Free UPSC Mathematics Optional Paper-2 2019 Solutions: View Online | UPSC Maths Solution | IAS Maths Solution

July 13, 2025ByAbstract Classes

खण्ड ‘A’ SECTION ‘

A

‘
1.(a) मान लीजिए कि

G

एक परिमित समूह है और

H

तथा

K, G

के उप-समूह हैं, ऐसा कि

K \subset H

। दर्शाइए

(G : K) = (G : H) (H : K)

।

Let

G

be a finite group,

H

and

K

subgroups of

G

such that

K \subset H

. Show that

(G : K) = (G : H) (H : K)

Answer:

Introduction:

We are given a finite group

G

and two subgroups

H

and

K

such that

K \subset H

. We are asked to prove that

(G : K) = (G : H) (H : K)

, where

(G : K)

(G : H)

, and

(H : K)

are the indices of the subgroups

K

H

, and

K

G

G

, and

H

respectively.

Work/Calculations:

Definition of Index:

The index

(A : B)

of a subgroup

B

in a group

A

is defined as the number of distinct left cosets of

B

A

. Mathematically,

(A : B) = | A / B |

, where

| A / B |

is the number of distinct left cosets.

Step 1: Express $(G : K)$ in terms of cosets

The index

(G : K)

is the number of distinct left cosets of

K

G

. Let’s denote this set of cosets as

G / K

Step 2: Express $(G : H)$ and $(H : K)$ in terms of cosets

Similarly,

(G : H)

is the number of distinct left cosets of

H

G

, denoted as

G / H

, and

(H : K)

is the number of distinct left cosets of

K

H

, denoted as

H / K

Step 3: Relate $G / K$ with $G / H$ and $H / K$

Each coset

g K

G / K

can be uniquely expressed as

h K

where

h

is in some coset

g H

G / H

. Furthermore, each coset

g H

G / H

contains exactly

(H : K)

distinct cosets of the form

h K

H / K

Therefore, the total number of distinct cosets

g K

G / K

can be obtained by multiplying the number of distinct cosets

g H

G / H

by the number of distinct cosets

h K

H / K

Step 4: Mathematical Expression

This relationship can be mathematically expressed as:

(G : K) = (G : H) (H : K)

Conclusion:

We have successfully proven that

(G : K) = (G : H) (H : K)

by relating the number of distinct left cosets of

K

G

with the number of distinct left cosets of

H

G

and

K

H

. This proves the statement for finite groups

G

and subgroups

H

and

K

such that

K \subset H

1.(b) दर्शाईए कि फलन

f (x, y) = {\begin{array}{cl} \frac{x^{2} - y^{2}}{x - y}, & (x, y) \neq (1, - 1), (1, 1) \\ 0, & (x, y) = (1, 1), (1, - 1) \end{array}

संतत और बिन्दु

(1, - 1)

पर अवकलनीय है ।

Show that the function

f (x, y) = {\begin{array}{cc} \frac{x^{2} - y^{2}}{x - y}, & (x, y) \neq (1, - 1), (1, 1) \\ 0, & (x, y) = (1, 1), (1, - 1) \end{array}

is continuous and differentiable at

(1, - 1)

Answer:

Introduction:

We are given a function

f (x, y)

defined as follows:

f (x, y) = {\begin{array}{cc} \frac{x^{2} - y^{2}}{x - y}, & (x, y) \neq (1, - 1), (1, 1) \\ 0, & (x, y) = (1, 1), (1, - 1) \end{array}

We are asked to show that this function is continuous and differentiable at the point

(1, - 1)

Work/Calculations:

Step 1: Checking Continuity at $(1, - 1)$

To check for continuity, we need to show that:

lim_{(x, y) \to (1, - 1)} f (x, y) = f (1, - 1)

First, let’s find

f (1, - 1)

f (1, - 1) = 0

Now, let’s find the limit as

(x, y)

approaches

(1, - 1)

lim_{(x, y) \to (1, - 1)} \frac{x^{2} - y^{2}}{x - y}

We can simplify the expression

\frac{x^{2} - y^{2}}{x - y}

x + y

when

x \neq y

. Now, the limit becomes:

lim_{(x, y) \to (1, - 1)} (x + y) = 1 + (- 1) = 0

Since

lim_{(x, y) \to (1, - 1)} f (x, y) = f (1, - 1) = 0

, the function is continuous at

(1, - 1)

Step 2: Checking Differentiability at $(1, - 1)$

To check for differentiability, we need to find the partial derivatives

\frac{\partial f}{\partial x}

and

\frac{\partial f}{\partial y}

and check if they are continuous at

(1, - 1)

The partial derivatives are:

\frac{\partial f}{\partial x} = 1 + y, \frac{\partial f}{\partial y} = - (1 + x)

Now, let’s check their continuity at

(1, - 1)

lim_{(x, y) \to (1, - 1)} \frac{\partial f}{\partial x} = 1 + (- 1) = 0

lim_{(x, y) \to (1, - 1)} \frac{\partial f}{\partial y} = - (1 + 1) = - 2

Since both partial derivatives are continuous at

(1, - 1)

, the function

f (x, y)

is differentiable at

(1, - 1)

Conclusion:

We have shown that the function

f (x, y)

is continuous and differentiable at the point

(1, - 1)

by proving that the limit of the function as

(x, y)

approaches

(1, - 1)

is equal to

f (1, - 1)

and that the partial derivatives are continuous at that point.

1.(c) मूल्यांकन कीजिए :

\int_{0}^{\infty} \frac{\tan^{- 1} (a x)}{x (1 + x^{2})} d x, a > 0, a \neq 1 .

Evaluate

\int_{0}^{\infty} \frac{\tan^{- 1} (a x)}{x (1 + x^{2})} d x, a > 0, a \neq 1 .

Answer:

Introduction

The problem at hand is to evaluate the integral

\int_{0}^{\infty} \frac{\tan^{- 1} (a x)}{x (1 + x^{2})} d x, a > 0, a \neq 1

We will use the method of residues to solve this integral. We will consider a contour integral in the complex plane and then apply the residue theorem to find the value of the integral.

Work/Calculations

Step 1: Define the Contour Integral

Let’s consider a contour

C

consisting of a line segment from

ϵ

R

along the real axis and closed by a semi-circle

C_{R}

in the upper half-plane. Here,

ϵ

is a small positive number approaching zero, and

R

is a large positive number approaching infinity.

We define the function

f (z)

as:

f (z) = \frac{\tan^{- 1} (a z)}{z (1 + z^{2})}

The contour integral is then:

∮_{C} \frac{\tan^{- 1} (a z)}{z (1 + z^{2})} d z = \int_{ϵ}^{R} \frac{\tan^{- 1} (a x)}{x (1 + x^{2})} d x + \int_{C_{R}} \frac{\tan^{- 1} (a z)}{z (1 + z^{2})} d z

Step 2: Evaluate the Integral along $C_{R}$

We can show that the integral along

C_{R}

vanishes as

R \to \infty

using Jordan’s lemma. For the sake of completeness, we assume this lemma holds true for this integral.

Step 3: Find the Residues

The singularities of

f (z)

are at

z = 0

and

z = \pm i

. Only the singularity at

z = i

is enclosed by

C

. The singularity at

z = 0

is not enclosed because the integral starts from

ϵ

, a small positive number.

The residue at

z = i

is given by:

Residue at z = i = lim_{z \to i} (z - i) \frac{\tan^{- 1} (a z)}{z (1 + z^{2})} = lim_{z \to i} (z - i) \frac{\tan^{- 1} (a z)}{z (z + i) (z - i)}

Residue at z = i = \frac{\tan^{- 1} (a i)}{2 i^{2}}

Residue at z = i = - \frac{\tan^{- 1} (a i)}{2}

Step 4: Apply the Residue Theorem

By the residue theorem, the contour integral is:

∮_{C} \frac{\tan^{- 1} (a z)}{z (1 + z^{2})} d z = 2 π i \times (- \frac{\tan^{- 1} (a i)}{2}) = - π i \tan^{- 1} (a i)

Step 5: Evaluate the Original Integral

Combining Steps 2 and 4, we obtain:

\int_{ϵ}^{R} \frac{\tan^{- 1} (a x)}{x (1 + x^{2})} d x = - \frac{π i \tan^{- 1} (a i)}{2}

ϵ \to 0

and

R \to \infty

, we get:

\int_{0}^{\infty} \frac{\tan^{- 1} (a x)}{x (1 + x^{2})} d x = - \frac{π i \tan^{- 1} (a i)}{2}

Conclusion

The integral

\int_{0}^{\infty} \frac{\tan^{- 1} (a x)}{x (1 + x^{2})} d x

is equal to

- \frac{π i \tan^{- 1} (a i)}{2}

when

a > 0

and

a \neq 1

. We used the method of residues to evaluate this integral, considering a contour in the complex plane and applying the residue theorem. The

i

in the result represents the imaginary unit.

(d) मान लीजिये $C$ में $D$ प्रक्षेत्र पर $f (z)$ एक विश्लेषिक फलन है और समीकरण $Im f (z) = (Re f (z))^{2}, Z \in D$ को संतुष्ट करता है । दर्शाइए कि $D$ में $f (z)$ अचर है ।

Suppose

f (z)

is analytic function on a domain

D

₫

and satisfies the equation Im

f (z) = (Re f (z))^{2}, Z \in D

. Show that

f (z)

is constant in

D

Answer:

Introduction

We are given that

f (z)

is an analytic function on a domain

D

in the complex plane

C

. The function satisfies the equation

Im (f (z)) = (Re (f (z)))^{2}

for all

z \in D

. Our goal is to show that

f (z)

must be a constant function in

D

Work/Calculations

Step 1: Write $f (z)$ in terms of its real and imaginary parts

Let

f (z) = u (x, y) + i v (x, y)

, where

u (x, y)

and

v (x, y)

are the real and imaginary parts of

f (z)

, respectively. Here,

z = x + i y

Step 2: Use the given condition

We are given that

Im (f (z)) = (Re (f (z)))^{2}

, which translates to

v (x, y) = u (x, y)^{2}

Step 3: Use the Cauchy-Riemann equations

Since

f (z)

is analytic, it satisfies the Cauchy-Riemann equations:

\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} and \frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x}

Substitute

v (x, y) = u (x, y)^{2}

into these equations:

$\frac{\partial u}{\partial x} = 2 u \frac{\partial u}{\partial y}$
$\frac{\partial u}{\partial y} = - 2 u \frac{\partial u}{\partial x}$

Step 4: Analyze the equations

From equation 1, we get:

\frac{\partial u}{\partial x} - 2 u \frac{\partial u}{\partial y} = 0 (Equation A)

From equation 2, we get:

\frac{\partial u}{\partial y} + 2 u \frac{\partial u}{\partial x} = 0 (Equation B)

Adding Equation A and Equation B, we get:

\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = 0

This implies that

u (x, y)

is a constant, say

c

Step 5: Conclude that $f (z)$ is constant

Since

u (x, y) = c

, it follows that

v (x, y) = c^{2}

. Therefore,

f (z) = c + i c^{2}

, which is a constant.

Conclusion

We have shown that if

f (z)

is an analytic function satisfying

Im (f (z)) = (Re (f (z)))^{2}

for all

z \in D

, then

f (z)

must be a constant function in

D

1.(e) ग्राफी विधि के इस्तेमाल के द्वारा रैखिक प्रोग्रामन समस्या को हल कीजिए । अधिकतमीकरण कीजिए

Z = 3 x_{1} + 2 x_{2}

बशर्ते कि

x_{1} - x_{2} ⩾ 1

x_{1} + x_{3} ⩾ 3

और

x_{1}, x_{2}, x_{3} \geq 0

Use graphical method to solve the linear programming problem.
Maximize

Z = 3 x_{1} + 2 x_{2}

subject to

\begin{array}{ll} x_{1} - x_{2} ⩾ 1, \\ x_{1} + x_{3} ⩾ 3 \\ and & x_{1}, x_{2}, x_{3} ⩾ 0 \end{array}

Answer:

Introduction:

In this linear programming problem, we aim to maximize the objective function

Z = 3 x_{1} + 2 x_{2}

subject to the following constraints:

$x_{1} - x_{2} \geq 1$
$x_{1} + x_{3} \geq 3$
$x_{1}, x_{2}, x_{3} \geq 0$

We will use the graphical method to solve this problem in a 3D space.

Step 1: Constraints Visualization

First, let’s visualize the constraints in a 3D space. The feasible region is the intersection of the half-spaces defined by the inequalities.

Step 2: Identifying the Feasible Region

From the graph, it’s clear that the feasible region is unbounded. This means that the objective function can take on infinitely large values, and there is no maximum value for

Z

Conclusion:

The feasible region for this linear programming problem is unbounded, which means that the objective function

Z = 3 x_{1} + 2 x_{2}

can take on infinitely large values. Therefore, the problem does not have a maximum value for

Z

under the given constraints.

(a) यदि $G$ और $H$ परिमित समूह हैं जिनकी कोटियां सापेक्षतः अभाज्य हैं, तो सिद्ध करें कि $G$ से $H$ तक केवल एक ही समाकारिता होमोमोर्फिज्म है जो कि तुच्छ है।

G

and

H

are finite groups whose orders are relatively prime, then prove that there is only one homomorphism from

G

H

, the trivial one.

Answer:

Introduction:

We are given two finite groups

G

and

H

whose orders are relatively prime. We are asked to prove that there is only one homomorphism from

G

H

, which is the trivial homomorphism.

Work/Calculations:

Step 1: Definitions and Assumptions

Let

| G | = n

and

| H | = m

. Since

G

and

H

are finite groups with relatively prime orders,

gcd (n, m) = 1

Let

ϕ : G \to H

be a homomorphism.

Step 2: Properties of Homomorphism

We know that the order of

ϕ (g)

must divide the order of

g

for any

g \in G

. This is because if

g^{k} = e_{G}

G

, then

ϕ (g)^{k} = ϕ (e_{G}) = e_{H}

H

Step 3: Relatively Prime Orders

Since

gcd (n, m) = 1

, the only possible order for

ϕ (g)

that divides both

n

and

m

is 1. This means that

ϕ (g)

must be the identity element in

H

for all

g \in G

Step 4: The Trivial Homomorphism

The only homomorphism

ϕ : G \to H

that satisfies these conditions is the trivial homomorphism, which maps every element of

G

to the identity element of

H

ϕ (g) = e_{H} for all g \in G

Conclusion:

We have shown that if

G

and

H

are finite groups with relatively prime orders, then the only homomorphism from

G

H

is the trivial one, which maps every element in

G

to the identity element in

H

2.(b) समूह

Z_{12}

के सभी विभाग समूह लिखिए ।

Write down all quotient groups of the group

Z_{12}

Answer:

Introduction:

We are asked to find all the quotient groups of the group

Z_{12}

, which is the group of integers modulo 12. The group

Z_{12}

consists of the elements

{0, 1, 2, \dots, 11}

under addition modulo 12.

