Free UPSC Mathematics Optional Paper-2 2021 Solutions: View Online | UPSC Maths Solution | IAS Maths Solution

July 13, 2025ByAbstract Classes

खण्ड-A / SECTION-A

(a) मान लीजिए कि $m_{1}, m_{2}, \dots, m_{k}$ धनात्मक पूर्णांक हैं तथा $d > 0, m_{1}, m_{2}, \dots, m_{k}$ का महत्तम समापवर्तक है। दर्शाइए कि ऐसे पूर्णांक $x_{1}, x_{2}, \dots, x_{k}$ अस्तित्व में हैं ताकि

d = x_{1} m_{1} + x_{2} m_{2} + \dots + x_{k} m_{k}

Let

m_{1}, m_{2}, \dots, m_{k}

be positive integers and

d > 0

the greatest common divisor of

m_{1}, m_{2}, \dots, m_{k}

. Show that there exist integers

x_{1}, x_{2}, \dots, x_{k}

such that

d = x_{1} m_{1} + x_{2} m_{2} + \dots + x_{k} m_{k}

Answer:

Step 1: Defining a Set S

Let

s = {a u + b v | u, v

are integers and

a u + b v > 0}

Step 2: Investigating Cases for $a$ and $b$

a > 0

, then

a = a 1 + b (0) > 0

, which implies that

a > 0

a < 0

, then

- a = a (- 1) + b (0) > 0

, which implies that

- a \in S

Similarly, if

b > 0

, then

b \in S

b < 0

, then

- b \in S

Therefore,

S \neq ϕ

and

S

contains positive integers.

By the well-ordering principle,

S

has a least element, say

d

Step 3: Proving that $d$ is the GCD of $a$ and $b$

Now we have

d \in S

such that

d = a x + b y - - - - (1)

for some integers

x, y

Also,

d > 0

To prove that

d

is the greatest common divisor (GCD) of

a

and

b

, consider the following:

Let

a = d q + r - - - - (2)

(where

0 ⩽ r < d

) be a division of

a

d

r \neq 0

, then

r = a - d q

\begin{aligned} = a - (a x + b y) q (as d = a x + b y) \\ = a (1 - λ q) + b (- y z) > 0 (since 1 - λ q, - y q are integers) \\ r > 0, r \in S if r < d and r \in S \end{aligned}

This leads to a contradiction since

d

is the least element of

S

Therefore,

r = 0

, and we have

a = d q

This implies that

\frac{a}{d} = q

and

\frac{d}{a}

are integers.

Similarly,

\frac{d}{b}

is also an integer.

Step 4: Proving Uniqueness of GCD

Suppose

\frac{C}{c}, \frac{c}{b} \Rightarrow \frac{c}{a x + b y} \Rightarrow \frac{c}{d}

Therefore

\frac{d}{a}, \frac{d}{b}

also if

\frac{c}{a}, \frac{c}{b} \Rightarrow \frac{c}{d}

d

is gcd of

a

and

b

If possible

d^{'}

is also gcd of

a

and

b

Then

\frac{d^{'}}{a}, \frac{d^{'}}{b} \Rightarrow \frac{d}{d^{'}} \to

(3)
Similarly,

\frac{d}{c}, \frac{d}{b} \Rightarrow \frac{d^{'}}{d} \to

(4)

This implies

\frac{d}{d^{'}}

(from the divisibility relation) which, in turn, implies

d = d^{'}

(from the uniqueness of GCD).

Step 5: Extending the Argument

The above argument can be extended to more than two integers.

d = \gcd (m_{1}, m_{2}, \dots, m_{k})

, there exist integers

λ_{1}, λ_{2}, \dots, λ_{k}

such that

\begin{aligned} d = λ_{1} m_{1} + λ_{2} m_{2} + \dots + m_{k} λ_{k} \\ \Rightarrow d = λ_{1} m_{1} + λ_{2} m_{2} + \dots + λ_{k} m_{k} \end{aligned}

This concludes the proof.

(b) श्रेणी

x^{4} + \frac{x^{4}}{1 + x^{4}} + \frac{x^{4}}{{(1 + x^{4})}^{2}} + \frac{x^{4}}{{(1 + x^{4})}^{3}} + \dots

के

[0, 1]

पर एकसमान अभिसरण की जाँच कीजिए।

Test the uniform convergence of the series

x^{4} + \frac{x^{4}}{1 + x^{4}} + \frac{x^{4}}{{(1 + x^{4})}^{2}} + \frac{x^{4}}{{(1 + x^{4})}^{3}} + \dots

[0, 1]

Answer:

Preliminaries

Definition: Uniform Convergence

A series

\sum_{n = 0}^{\infty} f_{n} (x)

is said to be uniformly convergent on a set

S

if for every

ϵ > 0

, there exists an

N \in N

such that for all

m > N

and for all

x \in S

| \sum_{n = m + 1}^{\infty} f_{n} (x) | < ϵ

Approach

Step 1: Recognize the Series as a Geometric Series

Explanation

For each fixed

x

, the series is a geometric series. A geometric series is a series of the form

\sum_{n = 0}^{\infty} a \cdot r^{n}

, where

a

is the first term and

r

is the common ratio.

Application to Our Series

In our case, the first term

a = x^{4}

and the common ratio

r = \frac{1}{1 + x^{4}}

Step 2: Find the $n^{t h}$ Partial Sum $s_{n} (x)$

Explanation

The

n^{t h}

partial sum of a geometric series is given by:

s_{n} = \frac{a (1 - r^{n})}{1 - r}

r \neq 1

Application to Our Series

For our series, the

n^{t h}

partial sum

s_{n} (x)

becomes:

s_{n} (x) = \frac{x^{4} (1 - {(\frac{1}{1 + x^{4}})}^{n})}{1 - \frac{1}{1 + x^{4}}} = 1 + x^{4} - \frac{1}{(1 + x^{4})^{n - 1}}

Step 3: Determine the Limit Function $S (x)$

Explanation

The limit function

S (x)

is defined as:

S (x) = lim_{n \to \infty} s_{n} (x)

Application to Our Series

n \to \infty

s_{n} (x)

approaches:

S (x) = {\begin{cases} 1 + x^{4}, & if x \neq 0 \\ 0, & if x = 0 \end{cases}

Step 4: Check for Discontinuity

Explanation

A uniformly convergent series of continuous functions must converge to a continuous limit function.

Application to Our Series

The limit function

S (x)

is discontinuous at

x = 0

. This implies that the original series cannot be uniformly convergent, as a uniformly convergent series of continuous functions would converge to a continuous function.

Conclusion

By the definition of uniform convergence and the properties of geometric series, we conclude that the series

\sum_{n = 0}^{\infty} \frac{x^{4}}{(1 + x^{4})^{n}}

is not uniformly convergent on the interval

[0, 1]

because its limit function

S (x)

is discontinuous at

x = 0

f

, अन्तराल

[a, b]

में एकदिष्ट है, तब सिद्ध कीजिए कि

f, [a, b]

में रीमान समाकलनीय है।

If a function

f

is monotonic in the interval

[a, b]

, then prove that

f

is Riemann integrable in

[a, b]

Answer:

To prove that a monotonic function

f

is Riemann integrable on the interval

[a, b]

, we need to show that for any given

ϵ > 0

, there exists a partition

P

[a, b]

such that the upper sum

U (f, P)

and the lower sum

L (f, P)

satisfy:

U (f, P) - L (f, P) < ϵ .

Definitions:

Monotonic Function: A function $f$ is said to be monotonic on $[a, b]$ if it is either non-decreasing or non-increasing on that interval.
Partition $P$ : A partition $P$ of $[a, b]$ is a finite sequence $a = x_{0} < x_{1} < \dots < x_{n} = b$ .
Upper Sum $U (f, P)$ : Given a partition $P$ , the upper sum is defined as:

U (f, P) = \sum_{i = 1}^{n} M_{i} Δ x_{i},

where

M_{i} = sup {f (x) : x \in [x_{i - 1}, x_{i}]}

and

Δ x_{i} = x_{i} - x_{i - 1}

Lower Sum $L (f, P)$ : Given a partition $P$ , the lower sum is defined as:

L (f, P) = \sum_{i = 1}^{n} m_{i} Δ x_{i},

where

m_{i} = inf {f (x) : x \in [x_{i - 1}, x_{i}]}

and

Δ x_{i} = x_{i} - x_{i - 1}

Proof:

Choose $ϵ > 0$ : Let $ϵ > 0$ be given.
Partition Size: Since $f$ is monotonic, it is bounded on $[a, b]$ . Let $M$ and $m$ be the upper and lower bounds of $f$ on $[a, b]$ , respectively. Choose a partition size $Δ x$ such that:

Δ x < \frac{ϵ}{M - m} .

Construct the Partition $P$ : Divide the interval $[a, b]$ into subintervals of length $Δ x$ (the last interval may be smaller if $[b - a]$ is not a multiple of $Δ x$ ).
Upper and Lower Sums: For each subinterval $[x_{i - 1}, x_{i}]$ , the function $f$ will attain its maximum and minimum at the endpoints $x_{i - 1}$ and $x_{i}$ (or vice versa) because $f$ is monotonic. Therefore, $M_{i} - m_{i} = f (x_{i}) - f (x_{i - 1})$ .
Difference Between Upper and Lower Sums:

\begin{aligned} U (f, P) - L (f, P) & = \sum_{i = 1}^{n} (M_{i} - m_{i}) Δ x_{i} \\ = \sum_{i = 1}^{n} (f (x_{i}) - f (x_{i - 1})) Δ x_{i} \\ = (M - m) Δ x \\ < ϵ . \end{aligned}

Conclusion:

Since we can find a partition

P

such that

U (f, P) - L (f, P) < ϵ

for any given

ϵ > 0

, the function

f

is Riemann integrable on

[a, b]

This completes the proof.

(d) मान लीजिए कि

c : [0, 1] \to C, c (t) = e^{4 π i t}, 0 \leq t \leq 1

के द्वारा परिभाषित एक वक्र है। कन्दूर समाकल

\int_{c} \frac{d z}{2 z^{2} - 5 z + 2}

का मान निकालिए।

Let

c : [0, 1] \to C

be the curve, where

c (t) = e^{4 π i t}, 0 \leq t \leq 1

. Evaluate the contour integral

\int_{c} \frac{d z}{2 z^{2} - 5 z + 2}

Answer:

Given Function and Circle

Let

z (t) = e^{4 π i t}, 0 ⩽ t ⩽ 1

. Then,

| Z (t) | = 1

. Therefore, the curve

C

can be defined as the circle with the equation

| z - 0 | = 1

. This means that

C

is a circle of unit radius with its center at the origin

(0, 0)

denoted as

O

Integral Setup

Let’s find the integral:

\int_{C : | z - 0 | = 1} \frac{1}{2 z^{2} - 5 z + 2} d z = \int_{C : | z | = 1} f (z) d z

Here,

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

Partial Fraction Decomposition

We can perform partial fraction decomposition to simplify

f (z)

\begin{aligned} = \frac{1}{(2 z - 1) (z - 2)} \\ = \frac{1}{2 (z - \frac{1}{2}) (z - 2)} \end{aligned}

Clearly,

z = \frac{1}{2}

lies on the curve

| z | = 1

Residue Integration

Now, we can proceed with the integral:

\begin{aligned} \int_{C : | z - 0 | = 1} f (z) d z = \int_{C} \frac{\frac{1}{2 (z - 2)}}{(z - \frac{1}{2})} d z \\ = \int_{C} \frac{f (z)}{z - \frac{1}{2}} d z (2) \end{aligned}

Where

f_{1} (z) = \frac{1}{2 (z - 2)}

By Cauchy’s First Integral Formula,

\begin{aligned} \int_{C} \frac{f_{1} (z)}{z - \frac{1}{2}} d z = 2 π i f_{1} (\frac{1}{2}) \\ \Rightarrow \int_{C} \frac{\frac{1}{2 (z - 2)}}{z - \frac{1}{2}} d z = 2 π i [\frac{1}{2 (\frac{1}{2} - 2)}] = \frac{\frac{π i}{1 - 4}}{2} = \frac{2 π i}{- 3} = - \frac{2 π i}{3} \\ \int_{C} f (z) d z = - \frac{2 π i}{3} \end{aligned}

Winding Number Calculation

Since

C (t) = e^{4 π t}, 0 ⩽ t ⩽ 1

, it is clear that curve

C

winds up two times about the origin with unit radius.

Therefore, the final result is:

\int_{C} f (z) d z = - \frac{4 π i}{3}

(e) एक कम्पनी के एक विभाग के पाँच कर्मचारियों को पाँच कार्य सम्पन्न करने हैं। जितना समय (घंटों में) एक व्यक्ति एक कार्य को सम्पन्न करने के लिए लेता है, वह प्रभाविता आव्यूह में दिया गया है। इन पाँच कर्मचारियों को इन सभी कार्यों को इस तरह निर्धारित कीजिए जिससे कि समस्त कार्य सम्पन्न करने का समय न्यूनतम हो :

A department of a company has five employees with five jobs to be performed. The time (in hours) that each man takes to perform each job is given in the effectiveness matrix. Assign all the jobs to these five employees to minimize the total processing time :

\begin{array}{ccccccc} 1 & 2 & 3 & 4 & 5 \\ A & 10 & 5 & 13 & 15 & 16 \\ B & 3 & 9 & 18 & 13 & 6 \\ C & 10 & 7 & 2 & 2 & 2 \\ D & 7 & 11 & 9 & 7 & 12 \\ E & 7 & 9 & 10 & 4 & 12 \end{array}

Answer:

Step 1: Original Cost Matrix

This is the original cost matrix:

\begin{array}{ccccccc} 1 & 2 & 3 & 4 & 5 \\ A & 10 & 5 & 13 & 15 & 16 \\ B & 3 & 9 & 18 & 13 & 6 \\ C & 10 & 7 & 2 & 2 & 2 \\ D & 7 & 11 & 9 & 7 & 12 \\ E & 7 & 9 & 10 & 4 & 12 \end{array}

Step 2: Subtract Row Minima

Subtract the row minimum from each row:

\begin{array}{rrrrrr} 5 & 0 & 8 & 10 & 11 & (- 5) \\ 0 & 6 & 15 & 10 & 3 & (- 3) \\ 8 & 5 & 0 & 0 & 0 & (- 2) \\ 0 & 4 & 2 & 0 & 5 & (- 7) \\ 3 & 5 & 6 & 0 & 8 & (- 4) \end{array}

Step 3: Subtract Column Minima

Because each column contains a zero, subtracting column minima has no effect.

