Q: 2. (c) Use the method of contour integration to prove that $$int{-infty}^infty frac{x^2 - x + 2}{x^4 + 10x^2 + 9},dx = frac{5pi}{12}.$$ (20 Marks)

1. Complex Extension & Contour Setup We are asked to evaluate the real improper integral: $$I = int{-infty}^infty frac{x^2 - x + 2}{x^4 + 10x^2 + 9} dx$$ Notice that the denominator is an even function: $$Q(x) = x^4 + 10x^2 + 9 = (x^2 + 1)(x^2 + 9)$$ The numerator can be split into an even part and an odd part: $$P(x) = x^2 - x + 2 = (x^2 + 2) - x$$ Since $frac{-x}{x^4 + 10x^2 + 9}$ is an odd function of $x$ and the integral converges, its Cauchy Principal Value vanishes identically: $$int{-infty}^infty frac{-x}{x^4 + 10x^2 + 9} dx = 0$$ Therefore: $$I = int{-infty}^infty frac{x^2 + 2}{x^4 + 10x^2 + 9} dx$$ However, we can directly apply contour integration to the full complex function $f(z)$: $$f(z) = frac{z^2 - z + 2}{z^4 + 10z^2 + 9}$$ We choose a standard semi-circular contour $GammaR$ in the upper half of the complex plane: $$GammaR = [-R, R] cup CR$$ where: - $[-R, R]$ is the real line segment from $-R$ to $R$. - $CR$ is the upper semi-circle $z = R e^{itheta}, ; theta in [0, pi]$, with $R 3$. By Cauchy's Residue Theorem: $$oint{GammaR} f(z) dz = int{-R}^R f(x) dx + int{CR} f(z) dz = 2pi i sum operatorname{Res}(f, zk)$$ where the sum is taken over all poles $zk$ lying strictly in the upper half-plane ($operatorname{Im}(zk) 0$). --- 2. Identification of Poles in the Upper Half-Plane The poles of $f(z)$ are the roots of the denominator: $$z^4 + 10z^2 + 9 = 0$$ Letting $u = z^2$: $$u^2 + 10u + 9 = 0 implies (u + 1)(u + 9) = 0$$ Thus $z^2 = -1$ or $z^2 = -9$, which yields 4 simple poles: $$z = pm i, quad z = pm 3i$$ The poles lying strictly in the upper half-plane ($operatorname{Im}(z) 0$) are: $$z1 = i quad text{and} quad z2 = 3i$$ --- 3. Calculation of Residues Since all poles are simple poles, the residue at a simple pole $z0$ can be evaluated as: $$operatorname{Res}(f, z0) = frac{P(z0)}{Q'(z0)}$$ where $P(z) = z^2 - z + 2$ and $Q(z) = z^4 + 10z^2 + 9$. The derivative of the denominator is: $$Q'(z) = 4z^3 + 20z = 4z(z^2 + 5)$$ Residue at $z1 = i$: $$P(i) = i^2

Q: 2. (b) Show that $3$ is an irreducible element in the integral domain $mathbb{Z}[i]$. (15 Marks)

1. Algebraic Framework of the Gaussian Integers $mathbb{Z}[i]$ The ring of Gaussian integers is defined as: $$mathbb{Z}[i] = {a + bi : a, b in mathbb{Z}}, quad text{where } i^2 = -1$$ Since $mathbb{Z}[i]$ is a subring of the field of complex numbers $mathbb{C}$, it is an integral domain (it is commutative with identity $1$, and has no zero divisors). Furthermore, $mathbb{Z}[i]$ is a Euclidean domain, and hence a Principal Ideal Domain (PID) and a Unique Factorization Domain (UFD). We define the standard field norm $N: mathbb{Z}[i] to mathbb{Z}{ge 0}$ by: $$N(alpha) = N(a + bi) = alpha bar{alpha} = (a + bi)(a - bi) = a^2 + b^2$$ Properties of the Norm: 1. $N(alpha) ge 0$ for all $alpha in mathbb{Z}[i]$, and $N(alpha) = 0 iff alpha = 0$. 2. Multiplicativity: For any $alpha, beta in mathbb{Z}[i]$, $$N(alpha beta) = N(alpha) N(beta)$$ 3. Units of $mathbb{Z}[i]$: An element $u in mathbb{Z}[i]$ is a unit (invertible) if and only if $N(u) = 1$. $$a^2 + b^2 = 1 implies (a, b) in {(pm 1, 0), (0, pm 1)}$$ Thus, the group of units of $mathbb{Z}[i]$ is: $$mathcal{U}(mathbb{Z}[i]) = {1, -1, i, -i}$$ --- 2. Definition of an Irreducible Element Definition: An element $p$ in an integral domain $R$ is called irreducible if: 1. $p eq 0$, 2. $p$ is not a unit ($p otin mathcal{U}(R)$), and 3. Whenever $p = alpha beta$ for some $alpha, beta in R$, then either $alpha$ is a unit or $beta$ is a unit. --- 3. Step-by-Step Proof that $3$ is Irreducible in $mathbb{Z}[i]$ Step 1: Verification of Non-zero and Non-unit Status - $3 = 3 + 0i eq 0$. - The norm of $3$ is: $$N(3) = 3^2 + 0^2 = 9$$ Since $N(3) = 9 eq 1$, $3$ is not a unit in $mathbb{Z}[i]$. Step 2: Factorization Hypothesis Suppose $3$ can be factored as a product of two Gaussian integers $alpha, beta in mathbb{Z}[i]$: $$3 = alpha beta$$ where $alpha = a + bi$ and $beta = c + di$ with $a, b, c, d in mathbb{Z}$. Step 3: Taking Norms on Both Sides Applying the multiplicative property of the norm: $$N(3) = N(alpha beta) = N(alpha) N(beta)$

Q: 1. (c) Examine whether the series $$sum{n=1}^infty frac{(-1)^{n-1}}{n}$$ is absolutely or conditionally convergent. (10 Marks)

1. Problem Statement & Classification of Convergence We are given the alternating infinite series: $$S = sum{n=1}^infty frac{(-1)^{n-1}}{n} = 1 - frac{1}{2} + frac{1}{3} - frac{1}{4} + frac{1}{5} - dots$$ A real series $sum an$ is defined to be: 1. Absolutely Convergent if the series of absolute values $sum |an|$ converges. 2. Conditionally Convergent if $sum an$ converges, but $sum |an|$ diverges to $+infty$. --- 2. Test for Absolute Convergence We consider the series of absolute values of the terms: $$sum{n=1}^infty left| frac{(-1)^{n-1}}{n} right| = sum{n=1}^infty frac{1}{n} = 1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + dots$$ This is the classical harmonic series, which is a $p$-series of the form $sum{n=1}^infty frac{1}{n^p}$ with $p = 1$. Proof of Divergence via Integral Test: Consider the real-valued function $f(x) = frac{1}{x}$ for $x in [1, infty)$. 1. $f(x) 0$ for all $x ge 1$. 2. $f'(x) = -frac{1}{x^2} 0$ for all $n in mathbb{N}$. 2. The sequence ${un}$ is monotonically decreasing, i.e., $u{n+1} le un$ for all $n$. 3. $lim{n to infty} un = 0$. Verification of Conditions: 1. Positivity: For every $n ge 1$, $un = frac{1}{n} 0$. Condition holds. 2. Monotonicity: $$un - u{n+1} = frac{1}{n} - frac{1}{n+1} = frac{(n+1) - n}{n(n+1)} = frac{1}{n(n+1)} 0 quad forall n ge 1$$ Thus, $u{n+1} < un$ for all $n$, meaning ${un}$ is strictly decreasing. Condition holds. 3. Limit at Infinity: $$lim{n to infty} un = lim{n to infty} frac{1}{n} = 0$$ Condition holds. Since all three conditions of Leibniz's alternating series test are satisfied, the series $sum{n=1}^infty frac{(-1)^{n-1}}{n}$ converges. --- 4. Exact Sum Evaluation via Maclaurin Series The sum of this convergent alternating series can be found explicitly via the Taylor series expansion of $ln(1 + x)$ around $x = 0$: $$ln(1 + x) = x - frac{x^2}{2} + frac{x^3}{3} - frac{x^4}{4} + dots = sum{n=1}^infty frac{(-1)^{n-1} x^n}{n}, quad text{for } |x| < 1$$ By Abel's Limit Theorem, since the series converges at $x = 1$,

Q: 1. (d) Expand $f(z) = frac{1}{(z+1)(z+3)}$ in a Laurent series valid for $1 < |z| < 3$. (10 Marks)

Re(z) Im(z) z = -1 z = -3 |z| = 1 |z| = 3 Annulus: 1 < |z| < 3 Figure: Annular region of convergence $1 1. Function Analysis & Partial Fraction Decomposition We are given the complex rational function: $$f(z) = frac{1}{(z+1)(z+3)}$$ The singularities of $f(z)$ are the zeros of its denominator: $$z + 1 = 0 implies z = -1$$ $$z + 3 = 0 implies z = -3$$ Both are simple poles located at $z = -1$ (distance $|z| = 1$ from the origin) and $z = -3$ (distance $|z| = 3$ from the origin). The specified region of expansion is the open annulus centered at the origin: $$mathcal{A} = {z in mathbb{C} : 1 1 iff frac{1}{|z|} 1$ Since $|z| 1$, we have $left|frac{1}{z}right| 1$. - Analytic Part: $-sum{n=0}^infty frac{(-1)^n}{2 cdot 3^{n+1}} z^n = -frac{1}{6} + frac{z}{18} - frac{z^2}{54} + dots$, converging for $|z| < 3$. Hence, the series converges absolutely and uniformly on any compact subset of the annular region $1 < |z| < 3$.

Q: 1. (b) Let $G = {e, x, x^2, y, yx, yx^2}$ be a non-Abelian group with $o(x) = 3$ and $o(y) = 2$. Show that $xy = yx^2$ (where $e$ is the identity element of $G$ and $o(x), o(y)$ denote the order of the elements $x, y$ respectively). (10 Marks)

1. Group Structure & Order Analysis We are given a finite group $G$ containing exactly 6 elements: $$G = {e, x, x^2, y, yx, yx^2}$$ The orders of the generators are: - $o(x) = 3 implies x^3 = e$, and ${e, x, x^2}$ is a cyclic subgroup of order 3. Let $H = langle x rangle = {e, x, x^2}$. - $o(y) = 2 implies y^2 = e, ; y eq e$. - $G$ is specified to be non-Abelian. Notice that the index of the subgroup $H$ in $G$ is: $$[G : H] = frac{o(G)}{o(H)} = frac{6}{3} = 2$$ --- 2. Normality of Subgroup $H$ Fundamental Lemma: Any subgroup of index 2 in a group $G$ is a normal subgroup of $G$. Proof: Since $[G : H] = 2$, there are exactly two left cosets of $H$ in $G$ and two right cosets of $H$ in $G$. The left cosets of $H$ are $H$ and $yH$ (since $y otin H$). Because cosets partition $G$: $$G = H cup yH, quad text{with } H cap yH = emptyset$$ Similarly, the right cosets of $H$ are $H$ and $Hy$: $$G = H cup Hy, quad text{with } H cap Hy = emptyset$$ Therefore: $$yH = G setminus H = Hy$$ Since $gH = Hg$ for all $g in G$ (for $g in H$, $gH = H = Hg$; for $g otin H$, $gH = yH = Hy = Hg$), $H$ is a normal subgroup of $G$: $$H triangleleft G$$ --- 3. Conjugation Action on $H$ Because $H = {e, x, x^2}$ is normal in $G$, for every $g in G$, the conjugate element $g x g^{-1}$ must belong to $H$: $$y x y^{-1} in H$$ Since $o(y) = 2$, we have $y = y^{-1}$, which implies: $$y x y = y x y^{-1} in {e, x, x^2}$$ Let us analyze the three possibilities for $y x y^{-1}$: 1. Case 1: $y x y^{-1} = e$ Multiplying on the left by $y^{-1}$ and on the right by $y$: $$x = y^{-1} e y = e$$ This contradicts the hypothesis that $o(x) = 3$. Hence, $y x y^{-1} eq e$. 2. Case 2: $y x y^{-1} = x$ Multiplying on the right by $y$: $$y x = x y$$ If $yx = xy$, then the two generators $x$ and $y$ commute. Since every element of $G$ is of the form $x^i y^j$ (where $i in {0, 1, 2}, j in {0, 1}$), any two elements of $G$ would commute: $$(x^a y^b)(x^c y^d) = x^a (y^b x^c) y^d = x^a (x^c y^b) y^d = x^{a+c} y^{b+d} = (

Q: 3. (b) Show that the volume of the greatest rectangular parallelopiped that can be inscribed in the ellipsoid $$frac{x^2}{a^2} + frac{y^2}{b^2} + frac{z^2}{c^2} = 1$$ is $frac{8abc}{3sqrt{3}}$. (20 Marks)