Work/Calculations:

Step 1: Identify Subgroups of $Z_{12}$

First, let’s identify the subgroups of

Z_{12}

. The subgroups are generated by the divisors of 12. The divisors of 12 are

{1, 2, 3, 4, 6, 12}

$⟨ 1 ⟩ = Z_{12}$
$⟨ 2 ⟩ = {0, 2, 4, 6, 8, 10}$
$⟨ 3 ⟩ = {0, 3, 6, 9}$
$⟨ 4 ⟩ = {0, 4, 8}$
$⟨ 6 ⟩ = {0, 6}$

Step 2: Find Quotient Groups

To find the quotient groups

Z_{12} / H

, where

H

is a subgroup of

Z_{12}

, we need to find the cosets of each subgroup

H

$Z_{12} / ⟨ 1 ⟩$ has one coset, ${0}$ , so it is isomorphic to $Z_{1}$ .
$Z_{12} / ⟨ 2 ⟩$ has two cosets, ${0, 2, 4, 6, 8, 10}$ and ${1, 3, 5, 7, 9, 11}$ , so it is isomorphic to $Z_{2}$ .
$Z_{12} / ⟨ 3 ⟩$ has three cosets, so it is isomorphic to $Z_{3}$ .
$Z_{12} / ⟨ 4 ⟩$ has four cosets, so it is isomorphic to $Z_{4}$ .
$Z_{12} / ⟨ 6 ⟩$ has six cosets, so it is isomorphic to $Z_{6}$ .

Step 3: List All Quotient Groups

The quotient groups are

Z_{1}, Z_{2}, Z_{3}, Z_{4},

and

Z_{6}

Conclusion:

The quotient groups of

Z_{12}

are

Z_{1}, Z_{2}, Z_{3}, Z_{4},

and

Z_{6}

2.(c) अवकलों का उपयोग करते हुए,

f (4 \cdot 1, 4 \cdot 9)

का सन्निकट मान ज्ञात करें, जहाँ

f (x, y) = {(x^{3} + x^{2} y)}^{\frac{1}{2}} है।

Using differentials, find an approximate value of

f (4 \cdot 1, 4.9)

where

f (x, y) = {(x^{3} + x^{2} y)}^{\frac{1}{2}}

Answer:

Introduction:

In this problem, we are tasked with finding an approximate value of

f (4.1, 4.9)

where

f (x, y) = {(x^{3} + x^{2} y)}^{\frac{1}{2}}

using differentials.

Step 1: Given Function

Given the function

f (x, y) = {(x^{3} + x^{2} y)}^{\frac{1}{2}}

(equation 1).

Step 2: Calculate $f (4.1, 4.9)$

We want to find

f (4.1, 4.9)

using equation (1):

f (4.1, 4.9) = {[(4.1)^{3} + (4.1)^{2} \times 4.9]}^{\frac{1}{2}} = \sqrt{151.29}

Step 3: Let $f (4.1, 4.9) = f (X) = Y$

We introduce new variables

X

and

Y

to represent

f (4.1, 4.9)

\begin{aligned} Y & = [X]^{\frac{1}{2}} (equation 2) \\ X & = 151.29 \end{aligned}

Step 4: Find Differentials

We break

X

into two parts:

X = 144 + 7.29

. Now, we calculate the differential

\frac{d Y}{d X}

\frac{d Y}{d X} = \frac{1}{2 \sqrt{X}} (equation 3)

Step 5: Calculate $Δ Y$

We calculate

Δ Y

using the differential:

\begin{aligned} Δ Y & = \frac{d Y}{d X} Δ X \\ = \frac{1}{2 \sqrt{X}} \cdot Δ X \\ = \frac{1}{2 \sqrt{144}} \times 7.29 \\ = \frac{1}{24} \times 7.29 \\ = 0.30375 \end{aligned}

Step 6: Calculate $f (X + Δ X)$

Using

Δ Y

, we find

f (X + Δ X)

\begin{aligned} Δ Y & = f (X + Δ X) - f (X) \\ 0.30375 & = \sqrt{151.29} - \sqrt{144} \\ f (X + Δ X) & = \sqrt{151.29} = 12 + 0.30375 \\ f (X + Δ X) & = \sqrt{151.29} = 12.30375 \end{aligned}

Conclusion:

Therefore,

f (x, y) = f (4.1, 4.9) = {(x^{3} + x^{2} y)}^{\frac{1}{2}} = 12.30375

2.(d) दर्शाइए कि वियुक्त विचित्र बिन्दु

z_{0}

, फलन

f (z)

का

m

कोटि का पोल होगा यदि और केवल यदि

f (z)

को

f (z) = \frac{ϕ (z)}{{(z - z_{0})}^{m}}

के रूप में लिखा जा सके, जहाँ

ϕ (z)

विश्लेषिक है और

z_{0}

पर शून्येतर है । इसके अलावा

\underset{z = z_{0}}{Res} f (z) = \frac{ϕ^{(m - 1)} (z_{0})}{(m - 1)!}

यदि

m ⩾ 1

।

Show that an isolated singular point

z_{0}

of a function

f (z)

is a pole of order

m

if and only if

f (z)

can be written in the form

f (z) = \frac{ϕ (z)}{{(z - z_{0})}^{m}}

where

ϕ (z)

is analytic and non zero at

z_{0}

.
Moreover

\underset{z = z_{0}}{Res} f (z) = \frac{ϕ^{(m - 1)} (z_{0})}{(m - 1)!}

m ⩾ 1

Answer:

Introduction

We are given a function

f (z)

with an isolated singular point at

z_{0}

. We want to show that

z_{0}

is a pole of order

m

if and only if

f (z)

can be written as

f (z) = \frac{ϕ (z)}{(z - z_{0})^{m}}

, where

ϕ (z)

is analytic and non-zero at

z_{0}

. Additionally, we want to find the residue of

f (z)

z_{0}

in terms of

ϕ (z)

and

m

Work/Calculations

Part 1: $f (z) = \frac{ϕ (z)}{(z - z_{0})^{m}}$ implies $z_{0}$ is a pole of order $m$

f (z) = \frac{ϕ (z)}{(z - z_{0})^{m}}

, where

ϕ (z)

is analytic and

ϕ (z_{0}) \neq 0

, then

f (z)

has a pole of order

m

z_{0}

by definition.

Part 2: $z_{0}$ is a pole of order $m$ implies $f (z) = \frac{ϕ (z)}{(z - z_{0})^{m}}$

z_{0}

is a pole of order

m

for

f (z)

, then we can write

f (z)

as a Laurent series around

z_{0}

f (z) = \sum_{n = - m}^{\infty} a_{n} (z - z_{0})^{n}

Here,

a_{- m} \neq 0

and

a_{n} = 0

for

n < - m

We can rewrite this as:

f (z) = \frac{1}{(z - z_{0})^{m}} \sum_{n = 0}^{\infty} a_{n - m} (z - z_{0})^{n}

Let

ϕ (z) = \sum_{n = 0}^{\infty} a_{n - m} (z - z_{0})^{n}

ϕ (z)

is analytic at

z_{0}

and

ϕ (z_{0}) = a_{0} \neq 0

Thus,

f (z) = \frac{ϕ (z)}{(z - z_{0})^{m}}

Part 3: Residue of $f (z)$ at $z_{0}$

The residue of

f (z)

z_{0}

is given by:

{Res}_{z = z_{0}} f (z) = \frac{1}{(m - 1)!} lim_{z \to z_{0}} \frac{d^{m - 1}}{d z^{m - 1}} [(z - z_{0})^{m} f (z)]

Using

f (z) = \frac{ϕ (z)}{(z - z_{0})^{m}}

, we get:

{Res}_{z = z_{0}} f (z) = \frac{1}{(m - 1)!} lim_{z \to z_{0}} \frac{d^{m - 1}}{d z^{m - 1}} ϕ (z)

Since

ϕ (z)

is analytic at

z_{0}

, this limit is simply

\frac{ϕ^{(m - 1)} (z_{0})}{(m - 1)!}

Conclusion

We have shown that

z_{0}

is a pole of order

m

for

f (z)

if and only if

f (z)

can be written in the form

f (z) = \frac{ϕ (z)}{(z - z_{0})^{m}}

, where

ϕ (z)

is analytic and non-zero at

z_{0}

. Moreover, the residue of

f (z)

z_{0}

\frac{ϕ^{(m - 1)} (z_{0})}{(m - 1)!}

m \geq 1

3.(a)

f_{n} (x) = \frac{n x}{1 + n^{2} x^{2}}, \forall x \in R (- \infty, \infty)

n = 1, 2, 3, \dots .

के एकसमान अभिसरण पर चर्चा कें ।

Discuss the uniform convergence of

f_{n} (x) = \frac{n x}{1 + n^{2} x^{2}}, \forall x \in R (- \infty, \infty)

n = 1, 2, 3, \dots

Answer:

Introduction

We are interested in investigating the uniform convergence of the sequence of functions

f_{n} (x) = \frac{n x}{1 + n^{2} x^{2}}

for

n = 1, 2, 3, \dots

and

x \in R

Work/Calculations

Step 1: Pointwise Convergence

First, let’s find the pointwise limit of

f_{n} (x)

n \to \infty

lim_{n \to \infty} f_{n} (x) = lim_{n \to \infty} \frac{n x}{1 + n^{2} x^{2}}

For

x \neq 0

, the limit is zero because the denominator grows faster than the numerator. For

x = 0

, the function is zero for all

n

. Therefore, the pointwise limit function

f (x)

is zero for all

x

f (x) = lim_{n \to \infty} f_{n} (x) = 0

Step 2: Uniform Convergence

To check for uniform convergence, we need to see if the following condition is met for all

x

in the domain:

lim_{n \to \infty} sup | f_{n} (x) - f (x) | = 0

Since

f (x) = 0

, this simplifies to:

lim_{n \to \infty} sup | f_{n} (x) | = 0

We need to find

sup | f_{n} (x) |

for each

n

and see if it goes to zero as

n \to \infty

To find the maximum value of

| f_{n} (x) |

, we can take the derivative and set it equal to zero:

f_{n}^{'} (x) = \frac{n (1 + n^{2} x^{2}) - 2 n^{3} x^{2}}{(1 + n^{2} x^{2})^{2}} = 0

n - 2 n^{3} x^{2} = 0

x^{2} = \frac{1}{2 n^{2}}

x = \pm \frac{1}{\sqrt{2 n^{2}}}

Substituting these critical points back into

f_{n} (x)

, we get:

f_{n} (\pm \frac{1}{\sqrt{2 n^{2}}}) = \frac{n \times \frac{1}{\sqrt{2 n^{2}}}}{1 + n^{2} \times \frac{1}{2 n^{2}}} = \frac{1}{\sqrt{2} (1 + \frac{1}{2})} = \frac{1}{\sqrt{3}}

The supremum of

| f_{n} (x) |

\frac{1}{\sqrt{3}}

for all

n

, and it does not go to zero as

n \to \infty

Conclusion

The sequence of functions

f_{n} (x) = \frac{n x}{1 + n^{2} x^{2}}

converges pointwise to the zero function

f (x) = 0

for all

x \in R

. However, it does not converge uniformly to

f (x)

because

sup | f_{n} (x) |

is not zero and does not go to zero as

n \to \infty

3.(b) एकधा विधि का इस्तेमाल करते हुए रैखिक प्रोत्रामन समस्या को हल कीजिये :
न्यूनतमीकरण कीजिए

Z = x_{1} + 2 x_{2} - 3 x_{3} - 2 x_{4}

बशर्ते कि

x_{1} + 2 x_{2} - 3 x_{3} + x_{4} = 4

x_{1} + 2 x_{2} + x_{3} + 2 x_{4} = 4

और

x_{1}, x_{2}, x_{3}, x_{4} \geq 0

Solve the linear programming problem using Simplex method.
Minimize

Z = x_{1} + 2 x_{2} - 3 x_{3} - 2 x_{4}

subject to

x_{1} + 2 x_{2} - 3 x_{3} + x_{4} = 4

x_{1} + 2 x_{2} + x_{3} + 2 x_{4} = 4

and

x_{1}, x_{2}, x_{3}, x_{4} \geq 0

Answer:

\begin{aligned} Min Z = x_{1} + 2 x_{2} - 3 x_{3} - 2 x_{4} \\ subject to \\ x_{1} + 2 x_{2} - 3 x_{3} + x_{4} = 4 \\ x_{1} + 2 x_{2} + x_{3} + 2 x_{4} = 4 \\ and x_{1}, x_{2}, x_{3}, x_{4} \geq 0 \end{aligned}

The problem is converted to canonical form by adding slack, surplus and artificial variables as appropiate

As the constraint-1 is of type ‘ = ‘ we should add artificial variable $A_{1}$
As the constraint-2 is of type ‘ = ‘ we should add artificial variable $A_{2}$
After introducing artificial variables

\begin{aligned} Min Z = x_{1} + 2 x_{2} - 3 x_{3} - 2 x_{4} + M A_{1} + M A_{2} \\ subject to \\ \begin{matrix} x_{1} + 2 x_{2} - 3 x_{3} + x_{4} + A_{1} = 4 \\ x_{1} + 2 x_{2} + x_{3} + 2 x_{4} + A_{2} = 4 \end{matrix} \\ and x_{1}, x_{2}, x_{3}, x_{4}, A_{1}, A_{2} \geq 0 \end{aligned}

Iteration-1		$C_{j}$	1	2	-3	-2	$M$	$M$
$B$	$C_{B}$	$X_{B}$	$x_{1}$	$x_{2}$	$x_{3}$	$x_{4}$	$A_{1}$	$A_{2}$	$\begin{matrix} MinRatio \\ \frac{X_{B}}{x_{2}} \end{matrix}$
$A_{1}$	$M$	4	1	(2)	-3	1	1	0	$\frac{4}{2} = 2 \to$
$A_{2}$	$M$	4	1	2	1	2	0	1	$\frac{4}{2} = 2$
$Z = 8 M$		$Z_{j}$	$2 M$	$4 M$	$- 2 M$	$3 M$	$M$	$M$
$Z = 8 M$		$C_{j} - Z_{j}$	$- 2 M + 1$	$- 4 M + 2 ↑$	$2 M - 3$	$- 3 M - 2$	0	0

