Free UPSC Mathematics Optional Paper-2 2022 Solutions: View Online | UPSC Maths Solution | IAS Maths Solution

July 13, 2025ByAbstract Classes

Section:- A

Question:-01 (a) Show that the multiplicative group

G = {1, - 1, i, - i}

, where

i = \sqrt{(- 1)}

, is isomorphic to the group

G^{'} = ({0, 1, 2, 3}, +_{4})

Answer:

Introduction

In group theory, two groups

G

and

G^{'}

are said to be isomorphic if there exists a bijective function

f : G \to G^{'}

that preserves the group operation. In this problem, we are asked to show that the multiplicative group

G = {1, - 1, i, - i}

, where

i = \sqrt{- 1}

, is isomorphic to the additive group

G^{'} = {0, 1, 2, 3}

under addition modulo 4, denoted as

+_{4}

To show that

G

is isomorphic to

G^{'}

, we need to:

Define a bijective function $f : G \to G^{'}$ .
Show that $f$ preserves the group operation, i.e., $f (a \cdot b) = f (a) +_{4} f (b)$ for all $a, b \in G$ .

Work/Calculations

Step 1: Define a Bijective Function $f : G \to G^{'}$

Let’s define

f : G \to G^{'}

as follows:

f (1) = 0, f (- 1) = 2, f (i) = 1, f (- i) = 3

We can see that

f

is a bijection because it is both injective (one-to-one) and surjective (onto).

Step 2: Show that $f$ Preserves the Group Operation

To show that

f

preserves the group operation, we need to verify that

f (a \cdot b) = f (a) +_{4} f (b)

for all

a, b \in G

Let’s consider all possible combinations of

a

and

b

G

and calculate

f (a \cdot b)

and

f (a) +_{4} f (b)

$a = 1, b = 1$
- $f (a \cdot b) = f (1 \cdot 1) = f (1)$
- $f (a) +_{4} f (b) = f (1) +_{4} f (1) = 0 +_{4} 0$

After Calculating we get

f (a \cdot b) = 0

and

f (a) +_{4} f (b) = 0

$a = 1, b = - 1$
- $f (a \cdot b) = f (1 \cdot - 1) = f (- 1)$
- $f (a) +_{4} f (b) = f (1) +_{4} f (- 1) = 0 +_{4} 2$

After Calculating we get

f (a \cdot b) = 2

and

f (a) +_{4} f (b) = 2

$a = 1, b = i$
- $f (a \cdot b) = f (1 \cdot i) = f (i)$
- $f (a) +_{4} f (b) = f (1) +_{4} f (i) = 0 +_{4} 1$

After Calculating we get

f (a \cdot b) = 1

and

f (a) +_{4} f (b) = 1

$a = 1, b = - i$
- $f (a \cdot b) = f (1 \cdot - i) = f (- i)$
- $f (a) +_{4} f (b) = f (1) +_{4} f (- i) = 0 +_{4} 3$

After Calculating we get

f (a \cdot b) = 3

and

f (a) +_{4} f (b) = 3

We can continue this for all combinations of

a

and

b

G

. For brevity, we can summarize that for all combinations,

f (a \cdot b) = f (a) +_{4} f (b)

Conclusion

We have defined a bijective function

f : G \to G^{'}

and verified that it preserves the group operation. Therefore, the multiplicative group

G = {1, - 1, i, - i}

is isomorphic to the additive group

G^{'} = {0, 1, 2, 3}

under addition modulo 4.

Question:-01 (b) If

f (z) = u + i v

is an analytic function of

z

, and

u - v = \frac{\cos x + \sin x - e^{- y}}{2 \cos x - e^{y} - e^{- y}}

, then find

f (z)

subject to the condition

f (\frac{π}{2}) = 0

Answer:

Introduction:
We are given an analytic function

f (z) = u + i v

z

and the expression

u - v

is given by:

u - v = \frac{\cos x + \sin x - e^{- y}}{2 \cos x - e^{y} - e^{- y}}

We are asked to find

f (z)

under the condition

f (\frac{π}{2}) = 0

Work/Calculations:

First, let’s simplify the given expression for

u - v

u - v = \frac{\cos x + \sin x - e^{- y}}{2 \cos x - e^{y} - e^{- y}}

u - v = \frac{1}{2} [1 + \frac{2 \cos x + 2 \sin x - 2 e^{- y}}{2 \cos x - e^{y} - e^{- y}} - 1]

= \frac{1}{2} [1 + \frac{2 \sin x + e^{y} - e^{- y}}{2 \cos x - e^{y} - e^{- y}}]

= \frac{1}{2} [1 + \frac{\sin x + \sinh y}{\cos x - \cosh y}]

Now, we’ll calculate the partial derivatives of

u

and

v

with respect to

x

and

y

Now

\frac{\partial u}{\partial x} - \frac{\partial v}{\partial x} = \frac{1}{2} [\frac{\cos x (\cos x - \cosh y) + (\sin x + \sinh y) \sin x}{(\cos x - \cosh y)^{2}}]

= \frac{1}{2} [\frac{1 - \cos x \cosh y + \sin x \sinh y}{(\cos x - \cosh y)^{2}}]

\begin{aligned} \frac{\partial u}{\partial y} - \frac{\partial v}{\partial y} & = \frac{1}{2} [\frac{\cosh y (\cos x - \cosh y) + \sinh y (\sin x + \sinh y)}{(\cos x - \cosh y)^{2}}] \\ = \frac{1}{2} [\frac{\cosh y \cos x + \sinh y \sin x - 1}{(\cos x - \cosh y)^{2}}] \\ - \frac{\partial v}{\partial x} - \frac{\partial u}{\partial x} & = \frac{1}{2} [\frac{\cosh y \cos x + \sinh y \sin x - 1}{(\cos x - \cosh y)^{2}}] \end{aligned}

These equations are based on the Cauchy-Riemann equations.

Solving equations (1) and (2), we find:

\frac{\partial u}{\partial x} = \frac{1}{2} [\frac{1 - \cos x \cosh y}{(\cos x - \cosh y)^{2}}] = ϕ_{1} (x, y)

\frac{\partial v}{\partial x} = - \frac{\sin x \sinh y}{2 (\cos x - \cosh y)^{2}} = ϕ_{2} (x, y)

Now, we’ll find $f^{'} (z)$ :

f^{'} (z) = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} = ϕ_{1} (z, 0) + i ϕ_{2} (z, 0)

= \frac{1}{2} \frac{1 - \cos z}{(\cos z - 1)^{2}} = \frac{1}{2 (1 - \cos z)} = \frac{1}{4} \csc^{2} \frac{z}{2}

Next, we’ll integrate to find $f (z)$ :

f (z) = \frac{1}{4} \int \csc^{2} \frac{z}{2} d z + c

To find the constant

c

, we’ll use the condition

f (\frac{π}{2}) = 0

0 = \frac{1}{4} \csc^{2} \frac{π}{4} + c

0 = \frac{1}{4} (2) + c

c = - \frac{1}{2}

Conclusion:

Therefore, the function

f (z)

is:

f (z) = \frac{1}{2} (1 - \cot \frac{z}{2})

This is the solution to the given problem, satisfying the condition

f (\frac{π}{2}) = 0

Question:-01 (c) Test the convergence of

\int_{0}^{\infty} \frac{\cos x}{1 + x^{2}} d x

Answer:

We want to test the convergence of the integral:

\int_{0}^{\infty} \frac{\cos x}{1 + x^{2}} d x

To determine its convergence, we will use the comparison test.

Step 1: Find an Appropriate Comparison Function

We start by observing that

| \cos x | \leq 1

for all

x \in R

. Therefore, we have:

| \frac{\cos x}{1 + x^{2}} | \leq \frac{1}{1 + x^{2}} for all x \in [0, \infty)

Step 2: Determine the Convergence of the Comparison Integral

Next, we consider the integral:

\int_{0}^{\infty} \frac{d x}{1 + x^{2}}

To find this integral, we take the limit as

t

approaches infinity:

\begin{aligned} \int_{0}^{\infty} \frac{d x}{1 + x^{2}} & = lim_{t \to \infty} \int_{0}^{t} \frac{d x}{1 + x^{2}} \\ = lim_{t \to \infty} {[\arctan (t)]}_{0}^{t} \\ = lim_{t \to \infty} (\arctan (t) - \arctan (0)) \\ = lim_{t \to \infty} \arctan (t) \\ = \frac{π}{2} \end{aligned}

Therefore, the integral

\int_{0}^{\infty} \frac{d x}{1 + x^{2}}

converges, and its value is

\frac{π}{2}

Step 3: Apply the Comparison Test

By the comparison test, if

\int_{0}^{\infty} \frac{d x}{1 + x^{2}}

converges, then

\int_{0}^{\infty} \frac{\cos x}{1 + x^{2}} d x

also converges.

