$ and $n$ are prime numbers with $m n$. Show that $G$ has at most one subgroup of order $m$.

Question 7: Q2. (c) Find the function which is analytic inside and on the circle $C : z = e^{itheta}, 0 le theta le 2pi$ and has the value $$frac{(a^2 - 1)costhet

Question 8: Q2. (b) Show that every homomorphic image of an abelian group is abelian, but the converse is not necessarily true.

Question 9: Q3. (a) Locate the poles and their order for the function $$f(z) = frac{1}{z (sin pi z)left(z + frac{1}{2}right)}.$$ Also, find the residue of $f(z)$

Question 10: Q3. (b) Consider the series $sum{n=1}^infty Un(x)$, $0 le x le 1$, the sum of whose first $n$ terms is given by $Sn(x) = frac{1}{2n^2}log(1+n^4 x^2)$,

Question 11: Q3. (c) Using duality principle, solve the following linear programming problem: Minimize $$z = 4x1 + 3x2 + x3$$ subject to $$begin{aligned} x1 + 2x2

Question 12: Q4. (a) Consider the polynomial ring $mathbb{Z}[x]$ over the ring $mathbb{Z}$ of integers. Let $S$ be an ideal of $mathbb{Z}[x]$ generated by $x$. Sho

Question 13: Q4. (b) Find the upper and lower Riemann integrals for the function $f$ defined on $[0, 1]$ as follows: $$f(x) = begin{cases} (1 - x^2)^{1/2}, & text{

Question 14: Q4. (c) The personnel manager of a company wants to assign three officers $A$, $B$ and $C$ to regional offices at Delhi, Mumbai, Kolkata and Chennai.

Question 15: Q5. (a) Show that if $f$ and $g$ are arbitrary functions of their respective arguments, then $u = f(x - kt + ialpha y) + g(x - kt - ialpha y)$, is a s

Question 16: Q6. (b) Draw the logical circuit for the Boolean expression $$Y = ABoverline{C} + Boverline{C} + overline{A}B.$$ Also, obtain the output $Y$ (truth ta

Question 17: Q6. (a) Show that the solution of the two-dimensional Laplace's equation $$frac{partial^2 phi(x, y)}{partial x^2} + frac{partial^2 phi(x, y)}{partial

Question 18: Q5. (e) The velocity potential $phi$ of a flow is given by $$phi = frac{1}{2}(x^2 + y^2 - 2z^2).$$ Determine the streamlines.

Question 19: Q5. (d) A rough uniform board of mass $m$ and length $2a$ rests on a smooth horizontal plane and a man of mass $M$ walks on it from one end to the oth

Question 20: Q5. (c) (i) Determine the decimal equivalent in sign magnitude form of $(8D){16}$ and $(FF){16}$. (ii) Determine the decimal equivalent of $(9B2.1A){1

Question 21: Q5. (b) Solve the following system of linear equations by Gauss-Jordan method: $$begin{aligned} 2x + 3y - z &= 5 4x + 4y - 3z &= 3 2x - 3y + 2z &= 2 e

Question 22: Q7. (a) Find the integral surface of the following quasi-linear equation $$(y - phi)frac{partial phi}{partial x} + (phi - x)frac{partial phi}{partial

Question 23: Q7. (b) Integrate $f(x) = 5x^3 - 3x^2 + 2x + 1$ from $x = -2$ to $x = 4$ using (i) Simpson's $frac{3}{8}$ rule with width $h = 1$, and (ii) Trapezoida

Question 24: Q7. (c) Let the velocity field $$u(x, y) = frac{B(x^2 - y^2)}{(x^2 + y^2)^2}, quad v(x, y) = frac{2Bxy}{(x^2 + y^2)^2}, quad w(x, y) = 0$$ satisfy the

Question 25: Q8. (a) Solve the partial differential equation $$frac{partial}{partial y}left(frac{partial phi}{partial x} + phiright) + 2x^2 yleft(frac{partial phi}

Question 26: Q8. (b) Using Newton's forward difference formula for interpolation, estimate the value of $f(2.5)$ from the following data: | x | 1 | 2 | 3 | 4 | 5 |

Question 27: Q8. (c) Suppose an infinite liquid contains two parallel, equal and opposite rectilinear vortices at a distance $2a$. Show that the streamlines relati

Question 28: Q6. (c) Find the moment of inertia of a quadrant of an elliptic disk $frac{x^2}{a^2} + frac{y^2}{b^2} = 1$, of mass $M$ about the line passing through