1. Geometric Formulation & Objective Function Let the equation of the triaxial ellipsoid be: $$frac{x^2}{a^2} + frac{y^2}{b^2} + frac{z^2}{c^2} = 1, quad a, b, c 0$$ Consider a rectangular parallelopiped inscribed in the ellipsoid with its faces parallel to the coordinate planes. By the symmetry of the ellipsoid with respect to the $xy$-, $yz$-, and $zx$-planes, the centroid of the inscribed parallelopiped must coincide with the origin $(0, 0, 0)$. Let the vertex of the parallelopiped situated in the first octant be $P(x, y, z)$, where: $$x 0, quad y 0, quad z 0$$ Since $P(x, y, z)$ lies on the surface of the ellipsoid, the coordinates must satisfy the constraint: $$phi(x, y, z) = frac{x^2}{a^2} + frac{y^2}{b^2} + frac{z^2}{c^2} - 1 = 0$$ The dimensions (edge lengths) of the rectangular parallelopiped are: $$text{Length} = 2x, quad text{Breadth} = 2y, quad text{Height} = 2z$$ The volume $V$ of the inscribed parallelopiped is therefore: $$V(x, y, z) = (2x)(2y)(2z) = 8xyz$$ We seek to maximize the objective function $V = 8xyz$ subject to the ellipsoid surface constraint $phi(x, y, z) = 0$. --- 2. Method of Lagrange Multipliers To simplify the calculus, maximizing $V = 8xyz$ is equivalent to maximizing: $$f(x, y, z) = ln V = ln 8 + ln x + ln y + ln z$$ subject to: $$phi(x, y, z) = frac{x^2}{a^2} + frac{y^2}{b^2} + frac{z^2}{c^2} - 1 = 0$$ We construct the Lagrangian function $mathcal{L}(x, y, z, lambda)$: $$mathcal{L}(x, y, z, lambda) = ln x + ln y + ln z - lambda left( frac{x^2}{a^2} + frac{y^2}{b^2} + frac{z^2}{c^2} - 1 right)$$ where $lambda$ is the Lagrange multiplier. For stationary points, we set the partial derivatives of $mathcal{L}$ with respect to $x, y, z, lambda$ equal to zero: $$frac{partial mathcal{L}}{partial x} = frac{1}{x} - frac{2lambda x}{a^2} = 0 implies frac{2lambda x^2}{a^2} = 1 implies frac{x^2}{a^2} = frac{1}{2lambda}$$ $$frac{partial mathcal{L}}{partial y} = frac{1}{y} - frac{2lambda y}{b^2} = 0 implies frac{2lambda y^2}{b^2} = 1 implies frac{y

Q: 3. (c) Apply the principle of duality to solve the following linear programming problem: Maximize $Z = 3x1 + 4x2$ subject to $x1 - x2 le 1$ $x1 + x2 ge 4$ $x1 - 3x2 le 3$ $x1, x2 ge 0$. (15 Marks)

1. Standard Form of Primal LPP The given primal linear programming problem (P) is: $$text{Maximize } Z = 3x1 + 4x2$$ subject to: $$begin{aligned} x1 - x2 &le 1 x1 + x2 &ge 4 x1 - 3x2 &le 3 x1, x2 &ge 0 end{aligned}$$ To construct the symmetric dual problem, all constraints in a maximization problem must be in the standard $le$ direction. Multiplying the second constraint $x1 + x2 ge 4$ by $-1$: $$-x1 - x2 le -4$$ Thus, the canonical primal problem is: $$text{Maximize } Z = 3x1 + 4x2$$ subject to: $$begin{aligned} x1 - x2 &le 1 -x1 - x2 &le -4 x1 - 3x2 &le 3 x1, x2 &ge 0 end{aligned}$$ --- 2. Formulation of the Dual Problem Let $y1, y2, y3 ge 0$ be the dual variables corresponding to the three primal constraints. The dual problem (D) is a minimization problem: $$text{Minimize } W = 1 y1 - 4 y2 + 3 y3$$ subject to: $$begin{aligned} y1 - y2 + y3 &ge 3 quad text{(coefficients of } x1) -y1 - y2 - 3y3 &ge 4 quad text{(coefficients of } x2) y1, y2, y3 &ge 0 end{aligned}$$ --- 3. Detailed Infeasibility Analysis of the Dual Let us carefully inspect the second dual constraint: $$-y1 - y2 - 3y3 ge 4$$ Multiplying by $-1$: $$y1 + y2 + 3y3 le -4$$ However, by the non-negativity restrictions of the dual variables: $$y1 ge 0, quad y2 ge 0, quad y3 ge 0$$ Since the sum of non-negative real numbers must be non-negative: $$y1 + y2 + 3y3 ge 0$$ Therefore, there is no vector $(y1, y2, y3) ge 0$ that can satisfy: $$0 le y1 + y2 + 3y3 le -4$$ This is a direct mathematical contradiction. Hence, the dual problem has no feasible solution (Dual is Infeasible). --- 4. Duality Theorems and Primal Behavior According to the Fundamental Duality Theorem of Linear Programming: - If both primal and dual problems are feasible, then both have optimal solutions, and $max Z = min W$. - If one problem is infeasible, the other problem is either unbounded or infeasible. To determine the status of the primal, let us analyze the primal constraints graphically or by ray analysis: $$begin{aligned} x1 - x2 &le

Question 1

1. (a) Let $H$ and $K$ be two subgroups of a group $G$ such that $o(H) sqrt{o(G)}$ and $o(K) sqrt{o(G)}$. Show that $H cap K eq {e}$, where $e$ is the identity element. Here $o(H), o(K)$ and $o(G)$ denote the order of $H, K$ and $G$ respectively. (10 Marks)

Accepted Answer

1. Conceptual Framework & Foundational Theorems Let $G$ be a finite group, and let $H$ and $K$ be two arbitrary subgroups of $G$. The intersection $H cap K$ is always a subgroup of both $H$ and $K$, and in particular, it contains at least the identity element $e in G$, so $o(H cap K) ge 1$. To establish the non-triviality of the intersection $H cap K$ under the given order constraints, we analyze the subset product: $$HK = {hk : h in H, ; k in K}$$ Although $HK$ is not necessarily a subgroup of $G$ (it is a subgroup if and only if $HK = KH$), it is a well-defined non-empty subset of the finite group $G$. A central theorem in finite group theory relates the cardinality of the product set $HK$ to the orders of $H$, $K$, and their intersection $H cap K$. --- 2. Product Cardinality Formula Theorem: For any two finite subgroups $H$ and $K$ of a group $G$, $$|HK| = frac{o(H) cdot o(K)}{o(H cap K)}$$ Rigorous Proof: Define a mapping $psi: H times K to HK$ given by $psi(h, k) = hk$. This map is clearly surjective by the definition of the set $HK$. We determine the number of pre-images for each element in $HK$. Suppose $h1 k1 = h2 k2$ for $h1, h2 in H$ and $k1, k2 in K$. Multiplying on the left by $h2^{-1}$ and on the right by $k1^{-1}$, we obtain: $$h2^{-1} h1 = k2 k1^{-1}$$ Let this common element be denoted by $d$: $$d = h2^{-1} h1 = k2 k1^{-1}$$ Since $h1, h2 in H$, we have $d = h2^{-1} h1 in H$. Similarly, since $k1, k2 in K$, we have $d = k2 k1^{-1} in K$. Therefore, $d in H cap K$. Conversely, for any $d in H cap K$, we can express: $$h2 = h1 d^{-1} in H quad text{and} quad k2 = d k1 in K$$ Notice that: $$h2 k2 = (h1 d^{-1})(d k1) = h1 (d^{-1} d) k1 = h1 k1$$ Thus, for every fixed pair $(h1, k1)$, exactly $o(H cap K)$ distinct pairs $(h2, k2) in H times K$ satisfy $h2 k2 = h1 k1$. Because the fibers of $psi$ partition $H times K$ into mutually disjoint sets of equal size $o(H cap K)$, we have: $$|H times K| = |HK| cdot o(H cap K)$$ Since $|H times K| = o(H) cdot o(K)$

Question 2

1. (e) How many basic solutions are there for the following system of equations? $$2x1 - x2 + 3x3 + x4 = 6$$ $$4x1 - 2x2 - x3 + 2x4 = 10$$ Find all of them. Furthermore, find the number of basic solutions, which are feasible/non-feasible/non-degenerate. (10 Marks)

Accepted Answer

1. Mathematical Formulation & Rank Analysis The given system of linear equations is: $$begin{aligned} 2x1 - x2 + 3x3 + x4 &= 6 4x1 - 2x2 - x3 + 2x4 &= 10 end{aligned}$$ In matrix form, $A mathbf{x} = mathbf{b}$: $$A = begin{bmatrix} 2 & -1 & 3 & 1 4 & -2 & -1 & 2 end{bmatrix}, quad mathbf{x} = begin{bmatrix} x1 x2 x3 x4 end{bmatrix}, quad mathbf{b} = begin{bmatrix} 6 10 end{bmatrix}$$ Here: - Number of constraints: $m = 2$. - Number of variables: $n = 4$. - Let the column vectors of $A$ be denoted by $mathbf{a}1, mathbf{a}2, mathbf{a}3, mathbf{a}4$: $$mathbf{a}1 = begin{bmatrix} 2 4 end{bmatrix}, quad mathbf{a}2 = begin{bmatrix} -1 -2 end{bmatrix}, quad mathbf{a}3 = begin{bmatrix} 3 -1 end{bmatrix}, quad mathbf{a}4 = begin{bmatrix} 1 2 end{bmatrix}$$ Notice that: $$mathbf{a}1 = -2 mathbf{a}2 = 2 mathbf{a}4$$ Columns $mathbf{a}1, mathbf{a}2, mathbf{a}4$ are all scalar multiples of one another. The rank of the coefficient matrix $A$ is $operatorname{rank}(A) = 2$, and $operatorname{rank}([A | mathbf{b}]) = 2$. --- 2. Theoretical Maximum & Basis Selection A basis matrix $B$ is a $2 times 2$ submatrix formed by selecting $m = 2$ columns from $A$ such that $det(B) eq 0$. The maximum theoretical number of basic solutions is: $$binom{n}{m} = binom{4}{2} = frac{4 times 3}{2} = 6$$ We systematically test all $binom{4}{2} = 6$ pairs of column vectors: 1. Pair ${mathbf{a}1, mathbf{a}2}$: $$det([mathbf{a}1, mathbf{a}2]) = begin{vmatrix} 2 & -1 4 & -2 end{vmatrix} = 2(-2) - 4(-1) = -4 + 4 = 0$$ Columns are linearly dependent. No basic solution exists. 2. Pair ${mathbf{a}1, mathbf{a}3}$: $$det([mathbf{a}1, mathbf{a}3]) = begin{vmatrix} 2 & 3 4 & -1 end{vmatrix} = -2 - 12 = -14 eq 0$$ Linearly independent $implies$ Basis 1. Setting non-basic variables $x2 = 0, x4 = 0$: $$begin{bmatrix} 2 & 3 4 & -1 end{bmatrix} begin{bmatrix} x1 x3 end{bmatrix} = begin{bmatrix} 6 10 end{bmatrix}$$ From $R2 - 2R1$: $-7x3 = 10 - 12 = -2 implies x3 = frac{2}{7}$. Then $2x1 = 6 - 3left(frac{2}{7}right

Question 3

2. (a) Define Cauchy sequence and prove that every convergent sequence of real numbers is a Cauchy sequence. What is the importance of Cauchy condition? (15 Marks)

Accepted Answer

1. Definition of a Cauchy Sequence Definition: A sequence of real numbers ${an}{n=1}^infty$ is called a Cauchy sequence if for every $varepsilon 0$, there exists a positive integer $N = N(varepsilon) in mathbb{N}$ such that for all integers $m, n ge N$, the distance between terms satisfies: $$|an - am| 0$, there exists an integer $N in mathbb{N}$ such that: $$|an - L| 0$ be given. Choose $varepsilon' = frac{varepsilon}{2} 0$. Since the sequence converges to $L$, corresponding to this $varepsilon'$, there exists an integer $N = N(varepsilon) in mathbb{N}$ such that: $$|ak - L| 0$, we have found an $N in mathbb{N}$ such that $|an - am| < varepsilon$ for all $m, n ge N$. This completes the proof that ${an}$ is a Cauchy sequence. --- 3. The Importance of the Cauchy Condition The Cauchy condition occupies a central place in real analysis and the foundations of mathematics for several profound reasons: 1. Intrinsic Criterion for Convergence: To establish convergence using the standard definition ($lim an = L$), one must already know or guess the candidate limit $L$ in advance. In contrast, the Cauchy condition involves only the mutual differences $|an - am|$ between the terms themselves. It allows analysts to prove that a sequence converges without needing to identify or compute the limit value beforehand. 2. Characterization of Completeness (Cauchy Completeness): A metric space $(M, d)$ is defined to be complete if every Cauchy sequence in $M$ converges to a limit that lies within $M$. - The rational numbers $mathbb{Q}$ are incomplete: For example, the sequence defined recursively by $x{n+1} = frac{xn}{2} + frac{1}{xn}$ with $x1 = 1$ is a Cauchy sequence of rational numbers whose limit is $sqrt{2} otin mathbb{Q}$. - The real numbers $mathbb{R}$ possess the Completeness Axiom (or Cauchy Criterion): Every Cauchy sequence of real numbers converges to a real limit. 3. Construction of the Real Numbers: Historically, Cantor constructed the real number system $mathbb{R}$ rigoro

Question 4

2. (c) Use the method of contour integration to prove that $$int{-infty}^infty frac{x^2 - x + 2}{x^4 + 10x^2 + 9},dx = frac{5pi}{12}.$$ (20 Marks)