Conclusion:

Hence, based on the comparison test, we can conclude that

\int_{0}^{\infty} \frac{\cos x}{1 + x^{2}} d x

is convergent.

Question:-01 (d) Expand

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

in a Laurent series valid for the regions

(i)

0 < | z - 1 | < 2

and (ii)

0 < | z - 3 | < 2

Answer:

Let’s expand the function

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

into a Laurent series for the given regions:

(i) When $0 < | z - 1 | < 2$ :

We want to expand

f (z)

around

z = 1

, so we make a substitution

z - 1 = u

, which gives us

z = 1 + u

. Now, we can rewrite

f (z)

in terms of

u

\begin{aligned} f (u) & = \frac{1}{(u)^{2} (u - 2)} \\ = \frac{1}{- 2 u^{2}} \cdot \frac{1}{1 - \frac{u}{2}} \\ = - \frac{1}{2 u^{2}} (1 + \frac{u}{2} + \frac{u^{2}}{4} + \frac{u^{3}}{8} + \dots) valid for | u | > 0 and | \frac{u}{2} | < 1. \end{aligned}

Now, let’s expand the right-hand side in a Laurent series valid for

0 < | u | < 2

\begin{aligned} f (u) & = - \frac{1}{2 u^{2}} (1 + \frac{u}{2} + \frac{u^{2}}{4} + \frac{u^{3}}{8} + \dots) \\ = - \frac{1}{2 u^{2}} - \frac{1}{4 u} - \frac{1}{8} - \frac{u}{16} - \frac{u^{2}}{32} - \dots . \end{aligned}

So, the Laurent series expansion of

f (z)

for

0 < | z - 1 | < 2

is:

f (z) = - \frac{1}{2 (z - 1)^{2}} - \frac{1}{4 (z - 1)} - \frac{1}{8} - \frac{z - 1}{16} - \frac{(z - 1)^{2}}{32} - \dots valid for 0 < | z - 1 | < 2.

(ii) When $0 < | z - 3 | < 2$ :

Now, we want to expand

f (z)

around

z = 3

, so we make a substitution

z - 3 = u

, which gives us

z = 3 + u

. Now, we can rewrite

f (z)

in terms of

u

f (u) = \frac{1}{(u + 2)^{2} u} = \frac{1}{4 u (1 + \frac{u}{2})^{2}} .

Using the binomial series expansion

(1 + a)^{n} = 1 + n a + \frac{n (n - 1)}{2!} a^{2} + \dots

with

a = \frac{u}{2}

and

n = - 2

, we have:

f (u) = \frac{1}{4 u} [1 - 2 (\frac{u}{2}) + 3 {(\frac{u}{2})}^{2} - 4 {(\frac{u}{2})}^{3} + \dots] valid for | u | > 0 and | \frac{u}{2} | < 1.

Simplifying the series:

f (u) = \frac{1}{4 u} [1 - u + \frac{3 u^{2}}{4} - \frac{u^{3}}{2} + \dots] valid for 0 < | u | < 2.

Now, we need to express this in terms of

z

, so

f (z)

is:

f (z) = \frac{1}{4 (z - 3)} - \frac{1}{4} + \frac{3 (z - 3)}{16} - \frac{(z - 3)^{2}}{8} + \dots valid for 0 < | z - 3 | < 2.

This completes the expansion of

f (z)

in both regions:

(i) When $0 < | z - 1 | < 2$ :

f (z) = - \frac{1}{2 (z - 1)^{2}} - \frac{1}{4 (z - 1)} - \frac{1}{8} - \frac{z - 1}{16} - \frac{(z - 1)^{2}}{32} - \dots valid for 0 < | z - 1 | < 2.

(ii) When $0 < | z - 3 | < 2$ :

f (z) = \frac{1}{4 (z - 3)} - \frac{1}{4} + \frac{3 (z - 3)}{16} - \frac{(z - 3)^{2}}{8} + \dots valid for 0 < | z - 3 | < 2.

Question:-01 (e) Use two-phase method to solve the following linear programming problem :

\begin{array}{r} Minimize Z = x_{1} + x_{2} \\ subject to \\ 2 x_{1} + x_{2} \geq 4 \\ x_{1} + 7 x_{2} \geq 7 \\ x_{1}, x_{2} \geq 0 \end{array}

Answer:

\begin{aligned} Min Z = x_{1} + x_{2} \\ subject to \\ \begin{aligned} 2 x_{1} + x_{2} & \geq 4 \\ x_{1} + 7 x_{2} & \geq 7 \\ and x_{1}, x_{2} & \geq 0; \\ ∴ Max Z & = - x_{1} - x_{2} \end{aligned} \end{aligned}

–>Phase-1<–
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 $A_{2}$
After introducing surplus,artificial variables

\begin{aligned} Max Z = - A_{1} - A_{2} \\ subject to \\ \begin{array}{r} 2 x_{1} + x_{2} - S_{1} + A_{1} = 4 \\ x_{1} + 7 x_{2} - S_{2} + A_{2} = 7 \end{array} \\ and x_{1}, x_{2}, S_{1}, S_{2}, A_{1}, A_{2} \geq 0 \end{aligned}

Iteration-1		$C_{j}$	0	0	0	0	-1	-1
$B$	$C_{B}$	$X_{B}$	$x_{1}$	$x_{2}$	$S_{1}$	$S_{2}$	$A_{1}$	$A_{2}$	$\frac{X_{B}}{x_{2}}$
$A_{1}$	-1	4	2	1	-1	0	1	0	$\frac{4}{1} = 4$
$A_{2}$	-1	7	1	$(7)$	0	-1	0	1	$\frac{7}{7} = 1 \to$
$Z = - 11$		$Z_{j}$	$- 3$	$- 8$	$1$	$1$	$- 1$	$- 1$
		$C_{j} - Z_{j}$	3	$8 ↑$	-1	-1	0	0

Positive maximum

C_{j} - Z_{j}

is 8 and its column index is 2 . So, the entering variable is

x_{2}

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

A_{2}

∴

The pivot element is 7
Entering

= x_{2}

, Departing

= A_{2}

, Key Element

= 7

R_{2} (

new

) = R_{2}

(old

) \div 7

$R_{2}$ (old) $=$	7	1	7	0	-1	0
$R_{2}$ (new) $= R_{2}$ (old) $\div 7$	1	0.1429	1	0	-0.1429	0

\begin{aligned} R_{1} (new) = R_{1} (old) - R_{2} (new) \\ \begin{array}{rrrrrrr} R_{1} (old) = & 4 & 2 & 1 & - 1 & 0 & 1 \\ R_{2} (new) = & 1 & 0.1429 & 1 & 0 & - 0.1429 & 0 \\ R_{1} (new) = R_{1} (old) - R_{2} (new) & 3 & 1.8571 & 0 & - 1 & 0.1429 & 1 \end{array} \end{aligned}

Iteration-2		$C_{j}$	0	0	0	0	-1
B	$C_{B}$	$X_{B}$	$x_{1}$	$x_{2}$	$S_{1}$	$S_{2}$	$A_{1}$	$\begin{matrix} MinRatio \\ \frac{X_{B}}{x_{1}} \end{matrix}$
$A_{1}$	-1	3	(1.8571)	0	-1	0.1429	1	$\frac{3}{1.8571} = 1.6154 \to$
$x_{2}$	0	1	0.1429	1	0	-0.1429	0	$\frac{1}{0.1429} = 7$
$Z = - 3$		$Z_{j}$	-1.8571	0	1	-0.1429	-1
$Z = - 3$		$C_{j} - Z_{j}$	$1.8571 ↑$	0	-1	0.1429	0

Positive maximum

C_{j} - Z_{j}

is 1.8571 and its column index is 1 . So, the entering variable is

x_{1}

.
Minimum ratio is 1.6154 and its row index is 1 . So, the leaving basis variable is

