Free UPSC Mathematics Optional Paper-2 2018 Solutions: View Online | UPSC Maths Solution | IAS Maths Solution

July 13, 2025ByAbstract Classes

खण्ड ‘A’ SECTION ‘

A

‘
1.(a) मान लीजिए

R

तत्समक अवयव सहित एक पूर्णांकीय प्रांत है । दर्शाइए कि

R [x]

में कोई भी एवक

R

में एक एकक है।

Let

R

be an integral domain with unit element. Show that any unit in

R [x]

is a unit in

R

Answer:

Introduction

In this problem, we are dealing with the concept of units in the context of integral domains and polynomial rings. Specifically, we are given an integral domain

R

with a unit element and are asked to prove that any unit in

R [x]

(the polynomial ring over

R

) must also be a unit in

R

Definitions

Integral Domain: An integral domain is a commutative ring $R$ with a multiplicative identity (unit element) such that the product of any two non-zero elements is non-zero.
Unit in a Ring: An element $a$ in a ring $R$ is called a unit if there exists an element $b$ in $R$ such that $a \cdot b = b \cdot a = 1$ , where $1$ is the multiplicative identity in $R$ .
Polynomial Ring $R [x]$ : The set of all polynomials with coefficients in $R$ .

Work/Calculations

Step 1: Assume $f (x)$ is a Unit in $R [x]$

Let’s assume that

f (x)

is a unit in

R [x]

. This means there exists a polynomial

g (x)

R [x]

such that:

f (x) \cdot g (x) = 1

Step 2: Examine the Degrees of $f (x)$ and $g (x)$

The degree of the polynomial

f (x) \cdot g (x)

is the sum of the degrees of

f (x)

and

g (x)

. Since

f (x) \cdot g (x) = 1

, a constant polynomial, the degree of

f (x) \cdot g (x)

is 0.

Let

deg (f (x)) = m

and

deg (g (x)) = n

deg (f (x) \cdot g (x)) = m + n = 0

Step 3: Conclude $m = n = 0$

Since

m + n = 0

and

m, n \geq 0

, it must be the case that

m = n = 0

Step 4: Show $f (x)$ is a Unit in $R$

Since

m = 0

f (x)

is a constant polynomial, say

f (x) = a

, where

a

is in

R

. Similarly,

g (x) = b

, where

b

is in

R

From

f (x) \cdot g (x) = 1

, we have

a \cdot b = 1

This shows that

a

is a unit in

R

, as

a \cdot b = 1

Conclusion

We have shown that if

f (x)

is a unit in

R [x]

, then it must be a constant polynomial and also a unit in

R

. Therefore, any unit in

R [x]

is a unit in

R

1.(b) असमिका :

\frac{π^{2}}{9} < \int_{\frac{π}{6}}^{\frac{π}{2}} \frac{x}{\sin x} d x < \frac{2 π^{2}}{9}

को सिद्ध कीजिए ।

Prove the inequality :

\frac{π^{2}}{9} < \int_{\frac{π}{6}}^{\frac{π}{2}} \frac{x}{\sin x} d x < \frac{2 π^{2}}{9}

Answer:

Introduction:
We have the inequality:

1 \leq \frac{1}{\sin x} \leq 2 for all x \in [\frac{π}{6}, \frac{π}{2}] .

Multiply by

x

Therefore, it follows that:

x \leq \frac{x}{\sin x} \leq 2 x for all x \in [\frac{π}{6}, \frac{π}{2}]

Step 1: Defining $f (x)$ :
Let’s define a function

f (x) = \frac{x}{\sin x}

Step 2: Defining $ϕ (x)$ and $ψ (x)$ :
We define two more functions as follows:

\begin{aligned} ϕ (x) = x \\ ψ (x) = 2 x, x \in [π / 6, π / 2] \end{aligned}

Step 3: Boundedness and Integrability:
Both

f

and

ϕ

are bounded and integrable on

[π / 6, π / 2]

, and

f (x) \geq ϕ (x)

for all

x \in [π / 6, π / 2]

Step 4: Continuity at $π / 3$ :
Furthermore,

f

and

ϕ

are both continuous at

x = π / 3

, and

f (π / 3) > ϕ (π / 3)

Step 5: Integral Comparison:
Hence, we can compare the integrals:

\begin{aligned} \int_{π / 6}^{π / 2} f (x) d x > \int_{π / 6}^{π / 2} ϕ (x) d x \\ = \int_{π / 6}^{π / 2} x d x \\ = \frac{π^{2}}{9} \end{aligned}

Step 6: Comparing with $ψ$ :
Similarly, we have:

\begin{aligned} \int_{π / 6}^{π / 2} f (x) d x & < \int_{π / 6}^{π / 2} ψ (x) d x \\ = 2 \int_{π / 6}^{π / 2} x d x \\ = \frac{2 π^{2}}{9} \end{aligned}

Conclusion:
Consequently, we can conclude that:

\frac{π^{2}}{9} < \int_{π / 6}^{π / 2} \frac{x}{\sin x} d x < \frac{2 π^{2}}{9}

1.(c) सिद्ध कीजिए कि फलन :

u (x, y) = (x - 1)^{3} - 3 x y^{2} + 3 y^{2}

प्रसंवादी है और इसके प्रसंबादी संयुग्मी को और संगत विश्लेषिक फलन

f (z)

को,

z

के ल्प में ज्ञात कीजिए ।

Prove that the function:

u (x, y) = (x - 1)^{3} - 3 x y^{2} + 3 y^{2}

is harmonic and find its harmonic conjugate and the corresponding analytic function

f (z)

in terms of

z

Answer:

Introduction:

In this problem, we are tasked with proving that the function

u (x, y) = (x - 1)^{3} - 3 x y^{2} + 3 y^{2}

is harmonic. Furthermore, we need to find its harmonic conjugate and express the corresponding analytic function

f (z)

in terms of

z

Step 1: Verification of Harmonicity:

To begin, we need to determine whether the given function

u (x, y)

is harmonic. We calculate its partial derivatives with respect to

x

and

y

\begin{aligned} \frac{\partial u}{\partial x} & = 3 (x - 1)^{2} - 3 y^{2} \\ \frac{\partial^{2} u}{\partial x^{2}} & = 6 (x - 1) \\ \frac{\partial u}{\partial y} & = - 6 x y + 6 y \\ \frac{\partial^{2} u}{\partial y^{2}} & = - 6 (x - 1) \end{aligned}

We notice that

\frac{\partial^{2} u}{\partial x^{2}} + \frac{\partial^{2} u}{\partial y^{2}} = 6 (x - 1) - 6 (x - 1) = 0

. This confirms that

u

is a harmonic function.

Step 2: Harmonic Conjugate and Analytic Function $f (z)$ :

Next, we aim to find the harmonic conjugate of

u

and express the corresponding analytic function

f (z)

in terms of

z

We recognize that

u (x, y)

is the real part of the analytic function

f (z)

, where

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

We utilize the Cauchy-Riemann equations, which state that

u_{x} = v_{y}

and

u_{y} = - v_{x}

, to determine the imaginary part

v (x, y)

f (z)

\begin{aligned} v_{x} & = - u_{y} = - \frac{\partial u}{\partial y} = - (- 6 x y + 6 y) = 6 x y - 6 y \\ v (x, y) & = \int (6 x y - 6 y) d y = 3 x^{2} y - 6 x y + f_{1} (y) \end{aligned}

Here,

f_{1} (y)

is an arbitrary function of

y

Now, we differentiate

v (x, y)

with respect to

y

to find

f_{1} (y)

\begin{aligned} \frac{\partial v}{\partial y} & = 3 x^{2} - 6 x + f_{1}^{'} (y) \\ \Rightarrow \frac{\partial u}{\partial x} & = 3 x^{2} - 6 x + f_{1}^{'} (y) (by the Cauchy-Riemann equations) \\ 3 (x - 1)^{2} - 3 y^{2} & = 3 x^{2} - 6 x + f_{1}^{'} (y) \\ f_{1}^{'} (y) & = 3 - 3 y^{2} \\ f_{1} (y) & = 3 y - y^{3} + C (where C is an arbitrary constant) \end{aligned}

Thus, the imaginary part

v (x, y)

is given by

v (x, y) = 3 x^{2} y - 6 x y + 3 y - y^{3} + C

Step 3: Expressing $f (z)$ in terms of $z$ :

\begin{aligned} f^{'} (z) & = ϕ_{1} (z, 0) - i ϕ_{2} (z, 0) \\ f^{'} (z) & = 3 (z - 1)^{2} - 3 (0)^{2} - i [6 \cdot z \cdot (0) + 6 (0)] \\ f^{'} (z) & = 3 (z - 1)^{2} \\ ∴ & f (z) = (z - 1)^{3} + C \end{aligned}

Finally, we express the analytic function

f (z)

in terms of

z

. We use the result that

f (z) = (z - 1)^{3} + C

Conclusion:

In conclusion, the given function

u (x, y)

is confirmed to be harmonic. We have found its harmonic conjugate,

v (x, y)

, and expressed the corresponding analytic function

f (z)

f (z) = (z - 1)^{3} + C

, where

C

is an arbitrary constant.

1.(d)

p (> 0)

का वह परास ज्ञात कीजिए, जिसके लिए श्रेणी:

\frac{1}{(1 + a)^{p}} - \frac{1}{(2 + a)^{p}} + \frac{1}{(3 + a)^{p}} - \dots, a > 0

(i) निरपेक्षतः अभिसारी तथा (ii) सापेक्ष अभिसारी है ।

Find the range of

p (> 0)

for which the series:

\frac{1}{(1 + a)^{p}} - \frac{1}{(2 + a)^{p}} + \frac{1}{(3 + a)^{p}} - \dots, a > 0

, is
(i) absolutely convergent and (ii) conditionally convergent.

Answer:

Introduction:

We are tasked with finding the range of

p (> 0)

for which the series given by:

\frac{1}{(1 + a)^{p}} - \frac{1}{(2 + a)^{p}} + \frac{1}{(3 + a)^{p}} - \dots, where a > 0

converges absolutely and conditionally. To do this, we will use various convergence tests and mathematical techniques.

Step 1: Definition of $v_{n}$ and $ω_{n}$ :

Let

\sum u_{n}

be the given series, and define a new series

v_{n}

as follows:

v_{n} = \frac{1}{(n + a)^{p}}

Additionally, define

ω_{n}

as:

ω_{n} = \frac{1}{n^{p}}

Step 2: Checking Convergence of $\sum v_{n}$ for $p > 1$ :

By comparing

v_{n}

with

ω_{n}

, we find that:

lim_{n \to \infty} \frac{v_{n}}{ω_{n}} = 1

Using the comparison test, we conclude that

\sum v_{n}

is convergent when

p > 1

Step 3: Case 1 – Absolute Convergence for $P > 1$ :

\sum u_{n}

is an alternating series and

\sum | u_{n} |

is convergent. Therefore,

\sum u_{n}

is absolutely convergent.

Step 4: Case 2 – Conditional Convergence for $0 < p \leq 1$ :

When

0 < p \leq 1

, the sequence

{v_{n}}

is a monotone decreasing sequence of positive real numbers with

lim v_{n} = 0

. By Leibniz’s test,

\sum (- 1)^{n + 1} v_{n}

, which is equivalent to

\sum u_{n}

, is convergent.

Since

\sum | u_{n} |

is divergent in this case,

\sum u_{n}

is conditionally convergent.

Step 5: Decomposition of $u_{n}$ into $P_{n}$ and $q_{n}$ :

Let

Σ u_{n}

be a series of positive real numbers. We can decompose it into two new series:

\begin{aligned} P_{n} & = {\begin{cases} u_{n} & if u_{n} > 0 \\ 0 & if u_{n} \leq 0 \end{cases} \\ q_{n} & = {\begin{cases} 0 & if u_{n} \geq 0 \\ u_{n} & if u_{n} < 0 \end{cases} \end{aligned}

Conclusion:

In summary, the given series

\sum u_{n}

is absolutely convergent when

P > 1

, and it is conditionally convergent when

0 < p \leq 1

. This analysis provides a comprehensive understanding of the convergence behavior of the series for different values of

p

in the specified range.

1.(e) एक कृषि फर्म के पास 180 टन नाइट्रोजन उर्बरक, 250 टन फॉस्फेट तथा 220 टन पोटाश है । फ़र्म इन पदार्थों के क्रमशः

3 : 3 : 4

के अनुपात में मिश्रण को 1500 रुपये प्रति टन के मुनाफे से तथा

2 : 4 : 2

के अनुपात में मिश्रण को 1200 रुपये प्रति टन के मुनाफे से बेच पायेगी। एक रैखिक-प्रोग्रामन समस्या प्रस्तुत कीजिए, जो यह दर्शाए कि अधिकतम मुनाफा प्राप्त करने के लिए, इन मिश्रणों की कितने टन मात्रा तैयार की जानी चाहिए।

An agricultural firm has 180 tons of nitrogen fertilizer, 250 tons of phosphate and 220 tons of potash. It will be able to sell a mixture of these substances in their respective ratio

3 : 3 : 4

at a profit of Rs. 1500 per ton and a mixture in the ratio

2 : 4 : 2

at a profit of Rs. 1200 per ton. Pose a linear programming problem to show how many tons of these two mixtures should be prepared to obtain the maximum profit.

Answer:

To formulate a Linear Programming (LP) problem, we need to define the decision variables, the objective function, and the constraints.