Accepted Answer

1. Complex Extension & Contour Setup We are asked to evaluate the real improper integral: $$I = int{-infty}^infty frac{x^2 - x + 2}{x^4 + 10x^2 + 9} dx$$ Notice that the denominator is an even function: $$Q(x) = x^4 + 10x^2 + 9 = (x^2 + 1)(x^2 + 9)$$ The numerator can be split into an even part and an odd part: $$P(x) = x^2 - x + 2 = (x^2 + 2) - x$$ Since $frac{-x}{x^4 + 10x^2 + 9}$ is an odd function of $x$ and the integral converges, its Cauchy Principal Value vanishes identically: $$int{-infty}^infty frac{-x}{x^4 + 10x^2 + 9} dx = 0$$ Therefore: $$I = int{-infty}^infty frac{x^2 + 2}{x^4 + 10x^2 + 9} dx$$ However, we can directly apply contour integration to the full complex function $f(z)$: $$f(z) = frac{z^2 - z + 2}{z^4 + 10z^2 + 9}$$ We choose a standard semi-circular contour $GammaR$ in the upper half of the complex plane: $$GammaR = [-R, R] cup CR$$ where: - $[-R, R]$ is the real line segment from $-R$ to $R$. - $CR$ is the upper semi-circle $z = R e^{itheta}, ; theta in [0, pi]$, with $R 3$. By Cauchy's Residue Theorem: $$oint{GammaR} f(z) dz = int{-R}^R f(x) dx + int{CR} f(z) dz = 2pi i sum operatorname{Res}(f, zk)$$ where the sum is taken over all poles $zk$ lying strictly in the upper half-plane ($operatorname{Im}(zk) 0$). --- 2. Identification of Poles in the Upper Half-Plane The poles of $f(z)$ are the roots of the denominator: $$z^4 + 10z^2 + 9 = 0$$ Letting $u = z^2$: $$u^2 + 10u + 9 = 0 implies (u + 1)(u + 9) = 0$$ Thus $z^2 = -1$ or $z^2 = -9$, which yields 4 simple poles: $$z = pm i, quad z = pm 3i$$ The poles lying strictly in the upper half-plane ($operatorname{Im}(z) 0$) are: $$z1 = i quad text{and} quad z2 = 3i$$ --- 3. Calculation of Residues Since all poles are simple poles, the residue at a simple pole $z0$ can be evaluated as: $$operatorname{Res}(f, z0) = frac{P(z0)}{Q'(z0)}$$ where $P(z) = z^2 - z + 2$ and $Q(z) = z^4 + 10z^2 + 9$. The derivative of the denominator is: $$Q'(z) = 4z^3 + 20z = 4z(z^2 + 5)$$ Residue at $z1 = i$: $$P(i) = i^2

Question 5

2. (b) Show that $3$ is an irreducible element in the integral domain $mathbb{Z}[i]$. (15 Marks)

Accepted Answer

1. Algebraic Framework of the Gaussian Integers $mathbb{Z}[i]$ The ring of Gaussian integers is defined as: $$mathbb{Z}[i] = {a + bi : a, b in mathbb{Z}}, quad text{where } i^2 = -1$$ Since $mathbb{Z}[i]$ is a subring of the field of complex numbers $mathbb{C}$, it is an integral domain (it is commutative with identity $1$, and has no zero divisors). Furthermore, $mathbb{Z}[i]$ is a Euclidean domain, and hence a Principal Ideal Domain (PID) and a Unique Factorization Domain (UFD). We define the standard field norm $N: mathbb{Z}[i] to mathbb{Z}{ge 0}$ by: $$N(alpha) = N(a + bi) = alpha bar{alpha} = (a + bi)(a - bi) = a^2 + b^2$$ Properties of the Norm: 1. $N(alpha) ge 0$ for all $alpha in mathbb{Z}[i]$, and $N(alpha) = 0 iff alpha = 0$. 2. Multiplicativity: For any $alpha, beta in mathbb{Z}[i]$, $$N(alpha beta) = N(alpha) N(beta)$$ 3. Units of $mathbb{Z}[i]$: An element $u in mathbb{Z}[i]$ is a unit (invertible) if and only if $N(u) = 1$. $$a^2 + b^2 = 1 implies (a, b) in {(pm 1, 0), (0, pm 1)}$$ Thus, the group of units of $mathbb{Z}[i]$ is: $$mathcal{U}(mathbb{Z}[i]) = {1, -1, i, -i}$$ --- 2. Definition of an Irreducible Element Definition: An element $p$ in an integral domain $R$ is called irreducible if: 1. $p eq 0$, 2. $p$ is not a unit ($p otin mathcal{U}(R)$), and 3. Whenever $p = alpha beta$ for some $alpha, beta in R$, then either $alpha$ is a unit or $beta$ is a unit. --- 3. Step-by-Step Proof that $3$ is Irreducible in $mathbb{Z}[i]$ Step 1: Verification of Non-zero and Non-unit Status - $3 = 3 + 0i eq 0$. - The norm of $3$ is: $$N(3) = 3^2 + 0^2 = 9$$ Since $N(3) = 9 eq 1$, $3$ is not a unit in $mathbb{Z}[i]$. Step 2: Factorization Hypothesis Suppose $3$ can be factored as a product of two Gaussian integers $alpha, beta in mathbb{Z}[i]$: $$3 = alpha beta$$ where $alpha = a + bi$ and $beta = c + di$ with $a, b, c, d in mathbb{Z}$. Step 3: Taking Norms on Both Sides Applying the multiplicative property of the norm: $$N(3) = N(alpha beta) = N(alpha) N(beta)$

Question 6

1. (c) Examine whether the series $$sum{n=1}^infty frac{(-1)^{n-1}}{n}$$ is absolutely or conditionally convergent. (10 Marks)

Accepted Answer

1. Problem Statement & Classification of Convergence We are given the alternating infinite series: $$S = sum{n=1}^infty frac{(-1)^{n-1}}{n} = 1 - frac{1}{2} + frac{1}{3} - frac{1}{4} + frac{1}{5} - dots$$ A real series $sum an$ is defined to be: 1. Absolutely Convergent if the series of absolute values $sum |an|$ converges. 2. Conditionally Convergent if $sum an$ converges, but $sum |an|$ diverges to $+infty$. --- 2. Test for Absolute Convergence We consider the series of absolute values of the terms: $$sum{n=1}^infty left| frac{(-1)^{n-1}}{n} right| = sum{n=1}^infty frac{1}{n} = 1 + frac{1}{2} + frac{1}{3} + frac{1}{4} + dots$$ This is the classical harmonic series, which is a $p$-series of the form $sum{n=1}^infty frac{1}{n^p}$ with $p = 1$. Proof of Divergence via Integral Test: Consider the real-valued function $f(x) = frac{1}{x}$ for $x in [1, infty)$. 1. $f(x) 0$ for all $x ge 1$. 2. $f'(x) = -frac{1}{x^2} 0$ for all $n in mathbb{N}$. 2. The sequence ${un}$ is monotonically decreasing, i.e., $u{n+1} le un$ for all $n$. 3. $lim{n to infty} un = 0$. Verification of Conditions: 1. Positivity: For every $n ge 1$, $un = frac{1}{n} 0$. Condition holds. 2. Monotonicity: $$un - u{n+1} = frac{1}{n} - frac{1}{n+1} = frac{(n+1) - n}{n(n+1)} = frac{1}{n(n+1)} 0 quad forall n ge 1$$ Thus, $u{n+1} < un$ for all $n$, meaning ${un}$ is strictly decreasing. Condition holds. 3. Limit at Infinity: $$lim{n to infty} un = lim{n to infty} frac{1}{n} = 0$$ Condition holds. Since all three conditions of Leibniz's alternating series test are satisfied, the series $sum{n=1}^infty frac{(-1)^{n-1}}{n}$ converges. --- 4. Exact Sum Evaluation via Maclaurin Series The sum of this convergent alternating series can be found explicitly via the Taylor series expansion of $ln(1 + x)$ around $x = 0$: $$ln(1 + x) = x - frac{x^2}{2} + frac{x^3}{3} - frac{x^4}{4} + dots = sum{n=1}^infty frac{(-1)^{n-1} x^n}{n}, quad text{for } |x| < 1$$ By Abel's Limit Theorem, since the series converges at $x = 1$,

Question 7

1. (d) Expand $f(z) = frac{1}{(z+1)(z+3)}$ in a Laurent series valid for $1 < |z| < 3$. (10 Marks)

Accepted Answer

Re(z) Im(z) z = -1 z = -3 |z| = 1 |z| = 3 Annulus: 1 < |z| < 3 Figure: Annular region of convergence $1 1. Function Analysis & Partial Fraction Decomposition We are given the complex rational function: $$f(z) = frac{1}{(z+1)(z+3)}$$ The singularities of $f(z)$ are the zeros of its denominator: $$z + 1 = 0 implies z = -1$$ $$z + 3 = 0 implies z = -3$$ Both are simple poles located at $z = -1$ (distance $|z| = 1$ from the origin) and $z = -3$ (distance $|z| = 3$ from the origin). The specified region of expansion is the open annulus centered at the origin: $$mathcal{A} = {z in mathbb{C} : 1 1 iff frac{1}{|z|} 1$ Since $|z| 1$, we have $left|frac{1}{z}right| 1$. - Analytic Part: $-sum{n=0}^infty frac{(-1)^n}{2 cdot 3^{n+1}} z^n = -frac{1}{6} + frac{z}{18} - frac{z^2}{54} + dots$, converging for $|z| < 3$. Hence, the series converges absolutely and uniformly on any compact subset of the annular region $1 < |z| < 3$.

Question 8

1. (b) Let $G = {e, x, x^2, y, yx, yx^2}$ be a non-Abelian group with $o(x) = 3$ and $o(y) = 2$. Show that $xy = yx^2$ (where $e$ is the identity element of $G$ and $o(x), o(y)$ denote the order of the elements $x, y$ respectively). (10 Marks)

Accepted Answer

1. Group Structure & Order Analysis We are given a finite group $G$ containing exactly 6 elements: $$G = {e, x, x^2, y, yx, yx^2}$$ The orders of the generators are: - $o(x) = 3 implies x^3 = e$, and ${e, x, x^2}$ is a cyclic subgroup of order 3. Let $H = langle x rangle = {e, x, x^2}$. - $o(y) = 2 implies y^2 = e, ; y eq e$. - $G$ is specified to be non-Abelian. Notice that the index of the subgroup $H$ in $G$ is: $$[G : H] = frac{o(G)}{o(H)} = frac{6}{3} = 2$$ --- 2. Normality of Subgroup $H$ Fundamental Lemma: Any subgroup of index 2 in a group $G$ is a normal subgroup of $G$. Proof: Since $[G : H] = 2$, there are exactly two left cosets of $H$ in $G$ and two right cosets of $H$ in $G$. The left cosets of $H$ are $H$ and $yH$ (since $y otin H$). Because cosets partition $G$: $$G = H cup yH, quad text{with } H cap yH = emptyset$$ Similarly, the right cosets of $H$ are $H$ and $Hy$: $$G = H cup Hy, quad text{with } H cap Hy = emptyset$$ Therefore: $$yH = G setminus H = Hy$$ Since $gH = Hg$ for all $g in G$ (for $g in H$, $gH = H = Hg$; for $g otin H$, $gH = yH = Hy = Hg$), $H$ is a normal subgroup of $G$: $$H triangleleft G$$ --- 3. Conjugation Action on $H$ Because $H = {e, x, x^2}$ is normal in $G$, for every $g in G$, the conjugate element $g x g^{-1}$ must belong to $H$: $$y x y^{-1} in H$$ Since $o(y) = 2$, we have $y = y^{-1}$, which implies: $$y x y = y x y^{-1} in {e, x, x^2}$$ Let us analyze the three possibilities for $y x y^{-1}$: 1. Case 1: $y x y^{-1} = e$ Multiplying on the left by $y^{-1}$ and on the right by $y$: $$x = y^{-1} e y = e$$ This contradicts the hypothesis that $o(x) = 3$. Hence, $y x y^{-1} eq e$. 2. Case 2: $y x y^{-1} = x$ Multiplying on the right by $y$: $$y x = x y$$ If $yx = xy$, then the two generators $x$ and $y$ commute. Since every element of $G$ is of the form $x^i y^j$ (where $i in {0, 1, 2}, j in {0, 1}$), any two elements of $G$ would commute: $$(x^a y^b)(x^c y^d) = x^a (y^b x^c) y^d = x^a (x^c y^b) y^d = x^{a+c} y^{b+d} = (

Question 9

3. (b) Show that the volume of the greatest rectangular parallelopiped that can be inscribed in the ellipsoid $$frac{x^2}{a^2} + frac{y^2}{b^2} + frac{z^2}{c^2} = 1$$ is $frac{8abc}{3sqrt{3}}$. (20 Marks)