Step 4: Cover All Zeros with Lines

To cover all zeros, we need 4 lines:

\begin{array}{rrrrrr} 5 & 0 & 8 & 10 & 11 & (x) \\ 0 & 6 & 15 & 10 & 3 \\ 8 & 5 & 0 & 0 & 0 & (x) \\ 0 & 4 & 2 & 0 & 5 \\ 3 & 5 & 6 & 0 & 8 & (x) \\ x & x \end{array}

Step 5: Create Additional Zeros

The number of lines is smaller than 5, and the smallest uncovered number is 2. Subtract this number from all uncovered elements and add it to all elements that are covered twice:

\begin{array}{rrrrr} 7 & 0 & 8 & 12 & 11 \\ 0 & 4 & 13 & 10 & 1 \\ 10 & 5 & 0 & 2 & 0 \\ 0 & 2 & 0 & 0 & 3 \\ 3 & 3 & 4 & 0 & 6 \end{array}

Step 6: Cover All Zeros with Lines Again

Now, 5 lines are required to cover all zeros:

\begin{array}{ccccc} 7 & 0 & 8 & 12 & 11 \\ 0 & 4 & 13 & 10 & 1 \\ 10 & 5 & 0 & 2 & 0 \\ 0 & 2 & 0 & 0 & 3 \\ 3 & 3 & 4 & 0 & 6 \end{array}

Step 7: The Optimal Assignment

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

\begin{array}{rrrrr} 7 & 0 & 8 & 12 & 11 \\ 0 & 4 & 13 & 10 & 1 \\ 10 & 5 & 0 & 2 & 0 \\ 0 & 2 & 0 & 0 & 3 \\ 3 & 3 & 4 & 0 & 6 \end{array}

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

\begin{array}{ccccc} 10 & 5 & 13 & 15 & 16 \\ 3 & 9 & 18 & 13 & 6 \\ 10 & 7 & 2 & 2 & 2 \\ 7 & 11 & 9 & 7 & 12 \\ 7 & 9 & 10 & 4 & 12 \end{array}

Step 8: Optimal Value

The optimal value equals 23.

Conclusion:

In this optimization problem, we aimed to assign five jobs to five employees in a department of a company in a way that minimizes the total processing time. We began with an original cost matrix representing the time each employee takes to perform each job. Through a series of steps involving the subtraction of row minima, column minima, and the creation of additional zeros, we arrived at an optimal assignment.

The optimal assignment, shown in the original cost matrix, assigns each job to an employee in such a way that the total processing time is minimized. The optimal value for this assignment is 23 hours.

This optimized job assignment can help the department efficiently utilize its workforce and minimize the time required to complete the tasks, ultimately leading to increased productivity and cost savings.

(a) $f (x) = x^{3} - 9 x^{2} + 26 x - 24$ का, $0 \leq x \leq 1$ के लिए, अधिकतम तथा न्यूनतम मान निकालिए।

Find the maximum and minimum values of

f (x) = x^{3} - 9 x^{2} + 26 x - 24

for

0 \leq x \leq 1

Answer:

Step 1: Objective Function

Given the function

f (x) = x^{3} - 9 x^{2} + 26 x - 24

for

0 \leq x \leq 1

f (x) = x^{3} - 9 x^{2} + 26 x - 24 for \forall x \in [0, 1] (1)

Step 2: Calculate the Derivative

Calculate the derivative of

f (x)

f^{'} (x) = 3 x^{2} - 18 x + 26 (2)

Step 3: Find Critical Points

To find critical points, set

f^{'} (x) = 0

3 x^{2} - 18 x + 26 = 0

Solving for

x

x = \frac{18 \pm \sqrt{324 - 4 (3) (26)}}{6} = \frac{18 \pm \sqrt{12}}{6} = \frac{9 \pm \sqrt{3}}{3}

These critical points do not fall within the interval

[0, 1]

Step 4: Determine Monotonicity

Check if

f (x)

is monotonically increasing or decreasing on

[0, 1]

We have:

f^{'} (0) = 3 (0)^{2} - 18 (0) + 26 = 26 > 0

and

f^{'} (1) = 3 (1)^{2} - 18 (1) + 26 = 11 > 0

Therefore,

f^{'} (x) > 0

for all

x \in [0, 1]

, indicating that

f (x)

is monotonically increasing on

[0, 1]

Step 5: Find Maximum and Minimum Values

Calculate the maximum and minimum values of

f (x)

within the given interval:

\begin{aligned} f_{max} = {f (x) |}_{x = 1} = 1 - 9 + 26 - 24 = - 6 \\ f_{max} = - 6 at x = 1 \\ f_{min} = {f (x) |}_{x = 0} = 0 - 0 + 0 - 24 = - 24 \\ f_{min} (0) = - 24; f_{max} (1) = - 6 \end{aligned}

Step 6: Conclusion

In the given interval

[0, 1]

, the function

f (x) = x^{3} - 9 x^{2} + 26 x - 24

reaches its maximum value of -6 at

x = 1

and its minimum value of -24 at

x = 0

(b) मान लीजिए कि

F

एक क्षेत्र है तथा

f (x) \in F [x]

, क्षेत्र

F

के ऊपर घात

> 0

का एक बहुपद है। दर्शाइए कि एक क्षेत्र

F^{'}

तथा एक अंतःस्थापन

q : F \to F^{'}

इस प्रकार से अस्तित्व में हैं कि बहुपद

f^{q} \in F^{'} [x]

का एक मूल

F^{'}

में है, जहाँ

f^{q}, f

के प्रत्येक गुणांक

a

को

q (a)

द्वारा प्रतिस्थापित करने से प्राप्त होता है।

Let

F

be a field and

f (x) \in F [x]

a polynomial of degree

> 0

over

F

. Show that there is a field

F^{'}

and an imbedding

q : F \to F^{'}

s.t. the polynomial

f^{q} \in F^{'} [x]

has a root in

F^{'}

, where

f^{q}

is obtained by replacing each coefficient

a

f

q (a)

Answer:

Step 1: Problem Statement

We are given a field

F

and a polynomial

f (x) \in F [x]

of degree greater than 0. The goal is to show the existence of a field

F^{'}

and an embedding

q : F \to F^{'}

such that the polynomial

f^{q} \in F^{'} [x]

has a root in

F^{'}

. Here,

f^{q}

is obtained by replacing each coefficient

a

f

q (a)

Step 2: Maximal Ideal and Field Extension

Suppose

f (x) \in F [x]

is a polynomial of degree greater than 0 over a field

F

. We consider the ideal

M = ⟨ f (x) ⟩

F [x]

. Since

f (x)

is irreducible,

M

is a maximal ideal in

F [x]

Step 3: Field Extension

We create a field extension by taking

F

modulo this maximal ideal, denoted as

\frac{F [x]}{M}

, which is itself a field.

Step 4: Define Embedding $q$

Define the embedding

q : F \to \frac{F [x]}{M}

q (a) = a + M

. This embedding

q

is a homomorphism.

Step 5: Ker $q$ and Injectiveness

We consider the kernel of

q

, i.e., elements

a \in F

such that

q (a) = 0 + M

. This implies

a + M = M

, which further implies

a \in M = ⟨ f (x) ⟩

. Therefore,

a = f (x) q (x)

for some

q (x) \in F [x]

. Since

f (x)

is irreducible and

a

is a polynomial of degree greater than 0, this relation holds only when

a = 0

. Consequently,

ker q = {0}

q

is injective.

Step 6: Isomorphism and Subfield

Hence,

F

is isomorphic to

q (F)

. We can view

F^{'}

as containing

F

by identifying

a \in F

with

q (a)

and vice versa.

Step 7: Construct $Ψ$

Define

Ψ : F [x] \to \frac{F [x]}{M}

such that

Ψ (f (x)) = f (x) + M

Ψ

is a natural homomorphism.

Step 8: Show $α$ as a Root

Let

α = Ψ (x) = x + M

. We claim that

α

is a root of

f^{q}

F

Step 9: Evaluate $f^{q}$

Evaluate

f^{q}

as follows:

\begin{aligned} Ψ (f^{q}) & = Ψ (α_{0} + α_{1} x + α_{2} x^{2} + \dots + α_{n} x^{n}) \\ = Ψ (α_{0}) + Ψ (α_{1}) Ψ (x) + \dots + Ψ (α_{n}) Ψ (x^{n}) \end{aligned}

Step 10: Use $p^{(x)}$

Note that

Ψ (p^{(x)}) = p^{(x)} + M = p^{(x)} + ⟨ p^{(x)} ⟩ = M =

zero of

F

Step 11: Zero of $F^{'}$

Thus, the zero of

F^{'}

is:

\begin{aligned} Zero of F^{'} = Ψ (α_{0}) + Ψ (α_{1}) Ψ (x) + \dots + Ψ (α_{n}) Ψ (x^{n}) \\ = q (a_{0}) + q (a_{1}) Ψ (x) + \dots + q (a_{n}) Ψ (x^{n}) \end{aligned}

Step 12: Zero is a Root

Since

F ≅ q (F)

, the zero is also

a_{0} + a_{1} α + a_{2} α^{2} + \dots + a_{n} α^{n}

, which is equal to

f (a) = f^{q}

Step 13: Conclusion

Hence,

a

is a root of

f (a)

F

. By replacing each coefficient

a

f

q (a)

, we obtain

f^{q}

Therefore, there exists a field

F^{'}

and an embedding

q : F \to F^{'}

such that the polynomial

f^{q} \in F^{'} [x]

has a root in

F^{'}

, where

f^{q}

is obtained by replacing each coefficient

a

f

q (a)

| z + 1 | > 3

में

f (z) = \frac{z^{2} - z + 1}{z (z^{2} - 3 z + 2)}

का लौराँ श्रेणी प्रसार,

(z + 1)

की घातों में ज्ञात कीजिए।

Find the Laurent series expansion of

f (z) = \frac{z^{2} - z + 1}{z (z^{2} - 3 z + 2)}

in the powers of

(z + 1)

in the region

| z + 1 | > 3

Answer:

Step 1: Problem Statement

Find the Laurent series expansion of

f (z) = \frac{z^{2} - z + 1}{z (z^{2} - 3 z + 2)}

in the powers of

(z + 1)

in the region

| z + 1 | > 3

Step 2: Rewrite $f (z)$ in terms of $u$

Let

f (z) = \frac{z^{2} - z + 1}{z (z^{2} - 3 z + 2)}

. Now, let

z + 1 = u

, which implies

z = u - 1

\begin{aligned} f (u) = \frac{(u - 1)^{2} - (u - 1) + 1}{(u - 1) [(u - 1)^{2} - 3 (u - 1) + 2]} \\ \Rightarrow f (u) = \frac{u^{2} + 1 - 2 u - u + 1 + 1}{(u - 1) [u^{2} + 1 - 2 u - 3 u + 3 + 2]} \\ \Rightarrow f (u) = \frac{u^{2} - 3 u + 3}{(u - 1) (u^{2} - 5 u + 6)} \\ f (u) = \frac{u^{2} - 3 u + 3}{(u - 1) (u - 2) (u - 3)} \\ = \frac{A}{u - 1} + \frac{B}{u - 2} + \frac{C}{u - 3} \to (1) \\ \Rightarrow u^{2} - 3 u + 3 = A (u - 2) (u - 3) + B (u - 1) (u - 3) + C (u - 1) (u - 2) \end{aligned}

Step 3: Solve for Constants $A$ , $B$ , and $C$

u = 2

then

4 - 6 + 3 = B (1) (- 1) \Rightarrow B = - 1

u = 2

then

1 - 3 + 3 = A (- 1) (- 2) \Rightarrow A = \frac{1}{2}

u = 3

then

9 - 9 + 3 = C (2) (1) \Rightarrow C = \frac{3}{2}

Step 4: Rewrite $f (u)$ in Partial Fractions

From (1), we have:

f (u) = \frac{\frac{1}{2}}{u - 1} + \frac{1}{u - 2} + \frac{\frac{3}{2}}{u - 3}

Step 5: Express Fractions in Terms of $u$

To express

\frac{1}{u - 1}

\frac{1}{u - 1} = \frac{1}{u} {(1 - \frac{1}{u})}^{- 1} = \frac{1}{u} (1 + \frac{1}{u} + \frac{1}{u^{2}} + \dots)

This is valid for

| \frac{1}{u} | < 1

and

| u | > 0

\frac{1}{u - 1} = \frac{1}{u} + \frac{1}{u^{2}} + \frac{1}{u^{3}} + \dots

is valid for

| u | > 1

(2).