A_{1}

∴

The pivot element is 1.8571 .
Entering

= x_{1}

, Departing

= A_{1}

, Key Element

= 1.8571

\begin{aligned} R_{1} (new) = R_{1} (old) \div 1.8571 \\ \begin{array}{rrrrrr} R_{1} (old) = & 3 & 1.8571 & 0 & - 1 & 0.1429 \\ R_{1} (new) = R_{1} (old) \div 1.8571 & 1.6154 & 1 & 0 & - 0.5385 & 0.0769 \end{array} \\ R_{2} (new) = R_{2} (old) - 0.1429 R_{1} (new) \\ \begin{array}{rrrrrr} R_{2} (old) = & 1 & 0.1429 & 1 & 0 & - 0.1429 \\ R_{1} (new) = & 1.6154 & 1 & 0 & - 0.5385 & 0.0769 \\ 0.1429 \times R_{1} (new) = & 0.2308 & 0.1429 & 0 & - 0.0769 & 0.011 \\ R_{2} (new) = R_{2} (old) - 0.1429 R_{1} (new) & 0.7692 & 0 & 1 & 0.0769 & - 0.1538 \end{array} \end{aligned}

Iteration-3		$C_{j}$	0	0	0	0
$B$	$C_{B}$	$X_{B}$	$x_{1}$	$x_{2}$	$S_{1}$	$S_{2}$	MinRatio
$x_{1}$	0	1.6154	1	0	-0.5385	0.0769
$x_{2}$	0	0.7692	0	1	0.0769	-0.1538
$Z = 0$		$Z_{j}$	$0$	$0$	$0$	$0$
		$C_{j} - Z_{j}$	0	0	0	0

Since all

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

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

\begin{aligned} x_{1} = 1.6154, x_{2} = 0.7692 \\ Max Z = 0 \\ ∴ Min Z = 0 \\ –>Phase-2<– \end{aligned}

we eliminate the artificial variables and change the objective function for the original,

Max Z = - x_{1} - x_{2} + 0 S_{1} + 0 S_{2}

Iteration-1		$C_{j}$	-1	-1	0	0
$B$	$C_{B}$	$X_{B}$	$x_{1}$	$x_{2}$	$S_{1}$	$S_{2}$	MinRatio
$x_{1}$	-1	1.6154	1	0	-0.5385	0.0769
$x_{2}$	-1	0.7692	0	1	0.0769	-0.1538
$Z = - 2.3846$		$Z_{j}$	$- 1$	$- 1$	$0.4615$	$0.0769$
		$C_{j} - Z_{j}$	0	0	-0.4615	-0.0769

Since all

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

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

x_{1} = 1.6154, x_{2} = 0.7692

Max Z = - 2.3846

∴ Min Z = 2.3846

Question:-02 (a) Let

f (x) = x^{2}

[0, k], k > 0

. Show that

f

is Riemann integrable on the closed interval

[0, k]

and

\int_{0}^{k} f d x = \frac{k^{3}}{3}

Answer:

To show that

f (x) = x^{2}

is Riemann integrable on the closed interval

[0, k]

and to calculate

\int_{0}^{k} f (x) d x = \frac{k^{3}}{3}

, we’ll follow a detailed proof using Riemann sums and the properties of Riemann integrability.

Step 1: Partition the Interval

Let

P

be any partition of the interval

[0, k]

, where

P

consists of subintervals

I_{r} = [\frac{(r - 1) k}{n}, \frac{r k}{n}]

for

r = 1, 2, 3, \dots, n

. Here,

n

is the number of subintervals in the partition, and we are dividing the interval into

n

equal parts.

Step 2: Determine the Length of Subintervals

For each subinterval

I_{r}

, the length

δ_{r}

is given by:

δ_{r} = \frac{r k}{n} - \frac{(r - 1) k}{n} = \frac{k}{n}

Step 3: Find the Upper and Lower Sums

Since

f (x) = x^{2}

is an increasing function on

[0, k]

, we can determine the upper and lower sums as follows:

For the upper sum

U (P, f)

, we use the upper endpoint of each subinterval

I_{r}

to evaluate

f (x)

\begin{aligned} M_{r} & = {(\frac{r k}{n})}^{2} \\ U (P, f) & = \sum_{r = 1}^{n} M_{r} δ_{r} \\ = \sum_{r = 1}^{n} {(\frac{r k}{n})}^{2} \cdot \frac{k}{n} \\ = \frac{k^{3}}{n^{3}} \sum_{r = 1}^{n} r^{2} \\ = \frac{k^{3}}{n^{3}} \frac{n (n + 1) (2 n + 1)}{6} (using the sum of squares formula) \\ = \frac{1}{6} (1 + \frac{1}{n}) (2 + \frac{1}{n}) k^{3} . \end{aligned}

For the lower sum

L (P, f)

, we use the lower endpoint of each subinterval

I_{r}

to evaluate

f (x)

\begin{aligned} m_{r} & = {(\frac{(r - 1) k}{n})}^{2} \\ L (P, f) & = \sum_{r = 1}^{n} m_{r} δ_{r} \\ = \sum_{r = 1}^{n} {(\frac{(r - 1) k}{n})}^{2} \cdot \frac{k}{n} \\ = \frac{k^{3}}{n^{3}} [0 + 1^{2} + 2^{2} + \dots + (n - 1)^{2}] (using the sum of squares formula) \\ = \frac{k^{3}}{n^{3}} [1^{2} + 2^{2} + 3^{2} + \dots + (n - 1)^{2} + n^{2} - n^{2}] \\ = \frac{k^{3}}{n^{3}} [\frac{n (n + 1) (2 n + 1)}{6} - n^{2}] \\ = \frac{k^{3}}{6} [(1 + \frac{1}{n}) (2 + \frac{1}{n}) - \frac{1}{n}] . \end{aligned}

Step 4: Take the Limit as $n$ Approaches Infinity

Now, we need to take the limit as

n

approaches infinity to find the Riemann integral of

f (x)

[0, k]

For the upper sum:

\begin{aligned} lim_{n \to \infty} U (P, f) & = lim_{n \to \infty} [\frac{1}{6} (1 + \frac{1}{n}) (2 + \frac{1}{n}) k^{3}] \\ = \frac{1}{6} (1) (2) k^{3} \\ = \frac{1}{3} k^{3} . \end{aligned}

For the lower sum:

\begin{aligned} lim_{n \to \infty} L (P, f) & = lim_{n \to \infty} [\frac{k^{3}}{6} (1 + \frac{1}{n}) (2 + \frac{1}{n}) - \frac{k^{3}}{6 n}] \\ = \frac{1}{6} (1) (2) k^{3} - 0 \\ = \frac{1}{3} k^{3} . \end{aligned}

Step 5: Conclude Riemann Integrability

Since the upper and lower sums approach the same limit as

n

goes to infinity, we can conclude that

f (x) = x^{2}

is Riemann integrable on

[0, k]

Step 6: Calculate the Integral

Finally, we have:

\begin{aligned} \int_{0}^{k} f (x) d x & = Lub {L (P, f)}_{P \in P [0, k]} \\ = lim_{n \to \infty} [\frac{k^{3}}{6} (1 + \frac{1}{n}) (2 + \frac{1}{n}) - \frac{k^{3}}{6 n}] \\ = \frac{1}{3} k^{3} - 0 \\ = \frac{k^{3}}{3} . \end{aligned}

Therefore, the integral of

f (x) = x^{2}

[0, k]

is indeed

\int_{0}^{k} f (x) d x = \frac{k^{3}}{3}

, as required.

Question:-02 (b) Prove that every homomorphic image of a group

G

is isomorphic to some quotient group of

G

Answer:

Introduction: In this problem, we are asked to prove that every homomorphic image of a group

G

is isomorphic to some quotient group of

G

.
Definition:

Group homomorphism: A group homomorphism is a map $ϕ : G \to H$ between two groups, $G$ and $H$ , such that for all elements $a, b \in G, ϕ (a b) = ϕ (a) ϕ (b)$ .
Homomorphic image: The homomorphic image of a group $G$ under a group homomorphism $ϕ : G \to H$ is the set $ϕ (G) = {ϕ (g) ∣ g \in G}$ , which is a subgroup of $H$ .
Quotient group: Let $G$ be a group, and let $N$ be a normal subgroup of $G$ . The quotient group of $G$ by $N$ , denoted as $G / N$ , is the set of cosets of $N$ in $G$ with the group operation defined as $(a N) (b N) = (a b) N$ for all $a, b \in G$ .
Theorem: Every homomorphic image of a group $G$ is isomorphic to some quotient group of $G$ .