Decision Variables

Let:

$x$ be the number of tons of the mixture in the ratio $3 : 3 : 4$ to be prepared.
$y$ be the number of tons of the mixture in the ratio $2 : 4 : 2$ to be prepared.

Objective Function

The firm wants to maximize its profit. The profit for the mixture in the ratio

3 : 3 : 4

is Rs. 1500 per ton, and for the mixture in the ratio

2 : 4 : 2

, it is Rs. 1200 per ton. So, the objective function to maximize is:

P = 1500 x + 1200 y

Constraints

The firm has limited amounts of nitrogen fertilizer, phosphate, and potash, so we need to set up constraints based on the availability of these substances.

Certainly! Here is the corrected version with

x

and

y

replaced by

x_{1}

and

x_{2}

Nitrogen Fertilizer Constraint

For the mixture in the ratio

3 : 3 : 4

, 3 parts out of 10 are nitrogen fertilizer. So, for every ton of this mixture,

\frac{3}{10}

tons of nitrogen fertilizer are used. Similarly, for the mixture in the ratio

2 : 4 : 2

, 2 parts out of 8 are nitrogen fertilizer, i.e.,

\frac{2}{8} = \frac{1}{4}

tons of nitrogen fertilizer per ton of mixture. The firm has 180 tons of nitrogen fertilizer available, so:

\frac{3}{10} x_{1} + \frac{1}{4} x_{2} \leq 180

6 x_{1} + 5 x_{2} \leq 3600

Phosphate Constraint

For the mixture in the ratio

3 : 3 : 4

, 3 parts out of 10 are phosphate. So, for every ton of this mixture,

\frac{3}{10}

tons of phosphate are used. For the mixture in the ratio

2 : 4 : 2

, 4 parts out of 8 are phosphate, i.e.,

\frac{4}{8} = \frac{1}{2}

tons of phosphate per ton of mixture. The firm has 250 tons of phosphate available, so:

\frac{3}{10} x_{1} + \frac{1}{2} x_{2} \leq 250

3 x_{1} + 5 x_{2} \leq 2500

Potash Constraint

For the mixture in the ratio

3 : 3 : 4

, 4 parts out of 10 are potash. So, for every ton of this mixture,

\frac{4}{10}

tons of potash are used. For the mixture in the ratio

2 : 4 : 2

, 2 parts out of 8 are potash, i.e.,

\frac{2}{8} = \frac{1}{4}

tons of potash per ton of mixture. The firm has 220 tons of potash available, so:

\frac{4}{10} x_{1} + \frac{1}{4} x_{2} \leq 220

8 x_{1} + 5 x_{2} \leq 4400

Non-negativity Constraint

The number of tons of the mixtures cannot be negative:

x_{1} \geq 0

x_{2} \geq 0

Linear Programming Problem

To summarize, the Linear Programming problem is as follows:

Maximize

Z = 1500 x_{1} + 1200 x_{2}

Subject to

\begin{aligned} 6 x_{1} + 5 x_{2} \leq 3600 \\ 3 x_{1} + 5 x_{2} \leq 2500 \\ 8 x_{1} + 5 x_{2} \leq 4400 \\ x_{1}, x_{2} \geq 0 \end{aligned}

Now, let’s solve this Linear Programming problem.

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

As the constraint-1 is of type ‘ $\leq$ ‘ we should add slack variable $S_{1}$
As the constraint-2 is of type ‘ $\leq$ ‘ we should add slack variable $S_{2}$
As the constraint-3 is of type ‘ $\leq$ ‘ we should add slack variable $S_{3}$

After introducing slack variables

Max Z = 1500 x_{1} + 1200 x_{2} + 0 S_{1} + 0 S_{2} + 0 S_{3}

subject to

\begin{aligned} 6 x_{1} + 5 x_{2} + S_{1} & = 3600 \\ 3 x_{1} + 5 x_{2} + S_{2} & = 2500 \\ 8 x_{1} + 5 x_{2} + S_{3} & = 4400 \end{aligned}

and

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

\begin{array}{ccccccccc} Iteration-1 & C_{j} & 1500 & 1200 & 0 & 0 & 0 \\ B & C_{B} & X_{B} & x_{1} & x_{2} & S_{1} & S_{2} & S_{3} & \begin{matrix} \frac{X_{B}}{x_{1}} \\ S_{1} \end{matrix} \\ S_{2} & 0 & 2500 & 3 & 5 & 0 & 1 & 0 & \frac{2500}{3} = 833.3333 \\ S_{3} & 0 & 4400 & (8) & 5 & 0 & 0 & 1 & \frac{4400}{8} = 550 \to \\ Z = 0 & Z_{j} & 0 & 0 & 0 & 0 & 0 \\ C_{j} - Z_{j} & 1500 ↑ & 1200 & 0 & 0 & 0 \end{array}

Positive maximum

C_{j} - Z_{j}

is 1500 and its column index is 1 . So, the entering variable is

x_{1}

.
Minimum ratio is 550 and its row index is 3 . So, the leaving basis variable is

S_{3}

∴

The pivot element is 8 .
Entering

= x_{1}

, Departing

= S_{3}

, Key Element

= 8

R_{3} (

new

) = R_{3} (

old

) \div 8

R_{1} (

new

) = R_{1} (

old

) - 6 R_{3} (

new

)

R_{2} (

new

) = R_{2} (

old

) - 3 R_{3} (

new

)

Iteration-2		$C_{j}$	1500	1200	0	0	0
$B$	$C_{B}$	$X_{B}$	$x_{1}$	$x_{2}$	$s_{1}$	$s_{2}$	$S_{3}$	$\begin{matrix} MinRatio \\ \frac{X_{B}}{x_{2}} \end{matrix}$
$s_{1}$	0	300	0	(1.25)	1	0	-0.75	$\frac{300}{1.25} = 240 \to$
$S_{2}$	0	850	0	3.125	0	1	-0.375	$\frac{850}{3.125} = 272$
$x_{1}$	1500	550	1	0.625	0	0	0.125	$\frac{550}{0.625} = 880$
$Z = 825000$		$Z_{j}$	1500	937.5	0	0	187.5
$Z = 825000$		$C_{j} - Z_{j}$	0	$262.5 ↑$	0	0	-187.5

Positive maximum

C_{j} - Z_{j}

is 262.5 and its column index is 2 . So, the entering variable is

x_{2}

.
Minimum ratio is 240 and its row index is 1 . So, the leaving basis variable is

S_{1}

∴

The pivot element is 1.25 .
Entering

= x_{2}

, Departing

= S_{1}

, Key Element

= 1.25

\begin{aligned} R_{1} (new) = R_{1} (old) \div 1.25 \\ R_{2} (new) = R_{2} (old) - 3.125 R_{1} (new) \\ R_{3} (new) = R_{3} (old) - 0.625 R_{1} (new) \end{aligned}

Iteration-3		$C_{j}$	1500	1200	0	0	0
$B$	$C_{B}$	$X_{B}$	$x_{1}$	$x_{2}$	$S_{1}$	$S_{2}$	$S_{3}$	MinRatio
$x_{2}$	1200	240	0	1	0.8	0	-0.6
$S_{2}$	0	100	0	0	-2.5	1	1.5
$x_{1}$	1500	400	1	0	-0.5	0	0.5
$Z = 888000$		$Z_{j}$	$1500$	$1200$	$210$	$0$	$30$
		$C_{j} - Z_{j}$	0	0	-210	0	-30

Since all

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

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

x_{1} = 400, x_{2} = 240

Max Z = 888000

Conclusion

Through iterative calculations and optimizations, the LP problem reached an optimal solution where:

$x_{1} = 400$ tons (mixture in the ratio $3 : 3 : 4$ )
$x_{2} = 240$ tons (mixture in the ratio $2 : 4 : 2$ )
Maximum Profit:
The maximum profit that the firm can achieve with this optimal solution is: $Max Z = 888, 000$ (in Rs.)

2.(a) दर्शाइए कि

(R, +)

मोड्यूलो

Z

का विभाग समूह, सम्मिश्र तल में एकांक वृत्त पर सम्मिश्र संख्याओं के गुणनात्मक समूह से तुल्यकारी होता है । यहाँ पर

R R

, वास्तबिक संख्याओं का समुच्चय है तथा

Z

पूर्णांकों का समुच्चय है ।

Show that the quotient group of

(R, +)

modulo

Z

is isomorphic to the multiplicative group of complex numbers on the unit circle in the complex plane. Here

R

is the set of real numbers and

Z

is the set of integers.

Answer:

To show that the quotient group of

(R, +)

modulo

Z

is isomorphic to the multiplicative group of complex numbers on the unit circle, we will define an isomorphism between these two groups and then prove that this mapping is indeed an isomorphism.

Quotient Group Definition

The quotient group of

(R, +)

modulo

Z

is denoted as

R / Z

and consists of all cosets of the form

r + Z

where

r \in R

. Here,

r + Z

represents the set of all real numbers differing by an integer from

r

Multiplicative Group of Complex Numbers on the Unit Circle

The multiplicative group of complex numbers on the unit circle is the set of all complex numbers of the form

e^{2 π i θ}

where

θ \in R

and

i

is the imaginary unit. This group is usually denoted as

U (1)

S^{1}

Define an Isomorphism

Let’s define a function

f : R / Z \to U (1)

as follows:

f (r + Z) = e^{2 π i r}

Prove Isomorphism

To prove that

f

is an isomorphism, we need to show that

f

is a bijective homomorphism.

1. Homomorphism

A function

f

is a homomorphism if for all

a, b \in R / Z

f (a + b) = f (a) \cdot f (b)

Let’s verify this property for our function

f

f ((r_{1} + Z) + (r_{2} + Z)) = f (r_{1} + r_{2} + Z) = e^{2 π i (r_{1} + r_{2})} = e^{2 π i r_{1}} \cdot e^{2 π i r_{2}} = f (r_{1} + Z) \cdot f (r_{2} + Z)

2. Injective (One-to-One)

A function

f

is injective if different elements in the domain map to different elements in the codomain.

f (r_{1} + Z) = f (r_{2} + Z)

, then:

e^{2 π i r_{1}} = e^{2 π i r_{2}} ⟹ r_{1} - r_{2} \in Z ⟹ r_{1} + Z = r_{2} + Z

3. Surjective (Onto)

A function

f

is surjective if every element in the codomain has a preimage in the domain.

For any

e^{2 π i θ} \in U (1)

θ + Z \in R / Z

is a preimage, since:

f (θ + Z) = e^{2 π i θ}

Conclusion

Since

f

is a bijective homomorphism, it is an isomorphism. Thus, the quotient group

R / Z

is isomorphic to the multiplicative group of complex numbers on the unit circle,

U (1)

S^{1}

2.(b) निम्नलिखित रैख्रिक प्रोग्रामन समस्या को Big

M

विधि से हल कीजिए तथा दर्शाइए कि समस्या के परिमित इष्टतम हल हैं। साथ ही उद्देश्य फलन का मान भी ज्ञात कीजिए :
न्यूनतमीकरण कीजिए

z = 3 x_{1} + 5 x_{2}

बशाते कि

x_{1} + 2 x_{2} ⩾ 8

\begin{aligned} 3 x_{1} + 2 x_{2} ⩾ 12 \\ 5 x_{1} + 6 x_{2} ⩽ 60, \\ x_{1}, x_{2} ⩾ 0 . \end{aligned}

Solve the following linear programming problem by Big M-method and show that the problem has finite optimal solutions. Also find the value of the objective function :
Minimize

z = 3 x_{1} + 5 x_{2}

subject to

x_{1} + 2 x_{2} ⩾ 8

\begin{aligned} 3 x_{1} + 2 x_{2} ⩾ 12 \\ 5 x_{1} + 6 x_{2} ⩽ 60, \\ x_{1}, x_{2} ⩾ 0 . \end{aligned}

Answer:

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 ‘ $\leq$ ‘ we should add slack variable $S_{3}$
After introducing slack,surplus, artificial variables

\begin{aligned} Min Z = 3 x_{1} + 5 x_{2} + 0 S_{1} + 0 S_{2} + 0 S_{3} + M A_{1} + M A_{2} \\ subject to \\ x_{1} + 2 x_{2} - S_{1} + A_{1} = 8 \\ 3 x_{1} + 2 x_{2} - S_{2} + A_{2} = 12 \\ 5 x_{1} + 6 x_{2} + S_{3} = 60 \\ and x_{1}, x_{2}, S_{1}, S_{2}, S_{3}, A_{1}, A_{2} \geq 0 \end{aligned}

Negative minimum

C_{j} - Z_{j}

- 4 M + 3

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

x_{1}

.
Minimum ratio is 4 and its row index is 2 . So, the leaving basis variable is

A_{2}

∴

The pivot element is 3 .
Entering

= x_{1}

, Departing

= A_{2}

, Key Element

= 3

R_{2} (

new

) = R_{2} (

old

) \div 3

R_{1} (

new

) = R_{1}

(old)

- R_{2}

(new)

R_{3} (

new

) = R_{3} (

old

) - 5 R_{2} (

new

)