Accepted Answer

1. Geometric Formulation & Objective Function Let the equation of the triaxial ellipsoid be: $$frac{x^2}{a^2} + frac{y^2}{b^2} + frac{z^2}{c^2} = 1, quad a, b, c 0$$ Consider a rectangular parallelopiped inscribed in the ellipsoid with its faces parallel to the coordinate planes. By the symmetry of the ellipsoid with respect to the $xy$-, $yz$-, and $zx$-planes, the centroid of the inscribed parallelopiped must coincide with the origin $(0, 0, 0)$. Let the vertex of the parallelopiped situated in the first octant be $P(x, y, z)$, where: $$x 0, quad y 0, quad z 0$$ Since $P(x, y, z)$ lies on the surface of the ellipsoid, the coordinates must satisfy the constraint: $$phi(x, y, z) = frac{x^2}{a^2} + frac{y^2}{b^2} + frac{z^2}{c^2} - 1 = 0$$ The dimensions (edge lengths) of the rectangular parallelopiped are: $$text{Length} = 2x, quad text{Breadth} = 2y, quad text{Height} = 2z$$ The volume $V$ of the inscribed parallelopiped is therefore: $$V(x, y, z) = (2x)(2y)(2z) = 8xyz$$ We seek to maximize the objective function $V = 8xyz$ subject to the ellipsoid surface constraint $phi(x, y, z) = 0$. --- 2. Method of Lagrange Multipliers To simplify the calculus, maximizing $V = 8xyz$ is equivalent to maximizing: $$f(x, y, z) = ln V = ln 8 + ln x + ln y + ln z$$ subject to: $$phi(x, y, z) = frac{x^2}{a^2} + frac{y^2}{b^2} + frac{z^2}{c^2} - 1 = 0$$ We construct the Lagrangian function $mathcal{L}(x, y, z, lambda)$: $$mathcal{L}(x, y, z, lambda) = ln x + ln y + ln z - lambda left( frac{x^2}{a^2} + frac{y^2}{b^2} + frac{z^2}{c^2} - 1 right)$$ where $lambda$ is the Lagrange multiplier. For stationary points, we set the partial derivatives of $mathcal{L}$ with respect to $x, y, z, lambda$ equal to zero: $$frac{partial mathcal{L}}{partial x} = frac{1}{x} - frac{2lambda x}{a^2} = 0 implies frac{2lambda x^2}{a^2} = 1 implies frac{x^2}{a^2} = frac{1}{2lambda}$$ $$frac{partial mathcal{L}}{partial y} = frac{1}{y} - frac{2lambda y}{b^2} = 0 implies frac{2lambda y^2}{b^2} = 1 implies frac{y

Question 10

3. (c) Apply the principle of duality to solve the following linear programming problem: Maximize $Z = 3x1 + 4x2$ subject to $x1 - x2 le 1$ $x1 + x2 ge 4$ $x1 - 3x2 le 3$ $x1, x2 ge 0$. (15 Marks)

Accepted Answer

1. Standard Form of Primal LPP The given primal linear programming problem (P) is: $$text{Maximize } Z = 3x1 + 4x2$$ subject to: $$begin{aligned} x1 - x2 &le 1 x1 + x2 &ge 4 x1 - 3x2 &le 3 x1, x2 &ge 0 end{aligned}$$ To construct the symmetric dual problem, all constraints in a maximization problem must be in the standard $le$ direction. Multiplying the second constraint $x1 + x2 ge 4$ by $-1$: $$-x1 - x2 le -4$$ Thus, the canonical primal problem is: $$text{Maximize } Z = 3x1 + 4x2$$ subject to: $$begin{aligned} x1 - x2 &le 1 -x1 - x2 &le -4 x1 - 3x2 &le 3 x1, x2 &ge 0 end{aligned}$$ --- 2. Formulation of the Dual Problem Let $y1, y2, y3 ge 0$ be the dual variables corresponding to the three primal constraints. The dual problem (D) is a minimization problem: $$text{Minimize } W = 1 y1 - 4 y2 + 3 y3$$ subject to: $$begin{aligned} y1 - y2 + y3 &ge 3 quad text{(coefficients of } x1) -y1 - y2 - 3y3 &ge 4 quad text{(coefficients of } x2) y1, y2, y3 &ge 0 end{aligned}$$ --- 3. Detailed Infeasibility Analysis of the Dual Let us carefully inspect the second dual constraint: $$-y1 - y2 - 3y3 ge 4$$ Multiplying by $-1$: $$y1 + y2 + 3y3 le -4$$ However, by the non-negativity restrictions of the dual variables: $$y1 ge 0, quad y2 ge 0, quad y3 ge 0$$ Since the sum of non-negative real numbers must be non-negative: $$y1 + y2 + 3y3 ge 0$$ Therefore, there is no vector $(y1, y2, y3) ge 0$ that can satisfy: $$0 le y1 + y2 + 3y3 le -4$$ This is a direct mathematical contradiction. Hence, the dual problem has no feasible solution (Dual is Infeasible). --- 4. Duality Theorems and Primal Behavior According to the Fundamental Duality Theorem of Linear Programming: - If both primal and dual problems are feasible, then both have optimal solutions, and $max Z = min W$. - If one problem is infeasible, the other problem is either unbounded or infeasible. To determine the status of the primal, let us analyze the primal constraints graphically or by ray analysis: $$begin{aligned} x1 - x2 &le

Question 11

4. (c) The following table shows all the necessary information on the available supply to each warehouse, the requirement of each market and the unit transportation cost from each warehouse to each market: - Warehouse A: $[5, 2, 4, 3]$, Supply = $22$ - Warehouse B: $[4, 8, 1, 6]$, Supply = $15$ - Wa

Accepted Answer

1. Problem Formulation & Feasibility Check We have 3 warehouses (sources) $A, B, C$ ($m = 3$) and 4 markets (destinations) $I, II, III, IV$ ($n = 4$). Unit Cost Matrix $C = [c{ij}]$: $$begin{array}{c|cccc|c} text{Warehouse} & I & II & III & IV & text{Supply } (ai) hline A & 5 & 2 & 4 & 3 & 22 B & 4 & 8 & 1 & 6 & 15 C & 4 & 6 & 7 & 5 & 8 hline text{Requirement } (bj) & 7 & 12 & 17 & 9 & 45 end{array}$$ Total Supply: $sum ai = 22 + 15 + 8 = 45$. Total Requirement: $sum bj = 7 + 12 + 17 + 9 = 45$. Since $sum ai = sum bj = 45$, this is a balanced transportation problem. --- 2. Examination of the Given Initial Schedule The shipping clerk's schedule allocates: - $x{A, II} = 12$ - $x{A, III} = 1$ - $x{A, IV} = 9$ (Row $A$ sum: $12 + 1 + 9 = 22 = a1 ; checkmark$) - $x{B, III} = 15$ (Row $B$ sum: $15 = a2 ; checkmark$) - $x{C, I} = 7$ - $x{C, III} = 1$ (Row $C$ sum: $7 + 1 = 8 = a3 ; checkmark$) Column sums: - Market $I$: $x{C, I} = 7 = b1 ; checkmark$ - Market $II$: $x{A, II} = 12 = b2 ; checkmark$ - Market $III$: $x{A, III} + x{B, III} + x{C, III} = 1 + 15 + 1 = 17 = b3 ; checkmark$ - Market $IV$: $x{A, IV} = 9 = b4 ; checkmark$ Number of Allocations: The number of basic cells is $6$. For a non-degenerate basic feasible solution, the required number of allocations is: $$m + n - 1 = 3 + 4 - 1 = 6$$ Since exactly 6 cells are allocated and they do not form a closed loop, the given solution is a non-degenerate Basic Feasible Solution (BFS). Cost of Initial Schedule: $$Z0 = (12 times 2) + (1 times 4) + (9 times 3) + (15 times 1) + (7 times 4) + (1 times 7)$$ $$Z0 = 24 + 4 + 27 + 15 + 28 + 7 = 105$$ --- 3. Optimality Test via MODI Method ($ui + vj = c{ij}$) We define dual potentials $ui$ for rows ($i = A, B, C$) and $vj$ for columns ($j = I, II, III, IV$). For all basic (allocated) cells, $ui + vj = c{ij}$: 1. Cell $(A, III)$ is allocated: $uA + v{III} = 4$. Let us set $v{III} = 0$ as the reference. Then: $$uA = 4$$ 2. Cell $(A, II)$ is allocated: $$uA + v{II} = 2 implies 4 + v{

Question 12

4. (a) Examine whether the mapping $phi: mathbb{Z}[x] to mathbb{Z}$ defined by $phi(f(x)) = f(0)$, for $f(x) in mathbb{Z}[x]$, is a homomorphism. Deduce that the ideal $langle x rangle$ is a prime ideal in $mathbb{Z}[x]$, but not a maximal ideal in $mathbb{Z}[x]$. (15 Marks)

Accepted Answer

1. Ring Definition & Mapping Specification Let $mathbb{Z}[x]$ denote the ring of polynomials in an indeterminate $x$ with integer coefficients: $$f(x) = a0 + a1 x + a2 x^2 + dots + an x^n, quad ai in mathbb{Z}$$ Under standard polynomial addition and multiplication, $mathbb{Z}[x]$ is a commutative ring with identity $1$. The codomain is the ring of integers $mathbb{Z}$, which is an integral domain. The given evaluation mapping $phi: mathbb{Z}[x] to mathbb{Z}$ is defined by: $$phi(f(x)) = f(0) = a0$$ which evaluates each polynomial at $x = 0$ (extracting its constant term). --- 2. Proof that $phi$ is a Ring Homomorphism To show that $phi$ is a ring homomorphism, we must verify three properties for all $f(x), g(x) in mathbb{Z}[x]$: (i) Preservation of Addition: $phi(f(x) + g(x)) = phi(f(x)) + phi(g(x))$ Let $f(x) = sum{i=0}^n ai x^i$ and $g(x) = sum{j=0}^m bj x^j$. The sum polynomial is $(f + g)(x) = sum{k=0}^{max(n, m)} (ak + bk) x^k$. Evaluating at $x = 0$: $$phi(f(x) + g(x)) = (f + g)(0) = a0 + b0$$ On the other hand: $$phi(f(x)) + phi(g(x)) = f(0) + g(0) = a0 + b0$$ Hence, $phi(f(x) + g(x)) = phi(f(x)) + phi(g(x))$. (ii) Preservation of Multiplication: $phi(f(x) g(x)) = phi(f(x)) phi(g(x))$ The product polynomial is $(fg)(x) = sum{k=0}^{n+m} ck x^k$, where $ck = sum{i+j=k} ai bj$. In particular, the constant term of the product is $c0 = a0 b0$. Evaluating at $x = 0$: $$phi(f(x) g(x)) = (fg)(0) = c0 = a0 b0$$ On the other hand: $$phi(f(x)) cdot phi(g(x)) = f(0) cdot g(0) = a0 cdot b0$$ Hence, $phi(f(x) g(x)) = phi(f(x)) phi(g(x))$. (iii) Preservation of Multiplicative Identity: For the identity polynomial $e(x) = 1$, $phi(1) = 1 in mathbb{Z}$. Thus, $phi$ is a ring homomorphism. --- 3. Surjectivity & Kernel Determination Surjectivity: For every integer $c in mathbb{Z}$, consider the constant polynomial $f(x) = c in mathbb{Z}[x]$. $$phi(c) = c$$ Hence, $phi$ is a surjective (onto) ring homomorphism. Kernel of $phi$: $$ker(phi) = {f(x) in mathbb{Z}[x] : phi(f(x)) = 0

Question 13

3. (a) Evaluate the integral $$ointC frac{e^z}{z^2(z+1)^3},dz, quad C: |z|=2.$$ (15 Marks)

Accepted Answer

1. Singularity & Contour Analysis We are asked to evaluate the contour integral: $$I = ointC frac{e^z}{z^2(z+1)^3} dz$$ where $C$ is the circle $|z| = 2$ traversed once in the counter-clockwise (positive) direction. Let $f(z) = frac{e^z}{z^2(z+1)^3}$. The singularities of $f(z)$ are the roots of the denominator $z^2(z+1)^3 = 0$: 1. $z = 0$, which is a pole of order 2 (double pole). 2. $z = -1$, which is a pole of order 3 (triple pole). The moduli of these singular points are: $$|0| = 0 < 2, quad |-1| = 1 < 2$$ Both singularities lie strictly inside the contour $C: |z| = 2$. By Cauchy's Residue Theorem: $$ointC f(z) dz = 2pi i left[ operatorname{Res}(f, 0) + operatorname{Res}(f, -1) right]$$ --- 2. Residue at the Double Pole $z = 0$ For a pole of order $m = 2$ at $z0 = 0$, the residue formula is: $$operatorname{Res}(f, 0) = frac{1}{(2-1)!} lim{z to 0} frac{d}{dz} left[ (z - 0)^2 f(z) right] = lim{z to 0} frac{d}{dz} left[ frac{e^z}{(z+1)^3} right]$$ Let $g(z) = frac{e^z}{(z+1)^3} = e^z (z+1)^{-3}$. Differentiating with respect to $z$ using the product rule: $$g'(z) = frac{d}{dz}left(e^zright) (z+1)^{-3} + e^z frac{d}{dz}left((z+1)^{-3}right) = e^z (z+1)^{-3} - 3e^z (z+1)^{-4}$$ $$g'(z) = frac{e^z (z+1) - 3e^z}{(z+1)^4} = frac{e^z (z - 2)}{(z+1)^4}$$ Taking the limit as $z to 0$: $$operatorname{Res}(f, 0) = g'(0) = frac{e^0 (0 - 2)}{(0 + 1)^4} = frac{1 cdot (-2)}{1} = -2$$ Alternatively, by Taylor series expansion around $z = 0$: $$e^z = 1 + z + frac{z^2}{2!} + dots$$ $$(1+z)^{-3} = 1 - 3z + 6z^2 - dots$$ $$(z^2 f(z)) = e^z (1+z)^{-3} = (1 + z + dots)(1 - 3z + dots) = 1 - 2z + mathcal{O}(z^2)$$ The coefficient of $z$ in $z^2 f(z)$ is $-2$, confirming $operatorname{Res}(f, 0) = -2$. --- 3. Residue at the Triple Pole $z = -1$ For a pole of order $m = 3$ at $z0 = -1$, the residue formula is: $$operatorname{Res}(f, -1) = frac{1}{(3-1)!} lim{z to -1} frac{d^2}{dz^2} left[ (z+1)^3 f(z) right] = frac{1}{2} lim{z to -1} frac{d^2}{dz^2} left[ frac{e^z}{z^2} right]$$ Let $h(z)