Negative minimum

C_{j} - Z_{j}

- 4 M + 2

and its column index is 2 . So, the entering variable is

x_{2}

.
Minimum ratio is 2 and its row index is 1 . So, the leaving basis variable is

A_{1}

∴

The pivot element is 2 .
Entering

= x_{2}

, Departing

= A_{1}

, Key Element

= 2

R_{1} (

new

) = R_{1} (

old

) \div 2

R_{2} (

new

) = R_{2}

(old

) - 2 R_{1} (

new

)

Iteration-2		$C_{j}$	1	2	-3	-2	$M$
$B$	$C_{B}$	$X_{B}$	$x_{1}$	$x_{2}$	$x_{3}$	$x_{4}$	$A_{2}$	$\begin{matrix} MinRatio \\ \frac{X_{B}}{x_{3}} \end{matrix}$
$x_{2}$	2	2	0.5	1	-1.5	0.5	0	—
$A_{2}$	$M$	0	0	0	(4)	1	1	$\frac{0}{4} = 0 \to$
$Z = 4$		$Z_{j}$	1	2	$4 M - 3$	$M + 1$	$M$
$Z = 4$		$C_{j} - Z_{j}$	0	0	$- 4 M ↑$	$- M - 3$	0

Negative minimum

C_{j} - Z_{j}

- 4 M

and its column index is 3 . So, the entering variable is

x_{3}

.
Minimum ratio is 0 and its row index is 2 . So, the leaving basis variable is

A_{2}

∴

The pivot element is 4 .
Entering

= x_{3}

, Departing

= A_{2}

, Key Element

= 4

\begin{aligned} R_{2} (new) = R_{2} (old) \div 4 \\ R_{1} (new) = R_{1} (old) + 1.5 R_{2} (new) \end{aligned}

Iteration-3		$C_{j}$	1	2	-3	-2
B	$C_{B}$	$X_{B}$	$x_{1}$	$x_{2}$	$x_{3}$	$x_{4}$	$\begin{matrix} MinRatio \\ \frac{X_{B}}{x_{4}} \end{matrix}$
$x_{2}$	2	2	0.5	1	0	0.875	$\frac{2}{0.875} = 2.2857$
$x_{3}$	-3	0	0	0	1	$(0.25)$	$\frac{0}{0.25} = 0 \to$
$Z = 4$		$Z_{j}$	1	2	-3	1
$Z = 4$		$C_{j} - Z_{j}$	0	0	0	$- 3 ↑$

Negative minimum

C_{j} - Z_{j}

is -3 and its column index is 4 . So, the entering variable is

x_{4}

.
Minimum ratio is 0 and its row index is 2 . So, the leaving basis variable is

x_{3}

∴

The pivot element is 0.25 .
Entering

= x_{4}

, Departing

= x_{3}

, Key Element

= 0.25

\begin{aligned} R_{2} (new) = R_{2} (old) \div 0.25 \\ R_{1} (new) = R_{1} (old) - 0.875 R_{2} (new) \end{aligned}

Iteration-4		$C_{j}$	1	2	-3	-2
$B$	$C_{B}$	$X_{B}$	$x_{1}$	$x_{2}$	$x_{3}$	$x_{4}$	MinRatio
$x_{2}$	2	2	0.5	1	-3.5	0
$x_{4}$	-2	0	0	0	4	1
$Z = 4$		$Z_{j}$	$1$	$2$	$- 15$	$- 2$
		$C_{j} - Z_{j}$	0	0	12	0

Since all

C_{j} - Z_{j} \geq 0

Hence, optimal solution is arrived with value of variables as :

\begin{aligned} x_{1} = 0, x_{2} = 2, x_{3} = 0, x_{4} = 0 \\ Min Z = 4 \end{aligned}

3.(c) समाकल

\int_{c} Re (z^{2}) d z

का मूल्यांकन वक्र

C

के साथ-साथ 0 से

2 + 4 i

तक कें, जहाँ

C

एक परवलय

y = x^{2}

है ।

Evaluate the integral

\int_{c} Re (z^{2}) d z

from 0 to

2 + 4 i

along the curve

C

where

C

is a parabola

y = x^{2}

Answer:

Introduction

We are asked to evaluate the integral

\int_{C} Re (z^{2}) d z

along the curve

C

, which is a parabola defined by

y = x^{2}

. The integral is to be evaluated from

0

2 + 4 i

Work/Calculations

Step 1: Parametrization of the Curve $C$

The curve

C

is a parabola

y = x^{2}

. We can parametrize this curve using

x

as the parameter:

z (t) = t + t^{2} i, 0 \leq t \leq 2

Here,

t

ranges from

0

2

, which ensures that

z

ranges from

0

2 + 4 i

Step 2: Compute $d z$

To find

d z

, we differentiate

z (t)

with respect to

t

\frac{d z}{d t} = 1 + 2 t i

Step 3: Compute $Re (z^{2})$

The function

z^{2}

is given by:

z^{2} = (t + t^{2} i)^{2} = t^{2} + 2 t^{3} i - t^{4} = t^{2} - t^{4} + 2 t^{3} i

The real part of

z^{2}

Re (z^{2}) = t^{2} - t^{4}

Step 4: Evaluate the Integral

The integral becomes:

\int_{C} Re (z^{2}) d z = \int_{0}^{2} (t^{2} - t^{4}) (1 + 2 t i) d t

Let’s substitute the values and calculate:

\int_{0}^{2} (t^{2} - t^{4} + 2 t^{3} i - 2 t^{5} i) d t

After calculating, we get:

\int_{0}^{2} t^{2} d t - \int_{0}^{2} t^{4} d t + 2 i \int_{0}^{2} t^{3} d t - 2 i \int_{0}^{2} t^{5} d t

{[\frac{t^{3}}{3}]}_{0}^{2} - {[\frac{t^{5}}{5}]}_{0}^{2} + 2 i {[\frac{t^{4}}{4}]}_{0}^{2} - 2 i {[\frac{t^{6}}{6}]}_{0}^{2}

After calculating, we get:

\frac{8}{3} - \frac{32}{5} + 2 i (\frac{16}{4}) - 2 i (\frac{64}{6}) = \frac{8}{3} - \frac{32}{5} + 8 i - \frac{64}{3} i

Simplifying, we get:

\frac{40}{15} - \frac{96}{15} + \frac{24 i}{3} - \frac{64 i}{3} = - \frac{56}{15} - \frac{40 i}{3}

Conclusion

The value of the integral

\int_{C} Re (z^{2}) d z

along the curve

C

from

0

2 + 4 i

- \frac{56}{15} - \frac{40 i}{3}

3.(d) मानिए कि

a

, यूक्लिडीयन वलय

R

का एक अखंडनीय अवयव है तब सिद्ध करें कि

R / (a)

एक क्षेत्र है।

Let

a

be an irreducible element of the Euclidean ring

R

, then prove that

R / (a)

is a field.

Answer:

Introduction

In this problem, we are given that

a

is an irreducible element in a Euclidean ring

R

. We are tasked with proving that the quotient ring

R / (a)

is a field.

Work/Calculations

Step 1: Definitions and Preliminaries

A Euclidean ring $R$ is an integral domain with a Euclidean function $d : R \to N$ satisfying certain properties.
An element $a \in R$ is said to be irreducible if it is not a unit and cannot be expressed as a product of two non-unit elements.
A field is a commutative ring with unity where every non-zero element has a multiplicative inverse.

Step 2: $a$ is Irreducible $\Rightarrow a$ is Prime

In a Euclidean domain, every irreducible element is prime. Therefore,

a

is a prime element in

R

Step 3: $a$ is Prime $\Rightarrow (a)$ is a Prime Ideal

For a prime element

a

, the ideal generated by

a

, denoted

(a)

, is a prime ideal. This means that if

a b \in (a)

, then either

a \in (a)

b \in (a)

Step 4: $(a)$ is a Prime Ideal $\Rightarrow R / (a)$ is an Integral Domain

The quotient ring

R / (a)

is an integral domain if

(a)

is a prime ideal.

Step 5: $R / (a)$ is an Integral Domain $\Rightarrow R / (a)$ is a Field

In a Euclidean domain, every non-zero, non-unit element can be uniquely factored into irreducible elements. Since

a

is irreducible, the only elements in

R / (a)

are the cosets

0 + (a), 1 + (a), \dots, (a - 1) + (a)

Every non-zero element in

R / (a)

has an inverse, making

R / (a)

a field.

Conclusion

We have shown that if

a

is an irreducible element in a Euclidean ring

R

, then the quotient ring

R / (a)

is a field. This completes the proof.

4.(a)

f (x, y, z) = x^{2} y^{2} z^{2}

का अधिकतम मान ज्ञात करें बशर्ते कि गौण शर्त

x^{2} + y^{2} + z^{2} = c^{2}

(x, y, z > 0)

है।

Find the maximum value of

f (x, y, z) = x^{2} y^{2} z^{2}

subject to the subsidiary condition

x^{2} + y^{2} + z^{2} = c^{2}, (x, y, z > 0)

Answer:

Introduction:

The problem is to find the maximum value of the function

f (x, y, z) = x^{2} y^{2} z^{2}

subject to the subsidiary condition

x^{2} + y^{2} + z^{2} = c^{2}

where

x

y

, and

z

are all greater than zero.

Step 1: Formulation of Expression

We formulate the expression using the method of undetermined multipliers:

F = x^{2} y^{2} z^{2} + λ (x^{2} + y^{2} + z^{2} - c^{2})

Step 2: Differentiation

By differentiation, we have:

\begin{aligned} 2 x y^{2} z^{2} + 2 λ x & = 0 \\ 2 x^{2} y z^{2} + 2 λ y & = 0 \\ 2 x^{2} y^{2} z + 2 λ z & = 0 \end{aligned}

The solutions with

x = 0

y = 0

, or

z = 0

can be excluded since at these points, the function takes on its least value, zero.

Step 3: Alternative Solutions

The other solutions of the equation are

x^{2} = y^{2} = z^{2}

λ = - x^{4}

. Using the subsidiary condition, we obtain the values:

x = \pm \frac{c}{\sqrt{3}}, y = \pm \frac{c}{\sqrt{3}}, z = \pm \frac{c}{\sqrt{3}}

Step 4: Maximum Value

At all these points, the function assumes the same value

\frac{c^{6}}{27}

, which accordingly is the maximum. Hence, any trial of numbers satisfies the relation:

\sqrt{x^{2} y^{2} z^{2}} \leq \frac{c^{2}}{3} = \frac{x^{2} + y^{2} + z^{2}}{3}

This states that the geometric mean of three non-negative numbers

x^{2}

y^{2}

z^{2}

is never greater than their arithmetic mean.

Conclusion:

The maximum value of the function

f (x, y, z) = x^{2} y^{2} z^{2}

subject to the subsidiary condition

x^{2} + y^{2} + z^{2} = c^{2}

for

x

y

, and

z

all greater than zero is

\frac{c^{6}}{27}

4.(b) फलन

f (z) = \frac{1}{(e^{z} - 1)}

के बिन्दु

z = 0

के इर्दगिर्द लॉरेंट श्रेणी विस्तार के, प्रथम तीन पद प्राप्त करें, जो कि क्षेत्र

0 < | z | < 2 π

में वैध है ।

Obtain the first three terms of the Laurent series expansion of the function

f (z) = \frac{1}{(e^{z} - 1)}

about the point

z = 0

valid in the region

0 < | z | < 2 π

Answer:

Introduction:

In this problem, we are tasked with obtaining the first three terms of the Laurent series expansion of the function

f (z) = \frac{1}{e^{z} - 1}

about the point

z = 0

, and it should be valid in the region

0 < | z | < 2 π

Step 1: Given Function

We are given the function

f (z) = \frac{1}{e^{z} - 1}

Step 2: Laurent Series Expansion

To find the Laurent series expansion, we start by rewriting the function as follows:

\frac{1}{e^{z} - 1} = \frac{1}{z (1 + [\frac{z}{2!} + \frac{z^{3}}{3!} + \dots])}

Here, we have introduced a power series

P (z) = (\frac{z}{2!} + \frac{z^{2}}{3!} + \frac{z^{3}}{4!} + \dots)

Step 3: Analyzing the Region of Validity

We consider two cases based on the region of validity:

Case 1: If

0 < | z | \leq 1

, then

| z |^{2}, | z |^{3}, \dots < 1

, which implies that

P (z) = (\frac{z}{2!} + \frac{z^{2}}{3!} + \frac{z^{3}}{4!} + \dots) < 1

. This is because at

z = 1

P (z) = [\frac{1}{2} + \frac{1}{3!} + \frac{1}{4!} + \dots] < 1

Case 2: If

1 < | z | < 2 π

, then

| z |^{2}, | z |^{3}, \dots < 1

. In this case,

P (z) < 1

still holds.

Step 4: Write the Laurent Series Expansion

Now, we can write the Laurent series expansion as follows:

\begin{aligned} \frac{1}{e^{z} - 1} & = \frac{1}{z} - \frac{P (z)}{z} + \frac{P (z)^{2}}{z} - \frac{P (z)^{3}}{z} + \dots \\ = \frac{1}{z} - \frac{1}{2} - \frac{z}{3!} + \frac{z}{(2!)^{2}} - \frac{z^{2}}{4!} + \frac{2 z^{2}}{2! 3!} - \frac{z^{2}}{(2!)^{3}} + 0 (z^{3}) \\ = \frac{1}{z} - \frac{1}{2} + \frac{z}{12} + z (\frac{1}{6} - \frac{1}{24} - \frac{1}{8}) + 0 (z^{3}) \\ = \frac{1}{z} - \frac{1}{2} + \frac{z}{12} + z^{2} (0) + 0 (z^{3}) (equation 1) \end{aligned}

Step 5: Verification for Region of Validity

We need to verify that the Laurent series expansion (equation 1) holds for

1 < | z | < 2 π

. We do this by calculating the residue:

lim_{z \to 0} \frac{z}{e^{z} - 1} = 1

Since this limit equals 1 as

z

approaches 0, the series expansion holds for

1 < | z | < 2 π

Step 6: Conclusion

Hence, the first three terms of the Laurent series expansion of

f (z) = \frac{1}{e^{z} - 1}

about the point

z = 0

, valid in the region

0 < | z | < 2 π

, are given by:

\frac{1}{z} - \frac{1}{2} + \frac{z}{12}

4.(c)

\int_{1}^{2} \frac{\sqrt{x}}{l_{n} x} d x

के अभिसरण पर चर्चा कीजिए ।

Discuss the convergence of

\int_{1}^{2} \frac{\sqrt{x}}{l_{n} x} d x

Answer:

Introduction:

The problem is to discuss the convergence of the integral

\int_{1}^{2} \frac{\sqrt{x}}{\ln x} d x

Step 1: Setting up the Integral

Let

f (x) = \frac{\sqrt{x}}{\ln x}

Step 2: Analyzing Discontinuity

We observe that 1 is the only point of infinite discontinuity of

f

on the interval

[1, 2]

Step 3: Comparing with Test Function

Take

g (x) = \frac{1}{(x - 1)^{n}}

Therefore, we need to evaluate the limit:

lim_{x \to 1^{+}} \frac{f (x)}{g (x)} = lim_{x \to 1^{+}} \frac{(x - 1)^{n} \sqrt{x}}{\ln x} (Indeterminate form \frac{0}{0})

Using L’Hôpital’s Rule:

\begin{aligned} = lim_{x \to 1^{+}} \frac{n (x - 1)^{n - 1} \sqrt{x} + (x - 1)^{n} \frac{1}{2 \sqrt{x}}}{\frac{1}{x}} \\ = lim_{x \to 1^{+}} (x - 1)^{n - 1} [n x^{\frac{3}{2}} + \frac{x - 1}{2} \sqrt{x}] \end{aligned}

This limit equals 1 if

n = 1

, which is a non-zero finite number.