Step 6: Express $\frac{1}{u - 2}$

To express

\frac{1}{u - 2}

\frac{1}{u - 2} = \frac{1}{u} {(1 - \frac{2}{u})}^{- 1} = \frac{1}{u} (1 + \frac{2}{4} + \frac{4}{u^{2}} + \dots)

This is valid for

| \frac{2}{u} | < 1

and

| u | > 0

\frac{1}{u - 2} = \frac{1}{4} + \frac{2}{u^{2}} + \frac{4}{u^{3}} + \dots

is valid for

| u | > 2

(3).

Step 7: Express $\frac{1}{u - 3}$

To express

\frac{1}{u - 3}

\frac{1}{u - 3} = \frac{1}{u} {(1 - \frac{3}{u})}^{- 1} = \frac{1}{u} (1 + \frac{3}{u} + \frac{9}{u^{2}} + \dots)

This is valid for

| u | > 0

and

| \frac{3}{u} | < 1

\frac{1}{u - 3} = \frac{1}{u} + \frac{3}{u^{2}} + \frac{9}{u^{3}} + \dots

is valid for

| u | > 3

(4).

Step 8: Combine Expressions for $f (u)$

From (2), we have:

f (u) = \frac{1}{2} (\frac{1}{u} + \frac{1}{u^{2}} + \frac{1}{u^{3}} + \dots) - (\frac{1}{u} + \frac{2}{u^{2}} + \frac{4}{u^{3}} + \dots) + \frac{3}{2} (\frac{1}{u} + \frac{3}{u^{2}} + \frac{9}{u^{3}} + \dots)

This is valid for

| u | > 3

Step 9: Simplify the Expression

Simplifying the expression gives:

f (u) = \frac{1}{u} + \frac{3}{u^{2}} + \frac{11}{u^{3}} + \dots

This is valid for

| u | > 3

Step 10: Substitute Back for $z$

Therefore,

f (z) = \frac{1}{z + 1} + \frac{3}{(z + 1)^{2}} + \frac{11}{(z + 1)^{3}} + \dots

is valid for

| z + 1 | > 3

(a) मान लीजिए कि $f$ एक सर्वत्र वैश्लेषिक फलन है जिसके केन्द्र $z = 0$ पर टेलर श्रेणी प्रसार में अपरिमित रूप से अनेक पद हैं। दर्शाइए कि $f (\frac{1}{z})$ की $z = 0$ एक अनिवार्य विचित्रता है।

Let

f

be an entire function whose Taylor series expansion with centre

z = 0

has infinitely many terms. Show that

z = 0

is an essential singularity of (

f (\frac{1}{z}))

Answer:

Step 1: Problem Statement

Show that

z = 0

is an essential singularity of

(f (\frac{1}{z}))

, where

f

is an entire function with a Taylor series expansion centered at

z = 0

containing infinitely many terms.

Step 2: Introduction

Let

f (z)

be an entire function with a Taylor series expansion centered at

z = 0

that contains infinitely many terms.

Step 3: Example Function

Consider the example function

f (z) = e^{z}

, which is an entire function. Its Taylor series expansion centered at

z = 0

is given by:

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

Step 4: Transformation $f (\frac{1}{z})$

Now, let’s consider the function

f (\frac{1}{z})

f (\frac{1}{z}) = 1 + \frac{1}{z} + \frac{1}{2!} \frac{1}{z^{2}} + \frac{1}{3!} \frac{1}{z^{3}} + \dots

Step 5: Principal Part of Laurent Series

Notice that the Laurent series of

f (\frac{1}{z})

contains infinitely many terms in the principal part.

Step 6: Conclusion

Since the Laurent series expansion of

f (\frac{1}{z})

contains infinitely many terms in the principal part, it implies that

z = 0

is an essential singular point for

f (\frac{1}{z})

Therefore,

z = 0

is indeed an essential singularity of

(f (\frac{1}{z}))

(b) शर्तों

a x^{2} + b y^{2} + c z^{2} = 1

तथा

l x + m y + n z = 0

से प्रतिबन्धित

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

के स्तब्ध (अचर) मान निकालिए। परिणाम की ज्यामितीय व्याख्या कीजिए।

Find the stationary values of

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

subject to the conditions

a x^{2} + b y^{2} + c z^{2} = 1

and

l x + m y + n z = 0

. Interpret the result.

Answer:

Step 1: Introduction

Let’s define a function

F

as follows:

F = x^{2} + y^{2} + z^{2} + λ_{1} (a x^{2} + b y^{2} + c z^{2} - 1) + λ_{2} (l x + m y + n z) - - - - (1)

This function represents the problem statement and incorporates Lagrange multipliers

λ_{1}

and

λ_{2}

for the given constraints.

Step 2: Deriving Stationary Conditions

For the maxima and minima of

F

, we need to find the stationary points. We calculate the partial derivatives with respect to

x

y

, and

z

and set them to zero:

\begin{aligned} \frac{\partial F}{\partial x} = 2 x + 2 a x λ_{1} + l λ_{2} = 0 - - - - (2) \\ \frac{\partial F}{\partial y} = 2 y + 2 b y λ_{1} + m λ_{2} = 0 - - - - (3) \\ \frac{\partial F}{\partial z} = 2 z + 2 c z λ_{1} + n λ_{2} = 0 - - - - (4) \end{aligned}

Step 3: Combining Equations

Multiply equations (2), (3), and (4) by

x

y

, and

z

respectively and then add them together:

2 (x^{2} + y^{2} + z^{2}) + 2 λ_{1} (a x^{2} + b y^{2} + c z^{2}) + λ_{2} (l x + m y + n z) = 0

Step 4: Solving for $λ_{1}$

From the equation above, we can solve for

λ_{1}

\Rightarrow 2 u + 2 λ_{1} = 0 \Rightarrow λ_{1} = - u

Step 5: Expressing $x, y, z$ in Terms of $λ_{2}$

Using the values of

λ_{1}

obtained, we can express

x

y

, and

z

in terms of

λ_{2}

x = \frac{l λ_{2}}{2 (a u - 1)}, y = \frac{m λ_{2}}{2 (b u - 1)}, z = \frac{n λ_{2}}{2 (c u - 1)}

Step 6: Substituting into Constraint Equation

Substituting these expressions for

x, y, z

into the constraint equation

l x + m y + n z = 0

, we get:

l \cdot \frac{l λ_{2}}{2 (a u - 1)} + m \cdot \frac{m λ_{2}}{2 (b u - 1)} + n \cdot \frac{n λ_{2}}{2 (c u - 1)} = 0

\Rightarrow \frac{l^{2}}{a u - 1} + \frac{m^{2}}{b u - 1} + \frac{n^{2}}{c u - 1} = 0

Stationary Values:

The stationary values of

u

are the solutions to the above equation. These values represent the square of the distance from the origin to the points on the conicoid

a x^{2} + b y^{2} + c z^{2} = 1

that also lie on the plane

l x + m y + n z = 0

Geometrical Interpretation

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

is the square of the distance of any point

(x, y, z)

from the origin

(0, 0, 0)

. Hence in this problem, we have found the maximum and minimum values of the square of distance of the origin from the point of intersection of the central conicoid

a x^{2} + b y^{2} + c z^{2} = 1

by the central plane

l x + m y + n z = 0

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

बशर्ते कि

\begin{aligned} 3 x_{1} - x_{2} + 2 x_{3} & \leq 7 \\ 2 x_{1} - 4 x_{2} & \geq 12 \\ - 4 x_{1} + 3 x_{2} + 8 x_{3} & = 10 \end{aligned}

जहाँ

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

तथा

x_{3}

का चिह्न अप्रतिबंधित है।

Convert the following LPP into dual LPP :
Minimize

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

subject to

\begin{aligned} 3 x_{1} - x_{2} + 2 x_{3} & \leq 7 \\ 2 x_{1} - 4 x_{2} & \geq 12 \\ - 4 x_{1} + 3 x_{2} + 8 x_{3} & = 10 \end{aligned}

where

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

and

x_{3}

is unrestricted in sign.

Answer:

Step 1: Formulating the Primal Linear Programming Problem (LPP)

The primal LPP is defined as follows:
Minimize

Z_{x} = x_{1} - 3 x_{2} - 2 x_{3}

Subject to:

$3 x_{1} - x_{2} + 2 x_{3} \leq 7$
$2 x_{1} - 4 x_{2} \geq 12$
$- 4 x_{1} + 3 x_{2} + 8 x_{3} = 10$
$x_{1}, x_{2} \geq 0$
$x_{3}$ is unrestricted in sign.

Step 2: Converting Primal Constraints

In this step, the answer converts the primal constraints from

\leq

type to

\geq

type because the primal objective function is minimizing. This leads to the following primal LPP:

Minimize

Z_{x} = x_{1} - 3 x_{2} - 2 x_{3}

Subject to:

$- 3 x_{1} + x_{2} - 2 x_{3} \geq - 7$
$2 x_{1} - 4 x_{2} \geq 12$
$- 4 x_{1} + 3 x_{2} + 8 x_{3} = 10$
$x_{1}, x_{2} \geq 0$
$x_{3}$ is unrestricted in sign.

Step 3: Identifying Dual Problem Characteristics

Here, the answer identifies the characteristics of the dual problem based on the properties of the primal problem:

There must be 3 constraints and 3 variables in the dual problem.
The coefficients of the objective function in the primal ( $c_{1} = 1, c_{2} = - 3, c_{3} = - 2$ ) become the right-hand side constants in the dual.
The right-hand side constants in the primal ( $b_{1} = - 7, b_{2} = 12, b_{3} = 10$ ) become the coefficients of the objective function in the dual.
Since the primal is a minimization problem, the dual must be a maximization problem.
The presence of an unrestricted variable ( $x_{3}$ ) in the primal results in an equality constraint in the dual, and the corresponding dual variable ( $y_{3}$ ) is unrestricted.

Step 4: Formulating the Dual Linear Programming Problem (LPP)

Based on the characteristics identified in Step 3, the dual LPP is formulated as follows:

Maximize

Z_{y} = - 7 y_{1} + 12 y_{2} + 10 y_{3}

Subject to:

$- 3 y_{1} + 2 y_{2} - 4 y_{3} \leq 1$
$y_{1} - 4 y_{2} + 3 y_{3} \leq - 3$
$- 2 y_{1} + 8 y_{3} = - 2$
$y_{1}, y_{2} \geq 0$
$y_{3}$ is unrestricted in sign.

(a) दर्शाइए कि परिमेय संख्याओं के योज्य समूह $Q$ के अपरिमित रूप से अनेक उपसमूह हैं।

Show that there are infinitely many subgroups of the additive group

Q

of rational numbers.

Answer:

The additive group of rational numbers, denoted by

Q

, consists of all rational numbers under the operation of addition. To show that there are infinitely many subgroups of

Q

, we can construct an infinite family of subgroups.

Construction of Subgroups:

Consider the set

n Q

, where

n

is any positive integer. This set consists of all rational numbers that can be written in the form

n q

, where

q

is any rational number. Formally,

n Q = {n q ∣ q \in Q}

Properties:

Identity Element: $0$ is in $n Q$ because $0 = n \times 0$ .
Closure under Addition: If $a, b \in n Q$ , then $a + b = n q_{1} + n q_{2} = n (q_{1} + q_{2})$ is also in $n Q$ .
Closure under Inverse: If $a \in n Q$ , then $- a = - n q = n (- q)$ is also in $n Q$ .

Therefore,

n Q

is a subgroup of

Q

for each positive integer

n

Infinitude of Subgroups:

Since

n

can be any positive integer, and there are infinitely many positive integers, we have constructed an infinite family of subgroups

n Q

Q

Conclusion:

Hence, we have shown that there are infinitely many subgroups of the additive group

Q

of rational numbers.

(b) कन्टूर समाकलन का उपयोग कर समाकल

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

का मान ज्ञात कीजिए।

Using contour integration, evaluate the integral

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

Answer:

Introduction:

We aim to evaluate the integral

\int_{- \infty}^{\infty} \frac{\sin x d x}{x (x^{2} + a^{2})}

using contour integration techniques. Here,

a

is a positive constant.

Contour Integration Approach:

Consider the integral

\int_{- \infty}^{\infty} e^{i z} \frac{d z}{z (z^{2} + a^{2})}

taken around a contour

C

consisting of the following segments:

i. The real axis from point

P

R

.
ii. A large semi-circle

T

in the upper half plane defined by

| z | = R

.
iii. The real axis from

- R

- P

.
iv. A small semi-circle

γ

defined by

| z | = P

Now, the function

f (z)

has simple poles at

z = 0, \pm a i

, but only

z = a i

lies within the contour

C

The residue of

f (z)

z = a i

is given by

lim_{z \to a i} (z - a i) f (z) = lim_{z \to a i} \frac{e^{i z}}{z (z + a i)} = \frac{e^{- a}}{- 2 a^{2}}

By the residue theorem, we have

\begin{aligned} \int_{C} f (z) d z = \int_{μ}^{R} f (z) d z + \int_{T} f (z) d z + \int_{- R}^{- P} f (z) d z + \int_{γ} f (z) d z \\ = 2 π i \cdot \frac{e^{- a}}{- 2 a^{2}} = - \frac{π i}{a^{2}} e^{- a} (1) \end{aligned}

Using Jordan’s lemma, we can show that

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

Also, since

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

, we have

\int_{γ} f (z) d z = \frac{i}{a^{2}} (0 - π) = - \frac{π i}{a^{2}}

P

approaches 0 and

R

approaches infinity, we can derive from equation (1):

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

Conclusion:

Equating the imaginary parts, we obtain the final result:

\int_{- \infty}^{\infty} \frac{\sin x d x}{x (x^{2} + a^{2})} = \frac{π}{a^{2}} (1 - e^{- a})

This is the value of the given integral using contour integration.