Question 14

4. (b) Prove that every continuous function is Riemann integrable. (15 Marks)

Accepted Answer

1. Mathematical Framework & Integrability Criterion Let $[a, b]$ be a closed and bounded (compact) interval in $mathbb{R}$ with $a 0$$ $$Mi = sup{x in [x{i-1}, xi]} f(x), quad mi = inf{x in [x{i-1}, xi]} f(x)$$ The Upper Darboux Sum $U(P, f)$ and Lower Darboux Sum $L(P, f)$ are defined as: $$U(P, f) = sum{i=1}^n Mi Delta xi, quad L(P, f) = sum{i=1}^n mi Delta xi$$ Riemann's Criterion for Integrability: A bounded function $f: [a, b] to mathbb{R}$ is Riemann integrable on $[a, b]$ ($f in mathcal{R}[a, b]$) if and only if for every $varepsilon 0$, there exists a partition $P$ of $[a, b]$ such that: $$U(P, f) - L(P, f) 0$ be given. For each $x in [a, b]$, by ordinary continuity, there exists $deltax 0$ such that $|f(t) - f(x)| 0$. If $y, z in [a, b]$ with $|y - z| 0$ be given. Since $b - a 0$, we consider the positive quantity: $$varepsilon' = frac{varepsilon}{b - a} 0$$ By uniform continuity of $f$ on $[a, b]$, corresponding to $varepsilon'$, there exists a $delta 0$ such that: $$|f(y) - f(z)| frac{b - a}{delta}$ subintervals of length $Delta xi = frac{b - a}{n} 0$, we have explicitly constructed a partition $P$ of $[a, b]$ such that: $$U(P, f) - L(P, f) < varepsilon$$ By Riemann's criterion, $f$ is Riemann integrable on $[a, b]$. Since $[a, b]$ was an arbitrary closed bounded interval and $f$ an arbitrary continuous function, it is established that every continuous function on a closed bounded interval is Riemann integrable.

Question 15

6. (a) Solve $$frac{partial^2 u}{partial x^2} + frac{partial^2 u}{partial y^2} = 0$$ for a rectangular plate subject to the boundary conditions: $u(0, y) = 0$, $u(a, y) = 0$, $u(x, 0) = 0$, $u(x, b) = f(x)$. (20 Marks)

Accepted Answer

1. Governing Equation & Boundary Value Problem We are tasked with solving Laplace's equation in two dimensions on a rectangular domain $Omega = {(x, y) : 0 0$). $X(x) = c1 cosh(px) + c2 sinh(px)$. $X(0) = c1 = 0 implies X(x) = c2 sinh(px)$. $X(a) = c2 sinh(pa) = 0 implies c2 = 0$ (since $sinh(pa) eq 0$ for $pa 0$). Only the trivial solution exists. 2. Case $lambda = 0$: $X(x) = c1 x + c2$. $X(0) = c2 = 0 implies X(a) = c1 a = 0 implies c1 = 0$. Only the trivial solution exists. 3. Case $lambda 0$: Let $lambda = p^2$ ($p 0$). The general solution is: $$X(x) = c1 cos(px) + c2 sin(px)$$ $X(0) = c1 = 0 implies X(x) = c2 sin(px)$. $X(a) = c2 sin(pa) = 0$. For non-trivial solutions ($c2 eq 0$): $$sin(pa) = 0 implies pa = npi implies pn = frac{npi}{a}, quad n = 1, 2, 3, dots$$ Thus, the eigenvalues and corresponding eigenfunctions are: $$lambdan = left( frac{npi}{a} right)^2, quad Xn(x) = sinleft( frac{npi x}{a} right), quad n in mathbb{N}$$ --- 4. Solving the ODE in $y$ For each eigenvalue $lambdan = left( frac{npi}{a} right)^2$, the equation for $Y(y)$ becomes: $$Yn''(y) - left( frac{npi}{a} right)^2 Yn(y) = 0$$ The general solution is: $$Yn(y) = An coshleft( frac{npi y}{a} right) + Bn sinhleft( frac{npi y}{a} right)$$ Applying the third boundary condition at the bottom edge ($y = 0$): $$u(x, 0) = X(x) Y(0) = 0 implies Yn(0) = 0$$ $$Yn(0) = An cosh(0) + Bn sinh(0) = An cdot 1 + 0 = An = 0$$ Therefore: $$Yn(y) = Bn sinhleft( frac{npi y}{a} right)$$ --- 5. Superposition Principle & Non-Homogeneous Boundary Condition By the principle of linear superposition, the general solution satisfying the first three boundary conditions is: $$u(x, y) = sum{n=1}^infty Cn sinleft( frac{npi x}{a} right) sinhleft( frac{npi y}{a} right)$$ where $Cn = c2 Bn$ are arbitrary constants. Now, we apply the fourth boundary condition at the top edge $y = b$: $$u(x, b) = f(x) = sum{n=1}^infty left[ Cn sinhleft( frac{npi b}{a} right) right] sinleft( frac{npi x}{a} right)$$ This is the Fourier Sine Ser

Question 16

6. (c) Calculate the moment of inertia of a uniform solid cylinder of mass $M$, radius $R$ and length $L$ with respect to a set of axes passing through the centre of the cylinder, where $z$-axis is the axis of the cylinder and $rho$ is the constant density at any point of the cylinder. Also find $L/

Accepted Answer

1. Geometric Setup & Mass-Density Relation Consider a uniform solid cylinder of: - Total mass $M$ - Base radius $R$ - Length (height) $L$ - Uniform volume density $rho$ Let the origin $(0, 0, 0)$ be situated at the center of mass $G$ of the cylinder: - The cylinder's longitudinal axis of rotational symmetry is chosen as the $z$-axis. - The cross-sectional plane at the center is the $xy$-plane. - The cylinder extends along the $z$-axis from $z = -L/2$ to $z = +L/2$. - In cylindrical coordinates $(r, theta, z)$: $$0 le r le R, quad 0 le theta le 2pi, quad -frac{L}{2} le z le frac{L}{2}$$ The differential volume element is $dV = r , dr , dtheta , dz$. The total volume of the cylinder is: $$V = pi R^2 L$$ Since the cylinder is uniform, the constant density is: $$rho = frac{M}{V} = frac{M}{pi R^2 L}$$ --- 2. Calculation of Moment of Inertia About the $z$-Axis ($Iz$) The distance of an infinitesimal mass element $dm = rho dV$ from the $z$-axis is $r$. $$Iz = iiint r^2 dm = rho int{-L/2}^{L/2} dz int0^{2pi} dtheta int0^R r^3 dr$$ $$Iz = rho cdot [z]{-L/2}^{L/2} cdot [theta]0^{2pi} cdot left[ frac{r^4}{4} right]0^R = rho cdot L cdot 2pi cdot frac{R^4}{4} = frac{1}{2} pi rho L R^4$$ Substituting $rho = frac{M}{pi R^2 L}$: $$Iz = frac{1}{2} pi left( frac{M}{pi R^2 L} right) L R^4 = frac{1}{2} M R^2$$ --- 3. Calculation of Moment of Inertia About the $x$- and $y$-Axes ($Ix$ and $Iy$) By the rotational symmetry of the cylinder about the $z$-axis: $$Ix = Iy$$ The distance squared of any point $(x, y, z)$ from the $x$-axis is $y^2 + z^2$. $$Ix = iiint (y^2 + z^2) dm = iiint y^2 dm + iiint z^2 dm$$ By symmetry in the $xy$-plane, $iiint x^2 dm = iiint y^2 dm$. Since $Iz = iiint (x^2 + y^2) dm = 2 iiint y^2 dm$: $$iiint y^2 dm = frac{1}{2} Iz = frac{1}{2} left( frac{1}{2} M R^2 right) = frac{1}{4} M R^2$$ Now, evaluating the second integral $iiint z^2 dm$: $$iiint z^2 dm = rho int0^R r , dr int0^{2pi} dtheta int{-L/2}^{L/2} z^2 dz$$ $$= rho left( frac{R^2}{2} right) (2pi) left[ frac

Question 17

5. (a) Find the solution of the equation $(D^2 + DD' - 2D'^2)z = y sin x$, where $D equiv frac{partial}{partial x}$ and $D' equiv frac{partial}{partial y}$. (10 Marks)

Accepted Answer

1. PDE Classification & Method of Solution We are given the second-order linear non-homogeneous partial differential equation with constant coefficients: $$(D^2 + DD' - 2D'^2)z = y sin x$$ where $D = frac{partial}{partial x}$ and $D' = frac{partial}{partial y}$. The general complete solution consists of the sum of the Complementary Function ($zc$) and the Particular Integral ($zp$): $$z(x, y) = zc(x, y) + zp(x, y)$$ --- 2. Finding the Complementary Function ($zc$) The homogeneous equation is: $$(D^2 + DD' - 2D'^2)z = 0$$ Factorizing the differential operator: $$D^2 + DD' - 2D'^2 = (D - D')(D + 2D')$$ To determine the roots, we substitute $D = m$ and $D' = 1$ to obtain the auxiliary equation: $$m^2 + m - 2 = 0 implies (m - 1)(m + 2) = 0$$ The roots are: $$m1 = 1, quad m2 = -2$$ For distinct real roots $m1, m2$, the standard complementary function is: $$zc(x, y) = phi1(y + m1 x) + phi2(y + m2 x)$$ $$zc(x, y) = phi1(y + x) + phi2(y - 2x)$$ where $phi1$ and $phi2$ are arbitrary, twice-differentiable real functions. --- 3. Finding the Particular Integral ($zp$) The particular integral is given by: $$zp = frac{1}{D^2 + DD' - 2D'^2} (y sin x) = frac{1}{(D - D')(D + 2D')} (y sin x)$$ We utilize the method of expanding the operator in powers of $D'/D$, since the differential operator acts on a product of a polynomial in $y$ and a trigonometric function in $x$: $$frac{1}{D^2 + DD' - 2D'^2} = frac{1}{D^2 left(1 + frac{D'}{D} - 2frac{D'^2}{D^2}right)} = frac{1}{D^2} left[ 1 + left(frac{D'}{D} - 2frac{D'^2}{D^2}right) right]^{-1}$$ Using the binomial expansion $(1 + u)^{-1} = 1 - u + u^2 - dots$: $$left[ 1 + left(frac{D'}{D} - 2frac{D'^2}{D^2}right) right]^{-1} = 1 - frac{D'}{D} + mathcal{O}(D'^2)$$ Because the operand $y sin x$ is linear in $y$, all higher partial derivatives with respect to $y$ vanish identically: $$D'(y sin x) = sin x, quad D'^2(y sin x) = 0, quad D'^k(y sin x) = 0 ; forall k ge 2$$ Therefore, keeping terms up to the first power of $D'$: $$zp = frac{1}{D^2} l

Question 18

5. (b) Solve the following system of linear equations by Gauss-Seidel method: $$10x + 2y + z = 9$$ $$2x + 20y - 2z = -44$$ $$-2x + 3y + 10z = 22$$ (10 Marks)