Iteration-2		$C_{j}$	3	5	0	0	0	$M$
$B$	$C_{B}$	$X_{B}$	$x_{1}$	$x_{2}$	$s_{1}$	$S_{2}$	$S_{3}$	$A_{1}$	$\begin{matrix} MinRatio \\ \frac{X_{B}}{x_{2}} \end{matrix}$
$A_{1}$	$M$	4	0	(1.3333)	-1	0.3333	0	1	$\frac{4}{1.3333} = 3 \to$
$x_{1}$	3	4	1	0.6667	0	-0.3333	0	0	$\frac{4}{0.6667} = 6$
$S_{3}$	0	40	0	2.6667	0	1.6667	1	0	$\frac{40}{2.6667} = 15$
$Z = 4 M + 12$		$Z_{j}$	3	$1.3333 M + 2$	$- M$	$0.3333 M - 1$	$0$	$M$
$Z = 4 M + 12$		$C_{j} - Z_{j}$	0	$- 1.3333 M + 3 ↑$	$M$	$- 0.3333 M + 1$	0	0

Negative minimum

C_{j} - Z_{j}

- 1.3333 M + 3

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

x_{2}

.
Minimum ratio is 3 and its row index is 1 . So, the leaving basis variable is

A_{1}

∴

The pivot element is 1.3333 .
Entering

= x_{2}

, Departing

= A_{1}

, Key Element

= 1.3333

\begin{aligned} R_{1} (new) = R_{1} (old) \div 1.3333 \\ R_{2} (new) = R_{2} (old) - 0.6667 R_{1} (new) \\ R_{3} (new) = R_{3} (old) - 2.6667 R_{1} (new) \end{aligned}

Iteration-3		$C_{j}$	3	5	0	0	0
$B$	$C_{B}$	$X_{B}$	$x_{1}$	$x_{2}$	$S_{1}$	$S_{2}$	$S_{3}$	MinRatio
$x_{2}$	5	3	0	1	-0.75	0.25	0
$x_{1}$	3	2	1	0	0.5	-0.5	0
$S_{3}$	0	32	0	0	2	1	1
$Z = 21$		$Z_{j}$	$3$	$5$	$- 2.25$	$- 0.25$	$0$
		$C_{j} - Z_{j}$	0	0	2.25	0.25	0

Since all

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

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

x_{1} = 2, x_{2} = 3

Min Z = 21

2.(c) दर्शाइए कि यदि

R

के विदृत अन्तराल

(a, b)

पर परिभाषित फलन

f

अवमुख हो, तो बह संतत है । उदाहरण के द्वारा दर्शाइए कि यदि विवृत अन्तराल होने की शर्त न हो, बब अवमुख फलन का संतत होना आवश्यक नहीं है ।

Show that if a function

f

defined on an open interval

(a, b)

R

is convex, then

f

is continuous. Show, by example, if the condition of open interval is dropped, then the convex function need not be continuous.

Answer:

Proof that a Convex Function on an Open Interval is Continuous

A function

f : (a, b) \to R

is convex if, for any

x, y \in (a, b)

and any

λ \in (0, 1)

, the following inequality holds:

f (λ x + (1 - λ) y) \leq λ f (x) + (1 - λ) f (y)

To show that

f

is continuous on

(a, b)

, we need to show that for every point

c \in (a, b)

, for every

ϵ > 0

, there exists a

δ > 0

such that if

| x - c | < δ

, then

| f (x) - f (c) | < ϵ

Proof:

Let

c \in (a, b)

and let

x \in (a, b)

be such that

x < c

. Since

f

is convex, for any

t \in (0, 1)

, we have:

f ((1 - t) c + t x) \leq (1 - t) f (c) + t f (x)

Let’s rearrange terms and isolate

f (x)

f (x) \geq \frac{f ((1 - t) c + t x) - (1 - t) f (c)}{t}

Now, let’s take the limit as

t \to 0^{+}

f (x) \geq lim_{t \to 0^{+}} \frac{f ((1 - t) c + t x) - (1 - t) f (c)}{t} = f (c) + lim_{t \to 0^{+}} \frac{f ((1 - t) c + t x) - f (c)}{t}

The limit on the right side exists because

f

is convex, and it represents the right-hand derivative of

f

c

. Thus,

f

has a right-hand limit at every point in

(a, b)

Similarly, by considering points

x > c

, we can show that

f

has a left-hand limit at every point in

(a, b)

Since

f

has both left-hand and right-hand limits at every point in

(a, b)

, and these limits are equal,

f

is continuous on

(a, b)

Counterexample when the Open Interval Condition is Dropped

Consider the function

f : [0, 1] \to R

defined as follows:

f (x) = {\begin{cases} 0 & if x = 0 \\ 1 & if 0 < x \leq 1 \end{cases}

This function is convex because for any

x, y \in [0, 1]

and any

λ \in (0, 1)

, the inequality

f (λ x + (1 - λ) y) \leq λ f (x) + (1 - λ) f (y)

holds. However, this function is not continuous at

x = 0

, as the left-hand limit does not exist, and the right-hand limit at

x = 0

1

, which is not equal to

f (0) = 0

(a) क्षेत्र $(Z_{13}, +_{13}, x_{13})$ , जहाँ पर $+_{13}$ तथा $x_{13}$ क्रमशः योग मोडयूलो 13 व गुणन मोडयूलो 13 निरूपित करते है, के गुणनात्मक समूह के सभी उचित उपसमूहों को ज्ञात कीजिए।

Find all the proper subgroups of the multiplicative group of the field

(Z_{13}, +_{13}, \times_{13})

, where

+_{13}

and

x_{13}

represent addition modulo 13 and multiplication modulo 13 respectively.

Answer:

To find all the proper subgroups of the multiplicative group of the field

(Z_{13}, +_{13}, \times_{13})

, we need to consider the elements and their orders in this group.

The multiplicative group of a finite field is formed by the nonzero elements of that field under multiplication. In this case, we are working with

Z_{13}

, so the nonzero elements are

{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}

Finding the Generator

The generator of the multiplicative group

Z_{13}^{*}

2

. This means that every element in the multiplicative group

Z_{13}^{*}

can be represented as a power of

2

modulo

13

, confirming that

Z_{13}^{*}

is indeed a cyclic group.

The cyclic group generated by

2

is as follows:

{2, 4, 8, 3, 6, 12, 11, 9, 5, 10, 7, 1}

Finding the Proper Subgroups

Now, let’s find all the proper subgroups of this cyclic group. A proper subgroup is a subgroup that is not equal to the entire group. The proper subgroups of the cyclic group generated by

2

are as follows:

Subgroup generated by $2$ : ${1, 2, 4, 8, 3, 6, 12, 11, 9, 5, 10, 7}$ (This is the entire group, so it is not a proper subgroup)
Subgroup generated by powers of $2$ with step $2$ : ${1, 4, 3, 12, 9, 10}$ (Order $6$ )
Subgroup generated by powers of $2$ with step $3$ : ${1, 8, 12, 5}$ (Order $4$ )
Subgroup generated by powers of $2$ with step $4$ : ${1, 3, 9}$ (Order $3$ )
Subgroup generated by powers of $2$ with step $6$ : ${1, 12}$ (Order $2$ )
Subgroup generated by powers of $2$ with step $12$ : ${1}$ (Trivial subgroup)

Conclusion

The proper subgroups of the multiplicative group of the field

(Z_{13}, +_{13}, \times_{13})

are:

${1, 4, 3, 12, 9, 10}$ (Order $6$ )
${1, 8, 12, 5}$ (Order $4$ )
${1, 3, 9}$ (Order $3$ )
${1, 12}$ (Order $2$ )
${1}$ (Trivial subgroup)

Each of these subgroups is closed under multiplication modulo

13

and contains the identity element

1

3.(b) अवशेष प्रमेय के अनुप्रयोग के द्वारा दर्शाइए कि

\int_{0}^{\infty} \frac{d x}{{(x^{2} + a^{2})}^{2}} = \frac{π}{4 a^{3}}, a > 0

Show by applying the residue theorem that

\int_{0}^{\infty} \frac{d x}{{(x^{2} + a^{2})}^{2}} = \frac{π}{4 a^{3}}, a > 0

Answer:

Introduction:

In this problem, we are asked to prove the integral identity

\int_{0}^{\infty} \frac{d x}{(x^{2} + a^{2})^{2}} = \frac{π}{4 a^{3}}

where

a > 0

. We will use the residue theorem to evaluate this integral.

Step 1: Setting up the Contour Integral:

We consider the contour integral

\int_{c} \frac{d z}{(z^{2} + a^{2})^{2}} = \int_{c} f (z) d z

where

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

, and

c

is a contour consisting of a large semicircle

T

of radius

R

together with a part of the real axis from

x = - R

R

Step 2: Applying the Residue Theorem:

By Cauchy’s residue theorem, we have:

\int_{c} f (z) d z = 2 π i \sum Res

where the sum is taken over the residues of

f (z)

inside the contour

c

We first consider the behavior of

f (z)

z

approaches infinity:

lim_{z \to \infty} z f (z) = lim_{z \to \infty} \frac{z}{(z^{2} + a^{2})^{2}} = lim_{z \to \infty} \frac{z}{z^{4} {(\frac{a^{2}}{z^{2}} + 1)}^{2}} = 0

Therefore, as

R \to \infty

, the integral over the semicircle

T

lim_{R \to \infty} \int_{T} f (z) d z

, approaches zero.

Hence, we have:

lim_{R \to \infty} \int_{T} f (z) d z = 0

Now, taking

R \to \infty

in the original contour integral equation, we get:

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

Step 3: Finding the Residue:

The poles of

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

are given by

(z^{2} + a^{2})^{2} = 0

, which implies

z = \pm a i

(twice). However, only the pole at

z = a i

(of order 2) lies inside the contour

c

To find the residue at

a i

, we calculate:

\begin{aligned} ϕ (z) & = (z - a i)^{2} f (z) \\ = (z - a i)^{2} \frac{1}{(z^{2} + a^{2})^{2}} \\ = \frac{1}{(z + a i)^{2}} \end{aligned}

Taking the derivative of

ϕ (z)

, we get:

ϕ^{'} (z) = - \frac{2}{(z + a i)^{3}}

Therefore, the residue at

a i

is given by:

Residue at a i = ϕ^{'} (a i) = - \frac{2}{(a i + a i)^{3}} = \frac{1}{4 a^{3} i}

Conclusion:

Putting all the pieces together, we have:

\int_{- \infty}^{\infty} \frac{d x}{(a^{2} + x^{2})^{2}} + 0 = 2 π i \times \frac{1}{4 a^{3} i}

Simplifying, we find:

\int_{- \infty}^{\infty} \frac{d x}{(a^{2} + x^{2})^{2}} = \frac{π}{2 a^{3}}

\frac{d x}{(a^{2} + x^{2})^{2}}

is an even function.

So,

2 \times \int_{0}^{\infty} \frac{d x}{(x^{2} + a^{2})^{2}} = \frac{π}{2 a^{3}}, a > 0

\int_{0}^{\infty} \frac{d x}{(x^{2} + a^{2})^{2}} = \frac{π}{4 a^{3}}, a > 0

\begin{aligned} 2 x_{1} - x_{2} + 3 x_{3} + x_{4} = 6 \\ 4 x_{1} - 2 x_{2} - x_{3} + 2 x_{4} = 10 . \end{aligned}

How many basic solutions are there in the following linearly independent set of equations? Find all of them.

\begin{aligned} 2 x_{1} - x_{2} + 3 x_{3} + x_{4} = 6 \\ 4 x_{1} - 2 x_{2} - x_{3} + 2 x_{4} = 10 . \end{aligned}

Answer:

Given System of Equations:

\begin{aligned} 2 x_{1} - x_{2} + 3 x_{3} + x_{4} = 6 \\ 4 x_{1} - 2 x_{2} - x_{3} + 2 x_{4} = 10 . \end{aligned}

Method:

To find the basic solutions, we will express the non-pivot variables in terms of the pivot variables and then set each pivot variable to 1 (and the others to 0) one at a time to find the basic solutions.

Let’s write the augmented matrix for the given system of equations and then perform row operations to get it to reduced row echelon form (RREF):

Augmented Matrix:

[\begin{array}{ccccc} 2 & - 1 & 3 & 1 & 6 \\ 4 & - 2 & - 1 & 2 & 10 \end{array}]

Row Operations to get RREF:

Swap $R_{1}$ and $R_{2}$ :

[\begin{array}{ccccc} 4 & - 2 & - 1 & 2 & 10 \\ 2 & - 1 & 3 & 1 & 6 \end{array}]

$R_{2} \leftarrow R_{2} - \frac{1}{2} R_{1}$ :

[\begin{array}{ccccc} 4 & - 2 & - 1 & 2 & 10 \\ 0 & 0 & \frac{7}{2} & 0 & 1 \end{array}]

$R_{1} \leftarrow R_{1} + \frac{1}{7} R_{2}$ :

[\begin{array}{ccccc} 4 & - 2 & 0 & 2 & \frac{17}{7} \\ 0 & 0 & 1 & 0 & \frac{2}{7} \end{array}]

Expressing Variables:

From the RREF, we have:

Pivot variables: $x_{3}$
Non-pivot variables: $x_{1}$ , $x_{2}$ , and $x_{4}$

Expressing the non-pivot variables in terms of the pivot variable:

From the second row: $x_{3} = \frac{2}{7}$
From the first row: $x_{1} = \frac{1}{4} (\frac{17}{7} - 2 x_{4} + 2 x_{2}) = \frac{17}{28} + \frac{1}{2} x_{2} - \frac{1}{2} x_{4}$

Finding Basic Solutions:

Set $x_{2} = 1$ , $x_{4} = 0$ :
- $x_{1} = \frac{17}{28} + \frac{1}{2} = \frac{20}{7}$
- $x_{3} = \frac{2}{7}$
- Basic Solution: $(\frac{20}{7}, 1, \frac{2}{7}, 0)$
Set $x_{2} = 0$ , $x_{4} = 1$ :
- $x_{1} = \frac{17}{28} - \frac{1}{2} = \frac{19}{7}$
- $x_{3} = \frac{2}{7}$
- Basic Solution: $(\frac{19}{7}, 0, \frac{2}{7}, 1)$

Conclusion:

The basic solutions for the given system of linearly independent equations are:

$(\frac{20}{7}, 1, \frac{2}{7}, 0)$
$(\frac{19}{7}, 0, \frac{2}{7}, 1)$

4.(a) मान लीजिए कि

R

सभी वास्तविक संख्याओं का समुच्चय है तथा

f : R \to R

ऐसा फलन है कि समी

x, y \in R

के लिए निम्नलिखित समीकरण लागू होते हैं :
(i)

f (x + y) = f (x) + f (y)

(ii)

f (x y) = f (x) f (y)

दर्शाइए कि सभी

\forall x \in R

के लिए या तो

f (x) = 0

या

f (x) = x

है।

Suppose

R

be the set of all real numbers and

f : R \to R

is a function such that the following equations hold for all

x, y \in R

:
(i)

f (x + y) = f (x) + f (y)

(ii)

f (x y) = f (x) f (y)

Show that

\forall x \in R

either

f (x) = 0

, or,

f (x) = x

Answer:

To solve this problem, we will use the given properties of the function

f : R \to R

and try to deduce the possible forms of the function.