M

(बिग

M

) विधि का उपयोग करके निम्नलिखित रैखिक प्रोग्रामन समस्या को हल कीजिए :
अधिकतमीकरण कीजिए

Z = 4 x_{1} + 5 x_{2} + 2 x_{3}

बशर्ते कि

\begin{aligned} 2 x_{1} + x_{2} + x_{3} & \geq 10 \\ x_{1} + 3 x_{2} + x_{3} & \leq 12 \\ x_{1} + x_{2} + x_{3} & = 6 \\ x_{1}, x_{2}, x_{3} & \geq 0 \end{aligned}

Solve the following linear programming problem using Big M method :
Maximize

Z = 4 x_{1} + 5 x_{2} + 2 x_{3}

subject to

\begin{aligned} 2 x_{1} + x_{2} + x_{3} & \geq 10 \\ x_{1} + 3 x_{2} + x_{3} & \leq 12 \\ x_{1} + x_{2} + x_{3} & = 6 \\ x_{1}, x_{2}, x_{3} & \geq 0 \end{aligned}

Answer:

Problem is

\begin{aligned} Max Z = 4 x_{1} + 5 x_{2} + 2 x_{3} \\ subject to \\ \begin{array}{c} 2 x_{1} + x_{2} + x_{3} \geq 10 \\ x_{1} + 3 x_{2} + x_{3} \leq 12 \\ x_{1} + x_{2} + x_{3} = 6 \end{array} \\ and x_{1}, x_{2}, x_{3} \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 ‘ $\leq$ ‘ we should add slack variable $S_{2}$
As the constraint- 3 is of type ‘ $=$ ‘ we should add artificial variable $A_{2}$

After introducing slack,surplus, artificial variables

Max Z = 4 x_{1} + 5 x_{2} + 2 x_{3} + 0 S_{1} + 0 S_{2} - M A_{1} - M A_{2}

subject to

\begin{aligned} 2 x_{1} + x_{2} + x_{3} - S_{1} + A_{1} & = 10 \\ x_{1} + 3 x_{2} + x_{3} + S_{2} & = 12 \\ x_{1} + x_{2} + x_{3} & + A_{2} = 6 \\ and x_{1}, x_{2}, x_{3}, S_{1}, S_{2}, A_{1}, A_{2} \geq 0 \end{aligned}

Iteration-1		$C_{j}$	4	5	2	0	0	$- M$	$- M$
$B$	$C_{B}$	$X_{B}$	$x_{1}$	$x_{2}$	$x_{3}$	$s_{1}$	$S_{2}$	$A_{1}$	$A_{2}$	$\begin{matrix} MinRatio \\ \frac{X_{B}}{x_{1}} \end{matrix}$
$A_{1}$	$- M$	10	(2)	1	1	-1	0	1	0	$\frac{10}{2} = 5 \to$
$S_{2}$	0	12	1	3	1	0	1	0	0	$\frac{12}{1} = 12$
$A_{2}$	$- M$	6	1	1	1	0	0	0	1	$\frac{6}{1} = 6$
$Z = - 16 M$		$Z_{j}$	$- 3 M$	$- 2 M$	$- 2 M$	$M$	0	$- M$	$- M$
$Z = - 16 M$		$C_{j} - Z_{j}$	$3 M + 4 ↑$	$2 M + 5$	$2 M + 2$	$- M$	0	0	0

Positive maximum

C_{j} - Z_{j}

3 M + 4

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

x_{1}

.
Minimum ratio is 5 and its row index is 1 . So, the leaving basis variable is

A_{1}

∴

The pivot element is 2 .
Entering

= x_{1}

, Departing

= A_{1}

, Key Element

= 2

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

Iteration-2		$C_{j}$	4	5	2	0	0	$- M$
$B$	$C_{B}$	$X_{B}$	$x_{1}$	$x_{2}$	$x_{3}$	$s_{1}$	$S_{2}$	$A_{2}$	$\begin{matrix} MinRatio \\ \frac{X_{B}}{x_{2}} \end{matrix}$
$x_{1}$	4	5	1	0.5	0.5	-0.5	0	0	$\frac{5}{0.5} = 10$
$S_{2}$	0	7	0	2.5	0.5	0.5	1	0	$\frac{7}{2.5} = 2.8$
$A_{2}$	$- M$	1	0	$(0.5)$	0.5	0.5	0	1	$\frac{1}{0.5} = 2 \to$
$Z = - M + 20$		$Z_{j}$	4	$- 0.5 M + 2$	$- 0.5 M + 2$	$- 0.5 M - 2$	0	$- M$
$Z = - M + 20$		$C_{j} - Z_{j}$	0	$0.5 M + 3 ↑$	$0.5 M$	$0.5 M + 2$	0	0

Positive maximum

C_{j} - Z_{j}

0.5 M + 3

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

x_{2}

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

A_{2}

∴

The pivot element is 0.5 .
Entering

= x_{2}

, Departing

= A_{2}

, Key Element

= 0.5

R_{3} (

new

) = R_{3}

(old)

\div 0.5

R_{1} (

new

) = R_{1} (

old

) - 0.5 R_{3}

(new)

R_{2} (

new

) = R_{2} (

old

) - 2.5 R_{3} (

new

)

Iteration-3		$C_{j}$	4	5	2	0	0
$B$	$C_{B}$	$X_{B}$	$x_{1}$	$x_{2}$	$x_{3}$	$S_{1}$	$S_{2}$	MinRatio
$x_{1}$	4	4	1	0	0	-1	0
$S_{2}$	0	2	0	0	-2	-2	1
$x_{2}$	5	2	0	1	1	1	0
$Z = 26$		$Z_{j}$	$4$	$5$	$5$	$1$	$0$
		$C_{j} - Z_{j}$	0	0	-3	-1	0

Since all

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

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

x_{1} = 4, x_{2} = 2, x_{3} = 0

Max Z = 26

खण्ड-B / SECTION-B

(a) समीकरण $f (x + y + z, x^{2} + y^{2} + z^{2}) = 0$ से स्वैच्छिक फलन $f$ का विलोपन कर आंशिक अवकल समीकरण को प्राप्त कीजिए।

Obtain the partial differential equation by eliminating arbitrary function

f

from the equation

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

Answer:

Introduction:

The objective is to obtain a partial differential equation (PDE) by eliminating the arbitrary function

f

from the given equation:

f (x + y + z, x^{2} + y^{2} + z^{2}) = 0 \to (1)

Elimination of Arbitrary Function:

Let’s proceed to eliminate

f

by introducing new variables

u

and

v

as follows:

u = x + y + z, v = x^{2} + y^{2} + z^{2} \to (2)

The equation (1) can now be expressed as:

ϕ (u, v) = 0 \to (3)

Partial Differentiation:

Now, we will differentiate equation (3) with respect to

x

partially:

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

From the relationships in equation (2), we have:

\frac{\partial u}{\partial x} = 1, \frac{\partial u}{\partial z} = 1, \frac{\partial v}{\partial x} = 2 x, \frac{\partial v}{\partial z} = 2 z, \frac{\partial u}{\partial y} = 1, \frac{\partial v}{\partial y} = 2 y \to (5)

Using equations (4) and (5), we can simplify as follows:

\begin{aligned} \frac{\partial ϕ}{\partial u} (1 + p) + 2 \frac{\partial ϕ}{\partial v} (x + p z) = 0 \\ \Rightarrow \frac{\frac{\partial ϕ}{\partial u}}{\frac{\partial ϕ}{\partial v}} = - \frac{2 (x + p z)}{1 + p} \to (6) \end{aligned}

Partial Differentiation (w.r.t. $y$ ):

Next, we differentiate equation (3) partially with respect to

y

\begin{aligned} \frac{\partial ϕ}{\partial u} (\frac{\partial u}{\partial y} + q \frac{\partial u}{\partial z}) + \frac{\partial ϕ}{\partial v} (\frac{\partial v}{\partial y} + q \frac{\partial v}{\partial z}) = 0 \\ \Rightarrow \frac{\partial ϕ}{\partial u} (1 + q) + 2 \frac{\partial ϕ}{\partial v} (y + z q) = 0 \\ \Rightarrow \frac{\frac{\partial ϕ}{\partial u}}{\frac{\partial ϕ}{\partial v}} = - \frac{2 (y + z q)}{1 + q} \to (7) \end{aligned}

Eliminating $ϕ$ :

Now, by eliminating

ϕ

from equations (6) and (7), we obtain:

\begin{aligned} \frac{x + p z}{1 + p} = \frac{y + q z}{1 + q} \\ \Rightarrow (1 + q) (x + p z) = (1 + p) (y + q z) \\ \Rightarrow x + p z + q x + p q z = y + q z + p y + p q z \\ \Rightarrow x - y = p (y - z) + q (z - x) \\ \Rightarrow (y - z) p + (z - x) q = x - y \end{aligned}

This final expression represents the desired partial differential equation (PDE) of the first order.

Conclusion:

The elimination of the arbitrary function

f

from the given equation results in the first-order PDE:

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

This is the partial differential equation obtained by eliminating

f

from the initial equation.

(b) प्रारंभिक मानों

0, \frac{π}{2}

का उपयोग करके एक संख्यात्मक तकनीक के द्वारा समीकरण

3 x = 1 + \cos x

का एक धनात्मक मूल ज्ञात कीजिए, तथा न्यूटन-राफ्सन विधि के द्वारा परिणाम को 8 सार्थक अंकों तक और शुद्ध मान के निकट लाइए।

Find a positive root of the equation

3 x = 1 + \cos x

by a numerical technique using initial values

0, \frac{π}{2}

; and further improve the result using Newton-Raphson method correct to 8 significant figures.

Answer:

Introduction:

The goal is to find a positive root of the equation

3 x = 1 + \cos (x)

using a numerical technique. Initially, we use the interval

[0, \frac{π}{2}]

as our guess range, and then we further improve the result using the Newton-Raphson method with an accuracy of 8 significant figures.

Finding Initial Interval:

Given the equation

3 x = 1 + \cos (x)

, we can rearrange it as:

3 x - \cos (x) - 1 = 0

Now, let

f (x) = 3 x - \cos (x) - 1

We also find the derivative of

f (x)

f^{'} (x) = 3 + \sin (x)

Now, we evaluate

f (x)

and

f^{'} (x)

at two points:

x = 0

and

x = 1

$x$	0	1
$f (x)$	-2	1.45969769

From the values, we see that

f (0) < 0

and

f (1) > 0

. Hence, the root lies between 0 and 1.

Let’s start with an initial guess,

x_{0} = \frac{0 + 1}{2} = 0.5

Newton-Raphson Method:

1st iteration:

Evaluate

f (x_{0})

and

f^{'} (x_{0})

\begin{aligned} f (x_{0}) = f (0.5) = 3 \cdot 0.5 - \cos (0.5) - 1 = - 0.37758256 \\ f^{'} (x_{0}) = f^{'} (0.5) = 3 + \sin (0.5) = 3.47942554 \end{aligned}

Now, update

x_{1}

x_{1} = x_{0} - \frac{f (x_{0})}{f^{'} (x_{0})} = 0.5 - \frac{- 0.37758256}{3.47942554} = 0.60851865

2nd iteration:

Evaluate

f (x_{1})

and

f^{'} (x_{1})

\begin{aligned} f (x_{1}) = f (0.60851865) = 3 \cdot 0.60851865 - \cos (0.60851865) - 1 = 0.00506021 \\ f^{'} (x_{1}) = f^{'} (0.60851865) = 3 + \sin (0.60851865) = 3.57165265 \end{aligned}

Now, update

x_{2}

x_{2} = x_{1} - \frac{f (x_{1})}{f^{'} (x_{1})} = 0.60851865 - \frac{0.00506021}{3.57165265} = 0.60710188

3rd iteration:

Evaluate

f (x_{2})

and

f^{'} (x_{2})

\begin{aligned} f (x_{2}) = f (0.60710188) = 3 \cdot 0.60710188 - \cos (0.60710188) - 1 = 0.00000082 \\ f^{'} (x_{2}) = f^{'} (0.60710188) = 3 + \sin (0.60710188) = 3.57048962 \end{aligned}

Now, update

x_{3}

x_{3} = x_{2} - \frac{f (x_{2})}{f^{'} (x_{2})} = 0.60710188 - \frac{0.00000082}{3.57048962} = 0.60710165

4th iteration:

Evaluate

f (x_{3})

and

f^{'} (x_{3})

\begin{aligned} f (x_{3}) = f (0.60710165) = 3 \cdot 0.60710165 - \cos (0.60710165) - 1 = 0 \\ f^{'} (x_{3}) = f^{'} (0.60710165) = 3 + \sin (0.60710165) = 3.57048943 \end{aligned}

Now, update

x_{4}

x_{4} = x_{3} - \frac{f (x_{3})}{f^{'} (x_{3})} = 0.60710165 - \frac{0}{3.57048943} = 0.60710165

\begin{array}{cccccc} n & x_{0} & f (x_{0}) & f^{'} (x_{0}) & x_{1} & Update \\ 1 & 0.5 & - 0.37758256 & 3.47942554 & 0.60851865 & x_{0} = x_{1} \\ 2 & 0.60851865 & 0.00506021 & 3.57165265 & 0.60710188 & x_{0} = x_{1} \\ 3 & 0.60710188 & 0.00000082 & 3.57048962 & 0.60710165 & x_{0} = x_{1} \\ 4 & 0.60710165 & 0 & 3.57048943 & 0.60710165 & x_{0} = x_{1} \end{array}

Conclusion:

The approximate root of the equation

3 x - \cos (x) - 1 = 0

obtained using the Newton-Raphson method after 4 iterations is

x \approx 0.60710165

(correct to 8 significant figures).