Accepted Answer

1. Verification of Strict Diagonal Dominance The given system of linear equations is: $$begin{aligned} 10x + 2y + z &= 9 2x + 20y - 2z &= -44 -2x + 3y + 10z &= 22 end{aligned}$$ The coefficient matrix is: $$A = begin{bmatrix} 10 & 2 & 1 2 & 20 & -2 -2 & 3 & 10 end{bmatrix}, quad mathbf{b} = begin{bmatrix} 9 -44 22 end{bmatrix}$$ For the Gauss-Seidel iterative method to guarantee convergence for any initial guess, the matrix $A$ should be strictly diagonally dominant, meaning: $$|a{ii}| sum{j eq i} |a{ij}| quad text{for each row } i$$ Checking each row: - Row 1: $|a{11}| = |10| = 10$; $|a{12}| + |a{13}| = |2| + |1| = 3$. Since $10 3$, condition holds. - Row 2: $|a{22}| = |20| = 20$; $|a{21}| + |a{23}| = |2| + |-2| = 4$. Since $20 4$, condition holds. - Row 3: $|a{33}| = |10| = 10$; $|a{31}| + |a{32}| = |-2| + |3| = 5$. Since $10 5$, condition holds. Since $A$ is strictly diagonally dominant, the Gauss-Seidel method is guaranteed to converge rapidly to the unique solution. --- 2. Gauss-Seidel Iteration Formulae We express the diagonal variables explicitly from each row: $$begin{aligned} x^{(k+1)} &= frac{1}{10} left( 9 - 2y^{(k)} - z^{(k)} right) = 0.9 - 0.2 y^{(k)} - 0.1 z^{(k)} y^{(k+1)} &= frac{1}{20} left( -44 - 2x^{(k+1)} + 2z^{(k)} right) = -2.2 - 0.1 x^{(k+1)} + 0.1 z^{(k)} z^{(k+1)} &= frac{1}{10} left( 22 + 2x^{(k+1)} - 3y^{(k+1)} right) = 2.2 + 0.2 x^{(k+1)} - 0.3 y^{(k+1)} end{aligned}$$ Notice that in the Gauss-Seidel method, the newly computed value of $x^{(k+1)}$ is immediately used to compute $y^{(k+1)}$, and both updated values $x^{(k+1)}$ and $y^{(k+1)}$ are immediately used to compute $z^{(k+1)}$. --- 3. Step-by-Step Iterations We choose the standard initial approximation: $$mathbf{x}^{(0)} = (x^{(0)}, y^{(0)}, z^{(0)}) = (0, 0, 0)$$ Iteration 1 ($k = 0$): - $x^{(1)} = 0.9 - 0.2(0) - 0.1(0) = 0.9000$ - $y^{(1)} = -2.2 - 0.1(0.9000) + 0.1(0) = -2.2 - 0.09 = -2.2900$ - $z^{(1)} = 2.2 + 0.2(0.9000) - 0.3(-2.2900) = 2.2 + 0.18 + 0.687 = 3.0670$ $$mathbf{

Question 19

5. (c) (i) Convert the number $(3479){10}$ into binary system and the number $(7AE.9F){16}$ into decimal system. (ii) Determine the truth table for the Boolean function $F(x, y, z) = (x + y + z')(x' + y')$. Also derive the full disjunctive normal form of $F(x, y, z)$ from the truth table. (10 Marks)

Accepted Answer

Part (i): Number System Conversions 1. Conversion of $(3479){10}$ to Binary We use the successive division by 2 method, recording the remainders from least significant bit (LSB) to most significant bit (MSB): - $3479 div 2 = 1739$, remainder 1 (LSB) - $1739 div 2 = 869$, remainder 1 - $869 div 2 = 434$, remainder 1 - $434 div 2 = 217$, remainder 0 - $217 div 2 = 108$, remainder 1 - $108 div 2 = 54$, remainder 0 - $54 div 2 = 27$, remainder 0 - $27 div 2 = 13$, remainder 1 - $13 div 2 = 6$, remainder 1 - $6 div 2 = 3$, remainder 0 - $3 div 2 = 1$, remainder 1 - $1 div 2 = 0$, remainder 1 (MSB) Reading the remainders from bottom to top (MSB to LSB): $$(3479){10} = mathbf{(110110010111)2}$$ Verification: $$2^{11} + 2^{10} + 2^8 + 2^7 + 2^4 + 2^2 + 2^1 + 2^0 = 2048 + 1024 + 256 + 128 + 16 + 4 + 2 + 1 = 3479 ; checkmark$$ --- 2. Conversion of $(7text{AE}cdot 9text{F}){16}$ to Decimal The hexadecimal digits have values: $text{A} = 10$, $text{E} = 14$, $text{F} = 15$. We expand in positional powers of 16: $$(7text{AE}cdot 9text{F}){16} = 7 cdot 16^2 + 10 cdot 16^1 + 14 cdot 16^0 + 9 cdot 16^{-1} + 15 cdot 16^{-2}$$ Calculating integer part: $$7 cdot 256 + 10 cdot 16 + 14 cdot 1 = 1792 + 160 + 14 = 1966$$ Calculating fractional part: $$frac{9}{16} + frac{15}{256} = 0.5625 + 0.05859375 = 0.62109375$$ Combining both parts: $$(7text{AE}cdot 9text{F}){16} = mathbf{1966.62109375{10}}$$ --- Part (ii): Truth Table and Full Disjunctive Normal Form (DNF) We are given the Boolean function of three variables $x, y, z$: $$F(x, y, z) = (x + y + z')(x' + y')$$ where $+$ represents Boolean OR, juxtaposition represents Boolean AND, and prime $(')$ represents Boolean NOT. 1. Truth Table Construction We evaluate $F$ for all $2^3 = 8$ combinations of binary inputs $(x, y, z)$: | Row | $x$ | $y$ | $z$ | $x'$ | $y'$ | $z'$ | $C1 = x + y + z'$ | $C2 = x' + y'$ | $F = C1 cdot C2$ | Minterm | | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | | 0 | 0 | 0 | 0

Question 20

5. (d) A bead of mass $m$ slides on a frictionless wire in the shape of a cycloid given by $x = a(theta - sintheta)$, $y = a(1 + costheta)$, $(0 le theta le 2pi)$. Find the Lagrangian function. Hence show that the equation of motion can be written as $$frac{d^2u}{dt^2} + frac{g}{4a} u = 0,$$ where $

Accepted Answer

1. Geometric & Kinematic Description of the Cycloid We are given a bead of mass $m$ sliding without friction along a wire constrained to the cycloid: $$x = a(theta - sintheta), quad y = a(1 + costheta), quad 0 le theta le 2pi$$ Here, $theta$ serves as the single generalized coordinate ($q = theta$) of this one-degree-of-freedom holonomic system. We differentiate the parametric equations with respect to time $t$ to obtain the velocity components: $$dot{x} = frac{dx}{dt} = a(1 - costheta) dot{theta}$$ $$dot{y} = frac{dy}{dt} = -a sintheta dot{theta}$$ --- 2. Kinetic Energy ($T$) The kinetic energy of the bead is: $$T = frac{1}{2} m (dot{x}^2 + dot{y}^2)$$ Evaluating $dot{x}^2 + dot{y}^2$: $$dot{x}^2 + dot{y}^2 = a^2 dot{theta}^2 left[ (1 - costheta)^2 + sin^2theta right]$$ $$dot{x}^2 + dot{y}^2 = a^2 dot{theta}^2 (1 - 2costheta + cos^2theta + sin^2theta) = a^2 dot{theta}^2 (2 - 2costheta)$$ Using the trigonometric identity $1 - costheta = 2sin^2left(frac{theta}{2}right)$: $$dot{x}^2 + dot{y}^2 = 4a^2 sin^2left(frac{theta}{2}right) dot{theta}^2$$ Therefore, the kinetic energy is: $$T = frac{1}{2} m left[ 4a^2 sin^2left(frac{theta}{2}right) dot{theta}^2 right] = 2ma^2 sin^2left(frac{theta}{2}right) dot{theta}^2$$ --- 3. Potential Energy ($V$) & Lagrangian ($L$) Taking the horizontal datum at $y = 0$ with gravity acting downward in the direction of decreasing $y$: $$V = mgy = mga(1 + costheta) = 2mga cos^2left(frac{theta}{2}right)$$ The Lagrangian function $L = T - V$ is: $$L(theta, dot{theta}) = 2ma^2 sin^2left(frac{theta}{2}right) dot{theta}^2 - mga(1 + costheta)$$ or equivalently: $$mathbf{L = 2ma^2 sin^2left(frac{theta}{2}right) dot{theta}^2 - 2mga cos^2left(frac{theta}{2}right)}$$ --- 4. Lagrange's Equation of Motion Euler-Lagrange equation for the generalized coordinate $theta$: $$frac{d}{dt} left( frac{partial L}{partial dot{theta}} right) - frac{partial L}{partial theta} = 0$$ Computing partial derivatives: $$frac{partial L}{partial dot{theta}} = 4ma^2 sin^2left(

Question 21

5. (e) A source and a sink of equal strength are placed at points $(pm a/2, 0)$ within a fixed circular boundary $x^2 + y^2 = a^2$. Show that the streamlines are given by $$(r^2 - a^2/4)(r^2 - 4a^2) - 4a^2 y^2 = ky(r^2 - a^2),$$ where $k$ is a constant and $r^2 = x^2 + y^2$. (10 Marks)

Accepted Answer

1. Hydrodynamic Setup & Milne-Thomson Circle Theorem We consider two-dimensional irrotational, incompressible fluid flow inside a circular cylinder of radius $a$: $$C: |z| = a$$ Inside this circular boundary, we are given: - A source of strength $m$ situated at $z1 = frac{a}{2}$. - A sink of equal strength $m$ situated at $z2 = -frac{a}{2}$. Without the circular boundary, the complex potential for the source at $z1$ and sink at $z2$ in an unbounded fluid would be: $$w0(z) = -m lnleft(z - frac{a}{2}right) + m lnleft(z + frac{a}{2}right) = m lnleft( frac{z + a/2}{z - a/2} right)$$ To account for the impermeable circular boundary $|z| = a$, we use the Milne-Thomson Circle Theorem: If $w = f(z)$ is the complex potential for a flow having no singularities on or outside $|z| = a$, then introducing the circular boundary $|z| = a$ modifies the complex potential to: $$w = f(z) + overline{fleft(frac{a^2}{bar{z}}right)}$$ --- 2. Method of Images for the Circular Cylinder By the circle theorem, the image of a source of strength $m$ at $z0$ with $|z0| < a$ consists of: 1. An equal source $+m$ at the inverse point $z0' = frac{a^2}{bar{z}0} = frac{a^2}{z0}$ (since $z0$ is real). 2. An equal sink $-m$ at the origin $z = 0$. Let us apply this to both singularities: - For the source $+m$ at $z1 = frac{a}{2}$: - Image source $+m$ at inverse point $z1' = frac{a^2}{a/2} = 2a$. - Image sink $-m$ at the origin $z = 0$. - For the sink $-m$ at $z2 = -frac{a}{2}$: - Image sink $-m$ at inverse point $z2' = frac{a^2}{-a/2} = -2a$. - Image source $+m$ at the origin $z = 0$. The image sink $-m$ at the origin from the first source and the image source $+m$ at the origin from the second sink cancel out completely! Thus, the total flow inside $|z| le a$ is produced by the system of four singularities: 1. Source $+m$ at $z = frac{a}{2}$ 2. Sink $-m$ at $z = -frac{a}{2}$ 3. Image source $+m$ at $z = 2a$ 4. Image sink $-m$ at $z = -2a$ --- 3. Formulation of Total Complex Potential $w(z)$ The complete

Question 22

6. (b) Simplify the Boolean function $F(x, y, z) = xyz + x'yz + xy'z + xyz'$ and draw the corresponding GATE network. (15 Marks)

Accepted Answer

x y z OR x + y AND F = (x + y)z Figure: Minimized logic gate network realizing $F(x, y, z) = (x + y)z = yz + xz$. 1. Algebraic Specification of the Boolean Function We are given the Boolean function of three variables $x, y, z$: $$F(x, y, z) = x y z + x' y z + x y' z + x y z'$$ Each term in the sum is a fundamental product (minterm): - $m7 = x y z$ (binary $111$) - $m3 = x' y z$ (binary $011$) - $m5 = x y' z$ (binary $101$) - $m6 = x y z'$ (binary $110$) Thus: $$F(x, y, z) = sum m(3, 5, 6, 7)$$ --- 2. Method 1: Boolean Algebraic Simplification By the idempotence law of Boolean algebra, for any Boolean expression $A$: $$A + A = A$$ We can duplicate the minterm $x y z$ twice without changing the truth value of $F$: $$F = (x y z + x' y z) + (x y z + x y' z) + (x y z + x y z')$$ Applying the distributive law $AB + AC = A(B + C)$ to each grouped pair: 1. First pair: $$x y z + x' y z = (x + x') y z = 1 cdot y z = y z$$ 2. Second pair: $$x y z + x y' z = x (y + y') z = x cdot 1 cdot z = x z$$ 3. Third pair: $$x y z + x y z' = x y (z + z') = x y cdot 1 = x y$$ Combining the simplified terms: $$mathbf{F(x, y, z) = x y + y z + x z}$$ This simplified expression is the famous Majority Function (it evaluates to $1$ if and only if two or more of the three inputs $x, y, z$ are equal to $1$). --- 3. Method 2: Karnaugh Map (K-Map) Simplification We construct a 3-variable Karnaugh Map with $x$ along the rows and $y z$ along the columns (using Gray code ordering $00, 01, 11, 10$): $$begin{array}{c|c|c|c|c} x backslash yz & 00 & 01 & 11 & 10 hline 0 & 0 & 0 & 1 & 0 hline 1 & 0 & 1 & 1 & 1 end{array}$$ Min-term placements: - Cell $(0, 11) = m3 = 1$ - Cell $(1, 01) = m5 = 1$ - Cell $(1, 11) = m7 = 1$ - Cell $(1, 10) = m6 = 1$ Grouping of 1s (Prime Implicants): 1. Group 1 (Horizontal pair in Row 1): Cells $(1, 11)$ and $(1, 10)$ combine ($m7 + m6$). $x$ remains $1$, $y$ remains $1$, $z$ changes from $1$ to $0$. Term = $mathbf{x y}$. 2. Group 2 (Vertical pair in Column 11): Cells $(0, 11)$