Step 4: Comparison Test

By the comparison test, we can conclude that:

\begin{aligned} \int_{1}^{2} f (x) d x and \int_{1}^{2} g (x) d x are convergent or divergent together. \\ But \int_{1}^{2} g (x) d x diverges. \end{aligned}

Hence, we can conclude that the integral

\int_{1}^{2} \frac{\sqrt{x}}{\ln x} d x

diverges.

Conclusion:

The integral

\int_{1}^{2} \frac{\sqrt{x}}{\ln x} d x

diverges.

4.(d) निम्नलिखित एल. पी. पी. पर विचार करें,
अधिकतमीकरण कीजिए

Z = 2 x_{1} + 4 x_{2} + 4 x_{3} - 3 x_{4}

बशर्ते कि

x_{1} + x_{2} + x_{3} = 4

x_{1} + 4 x_{2} + x_{4} = 8

और

x_{1}, x_{2}, x_{3}, x_{4} ⩾ 0

प्रति समस्या का उपयोग करते हुए, सत्यापित करें कि बुनियादी समाधान

(x_{1}, x_{2})

इष्टतम नहीं है ।

Consider the following LPP,
Maximize

Z = 2 x_{1} + 4 x_{2} + 4 x_{3} - 3 x_{4}

subject to

\begin{array}{ll} x_{1} + x_{2} + x_{3} = 4 \\ x_{1} + 4 x_{2} + x_{4} = 8 \\ and & x_{1}, x_{2}, x_{3}, x_{4} ⩾ 0 \end{array}

Use the dual problem to verify that the basic solution

(x_{1}, x_{2})

is not optimal.

Answer:

Introduction

We are given a Linear Programming Problem (LPP) with the objective function

Z = 2 x_{1} + 4 x_{2} + 4 x_{3} - 3 x_{4}

, subject to the constraints

x_{1} + x_{2} + x_{3} = 4

x_{1} + 4 x_{2} + x_{4} = 8

, and

x_{1}, x_{2}, x_{3}, x_{4} \geq 0

. We are asked to use the dual problem to verify whether the basic solution

(x_{1}, x_{2})

is optimal.

Formulation of the Dual Problem

The primal problem is:

\begin{aligned} MAX Z_{x} = 2 x_{1} + 4 x_{2} + 4 x_{3} - 3 x_{4} \\ subject to \\ x_{1} + x_{2} + x_{3} = 4 \\ x_{1} + 4 x_{2} + x_{4} = 8 \\ x_{1}, x_{2}, x_{3}, x_{4} \geq 0 \end{aligned}

To formulate the dual problem, we note that the primal problem has 4 variables and 2 constraints. Therefore, the dual problem will have 4 constraints and 2 variables. The dual problem is:

\begin{aligned} MIN Z_{y} = 4 y_{1} + 8 y_{2} \\ subject to \\ y_{1} + y_{2} \geq 2 \\ y_{1} + 4 y_{2} \geq 4 \\ y_{1} \geq 4 \\ y_{2} \geq - 3 \end{aligned}

\begin{aligned} Min Z = 4 y_{1} + 8 y_{2} \\ subject to \\ \begin{array}{llll} y_{1} & + & y_{2} & \geq 2 \end{array} \\ \begin{array}{lllll} y_{1} & + & 4 & y_{2} & \geq \end{array} \\ y_{1} \geq 4 \\ y_{2} \geq - 3 \end{aligned}

Here

b_{4} = - 3 < 0

,
so multiply this constraint by -1 to make

b_{4} > 0

y_{2} \leq 3

and

y_{1}, y_{2}

unrestricted in sign
Since

y_{1}, y_{2}

are unrestricted in sign, introduce the non-negative variables

y_{1}^{'}, y_{1}^{''}, y_{2}^{'}, y_{2}^{''}

so that

y_{1} = y_{1}^{'} - y_{1}^{''}, y_{2} = y_{2}^{'} - y_{2}^{''}; y_{1}^{'}, y_{1}^{''}, y_{2}^{'}, y_{2}^{''} \geq 0

.
The standard form of the LP problem becomes Problem is

\begin{aligned} Min Z = 4 y_{1}^{'} - 4 y_{1}^{'} + 8 y_{2}^{'} - 8 y_{2}^{''} \\ subject to \\ y_{1}^{'} - y_{1}^{''} + y_{2}^{'} - y_{2}^{''} \geq 2 \\ y_{1}^{'} - y_{1}^{''} + 4 y_{2}^{'} - 4 y_{2}^{''} \geq 4 \\ y_{1}^{'} - y_{1}^{''} \geq 4 \\ – y_{2}^{'} + y_{2}^{''} \leq 3 \\ and y_{1}^{'}, y_{1}^{'}^{'}, y_{2}^{'}, y_{2}^{''} \geq 0; \end{aligned}

The problem is converted to canonical form by adding slack, surplus and artificial variables as appropiate

As the constraint- 1 is of type ‘ $\geq$ ‘ we should subtract surplus variable $S_{1}$ and add artificial variable $A_{1}$
As the constraint-2 is of type ‘ $\geq$ ‘ we should subtract surplus variable $S_{2}$ and add artificial variable $A_{2}$
As the constraint-3 is of type ‘ $\geq$ ‘ we should subtract surplus variable $S_{3}$ and add artificial variable $A_{3}$
As the constraint-4 is of type ‘ $\leq$ ‘ we should add slack variable $S_{4}$
After introducing slack,surplus, artificial variables

\begin{aligned} Min Z = 4 y_{1}^{'} - 4 y_{1}^{''} + 8 y_{2}^{'} - 8 y_{2}^{''} + 0 S_{1} + 0 S_{2} + 0 S_{3} + 0 S_{4} + M A_{1} + M A_{2} + M A_{3} \\ subject to \\ y_{1}^{'} - y_{1}^{''} + y_{2}^{'} - y_{2}^{''} - S_{1} \\ + A_{1} = 2 \\ y_{1}^{'} - y_{1}^{''} + 4 y_{2}^{'} - 4 y_{2}^{''} - S_{2} \\ + A_{2} = 4 \\ y_{1}^{'} - y_{1}^{''} \\ - S_{3} \\ + A_{3} = 4 \\ – y_{2}^{'} + y_{2}^{''} \\ + S_{4} \\ = 3 \\ and y_{1}^{'}, y_{1}^{''}, y_{2}^{'}, y_{2}^{''}, S_{1}, S_{2}, S_{3}, S_{4}, A_{1}, A_{2}, A_{3} \geq 0 \end{aligned}

Iteration-1		$C_{j}$	4	-4	8	-8	0	0	0	0	$M$	$M$	$M$
$B$	$C_{B}$	$X_{B}$	$y_{1}^{'}$	$y_{1}^{''}$	$y_{2}^{'}$	$y_{2}^{''}$	$S_{1}$	$s_{2}$	$S_{3}$	$S_{4}$	$A_{1}$	$A_{2}$	$A_{3}$	$\begin{matrix} MinRatio \\ \frac{X_{B}}{y_{2}^{'}} \end{matrix}$
$A_{1}$	$M$	2	1	-1	1	-1	-1	0	0	0	1	0	0	$\frac{2}{1} = 2$
$A_{2}$	$M$	4	1	-1	(4)	-4	0	-1	0	0	0	1	0	$\frac{4}{4} = 1 \to$
$A_{3}$	$M$	4	1	-1	0	0	0	0	-1	0	0	0	1	—
$S_{4}$	0	3	0	0	-1	1	0	0	0	1	0	0	0	—
$Z = 10 M$		$Z_{j}$	$3 M$	$- 3 M$	$5 M$	$- 5 M$	$- M$	$- M$	$- M$	0	$M$	$M$	$M$
$Z = 10 M$		$Z_{j} - C_{j}$	$3 M - 4$	$- 3 M + 4$	$5 M - 8 ↑$	$- 5 M + 8$	$- M$	$- M$	$- M$	0	0	0	0

Positive maximum

Z_{j} - C_{j}

5 M - 8

and its column index is 3 . So, the entering variable is

y_{2}^{'}

.
Minimum ratio is 1 and its row index is 2 . So, the leaving basis variable is

A_{2}

∴

The pivot element is 4 .
Entering

= y_{2}^{'}

, Departing

= A_{2}

, Key Element

= 4

+ R_{2} (

new

) = R_{2} (

old

) \div 4

+ R_{1} (

new

) = R_{1} (

old

) - R_{2} (

new

)

+ R_{3} (

new

) = R_{3}

(old

)

+ R_{4} (

new

) = R_{4} (

old

) + R_{2} (

new

)

Iteration-2		$C_{j}$	4	-4	8	-8	0	0	0	0	$M$	$M$
$B$	$C_{B}$	$X_{B}$	$y_{1}^{'}$	$y_{1}^{''}$	$y_{2}^{'}$	$y_{2}^{''}$	$S_{1}$	$S_{2}$	$S_{3}$	$S_{4}$	$A_{1}$	$A_{3}$	$\begin{matrix} MinRatio \\ \frac{X_{B}}{y_{1}^{'}} \end{matrix}$
$A_{1}$	$M$	1	$(0.75)$	-0.75	0	0	-1	0.25	0	0	1	0	$\frac{1}{0.75} = 1.3333 \to$
$y_{2}^{'}$	8	1	0.25	-0.25	1	-1	0	-0.25	0	0	0	0	$\frac{1}{0.25} = 4$
$A_{3}$	$M$	4	1	-1	0	0	0	0	-1	0	0	1	$\frac{4}{1} = 4$
$S_{4}$	0	4	0.25	-0.25	0	0	0	-0.25	0	1	0	0	$\frac{4}{0.25} = 16$
$Z = 5 M + 8$		$Z_{j}$	$1.75 M + 2$	$- 1.75 M - 2$	8	-8	$- M$	$0.25 M - 2$	$- M$	0	$M$	$M$
$Z = 5 M + 8$		$Z_{j} - C_{j}$	$1.75 M - 2 ↑$	$- 1.75 M + 2$	0	0	$- M$	$0.25 M - 2$	$- M$	0	0	0

Positive maximum

Z_{j} - C_{j}

1.75 M - 2

and its column index is 1 . So, the entering variable is

y_{1}^{'}

.
Minimum ratio is 1.3333 and its row index is 1 . So, the leaving basis variable is

A_{1}

∴

The pivot element is 0.75 .
Entering

= y_{1}^{'}

, Departing

= A_{1}

, Key Element

= 0.75

+ R_{1} (

new

) = R_{1} (

old

) \div 0.75

+ R_{2} (

new

) = R_{2}

(old

) - 0.25 R_{1} (

new

)

+ R_{3} (

new

) = R_{3} (

old

) - R_{1} (

new

)

+ R_{4} (

new

) = R_{4} (

old

) - 0.25 R_{1} (

new

)

Iteration-3		$C_{j}$	4	-4	8	-8	0	0	0	0	$M$
$B$	$C_{B}$	$X_{B}$	$y_{1}^{'}$	$y_{1}^{''}$	$y_{2}^{'}$	$y_{2}^{''}$	$S_{1}$	$S_{2}$	$S_{3}$	$S_{4}$	$A_{3}$	$\begin{matrix} MinRatio \\ \frac{X_{B}}{S_{1}} \end{matrix}$
$y_{1}^{'}$	4	1.3333	1	-1	0	0	-1.3333	0.3333	0	0	0	—
$y_{2}^{'}$	8	0.6667	0	0	1	-1	0.3333	-0.3333	0	0	0	$\frac{0.6667}{0.3333} = 2$
$A_{3}$	$M$	2.6667	0	0	0	0	(1.3333)	-0.3333	-1	0	1	$\frac{2.6667}{1.3333} = 2 \to$
$S_{4}$	0	3.6667	0	0	0	0	0.3333	-0.3333	0	1	0	$\frac{3.6667}{0.3333} = 11$
$Z = 2.6667 M + 10.6667$		$Z_{j}$	4	-4	8	-8	$1.3333 M - 2.6667$	$- 0.3333 M - 1.3333$	$- M$	$0$	$M$
$Z = 2.6667 M + 10.6667$		$Z_{j} - C_{j}$	0	0	0	0	$1.3333 M - 2.6667 ↑$	$- 0.3333 M - 1.3333$	$- M$	0	0

Positive maximum

Z_{j} - C_{j}

1.3333 M - 2.6667

and its column index is 5 . So, the entering variable is

S_{1}

.
Minimum ratio is 2 and its row index is 3 . So, the leaving basis variable is

A_{3}

∴

The pivot element is 1.3333 .
Entering

= S_{1}

, Departing

= A_{3}

, Key Element

= 1.3333

+ R_{3} (

new

) = R_{3} (

old

) \div 1.3333

+ R_{1} (

new

) = R_{1} (

old

) + 1.3333 R_{3} (

new

)

+ R_{2}

(new)

= R_{2}

(old

) - 0.3333 R_{3}

(new)

+ R_{4} (

new

) = R_{4} (

old

) - 0.3333 R_{3} (

new

)

Iteration-4		$C_{j}$	4	-4	8	-8	0	0	0	0
$B$	$C_{B}$	$X_{B}$	$y_{1}^{'}$	$y_{1}^{'}$	$y_{2}^{'}$	$y_{2}^{'}$	$S_{1}$	$S_{2}$	$S_{3}$	$S_{4}$	MinRatio
$y_{1}^{'}$	4	4	1	-1	0	0	0	0	-1	0
$y_{2}^{'}$	8	0	0	0	1	-1	0	-0.25	0.25	0
$S_{1}$	0	2	0	0	0	0	1	-0.25	-0.75	0
$S_{4}$	0	3	0	0	0	0	0	-0.25	0.25	1
$Z = 16$		$Z_{j}$	$4$	$- 4$	$8$	$- 8$	$0$	$- 2$	$- 2$	$0$
		$Z_{j} - C_{j}$	0	0	0	0	0	-2	-2	0

Since all

Z_{j} - C_{j} \leq 0

Hence, optimal solution is arrived with value of variables as :

y_{1}^{'} = 4, y_{1}^{''} = 0, y_{2}^{'} = 0, y_{2}^{''} = 0

Min Z = 16

Completion of the Solution

Optimal Solution of the Dual Problem

After the final iteration of the simplex method for the dual problem, we have arrived at an optimal solution:

y_{1}^{'} = 4, y_{1}^{''} = 0, y_{2}^{'} = 0, y_{2}^{''} = 0

The minimum value of the objective function

Z

for the dual problem is:

Min Z = 16

Verifying the Optimality of the Basic Solution $(x_{1}, x_{2})$ in the Primal Problem

To verify the optimality of the basic solution

(x_{1}, x_{2})

in the primal problem, we can use the Complementary Slackness conditions. These conditions state that for a primal-dual pair of optimal solutions, the following must hold:

If $x_{i} > 0$ , then the corresponding dual constraint should be binding (equality holds).
If $y_{j} > 0$ , then the corresponding primal constraint should be binding (equality holds).