(3798 \cdot 3875)_{10}

को अष्टधारी तथा षोडशाधारी तुल्यमानों में बदलिए।
(ii) (\rceil

P \to R) \land (Q ⇄ P)

का मुख्य संयोजक सामान्य रूप (प्रिंसिपल कंजंक्टिव नॉर्मल फॉर्म) प्राप्त कीजिए।

(i) Convert

(3798 \cdot 3875)_{10}

into octal and hexadecimal equivalents.

Answer:

Introduction:

We are tasked with converting the decimal number

(3798 \cdot 3875)_{10}

into its octal and hexadecimal equivalents.

Conversion to Octal:

To convert

(3798)_{10}

into octal, we perform successive divisions by 8, recording remainders at each step until the quotient becomes zero.

Divide 3798 by 8: Quotient = 474, Remainder = 6.
Divide 474 by 8: Quotient = 59, Remainder = 2.
Divide 59 by 8: Quotient = 7, Remainder = 3.
Divide 7 by 8: Quotient = 0, Remainder = 7.

The remainders, when read from bottom to top, give us the octal equivalent:

(3798)_{10} = (\underset{―}{7326})_{8}

Conversion of Decimal Fraction to Octal Fraction:

To convert the decimal fraction

(0.3875)_{10}

into its octal equivalent, we can multiply it by 8 successively, noting the integer parts.

$0.3875 \times 8 = 3.1000$ (integer part = 3).
$0.1000 \times 8 = 0.8000$ (integer part = 0).
$0.8000 \times 8 = 6.4000$ (integer part = 6).
$0.4000 \times 8 = 3.2000$ (integer part = 3).
$0.2000 \times 8 = 1.6000$ (integer part = 1).
$0.6000 \times 8 = 4.8000$ (integer part = 4).

Putting the integer parts together, we get the octal fraction:

(0.3875)_{10} = (. \underset{―}{306314})_{8}

Octal Representation:

Therefore,

(3798.3875)_{10} = (\underset{―}{7326.306314})_{8}

Conversion to Hexadecimal:

To convert

(3798)_{10}

into hexadecimal, we perform successive divisions by 16, recording remainders at each step until the quotient becomes zero.

Divide 3798 by 16: Quotient = 237, Remainder = 6.
Divide 237 by 16: Quotient = 14, Remainder = D (in hexadecimal).
Divide 14 by 16: Quotient = 0, Remainder = E (in hexadecimal).

The remainders, when read from bottom to top, give us the hexadecimal equivalent:

(3798)_{10} = (\underset{―}{E D 6})_{16}

Conversion of Decimal Fraction to Hexadecimal Fraction:

To convert the decimal fraction

(0.3875)_{10}

into its hexadecimal equivalent, we can multiply it by 16 successively, noting the integer parts.

$0.3875 \times 16 = 6.2000$ (integer part = 6).
$0.2000 \times 16 = 3.2000$ (integer part = 3).
$0.2000 \times 16 = 3.2000$ (integer part = 3).
$0.2000 \times 16 = 3.2000$ (integer part = 3).
$0.2000 \times 16 = 3.2000$ (integer part = 3).
$0.2000 \times 16 = 3.2000$ (integer part = 3).

Putting the integer parts together, we get the hexadecimal fraction:

(0.3875)_{10} = (. \underset{―}{633333})_{16}

Hexadecimal Representation:

Therefore,

(3798.3875)_{10} = (\underset{―}{E D 6.633333})_{16}

Conclusion:

The octal equivalent of

(3798 \cdot 3875)_{10}

(7326.306314)_{8}

, and the hexadecimal equivalent is

(E D 6.633333)_{16}

(ii) Obtain the principal conjunctive normal form of

(ד P \to R) \land (Q ⇋ P = X^{Y})

Answer:

Introduction:

We are tasked with obtaining the principal conjunctive normal form (PCNF) of the given logical expression:

(\neg P \to R) \land (Q ⇋ P = X^{Y})

Conversion Steps:

Transform the Expression:

Consider the expression $(\neg P \to R) \land (Q ⇋ P = X^{Y})$ and convert it to its equivalent form with negations and biconditionals expanded.
Create the Truth Table:

Build a truth table for the given expression with all relevant variables (P, Q, R, X, Y) and their respective negations.
Calculate Intermediate Results:

Calculate intermediate results for the implications and biconditionals in the truth table.
Identify False Rows:

Identify the rows in the truth table where the final result is false. These rows represent cases where the original expression is false.
Apply PCNF:

Use the rows identified in step 4 to construct the PCNF by combining the variables and their negations using disjunction (OR) to ensure that each row with a false result is included.

Truth Table:

Here’s the truth table for the given expression:

$P$	$Q$	$R$	$\sim P$	$\sim P \to R$	$Q ⇋ P$	$X \land Y$
$T$	$T$	$T$	$F$	$T$	$T$	$T$
$T$	$T$	$F$	$F$	$T$	$T$	$T$
$T$	$F$	$T$	$F$	$T$	$F$	$F$
$T$	$F$	$F$	$F$	$T$	$F$	$F$
$F$	$T$	$T$	$T$	$T$	$F$	$F$
$F$	$T$	$F$	$T$	$F$	$F$	$F$
$F$	$F$	$T$	$T$	$T$	$T$	$T$
$F$	$F$	$F$	$T$	$F$	$T$	$F$

Identify False Rows:

The rows with a false result are:

Row 6
Row 8

Apply PCNF:

Using the false rows identified above, we construct the PCNF by combining the variables and their negations using disjunction (OR):

(\sim P \lor Q \lor \sim R) \land (\sim P \lor \sim Q \lor R) \land (P \lor \sim Q \lor \sim R) \land (P \lor \sim Q \lor R) \land (P \lor Q \lor R)

Conclusion:

The principal conjunctive normal form (PCNF) of the given expression

(\neg P \to R) \land (Q ⇋ P = X^{Y})

is:

(\sim P \lor Q \lor \sim R) \land (\sim P \lor \sim Q \lor R) \land (P \lor \sim Q \lor \sim R) \land (P \lor \sim Q \lor R) \land (P \lor Q \lor R)

(d) ऊर्ध्वर्धर

x y

-तल में स्थित एक वृत्त के अनुदिश एक कण गति के लिए व्यवरुद्ध है। डी’एलम्बर्ट के नियम की सहायता से दर्शाइए कि इसकी गति का समीकरण

\ddot{x} y - \ddot{y} x - g x = 0

है, जहाँ

g

गुरुत्वीय त्वरण है।

A particle is constrained to move along a circle lying in the vertical

x y

-plane. With the help of the D’Alembert’s principle, show that its equation of motion is

\ddot{x} y - \ddot{y} x - g x = 0

, where

g

is the acceleration due to gravity.

Answer:

Introduction:

In this problem, we are considering a particle constrained to move along a circle lying in the vertical XY-plane. We need to show, with the help of D’Alembert’s principle, that its equation of motion is

\ddot{x} y - \ddot{y} x - g x = 0

, where

g

is the acceleration due to gravity.

Solution Steps:

Particle on a Circular Path:

Consider a particle of mass $m$ moving along a circle of radius $r$ in the XY-plane. Let $(x, y)$ represent the position of the particle at any instant $t$ with respect to the fixed point $O$ . The constraint on the motion of the particle is that its position coordinates always lie on the circle. Hence, the equation of the constraint is given by:

$x^{2} + y^{2} = r^{2} (1)$
Deriving Displacement Relations:

Differentiate equation (1) to obtain:

$2 x d x + 2 y d y = 0 (2)$

From equation (2), we derive the relation $δ x = - \frac{y}{x} δ y$ , where $δ x$ and $δ y$ are displacements in $x$ and $y$ respectively.
D’Alembert’s Principle:

Using D’Alembert’s principle, we have:

$(F - m \ddot{r}) δ r = 0$

In component form, this equation is:

$(F_{x} - m \ddot{x}) δ x + (F_{y} - m \ddot{y}) δ y = 0 (3)$
Force Analysis:

The only force acting on the particle at any instant $t$ is its weight $mg$ in the downward direction. We can resolve this force into horizontal and vertical components:
- $F_{x} = 0$ (horizontal component)
- $F_{y} = - m g$ (vertical component)
Substitute Forces into Equation (3):

Substituting the expressions for $F_{x}$ and $F_{y}$ into equation (3), we get:

$- m \ddot{x} δ x - (m g + m \ddot{y}) δ y = 0$
Substitute Displacement Relation (2):

Substituting the displacement relation $δ x = - \frac{y}{x} δ y$ from step 2 into the equation, we have:

$m (- \ddot{x} y + \ddot{y} x + g x) δ x = 0$
Final Equation of Motion:

For $δ x \neq 0$ and $m \neq 0$ , the equation simplifies to:

$\ddot{x} y - \ddot{y} x - g x = 0 (4)$

Conclusion:

The equation of motion for the particle constrained to move along a circular path in the vertical XY-plane is

\ddot{x} y - \ddot{y} x - g x = 0

, where

g

represents the acceleration due to gravity.

(e) उद्गमों (स्रोतों) व अभिगमों (सिंकों) के किस विन्यास से वेग विभव

w = \log_{e} (z - \frac{a^{2}}{z})

हो सकता है? संगत धारा-रेखाओं का ख़ाका खींचिए और सिद्ध कीजिए कि उनमें से दो, वृत्त

r = a

तथा

y

-अक्ष में प्रविभाजित होती हैं।

What arrangements of sources and sinks can have the velocity potential

w = \log_{e} (z - \frac{a^{2}}{z})

? Draw the corresponding sketch of the streamlines and prove that two of them subdivide into the circle

r = a

and the axis of

y

Answer:

Introduction

The problem asks us to determine the arrangements of sources and sinks that can produce the given velocity potential function

w = \log_{e} (z - \frac{a^{2}}{z})

. Additionally, we are tasked with drawing the corresponding streamlines and proving that two of these streamlines subdivide into a circle with radius

r = a

and the

y

-axis.

Analysis and Solution

Complex Potential Decomposition

The complex potential is given by the equation:

w = \log_{e} (z - \frac{a^{2}}{z}) = \log_{e} {\frac{(z - a) (z + a)}{z}}

We can decompose this expression as follows:

w = \log_{e} (z - a) + \log_{e} (z + a) - \log_{e} (z) - - - - (1)

This decomposition reveals that there are two sinks of unit strength located at distances

z = a

and

z = - a

, and a source of unit strength at the origin.

Complex Potential in Terms of Real and Imaginary Parts

We can express equation (1) in terms of the real and imaginary parts

ϕ

and

Ψ

as follows:

ϕ + i Ψ = \log {(x - a) + i y} + \log {(x + a) + i y} - \log (x + i y)

Equating the imaginary parts, we obtain:

Ψ = \tan^{- 1} \frac{y}{x - a} + \tan^{- 1} \frac{y}{x + a} - \tan^{- 1} \frac{y}{x}

Simplifying further:

Ψ = \tan^{- 1} (\frac{2 x y}{x^{2} - y^{2} - a^{2}}) - \tan^{- 1} \frac{y}{x}

And finally:

Ψ = \tan^{- 1} (\frac{\frac{2 x y}{x^{2} - y^{2} - a^{2}} - \frac{y}{x}}{1 + \frac{2 x y}{x^{2} - y^{2} - a^{1}} \frac{y}{x}})

Streamlines

The streamlines are given by

Ψ =

constant. Therefore, we have the equation:

\frac{y (x^{2} + y^{2} + a^{2})}{x (x^{2} + y^{2} - a^{2})} = k - - - - (2)

Case 1: When $k$ is infinite

If the constant

k

is infinite, then:

\frac{y (x^{2} + y^{2} + a^{2})}{x (x^{2} + y^{2} - a^{2})} = \infty

This implies:

x (x^{2} + y^{2} - a^{2}) = 0 \Rightarrow x = 0 and x^{2} + y^{2} = a^{2}

Hence,

x = 0

shows that the

y

-axis is a streamline, and the equation

x^{2} + y^{2} = a^{2}

represents a circle with radius

r = a

centered at the origin.

Case 2: When $k$ is zero

If the constant

k

is zero, then

y = 0

, which implies that the axis of

x

is a streamline.

Therefore, the rough sketch of the lines consists of a source of unit strength at the origin

O (0, 0)

and two sinks of unit strength at

A (a, 0)

and

B (- a, 0

Sketch

In the sketch, we would draw:

A source at $z = a$ and a sink at $z = - a$ .
A circle of radius $a$ centered at the origin, representing the streamline $r = a$ .
The $y$ -axis, representing another streamline.

The fluid would flow from the source to the sink, and the circle

r = a

and the

y

-axis would act as barriers that the fluid cannot cross.

Conclusion

In conclusion, we have determined the arrangements of sources and sinks that produce the given velocity potential function and shown that two of the resulting streamlines form a circle with radius

r = a

and the

y

-axis. This analysis provides insight into the fluid flow behavior in this configuration.