Given Properties:

$f (x + y) = f (x) + f (y)$ (Additive property)
$f (x \cdot y) = f (x) \cdot f (y)$ (Multiplicative property)

Step 1: Evaluate $f (0)$

Let’s start by finding the value of

f (0)

. We can use the additive property for this:

f (0) = f (0 + 0) = f (0) + f (0)

f (0) = 0

Step 2: Evaluate $f (1)$

Let’s find the value of

f (1)

using the multiplicative property:

f (1) = f (1 \cdot 1) = f (1) \cdot f (1)

Let’s assume

f (1) = a

, then:

a = a \cdot a

a^{2} = a

a (a - 1) = 0

So,

f (1) = a = 0

f (1) = a = 1

Step 3: Evaluate $f (x)$ for any $x$

Let’s use the additive property to find the value of

f (x)

for any

x

f (x) = f (1 \cdot x) = f (1) \cdot f (x)

f (1) = 0

, then

f (x) = 0 \cdot f (x) = 0

for all

x

.
If

f (1) = 1

, then

f (x) = 1 \cdot f (x) = f (x)

for all

x

Step 4: Conclusion

Based on the above steps, we can conclude that for all

x \in R

, either

f (x) = 0

f (x) = x

Detailed Conclusion:

If $f (1) = 0$ , then the function $f : R \to R$ is the zero function, i.e., $f (x) = 0$ for all $x \in R$ .
If $f (1) = 1$ , then the function $f : R \to R$ is the identity function, i.e., $f (x) = x$ for all $x \in R$ .

Thus, the function

f

is either the zero function or the identity function, satisfying the given conditions.

4.(b) फलन

\frac{1}{(1 + z^{2}) (z + 2)}

को निरूपित करने वाली लाँरेन्ज श्रेणी ज्ञात कीजिए जब
(i)

| z | < 1

(ii)

1 < | z | < 2

(iii)

| z | > 2

Find the Laurent’s series which represent the function

\frac{1}{(1 + z^{2}) (z + 2)}

when
(i)

| z | < 1

(ii)

1 < | z | < 2

(iii)

| z | > 2

Answer:

Let

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

Resolving into partial fractions ;

\begin{aligned} f (z) & = \frac{A z + B}{1 + z^{2}} + \frac{C}{z + 2} \\ 1 & = (A z + B) (z + 2) + C (1 + z^{2}) \\ 1 & = z^{2} (A + C) + z (B + 2 A) + (2 B + C) \end{aligned}

or
Comparing coefficients of

z^{2}

and

z

and constant on both the sides, we get

\begin{aligned} \begin{matrix} 1 = 2 B + C \\ 0 = A + C \\ 0 = B + 2 A \end{matrix}} \Rightarrow B = 2 C \\ ∴ \\ B = \frac{2}{5}, C = \frac{1}{5}, A = - \frac{1}{5} \\ f (z) = \frac{1}{(1 + z^{2}) (z + 2)} \end{aligned}

= - \frac{1}{5} \frac{(z - 2)}{1 + z^{2}} + \frac{1}{5 (z + 2)} .

(i)

| z | < 1

f (z) = \frac{1}{5} \cdot \frac{1}{2} \cdot {(1 + \frac{1}{2} \cdot z)}^{- 1} + \frac{2 - z}{5} {(1 + z^{2})}^{- 1}

Binomial expansion of

(1 + z)^{- 1}

is valid only when

| z | < 1

∴ f (z) = \frac{1}{10} \sum_{0}^{\infty} (- 1)^{n} {(\frac{z}{2})}^{n} + \frac{2 - z}{5} \sum_{0}^{\infty} (- 1)^{n} \cdot z^{2 n} .

This is a series in positive powers of

z

, so it is an expansion of

f (z)

in a Taylor’s series within the circle

| z | = 1

(ii)

1 < | z | < 2

\begin{aligned} f (z) & = \frac{1}{10} {(1 + \frac{1}{2} z)}^{- 1} + \frac{2 - z}{5} \cdot \frac{1}{z^{2}} {(1 + \frac{1}{z^{2}})}^{- 1} \\ = \frac{1}{10} \sum_{0}^{\infty} (- 1)^{n} {(\frac{z}{2})}^{n} + \frac{2 - z}{5 z^{2}} \sum_{0}^{\infty} (- 1)^{n} \cdot {(\frac{1}{z^{2}})}^{n} . \end{aligned}

This is a series in positive and negative powers of

z

, so it is an expansion of

f (z)

in a Laurent’s series within the region

1 < | z | < 2

(iii)

| z | > 2

\begin{aligned} f (z) & = \frac{1}{5} \cdot \frac{1}{z + 2} - \frac{1}{5} \cdot \frac{z - 2}{1 + z^{2}} \\ = \frac{1}{5 z} {(1 + \frac{2}{z})}^{- 1} - \frac{1}{5} (z - 2) \cdot \frac{1}{z^{2}} {(1 + \frac{1}{z^{2}})}^{- 1} \\ = \frac{1}{5 z} (\sum_{0}^{\infty} (- 1)^{n} {(\frac{2}{z})}^{n}) - \frac{1}{5} (\frac{1}{z} - \frac{2}{z^{2}}) \sum_{0}^{\infty} (- 1)^{n} {(\frac{1}{z^{2}})}^{n} \end{aligned}

This is Laurent’s series within the ring shaped region (or annulus)

2 < z < R

where

R

is large.

4.(c) एक फैक्ट्री में पाँच प्रचालक

O_{1}, O_{2}, O_{3}, O_{4}, O_{5}

तथा पाँच मशीनें

M_{1}, M_{2}, M_{3}, M_{4}, M_{5}

हैं । परिचालन लागत, जब कि

O_{i}

प्रचालक

M_{j} (i, j = 1, 2, \dots, 5)

मशीन को परिचालन करता है, दी गई हैं। लेकिन एक प्रतिबन्ध है कि

O_{3}

को तीसरी मशीन

M_{3}

का परिचालन करने तथा

O_{2}

को पाँचर्वीं मरीन

M_{5}

का परिचालन करने की इजाज़त नह्ही दी जा सकती है। लागत आव्यूह नीचे दी है । इएतम नियतन तथा इए्टम नियतन की लागत ज्ञात कीजिए।

\begin{array}{rrrrrr} M_{1} & M_{2} & M_{3} & M_{4} & M_{5} \\ O_{1} & 24 & 29 & 18 & 32 & 19 \\ O_{2} & 17 & 26 & 34 & 22 & 21 \\ O_{3} & 27 & 16 & 28 & 17 & 25 \\ O_{4} & 22 & 18 & 28 & 30 & 24 \\ O_{5} & 28 & 16 & 31 & 24 & 27 \end{array}

In a factory there are five operators

O_{1}, O_{2}, O_{3}, O_{4}, O_{5}

and five machines

M_{1}, M_{2}, M_{3}, M_{4}, M_{5}

. The operating costs are given when the

O_{1}

operator operates the

M_{j}

machine

(i, j = 1, 2, \dots, 5)

. But there is a restriction that

O_{3}

cannot be allowed to operate the third machine

M_{3}

and

O_{2}

cannot be allowed to operate the fifth machine

M_{5 +}

The cost matrix is given above. Find the optimal assignment and the optimal assignment cost also.

\begin{array}{rrrrrr} M_{1} & M_{2} & M_{3} & M_{4} & M_{5} \\ O_{1} & 24 & 29 & 18 & 32 & 19 \\ O_{2} & 17 & 26 & 34 & 22 & 21 \\ O_{3} & 27 & 16 & 28 & 17 & 25 \\ O_{4} & 22 & 18 & 28 & 30 & 24 \\ O_{5} & 28 & 16 & 31 & 24 & 27 \end{array}

Answer:

Introduction

In the realm of operational research, assignment problems are pivotal, especially in manufacturing settings where optimal assignments of jobs to resources are crucial for efficiency and cost-effectiveness. This problem involves a factory with five operators,

O_{1}, O_{2}, O_{3}, O_{4}, O_{5}

, and five machines,

M_{1}, M_{2}, M_{3}, M_{4}, M_{5}

, each with associated operating costs. The objective is to find the optimal assignment of operators to machines such that the total operating cost is minimized, considering certain restrictions. Specifically, operator

O_{3}

is not allowed to operate machine

M_{3}

, and operator

O_{2}

is not allowed to operate machine

M_{5}

This is the original cost matrix:

\begin{array}{lllll} 24 & 29 & 18 & 32 & 19 \\ 17 & 26 & 34 & 22 & 21 \\ 27 & 16 & 28 & 17 & 25 \\ 22 & 18 & 28 & 30 & 24 \\ 28 & 16 & 31 & 24 & 27 \end{array}

Subtract row minima

We subtract the row minimum from each row:

\begin{array}{rrrrrr} 6 & 11 & 0 & 14 & 1 & (- 18) \\ 0 & 9 & 17 & 5 & 4 & (- 17) \\ 11 & 0 & 12 & 1 & 9 & (- 16) \\ 4 & 0 & 10 & 12 & 6 & (- 18) \\ 12 & 0 & 15 & 8 & 11 & (- 16) \end{array}

Subtract column minima

We subtract the column minimum from each column:

\begin{array}{rrrrr} 6 & 11 & 0 & 13 & 0 \\ 0 & 9 & 17 & 4 & 3 \\ 11 & 0 & 12 & 0 & 8 \\ 4 & 0 & 10 & 11 & 5 \\ 12 & 0 & 15 & 7 & 10 \\ (- 1) & (- 1) \end{array}

Cover all zeros with a minimum number of lines

There are 4 lines required to cover all zeros:

$6$	$11$	$0$	$13$	$0$	$x$
$0$	$9$	$17$	$4$	$3$	$x$
$11$	$0$	$12$	$0$	$8$	$x$
$4$	$0$	10	11	5
12	$0$	15	7	10
	$x$

Create additional zeros

The number of lines is smaller than

5

. The smallest uncovered number is 4 . We subtract this number from all uncovered elements and add it to all elements that are covered twice:

\begin{array}{rrrrr} 6 & 15 & 0 & 13 & 0 \\ 0 & 13 & 17 & 4 & 3 \\ 11 & 4 & 12 & 0 & 8 \\ 0 & 0 & 6 & 7 & 1 \\ 8 & 0 & 11 & 3 & 6 \end{array}

Cover all zeros with a minimum number of lines

There are 4 lines required to cover all zeros:

\begin{array}{rrrrrr} 6 & 15 & 0 & 13 & 0 & x \\ 0 & 13 & 17 & 4 & 3 \\ 11 & 4 & 12 & 0 & 8 & x \\ 0 & 0 & 6 & 7 & 1 \\ 8 & 0 & 11 & 3 & 6 \\ x & x \end{array}

Create additional zeros

The number of lines is smaller than

5

. The smallest uncovered number is 1 . We subtract this number from all uncovered elements and add it to all elements that are covered twice:

\begin{array}{rrrrr} 7 & 16 & 0 & 13 & 0 \\ 0 & 13 & 16 & 3 & 2 \\ 12 & 5 & 12 & 0 & 8 \\ 0 & 0 & 5 & 6 & 0 \\ 8 & 0 & 10 & 2 & 5 \end{array}

Cover all zeros with a minimum number of lines

There are 5 lines required to cover all zeros:

\begin{array}{rrrrrr} 7 & 16 & 0 & 13 & 0 & x \\ 0 & 13 & 16 & 3 & 2 & x \\ 12 & 5 & 12 & 0 & 8 & x \\ 0 & 0 & 5 & 6 & 0 & x \\ 8 & 0 & 10 & 2 & 5 & x \end{array}

The optimal assignment

Because there are 5 lines required, the zeros cover an optimal assignment:

\begin{array}{rrrrr} 7 & 16 & 0 & 13 & 0 \\ 0 & 13 & 16 & 3 & 2 \\ 12 & 5 & 12 & 0 & 8 \\ 0 & 0 & 5 & 6 & 0 \\ 8 & 0 & 10 & 2 & 5 \end{array}

This corresponds to the following optimal assignment in the original cost matrix:

24	29	$18$	32	19
$17$	26	34	22	21
27	16	28	$17$	25
22	18	28	30	$24$
28	$16$	31	24	27

The optimal value equals 92 .