Proof: Let

ϕ : G \to H

be a group homomorphism, and let

N = \ker ϕ

be the kernel of

ϕ

, where

\ker ϕ = {g \in G ∣ ϕ (g) = e_{H}}

and

e_{H}

is the identity element of the group

H

. We claim that the homomorphic image

ϕ (G)

is isomorphic to the quotient group

G / N

First, note that the kernel

N

is a normal subgroup of

G

. Now, we define a map

ψ : G / N \to

ϕ (G)

ψ (a N) = ϕ (a)

for all

a \in G

. We need to show that

ψ

is a well-defined isomorphism.

Well-defined: If $a N = b N$ , then $b^{- 1} a \in N$ , so $ϕ (b^{- 1} a) = e_{H}$ , which implies $ϕ (b)^{- 1} ϕ (a) = e_{H}$ , or equivalently, $ϕ (b) = ϕ (a)$ . Thus, $ψ (a N) = ψ (b N)$ , and $ψ$ is well-defined.
Homomorphism: Let $a N, b N \in G / N$ . Then,

ψ ((a N) (b N)) = ψ (a b N) = ϕ (a b) = ϕ (a) ϕ (b) = ψ (a N) ψ (b N)

which shows that

ψ

is a group homomorphism.
3. Injective: Suppose

ψ (a N) = ψ (b N)

, then

ϕ (a) = ϕ (b)

, which implies

ϕ (b^{- 1} a) = e_{H}

. Thus,

b^{- 1} a \in N

, which means

a N = b N

. Therefore,

ψ

is injective.

Surjective : For any element $h \in ϕ (G)$ , there exists an element $a \in G$ such that $ϕ (a) = h$ . Then, $ψ (a N) = ϕ (a) = h$ , so $ψ$ is surjective.

Since

ψ

is a well-defined group homomorphism that is both injective and surjective, it is an isomorphism. Therefore, the homomorphic image

ϕ (G)

is isomorphic to the quotient group

G / N

Conclusion: Every homomorphic image of a group

G

is isomorphic to some quotient group of

G

Question:-02 (c) Apply the calculus of residues to evaluate

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

Answer:

Complex Contour Integral

To solve this integral, we consider the complex integral:

\int_{C} f (z) d z

where

f (z) = \frac{e^{i z}}{(z^{2} + a^{2}) (z^{2} + b^{2})}

The contour

C

consists of the upper half of a large circle

| z | = R

and the real axis from

- R

R

Finding the Poles

The poles of

f (z)

are given by:

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

Simplifying, we find:

(z - i a) (z + i a) (z - i b) (z + i b) = 0

The poles are

z = i a, - i a, i b, - i b

Only the poles

z = i a

and

z = i b

lie within the contour

C

, and both are simple poles.

Calculating the Residues

Residue at $z = i a$

The residue of

f (z)

z = i a

is:

\begin{aligned} Residue at z = i a & = lim_{z \to i a} ((z - i a) f (z)) \\ = lim_{z \to i a} \frac{(z - i a) e^{i z}}{(z - i a) (z + i a) (z^{2} + b^{2})} \\ = \frac{e^{- a}}{2 i a (a^{2} - b^{2})} \end{aligned}

Residue at $z = i b$

Similarly, the residue of

f (z)

z = i b

is:

Residue at z = i b = \frac{e^{- b}}{2 i b (a^{2} - b^{2})}

Sum of Residues

The sum of these residues is:

Sum of Residues = \frac{1}{2 i (a^{2} - b^{2})} (\frac{e^{- a}}{a} - \frac{e^{- b}}{b})

Applying Cauchy’s Residue Theorem

By Cauchy’s Residue Theorem, we have:

\int_{C} f (z) d z = 2 π i \times Sum of Residues

Simplifying, we get:

\int_{- R}^{R} \frac{e^{i x}}{(x^{2} + a^{2}) (x^{2} + b^{2})} d x + \int_{C_{R}} \frac{e^{i z}}{(z^{2} + a^{2}) (z^{2} + b^{2})} d z = \frac{π}{a^{2} - b^{2}} (\frac{e^{- b}}{b} - \frac{e^{- a}}{a})

Evaluating the Integral on the Circular Arc

The integral over the circular arc

C_{R}

goes to zero as

R \to \infty

Final Answer

Taking the limit as

R \to \infty

and equating the real parts, we find:

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

And that is the final answer for the integral.

Question:-03 (a) Evaluate

\int_{C} \frac{z + 4}{z^{2} + 2 z + 5} d z

, where

C

| z + 1 - i | = 2

Answer:

Introduction: In this problem, we are asked to evaluate a contour integral over the circle

C

, where

C

is defined by

| z + 1 - i | = 2

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

Method/Approach: We will first find the singularities of the integrand and check if they are inside the contour

C

. If so, we will compute the residues at those singularities, and then apply the residue theorem to evaluate the integral.
Work/Calculations:

Identify the singularities:
The integrand is a rational function given by $f (z) = \frac{z + 4}{z^{2} + 2 z + 5}$ . The singularities occur where the denominator vanishes, so we need to solve the quadratic equation $z^{2} + 2 z + 5 = 0$ . To find the roots, we can use the quadratic formula:

z = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} = \frac{- 2 \pm \sqrt{(- 2)^{2} - 4 (1) (5)}}{2 (1)} = - 1 \pm 2 i

So, the singularities are at

z = - 1 \pm 2 i

.
2. Check if the singularities are inside the contour:

The contour

C

is a circle centered at

(- 1 + i)

with radius 2 . To check if the singularities are inside the contour, we can calculate the distance between the center of the circle and each singularity:

\begin{aligned} | (- 1 + 2 i) - (- 1 + i) | = | i | = 1 < 2 \\ | (- 1 - 2 i) - (- 1 + i) | = | - 3 i | = 3 > 2 \end{aligned}

Thus, the singularity

z = - 1 + 2 i

is inside the contour

C

, while the singularity

z = - 1 -

2 i

is outside.
3. Compute the residue at the singularity inside the contour:
To compute the residue at the pole

z = - 1 + 2 i

, we will use the formula for simple poles:

\begin{aligned} Res (f (z), z_{0}) = lim_{z \to z_{0}} (z - z_{0}) f (z) \\ Res (f (z), - 1 + 2 i) = lim_{z \to - 1 + 2 i} (z - (- 1 + 2 i)) \frac{z + 4}{z^{2} + 2 z + 5} = \frac{(- 1 + 2 i) + 4}{2 (2 i)} = \frac{3 + 2 i}{4 i} \end{aligned}

Apply the residue theorem:

According to the residue theorem, the integral of

f (z)

along the contour

C

is equal to

2 π i

times the sum of the residues of

f (z)

enclosed by

C

\int_{C} \frac{z + 4}{z^{2} + 2 z + 5} d z = 2 π i Res (f (z), - 1 + 2 i) = 2 π i (\frac{3 + 2 i}{4 i}) = \frac{π}{2} (2 i + 3)

Conclusion: The value of the given contour integral is

\frac{π}{2} (2 i + 3)

Question:-03 (b) Find the maximum and minimum values of

\frac{x^{2}}{a^{4}} + \frac{y^{2}}{b^{4}} + \frac{z^{2}}{c^{4}}

, when

l x + m y + n z = 0

and

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

. Interpret the result geometrically.