(a) तरंग समीकरण

a^{2} \frac{\partial^{2} u}{\partial x^{2}} = \frac{\partial^{2} u}{\partial t^{2}}, 0 < x < L, t > 0

का शर्तों

\begin{matrix} u (0, t) = 0, u (L, t) = 0 \\ u (x, 0) = \frac{1}{4} x (L - x), {\frac{\partial u}{\partial t} |}_{t = 0} = 0 \end{matrix}

से प्रतिबन्धित हल ज्ञात कीजिए।

Solve the wave equation

a^{2} \frac{\partial^{2} u}{\partial x^{2}} = \frac{\partial^{2} u}{\partial t^{2}}, 0 < x < L, t > 0

subject to the conditions

\begin{matrix} u (0, t) = 0, u (L, t) = 0 \\ u (x, 0) = \frac{1}{4} x (L - x), {\frac{\partial u}{\partial t} |}_{t = 0} = 0 \end{matrix}

Answer:

Introduction

In this problem, we are tasked with solving the wave equation with specific boundary and initial conditions. The wave equation is given by:

a^{2} \frac{\partial^{2} u}{\partial x^{2}} = \frac{\partial^{2} u}{\partial t^{2}}, 0 < x < L, t > 0

Subject to the following conditions:

u (0, t) = 0, u (L, t) = 0, t > 0

u (x, 0) = \frac{1}{4} x (L - x), {\frac{\partial u}{\partial t} |}_{t = 0} = 0, 0 < x < L

We will solve this equation step by step, starting with the separation of variables method.

Step 1: Separation of Variables

We assume a separable solution of the form:

u (x, t) = X (x) \cdot T (t)

Substituting this into the wave equation (1) gives:

a^{2} X (x) T (t) = X (x) T^{″} (t)

This leads to two separate ordinary differential equations:

\frac{X^{″} (x)}{X (x)} = \frac{T^{″} (t)}{a^{2} T (t)} = - λ

Step 2: Solve for X(x)

For the spatial part, we have the equation:

X^{″} (x) + λ X (x) = 0

The solutions to this equation depend on the value of

λ

Case 1: $λ = 0$

For

λ = 0

, the equation becomes:

X^{″} (x) = 0

The general solution is

X (x) = c_{1} x + c_{2}

Case 2: $λ = - α^{2} (α > 0)$

For

λ = - α^{2}

, the equation becomes:

X^{″} (x) - α^{2} X (x) = 0

The general solution is

X (x) = C_{1} \cosh (α x) + C_{2} \sinh (α x)

Step 3: Solve for T(t)

For the temporal part, we have the equation:

T^{″} (t) + λ a^{2} T (t) = 0

Case 1: $λ = 0$

For

λ = 0

, the equation becomes:

T^{″} (t) = 0

The general solution is

T (t) = C_{3} t + C_{4}

Case 2: $λ = - α^{2} (α > 0)$

For

λ = - α^{2}

, the equation becomes:

T^{″} (t) + α^{2} a^{2} T (t) = 0

The general solution is

T (t) = C_{3} \cosh (α a t) + C_{4} \sinh (α a t)

Step 4: Combine X(x) and T(t)

The general solution for

u (x, t)

is a product of the solutions for X(x) and T(t) obtained in steps 2 and 3:

u (x, t) = X (x) \cdot T (t)

For

λ = 0

, the solution is:

u (x, t) = (c_{1} x + c_{2}) (C_{3} t + C_{4})

However, this leads to the trivial solution

u = 0

due to the boundary conditions.

For

λ = - α^{2}

, the solution is:

u (x, t) = (C_{1} \cosh (α x) + C_{2} \sinh (α x)) (C_{3} \cosh (α a t) + C_{4} \sinh (α a t))

Step 5: Apply Boundary Conditions

We apply the boundary conditions

u (0, t) = 0

and

u (L, t) = 0

to determine the constants

C_{1}

and

α

u (0, t) = 0 ⟹ C_{1} + 0 = 0 ⟹ C_{1} = 0

u (L, t) = 0 ⟹ C_{2} \sinh (α L) (C_{3} \cosh (α a t) + C_{4} \sinh (α a t)) = 0

For this equation to hold for all

t

, we must have

C_{2} = 0

\sinh (α L) = 0

. Since

\sinh (α L) = 0

implies

α = n π / L

for

n

being a positive integer, we have:

α = \frac{n π}{L}

Step 6: Apply Superposition Principle

The solution can now be written as a sum of solutions for different values of

n

using the superposition principle:

u (x, t) = \sum_{n = 1}^{\infty} u_{n}

u (x, t) = \sum_{n = 1}^{\infty} (A_{n} \cos (\frac{a π n}{L} t) + B_{n} \sin (\frac{a π n}{L} t)) \sin (\frac{n π}{L} x)

Step 7: Satisfy Initial Condition

To satisfy the initial condition

u (x, 0) = \frac{1}{4} x (L - x)

, we compare it with the series representation:

u (x, 0) = \sum_{n = 1}^{\infty} A_{n} \sin (\frac{n π}{L} x)

This implies that

A_{n}

can be evaluated using the integral:

A_{n} = \frac{2}{L} \int_{0}^{L} \frac{1}{4} x (L - x) \sin (\frac{n π}{L} x) d x

\begin{aligned} A_{n} = \frac{2}{L} \int_{0}^{L} \frac{1}{4} x (L - x) \cdot \sin \frac{n π}{L} x d x \\ = \frac{1}{2 L} \int_{0}^{L} L x \sin (\frac{n π}{L} x) d x - \frac{1}{2 L} \int_{0}^{L} x^{2} \sin (\frac{n π}{L} x) d x \\ = \frac{1}{2 L} {[- \frac{L^{2} x}{n π} \cos \frac{n π}{L} x + \frac{L^{3}}{n^{2} π^{2}} \sin \frac{n π}{L} x]}_{0}^{L} + \frac{1}{2 L} {[\frac{x^{2} L}{n π} \cos \frac{n π}{L} x]}_{0}^{L} - \frac{2 L}{n π} \int_{0}^{L} x \cos \frac{n π}{L} x d x \\ = - \frac{L^{2}}{2 n π} \cos n π + \frac{L^{2}}{2 n π} \cos n π - \frac{1}{n π} {[\frac{x L}{n π} \sin \frac{n π}{L} x + \frac{L^{2}}{n^{2} π^{2}} \cos \frac{n π}{L} x]}_{0}^{L} \\ = - \frac{L^{2}}{n^{3} π^{3}} [\cos n π - 1] \\ = \frac{L^{2} (1 - \cos n π)}{n^{3} π^{3}} \\ = \frac{L^{2} (1 - (- 1)^{n})}{n^{3} π^{3}} \end{aligned}

After evaluating the integral, we obtain:

A_{n} = \frac{L^{2} (1 - (- 1)^{n})}{n^{3} π^{3}}

Step 8: Satisfy ${\frac{\partial u}{\partial t} |}_{t = 0} = 0$

We apply the condition

{\frac{\partial u}{\partial t} |}_{t = 0} = 0

to the solution in step 6, which leads to:

0 = \sum_{n = 1}^{\infty} B_{n} \frac{a π n}{L} \sin (\frac{n π}{L} x)

This implies that

B_{n} = 0

Conclusion

The final solution to the wave equation with the given boundary and initial conditions is:

u (x, t) = \frac{L^{2}}{π^{3}} \sum_{n = 1}^{\infty} \frac{1 - (- 1)^{n}}{n^{3}} \sin (\frac{n π}{L} x) \cos (\frac{a π n}{L} t)

(b) नीचे दी गई सारणी पर आधारित बूलीय फलन

F (x, y, z)

को निकालिए और तब

F (x, y, z)

को सरल कीजिए तथा उसके अनुरूप GATE परिपथ खींचिए :

$x$	$y$	$z$	$F (x, y, z)$
1	1	1	1
1	1	0	1
1	0	1	1
1	0	0	0
0	1	1	1
0	1	0	0
0	0	1	0
0	0	0	0

Obtain the Boolean function

F (x, y, z)

based on the table given below. Then simplify

F (x, y, z)

and draw the corresponding GATE network :

$x$	$y$	$z$	$F (x, y, z)$
1	1	1	1
1	1	0	1
1	0	1	1
1	0	0	0
0	1	1	1
0	1	0	0
0	0	1	0
0	0	0	0

Answer:

Introduction

The problem involves obtaining a Boolean function

F (x, y, z)

based on the provided truth table and then simplifying this Boolean function. Additionally, the task includes drawing the corresponding GATE network based on the simplified Boolean function.

Step 1: Identify the Rows Where $F (x, y, z) = 1$

$(x, y, z) = (1, 1, 1)$
$(x, y, z) = (1, 1, 0)$
$(x, y, z) = (1, 0, 1)$
$(x, y, z) = (0, 1, 1)$

Step 2: Write the Minterms

For $(1, 1, 1)$ , the minterm is $x \land y \land z$
For $(1, 1, 0)$ , the minterm is $x \land y \land \neg z$
For $(1, 0, 1)$ , the minterm is $x \land \neg y \land z$
For $(0, 1, 1)$ , the minterm is $\neg x \land y \land z$

Step 3: Combine the Minterms

Combine all these minterms using the OR operation (

\lor

) to get the complete Boolean function:

F (x, y, z) = (x \land y \land z) \lor (x \land y \land \neg z) \lor (x \land \neg y \land z) \lor (\neg x \land y \land z)

F (x, y, z) = x y z + x y z^{'} + x y^{'} z + x^{'} y z

Step 4: Simplification

Now, simplify the obtained Boolean expression:

\begin{aligned} F (x, y, z) = x y z + x y z^{'} + x y^{'} z + x^{'} y z \\ = x y (z + z^{'}) + x y^{'} z + x^{'} y z \\ = x y (1) + x y + x y^{'} z \\ = x^{'} y z + x y + x y^{'} z \\ = x^{'} y z + x (y + y^{'} z) \\ = x^{'} y z + x [(y + y^{'}) (y + z)] \\ = x^{'} y z + x [1 (y + z)] \\ = x^{'} y z + x y + x z \\ = y (x^{'} z + x) + x z \\ = y [(x^{'} + x) (z + x)] + x z \\ = y [1 {x + z}] + x z \\ = x y + y z + x z \end{aligned}

The simplified Boolean function is

F (x, y, z) = x y + y z + x z

Step 5: Draw the Corresponding GATE Network

The gate network for the simplified Boolean function

F (x, y, z) = x y + y z + x z

consists of two gates:

OR gate with inputs $x$ and $y$ .
AND gate with inputs from the OR gate and $z$ .

Here is the corresponding gate network:

Conclusion

The Boolean function

F (x, y, z)

based on the provided truth table is

F (x, y, z) = x y + y z + x z

, and the corresponding gate network has been successfully drawn.

(c) एक छोटी चिकनी घिरनी के ऊपर से गुजरने वाली एक अवितान्य डोरी के सिरों से बंधे असमान संहति वाले दो कणों के निकाय की गति के लिए लग्रांजी समीकरण ज्ञात कीजिए।

Obtain the Lagrangian equation for the motion of a system of two particles of unequal masses connected by an inextensible string passing over a small smooth pulley.

Answer:

Introduction

The problem involves obtaining the Lagrangian equation for the motion of a system consisting of two particles with unequal masses (

m_{1}

and

m_{2}

) connected by an inextensible string passing over a small, smooth pulley with a radius of

R

. The system is subject to gravitational forces, and the goal is to derive the equation of motion for the system.

Step 1: System Configuration and Generalized Coordinate

The system consists of two particles with masses $m_{1}$ and $m_{2}$ connected by an inextensible string.
The string passes over a small, smooth pulley with radius $R$ , and both ends of the string are attached to the masses $m_{1}$ and $m_{2}$ respectively.
Let the length of the string between $m_{1}$ and $m_{2}$ be $l$ .

Step 2: Generalized Coordinate and Degrees of Freedom

The system has only one degree of freedom, and $x$ is chosen as the generalized coordinate.
The instantaneous configuration of the system is specified by $q = x$ .