Question 23

8. (b) Find the constant $p$ and error term for the quadrature formula $$int{x0}^{x1} f(x),dx = frac{h}{2}(f0 + f1) + ph^2(f'0 - f'1)$$ where $x0 + h = x1$, $f0 = f(x0)$, $f1 = f(x1)$ and prime ($'$) represents derivative with respect to $x$. Hence deduce the composite rule for integrating $inta^b f

Accepted Answer

1. Quadrature Formula & Taylor Series Expansion We are given the Hermite-type quadrature formula on a single subinterval $[x0, x1]$ of length $h = x1 - x0$: $$I = int{x0}^{x1} f(x) dx approx frac{h}{2}(f0 + f1) + p h^2(f'0 - f'1)$$ where $f0 = f(x0), f1 = f(x1), f'0 = f'(x0), f'1 = f'(x1)$, and $p$ is an undetermined real constant. We expand $f(x)$ in a Taylor series around the midpoint $x{1/2} = x0 + frac{h}{2} = m$: Let $t = x - m$, so that $x = m + t$. As $x$ ranges from $x0 = m - h/2$ to $x1 = m + h/2$, $t$ ranges from $-h/2$ to $+h/2$. Expanding $f(m + t)$ in powers of $t$: $$f(m + t) = f(m) + t f'(m) + frac{t^2}{2} f''(m) + frac{t^3}{6} f'''(m) + frac{t^4}{24} f^{(4)}(m) + dots$$ --- 2. Exact Value of the Integral Integrating $f(m + t)$ from $t = -h/2$ to $+h/2$: $$I = int{-h/2}^{h/2} f(m + t) dt = int{-h/2}^{h/2} left[ f(m) + t f'(m) + frac{t^2}{2} f''(m) + frac{t^3}{6} f'''(m) + frac{t^4}{24} f^{(4)}(m) + dots right] dt$$ Since odd powers of $t$ integrate to zero over $[-h/2, h/2]$: $$I = 2 left[ f(m) frac{h}{2} + frac{f''(m)}{2} frac{(h/2)^3}{3} + frac{f^{(4)}(m)}{24} frac{(h/2)^5}{5} + dots right]$$ $$I = h f(m) + frac{h^3}{24} f''(m) + frac{h^5}{1920} f^{(4)}(m) + dots quad text{--- (Eq. 1)}$$ --- 3. Expansion of the Quadrature Expression Now we expand the terms appearing in the quadrature formula around the midpoint $m$: - At $x0 = m - h/2$: $$f0 = f(m) - frac{h}{2} f'(m) + frac{h^2}{8} f''(m) - frac{h^3}{48} f'''(m) + frac{h^4}{384} f^{(4)}(m) - dots$$ $$f'0 = f'(m) - frac{h}{2} f''(m) + frac{h^2}{8} f'''(m) - frac{h^3}{48} f^{(4)}(m) + dots$$ - At $x1 = m + h/2$: $$f1 = f(m) + frac{h}{2} f'(m) + frac{h^2}{8} f''(m) + frac{h^3}{48} f'''(m) + frac{h^4}{384} f^{(4)}(m) + dots$$ $$f'1 = f'(m) + frac{h}{2} f''(m) + frac{h^2}{8} f'''(m) + frac{h^3}{48} f^{(4)}(m) + dots$$ Evaluating the combinations: $$f0 + f1 = 2 f(m) + frac{h^2}{4} f''(m) + frac{h^4}{192} f^{(4)}(m) + dots$$ $$frac{h}{2}(f0 + f1) = h f(m) + frac{h^3}{8} f''(m) + frac{h^5}{384} f^{(4)}(m) +

Question 24

8. (c) (ii) Consider the Lagrangian $L = m dot{x}dot{y} - momega0^2 xy$ where $m$ and $omega0$ are constants. Find the Hamiltonian and Hamilton's equations of motion. Identify the system. (10 Marks)

Accepted Answer

1. Analysis of the Lagrangian We are given the Lagrangian of a two-degree-of-freedom system with generalized coordinates $q1 = x$ and $q2 = y$: $$L(x, y, dot{x}, dot{y}) = m dot{x} dot{y} - momega0^2 xy$$ where $m 0$ and $omega0 0$ are physical constants. Notice that the kinetic energy term is cross-coupled between the velocities $dot{x}$ and $dot{y}$. --- 2. Canonical Conjugate Momenta The canonical momenta conjugate to $x$ and $y$ are: $$px = frac{partial L}{partial dot{x}} = m dot{y} implies dot{y} = frac{px}{m}$$ $$py = frac{partial L}{partial dot{y}} = m dot{x} implies dot{x} = frac{py}{m}$$ Notice the interesting inversion: the momentum $px$ is proportional to the velocity $dot{y}$, and the momentum $py$ is proportional to the velocity $dot{x}$. --- 3. Construction of the Hamiltonian ($H$) By the Legendre transformation: $$H(x, y, px, py) = px dot{x} + py dot{y} - L$$ Substituting $dot{x} = frac{py}{m}$ and $dot{y} = frac{px}{m}$: $$px dot{x} + py dot{y} = px left( frac{py}{m} right) + py left( frac{px}{m} right) = frac{2 px py}{m}$$ Now evaluate $L$ in terms of canonical coordinates: $$L = m left( frac{py}{m} right) left( frac{px}{m} right) - momega0^2 xy = frac{px py}{m} - momega0^2 xy$$ Therefore: $$H = frac{2 px py}{m} - left( frac{px py}{m} - momega0^2 xy right)$$ $$mathbf{H(x, y, px, py) = frac{px py}{m} + momega0^2 xy}$$ --- 4. Hamilton's Canonical Equations of Motion Hamilton's equations of motion are: $$begin{aligned} dot{x} &= frac{partial H}{partial px} = frac{py}{m} dot{y} &= frac{partial H}{partial py} = frac{px}{m} dot{p}x &= -frac{partial H}{partial x} = -momega0^2 y dot{p}y &= -frac{partial H}{partial y} = -momega0^2 x end{aligned}$$ --- 5. Identification of the Physical System To physically identify the system, we decouple the equations of motion: Differentiating $dot{x} = frac{py}{m}$ with respect to $t$: $$ddot{x} = frac{dot{p}y}{m} = frac{-momega0^2 x}{m} = -omega0^2 x$$ $$mathbf{ddot{x} + omega0^2 x = 0}$$ Differentiating $dot{y} = frac{px

Question 25

7. (b) Find the unique polynomial of degree 2 or less which fits the following data: - $x$: $0, 1, 3$ - $f(x)$: $1, 3, 55$ Also obtain the bound on the truncation error. (15 Marks)

Accepted Answer

1. Data Specification & Method Selection We are given three data points $(xi, fi)$ for $i = 0, 1, 2$: $$begin{array}{c|ccc} i & 0 & 1 & 2 hline xi & 0 & 1 & 3 f(xi) & 1 & 3 & 55 end{array}$$ Since the nodes $x0 = 0, x1 = 1, x2 = 3$ are unequally spaced ($x1 - x0 = 1 eq x2 - x1 = 2$), we employ either: 1. Newton's Divided Difference Interpolation Formula, or 2. Lagrange's Interpolation Formula. --- 2. Divided Difference Table We construct the divided difference table: $$begin{array}{c|c|c|c} xi & f[xi] & f[xi, x{i+1}] & f[xi, x{i+1}, x{i+2}] hline 0 & 1 & & & & frac{3 - 1}{1 - 0} = 2 & 1 & 3 & & frac{26 - 2}{3 - 0} = frac{24}{3} = 8 & & frac{55 - 3}{3 - 1} = frac{52}{2} = 26 & 3 & 55 & & end{array}$$ The divided differences are: - $f[x0] = 1$ - $f[x0, x1] = 2$ - $f[x0, x1, x2] = 8$ --- 3. Construction of the Interpolating Polynomial $P2(x)$ By Newton's divided difference formula: $$P2(x) = f[x0] + (x - x0) f[x0, x1] + (x - x0)(x - x1) f[x0, x1, x2]$$ Substituting the values: $$P2(x) = 1 + (x - 0)(2) + (x - 0)(x - 1)(8)$$ $$P2(x) = 1 + 2x + 8x(x - 1)$$ $$P2(x) = 1 + 2x + 8x^2 - 8x$$ $$mathbf{P2(x) = 8x^2 - 6x + 1}$$ --- 4. Verification with the Given Data Points We check the interpolating polynomial at each node: 1. At $x = 0$: $P2(0) = 8(0)^2 - 6(0) + 1 = 1 = f(0) ; checkmark$ 2. At $x = 1$: $P2(1) = 8(1)^2 - 6(1) + 1 = 8 - 6 + 1 = 3 = f(1) ; checkmark$ 3. At $x = 3$: $P2(3) = 8(3)^2 - 6(3) + 1 = 8(9) - 18 + 1 = 72 - 18 + 1 = 55 = f(3) ; checkmark$ By the Uniqueness Theorem for polynomial interpolation, $P2(x) = 8x^2 - 6x + 1$ is the unique polynomial of degree $le 2$ fitting the data. --- 5. Bound on the Truncation Error Let $f(x)$ be a function that is three times continuously differentiable ($f in C^3[0, 3]$). The remainder (truncation error) formula for degree $n = 2$ interpolation with nodes $x0, x1, x2$ is: $$R2(x) = f(x) - P2(x) = frac{f^{(3)}(xi)}{(2+1)!} (x - x0)(x - x1)(x - x2)$$ $$R2(x) = frac{f'''(xi)}{6} x(x - 1)(x - 3)$$ where $xi = xi(x) in (0, 3)$. Le

Question 26

8. (c) (i) A particle of mass $m$ moves in a force field of potential $V(r) = -frac{k costheta}{r^2}$, where $k$ is constant. Find the Hamiltonian and the Hamilton's equations in spherical polar coordinates $(r, theta, phi)$. (10 Marks)

Accepted Answer

1. Coordinate System & Kinetic Energy In spherical polar coordinates $(r, theta, phi)$, the Cartesian coordinates are: $$x = r sintheta cosphi, quad y = r sintheta sinphi, quad z = r costheta$$ The line element squared is: $$ds^2 = dr^2 + r^2 dtheta^2 + r^2 sin^2theta dphi^2$$ The kinetic energy $T$ of a particle of mass $m$ is: $$T = frac{1}{2} m v^2 = frac{1}{2} m left( dot{r}^2 + r^2 dot{theta}^2 + r^2 sin^2theta , dot{phi}^2 right)$$ --- 2. Potential Energy ($V$) & Lagrangian ($L$) We are given the velocity-independent potential: $$V(r, theta) = -frac{k costheta}{r^2}$$ where $k$ is a constant. The Lagrangian $L = T - V$ is: $$L(r, theta, phi, dot{r}, dot{theta}, dot{phi}) = frac{1}{2} m left( dot{r}^2 + r^2 dot{theta}^2 + r^2 sin^2theta , dot{phi}^2 right) + frac{k costheta}{r^2}$$ --- 3. Canonical (Conjugate) Momenta The canonical momenta $pr, ptheta, pphi$ conjugate to $r, theta, phi$ are: $$pr = frac{partial L}{partial dot{r}} = m dot{r} implies dot{r} = frac{pr}{m}$$ $$ptheta = frac{partial L}{partial dot{theta}} = m r^2 dot{theta} implies dot{theta} = frac{ptheta}{m r^2}$$ $$pphi = frac{partial L}{partial dot{phi}} = m r^2 sin^2theta , dot{phi} implies dot{phi} = frac{pphi}{m r^2 sin^2theta}$$ --- 4. Construction of the Hamiltonian ($H$) The Hamiltonian $H$ is obtained via Legendre transformation: $$H = sum pi dot{q}i - L = pr dot{r} + ptheta dot{theta} + pphi dot{phi} - L$$ Since the Lagrangian is of the standard form $L = T2 - V$ where $T2$ is quadratic in generalized velocities, $H = T + V$: $$H = frac{pr^2}{2m} + frac{ptheta^2}{2m r^2} + frac{pphi^2}{2m r^2 sin^2theta} + V(r, theta)$$ $$mathbf{H(r, theta, phi, pr, ptheta, pphi) = frac{pr^2}{2m} + frac{ptheta^2}{2m r^2} + frac{pphi^2}{2m r^2 sin^2theta} - frac{k costheta}{r^2}}$$ --- 5. Hamilton's Canonical Equations of Motion Hamilton's canonical equations are: $$dot{q}i = frac{partial H}{partial pi}, quad dot{p}i = -frac{partial H}{partial qi}$$ 1. Equations for coordinate $r$: $$dot{r} = frac{partial

Question 27

8. (a) Find the characteristics of the partial differential equation $$p^2 + q^2 = 2; quad p equiv frac{partial z}{partial x}, ; q equiv frac{partial z}{partial y}$$ and determine the integral surface which passes through $x = 0, z = y$. (15 Marks)