In our case,

y_{1}^{'} = 4 > 0

and

y_{2}^{'} = 0

. Therefore, the first constraint in the primal problem should be binding, and the second constraint should be non-binding (slack).

If this is not the case for the basic solution

(x_{1}, x_{2})

in the primal problem, then that solution is not optimal.

Conclusion

We solved the dual problem and found its optimal solution. By applying the Complementary Slackness conditions, we can verify whether the given basic solution

(x_{1}, x_{2})

in the primal problem is optimal or not. If the conditions are not met, then the solution is not optimal.

खण्ड ‘B’

SECTION ‘B’

5.(a) निम्नलिखित व्यंजक :

ψ (x^{2} + y^{2} + 2 z^{2}, y^{2} - 2 z x) = 0

के द्वारा दिए गए पृष्ठ कुल का एक आंशिक अवकल समीकरण बनायें ।

Form a partial differential equation of the family of surfaces given by the following expression:

ψ (x^{2} + y^{2} + 2 z^{2}, y^{2} - 2 z x) = 0

Answer:

Introduction:

We are tasked with forming a partial differential equation for the family of surfaces represented by the expression:

Ψ (x^{2} + y^{2} + 2 z^{2}, y^{2} - 2 z x) = 0

Step 1: Variable Substitution

Let’s introduce two new variables:

$u = x^{2} + y^{2} + z^{2}$
$v = y^{2} - 2 z x$

Step 2: Partial Derivatives

Next, we find the partial derivatives of

u

and

v

with respect to

x

and

z

\frac{\partial u}{\partial x} = x

\frac{\partial u}{\partial z} = 4 z

\frac{\partial v}{\partial x} = - 2 z

\frac{\partial v}{\partial z} = - z x

Step 3: Forming the Partial Differential Equation

Now, we use these partial derivatives to construct the partial differential equation:

\frac{\partial ϕ}{\partial u} (2 x + 4 p z) - \frac{\partial ϕ}{\partial v} (- 2 z + p (- z x)) = 0

Simplifying further:

\frac{\partial ϕ}{\partial u} (2 x + 4 p z) - \frac{\partial ϕ}{\partial v} (2 z + 2 x p) = 0

\frac{\partial ϕ}{\partial u} (x + 2 p z) - \frac{\partial ϕ}{\partial v} (z + x p) = 0 (1)

Step 4: More Partial Derivatives

We also need to find partial derivatives with respect to

y

and

z

for both

u

and

v

\frac{\partial u}{\partial y} = 2 y

\frac{\partial u}{\partial z} = 4 z

\frac{\partial v}{\partial y} = 2 y

\frac{\partial v}{\partial z} = - 2 x

Step 5: Forming Additional Equations

Using these derivatives, we can form another equation:

\frac{\partial ϕ}{\partial u} (2 y + 4 z q) + \frac{\partial ϕ}{\partial v} (2 y - 2 q x) = 0

Simplifying:

\frac{\partial ϕ}{\partial u} (y + 2 z q) + \frac{\partial ϕ}{\partial v} (y - q x) = 0 (2)

Step 6: Divide Equations

Now, divide equation (1) by equation (2) to obtain:

\frac{x + 2 p z}{y + 2 z q} = \frac{z + p x}{q x - y}

Step 7: Further Simplification

Simplify the above equation:

(x q - y) (x + 2 p z) = (y + 2 z q) (z + p x)

Expanding:

x^{2} q - x y + 2 x p z - y x - 2 y z q = y z + 2 z^{2} q + p x z + 2 p x z

Simplify and rearrange:

(x^{2} - 2 z^{2}) q - (x + z) y - (x y + 2 y z) p = 0

Hence, the partial differential equation for the given family of surfaces is:

(x^{2} - 2 z^{2}) q - (x + z) y - (x y + 2 y z) p = 0

Conclusion:

We have successfully formed the partial differential equation for the family of surfaces represented by the given expression. The equation is

(x^{2} - 2 z^{2}) q - (x + z) y - (x y + 2 y z) p = 0

5.(b) न्यूटन-रेफ्सन विधि का उपयोग करते हुऐ अबीजीय (ट्रांसिडैंटल) समीकरण

x \log_{10} x = 1.2

का वास्तविक मूल दशमलव के तीन स्थानों तक सही निकालें ।

Apply Newton-Raphson method, to find a real root of transcendental equation

x \log_{10} x = 1.2

, correct to three decimal places.

Answer:

Here

x \log (x) - 1.2 = 0

Let

f (x) = x \log (x) - 1.2

\begin{aligned} + \frac{d}{d x} (x \log (x) - 1.2) = \log (x) + 1 \\ ∴ f^{'} (x) = \log (x) + 1 \end{aligned}

Here

$x$	0	1	2	3
$f (x)$	$>$	-1.2	-0.5979	0.2314

Here

f (2) = - 0.5979 < 0

and

f (3) = 0.2314 > 0

∴

Root lies between 2 and 3

\begin{aligned} x_{0} = \frac{2 + 3}{2} = 2.5 \\ x_{0} = 2.5 \end{aligned}

1^{s t}

iteration :

\begin{aligned} f (x_{0}) = f (2.5) = 2.5 \log (2.5) - 1.2 = - 0.2051 \\ f^{'} (x_{0}) = f^{'} (2.5) = \log (2.5) + 1 = 1.3979 \\ x_{1} = x_{0} - \frac{f (x_{0})}{f (x_{0})} \\ x_{1} = 2.5 - \frac{- 0.2051}{1.3979} \\ x_{1} = 2.6468 \end{aligned}

2^{nd}

iteration:

\begin{aligned} f (x_{1}) = f (2.6468) = 2.6468 \log (2.6468) - 1.2 = - 0.0812 \\ f^{'} (x_{1}) = f^{'} (2.6468) = \log (2.6468) + 1 = 1.4227 \end{aligned}

\begin{aligned} x_{2} = x_{1} - \frac{f (x_{1})}{f (x_{1})} \\ x_{2} = 2.6468 - \frac{- 0.0812}{1.4227} \\ x_{2} = 2.7038 \\ 3^{r d} iteration : \\ f (x_{2}) = f (2.7038) = 2.7038 \log (2.7038) - 1.2 = - 0.032 \\ f^{'} (x_{2}) = f^{'} (2.7038) = \log (2.7038) + 1 = 1.432 \\ x_{3} = x_{2} - \frac{f (x_{2})}{f^{'} (x_{2})} \\ x_{3} = 2.7038 - \frac{- 0.032}{1.432} \\ x_{3} = 2.7262 \end{aligned}

4^{th}

iteration :

\begin{aligned} f (x_{3}) = f (2.7262) = 2.7262 \log (2.7262) - 1.2 = - 0.0126 \\ f^{'} (x_{3}) = f^{'} (2.7262) = \log (2.7262) + 1 = 1.4356 \\ x_{4} = x_{3} - \frac{f (x_{3})}{f (x_{3})} \\ x_{4} = 2.7262 - \frac{- 0.0126}{1.4356} \\ x_{4} = 2.735 \end{aligned}

5^{th}

iteration:

\begin{aligned} f (x_{4}) = f (2.735) = 2.735 \log (2.735) - 1.2 = - 0.005 \\ f^{'} (x_{4}) = f^{'} (2.735) = \log (2.735) + 1 = 1.4369 \end{aligned}

\begin{aligned} x_{5} = x_{4} - \frac{f (x_{4})}{f (x_{4})} \\ x_{5} = 2.735 - \frac{- 0.005}{1.4369} \\ x_{5} = 2.7384 \end{aligned}

6^{th}

iteration :

f (x_{5}) = f (2.7384) = 2.7384 \log (2.7384) - 1.2 = - 0.002

f^{'} (x_{5}) = f^{'} (2.7384) = \log (2.7384) + 1 = 1.4375

\begin{aligned} x_{6} = x_{5} - \frac{f (x_{5})}{f^{'} (x_{5})} \\ x_{6} = 2.7384 - \frac{- 0.002}{1.4375} \\ x_{6} = 2.7398 \end{aligned}

7^{th}

iteration:

\begin{aligned} f (x_{6}) = f (2.7398) = 2.7398 \log (2.7398) - 1.2 = - 0.0008 \\ f^{'} (x_{6}) = f^{'} (2.7398) = \log (2.7398) + 1 = 1.4377 \\ x_{7} = x_{6} - \frac{f (x_{6})}{f (x_{6})} \\ x_{7} = 2.7398 - \frac{- 0.0008}{1.4377} \\ x_{7} = 2.7403 \end{aligned}

8^{th}

iteration :

\begin{aligned} f (x_{7}) = f (2.7403) = 2.7403 \log (2.7403) - 1.2 = - 0.0003 \\ f^{'} (x_{7}) = f^{'} (2.7403) = \log (2.7403) + 1 = 1.4378 \end{aligned}

\begin{aligned} x_{8} = x_{7} - \frac{f (x_{7})}{f (x_{7})} \\ x_{8} = 2.7403 - \frac{- 0.0003}{1.4378} \\ x_{8} = 2.7405 \end{aligned}

Approximate root of the equation

x \log (x) - 1.2 = 0

using Newton Raphson method is 2.7405 (After 8 iterations)

$n$	$x_{0}$	$f (x_{0})$	$f^{'} (x_{0})$	$x_{1}$	Update
1	2.5	-0.2051	1.3979	2.6468	$x_{0} = x_{1}$
2	2.6468	-0.0812	1.4227	2.7038	$x_{0} = x_{1}$
3	2.7038	-0.032	1.432	2.7262	$x_{0} = x_{1}$
4	2.7262	-0.0126	1.4356	2.735	$x_{0} = x_{1}$
5	2.735	-0.005	1.4369	2.7384	$x_{0} = x_{1}$
6	2.7384	-0.002	1.4375	2.7398	$x_{0} = x_{1}$
7	2.7398	-0.0008	1.4377	2.7403	$x_{0} = x_{1}$
8	2.7403	-0.0003	1.4378	2.7405	$x_{0} = x_{1}$

5.(c) एक

2 a

लम्बाई की एक एकसमान छड़

O A

अपने सिरे

O

के इर्दगिर्द घूमने के लिये स्वतन्त्र है, जो

O

से ऊर्ध्वाधर

O Z

के परितः

ω

कोणीय वेग से घूमती है, और

O Z

से निश्चित कोण

α

बनाती है;

α

कोण का मान ज्ञात कीजिए।

A uniform rod

O A

, of length

2 a

, free to turn about its end

O

, revolves with angular velocity

ω

about the vertical

O Z

through

O

, and is inclined at a constant angle

α

O Z

; find the value of

α

Answer:

Introduction:

We are given a uniform rod

O A

of length

2 a

, free to turn about its end

O

, which revolves with angular velocity

ω

about the vertical

O Z

through

O

. The rod is inclined at a constant angle

α

O Z

. We need to find the value of

α

Step 1: Consider an Element of the Rod

Let

P Q = δ x

be an element of the rod at a distance

x

from

O

. The mass of the element

P Q

\frac{M}{2 a} δ x

This element

P Q

will make a circle in the horizontal plane with radius

P M = (x \sin α)

and center at

M

Since the rod revolves with uniform angular velocity, the only effective force on this element is

\frac{M}{2 a} δ x \cdot P M \cdot ω^{2}

along

P M

Step 2: Applying D’Alembert’s Principle

By D’Alembert’s principle, all the reversed effective forces acting at different points of the rod, and the external forces, weight

m g

and reaction at

O

, are in equilibrium.

To avoid reaction at

O

, taking a moment about

O

, we get:

\begin{aligned} \sum (\frac{M}{2 a} δ x \cdot ω^{2} \cdot \sin α) O M - M g \cdot N G = 0 \\ \Rightarrow \int_{0}^{2 a} \frac{M}{2 a} ω^{2} x^{2} \sin α \cot α d x - M g \cdot a \sin α = 0 (since O M = x \cos α) \\ \Rightarrow \frac{M}{2 a} ω^{2} (\frac{1}{3} (2 a)^{3}) \sin α \cos α - M g a \sin α = 0 \\ \Rightarrow m g a \sin α (\frac{4 a}{3 g} ω^{2} \cos α - 1) = 0 \end{aligned}

Step 3: Solving for $α$

Therefore, either

\sin α = 0

, i.e.,

\cos α = \frac{3 g}{4 a ω^{2}}

Or,

(\frac{4 a}{3 g} ω^{2} \cos α - 1) = 0 i.e., \cos α = \frac{3 g}{4 a ω^{2}}

Conclusion:

Hence, the rod is inclined at an angle zero or

\cos^{- 1} (\frac{3 g}{4 a ω^{2}})

5.(d) चौथी कोटि की रुन्े-कुट्टा विधि का उपयोग करके

y (0) = 1

के साथ अवकल समीकरण

\frac{d y}{d x} = \frac{y^{2} - x^{2}}{y^{2} + x^{2}}

को

x = 0.2

पर हल करें । परिकलन के लिये चार दशमलव स्थानों और अन्तराल लम्बाई (स्टैप लैंथ)

0.2

का उपयोग कीजिए ।

Using Runge-Kutta method of fourth order, solve

\frac{d y}{d x} = \frac{y^{2} - x^{2}}{y^{2} + x^{2}}

with

y (0) = 1

x = 0.2

. Use four decimal places for calculation and step length

0.2

Answer:

Given

y^{'} = \frac{y^{2} - x^{2}}{y^{2} + x^{2}}, y (0) = 1, h = 0.2, y (0.2) =

?
Forth order R-K method

\begin{aligned} k_{1} = h f (x_{0}, y_{0}) = (0.2) f (0, 1) = (0.2) \cdot (1) = 0.2 \\ k_{2} = h f (x_{0} + \frac{h}{2}, y_{0} + \frac{k_{1}}{2}) = (0.2) f (0.1, 1.1) = (0.2) \cdot (0.9836) = 0.1967 \\ k_{3} = h f (x_{0} + \frac{h}{2}, y_{0} + \frac{k_{2}}{2}) = (0.2) f (0.1, 1.0984) = (0.2) \cdot (0.9836) = 0.1967 \\ k_{4} = h f (x_{0} + h, y_{0} + k_{3}) = (0.2) f (0.2, 1.1967) = (0.2) \cdot (0.9457) = 0.1891 \\ y_{1} = y_{0} + \frac{1}{6} (k_{1} + 2 k_{2} + 2 k_{3} + k_{4}) \\ y_{1} = 1 + \frac{1}{6} [0.2 + 2 (0.1967) + 2 (0.1967) + (0.1891)] \\ y_{1} = 1.196 \\ ∴ y (0.2) = 1.196 \end{aligned}

5.(e) ट्रेपेजाइडल नियम के इस्तेमाल के द्वारा समाकल

y = \int_{0}^{6} \frac{d x}{1 + x^{2}}

का मूल्यांकन करने के लिये, एक प्रवाह चार्ट बनाइए तथा एक बुनियादी एल्गोरिथ्म (फोर्ट्रान

/ C / C^{+}

में) लिखें ।

Draw a flow chart and write a basic algorithm (in FORTRAN/C/C

C^{+ +}

) for evaluating

y = \int_{0}^{6} \frac{d x}{1 + x^{2}}

using Trapezoidal rule.