Step 3: Kinetic Energy

Assuming that the cord does not slip, the angular velocity of the pulley is $\frac{x}{R}$ .
The kinetic energy of the system is given by:

T = \frac{1}{2} m_{1} {\dot{x}}^{2} + \frac{1}{2} I \frac{x^{2}}{R^{2}}

Step 4: Potential Energy

The potential energy of the system is given by:

V = - m_{1} g x - m_{2} g (l - x)

Step 5: Lagrangian

The Lagrangian ( $L$ ) is the difference between the kinetic energy ( $T$ ) and potential energy ( $V$ ):

L = T - V = \frac{1}{2} (m_{1} + m_{2} + \frac{I}{R^{2}}) {\dot{x}}^{2} + (m_{1} - m_{2}) g x + m_{2} g l

Step 6: Partial Derivatives of Lagrangian

Calculate the partial derivatives of the Lagrangian with respect to the generalized coordinate $\dot{x}$ and $x$ :

\frac{\partial L}{\partial \dot{x}} = (m_{1} + m_{2} + \frac{I}{R^{2}}) \dot{x}

\frac{\partial L}{\partial x} = (m_{1} - m_{2}) g

Step 7: Lagrange’s Equation of Motion

Apply Lagrange’s equation of motion, which is given by:

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

Substituting the derivatives and partial derivatives:

\frac{d}{d t} [(m_{1} + m_{2} + \frac{I}{R^{2}}) \dot{x}] - (m_{1} - m_{2}) g = 0

Step 8: Solve for Acceleration ( $\ddot{x}$ )

Rearrange the equation to solve for acceleration ( $\ddot{x}$ ):

(m_{1} + m_{2} + \frac{I}{R^{2}}) \ddot{x} = (m_{1} - m_{2}) g

\Rightarrow \ddot{x} = \frac{(m_{1} - m_{2}) g}{m_{1} + m_{2} + \frac{I}{R^{2}}}

Conclusion

The Lagrangian equation of motion for the system of two particles of unequal masses connected by an inextensible string passing over a small, smooth pulley is given by:

\ddot{x} = \frac{(m_{1} - m_{2}) g}{m_{1} + m_{2} + \frac{I}{R^{2}}}

(a) आंशिक अवकल समीकरण

(D^{2} - D^{' 2} - 3 D + 3 D^{'}) z = x y + e^{x + 2 y}

का व्यापक हल ज्ञात कीजिए, जहाँ

D \equiv \frac{\partial}{\partial x}

तथा

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

हैं।

Find the general solution of the partial differential equation

(D^{2} - D^{' 2} - 3 D + 3 D^{'}) z = x y + e^{x + 2 y}

where

D \equiv \frac{\partial}{\partial x}

and

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

Answer:

Introduction

The problem involves finding the general solution of the partial differential equation with respect to the given derivatives

D

and

D^{'}

(D^{2} - D^{' 2} - 3 D + 3 D^{'}) z = x y + e^{x + 2 y}

Where

D \equiv \frac{\partial}{\partial x}

and

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

Step 1: Rearrange the Equation

The given equation is rearranged as follows:

\begin{aligned} (D - D^{'}) (D + D^{'} - 3) Z = x y + e^{x + 2 y} \\ C . F = ϕ_{1} (y + x) + e^{3 x} ϕ_{2} (y - x), \end{aligned}

Here,

ϕ_{1}

and

ϕ_{2}

are arbitrary functions representing the particular integral (P.I.) corresponding to

x y

Step 2: Calculate the Coefficients for $x y$ Term

\begin{aligned} = \frac{1}{(D - D^{'}) (D + D^{'} - 3)} x y \\ = - \frac{1}{3 D} {(1 - \frac{D^{'}}{D})}^{- 1} {(1 - \frac{D + D^{'}}{3})}^{- 1} x y \\ = - \frac{1}{3 D} (1 + \frac{D^{'}}{D} + \dots) [1 + \frac{D + D^{'}}{3} + {(\frac{D + D^{'}}{3})}^{2} + \dots] x y \\ = - \frac{1}{3 D} (1 + \frac{D^{'}}{D} + \dots) (1 + \frac{D + D^{'}}{3} + \frac{2 D D^{'}}{9} + \dots) x y \\ = - \frac{1}{3 D} (x y + \frac{y}{3} + \frac{2 x}{3} + \frac{1}{D} x + \frac{2}{9}) \\ = - \frac{1}{3} (\frac{x^{2} y}{2} + \frac{x y}{2} + \frac{x^{2}}{3} + \frac{x^{3}}{6} + \frac{2 x}{9}) \end{aligned}

Step 3: Calculate the Coefficients for $e^{x + 2 y}$ Term

\begin{aligned} P . I . corresponding to e^{x + 2 y} \\ = \frac{1}{(D + D^{'} - 3) (D - D^{'})} e^{x + 2 y} \\ = \frac{1}{(D + D^{'} - 3)} \frac{1}{(1 - z)} e^{x + 2 y} \\ = - \frac{1}{(D + D^{'} - 3)} e^{x + 2 y} \cdot 1 \\ = - e^{x + 2 y} \frac{1}{(D + 1) + (D^{'} + 2) - 3} 1 \\ = - e^{x + 2 y} \frac{1}{D + D^{'}} 0.1 \\ = - e^{x + 2 y} \frac{1}{D} {(1 + \frac{D^{'}}{D})}^{- 1} \cdot 1 \\ = - e^{x + 2 y} \frac{1}{D} (1 + \dots) \cdot 1 \\ = - x e^{x + 2 y} \end{aligned}

Step 4: Combine the Results

The required solution is obtained by combining the coefficients for the

x y

and

e^{x + 2 y}

terms:

\begin{aligned} Z = C . F + P . I \\ Z = ϕ_{1} (y + x) + e^{3 x} ϕ_{2} (y - x) - (\frac{1}{6}) x^{2} y - (\frac{1}{6}) x y - (\frac{1}{9}) x^{2} - (\frac{1}{18}) x^{3} - (\frac{2}{27}) x - x e^{x + 2 y} \end{aligned}

Conclusion

The general solution of the given partial differential equation is:

Z = ϕ_{1} (y + x) + e^{3 x} ϕ_{2} (y - x) - (\frac{1}{6}) x^{2} y - (\frac{1}{6}) x y - (\frac{1}{9}) x^{2} - (\frac{1}{18}) x^{3} - (\frac{2}{27}) x - x e^{x + 2 y}

(b) समीकरणों के निकाय

\begin{aligned} 3 x_{1} + 9 x_{2} - 2 x_{3} & = 11 \\ 4 x_{1} + 2 x_{2} + 13 x_{3} & = 24 \\ 4 x_{1} - 2 x_{2} + x_{3} & = - 8 \end{aligned}

का गाउस-सीडल विधि द्वारा 4 सार्थक अंकों तक सही हल ज्ञात कीजिए, यह सत्यापन करने के बाद कि क्या यह विधि आपके द्वारा निकाय के रूपांतरित रूप में अनुप्रयोज्य है।

Solve the system of equations

\begin{aligned} 3 x_{1} + 9 x_{2} - 2 x_{3} & = 11 \\ 4 x_{1} + 2 x_{2} + 13 x_{3} & = 24 \\ 4 x_{1} - 2 x_{2} + x_{3} & = - 8 \end{aligned}

correct up to 4 significant figures by using Gauss-Seidel method after verifying whether the method is applicable in your transformed form of the system.

Answer:

We have changed the variables, to

a, b, c, \dots

for calculations.
So new equations are

3 a + 9 b - 2 c = 11, 4 a + 2 b + 13 c = 24, 4 a - 2 b + c = - 8

Total Equations are 3

\begin{aligned} 3 a + 9 b - 2 c = 11 \\ 4 a + 2 b + 13 c = 24 \\ 4 a - 2 b + c = - 8 \end{aligned}

From the above equations

a_{k + 1} = \frac{1}{3} (11 - 9 b_{k} + 2 c_{k})

b_{k + 1} = \frac{1}{2} (24 - 4 a_{k + 1} - 13 c_{k})

c_{k + 1} = \frac{1}{1} (- 8 - 4 a_{k + 1} + 2 b_{k + 1})

Initial gauss

(a, b, c) = (0, 0, 0)

Solution steps are

1^{st}

Approximation

a_{1} = \frac{1}{3} [11 - 9 (0) + 2 (0)] = \frac{1}{3} [11] = 3.6667

b_{1} = \frac{1}{2} [24 - 4 (3.6667) - 13 (0)] = \frac{1}{2} [9.3333] = 4.6667

c_{1} = \frac{1}{1} [- 8 - 4 (3.6667) + 2 (4.6667)] = \frac{1}{1} [- 13.3333] = - 13.3333

2^{nd}

Approximation

\begin{aligned} a_{2} = \frac{1}{3} [11 - 9 (4.6667) + 2 (- 13.3333)] = \frac{1}{3} [- 57.6667] = - 19.2222 \\ b_{2} = \frac{1}{2} [24 - 4 (- 19.2222) - 13 (- 13.3333)] = \frac{1}{2} [274.2222] = 137.1111 \end{aligned}

c_{2} = \frac{1}{1} [- 8 - 4 (- 19.2222) + 2 (137.1111)] = \frac{1}{1} [343.1111] = 343.1111

3^{rd}

Approximation

a_{3} = \frac{1}{3} [11 - 9 (137.1111) + 2 (343.1111)] = \frac{1}{3} [- 536.7778] = - 178.9259

\begin{aligned} b_{3} = \frac{1}{2} [24 - 4 (- 178.9259) - 13 (343.1111)] = \frac{1}{2} [- 3720.7407] = - 1860.3704 \\ c_{3} = \frac{1}{1} [- 8 - 4 (- 178.9259) + 2 (- 1860.3704)] = \frac{1}{1} [- 3013.037] = - 3013.037 \end{aligned}

Equations are Divergent…

Iterations are tabulated as below

Iteration	$a$	$b$	$c$
1	3.6667	4.6667	-13.3333
2	-19.2222	137.1111	343.1111
3	-178.9259	-1860.3704	-3013.037

\vec{q} = \frac{λ (- y \hat{i} + x \hat{j})}{x^{2} + y^{2}}, (λ =

स्थिरांक) एक संभाव्य असंपीड्य तरल गति है। धारा-रेखाएँ निकालिए। क्या गति का प्रकार विभव है? यदि हाँ, तो वेग विभव निकालिए।
Show that

\vec{q} = \frac{λ (- y \hat{i} + x \hat{j})}{x^{2} + y^{2}}, (λ =

constant

)

is a possible incompressible fluid motion. Determine the streamlines. Is the kind of the motion potential? If yes, then find the velocity potential.

Answer:

Introduction

In this problem, we are asked to show that the given velocity field

\vec{q} = \frac{λ (- y \hat{i} + x \hat{j})}{x^{2} + y^{2}}

with

λ

as a constant represents possible incompressible fluid motion. We need to determine the streamlines and check if the motion is potential. If it is potential, we need to find the velocity potential.

Step 1: Check for Incompressibility

We know that for an incompressible fluid motion,

\nabla \cdot \vec{q} = 0

. Let’s verify this:

\nabla \cdot \vec{q} = λ {- \frac{\partial}{\partial x} (\frac{y}{x^{2} + y^{2}}) + \frac{\partial}{\partial y} (\frac{x}{x^{2} + y^{2}})}

Simplifying,

λ {\frac{2 x y}{{(x^{2} + y^{2})}^{2}} - \frac{2 x y}{{(x^{2} + y^{2})}^{2}}} = 0

Since

\nabla \cdot \vec{q} = 0

, the equation of continuity for an incompressible fluid is satisfied, indicating that it is a possible motion for an incompressible fluid.

Step 2: Determine Streamlines

The equation of streamlines is given by:

\frac{d x}{u} = \frac{d y}{v} = \frac{d z}{w}

For our velocity field

\vec{q} = \frac{λ (- y \hat{i} + x \hat{j})}{x^{2} + y^{2}}

, we have:

\frac{d x}{- \frac{λ y}{x^{2} + y^{2}}} = \frac{d y}{\frac{λ x}{x^{2} + y^{2}}} = \frac{d z}{0}

Simplifying,

x d x + y d y = 0, d z = 0

Integrating these equations, we get:

x^{2} + y^{2} = constant (2)

and

z = constant (3)

The streamlines are circular with centers on the Z-axis, and their planes are perpendicular to the axis.

Step 3: Determine if the Motion is Potential

To check if the motion is potential, we calculate

\nabla \times \vec{q}

. If

\nabla \times \vec{q} = 0

, the motion is potential.

\nabla \times \vec{q} = k λ [\frac{y^{2} - x^{2}}{{(x^{2} + y^{2})}^{2}} + \frac{x^{2} - y^{2}}{{(x^{2} + y^{2})}^{2}}] = 0

\nabla \times \vec{q} = 0

, the flow is of a potential kind.

Step 4: Find the Velocity Potential

If the flow is potential, we can determine the velocity potential

ϕ (x, y, z)

such that:

\vec{q} = - \nabla ϕ

We have the following relations:

\begin{aligned} \frac{\partial ϕ}{\partial x} = - u = \frac{λ y}{x^{2} + y^{2}} (4) \\ \frac{\partial ϕ}{\partial y} = - v = - \frac{λ x}{x^{2} + y^{2}} (5) \\ \frac{\partial ϕ}{\partial z} = - w = 0 (6) \end{aligned}

These equations indicate that

ϕ

is independent of

z

, and thus

ϕ = ϕ (x, y)

Integrating equation (4), we get:

ϕ (x, y) = λ \tan^{- 1} (\frac{x}{y}) + f (y)

Using equation (5), we find:

f^{'} (y) - \frac{λ x}{x^{2} + y^{2}} = 0 \Rightarrow f^{'} (y) = \frac{λ x}{x^{2} + y^{2}}

Integrating

f^{'} (y)

, we obtain

f (y)

as a constant.

Therefore, the velocity potential

ϕ (x, y)

is given by:

ϕ (x, y) = λ \tan^{- 1} (\frac{x}{y}) + C

Conclusion

The given velocity field represents possible incompressible fluid motion. The streamlines are circular with centers on the Z-axis, and the motion is of a potential kind. The velocity potential

ϕ (x, y)

is given by:

ϕ (x, y) = λ \tan^{- 1} (\frac{x}{y}) + C

(a) चार्पिट विधि का उपयोग करके आंशिक अवकल समीकरण $p = (z + q y)^{2}$ का पूर्ण समाकल ज्ञात कीजिए।

Find a complete integral of the partial differential equation

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

by using Charpit’s method.

Answer:

Introduction

In this problem, we are tasked with finding a complete integral of the partial differential equation

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

using Charpit’s method.