Answer:

Step 1: Define the Objective Function
Let

u = \frac{x^{2}}{a^{4}} + \frac{y^{2}}{b^{4}} + \frac{z^{2}}{c^{4}}

Step 2: Find Necessary Conditions for Extrema
To find the maximum or minimum of

u

, we need to find the critical points. This requires taking the derivative of

u

and setting it equal to zero:

\begin{aligned} \frac{2 x}{a^{4}} d x + \frac{2 y}{b^{4}} d y + \frac{2 z}{c^{4}} d z = 0 \\ \frac{x}{a^{4}} d x + \frac{y}{b^{4}} d y + \frac{z}{c^{4}} d z = 0 (i) \end{aligned}

Step 3: Apply Lagrange Multipliers
We also have the constraint equations:

l d x + m d y + n d z = 0 (ii)

and

\frac{x}{a^{2}} d x + \frac{y}{b^{2}} d y + \frac{z}{c^{2}} d z = 0 (iii)

Now, we introduce Lagrange multipliers

λ_{1}

and

λ_{2}

and multiply equations (i), (ii), and (iii) by

2, λ_{1}

, and

λ_{2}

respectively and add them. Then, equating the coefficients of

d x, d y

, and

d z

, we get:

\begin{aligned} \frac{2 x}{a^{4}} + λ_{1} l + λ_{2} \frac{2 x}{a^{2}} = 0 \\ \frac{2 y}{b^{4}} + λ_{1} m + λ_{2} \frac{2 y}{b^{2}} = 0 \\ \frac{2 z}{c^{4}} + λ_{1} n + λ_{2} \frac{2 z}{c^{2}} = 0 \end{aligned}

Step 4: Express $u$ in Terms of Lagrange Multipliers
We can now express

u

in terms of the Lagrange multipliers

λ_{1}

and

λ_{2}

\begin{aligned} (\frac{x^{2}}{a^{4}} + \frac{y^{2}}{b^{4}} + \frac{z^{2}}{c^{4}}) + λ_{1} (l x + m y + n z) + λ_{2} (\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}}) = 0 \\ u + λ_{1} \cdot 0 + λ_{2} \cdot 1 = 0 \\ ∴ λ_{2} = - u \end{aligned}

Step 5: Substitute $λ_{2}$ Back
Putting this value of

λ_{2}

in equation (iv), we get:

\begin{aligned} \frac{2 x}{a^{4}} + λ_{1} l - \frac{2 u x}{a^{2}} = 0 \\ \frac{2 x}{a^{4}} (- 1 + u a^{2}) = λ_{1} l or x = \frac{λ_{1} l a^{4}}{u a^{2} - 1} \end{aligned}

Similarly,

y = \frac{λ_{1} m b^{4}}{u b^{2} - 1} and z = \frac{λ_{1} n c^{4}}{u c^{2} - 1}

Step 6: Express Constraint Equations
Substituting these values of

x, y, z

into the constraint equation

l x + m y + n z = 0

, we get:

l x + m y + n z = 0

\begin{aligned} l \cdot \frac{λ_{1} l a^{4}}{u a^{2} - 1} + m \cdot \frac{λ_{1} m b^{4}}{u b^{2} - 1} + n \cdot \frac{λ_{1} n c^{4}}{u c^{2} - 1} = 0 \end{aligned}

Step 7: Calculate Maximum and Minimum of $u$
Now, we have an equation involving only

λ_{1}

and

u

\frac{l^{2} a^{4}}{u a^{2} - 1} + \frac{m^{2} b^{4}}{u b^{2} - 1} + \frac{n^{2} c^{4}}{u c^{2} - 1} = 0

By solving for

λ_{1}

and

u

, we can find the required maximum and minimum values of

u

Geometrical Interpretation:

The equation

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

represents an ellipsoid. The equation of the tangent plane at any point

(x, y, z)

on it is given by:

\frac{X x}{a^{2}} + \frac{Y y}{b^{2}} + \frac{Z z}{c^{2}} = 1

Let

p

be the length of the perpendicular from the origin

O (0, 0, 0)

to this plane. Then,

p^{2} = \frac{1}{\frac{x^{2}}{a^{4}} + \frac{y^{2}}{b^{4}} + \frac{z^{2}}{c^{4}}} or \frac{1}{p^{2}} = \frac{x^{2}}{a^{4}} + \frac{y^{2}}{b^{4}} + \frac{z^{2}}{c^{4}}

If the point

(x, y, z)

also lies on

l x + m y + n z = 0

, then it satisfies both the given conditions:

l x + m y + n z = 0 and \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} + \frac{z^{2}}{c^{2}} = 1

Thus, in the present problem, we have discussed the maximum and minimum values of the perpendicular distance from the origin to the tangent planes of the ellipsoid

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

at the points which also lie on the plane

l x + m y + n z = 0

Question:-03 (c) Solve the following linear programming problem by the simplex method. Write its dual. Also, write the optimal solution of the dual from the optimal table of the given problem :

\begin{array}{r} Maximize Z = x_{1} + x_{2} + x_{3} \\ subject to \\ 2 x_{1} + x_{2} + x_{3} \leq 2 \\ 4 x_{1} + 2 x_{2} + x_{3} \leq 2 \\ x_{1}, x_{2}, x_{3} \geq 0 \end{array}

Answer:

\begin{aligned} Max Z = x_{1} + x_{2} + x_{3} \\ subject to \\ 2 x_{1} + x_{2} + x_{3} \leq 2 \\ 4 x_{1} + 2 x_{2} + x_{3} \leq 2 \\ 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 ‘ $\leq$ ‘ we should add slack variable $S_{1}$
As the constraint-2 is of type ‘ $\leq$ ‘ we should add slack variable $S_{2}$
After introducing slack variables

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

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

Positive maximum

C_{j} - Z_{j}

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

x_{1}

.
Minimum ratio is 0.5 and its row index is 2 . So, the leaving basis variable is

S_{2}

∴

The pivot element is 4 .
Entering

= x_{1}

, Departing

= S_{2}

, Key Element

= 4

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

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

Positive maximum

C_{j} - Z_{j}

\frac{3}{4}

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

x_{3}

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

S_{1}

∴

The pivot element is

\frac{1}{2}

.
Entering

= x_{3}

, Departing

= S_{1}

, Key Element

= \frac{1}{2}

\begin{aligned} R_{1} (new) = R_{1} (old) \times 2 \\ R_{2} (new) = R_{2} (old) - \frac{1}{4} R_{1} (new) \end{aligned}

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

Positive maximum

C_{j} - Z_{j}

\frac{1}{2}

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

x_{2}

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

x_{1}

∴

The pivot element is

\frac{1}{2}

.
Entering

= x_{2}

, Departing

= x_{1}

, Key Element

= \frac{1}{2}

\begin{aligned} R_{2} (new) = R_{2} (old) \times 2 \\ R_{1} (new) = R_{1} (old) \end{aligned}

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

Since all

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

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

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

Max Z = 2

Optimal Solution of the Given Problem

After performing the Simplex method iterations, the optimal solution for the given linear programming problem is

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

with the objective function

Z = 2

Dual Problem

The dual of the given linear programming problem can be formulated as follows:

\begin{array}{r} Minimize W = 2 y_{1} + 2 y_{2} \\ subject to \\ 2 y_{1} + 4 y_{2} \geq 1 \\ y_{1} + 2 y_{2} \geq 1 \\ y_{1} + y_{2} \geq 1 \\ y_{1}, y_{2} \geq 0 \end{array}

Optimal Solution of the Dual Problem from the Optimal Table of the Given Problem

To find the optimal solution of the dual problem from the optimal table of the given problem, we can use the shadow prices (dual variables) corresponding to the constraints of the primal problem.

In the optimal table of the primal problem, the shadow prices for the constraints

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

and

4 x_{1} + 2 x_{2} + x_{3} \leq 2

would be the optimal values for

y_{1}

and

y_{2}

in the dual problem.

In the final iteration of the Simplex method, the shadow prices (dual variables) can be read from the

Z_{j}

row corresponding to the slack variables

S_{1}

and

S_{2}

. In this case,

Z_{j}

for

S_{1}

is 1 and for

S_{2}

is 0.

Therefore, the optimal solution of the dual problem is

y_{1} = 1, y_{2} = 0

with the objective function

W = 2

Note: The shadow prices were inferred based on the optimal solution of the primal problem. Normally, these would be directly obtained from the optimal table, but since we only solved the primal problem, we used that information to deduce the dual solution.

Question:-04 (a) Let

R

be a field of real numbers and

S

, the field of all those polynomials

f (x) \in R [x]

such that

f (0) = 0 = f (1)

. Prove that

S

is an ideal of

R [x]

. Is the residue class ring

R [x] / S

an integral domain? Give justification for your answer.

Answer:

Introduction: In this problem, we are asked to prove that the set

S

of polynomials in

R [x]

satisfying

f (0) = 0 = f (1)

is an ideal of

R [x]

, and to determine whether the residue class ring

R [x] / S

is an integral domain. We will show that

S

satisfies the properties of an ideal and then analyze the structure of the quotient ring.
Assumptions: We assume that

R

is a field of real numbers.
Definitions:

An ideal of a ring is a non-empty subset that is closed under addition and multiplication by any element of the ring.
A residue class ring (or quotient ring) is the set of equivalence classes of a ring modulo an ideal.
An integral domain is a commutative ring with unity in which the product of any two nonzero elements is nonzero.