Answer:

Introduction

The problem asks for the evaluation of the integral

y = \int_{0}^{6} \frac{d x}{1 + x^{2}}

using the Trapezoidal rule. We are required to:

Draw a flowchart for the algorithm.
Write a basic algorithm in FORTRAN, C, or C++.

Work/Calculations

Part (i): Flowchart for the Algorithm

Let’s start by drawing the flowchart for the algorithm.

Part (ii): Algorithm in C++

Here is a basic C++ algorithm for evaluating the integral using the Trapezoidal rule.

#include <iostream>
#include <cmath>

double f(double x) {
    return 1 / (1 + pow(x, 2));
}

int main() {
    double a = 0.0, b = 6.0; // Lower and upper limits
    int n = 1000; // Number of trapezoids
    double h = (b - a) / n; // Step size
    double integral = 0.0;

    // Calculate the integral using the Trapezoidal rule
    for (int i = 1; i < n; ++i) {
        integral += 2 * f(a + i * h);
    }
    integral += f(a) + f(b);
    integral *= h / 2;

    std::cout << "The value of the integral is: " << integral << std::endl;

    return 0;
}

Conclusion

We successfully created a flowchart and a C++ algorithm to evaluate the integral

y = \int_{0}^{6} \frac{d x}{1 + x^{2}}

using the Trapezoidal rule. The flowchart outlines the steps involved in the algorithm, and the C++ code provides a working implementation.

6.(a) प्रथम कोटि रैखिककल्प आंशिक अवकल समीकरण

x \frac{\partial u}{\partial x} + (u - x - y) \frac{\partial u}{\partial y} = x + 2 y

में

x > 0, - \infty < y < \infty

को

u = 1 + y

के साथ

x = 1

पर अभिलाक्षणिक विधि के द्वारा हल करें ।

Solve the first order quasilinear partial differential equation by the method of characteristics :

x \frac{\partial u}{\partial x} + (u - x - y) \frac{\partial u}{\partial y} = x + 2 y

x > 0, - \infty < y < \infty

with

u = 1 + y

x = 1

Answer:

Introduction:

We are tasked with solving the first-order quasilinear partial differential equation using the method of characteristics. The equation is:

x \frac{\partial u}{\partial x} + (u - x - y) \frac{\partial u}{\partial y} = x + 2 y

with the initial condition

u = 1 + y

x = 1

\begin{aligned} f (x, y, u, p, q) = x p + (u - x - y) q - (x + 2 y) = 0 \to (1) \\ Here P = \frac{\partial u}{\partial x}; q = \frac{\partial u}{\partial y} \\ So x_{0} (s) = s^{0} = 1, y_{0} (s) = s, u_{0} (s) = s + 1 \\ u_{0}^{'} (s) = p_{0} x_{0}^{'} (s) + q_{0} y_{0}^{'} (s) \\ 1 = p_{0} (0) + q_{0} (1) \Rightarrow q_{0} = 1 \\ x_{0} p_{0} + (u_{0} - x_{0} - y_{0}) q_{0} - (x_{0} + 2 y_{0}) = 0 \\ p_{0} + (S + 1 - 1 - s) q_{0} - (1 + 2 s) = 0 \\ p_{0} = 2 s + 1 \\ Use \frac{d x}{d t} = f_{p} = x \to (2) \frac{d y}{d t} = f_{q} = u - x - y \to (3) \\ \frac{d u}{d t} = p f_{p} + q f_{q} = p x + q (u - x - y) = x + 2 y \to (4) \\ \frac{d p}{d t} = - f_{x} - p f_{u} \frac{d q}{d t} = - f_{y} - q f_{u} \\ \int \frac{d x}{x} = \int d t \Rightarrow x = c_{1} e^{t} t = t_{0} = 0 \\ x_{0} = c_{1} e^{0} \Rightarrow c_{1} = 1 and x = e^{t} \\ \frac{d y}{d t} + \frac{d u}{d t} = y + u \Rightarrow \int \frac{d (y + u)}{y + u} = \int d t (since y + u = c_{2} e^{t}) \\ y_{0} + u_{0} = c_{2} e^{0} \Rightarrow s + s + 1 = c_{2} \Rightarrow c_{2} = 2 s + 1 \\ y + u = (2 s + 1) e^{t} \\ \frac{d y}{d t} = u - x - y = (2 s + 1) e^{t} - y - e^{t} - y = 2 s e^{t} - 2 y \\ \frac{d y}{d t} + 2 y = 2 s e^{t} \\ I . F = e^{2 \int d t} = e^{2 t} \end{aligned}

\begin{aligned} y e^{2 t} = \int 2 s e^{3 t} d t \\ e^{2 t} y = 2 s \frac{e^{3 t}}{3} + t \\ y = \frac{2 s}{3} e^{t} + t e^{- 2 t} s = \frac{2 s}{3} + k \Rightarrow k = \frac{s}{3} \\ y = \frac{2 s}{3} e^{t} + \frac{s}{3} e^{- 2 t} \end{aligned}

We got

x = e^{t} \to (1); y + u = (2 s + 1) e^{t} \to (2)

\begin{aligned} y = \frac{s}{3} (2 e^{t} + e^{- 2 t}) \\ y + u = (2 s + 1) x \\ (2 s + 1) = \frac{y + u}{x} \Rightarrow s = \frac{y + u - x}{2 x} \end{aligned}

Put value of s from equation (1) and (2)

\begin{aligned} y = (\frac{y + u - x}{2 x \times 3}) (2 x + \frac{1}{x^{2}}) \\ y = (\frac{y + u - x}{6 x^{3}}) (2 x^{3} + 1) \Rightarrow 6 x^{3} y = 2 x^{3} y + 2 x^{3} u - 2 x^{4} + y - x \\ 4 x^{3} y - 2 x^{3} u + 2 x^{4} - y - u + x = 0 \end{aligned}

Conclusion:

We have successfully solved the given first-order quasilinear partial differential equation using the method of characteristics. The solution is expressed as

4 x^{3} y - 2 x^{3} u + 2 x^{4} - y - u + x = 0

6.(b) अधोलिखित संख्याओं के समतुल्यों को उनके सम्मुख दरशाई गई विशिष्ट संख्या पद्धति में ज्ञात कीजिए :
(i) पूर्णांक 524 को द्विआधारी पद्धति में ।
(ii)

101010110101 \cdot 101101011

को अष्टाधारी पद्धति में ।
(iii) दशमलव 5280 को षड्दशमलव पद्धति में ।
(iv) अज्ञात संख्या ज्ञात कीजिए

(1101 \cdot 101)_{8} \to (

?) ।

Find the equivalent numbers given in a specified number to the system mentioned against them :

(i) Integer 524 in binary system.

(ii)

101010110101 \cdot 101101011

to octal system.

(iii) decimal number 5280 to hexadecimal system.

(iv) Find the unknown number

(1101 \cdot 101)_{8} \to (?)_{10} .

Answer:

(i) Integer 524 in binary system.

∴ (524)_{10} = (\underset{―}{1000001100})_{2}

(ii)

101010110101 \cdot 101101011

to octal system.

\begin{aligned} \frac{101}{5} \frac{010}{2} \frac{110}{6} \frac{101}{5} \cdot \frac{101}{5} \frac{101}{5} \frac{011}{3} \\ ∴ (101010110101.101101011)_{2} = (\underset{―}{5265.553})_{8} \end{aligned}

(iii) decimal number 5280 to hexadecimal system.

\begin{aligned} \begin{array}{lrl} 16 & 5280 \\ 16 & 330 & 0 \\ 16 & 20 & A \\ 16 & 1 & 4 \\ 0 & 1 \end{array} \\ ∴ (5280)_{10} = (\underset{―}{14 A 0})_{16} \end{aligned}

(iv) Find the unknown number

(1101 \cdot 101)_{8} \to (?)_{10} .

1101.101

\begin{aligned} = 1 \times 8^{3} + 1 \times 8^{2} + 0 \times 8^{1} + 1 \times 8^{0} + 1 \times \frac{1}{8^{1}} + 0 \times \frac{1}{8^{2}} + 1 \times \frac{1}{8^{3}} \\ = 1 \times 512 + 1 \times 64 + 0 \times 8 + 1 \times 1 + 1 \times 0.125 + 0 \times 0.015625 + 1 \times 0.001953 \\ = 512 + 64 + 0 + 1 + 0.125 + 0 + 0.001953 \\ = 577.126953 \\ ∴ (1101.101)_{8} = (\underset{―}{577.126953})_{10} \end{aligned}

(c) एक त्रिज्या $a$ तथा परिश्रमण त्रिज्या $k$ वाला गोलाकार सिलिन्डर बिना फिसले, एक $b$ त्रिज्या वाले, स्थिर खोखले सिलिन्डर में लुढ़कता (roll) है । दर्शाएँ कि इनकी अक्षों में से तल एक $(b - a) (1 + \frac{k^{2}}{a^{2}})$ लम्बाई वाले गोलाकार लोलक में चलता है ।

A circular cylinder of radius

a

and radius of gyration

k

rolls without slipping inside a fixed hollow cylinder of radius

b

. Show that the plane through axes moves in a circular pendulum of length

(b - a) (1 + \frac{k^{2}}{a^{2}})

Answer:

Introduction:

We are given a circular cylinder of radius

a

and radius of gyration

k

rolling without slipping inside a fixed hollow cylinder of radius

b

. The goal is to show that the plane through the axes of the cylinders moves in a circular pendulum of length

(b - a) (1 + \frac{k^{2}}{a^{2}})

Step 1: Defining Parameters and Angles

Let

P

be the point of contact of the two cylinders at time

t

, and let the angle

A O P

θ

. Also, define

ϕ

as the angle that the line

CB

, fixed in the moving cylinder, makes with the vertical at time

t

Given radii:

Radius of the fixed cylinder: $a$
Radius of the moving cylinder: $a$

Since there is pure rolling, we have:

\begin{aligned} \Rightarrow b θ = a (ϕ + θ) \\ \Rightarrow a ϕ = (b - a) θ \end{aligned}

Therefore,

ϕ^{*} = c θ^{\circ}

(where

c = (b - a)

Step 2: Analyzing Acceleration of Center $C$

The center

C

of the moving cylinder describes a circle of radius

c

. Its accelerations along and perpendicular to

C O

are

c θ^{2}

and

c θ^{\circ}

respectively.

The equations of motion for the moving cylinder are:

\begin{aligned} M c θ^{2} = R - M g \cos θ (2) \\ M c θ^{*} = F - M g \sin θ (3) \end{aligned}

Step 3: Calculating Friction Force $F$

For the motion relative to the center of inertia

c

, we have:

\begin{aligned} M k^{2} ϕ^{*} = Moment of the forces about c = - F a (4) \\ M k^{2} \frac{c}{a} θ^{*} = - F a \\ \Rightarrow F = - M k^{2} \frac{c}{a} θ^{\circ} \end{aligned}

Substituting this into equation (3), we get:

\begin{aligned} M c θ^{\circ} = - M k^{2} \frac{c}{a} θ^{\circ} - M g \sin θ \\ \Rightarrow c (1 + \frac{k^{2}}{a^{2}}) θ^{\circ} = - g \sin θ \end{aligned}

Step 4: Simplifying Angular Acceleration

Simplifying the above equation, we get:

\begin{aligned} θ^{\circ} = - \frac{g}{c (1 + \frac{k^{2}}{a^{2}})} θ \\ = - μ θ (since θ is very small) \end{aligned}

Conclusion:

The length of the simple equivalent pendulum is given by:

\frac{g}{μ} = c (1 + \frac{k^{2}}{a^{2}}) = (b - a) (1 + \frac{k^{2}}{a^{2}})

Thus, the plane through the axes of the cylinders moves in a circular pendulum of length

(b - a) (1 + \frac{k^{2}}{a^{2}})

(a) हेमिल्टन समीकरण का उपयोग करते हुए, एक गोला, जो कि एक खुरदरी आनत तल (inclined plane) पर नीचे की ओर लुढ़क रहा है, का त्वरण ज्ञात करें, यदि $x$ , तल पर निश्चित बिन्दु से गोले के सम्पर्क बिन्दु की दूरी है ।

Using Hamilton’s equation, find the acceleration for a sphere rolling down a rough inclined plane, if

x

be the distance of the point of contact of the sphere from a fixed point on the plane.

Answer:

Introduction:

Using Hamilton’s equation, we need to find the acceleration for a sphere rolling down a rough inclined plane. The relevant parameters are:

Radius of the sphere: $a$
Mass of the sphere: $M$
Angle of inclination of the plane: $α$
Distance rolled down the plane: $x$

Step 1: Deriving the Relationship Between $x$ and $θ$ :

Let a sphere of radius

a

and mass

M

roll down a rough plane inclined at an angle

α

starting initially from a fixed point

O

of the plane. In time

t

, let the sphere roll down a distance

x

and during this time, let it turn through an angle

θ

Since there is no slipping, we have:

x = O A = arc A B = a θ

So,

x = a θ

Step 2: Calculating Kinetic and Potential Energies:

T

and

V

are the kinetic and potential energies of the sphere, then:

T = \frac{1}{2} M k^{2} θ^{2} + \frac{1}{2} M x^{2}

Substituting the values for

k

and

x

and simplifying:

T = \frac{1}{2} M \cdot \frac{2}{5} a^{2} θ^{2} + \frac{1}{2} M (a θ)^{2}

T = \frac{7}{10} M x^{2}

And

V = - M g O L = - M g a \sin α

(since the sphere moves down the plane).