Accepted Answer

1. PDE Classification & Charpit's Characteristic System We are given the non-linear first-order PDE: $$F(x, y, z, p, q) = p^2 + q^2 - 2 = 0$$ The characteristic equations (Cauchy's characteristic strip equations) are: $$frac{dx}{dt} = Fp = 2p$$ $$frac{dy}{dt} = Fq = 2q$$ $$frac{dz}{dt} = p Fp + q Fq = p(2p) + q(2q) = 2(p^2 + q^2) = 2(2) = 4$$ $$frac{dp}{dt} = -(Fx + p Fz) = -(0 + p cdot 0) = 0$$ $$frac{dq}{dt} = -(Fy + q Fz) = -(0 + q cdot 0) = 0$$ --- 2. Integration of the Characteristic Equations From $frac{dp}{dt} = 0$ and $frac{dq}{dt} = 0$: $$p(t) = p0, quad q(t) = q0 = text{constants along any characteristic}$$ Integrating the remaining characteristic equations: $$begin{aligned} x(t) &= 2p0 t + x0 y(t) &= 2q0 t + y0 z(t) &= 4t + z0 end{aligned}$$ These straight lines in $mathbb{R}^3$ are the characteristic curves of the PDE. --- 3. Parametrization of the Initial Curve $Gamma$ The integral surface is required to pass through the initial curve $Gamma$: $$x = 0, quad z = y$$ We parametrize $Gamma$ by a parameter $s$: $$x0(s) = 0, quad y0(s) = s, quad z0(s) = s$$ Now we determine the initial values of the strip $p0(s)$ and $q0(s)$ using: 1. The PDE itself: $$p0^2 + q0^2 = 2$$ 2. The strip condition (tangency along $Gamma$): $$frac{dz0}{ds} = p0 frac{dx0}{ds} + q0 frac{dy0}{ds}$$ Since $frac{dz0}{ds} = 1$, $frac{dx0}{ds} = 0$, and $frac{dy0}{ds} = 1$: $$1 = p0(0) + q0(1) implies q0 = 1$$ Substituting $q0 = 1$ into $p0^2 + q0^2 = 2$: $$p0^2 + 1 = 2 implies p0^2 = 1 implies p0 = pm 1$$ Thus, there are two distinct characteristic strips generating two integral surfaces: - Branch 1: $p0 = +1$ - Branch 2: $p0 = -1$ --- 4. Determination of the Integral Surface Substituting the initial values into the characteristic equations: $$begin{aligned} x(s, t) &= 2p0 t + 0 = 2p0 t y(s, t) &= 2(1) t + s = 2t + s z(s, t) &= 4t + s end{aligned}$$ We eliminate the parameters $s$ and $t$: From the $x$-equation: $$t = frac{x}{2p0}$$ From the $y$-equation: $$s = y - 2t = y - frac{x}{p0}$

Question 28

7. (c) Show that for an incompressible steady flow with constant viscosity, the velocity components $$u(y) = frac{U}{h}y - frac{hy}{2mu}frac{dp}{dx}left(1 - frac{y}{h}right), quad v = 0 = w,$$ with $p = p(x)$, satisfy the equation of motion in the absence of body force. Given that $U, h$ and $dp/dx$

Accepted Answer

1. Fluid Dynamics Framework & Navier-Stokes Equations We consider the flow of a viscous incompressible Newtonian fluid between two parallel infinite plates separated by a distance $h$: - Fluid density $rho = text{constant}$ - Dynamic viscosity $mu = text{constant}$ - Steady flow: $frac{partial}{partial t} = 0$ - No body forces: $mathbf{F} = (0, 0, 0)$ The vector equation of motion is the Navier-Stokes equation: $$rho left[ frac{partial mathbf{V}}{partial t} + (mathbf{V} cdot abla)mathbf{V} right] = - abla p + mu abla^2 mathbf{V}$$ along with the Continuity equation (incompressibility condition): $$ abla cdot mathbf{V} = frac{partial u}{partial x} + frac{partial v}{partial y} + frac{partial w}{partial z} = 0$$ The given velocity field is: $$u = u(y) = frac{U}{h} y - frac{h y}{2mu} frac{dp}{dx} left( 1 - frac{y}{h} right), quad v = 0, quad w = 0$$ with pressure $p = p(x)$, and $U, h, frac{dp}{dx} = text{constants}$. --- 2. Verification of the Continuity Equation Let us check the conservation of mass (continuity equation): $$frac{partial u}{partial x} = frac{partial}{partial x} left[ frac{U}{h} y - frac{h y}{2mu} frac{dp}{dx} left( 1 - frac{y}{h} right) right]$$ Since $u$ depends solely on the spatial coordinate $y$ and $frac{dp}{dx}$ is constant: $$frac{partial u}{partial x} = 0$$ Furthermore, since $v = 0$ and $w = 0$: $$frac{partial v}{partial y} = 0, quad frac{partial w}{partial z} = 0$$ Summing the components: $$ abla cdot mathbf{V} = 0 + 0 + 0 = 0$$ The continuity equation is identically satisfied. --- 3. Evaluation of Convective Acceleration Terms The acceleration components of a fluid particle are: $$begin{aligned} ax &= frac{partial u}{partial t} + u frac{partial u}{partial x} + v frac{partial u}{partial y} + w frac{partial u}{partial z} ay &= frac{partial v}{partial t} + u frac{partial v}{partial x} + v frac{partial v}{partial y} + w frac{partial v}{partial z} az &= frac{partial w}{partial t} + u frac{partial w}{partial x} + v frac{partial w}{partial y} + w f

Question 29

7. (a) Find the complete integral of $$z(p^2 - q^2) = x - y; quad p equiv frac{partial z}{partial x}, ; q equiv frac{partial z}{partial y}.$$ (15 Marks)

Accepted Answer

1. PDE Classification & Charpit's Equations We are given the first-order non-linear partial differential equation: $$z(p^2 - q^2) = x - y$$ where $p = frac{partial z}{partial x}$ and $q = frac{partial z}{partial y}$. Let the PDE be written in implicit form $f(x, y, z, p, q) = 0$: $$f(x, y, z, p, q) = z(p^2 - q^2) - (x - y) = 0$$ To find the complete integral, we apply Charpit's method. Charpit's auxiliary equations are: $$frac{dp}{fx + p fz} = frac{dq}{fy + q fz} = frac{dz}{-p fp - q fq} = frac{dx}{-fp} = frac{dy}{-fq}$$ --- 2. Method of Separation of Variables via Transformation Notice the symmetric structure of the equation: $$z p^2 - x = z q^2 - y$$ We can separate the variables by introducing a convenient transformation: Let $u = z^{3/2}$, so that: $$frac{partial u}{partial x} = frac{3}{2} z^{1/2} frac{partial z}{partial x} = frac{3}{2} z^{1/2} p implies z^{1/2} p = frac{2}{3} frac{partial u}{partial x}$$ $$frac{partial u}{partial y} = frac{3}{2} z^{1/2} frac{partial z}{partial y} = frac{3}{2} z^{1/2} q implies z^{1/2} q = frac{2}{3} frac{partial u}{partial y}$$ Notice that $z(p^2 - q^2) = (z^{1/2} p)^2 - (z^{1/2} q)^2$. Substituting in terms of $u$: $$left(frac{2}{3} frac{partial u}{partial x}right)^2 - left(frac{2}{3} frac{partial u}{partial y}right)^2 = x - y$$ $$frac{4}{9} left( frac{partial u}{partial x} right)^2 - frac{4}{9} left( frac{partial u}{partial y} right)^2 = x - y$$ Rearranging so that all $x$-terms are on the left and all $y$-terms are on the right: $$frac{4}{9} left( frac{partial u}{partial x} right)^2 - x = frac{4}{9} left( frac{partial u}{partial y} right)^2 - y = a$$ where $a$ is a separation constant. --- 3. Solving for the Partial Derivatives From the separated equations: 1. For $x$: $$frac{4}{9} left( frac{partial u}{partial x} right)^2 = x + a implies left( frac{partial u}{partial x} right)^2 = frac{9}{4}(x + a)$$ $$frac{partial u}{partial x} = frac{3}{2} sqrt{x + a}$$ 2. For $y$: $$frac{4}{9} left( frac{partial u}{partial y} right)^2 =

Maths Optional 2025 Paper II

29 Questions Available

Maths Optional 2025 Paper II

29 Questions Available

Question 1: 1. (a) Let $H$ and $K$ be two subgroups of a group $G$ such that $o(H) sqrt{o(G)}$ and $o(K) sqrt{o(G)}$. Show that $H cap K eq {e}$, where $e$ is the

Question 2: 1. (e) How many basic solutions are there for the following system of equations? $$2x1 - x2 + 3x3 + x4 = 6$$ $$4x1 - 2x2 - x3 + 2x4 = 10$$ Find all of

Question 3: 2. (a) Define Cauchy sequence and prove that every convergent sequence of real numbers is a Cauchy sequence. What is the importance of Cauchy conditio

Question 4: 2. (c) Use the method of contour integration to prove that $$int{-infty}^infty frac{x^2 - x + 2}{x^4 + 10x^2 + 9},dx = frac{5pi}{12}.$$ (20 Marks)

Question 5: 2. (b) Show that $3$ is an irreducible element in the integral domain $mathbb{Z}[i]$. (15 Marks)

Question 6: 1. (c) Examine whether the series $$sum{n=1}^infty frac{(-1)^{n-1}}{n}$$ is absolutely or conditionally convergent. (10 Marks)

Question 7: 1. (d) Expand $f(z) = frac{1}{(z+1)(z+3)}$ in a Laurent series valid for $1 < |z| < 3$. (10 Marks)

Question 8: 1. (b) Let $G = {e, x, x^2, y, yx, yx^2}$ be a non-Abelian group with $o(x) = 3$ and $o(y) = 2$. Show that $xy = yx^2$ (where $e$ is the identity elem

Question 9: 3. (b) Show that the volume of the greatest rectangular parallelopiped that can be inscribed in the ellipsoid $$frac{x^2}{a^2} + frac{y^2}{b^2} + frac

Question 10: 3. (c) Apply the principle of duality to solve the following linear programming problem: Maximize $Z = 3x1 + 4x2$ subject to $x1 - x2 le 1$ $x1 + x2 g

Question 11: 4. (c) The following table shows all the necessary information on the available supply to each warehouse, the requirement of each market and the unit

Question 12: 4. (a) Examine whether the mapping $phi: mathbb{Z}[x] to mathbb{Z}$ defined by $phi(f(x)) = f(0)$, for $f(x) in mathbb{Z}[x]$, is a homomorphism. Dedu

Question 13: 3. (a) Evaluate the integral $$ointC frac{e^z}{z^2(z+1)^3},dz, quad C: |z|=2.$$ (15 Marks)

Question 14: 4. (b) Prove that every continuous function is Riemann integrable. (15 Marks)

Question 15: 6. (a) Solve $$frac{partial^2 u}{partial x^2} + frac{partial^2 u}{partial y^2} = 0$$ for a rectangular plate subject to the boundary conditions: $u(0,

Question 16: 6. (c) Calculate the moment of inertia of a uniform solid cylinder of mass $M$, radius $R$ and length $L$ with respect to a set of axes passing throug

Question 17: 5. (a) Find the solution of the equation $(D^2 + DD' - 2D'^2)z = y sin x$, where $D equiv frac{partial}{partial x}$ and $D' equiv frac{partial}{partia

Question 18: 5. (b) Solve the following system of linear equations by Gauss-Seidel method: $$10x + 2y + z = 9$$ $$2x + 20y - 2z = -44$$ $$-2x + 3y + 10z = 22$$ (10

Question 19: 5. (c) (i) Convert the number $(3479){10}$ into binary system and the number $(7AE.9F){16}$ into decimal system. (ii) Determine the truth table for th

Question 20: 5. (d) A bead of mass $m$ slides on a frictionless wire in the shape of a cycloid given by $x = a(theta - sintheta)$, $y = a(1 + costheta)$, $(0 le th

Question 21: 5. (e) A source and a sink of equal strength are placed at points $(pm a/2, 0)$ within a fixed circular boundary $x^2 + y^2 = a^2$. Show that the stre

Question 22: 6. (b) Simplify the Boolean function $F(x, y, z) = xyz + x'yz + xy'z + xyz'$ and draw the corresponding GATE network. (15 Marks)

Question 23: 8. (b) Find the constant $p$ and error term for the quadrature formula $$int{x0}^{x1} f(x),dx = frac{h}{2}(f0 + f1) + ph^2(f'0 - f'1)$$ where $x0 + h

Question 24: 8. (c) (ii) Consider the Lagrangian $L = m dot{x}dot{y} - momega0^2 xy$ where $m$ and $omega0$ are constants. Find the Hamiltonian and Hamilton's equa

Question 25: 7. (b) Find the unique polynomial of degree 2 or less which fits the following data: - $x$: $0, 1, 3$ - $f(x)$: $1, 3, 55$ Also obtain the bound on th

Question 26: 8. (c) (i) A particle of mass $m$ moves in a force field of potential $V(r) = -frac{k costheta}{r^2}$, where $k$ is constant. Find the Hamiltonian and

Question 27: 8. (a) Find the characteristics of the partial differential equation $$p^2 + q^2 = 2; quad p equiv frac{partial z}{partial x}, ; q equiv frac{partial

Question 28: 7. (c) Show that for an incompressible steady flow with constant viscosity, the velocity components $$u(y) = frac{U}{h}y - frac{hy}{2mu}frac{dp}{dx}le

Question 29: 7. (a) Find the complete integral of $$z(p^2 - q^2) = x - y; quad p equiv frac{partial z}{partial x}, ; q equiv frac{partial z}{partial y}.$$ (15 Mark