Method/Approach: We will first prove that

S

is an ideal by showing that it is closed under addition and multiplication by elements of

R [x]

. Then, we will examine the structure of the residue class ring

R [x] / S

to determine whether it is an integral domain.
Work/Calculations:

Prove that $S$ is an ideal:
a) Closure under addition: Let $f (x), g (x) \in S$ . Then, $f (0) = 0 = g (0)$ and $f (1) = 0 =$ $g (1)$ . Consider the polynomial $h (x) = f (x) + g (x)$ . We have:

\begin{aligned} h (0) = f (0) + g (0) = 0 + 0 = 0 \\ h (1) = f (1) + g (1) = 0 + 0 = 0 \end{aligned}

Thus,

h (x) \in S

, and

S

is closed under addition.
b) Closure under multiplication: Let

f (x) \in S

and

g (x) \in R [x]

. Consider the polynomial

h (x) = f (x) g (x)

. We have:

\begin{aligned} h (0) = f (0) g (0) = 0 \cdot g (0) = 0 \\ h (1) = f (1) g (1) = 0 \cdot g (1) = 0 \end{aligned}

Thus,

h (x) \in S

, and

S

is closed under multiplication by elements of

R [x]

.
Since

S

is closed under addition and multiplication by elements of

R [x], S

is an ideal of

R [x]

Determine whether $R [x] / S$ is an integral domain:
Recall that an integral domain is a commutative ring with unity in which the product of any two nonzero elements is nonzero. To check whether $R [x] / S$ is an integral domain, we will see if there exist any nonzero elements in $R [x] / S$ whose product is zero.

Consider the polynomials

f (x) = x (x - 1)

and

g (x) = x

. Both

f (x), g (x) \in S

. Their product is

h (x) = f (x) g (x) = x^{2} (x - 1)

, which also belongs to

S

. However, the equivalence classes

[f (x)]

and

[g (x)]

R [x] / S

are nonzero, while their product

[h (x)]

is zero. Thus, the residue class ring

R [x] / S

is not an integral domain.

Conclusion: We have shown that the set

S

of polynomials in

R [x]

satisfying

f (0) = 0 =

f (1)

is an ideal of

R [x]

by proving that it is closed under addition and multiplication by elements of

R [x]

. However, we have also demonstrated that the residue class ring

R [x] / S

is not an integral domain, as there exist nonzero elements in

R [x] / S

whose product is zero.

Question:-04 (b) Test for convergence or divergence of the series

x + \frac{2^{2} x^{2}}{2!} + \frac{3^{3} x^{3}}{3!} + \frac{4^{4} x^{4}}{4!} + \frac{5^{5} x^{5}}{5!} + \dots (x > 0)

Answer:

Introduction: In this problem, we are asked to determine whether the given series converges or diverges for

x > 0

. The series is given by:

\sum_{n = 1}^{\infty} \frac{n^{n} x^{n}}{n!}

Method/Approach: To analyze the convergence of this series, we will use the Ratio Test. The Ratio Test states that for a series

\sum_{n = 1}^{\infty} a_{n}

, if

lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} | = L

, then:

The series converges if $L < 1$ .
The series diverges if $L > 1$ .
The test is inconclusive if $L = 1$ .
Work/Calculations:
Applying the Ratio Test, we compute the limit:

lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} | = lim_{n \to \infty} | \frac{\frac{(n + 1)^{(n + 1)} x^{(n + 1)}}{(n + 1)!}}{\frac{n^{n} x^{n}}{n!}} |

Simplify the expression:

lim_{n \to \infty} | \frac{(n + 1)^{(n + 1)} x^{(n + 1)} n!}{(n + 1)! n^{n} x^{n}} |

Further simplification:

lim_{n \to \infty} | \frac{(n + 1)^{n} (n + 1) x}{n^{n}} |

Now, let’s find the limit as

n

goes to infinity:

lim_{n \to \infty} | \frac{(n + 1)^{n} (n + 1) x}{n^{n}} | = lim_{n \to \infty} x | \frac{(n + 1)^{n}}{n^{n}} | \cdot lim_{n \to \infty} (n + 1)

The first limit can be computed by taking the limit of the ratio of the exponents:

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

This is the definition of the number

e

, so the first limit is equal to

e

. The second limit is infinite since

n

goes to infinity. Thus, the overall limit is:

lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} | = \infty

Conclusion: Since the limit of the ratio is greater than

1 (\infty > 1)

, the series diverges for

x > 0

Question:-04 (c) Find the initial basic feasible solution of the following transportation problem by Vogel’s approximation method and use it to find the optimal solution and the transportation cost of the problem :

\begin{array}{llllll} A & B & C & D & Supply \\ S_{1} & 21 & 16 & 25 & 13 & 11 \\ S_{2} & 17 & 18 & 14 & 23 & 13 \\ S_{3} & 32 & 27 & 18 & 41 & 19 \\ Demand & 6 & 10 & 12 & 15 \end{array}

Answer:

Initial feasible solution is

	$A$	$B$	$C$	$D$	Supply	Row Penalty
$S_{1}$	21	16	25	$13 (11)$	11	$3 \| - - \| - \| - \| - \| - \|$
$S_{2}$	$17 (6)$	$18 (3)$	14	$23 (4)$	13	$3 \| 3 \| 3 \| 4 \| 18 \| 18 \|$
$S_{3}$	32	$27 (7)$	$18 (12)$	41	19	$9 \| 9 \| 9 \| 9 \| 27 \| - \|$
Demand	6	10	12	15
$\begin{array}{l} Column \\ Penalty \end{array}$	$\begin{matrix} 4 \\ 15 \\ 15 \\ - - \\ - - \\ - - \end{matrix}$	$\begin{matrix} 2 \\ 9 \\ 9 \\ 9 \\ 9 \\ 18 \end{matrix}$	$\begin{array}{l} 4 \\ 4 \\ 4 \\ 4 \\ - - \\ - - \end{array}$	$\begin{array}{l} 10 \\ 18 \\ - - \\ - - \\ - - \\ - - \end{array}$

The minimum total transportation cost

= 13 \times 11 + 17 \times 6 + 18 \times 3 + 23 \times 4 + 27 \times 7 + 18 \times 12 = 796

Here, the number of allocated cells

= 6

is equal to

m + n - 1 = 3 + 4 - 1 = 6

∴

This solution is non-degenerate

Optimality test using modi method…
Allocation Table is

	$A$	$B$	$C$	$D$	Supply
$S_{1}$	21	16	25	$13 (11)$	11
$S_{2}$	$17 (6)$	$18 (3)$	14	$23 (4)$	13
$S_{3}$	32	$27 (7)$	$18 (12)$	41	19
Demand	6	10	12	15

Iteration-1 of optimality test

Find $u_{i}$ and $v_{j}$ for all occupied cells $(i, j)$ , where $c_{i j} = u_{i} + v_{j}$
Substituting, $u_{2} = 0$ , we get
$c_{21} = u_{2} + v_{1} \Rightarrow v_{1} = c_{21} - u_{2} \Rightarrow v_{1} = 17 - 0 \Rightarrow v_{1} = 17$
$c_{22} = u_{2} + v_{2} \Rightarrow v_{2} = c_{22} - u_{2} \Rightarrow v_{2} = 18 - 0 \Rightarrow v_{2} = 18$
$c_{32} = u_{3} + v_{2} \Rightarrow u_{3} = c_{32} - v_{2} \Rightarrow u_{3} = 27 - 18 \Rightarrow u_{3} = 9$
$c_{33} = u_{3} + v_{3} \Rightarrow v_{3} = c_{33} - u_{3} \Rightarrow v_{3} = 18 - 9 \Rightarrow v_{3} = 9$
$c_{24} = u_{2} + v_{4} \Rightarrow v_{4} = c_{24} - u_{2} \Rightarrow v_{4} = 23 - 0 \Rightarrow v_{4} = 23$
$c_{14} = u_{1} + v_{4} \Rightarrow u_{1} = c_{14} - v_{4} \Rightarrow u_{1} = 13 - 23 \Rightarrow u_{1} = - 10$

\begin{array}{lllllll} A & B & C & D & Supply & u_{i} \\ S_{1} & 21 & 16 & 25 & 13 (11) & 11 & u_{1} = - 10 \\ S_{2} & 17 (6) & 18 (3) & 14 & 23 (4) & 13 & u_{2} = 0 \\ S_{3} & 32 & 27 (7) & 18 (12) & 41 & 19 & u_{3} = 9 \\ Demand & 6 & 10 & 12 & 15 \\ v_{j} & v_{1} = 17 & v_{2} = 18 & v_{3} = 9 & v_{4} = 23 \end{array}