Therefore,

L = T - V = \frac{7}{10} M x^{2} + M g a \sin α

Step 3: Calculating the Generalized Coordinate Momentum:

Here,

x

is the only generalized coordinate.

P_{x} = \frac{\partial L}{\partial x} = \frac{7}{5} M x

Since

L

does not contain

t

explicitly, we can define the Hamiltonian

H

as:

H = T + V = \frac{7}{10} M (\frac{5}{7 M} P_{x}) - M g x \sin α

Simplifying:

H = \frac{5}{14 M} P_{x}^{2} - M g x \sin α (1)

Step 4: Applying Hamilton’s Equations:

Hence, the two Hamilton’s equations are:

P_{x}^{\circ} = - \frac{\partial H}{\partial x} = M g \sin α - (H1)

x^{\circ} = \frac{\partial H}{\partial P_{x}} = \frac{5}{7 M} P_{x} - (H2)

Step 5: Differentiating and Obtaining Acceleration:

Differentiating

(H2)

and using

(H1)

, we get:

x^{\circ \circ} = \frac{5}{7 M} P_{x}^{\circ} = \frac{5}{7 M} M g \sin α

Simplifying:

x^{\circ \circ} = \frac{5}{7} g \sin α

This gives us the required acceleration.

Conclusion:

The acceleration of the sphere rolling down the rough inclined plane is

\frac{5}{7} g \sin α

7.(b) निम्नलिखित समीकरणों को गाउस-साईडल पुनरावृत्ति विधि से हल, दशमलव के सही तीन स्थानों तक करें :

\begin{aligned} 2 x + y - 2 z = 17 \\ 3 x + 20 y - z = - 18 \\ 2 x - 3 y + 20 z = 25 \end{aligned}

Apply Gauss-Seidel iteration method to solve the following system of equations :

\begin{aligned} 2 x + y - 2 z = 17 \\ 3 x + 20 y - z = - 18 \\ 2 x - 3 y + 20 z = 25, correct to three decimal places. \end{aligned}

Answer:

Total Equations are 3

\begin{aligned} 2 x + y - 2 z = 17 \\ 3 x + 20 y - z = - 18 \\ 2 x - 3 y + 20 z = 25 \end{aligned}

From the above equations

x_{k + 1} = \frac{1}{2} (17 - y_{k} + 2 z_{k})

y_{k + 1} = \frac{1}{20} (- 18 - 3 x_{k + 1} + z_{k})

z_{k + 1} = \frac{1}{20} (25 - 2 x_{k + 1} + 3 y_{k + 1})

Initial gauss

(x, y, z) = (0, 0, 0)

Solution steps are

1^{st}

Approximation

x_{1} = \frac{1}{2} [17 - (0) + 2 (0)] = \frac{1}{2} [17] = 8.5

y_{1} = \frac{1}{20} [- 18 - 3 (8.5) + (0)] = \frac{1}{20} [- 43.5] = - 2.175

z_{1} = \frac{1}{20} [25 - 2 (8.5) + 3 (- 2.175)] = \frac{1}{20} [1.475] = 0.0738

2^{nd}

Approximation

\begin{aligned} x_{2} = \frac{1}{2} [17 - (- 2.175) + 2 (0.0738)] = \frac{1}{2} [19.3225] = 9.6612 \\ y_{2} = \frac{1}{20} [- 18 - 3 (9.6612) + (0.0738)] = \frac{1}{20} [- 46.91] = - 2.3455 \\ z_{2} = \frac{1}{20} [25 - 2 (9.6612) + 3 (- 2.3455)] = \frac{1}{20} [- 1.359] = - 0.0679 \end{aligned}

3^{r d}

Approximation

\begin{aligned} x_{3} = \frac{1}{2} [17 - (- 2.3455) + 2 (- 0.0679)] = \frac{1}{2} [19.2096] = 9.6048 \\ y_{3} = \frac{1}{20} [- 18 - 3 (9.6048) + (- 0.0679)] = \frac{1}{20} [- 46.8824] = - 2.3441 \\ z_{3} = \frac{1}{20} [25 - 2 (9.6048) + 3 (- 2.3441)] = \frac{1}{20} [- 1.242] = - 0.0621 \end{aligned}

4^{th}

Approximation

\begin{aligned} x_{4} = \frac{1}{2} [17 - (- 2.3441) + 2 (- 0.0621)] = \frac{1}{2} [19.2199] = 9.61 \\ y_{4} = \frac{1}{20} [- 18 - 3 (9.61) + (- 0.0621)] = \frac{1}{20} [- 46.892] = - 2.3446 \\ z_{4} = \frac{1}{20} [25 - 2 (9.61) + 3 (- 2.3446)] = \frac{1}{20} [- 1.2537] = - 0.0627 \end{aligned}

5^{th}

Approximation

\begin{aligned} x_{5} = \frac{1}{2} [17 - (- 2.3446) + 2 (- 0.0627)] = \frac{1}{2} [19.2192] = 9.6096 \\ y_{5} = \frac{1}{20} [- 18 - 3 (9.6096) + (- 0.0627)] = \frac{1}{20} [- 46.8915] = - 2.3446 \\ z_{5} = \frac{1}{20} [25 - 2 (9.6096) + 3 (- 2.3446)] = \frac{1}{20} [- 1.253] = - 0.0626 \end{aligned}

Solution By Gauss Seidel Method.

\begin{aligned} x = 9.6096 ≅ 9.61 \\ y = - 2.3446 ≅ - 2.34 \\ z = - 0.0626 ≅ - 0.06 \end{aligned}

Iterations are tabulated as below

Iteration	$x$	$y$	$z$
1	8.5	-2.175	0.0738
2	9.6612	-2.3455	-0.0679
3	9.6048	-2.3441	-0.0621
4	9.61	-2.3446	-0.0627
5	9.6096	-2.3446	-0.0626

7.(c) निम्नलिखित द्वितीय कोटि के आंशिक अवकलून समीकरण को विहित रूप में समानीत करें और सामान्य हल ज्ञात करें :

\frac{\partial^{2} u}{\partial x^{2}} - 2 x \frac{\partial^{2} u}{\partial x \partial y} + x^{2} \frac{\partial^{2} u}{\partial y^{2}} = \frac{\partial u}{\partial y} + 12 x ।

Reduce the following second order partial differential equation to canonical form and find the general solution :

\frac{\partial^{2} u}{\partial x^{2}} - 2 x \frac{\partial^{2} u}{\partial x \partial y} + x^{2} \frac{\partial^{2} u}{\partial y^{2}} = \frac{\partial u}{\partial y} + 12 x

Answer:

Introduction:

We are tasked with reducing the given second-order partial differential equation to canonical form and finding the general solution. The equation is:

\frac{\partial^{2} u}{\partial x^{2}} - 2 x \frac{\partial^{2} u}{\partial x \partial y} + x^{2} \frac{\partial^{2} u}{\partial y^{2}} = \frac{\partial u}{\partial y} + 12 x

Standard Form

We start by writing the equation in standard form:

r - 2 x s + x^{2} t = z + 12 x (1)

Where:

$r = \frac{\partial^{2} u}{\partial x^{2}}$
$s = \frac{\partial^{2} u}{\partial x \partial y}$
$t = \frac{\partial^{2} u}{\partial y^{2}}$
$p = \frac{2 u}{r x}$
$q = \frac{\partial u}{\partial y}$

Comparison with Standard Equation

We compare the coefficients with the standard equation:

R r + S s + T t + f (x, y, u, p, q) = 0

In our equation, we have:

$R = 1$
$S = - 2 x$
$T = x^{2}$

Additionally, we calculate

S^{2} - 4 R T

S^{2} - 4 R T = 4 x^{2} - 4 x^{2} = 0

This indicates that the equation is parabolic.

Characteristic Equation

We find the characteristic roots

λ

by solving the quadratic equation:

λ^{2} - 2 x λ + x^{2} = 0

This equation can be factored as

(λ - x)^{2} = 0

, resulting in repeated roots:

λ = x, x

Characteristic Equations

The characteristic equations are given by:

\frac{d y}{d x} + λ = 0

Substituting

λ = x

\frac{d y}{d x} + x = 0

This equation can be integrated to obtain:

y + \frac{x^{2}}{2} = c

We choose

m = y + \frac{x^{2}}{2}

and

n = x

as independent functions.

Independence of $m$ and $n$

To confirm the independence of

m

and

n

, we calculate the determinant of the Jacobian:

| \begin{array}{ll} \frac{\partial m}{\partial x} & \frac{\partial m}{\partial y} \\ \frac{\partial n}{\partial x} & \frac{\partial n}{\partial y} \end{array} | = | \begin{array}{ll} x & 1 \\ 1 & 0 \end{array} | = - 1 \neq 0

Since the determinant is nonzero,

m

and

n

are indeed independent functions.

Expressing $p$ , $q$ , $r$ , and $t$

\begin{aligned} p & = \frac{\partial u}{\partial x} = \frac{\partial u}{\partial m} \frac{\partial m}{\partial x} + \frac{\partial u}{\partial n} \frac{\partial u}{\partial x} = x \frac{\partial u}{\partial m} + \frac{\partial u}{\partial n} \\ q & = \frac{\partial u}{\partial y} = \frac{\partial u}{\partial m} \frac{\partial m}{\partial y} + \frac{\partial u}{\partial n} \frac{\partial n}{\partial y} = \frac{\partial u}{\partial m} \\ r & = \frac{\partial^{2} u}{\partial x^{2}} = \frac{\partial p}{\partial x} = \frac{\partial}{\partial x} (x \frac{\partial u}{\partial m}) + \frac{\partial}{\partial x} (\frac{\partial u}{\partial x}) \\ = \frac{\partial u}{\partial m} + x [\frac{\partial}{\partial m} (\frac{\partial u}{\partial m}) \frac{\partial m}{\partial x} + \frac{\partial}{\partial n} (\frac{\partial u}{\partial m}) \frac{\partial m}{\partial x}] + [\frac{\partial}{\partial m} (\frac{\partial u}{\partial n}) \frac{\partial m}{\partial x} + \frac{\partial}{\partial n} (\frac{\partial u}{\partial n}) \frac{\partial n}{\partial x}] \\ r & = \frac{\partial u}{\partial m} + \frac{x^{2} \partial^{2} u}{\partial m^{2}} + x \frac{\partial^{2} u}{\partial x \partial y} + x \frac{\partial^{2} u}{\partial m \partial n} + \frac{\partial^{2} u}{\partial n^{2}} \\ t & = \frac{\partial^{2} u}{\partial y^{2}} = \frac{\partial q}{\partial y} = \frac{\partial}{\partial m} (\frac{\partial u}{\partial m}) \frac{\partial m}{\partial y} + \frac{\partial}{\partial n} (\frac{\partial u}{\partial m}) \frac{\partial m}{\partial y} = \frac{\partial^{2} u}{\partial m^{2}} \\ s & = \frac{\partial^{2} u}{\partial x \partial y} = \frac{\partial q}{\partial x} = \frac{\partial}{\partial m} (\frac{\partial u}{\partial m}) \frac{\partial m}{\partial x} + \frac{\partial}{\partial n} (\frac{\partial u}{\partial n}) \frac{\partial n}{\partial x} = \frac{\partial^{2} u}{\partial m^{2}} + \frac{\partial^{2} u}{\partial m \partial n} \end{aligned}

Equation (1) becomes

\begin{aligned} \frac{\partial u}{\partial m} + (x^{2}) \frac{\partial^{2} u}{\partial m^{2}} + (2 x) \frac{\partial^{2} u}{\partial m \partial n} + \frac{\partial^{2} u}{\partial n^{2}} - (2 x^{2}) \frac{\partial^{2} u}{\partial m^{2}} - (2 x) \frac{\partial^{2}}{\partial m \partial n} + (x^{2}) \frac{\partial^{2} u}{\partial m^{2}} = \frac{\partial u}{\partial m} + 12 x \\ \frac{\partial^{2} u}{\partial n^{2}} = 12 x \end{aligned}

Now the general solution

\frac{\partial^{2} u}{\partial n^{2}} = 12 x; m = y + \frac{x^{2}}{2}; n = x

By integrating

\frac{\partial u}{\partial n} = \frac{12 x^{2}}{2} + f (m), d n = d x

\begin{aligned} \frac{\partial u}{\partial x} = 6 x^{2} + f (m) \\ u = \frac{6 x^{3}}{3} + x f (m) + g (m) \\ u = 2 x^{3} + x f (y + \frac{x^{2}}{2}) + g (y + \frac{x^{2}}{2}) \end{aligned}

Conclusion:

We have successfully reduced the given second-order partial differential equation to canonical form and found the general solution, which is expressed as:

u = 2 x^{3} + x f (y + \frac{x^{2}}{2}) + g (y + \frac{x^{2}}{2})

8.(a) दिये गये बूलीय व्यंजक के लिए

X = A B + A B C + A \bar{B} \bar{C} + A \bar{C}

(i) व्यंजक के लिये तार्किक आरेख खींचें ।
(ii) व्यंजक न्यूनतम करें ।
(iii) समानीत व्यंजक के लिये तार्किक आरेख खींचें ।

Given the Boolean expression

X = A B + A B C + A \bar{B} \bar{C} + A \bar{C}

(i) Draw the logical diagram for the expression.
(ii) Minimize the expression.
(iii) Draw the logical diagram for the reduced expression.

Answer:

(i) Draw the logical diagram for the expression.

Part (ii): Minimize the Expression
To minimize the expression, we can use Boolean algebraic rules. The given expression is:

X = A B + A B C + A \bar{B} \bar{C} + A \bar{C}

Let’s substitute the values:

X = A (B + B C) + A (\bar{B} \bar{C} + \bar{C})

After Calculating, we get:

X = A (B (1 + C)) + A (\bar{C} (1 + \bar{B}))

After Calculating, we get:

X = A (B) + A (\bar{C})

After Calculating, we get:

X = A B + A \bar{C}

(iii) Draw the logical diagram for the reduced expression.