Conclusion

Through meticulous application of the Hungarian Algorithm, we have successfully determined the optimal assignment of operators to machines, adhering to the given restrictions and minimizing the total operating cost. The optimal assignments are as follows:

O_{1}

M_{3}

O_{2}

M_{1}

O_{3}

M_{4}

O_{4}

M_{5}

, and

O_{5}

M_{2}

, resulting in a minimized total operating cost of 92. This optimal assignment ensures that the factory operates at the highest possible efficiency, given the constraints, and incurs the least amount of operating cost, thereby maximizing the overall productivity and cost-effectiveness of the manufacturing process.

खण्ड ‘B’ SECTION ‘B’

5.(a) दीर्घवृत्तज :

x^{2} + 4 y^{2} + 4 z^{2} = 4

के उन समी स्पर्श-तलों के संकाय का आंशिक अवकल समीकरण ज्ञात कीजिए, जो

x y

समतल के लम्बवत नहीं हैं।

Find the partial differential equation of the family of all tangent planes to the ellipsoid:

x^{2} + 4 y^{2} + 4 z^{2} = 4

, which are not perpendicular to the

x y

plane.

Answer:

Given Ellipsoid

The given ellipsoid is represented by the equation:

x^{2} + 4 y^{2} + 4 z^{2} = 4 (1)

Tangent Plane to the Ellipsoid

The equation of the tangent plane to the ellipsoid at a point

P (x_{1}, y_{1}, z_{1})

is given by:

x x_{1} + 4 y y_{1} + 4 z z_{1} = 4 (2)

General Equation of a Plane

Let the general equation of a plane be represented as:

l x + m y + n z = P (3)

Coefficient Relations

Comparing the coefficients in equations (2) and (3), we have:

\frac{x_{1}}{l} = \frac{4 y_{1}}{m} = \frac{4 z_{1}}{n} = \frac{4}{P} (4)

From equation (4), we can express

x_{1}

y_{1}

, and

z_{1}

as:

x_{1} = \frac{4 l}{P} y_{1} = \frac{m}{P} z_{1} = \frac{n}{P} (5)

Substituting Back into the Ellipsoid Equation

Substituting equation (5) back into equation (1) gives:

\frac{16 l^{2}}{P^{2}} + \frac{4 m^{2}}{P^{2}} + \frac{4 n^{2}}{P^{2}} = 4 (6)

\Rightarrow 4 l^{2} + m^{2} + n^{2} = P^{2} (7)

Equation of the Tangent Plane

Substituting the values from equation (4) into equation (3) gives the equation of the tangent plane as:

l x + m y + n z = \pm \sqrt{4 l^{2} + m^{2} + n^{2}} (8)

Condition for the Plane not being Perpendicular to the $x y$ -plane

Since the plane is not perpendicular to the

x y

-plane, we can write equation (8) as:

\frac{l}{n} x + \frac{m}{n} y + z = \pm \sqrt{4 {(\frac{l}{n})}^{2} + {(\frac{m}{n})}^{2} + 1}

α x + β y + z = \pm \sqrt{4 α^{2} + β^{2} + 1} (9)

Partial Differential Equation

Differentiating equation (9) partially with respect to

x

and

y

, we get:

α + \frac{\partial z}{\partial x} = 0 \Rightarrow \frac{\partial z}{\partial x} = - α \Rightarrow α = - p (10)

β + \frac{\partial z}{\partial y} = 0 \Rightarrow \frac{\partial z}{\partial y} = - β \Rightarrow β = - q (11)

Substituting equations (10) and (11) back into equation (9) gives:

- p x + (- q) y + z = \pm \sqrt{4 p^{2} + q^{2} + 1} (12)

\Rightarrow (p x + q y - z)^{2} = 4 p^{2} + q^{2} + 1 (13)

Conclusion

Equation (13) is the required partial differential equation representing the family of all tangent planes to the given ellipsoid, which are not perpendicular to the

x y

-plane.

5.(b) न्यूटन के अग्रांतर फार्मूले से निम्नतम-घातीय बहुपद

u_{x}

ज्ञात कीजिए जब कि

u_{1} = 1, u_{2} = 9

u_{3} = 25, u_{4} = 55

तथा

u_{5} = 105

दिया गया है ।

Using Newton’s forward difference formula find the lowest degree polynomial

u_{x}

when it is given that

u_{1} = 1, u_{2} = 9, u_{3} = 25, u_{4} = 55

and

u_{5} = 105

Answer:

The value of table for

x

and

y

$x$	1	2	3	4	5
$y$	1	9	25	55	105

Newton’s forward difference interpolation method to find solution
Newton’s forward difference table is

$x$	$y$	$Δ y$	$Δ^{2} y$	$Δ^{3} y$	$Δ^{4}$
$1$	$1$
		$8$
$2$	9		$8$
		16		$6$
$3$	$25$		14		$0$
		30		6
$4$	$55$		20
		50
$5$	105

Newton’s forward difference interpolation formula is

\begin{aligned} y (x) = y_{0} + p Δ y_{0} + \frac{p (p - 1)}{2!} \cdot Δ^{2} y_{0} + \frac{p (p - 1) (p - 2)}{3!} \cdot Δ^{3} y_{0} + \frac{p (p - 1) (p - 2) (p - 3)}{4!} \cdot Δ^{4} y_{0} \\ p = \frac{x - x_{0}}{h} = \frac{x - 1}{1} = x - 1 \end{aligned}

Substituting these values, we get

\begin{aligned} y (x) = 1 + (x - 1) \times 8 + \frac{(x - 1) (x - 2)}{2} \times 8 + \frac{(x - 1) (x - 2) (x - 3)}{6} \times 6 + \frac{(x - 1) (x - 2) (x - 3) (x - 4)}{24} \times 0 \\ y (x) = 1 + 8 \times (x - 1) + 4 \times (x^{2} - 3 x + 2) + 1 \times (x^{3} - 6 x^{2} + 11 x - 6) \\ y (x) = 1 + (8 x - 8) + (4 x^{2} - 12 x + 8) + (x^{3} - 6 x^{2} + 11 x - 6) \\ y (x) = x^{3} - 2 x^{2} + 7 x - 5 \end{aligned}

5.(c) एक असंपीडय तरल प्रवाह के लिए वेग

(u, v, w)

के दो घटक

u = x^{2} + 2 y^{2} + 3 z^{2}

व

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

दिए गए हैं। वेग के तीसरे घटक

w

का निर्धारण कीजिए ताकि वे सांतत्य समीकरण को सन्तुष्ट करें। त्वरण के

z

-घटक को भी ज्ञात कीजिए।

For an incompressible fluid flow, two components of velocity

(u, v, w)

are given by

u = x^{2} + 2 y^{2} + 3 z^{2}, v = x^{2} y - y^{2} z + z x

. Determine the third component

w

so that they satisfy the equation of continuity. Also, find the z-component of acceleration.

Answer:

Introduction:

In this problem, we are given two components of velocity

(u, v, w)

for an incompressible fluid flow, specifically

u = x^{2} + 2 y^{2} + 3 z^{2}

and

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

. The task is to determine the third component

w

such that the velocity components satisfy the equation of continuity. Additionally, we need to find the

z

-component of acceleration.

Step 1: Continuity Equation and Solving for $w$ :

We start by applying the continuity equation for incompressible flow in Cartesian coordinates:

\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y} + \frac{\partial w}{\partial z} = 0

Plugging in the given expressions for

u

and

v

, we get:

2 x + x^{2} - 2 y z + \frac{\partial w}{\partial z} = 0

Now, isolate

\frac{\partial w}{\partial z}

by rearranging the equation:

\frac{\partial w}{\partial z} = 2 y z - 2 x - x^{2}

Step 2: Integrating to Find $w$ :

To find

w

, we integrate the above equation with respect to

z

. This will involve an arbitrary function

f (x, y)

since we are integrating with respect to only one variable. The equation becomes:

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

So,

w

is given by the above expression, including the arbitrary function

f (x, y)

Step 3: Finding the $z$ -Component of Acceleration:

The

z

-component of acceleration

a_{z}

is given by:

a_{z} = (q \cdot \nabla) w + \frac{\partial ω}{\partial t}

a_{z} = u \frac{\partial w}{\partial x} + v \frac{\partial w}{\partial y} + w \frac{\partial w}{\partial z}

Now, we need to substitute the given expressions for

u

v

, and

w

into this expression and calculate

a_{z}

Therefore,

\begin{aligned} a_{z} & = (x^{2} + 2 y^{2} + 3 z^{2}) (\frac{\partial f}{\partial x} - 2 z - 2 x z) \\ + (x^{2} y - y^{2} z + x z) (\frac{\partial f}{\partial y} + z^{2}) \\ + (y z^{2} - 2 x z - x^{2} z + f (x, y)) (2 y z - 2 z - x^{2}) \end{aligned}

This expression represents the

z

-component of acceleration

a_{z}

for the given velocity components

u

v

, and

w

Conclusion:

The third component

w

that satisfies the equation of continuity is given by

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

, where

f (x, y)

is an arbitrary function of

x

and

y

The

z

-component of acceleration

a_{z}

is given by the complex expression involving the derivatives of

f (x, y)

and the given velocity components.

5.(d) विराम अवस्था से प्रारम्भ हो कर एक रेलगाड़ी की रफतार (किमी/घं में) विभिन्न समयों (मिनट में) पर निम्न सारणी के द्वारा दी गई है :
सिम्पसन के

\frac{1}{3}

नियम के इस्तेमाल से प्रारंभ से 20 मिनटों में चली गई सन्तिकट दूरी (किमी. में) ज्ञात कीजिए ।

समय (मिनट) Time (Minutes)	2	4	6	8	10	12	14	16	18	20
रफ़तार (किमी /घं) Speed $($ Km/h $)$	10	18	25	29	32	20	11	5	2	$8.5$

Starting from rest in the beginning, the speed (in

Km / h

) of a train at different times (in minutes) is given by the above table:

Using Simpson’s

\frac{1}{3}

rd rule, find the approximate distance travelled (in

Km

) in 20 minutes from the beginning.

Answer:

To find the approximate distance traveled using Simpson’s

\frac{1}{3}

rd rule, we can use the formula:

Distance = \frac{h}{3} [y_{0} + 4 (y_{1} + y_{3} + \dots + y_{n - 1}) + 2 (y_{2} + y_{4} + \dots + y_{n - 2}) + y_{n}]

Here,

h

is the width of each interval,

y_{0}, y_{1}, \dots, y_{n}

are the function values at each interval, and

n

is the number of intervals.

Given Data

The given data points are in minutes and km/h. We have 10 intervals of 2 minutes each, so

h = 2

minutes

= \frac{1}{30}

hours.

The speed values

y_{0}, y_{1}, \dots, y_{10}

in km/h are given as:

$y_{0} = 0$ km/h (starting from rest)
$y_{1} = 10$ km/h
$y_{2} = 18$ km/h
$y_{3} = 25$ km/h
$y_{4} = 29$ km/h
$y_{5} = 32$ km/h
$y_{6} = 20$ km/h
$y_{7} = 11$ km/h
$y_{8} = 5$ km/h
$y_{9} = 2$ km/h
$y_{10} = 8.5$ km/h

Calculation

Let’s substitute the given values into Simpson’s

\frac{1}{3}

rd rule formula to find the approximate distance traveled in 20 minutes.

Substitution

Here is the substitution part with the given values:

Interval width, $h = 2$ minutes $= \frac{1}{30}$ hours.
Speed values, $y_{0} = 0$ , $y_{1} = 10$ , $y_{2} = 18$ , $y_{3} = 25$ , $y_{4} = 29$ , $y_{5} = 32$ , $y_{6} = 20$ , $y_{7} = 11$ , $y_{8} = 5$ , $y_{9} = 2$ , and $y_{10} = 8.5$ km/h.

Substituting these into Simpson’s

\frac{1}{3}

rd rule formula:

Distance = \frac{\frac{1}{30}}{3} [0 + 4 (10 + 25 + 32 + 11 + 2) + 2 (18 + 29 + 20 + 5) + 8.5]

Distance = 5.25 Km

This is the approximate distance traveled by the train in the first 20 minutes from the beginning.

5.(e) समीकरण :

x e^{x} - 1 = 0

को द्विभाजन-विधि के द्वारा, दशमलव के 4 अंकों तक, हल करने के लिए, आधारी ऐल्लोरिथ्म लिखिए।

Write down the basic algorithm for solving the equation :

x e^{x} - 1 = 0

by bisection method, correct to 4 decimal places.

Answer:

The Bisection Method is a root-finding algorithm that divides the interval into two subintervals and then selects the subinterval where the function changes sign, indicating the presence of a root. Here’s a basic algorithm to solve the equation

x e^{x} - 1 = 0

using the Bisection Method:

Algorithm for Bisection Method

Step 1: Define the Function

Define the function

f (x) = x e^{x} - 1

Step 2: Choose Interval [a, b]

Choose an interval

[a, b]

such that

f (a) \cdot f (b) < 0

, indicating that there is a root in the interval.