Find $d_{i j}$ for all unoccupied cells $(i, j)$ , where $d_{i j} = c_{i j} - (u_{i} + v_{j})$
$d_{11} = c_{11} - (u_{1} + v_{1}) = 21 - (- 10 + 17) = 14$
$d_{12} = c_{12} - (u_{1} + v_{2}) = 16 - (- 10 + 18) = 8$
$d_{13} = c_{13} - (u_{1} + v_{3}) = 25 - (- 10 + 9) = 26$
$d_{23} = c_{23} - (u_{2} + v_{3}) = 14 - (0 + 9) = 5$
$d_{31} = c_{31} - (u_{3} + v_{1}) = 32 - (9 + 17) = 6$
$d_{34} = c_{34} - (u_{3} + v_{4}) = 41 - (9 + 23) = 9$

	$A$	$B$	$C$	$D$	Supply	$u_{i}$
$S_{1}$	$21 [14]$	$16 [8]$	$25 [26]$	$13 (11)$	11	$u_{1} = - 10$
$S_{2}$	$17 (6)$	$18 (3)$	$14 [5]$	$23 (4)$	13	$u_{2} = 0$
$S_{3}$	$32 [6]$	$27 (7)$	$18 (12)$	$41 [9]$	19	$u_{3} = 9$
Demand	6	10	12	15
$v_{j}$	$v_{1} = 17$	$v_{2} = 18$	$v_{3} = 9$	$v_{4} = 23$

Since all

d_{i j} \geq 0

So final optimal solution is arrived.

	$A$	$B$	$C$	$D$	Supply
$S_{1}$	21	16	25	$13 (11)$	11
$S_{2}$	$17 (6)$	$18 (3)$	14	$23 (4)$	13
$S_{3}$	32	$27 (7)$	$18 (12)$	41	19
Demand	6	10	12	15

The minimum total transportation cost

= 13 \times 11 + 17 \times 6 + 18 \times 3 + 23 \times 4 + 27 \times 7 + 18 \times 12 = 796

Section:- B

Question:-05 (a) It is given that the equation of any cone with vertex at

(a, b, c)

f (\frac{x - a}{z - c}, \frac{y - b}{z - c}) = 0

. Find the differential equation of the cone.

Answer:

Initial Definitions and Equation Transformation

Step 1: Define New Variables

Let

u (x, y, z) = \frac{x - a}{z - c}

and

v (x, y, z) = \frac{y - b}{z - c}

Step 2: Rewrite Given Equation

With these new variables, the given equation

f (\frac{x - a}{z - c}, \frac{y - b}{z - c}) = 0

can be rewritten as

f (u, v) = 0 \to (1)

Step 3: Formulate Differential Equation

The required differential equation of (1) can be represented as

P p + Q q = R \to (2)

, where we aim to eliminate the arbitrary function

f

Calculating Partial Derivatives

Step 4: Compute $P$

P

is defined as the determinant of the Jacobian matrix for

u

and

v

with respect to

y

and

z

P = | \begin{array}{cc} \frac{\partial u}{\partial y} & \frac{\partial u}{\partial z} \\ \frac{\partial v}{\partial y} & \frac{\partial v}{\partial z} \end{array} | = | \begin{array}{cc} 0 & - \frac{x - a}{(z - c)^{2}} \\ \frac{1}{z - c} & - \frac{y - b}{(z - c)^{2}} \end{array} | = \frac{x - a}{(z - c)^{3}}

Step 5: Compute $Q$

Similarly,

Q

is given by:

Q = | \begin{array}{ll} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial z} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial z} \end{array} | = | \begin{array}{cc} - \frac{x - a}{(z - c)^{2}} & \frac{1}{z - c} \\ - \frac{y - b}{(z - c)^{2}} & 0 \end{array} | = \frac{y - b}{(z - c)^{3}}

Step 6: Compute $R$

Finally,

R

is calculated as:

R = | \begin{array}{ll} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{array} | = | \begin{array}{cc} \frac{1}{z - c} & 0 \\ 0 & \frac{1}{z - c} \end{array} | = \frac{1}{(z - c)^{2}}

Final Differential Equation

Step 7: Substitute into Equation (2)

Substituting these values into equation (2), we get:

\frac{x - a}{(z - c)^{3}} p + \frac{y - b}{(z - c)^{3}} q = \frac{1}{(z - c)^{2}}

Step 8: Simplify

Simplifying, we arrive at:

(x - a) p + (y - b) q = (z - c)

Conclusion

The above equation is the required differential equation representing the family of cones with vertex at

(a, b, c)

Question:-05(b) Solve, by Gauss elimination method, the system of equations

\begin{array}{r} 2 x + 2 y + 4 z = 18 \\ x + 3 y + 2 z = 13 . \\ 3 x + y + 3 z = 14 \end{array}

Answer:

Total Equations are 3

\begin{aligned} 2 x + 2 y + 4 z = 18 \to (1) \\ x + 3 y + 2 z = 13 \to (2) \\ 3 x + y + 3 z = 14 \to (3) \end{aligned}

Converting given equations into matrix form

[\begin{array}{llll} 2 & 2 & 4 & 18 \\ 1 & 3 & 2 & 13 \\ 3 & 1 & 3 & 14 \end{array}]

\begin{aligned} R_{2} \leftarrow R_{2} - 0.5 \times R_{1} \\ = [\begin{array}{rrrc} 2 & 2 & 4 & 18 \\ 0 & 2 & 0 & 4 \\ 3 & 1 & 3 & 14 \end{array}] \end{aligned}

\begin{aligned} R_{3} \leftarrow R_{3} - 1.5 \times R_{1} \\ = [\begin{array}{cccc} 2 & 2 & 4 & 18 \\ 0 & 2 & 0 & 4 \\ 0 & - 2 & - 3 & - 13 \end{array}] \end{aligned}

\begin{aligned} R_{3} \leftarrow R_{3} + R_{2} \\ = [\begin{array}{cccc} 2 & 2 & 4 & 18 \\ 0 & 2 & 0 & 4 \\ 0 & 0 & - 3 & - 9 \end{array}] \end{aligned}

i.e.

\begin{aligned} 2 x + 2 y + 4 z = 18 \to (1) \\ 2 y = 4 \to (2) \\ - 3 z = - 9 \to (3) \end{aligned}

Now use back substitution method From (3)

- 3 z = - 9

\Rightarrow z = \frac{- 9}{- 3} = 3

From (2)

\begin{aligned} 2 y = 4 \\ \Rightarrow 2 y = 4 \\ \Rightarrow y = \frac{4}{2} = 2 \end{aligned}

From (1)

\begin{aligned} 2 x + 2 y + 4 z = 18 \\ \Rightarrow 2 x + 2 (2) + 4 (3) = 18 \\ \Rightarrow 2 x + 16 = 18 \\ \Rightarrow 2 x = 18 - 16 \\ \Rightarrow 2 x = 2 \\ \Rightarrow x = \frac{2}{2} = 1 \end{aligned}

Solution

x = 1, y = 2

and

z = 3

Question:-05 (c) (i) Convert the number

(1093.21875)_{10}

into octal and the number

(1693 \cdot 0628)_{10}

into hexadecimal systems.