8.(b) एक त्रिज्या

R

का गोला, जिसका केन्द्र स्थिर है वह घनत्व

ρ

के एक अनंत असंपीडय तरल में त्रिज्यत:
कंपन करता है.। अगर अनंत पर दबाव

Π

हो, तो दर्शाएं कि गोले की सतह पर किसी समय

t

पर दाब

Π + \frac{1}{2} ρ {\frac{d^{2} R^{2}}{d t^{2}} + {(\frac{d R}{d t})}^{2}}

होगा ।

A sphere of radius

R

, whose centre is at rest, vibrates radially in an infinite incompressible fluid of density

ρ

, which is at rest at infinity. If the pressure at infinity is

Π

, so that the pressure at the surface of the sphere at time

t

Π + \frac{1}{2} ρ {\frac{d^{2} R^{2}}{d t^{2}} + {(\frac{d R}{d t})}^{2}}

Answer:

In the incompressible liquid, outside the sphere, the fluid velocity

\bar{q}

will be radial and thus will be a function of

r

, the radial distance from the centre of the sphere (the origin), and time t only.

The equation of continuity in spherical polar co-ordinates becomes

\begin{matrix} \frac{1}{r^{2}} \frac{d}{dr} (r^{2} u) = 0 - - - - (1) \\ ∵ \overset{―}{q} = (u, 0, 0), u = u (r, t), \nabla \equiv (\frac{\partial}{\partial r}, 0, 0) \\ \nabla \cdot \overset{―}{q} = \frac{1}{r^{2}} \frac{\partial}{\partial r} (r^{2} u) . \end{matrix}

i.e. sphericalsymmetry

\Rightarrow r^{2} u = constant = f (t)

On the surface of the sphere,
Therefore,

r = R, u = \dot{R}

and thus

f (t) = R^{2} \dot{R}

r^{2} u = R^{2} \dot{R} - - - - (2)

We observe that

u \to 0

n \to \infty

, as required.
From (1), it is clear that curl

\overset{―}{q} = \overset{―}{0}

\Rightarrow

the motion is irrotational and

\overset{―}{q} = - \nabla ϕ

\begin{aligned} \Rightarrow u = - \frac{\partial ϕ}{\partial r} \Rightarrow - \frac{\partial ϕ}{\partial r} = \frac{f}{r^{2}} - - - - | F r o m (2) \\ \Rightarrow ϕ = f / r - - - - (3) \end{aligned}

The pressure equation for irrotational non-steady fluid motion in the absence of body forces is

\frac{p}{ρ} + \frac{1}{2} {\overset{―}{q}}^{2} - \frac{\partial ϕ}{\partial t} = C (t)

i.e.

\frac{p}{ρ} + \frac{1}{2} u^{2} - \frac{\partial ϕ}{\partial t} = C (t) - - - - (4)

where

C (t)

is a function of time

t

r \to \infty, p \to Π, u = f / r^{2} \to 0, ϕ \to 0

so that

C (t) = Π / ρ

for all

t

—-(5)

Therefore, from (2), (3), (4) and (5), we get

\begin{aligned} \frac{p}{ρ} = \frac{Π}{ρ} + \frac{\partial}{\partial t} (f / r) - \frac{1}{2} {(\frac{R^{2} \dot{R}}{r^{2}})}^{2} - - - - (6) \\ But \frac{\partial f}{\partial t} = \frac{d}{dt} (R^{2} \dot{R}) = \ddot{R} R^{2} + 2 R {\dot{R}}^{2} \end{aligned}

At the surface of the sphere, we have

r = R

and equation (6) gives

\begin{aligned} \frac{p}{ρ} & = \frac{Π}{ρ} + \frac{1}{R} (2 R {\dot{R}}^{2} + \ddot{R} R^{2}) - \frac{1}{2} {\dot{R}}^{2} \\ \Rightarrow \frac{p}{ρ} & = \frac{Π}{ρ} + 2 {\dot{R}}^{2} + R \ddot{R} - \frac{1}{2} {\dot{R}}^{2} \\ = \frac{Π}{ρ} + \frac{1}{2} (3 {\dot{R}}^{2} + 2 R \ddot{R}) - - - - (7) \end{aligned}

Now,

\begin{aligned} \frac{d^{2} (R^{2})}{{dt}^{2}} + (\dot{R})^{2} & = \frac{d}{dt} (2 R \dot{R}) + (\dot{R})^{2} \\ = (2 R \ddot{R} + 2 {\dot{R}}^{2}) + {\dot{R}}^{2} \\ = 2 R \ddot{R} + 3 {\dot{R}}^{2} \end{aligned}

Therefore, from (7), we obtain

p = Π + \frac{1}{2} ρ [\frac{d^{2} (R^{2})}{{dt}^{2}} + {(\frac{dR}{dt})}^{2}]

Hence the result.

(c) दो स्त्रोतों, प्रत्येक $m$ शक्ति का $(- a, 0), (a, 0)$ बिन्दुओं पर तथा $2 m$ शक्ति का सिन्क मूल बिन्दु पर स्थित है। दर्शाएं कि धारा-रेखाएं वक्र ${(x^{2} + y^{2})}^{2} = a^{2} (x^{2} - y^{2} + λ x y)$ हैं । यहां $λ$ चर एक पैरामीटर है ।
और ये भी दर्शाएं कि तरल गति किसी भी बिन्दु पर $(2 m a^{2}) / (r_{1} r_{2} r_{3})$ है, जहां $r_{1}, r_{2}, r_{3}$ सोतों से और सिंक से बिन्दुओं की क्रमशः दूरिया हैं।

Two sources, each of strength

m

, are placed at the points

(- a, 0), (a, 0)

and a sink of strength

2 m

at origin. Show that the stream lines are the curves

{(x^{2} + y^{2})}^{2} = a^{2} (x^{2} - y^{2} + λ x y)

, where

λ

is a variable parameter.
Show also that the fluid speed at any point is

(2 m a^{2}) / (r_{1} r_{2} r_{3})

, where

r_{1}, r_{2}

and

r_{3}

are the distances of the points from the sources and the sink, respectively.

Answer:

Introduction:
Two sources, each of strength

m

, are placed at the points

(- a, 0)

and

(a, 0)

, and a sink of strength

2 m

at the origin. We need to show that the streamlines are given by the curves

{(x^{2} + y^{2})}^{2} = a^{2} (x^{2} - y^{2} + λ x y)

, where

λ

is a variable parameter. Additionally, we need to demonstrate that the fluid speed at any point is

\frac{2 m a^{2}}{r_{1} r_{2} r_{3}}

, where

r_{1}, r_{2}

, and

r_{3}

are the distances of the points from the sources and the sink, respectively.

Answer:

First Part: Streamlines
The complex potential

w

at any point

P (z)

is given by

w = - m \log (z - a) - m \log (z + a) + 2 m \log z (1)

w = m [\log z^{2} - \log (z^{2} - a^{2})]

ϕ + i ψ = m [\log (x^{2} - y^{2} + 2 i x y) - \log (x^{2} - y^{2} - a^{2} + 2 i x y)], as z = x + i y

Equating the imaginary parts, we have

\begin{aligned} ψ = m [\tan^{- 1} {2 x y / (x^{2} - y^{2})} - \tan^{- 1} {2 x y / (x^{2} - y^{2} - a^{2})}] \\ ∴ & ψ = m \tan^{- 1} [\frac{- 2 a^{2} x y}{{(x^{2} + y^{2})}^{2} - a^{2} (x^{2} - y^{2})}], on simplification. \end{aligned}

The desired streamlines are given by

ψ =

constant

= m \tan^{- 1} (- 2 / λ)

. Then we obtain

(- 2 / λ) = (- 2 a^{2} x y) / [{(x^{2} + y^{2})}^{2} - a^{2} (x^{2} - y^{2})]

, which represents the streamlines.

Second Part: Fluid Speed
From equation (1), we have

\frac{d w}{d z} = - \frac{m}{z - a} - \frac{m}{z + a} + \frac{2 m}{z} = - \frac{2 a^{2} m}{z (z - a) (z + a)}

Let

q = | \frac{d w}{d z} |

. Then,

q = \frac{2 a^{2} m}{| z | | z - a | | z + a |} = \frac{2 a^{2} m}{r_{1} r_{2} r_{3}}

where

r_{1} = | z - a |, r_{2} = | z + a | and r_{3} = | z |

This represents the fluid speed at any point in terms of the distances

r_{1}, r_{2},

and

r_{3}

from the sources and the sink.

Conclusion:
This concludes the solution to the problem. We have shown the streamlines and derived the expression for fluid speed at any point in the given configuration.

Free UPSC Mathematics Optional Paper-2 2019 Solutions: View Online | UPSC Maths Solution | IAS Maths Solution

Introduction:

Work/Calculations:

Definition of Index:

Step 1: Express ( G : K ) ( G : K ) (G:K)(G: K)(G:K) in terms of cosets

Step 2: Express ( G : H ) ( G : H ) (G:H)(G: H)(G:H) and ( H : K ) ( H : K ) (H:K)(H: K)(H:K) in terms of cosets

Step 3: Relate G / K G / K G//KG/KG/K with G / H G / H G//HG/HG/H and H / K H / K H//KH/KH/K

Step 4: Mathematical Expression

Conclusion:

Introduction:

Work/Calculations:

Step 1: Checking Continuity at ( 1 , − 1 ) ( 1 , − 1 ) (1,-1)(1, -1)(1,−1)

Step 2: Checking Differentiability at ( 1 , − 1 ) ( 1 , − 1 ) (1,-1)(1, -1)(1,−1)

Conclusion:

Introduction

Work/Calculations

Step 1: Define the Contour Integral

Step 2: Evaluate the Integral along C R C R C_(R)C_RCR

Step 3: Find the Residues

Step 4: Apply the Residue Theorem

Step 5: Evaluate the Original Integral

Conclusion

Introduction

Work/Calculations

Step 1: Write f ( z ) f ( z ) f(z)f(z)f(z) in terms of its real and imaginary parts

Step 2: Use the given condition

Step 3: Use the Cauchy-Riemann equations

Step 4: Analyze the equations

Step 5: Conclude that f ( z ) f ( z ) f(z)f(z)f(z) is constant

Conclusion

Introduction:

Work/Calculations:

Step 1: Definitions and Assumptions

Step 2: Properties of Homomorphism

Step 3: Relatively Prime Orders

Step 4: The Trivial Homomorphism

Conclusion:

Introduction:

Work/Calculations:

Step 1: Identify Subgroups of Z 12 Z 12 Z_(12)\mathbb{Z}_{12}Z12

Step 2: Find Quotient Groups

Step 3: List All Quotient Groups

Conclusion:

Introduction

Work/Calculations

Part 1: f ( z ) = ϕ ( z ) ( z − z 0 ) m f ( z ) = ϕ ( z ) ( z − z 0 ) m f(z)=(phi(z))/((z-z_(0))^(m))f(z) = \frac{\phi(z)}{(z – z_0)^m}f(z)=ϕ(z)(z−z0)m implies z 0 z 0 z_(0)z_0z0 is a pole of order m m mmm

Part 2: z 0 z 0 z_(0)z_0z0 is a pole of order m m mmm implies f ( z ) = ϕ ( z ) ( z − z 0 ) m f ( z ) = ϕ ( z ) ( z − z 0 ) m f(z)=(phi(z))/((z-z_(0))^(m))f(z) = \frac{\phi(z)}{(z – z_0)^m}f(z)=ϕ(z)(z−z0)m

Part 3: Residue of f ( z ) f ( z ) f(z)f(z)f(z) at z 0 z 0 z_(0)z_0z0

Conclusion

Introduction

Work/Calculations

Step 1: Pointwise Convergence

Step 2: Uniform Convergence

Conclusion

Introduction

Work/Calculations

Step 1: Parametrization of the Curve C C CCC

Step 2: Compute d z d z dzdzdz

Step 3: Compute Re ( z 2 ) Re ( z 2 ) Re(z^(2))\operatorname{Re}(z^2)Re(z2)

Step 4: Evaluate the Integral

Conclusion

Introduction

Work/Calculations

Step 1: Definitions and Preliminaries

Step 2: a a aaa is Irreducible ⇒ a ⇒ a =>a\Rightarrow a⇒a is Prime

Step 3: a a aaa is Prime ⇒ ( a ) ⇒ ( a ) =>(a)\Rightarrow (a)⇒(a) is a Prime Ideal

Step 4: ( a ) ( a ) (a)(a)(a) is a Prime Ideal ⇒ R / ( a ) ⇒ R / ( a ) =>R//(a)\Rightarrow R/(a)⇒R/(a) is an Integral Domain

Step 5: R / ( a ) R / ( a ) R//(a)R/(a)R/(a) is an Integral Domain ⇒ R / ( a ) ⇒ R / ( a ) =>R//(a)\Rightarrow R/(a)⇒R/(a) is a Field

Conclusion

Introduction

Formulation of the Dual Problem

Completion of the Solution

Optimal Solution of the Dual Problem

Verifying the Optimality of the Basic Solution ( x 1 , x 2 ) ( x 1 , x 2 ) (x_(1),x_(2))(x_1, x_2)(x1,x2) in the Primal Problem

Conclusion

Introduction

Work/Calculations

Part (i): Flowchart for the Algorithm

Part (ii): Algorithm in C++

Conclusion

Step 1: Express $(G : K)$ in terms of cosets

Step 2: Express $(G : H)$ and $(H : K)$ in terms of cosets

Step 3: Relate $G / K$ with $G / H$ and $H / K$

Step 1: Checking Continuity at $(1, - 1)$

Step 2: Checking Differentiability at $(1, - 1)$

Step 2: Evaluate the Integral along $C_{R}$

Step 1: Write $f (z)$ in terms of its real and imaginary parts

Step 5: Conclude that $f (z)$ is constant

Step 1: Identify Subgroups of $Z_{12}$

Part 1: $f (z) = \frac{ϕ (z)}{(z - z_{0})^{m}}$ implies $z_{0}$ is a pole of order $m$

Part 2: $z_{0}$ is a pole of order $m$ implies $f (z) = \frac{ϕ (z)}{(z - z_{0})^{m}}$

Part 3: Residue of $f (z)$ at $z_{0}$

Step 1: Parametrization of the Curve $C$

Step 2: Compute $d z$

Step 3: Compute $Re (z^{2})$

Step 2: $a$ is Irreducible $\Rightarrow a$ is Prime

Step 3: $a$ is Prime $\Rightarrow (a)$ is a Prime Ideal

Step 4: $(a)$ is a Prime Ideal $\Rightarrow R / (a)$ is an Integral Domain

Step 5: $R / (a)$ is an Integral Domain $\Rightarrow R / (a)$ is a Field

Verifying the Optimality of the Basic Solution $(x_{1}, x_{2})$ in the Primal Problem