Step 3: Calculate Midpoint

Calculate the midpoint

c

of the interval

[a, b]

as:

c = \frac{a + b}{2}

Step 4: Evaluate Function at Midpoint

Evaluate the function at the midpoint:

f (c) = c e^{c} - 1

Step 5: Update Interval

If $f (c) = 0$ , then $c$ is the root of the equation.
If $f (a) \cdot f (c) < 0$ , then the root lies in the interval $[a, c]$ , so update $b = c$ .
If $f (b) \cdot f (c) < 0$ , then the root lies in the interval $[c, b]$ , so update $a = c$ .

Step 6: Check Tolerance

Check if the width of the interval

[a, b]

is less than the desired tolerance,

10^{- 4}

for 4 decimal places. If the width is less than the tolerance, then stop the algorithm, and the midpoint

c

is the approximate root. Otherwise, go back to Step 3.

Step 7: Return the Root

Return the value of

c

as the approximate root of the equation, correct to 4 decimal places.

Its Iterations are; when we enter the values.

\begin{array}{cccccccc} n & a & f (a) & b & f (b) & c = \frac{a + b}{2} & f (c) & Update \\ 1 & 0 & - 1 & 1 & 1.7183 & 0.5 & - 0.1756 & a = c \\ 2 & 0.5 & - 0.1756 & 1 & 1.7183 & 0.75 & 0.5878 & b = c \\ 3 & 0.5 & - 0.1756 & 0.75 & 0.5878 & 0.625 & 0.1677 & b = c \\ 4 & 0.5 & - 0.1756 & 0.625 & 0.1677 & 0.5625 & - 0.0128 & a = c \\ 5 & 0.5625 & - 0.0128 & 0.625 & 0.1677 & 0.5938 & 0.0751 & b = c \\ 6 & 0.5625 & - 0.0128 & 0.5938 & 0.0751 & 0.5781 & 0.0306 & b = c \\ 7 & 0.5625 & - 0.0128 & 0.5781 & 0.0306 & 0.5703 & 0.0088 & b = c \\ 8 & 0.5625 & - 0.0128 & 0.5703 & 0.0088 & 0.5664 & - 0.002 & a = c \\ 9 & 0.5664 & - 0.002 & 0.5703 & 0.0088 & 0.5684 & 0.0034 & b = c \\ 10 & 0.5664 & - 0.002 & 0.5684 & 0.0034 & 0.5674 & 0.0007 & b = c \\ 11 & 0.5664 & - 0.002 & 0.5674 & 0.0007 & 0.5669 & - 0.0007 & a = c \\ 12 & 0.5669 & - 0.0007 & 0.5674 & 0.0007 & 0.5671 & 0 & a = c \end{array}

(a) आंशिक अवकल समीकरण :

(y^{3} x - 2 x^{4}) p + (2 y^{4} - x^{3} y) q = 9 z (x^{3} - y^{3}), का,

जहाँ

p = \frac{\partial z}{\partial x}, q = \frac{\partial z}{\partial y}

है, का व्यापक हल ज्ञात कीजिए, तथा इसके, वक्र:

x = t, y = t^{2}, z = 1

में से गुजरने वाले समाकल पृष्ठ को भी ज्ञात कीजिए।

Find the general solution of the partial differential equation:

(y^{3} x - 2 x^{4}) p + (2 y^{4} - x^{3} y) q = 9 z (x^{3} - y^{3}),

where

p = \frac{\partial z}{\partial x}, q = \frac{\partial z}{\partial y}

, and find its integral surface that passes through the curve:

x = t, y = t^{2}, z = 1 .

Answer:

Here Lagrange’s auxilliary equations are

\frac{d x}{y^{3} x - 2 x^{4}} = \frac{d y}{2 y^{4} - x^{3} y} = \frac{d z}{9 z (x^{3} - y^{3})} - - - - (1)

Choosing

\frac{1}{x}, \frac{1}{y}, \frac{1}{3 z}

as multipliers, we have

\begin{aligned} Each fraction & = \frac{\frac{1}{x} d x + \frac{1}{y} d y + \frac{1}{3 z} d z}{\frac{1}{x} (y^{3} x - 2 x^{4}) + \frac{1}{y} (2 y^{4} - x^{3} y) + \frac{1}{3 z} 9 z (x^{3} - y^{3})} \\ = \frac{\frac{1}{x} d x + \frac{1}{y} d y + \frac{1}{3 z} d z}{(y^{3} - 2 x^{3}) + (2 y^{3} - x^{3}) + (3 x^{3} - 3 y^{3})} \end{aligned}

Each fraction

= \frac{\frac{1}{x} d x + \frac{1}{y} d y + \frac{1}{3 z} d z}{0}

∴ \frac{1}{x} d x + \frac{1}{y} d y + \frac{1}{3 z} d z = 0

Integrating, we get

\log x + \log y + \frac{1}{3} \log z = \log a

\log ({xyz}^{1 / 3}) = \log a

{xyz}^{1 / 3} = a - - - - (2)

Again choosing

- 2 x y, x^{2}, 0

as multipliers, we have

\begin{aligned} each fraction = \frac{- 2 x y d x + x^{2} d y}{- 2 x y (y^{3} x - 2 x^{4}) + x^{2} (2 y^{4} - x^{3} y)} \\ = \frac{- 2 x y d x + x^{2} d y}{3 x^{5} y} - - - - (3) \end{aligned}

And taking

y^{2}, - 2 x y, 0

as multipliers, we have

\begin{aligned} each fraction & = \frac{y^{2} d x - 2 x y d y}{y^{2} (y^{3} x - 2 x^{4}) - 2 x y (2 y^{4} - x^{3} y)} \\ = \frac{y^{2} d x - 2 x y d y}{- 3 x y^{5}} - - - - (4) \end{aligned}

From (3) and (4), we get

\begin{aligned} - \frac{2 x y d x + x^{2} d y}{3 x^{5} y} = \frac{y^{2} d x - 2 x y d y}{- 3 x y^{5}} \\ or - \frac{2 x y d x + x^{2} d y}{x^{4}} = \frac{y^{2} d x - 2 x y d y}{- y^{4}} \\ or - \frac{2}{x^{3}} y d x + \frac{1}{x^{2}} d y = - [\frac{1}{y^{2}} d x - \frac{2}{y^{3}} x d y] \\ or d (\frac{y}{x^{2}}) = - d (\frac{x}{y^{2}}) \end{aligned}

Integrating, we get

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

\frac{x}{y^{2}} + \frac{y}{x^{2}} = b - - - - (5)

)

From (2) and (5), the general integral of given equation is

ϕ (x y z^{1 / 3}, \frac{x}{y^{2}} + \frac{y}{x^{2}}) = 0

where

ϕ

is an arbitrary function.

Given Equations:

The general integral of the given PDE is:

ϕ (x y z^{1 / 3}, \frac{x}{y^{2}} + \frac{y}{x^{2}}) = 0

The curve is defined as:

x = t

y = t^{2}

z = 1

Substitute the given curve equations into the general integral to find the form of the function

ϕ

Step 1: Substitute the Curve Equations

ϕ (t (t^{2}) (1)^{1 / 3}, \frac{t}{(t^{2})^{2}} + \frac{t^{2}}{t^{2}}) = 0

ϕ (t^{3}, \frac{1}{t^{3}} + 1) = 0

Step 2: Solve for $ϕ$

To find the integral surface that passes through the given curve, we need to determine the form of the function

ϕ

that satisfies the above equation for all values of

t

Let’s denote the arguments of

ϕ

as:

u = t^{3}

v = \frac{1}{t^{3}} + 1

So, the equation becomes:

ϕ (u, v) = 0

Since we are looking for an integral surface passing through the given curve, we can consider

ϕ

as a constant for this specific curve, say

C

ϕ (u, v) = C

Step 3: Formulate the Integral Surface

The integral surface of the given partial differential equation that passes through the curve

x = t

y = t^{2}

, and

z = 1

is represented by:

ϕ (x y z^{1 / 3}, \frac{x}{y^{2}} + \frac{y}{x^{2}}) = C

where

C

is a constant specific to this curve, and it represents the specific integral surface passing through the given curve in the family of integral surfaces represented by the general integral.

Conclusion:

The integral surface of the given partial differential equation that passes through the specified curve is characterized by the function

ϕ

with the constant

C

specific to this curve. The exact value of

C

can be determined if more information or specific conditions related to the curve or the integral surface are provided.

(b) अधोलिखित संख्याओं के समतुल्यों को उनके सम्मुख दर्शाई गई विशिष्ट संख्या पद्धति में, ज्ञात कीजिए।
(i) $(111011 \cdot 101)_{2}$ को दशमलव पद्धति में
(ii) $(1000111110000 - 00101100)_{2}$ को षड़दशमलव पद्धति में
(iii) $(C 4 F 2)_{16}$ को दशमलव पद्वति में
(iv) $(418)_{10}$ को द्विआधारी पद्धति में

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

(i)

(111011 \cdot 101)_{2}

to decimal system

(ii)

(1000111110000 \cdot 00101100)_{2}

to hexadecimal system

(iii)

(C 4 F 2)_{16}

to decimal system
(iv)

(418)_{10}

to binary system

Answer:

(i) (111011 \cdot 101)_{2} to decimal system

111011.101

\begin{aligned} = 1 \times 2^{5} + 1 \times 2^{4} + 1 \times 2^{3} + 0 \times 2^{2} + 1 \times 2^{1} + 1 \times 2^{0} + 1 \times \frac{1}{2^{1}} + 0 \times \frac{1}{2^{2}} + 1 \times \frac{1}{2^{3}} \\ = 1 \times 32 + 1 \times 16 + 1 \times 8 + 0 \times 4 + 1 \times 2 + 1 \times 1 + 1 \times 0.5 + 0 \times 0.25 + 1 \times 0.125 \\ = 32 + 16 + 8 + 0 + 2 + 1 + 0.5 + 0 + 0.125 \\ = 59.625 \end{aligned}

∴ (111011.101)_{2} = (\underset{―}{59.625})_{10}

(ii) (1000111110000.00101100)_{2} to hexadecimal system

\begin{aligned} \frac{0001}{1} \frac{0001}{1} \frac{1111}{15} \frac{0000}{0} \cdot \frac{0010}{2} \frac{1100}{12} \\ ∴ (1000111110000.00101100)_{2} = (\underset{―}{11 F 0.2 C})_{16} \end{aligned}

(iii) (C 4 F 2)_{16} to decimal system

C 4 F 2

\begin{aligned} = C \times 16^{3} + 4 \times 16^{2} + F \times 16^{1} + 2 \times 16^{0} \\ = 12 \times 4096 + 4 \times 256 + 15 \times 16 + 2 \times 1 \\ = 49152 + 1024 + 240 + 2 \\ = 50418 \end{aligned}

∴ (C 4 F 2)_{16} = (\underset{―}{50418})_{10}

(iv) (418)_{10} to binary system

\begin{aligned} \begin{array}{ccc} 418 \\ 2 & 209 & 0 ↑ \\ 2 & 104 & 1 ↑ \\ 2 & 52 & 0 ↑ \\ 2 & 26 & 0 ↑↑ \\ 2 & 13 & 0 ↑↑ \\ 2 & 6 & 1 ↑ \\ 2 & 3 & 0 ↑ \\ 2 & 1 & 1 ↑ \\ 0 & 1 ↑ \end{array} \\ ∴ (418)_{10} = (110100010)_{2} \end{aligned}

6.(c) मान लीजिए किसी यांत्रिक-निकाय का लेगरान्जियन :

L = \frac{1}{2} m (a {\dot{x}}^{2} + 2 b \dot{x} \dot{y} + c {\dot{y}}^{2}) - \frac{1}{2} k (a x^{2} + 2 b x y + c y^{2}),

के द्वारा द्योतित है जहाँ

a, b, c, m (> 0), k (> 0)

स्थिरांक हैं तथा

b^{2} \neq a c

लेगरान्जियन समीकरणों को लिखिए तथा निकाय को पहचानिए ।

Suppose the Lagrangian of a mechanical system is given by

L = \frac{1}{2} m (a {\dot{x}}^{2} + 2 b \dot{x} \dot{y} + c {\dot{y}}^{2}) - \frac{1}{2} k (a x^{2} + 2 b x y + c y^{2}),

where

a, b, c, m (> 0), k (> 0)

are constants and

b^{2} \neq a c

. Write down the Lagrangian equations of motion and identify the system.