Step 1: Formulate the Given Equation

The given equation is:

f (x, y, z, p, q) = (z + q y)^{2} - p (1)

Step 2: Charpit’s Auxiliary Equations

Charpit’s method involves deriving auxiliary equations. The auxiliary equations are:

\frac{d p}{\frac{\partial f}{\partial x} + p \frac{\partial f}{\partial z}} = \frac{d q}{\frac{\partial f}{\partial y} + q \frac{\partial f}{\partial z}} = \frac{d z}{- p \frac{\partial f}{\partial p} - q \frac{\partial f}{\partial q}} = \frac{d x}{- \frac{\partial f}{\partial p}} = \frac{d y}{- \frac{\partial f}{\partial q}}

Applying these equations to our problem:

\frac{d p}{2 p (z + q y)} = \frac{d q}{4 q (z + q y)} = \frac{d z}{p - 2 q y (z + q y)} = d x = \frac{d y}{- 2 y (z + q y)}

Step 3: Solve for $d p$ and $d y$

Taking the first and fifth fractions:

\frac{d p}{2 p (z + q y)} = \frac{d y}{- 2 y (z + q y)}

Integrate these expressions:

\begin{aligned} \frac{d p}{p} & = - \frac{d y}{y} \\ \log (p) & = \log (a) - \log (y) \\ p & = \frac{a}{y} (2) \end{aligned}

Step 4: Substitute into the Given Equation

Substituting (2) into (1):

(z + q y)^{2} = \frac{a}{y}

Solving for

q y

q y = \frac{\sqrt{a}}{\sqrt{y}} - z

Isolating

q

q = \frac{\sqrt{a}}{y \sqrt{y}} - \frac{z}{y} (3)

Step 5: Derive $d z$

Now, we need to find

d z

using our expressions for

p

and

q

\begin{aligned} d z & = p d x + q d y \\ = (\frac{a}{y}) d x + (\frac{\sqrt{a}}{y \sqrt{y}} - \frac{z}{y}) d y \\ = y d z + z d y = a d x + \sqrt{a} y^{- \frac{1}{2}} d y \end{aligned}

Step 6: Integrate to Find a Complete Integral

Integrating the above expression:

y d z + z d y = a d x + \sqrt{a} y^{- \frac{1}{2}} d y

We obtain:

y z = a x + 2 \sqrt{a y} + b

where

a

and

b

are arbitrary constants.

Hence, the complete integral of the partial differential equation

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

is:

y z = a x + 2 \sqrt{a y} + b

with

a

and

b

as arbitrary constants.

(b) न्यूटन के पश्चांतर अंतर्वेशन सूत्र की व्युत्पत्ति कीजिए तथा त्रुटि-विश्लेषण भी कीजिए।

Derive Newton’s backward difference interpolation formula and also do error analysis.

Answer:

Introduction

In this problem, we will derive Newton’s backward difference interpolation formula and conduct an error analysis.

Step 1: Formulate the Given Equation

Let

y = f (x)

be a function that takes values

y_{0}, y_{1}, y_{2}, \dots, y_{n}

corresponding to

x_{0}, x_{1}, x_{2}, \dots, x_{n}

with an interval of differencing

h

Step 2: Define $ϕ (x)$

Define

ϕ (x)

as a polynomial of degree

n

x

that represents

y = f (x)

such that

f (x_{r}) = ϕ (x_{r})

for

r = 0, 1, 2, \dots, n

Step 3: Write $ϕ (x)$ as a Polynomial

Express

ϕ (x)

as a polynomial:

ϕ (x) = a_{0} + a_{1} (x - x_{n}) + a_{2} (x - x_{n}) (x - x_{n - 1}) + \dots + a_{n} (x - x_{n}) (x - x_{n - 1}) \dots (x - x_{1})

Step 4: Determine Coefficients $a_{0}, a_{1}, \dots, a_{n}$

Calculate the coefficients

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

by matching the values of

ϕ (x)

and

f (x)

at the corresponding points:

For

r = 0

y_{n} = a_{0} ⟹ a_{0} = y_{n}

For

r = 1

y_{n - 1} = y_{n} + a_{1} (- h) ⟹ a_{1} h = y_{n} - y_{n - 1} = \nabla y_{n} ⟹ a_{1} = \frac{\nabla y_{n}}{1! h}

For

r = 2

y_{n - 2} = y_{n} - 2 y_{n} + 2 y_{n - 1} + (2 h)^{2} a_{2} ⟹ a_{2} = \frac{y_{n} - 2 y_{n - 1} + y_{n - 2}}{2 h^{2}} = \frac{\nabla^{2} y_{n}}{2! h^{2}}

Similarly, for

r = 3

n

a_{r} = \frac{\nabla^{r} y_{n}}{r! h^{r}}

Step 5: Express $ϕ (x)$ Using Coefficients

Writing

u = \frac{x - x_{n}}{h}

we get

x - x_{n} = u h

Therefore

x - x_{n - 1} = x - x_{n} + x_{n} - x_{n - 1} = (u h) + h = (u + 1) h

\Rightarrow x - x_{n - 2} = (u + 2) h, \dots, (x - x_{1}) = u + (n - 1) h

Substitute the calculated coefficients into the expression for

ϕ (x)

\begin{aligned} ϕ (x) & = y_{n} + \frac{\nabla y_{n}}{1! h} (x - x_{n}) + \frac{\nabla^{2} y_{n}}{2! h^{2}} (x - x_{n}) (x - x_{n - 1}) + \dots \\ + \frac{\nabla^{n} y_{n}}{n! h^{n}} (x - x_{n}) (x - x_{n - 1}) \dots (x - x_{1}) \\ = y_{n} + \frac{u \nabla y_{n}}{1!} + \frac{u (u + 1)}{2!} \nabla^{2} y_{n} + \frac{u (u + 1) (u + 2)}{3!} \nabla^{3} y_{n} + \dots \\ + \frac{u (u + 1) \dots (u + n - 1)}{n!} \nabla^{n} y_{n} \end{aligned}

Step 6: Error Analysis

The error in the Newton’s backward interpolation formula is given by:

R (x) = \frac{\nabla^{n + 1} y_{n}}{(n + 1)!} u (u + 1) \dots (u + n)

Here,

u = x - x_{n}

This error term quantifies the difference between the interpolated polynomial and the actual function

f (x

Conclusion

We have successfully derived Newton’s backward difference interpolation formula and conducted an error analysis to evaluate the interpolation error.

\tan^{- 1} z

के लिए धारा-रेखाएँ तथा समविभव वक्र, वृत्त हैं। किसी भी बिन्दु पर वेग निकालिए तथा

z = \pm i

पर विचित्रता जाँचिए।

Show that for the complex potential

\tan^{- 1} z

, the streamlines and equipotential curves are circles. Find the velocity at any point and check the singularities at

z = \pm i

Answer:

Introduction

In this problem, we will analyze the complex potential

\tan^{- 1} z

and show that the streamlines and equipotential curves are circles. Additionally, we will find the velocity at any point and examine the singularities at

z = \pm i

Step 1: Define the Complex Potential

The complex potential is given by:

w = ϕ + i Ψ = \tan^{- 1} z (1)

We will use this complex potential to derive various properties.

Step 2: Calculate $\bar{w}$

Calculate the complex conjugate of the potential:

\bar{w} = ϕ - i Ψ = \tan^{- 1} \bar{z} (2)

Step 3: Subtract $\bar{w}$ from $w$

Subtract equation (2) from equation (1):

2 i Ψ = \tan^{- 1} z - \tan^{- 1} \bar{z}

Simplify:

\tan 2 i Ψ = \frac{2 i y}{1 + x^{2} + y^{2}}

This leads to:

x^{2} + y^{2} + 1 = 2 y \coth 2 Ψ (3)

Step 4: Analyze Streamlines

The streamlines

Ψ = constant

represent circles, as described by equation (3).

Step 5: Add $w$ and $\bar{w}$

Add equation (1) and equation (2):

2 ϕ = \tan^{- 1} z + \tan^{- 1} \bar{z} = \tan^{- 1} \frac{z - \bar{z}}{1 - z \bar{z}}

Simplify:

1 - x^{2} - y^{2} = 2 x \cot 2 ϕ (4)

Step 6: Analyze Equipotential Curves

The equipotential curves

ϕ = constant

also represent circles, orthogonal to the streamlines

Ψ = constant

, forming a co-axial system. These circles have limit points at

z = \pm i

, as described by equation (4).

Step 7: Find Velocity Components

The velocity components

(u, v)

are calculated as:

\frac{\partial w}{\partial z} = - u + i v = \frac{1}{z^{2} + 1} (5)

Step 8: Identify Singularities

The denominator of equation (5) vanishes at

z = \pm i

, indicating singularities at these points.

Step 9: Analyze Singularities

Analyzing the singularity at

z = + i

Substitute

z = i + z_{1}

, where

| z_{1} |

is very small:

- u + i v = \frac{d w}{d z} = \frac{d w}{d z_{1}} = \frac{1}{1 + (- 1 + 2 i z_{1})} = \frac{1}{2 i z_{1}}

By integrating, we find:

w = - \frac{1}{2} i \log z_{1}

This implies that the singularity at

z = + i

is a vortex of strength

k = - \frac{1}{2}

with circulation

- π k

Similarly, the singularity at

z = - i

is a vortex of strength

k = \frac{1}{2}

with circulation

π k

Conclusion

We have shown that for the complex potential

\tan^{- 1} z

, the streamlines and equipotential curves are circles. We have also found the velocity at any point and examined the singularities at

z = \pm i

Free UPSC Mathematics Optional Paper-2 2021 Solutions: View Online | UPSC Maths Solution | IAS Maths Solution

Preliminaries

Definition: Uniform Convergence

Approach

Step 1: Recognize the Series as a Geometric Series

Explanation

Application to Our Series

Step 2: Find the n t h n t h n^(th)n^{th}nth Partial Sum s n ( x ) s n ( x ) s_(n)(x)s_n(x)sn(x)

Explanation

Application to Our Series

Step 3: Determine the Limit Function S ( x ) S ( x ) S(x)S(x)S(x)

Explanation

Application to Our Series

Step 4: Check for Discontinuity

Explanation

Application to Our Series

Conclusion

Definitions:

Proof:

Conclusion:

Construction of Subgroups:

Properties:

Infinitude of Subgroups:

Conclusion:

Introduction

Analysis and Solution

Complex Potential Decomposition

Complex Potential in Terms of Real and Imaginary Parts

Streamlines

Case 1: When k k kkk is infinite

Case 2: When k k kkk is zero

Sketch

Conclusion

Step 1: Identify the Rows Where F ( x , y , z ) = 1 F ( x , y , z ) = 1 F(x,y,z)=1F(x, y, z) = 1F(x,y,z)=1

Step 2: Write the Minterms

Step 3: Combine the Minterms

Step 4: Simplification

Step 5: Draw the Corresponding GATE Network

Step 1: System Configuration and Generalized Coordinate

Step 2: Generalized Coordinate and Degrees of Freedom

Step 3: Kinetic Energy

Step 4: Potential Energy

Step 5: Lagrangian

Step 6: Partial Derivatives of Lagrangian

Step 7: Lagrange’s Equation of Motion

Step 8: Solve for Acceleration ( x ¨ x ¨ x^(¨)\ddot{x}x¨)

Step 1: Rearrange the Equation

Step 2: Calculate the Coefficients for x y x y xyxyxy Term

Step 3: Calculate the Coefficients for e x + 2 y e x + 2 y e^(x+2y)e^{x+2y}ex+2y Term

Step 4: Combine the Results

Step 1: Check for Incompressibility

Step 2: Determine Streamlines

Step 3: Determine if the Motion is Potential

Step 4: Find the Velocity Potential

Step 1: Formulate the Given Equation

Step 2: Charpit’s Auxiliary Equations

Step 3: Solve for d p d p dpdpdp and d y d y dydydy

Step 4: Substitute into the Given Equation

Step 5: Derive d z d z dzdzdz

Step 6: Integrate to Find a Complete Integral

Step 1: Formulate the Given Equation

Step 2: Define ϕ ( x ) ϕ ( x ) phi(x)\phi(x)ϕ(x)

Step 3: Write ϕ ( x ) ϕ ( x ) phi(x)\phi(x)ϕ(x) as a Polynomial

Step 4: Determine Coefficients a 0 , a 1 , … , a n a 0 , a 1 , … , a n a_(0),a_(1),dots,a_(n)a_0, a_1, \ldots, a_na0,a1,…,an

Step 5: Express ϕ ( x ) ϕ ( x ) phi(x)\phi(x)ϕ(x) Using Coefficients

Step 6: Error Analysis

Step 1: Define the Complex Potential

Step 2: Calculate w ¯ w ¯ bar(w)\bar{w}w¯

Step 3: Subtract w ¯ w ¯ bar(w)\bar{w}w¯ from w w www

Step 4: Analyze Streamlines

Step 5: Add w w www and w ¯ w ¯ bar(w)\bar{w}w¯

Step 6: Analyze Equipotential Curves

Step 7: Find Velocity Components

Step 8: Identify Singularities

Step 9: Analyze Singularities

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Step 2: Find the $n^{t h}$ Partial Sum $s_{n} (x)$

Step 3: Determine the Limit Function $S (x)$

Case 1: When $k$ is infinite

Case 2: When $k$ is zero

Step 1: Identify the Rows Where $F (x, y, z) = 1$

Step 8: Solve for Acceleration ( $\ddot{x}$ )

Step 2: Calculate the Coefficients for $x y$ Term

Step 3: Calculate the Coefficients for $e^{x + 2 y}$ Term

Step 3: Solve for $d p$ and $d y$

Step 5: Derive $d z$

Step 2: Define $ϕ (x)$

Step 3: Write $ϕ (x)$ as a Polynomial

Step 4: Determine Coefficients $a_{0}, a_{1}, \dots, a_{n}$

Step 5: Express $ϕ (x)$ Using Coefficients

Step 2: Calculate $\bar{w}$

Step 3: Subtract $\bar{w}$ from $w$

Step 5: Add $w$ and $\bar{w}$