Answer:

Introduction:

Given the Lagrangian of a mechanical system, we need to find the Lagrangian equations of motion and identify the system. The Lagrangian is defined as:

L = \frac{1}{2} m (a {\dot{x}}^{2} + 2 b \dot{x} \dot{y} + c {\dot{y}}^{2}) - \frac{1}{2} k (a x^{2} + 2 b x y + c y^{2}),

where

a, b, c, m (> 0), k (> 0)

are constants, and

b^{2} \neq a c

Lagrangian Equations of Motion:

To find the Lagrangian equations of motion, we’ll calculate the partial derivatives of

L

with respect to

x

y

\dot{x}

, and

\dot{y}

\frac{\partial L}{\partial x} = - \frac{1}{2} k (2 a x + 2 b y) = - k (a x + b y) (1)

\frac{\partial L}{\partial \dot{x}} = \frac{1}{2} m (2 a \dot{x} + 2 b \dot{y}) = m (a \dot{x} + b \dot{y}) (2)

\frac{\partial L}{\partial y} = - \frac{1}{2} k (2 b x + 2 c y) = - k (b x + c y) (3)

\frac{\partial L}{\partial \dot{y}} = \frac{1}{2} m (2 b \dot{x} + 2 c \dot{y}) = m (b \dot{x} + c \dot{y}) (4)

Lagrangian Equations of Motion:

The Lagrangian equations of motion are given by:

\frac{d}{d t} (\frac{\partial L}{\partial \dot{x}}) - \frac{\partial L}{\partial x} = 0 (5)

and

\frac{d}{d t} (\frac{\partial L}{\partial \dot{y}}) - \frac{\partial L}{\partial y} = 0 (6)

Substituting the partial derivatives from equations (1)-(4) into equations (5) and (6), we get:

(a \ddot{x} + b \ddot{y}) + k (a x + b y) = 0 (7)

m (b \dot{x} + c \ddot{y}) + k (b x + c y) = 0 (8)

Solving for Equations of Motion:

Subtracting equation (8) multiplied by

b

from equation (7) multiplied by

a

gives:

m (a c - b^{2}) \ddot{x} + k (a c - b^{2}) x = 0

\Rightarrow \ddot{x} = - (k / m) x (A)

Similarly, subtracting equation (7) multiplied by

b

from equation (8) multiplied by

a

yields:

m (b^{2} - a c) \ddot{y} + k (b^{2} - a c) y = 0

\Rightarrow \ddot{y} = - (k / m) y (B)

Conclusion:

From equations (A) and (B), we have obtained the required equations of motion for the system. The system is identified as a 2-D Harmonic oscillator.

7.(a) आंशिक अवकल समीकरण :

(2 D^{2} - 5 D D^{'} + 2 D^{' 2}) z = 5 \sin (2 x + y) + 24 (y - x) + e^{3 x + 4 y}

को हल कीजिए
जहाँ

D \equiv \frac{\partial}{\partial x}, D^{'} \equiv \frac{\partial}{\partial y}

Solve the partial differential equation:

(2 D^{2} - 5 D D^{'} + 2 D^{' 2}) z = 5 \sin (2 x + y) + 24 (y - x) + e^{3 x + 4 y}

where

D = \frac{\partial}{\partial x}, D^{'} \equiv \frac{\partial}{\partial y}

Answer:

Introduction:

In this solution, we will solve a partial differential equation given by:

(2 D^{2} - 5 D D^{'} + 2 D^{' 2}) z = 5 \sin (2 x + y) + 24 (y - x) + e^{3 x + 4 y}

where

D = \frac{\partial}{\partial x}

and

D^{'} \equiv \frac{\partial}{\partial y}

Equation Analysis:

Given P.D.E is

(2 D^{2} - 5 D D^{'} + 5 D^{' 2}) z = 5 \sin (2 x + y) + 24 (y - x) + e^{3 x + 4 y}

The auxiliary equation is calculated as follows:

2 m^{2} - 5 m + 2 = 0 (Auxiliary Equation)

Solving this equation yields two values for

m

m = \frac{1}{2}

and

m = 1

Solution Process:

We start by finding the general solution using the values of

m

obtained from the auxiliary equation:

$ϕ_{1} (y + x / 2) + ϕ_{2} (y + 2 x)$ for $m = \frac{1}{2}$ .
$ϕ_{1} (2 y + x) + ϕ_{2} (y + 2 x)$ for $m = 1$ .

Intermediate Steps:

Now, let’s break down the solution into parts:

\begin{aligned} Now P . I . = \frac{1}{2 D^{2} - 5 D D^{'} + 2 D^{' 2}} [5 \sin (2 x + y) + 24 (y - x) + e^{3 x + 4 y}] \\ P . I = \frac{1}{2 D^{2} - 5 D D^{'} + 2 D^{' 2}} [5 \sin (2 x + y] + \frac{1}{2 D^{2} - 5 D D^{'} + 2 D^{' 2}} 24 (y - x) \\ + \frac{1}{2 D^{2} - 5 D D^{'} + 2 D^{' 2}} \cdot e^{3 x + 4 y} \\ P . I_{2} = \frac{1}{2 D^{2} - 5 D D^{'} + 2 D^{' 2}} [24 (y - x)] \\ = 24 \cdot \frac{1}{2 D^{2} - 5 D D^{'} + 2 D^{' 2}} (y - x) \\ = 24 \cdot \frac{1}{2 (- 1)^{2} - 5 \times 1 \times - 1 + 2 \times (1)^{2}} \iint v d u d v \\ = \frac{24}{20} \int \frac{v^{2}}{2} d v \end{aligned}

\begin{aligned} P \cdot I_{2} & = \frac{24}{20} \cdot (\frac{v^{3}}{6}) = \frac{1}{5} (y - x)^{3} \\ P \cdot I_{3} & = \frac{1}{2 D^{2} - 5 D D^{'} + 2 D^{' 2}} \cdot e^{2 x + 3 y} . \\ = e^{2 x + 3 y} \cdot \frac{1}{2 (2)^{2} - 5 \times 2 \times 3 + 2 \times 9} \\ = e^{2 x + 3 y} \cdot \frac{1}{8 - 30 + 18} = \frac{- e^{2 x + 3 y}}{4} \\ P \cdot I_{1} & = \frac{1}{2 D^{2} - 5 D D^{'} + 2 D^{' 2}} [5 \sin (2 x + y)] \\ = 5 [\frac{1}{2 D^{2} - 5 D D^{'} + 2 D^{2}} \sin (2 x + y)] \end{aligned}

\begin{aligned} = 5 \cdot \frac{1}{(D - 2 D) [(2 D - D^{'})} \sin (2 x + y)] \\ = 5 \cdot \frac{1}{(D - 2 D^{'})} [\frac{1}{2 \cdot 2 - 1} \int \sin v d v] where v = 2 x + y \\ = 5. \frac{1}{(D - 2 D^{'})} \times \frac{1}{3} (- \cos v) \\ P I_{1} = \frac{- 5}{3} \cdot \frac{1}{D - 2 D^{'}} \cos V = \frac{- 5}{3} \cdot \frac{x}{1! \cdot 1!} \cos (2 x + y) \\ ∴ P I = P I_{1} + P I_{2} + P I_{3} \\ P \cdot I = - \frac{5}{3} x \cos (2 x + y) + \frac{1}{5} (y - x)^{3} - \frac{e^{2 x + 3 y}}{4} \end{aligned}

Conclusion:

Combining the results of the partial integrals, we obtain the final expression for

P . I

P . I = - \frac{5}{3} x \cos (2 x + y) + \frac{1}{5} (y - x)^{3} - \frac{e^{2 x + 3 y}}{4}

The general solution for the given partial differential equation is then found by adding this

P . I

to the complementary function (C.F):

y = C . F + P . I = ϕ_{1} (2 y + x) + ϕ_{2} (y + 2 x) - \frac{5}{3} x \cos (2 x + y) + \frac{1}{5} (y - x)^{3} - \frac{e^{2 x + 3 y}}{4}

This completes the solution to the partial differential equation.

7.(b) स्थिराँकों

a, b, c

के मान निकालिए ताकि क्षेत्रकलन-सूत्र

\int_{o}^{h} f (x) d x = h [a f (o) + b f (\frac{h}{3}) + c f (h)]

अधिक से अधिक सम्भव घातीय बहुपदों के लिए सही हो । अताएव रुंडन-न्रुटि का क्रम भी ज्ञात कीजिए ।

Find the values of the constants

a, b, c

such that the quadrature formula

\int_{o}^{h} f (x) d x = h [a f (o) + b f (\frac{h}{3}) + c f (h)]

is exact for polynomials of as high degree as possible, and hence find the order of the truncation error.

Answer:

To find the values of the constants

a

b

, and

c

in the quadrature formula, we will use the method of undetermined coefficients. We will test the quadrature formula on polynomials of increasing degree until it is no longer exact. The quadrature formula is given as:

\int_{0}^{h} f (x) d x = h [a f (0) + b f (\frac{h}{3}) + c f (h)]

Step 1: Test with a Constant Function

Let’s start with a constant function,

f (x) = 1

\int_{0}^{h} 1 d x = h [a \cdot 1 + b \cdot 1 + c \cdot 1]

h = h (a + b + c)

For the formula to be exact for constant functions, we must have:

a + b + c = 1

(Equation 1)

Step 2: Test with a Linear Function

Next, let’s test with a linear function,

f (x) = x

\int_{0}^{h} x d x = h [a \cdot 0 + b \cdot (\frac{h}{3}) + c \cdot h]

\frac{1}{2} h^{2} = h [\frac{b h}{3} + c h]

\frac{1}{2} h = \frac{b h}{3} + c h

\frac{1}{2} = \frac{b}{3} + c

For the formula to be exact for linear functions, we must have:

\frac{b}{3} + c = \frac{1}{2}

(Equation 2)

Step 3: Test with a Quadratic Function

Next, let’s test with a quadratic function,

f (x) = x^{2}

\int_{0}^{h} x^{2} d x = h [a \cdot 0 + b \cdot {(\frac{h}{3})}^{2} + c \cdot h^{2}]

\frac{1}{3} h^{3} = h [\frac{b h^{2}}{9} + c h^{2}]

\frac{1}{3} h^{2} = \frac{b h^{2}}{9} + c h^{2}

\frac{1}{3} = \frac{b}{9} + c

For the formula to be exact for quadratic functions, we must have:

\frac{b}{9} + c = \frac{1}{3}

(Equation 3)

Step 4: Solve the System of Equations

Now, we have a system of three equations with three unknowns:

$a + b + c = 1$
$\frac{b}{3} + c = \frac{1}{2}$
$\frac{b}{9} + c = \frac{1}{3}$

Let’s solve this system of equations to find the values of

a

b

, and

c

The solution to the system of equations is:

a = 0

b = \frac{3}{4}

c = \frac{1}{4}

Order of the Truncation Error

To find the order of the truncation error, we need to determine the highest degree of polynomial for which the quadrature formula is exact. Since we have found the exact values of

a

b

, and

c

by testing the quadrature formula on polynomials up to degree 2 (quadratic), the formula is exact for polynomials of degree 2 or less.

Thus, the order of the truncation error is

O (h^{3})

, where

h

is the step size, as the error term will be proportional to the third derivative of the function being integrated.

Hence the required formula is

\int_{0}^{h} f (x) d x = h / 4 [3 f (h / 3) + f (h)]

The truncation error of the formula is given by

\begin{aligned} T . E . & = c / 3! f^{'''} (ξ), 0 < ξ < h \\ where C & = \int_{0}^{h} x^{3} d x - h [\frac{b h^{3}}{27} + c h^{3}] = - \frac{h^{4}}{36} \end{aligned}

Hence we have.

T E = \frac{- h^{4}}{216} f^{'''} (ξ) = O (h^{4}) .

7.(c) किसी यांत्रिक निकाय का हैमिल्टोनियन

H = p_{1} q_{1} - a q_{1}^{2} + b q_{2}^{2} - p_{2} q_{2}

के द्वारा द्योतित है, जहाँ

a, b

स्थिरांक हैं। हैमिल्टोनियन समीकरणों का हल निकालिए तथा दर्शाइए कि

\frac{p_{2} - b q_{2}}{q_{1}} =

स्थिराँक ।

The Hamiltonian of a mechanical system is given by,

H = p_{1} q_{1} - a q_{1}^{2} + b q_{2}^{2} - p_{2} q_{2}

, where a, b are the constants. Solve the Hamiltonian equations and show that

\frac{p_{2} - b q_{2}}{q_{1}} =

constant.

Answer:

The Hamiltonian of a mechanical system is given by,

H = p_{1} q_{1} - a q_{1}^{2} + b q_{2}^{2} - p_{2} q_{2}

, where

a

b

are constants. We are to solve the Hamiltonian equations and show that

\frac{p_{2} - b q_{2}}{q_{1}} =

constant.

Solution:

Let us consider the generalized coordinates

q_{1}, q_{2}

and the generalized components of momentum

p_{1}, p_{2}

. The Hamiltonian equations are given by:

{\dot{p}}_{i} = - \frac{\partial H}{\partial q_{i}} and {\dot{q}}_{i} = \frac{\partial H}{\partial p_{i}}

where

i = 1, 2

Applying the Hamiltonian equations to the given Hamiltonian, we have:

${\dot{q}}_{1} = q_{1} \Rightarrow \frac{d q_{1}}{d t} = q_{1} \Rightarrow q_{1} = c_{1} e^{t}$
${\dot{p}}_{1} = 2 a q_{1} - p_{1} \Rightarrow \frac{d p_{1}}{d t} + p_{1} = 2 a c_{1} e^{t} \Rightarrow p_{1} = a c_{1} e^{t} + c_{2} e^{- t}$ (using integrating factor $e^{t}$ )
${\dot{q}}_{2} = - q_{2} \Rightarrow q_{2} = c_{3} e^{- t}$
${\dot{p}}_{2} = p_{2} - 2 b q_{2} \Rightarrow \frac{d p_{2}}{d t} - p_{2} = - 2 b c_{3} e^{- t} \Rightarrow p_{2} = b c_{3} e^{- t} + c_{4} e^{t}$ (using integrating factor $e^{- t}$ )

Now, let’s consider the expression

\frac{p_{2} - b q_{2}}{q_{1}}

\frac{p_{2} - b q_{2}}{q_{1}} = \frac{b c_{3} e^{- t} + c_{4} e^{t} - b c_{3} e^{- t}}{c_{1} e^{t}} = \frac{c_{4} e^{t}}{c_{1} e^{t}} = \frac{c_{4}}{c_{1}} = constant

Conclusion:

The expression

\frac{p_{2} - b q_{2}}{q_{1}}

is indeed a constant, as shown above, under the solutions of the Hamiltonian equations for the given mechanical system.

8.(a) बूलीय व्यंजक :

(a + b) \cdot (\bar{b} + c) + b \cdot (\bar{a} + \vec{c})

को बूलीय-बीजगणित के नियमों का उपयोग करने के द्वारा सरल कीजिए। इस की सत्यता-सारणी से इसको मिनटर्म प्रसामान्य रूप में लिखिए।

Simplify the boolean expression:

(a + b) \cdot (\bar{b} + c) + b \cdot (\bar{a} + \bar{c})

by using the laws of boolean algebra. From its truth table write it in minterm normal form.

Answer:

To simplify the given Boolean expression, we will use the laws of Boolean algebra. The given expression is:

(a + b) \cdot (\overset{―}{b} + c) + b \cdot (\overset{―}{a} + \overset{―}{c})

Step 1: Simplification

Let’s simplify the expression step by step:

Distributive Law:

= a \cdot (\overset{―}{b} + c) + b \cdot (\overset{―}{b} + c) + b \cdot (\overset{―}{a} + \overset{―}{c})

Absorption Law:

= a \cdot (\overset{―}{b} + c) + b + b \cdot (\overset{―}{a} + \overset{―}{c})

Absorption Law:

= a \cdot (\overset{―}{b} + c) + b

Step 2: Truth Table and Minterm Normal Form

Now, let’s create a truth table for the simplified expression and then write the expression in minterm normal form. The minterm normal form is the OR of all the minterms where the output is 1.

$a$	$b$	$c$	$\overset{―}{b} + c$	$a \cdot (\overset{―}{b} + c)$	$a \cdot (\overset{―}{b} + c) + b$
0	0	0	1	0	0
0	0	1	1	0	0
0	1	0	0	0	1
0	1	1	1	0	1
1	0	0	1	1	1
1	0	1	1	1	1
1	1	0	0	0	1
1	1	1	1	1	1

Minterm Normal Form:

The minterm normal form is the OR of all the minterms where the output is 1. From the truth table, the minterms where the output is 1 are when

a, b, c

are:

$001$ (Minterm 1)
$011$ (Minterm 3)
$100$ (Minterm 4)
$101$ (Minterm 5)
$110$ (Minterm 6)
$111$ (Minterm 7)

So, the minterm normal form of the given Boolean expression is:

Σ m (1, 3, 4, 5, 6, 7)

From Truth Fable, Minter normal form is

\bar{a} b \bar{c} + \bar{a} b c + a \bar{b} \bar{c} + a \bar{b} c + a b \bar{c} + a b c

8.(b) एक द्विविमीय विभव-प्रवाह के लिए वेग विभव

ϕ = x^{2} y - x y^{2} + \frac{1}{3} (x^{3} - y^{3})

के द्वारा दिया गया है ।

x

व

y

दिशाओं के अनुदिश वेग घटकों का निर्धारण कीजिए । धारा-फलन

ψ

का भी निर्धारण कीजिए और जाँच कीजिए कि क्या

ϕ

एक सम्भव प्रवाह को निस्पित करता है अथवा नहीं।

For a two-dimensional potential flow, the velocity potential is given by

ϕ = x^{2} y - x y^{2} + \frac{1}{3} (x^{3} - y^{3})

. Determine the velocity components along the directions

x

and

y

. Also, determine the stream function

ψ

and check whether

ϕ

represents a possible case of flow or not.

Answer:

Velocity Components:

Let

\vec{q} = u \hat{i} + v \hat{ȷ}

. Then,

\begin{aligned} u = - \frac{\partial ϕ}{\partial x} \\ ⟹ u = - (2 x y - y^{2} + x^{2}) \\ ⟹ u = y^{2} - x^{2} - 2 x y \end{aligned}

Also,

\begin{aligned} v = - \frac{\partial ϕ}{\partial y} \\ \Rightarrow v = - (x^{2} - 2 x y - y^{2}) \\ ⟹ v = y^{2} - x^{2} + 2 x y \end{aligned}

Stream Function $ψ$ :

We know that

ϕ + i ψ

is an analytic function and satisfies Cauchy Riemann equations.

\begin{aligned} So, \frac{\partial ψ}{\partial y} = u and \frac{\partial ψ}{\partial x} = v \\ \Rightarrow \frac{\partial ψ}{\partial x} = y^{2} - x^{2} + 2 x y \end{aligned}

Integrating with respect to

x

ψ = x y^{2} - \frac{x^{3}}{3} + x^{2} y + f (y)

Now,

\begin{aligned} \frac{\partial ψ}{\partial y} = 2 x y + x^{2} + f^{'} (y) \\ \frac{\partial ψ}{\partial y} = - u \\ \Rightarrow x^{2} + 2 x y + f^{'} (y) = x^{2} + 2 x y - y^{2} \\ \Rightarrow f^{'} (y) = - y^{2} \\ \frac{d f}{d y} = - y^{2} \\ \Rightarrow f = \frac{- y^{3}}{3} + c \end{aligned}

So,

\begin{aligned} ψ = x y^{2} - \frac{(x^{3} + y^{3})}{3} + x^{2} y + c \\ ψ = 0 at origin ⟹ c = 0 \end{aligned}

So,

ψ = x y^{2} + x^{2} y - \frac{(x^{3} + y^{3})}{3}

Checking for Possible Flow:

Since

\frac{\partial u}{\partial x} + \frac{\partial y}{\partial y} = - 2 x - 2 y + 2 y + 2 x = 0

⟹

Equation of continuity is satisfied,

⟹

It is a possible flow.

(c) एक पतली वलयिका (एनुलस) क्षेत्र $0 < a ⩽ r ⩽ b, 0 ⩽ θ ⩽ 2 π$ को घेरती है । इसके तल तापअवरोधी हैं.। आन्तरिक किलारे के साथ-साथ ताप $0^{\circ}$ पर स्थिर रखा जाता है जबकि बाह्य किनारे का ताप $T = K \cos \frac{θ}{2}$ पर बनाए रखा जाता है, जहाँ $K$ एक स्थिरांक है । बलयिका में ताप-वितरण का निर्धारण कीजिए ।

A thin annulus occupies the region

0 < a ⩽ r ⩽ b, 0 ⩽ θ ⩽ 2 π

. The faces are insulated. Along the inner edge the temperature is maintained at

0^{\circ}

, while along the outer edge the temperature is held at

T = K \cos \frac{θ}{2}

, where

K

is a constant. Determine the temperature distribution in the annulus.

Answer:

Formulating the Differential Equation:

The problem is mathematically formulated as solving Laplace’s equation:

\nabla^{2} T = 0 for a \leq r \leq b, 0 \leq θ \leq 2 π

with the boundary conditions:

T (a, θ) = 0 and T (b, θ) = K \cos \frac{θ}{2}

General Solution of Laplace’s Equation:

The required general solution of Laplace’s equation in polar coordinates is given by:

T (r, θ) = (c_{1} r^{n} + c_{2} r^{- n}) (A \cos n θ + B \sin n θ)

The principle of superposition gives

T (r, θ) = \sum_{n = 1}^{\infty} [r^{n} - \frac{a^{2 n}}{λ^{n}}] (A_{n} \cos n θ + B_{n} \sin n θ) ∣

Now, using

B C

ii), we obtain

T (b, θ) = k \cos \frac{θ}{2} = \sum_{n = 1}^{\infty} [b^{n} - b^{- n} a^{2 n}] [A_{n} \cos n θ + B_{n} \sin n θ

Which is a full-range fourier series.

Hence,

\begin{aligned} {\dot{A}}_{n} (b^{n} - b^{- n} a^{2 n}) & = \frac{1}{π} \int_{0}^{2 π} k \cos \frac{θ}{2} \cos n θ d θ \\ = \frac{k}{2 π} \int_{0}^{2 π} [\cos [n + \frac{1}{2}] θ + \cos (n - \frac{1}{2}) θ] d θ \\ = \frac{k}{2 π} {[\frac{\sin (n + 1 / 2) θ}{(n + 1 / 2)} + \frac{\sin (n - 1 / 2) θ}{(n - 1 / 2)}]}_{0}^{2 π} \\ = 0 ie A_{n} = 0 \end{aligned}

Implying

A_{n} = 0

. Also

\begin{aligned} B n (b^{n} - b^{- n} a^{2 n}) = \frac{k}{\bar{n}} \int_{0}^{2 π} \cos θ / 2 \sin n θ d θ \\ = \frac{k}{2 π} \int_{0}^{2 π} [\sin (n + \frac{1}{2}) θ + \sin ((n - \frac{1}{2}) θ]] d θ \end{aligned}

\begin{aligned} = \frac{- k}{2 π} {[\frac{\cos (n + 1 / 2)}{n + 1 / 2} + \frac{\cos (n - 1 / 2) θ^{2}}{n - 1 / 2}]}_{0}^{2 π} \\ = \frac{- k}{2 π} [- \frac{1}{n + 1 / 2} - \frac{1}{n + 1 / 2} - \frac{1}{n - 1 / 2} - \frac{1}{n - 1 / 2}) \\ = \frac{k}{π} [\frac{1}{n + 1 / 2} + \frac{1}{n - 1 / 2}] = \frac{k}{π} [\frac{2 n}{n^{2} - 1 / 4}] . \\ ∴ B n (b^{n} - b^{- n} a^{2 n}) = \frac{8 k n}{π (4 n^{2} - 1)} \end{aligned}

Thus, the temperature distribution in annulus is given by.

T (r, θ) = \frac{δ k}{π} \sum_{n = 1}^{\infty} \frac{n}{4 n^{2} - 1} [\frac{(s / a)^{n} - (a / a)^{n}}{(b / a)^{n} - (a / b)^{n}}] \sin θ

Which is required result.

Free UPSC Mathematics Optional Paper-2 2018 Solutions: View Online | UPSC Maths Solution | IAS Maths Solution

Introduction

Definitions

Work/Calculations

Step 1: Assume f ( x ) f ( x ) f(x)f(x)f(x) is a Unit in R [ x ] R [ x ] R[x]R[x]R[x]

Step 2: Examine the Degrees of f ( x ) f ( x ) f(x)f(x)f(x) and g ( x ) g ( x ) g(x)g(x)g(x)

Step 3: Conclude m = n = 0 m = n = 0 m=n=0m = n = 0m=n=0

Step 4: Show f ( x ) f ( x ) f(x)f(x)f(x) is a Unit in R R RRR

Conclusion

Decision Variables

Objective Function

Constraints

Nitrogen Fertilizer Constraint

Phosphate Constraint

Potash Constraint

Non-negativity Constraint

Linear Programming Problem

Maximize

Subject to

Conclusion

Quotient Group Definition

Multiplicative Group of Complex Numbers on the Unit Circle

Define an Isomorphism

Prove Isomorphism

1. Homomorphism

2. Injective (One-to-One)

3. Surjective (Onto)

Conclusion

Proof that a Convex Function on an Open Interval is Continuous

Proof:

Counterexample when the Open Interval Condition is Dropped

Finding the Generator

Finding the Proper Subgroups

Conclusion

Given System of Equations:

Method:

Augmented Matrix:

Row Operations to get RREF:

Expressing Variables:

Finding Basic Solutions:

Conclusion:

Given Properties:

Step 1: Evaluate f ( 0 ) f ( 0 ) f(0)f(0)f(0)

Step 2: Evaluate f ( 1 ) f ( 1 ) f(1)f(1)f(1)

Step 3: Evaluate f ( x ) f ( x ) f(x)f(x)f(x) for any x x xxx

Step 4: Conclusion

Detailed Conclusion:

Introduction

Conclusion

Given Ellipsoid

Tangent Plane to the Ellipsoid

General Equation of a Plane

Coefficient Relations

Substituting Back into the Ellipsoid Equation

Equation of the Tangent Plane

Condition for the Plane not being Perpendicular to the x y x y xyxyxy-plane

Partial Differential Equation

Conclusion

Given Data

Calculation

Substitution

Algorithm for Bisection Method

Step 1: Define the Function

Step 2: Choose Interval [a, b]

Step 3: Calculate Midpoint

Step 4: Evaluate Function at Midpoint

Step 5: Update Interval

Step 6: Check Tolerance

Step 7: Return the Root

Given Equations:

Step 1: Substitute the Curve Equations

Step 2: Solve for ϕ ϕ phi\phiϕ

Step 3: Formulate the Integral Surface

Conclusion:

Step 1: Test with a Constant Function

Step 2: Test with a Linear Function

Step 3: Test with a Quadratic Function

Step 4: Solve the System of Equations

Order of the Truncation Error

Solution:

Step 1: Assume $f (x)$ is a Unit in $R [x]$

Step 2: Examine the Degrees of $f (x)$ and $g (x)$

Step 3: Conclude $m = n = 0$

Step 4: Show $f (x)$ is a Unit in $R$

Step 1: Evaluate $f (0)$

Step 2: Evaluate $f (1)$

Step 3: Evaluate $f (x)$ for any $x$

Condition for the Plane not being Perpendicular to the $x y$ -plane

Step 2: Solve for $ϕ$