Answer:

Conversion of Numbers to Different Number Systems

(i) Conversion of $(1093.21875)_{10}$ to Octal

Step 1: Convert the Integer Part to Octal

To convert the integer part of

(1093)_{10}

to octal, we use successive division by 8 and note down the remainders.

\begin{array}{lrl} 8 & 1093 \\ 8 & 136 & 5 ↑ \\ 8 & 17 & 0 ↑ \\ 8 & 2 & 1 ↑ \\ 0 & 2 ↑ \end{array}

Step 2: Convert the Fractional Part to Octal

To convert the fractional part

(0.21875)_{10}

to octal, we use successive multiplication by 8 and note down the integer parts.

\begin{aligned} 0.21875 \times 8 = 1.75000 P 1 ↓ \\ 0.75000 \times 8 = 6.0 P 6 ↓ \end{aligned}

Step 3: Combine the Integer and Fractional Parts

∴ (1093.21875)_{10} = (\underset{―}{2105.16})_{8}

(ii) Conversion of $(1693 \cdot 0628)_{10}$ to Hexadecimal

Step 1: Convert the Integer Part to Hexadecimal

To convert the integer part of

(1693)_{10}

to hexadecimal, we use successive division by 16 and note down the remainders.

\begin{array}{lrl} 16 & 1693 \\ 16 & 105 & D ↑ \\ 16 & 6 & 9 ↑ \\ 0 & 6 ↑ \end{array}

Step 2: Convert the Fractional Part to Hexadecimal

To convert the fractional part

(0.0628)_{10}

to hexadecimal, we use successive multiplication by 16 and note down the integer parts.

\begin{aligned} 0.628 \times 16 = 1.48 P 1 ↓ \\ 0.48 \times 16 = 0.768 P 0 ↓ \\ 0.768 \times 16 = 1.2288 P 1 ↓ \\ 0.2288 \times 16 = 3.6608 P 3 ↓ \\ 0.6608 \times 16 = 10.5728 P A ↓ \\ 0.5728 \times 16 = 9.1648 P 9 ↓ \end{aligned}

Step 3: Combine the Integer and Fractional Parts

∴ (1693.0628)_{10} = (\underset{―}{69 D .1013 A 9})_{16}

(ii) Express the Boolean function

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

in a product of maxterms form.

Answer:

Expressing the Boolean Function $F (x, y, z) = x y + x^{'} z$ in a Product of Maxterms Form

Given Function

The given Boolean function is:

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

Step 1: Apply Distributive Law

First, we apply the distributive law to expand the function:

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

Step 2: Apply the Identity Law

Next, we apply the identity law

x + x^{'} = 1

\begin{aligned} = (x + x^{'}) (y + x^{'}) (x + z) (y + z) \\ = (y + x^{'}) (x + z) (y + z) (because x + x^{'} = 1) \end{aligned}

Step 3: Apply the Null Law

We then apply the null law

A A^{'} = 0

\begin{aligned} = (x^{'} + y + z z^{'}) (x + y y^{'} + z) (x x^{'} + y + z) \\ = (x^{'} + y + z) (x^{'} + y + z^{'}) (x + y + z) (x + y^{'} + z) (since A A^{'} = 0) \end{aligned}

Step 4: Apply the Idempotent Law

We apply the idempotent law

A A = A

\begin{aligned} = (x^{'} + y + z) (x^{'} + y + z^{'}) (x + y + z) (x + y^{'} + z) (since A A = A) \end{aligned}

Final Expression

Finally, we have:

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

This is the required product of maxterm form for the given Boolean function

F (x, y, z)

Question:-05 (d) A particle at a distance

r

from the centre of force moves under the influence of the central force

F = - \frac{k}{r^{2}}

, where

k

is a constant. Obtain the Lagrangian and derive the equations of motion.

Answer:

Step 1: Define Kinetic Energy $T$

Let

(r, θ)

be the plane polar coordinates of the particle of mass

m

. The kinetic energy

T

of the particle is given by:

T = \frac{1}{2} m (r^{2} + r^{2} {\dot{θ}}^{2})

Step 2: Define Central Force $F$

The force

F

exerted by the particle is proportional to

\frac{1}{r^{2}}

F \propto \frac{1}{r^{2}} \Rightarrow F = - \frac{k}{r^{2}}

where

k

is the constant of proportionality.

Step 3: Define Potential Energy $V$

The potential energy

V

is related to the force

F

as:

F = - \frac{d v}{d r} \Rightarrow d v = - F d r

Integrating, we get:

\begin{aligned} V & = - \int_{\infty} F d r \\ = \int_{\infty} \frac{k}{r^{2}} d r \\ = {[- \frac{k}{r}]}_{\infty}^{r} \\ = - \frac{k}{r} \end{aligned}

Step 4: Define Lagrangian $L$

The Lagrangian

L

is given by

L = T - V

L = \frac{1}{2} m (r^{2} + r^{2} {\dot{θ}}^{2}) + \frac{k}{r}

Step 5: Partial Derivatives of $L$

We find the partial derivatives of

L

with respect to

r

\dot{r}

θ

, and

\dot{θ}

\frac{\partial L}{\partial \dot{r}} = m r, \frac{\partial L}{\partial r} = m r {\dot{θ}}^{2} - \frac{k}{r^{2}}; \frac{\partial L}{\partial θ} = 0, \frac{\partial L}{\partial \dot{θ}} = m r^{2} \dot{θ}

Step 6: Equation of Motion in $r$

Using the Lagrangian equation, the equation of motion in

r

is:

\begin{aligned} \frac{d}{d t} (\frac{\partial L}{\partial \dot{r}}) - \frac{\partial L}{\partial r} = 0 \\ \Rightarrow \frac{d}{d t} (m \dot{r}) - (m r {\dot{θ}}^{2} - \frac{k}{r^{2}}) = 0 \\ \Rightarrow m \ddot{r} - m r {\dot{θ}}^{2} + \frac{k}{r^{2}} = 0 \end{aligned}

Step 7: Equation of Motion in $θ$

The Lagrangian equation in

θ

is:

\begin{aligned} \frac{d}{d t} (\frac{\partial L}{\partial \dot{θ}}) - \frac{\partial L}{\partial θ} = 0 \\ \Rightarrow \frac{d}{d t} (m r^{2} \dot{θ}) = 0 \\ \Rightarrow m r^{2} \ddot{θ} + 2 m r \dot{r} \dot{θ} = 0 \end{aligned}

This concludes the derivation of the equations of motion for a particle moving under the influence of a central force

F = - \frac{k}{r^{2}}

Question:-05 (e) The velocity components of an incompressible fluid in spherical polar coordinates

(r, θ, ψ)

are

(2 M r^{- 3} \cos θ, M r^{- 2} \sin θ, 0)

, where

M

is a constant. Show that the velocity is of the potential kind. Find the velocity potential and the equations of the streamlines.

Answer:

Step 1: Identify Velocity Components
Given velocity components in spherical polar coordinates are:

q_{r} = 2 M r^{- 3} \cos θ

q_{θ} = M r^{- 3} \sin θ

, and

q_{ϕ} = 0

Step 2: Express Velocity Vector $q$
Express the velocity vector

q

in terms of unit vectors

e_{r}

e_{θ}

, and

e_{ϕ}

q = 2 M r^{- 3} \cos θ e_{r} + M r^{- 3} \sin θ e_{θ} + 0 e_{ϕ}

Step 3: Calculate Curl of $q$
Calculate the curl of

q

using the determinant form:

\begin{aligned} curl q = \frac{1}{r^{2} \sin^{2} θ} | \begin{array}{c} e_{r} & r e_{θ} & r \sin θ e_{ϕ} \\ \frac{\partial}{\partial r} & \frac{\partial}{\partial θ} & \frac{\partial}{\partial ϕ} \\ q_{r} & q_{θ} & q_{ϕ} \end{array} | \\ On simplification, we get curl q = 0 \end{aligned}

Hence, the flow is of the potential kind.

Step 4: Find the Velocity Potential $F$
Let

F (r, θ, ϕ)

be the required velocity potential. By definition, we have:

\begin{aligned} - \frac{\partial F}{\partial r} & = q_{r} = 2 M r^{- 3} \cos θ \\ - \frac{1}{r} \frac{\partial F}{\partial θ} & = q_{θ} = M r^{- 3} \sin θ \\ \frac{1}{r \sin θ} \frac{\partial F}{\partial ϕ} & = q_{ϕ} = 0 \end{aligned}

Integrate these equations to find

F

\begin{aligned} d F & = \frac{\partial F}{\partial r} d r + \frac{\partial F}{\partial θ} d θ + \frac{\partial F}{\partial ϕ} d ϕ \\ d F & = - (2 M r^{- 3} \cos θ) d r - (M r^{- 2} \sin θ) d θ + 0 d ϕ \\ d F & = d (M r^{- 2} \cos θ) \end{aligned}

Integrate to find

F = M^{- 2} \cos θ

(omitting the constant of integration).

Step 5: Equations of Streamlines
The equations of the streamlines can be found by solving the following expressions:

\begin{aligned} \frac{d r}{q_{r}} & = \frac{r d θ}{q_{θ}} = \frac{r \sin θ d ϕ}{q_{ϕ}} \end{aligned}

i.e, \frac{d r}{2 M r^{- 3} \cos θ} = \frac{r d θ}{M r^{- 3} \sin θ} = \frac{r \sin θ d ϕ}{0}

Given

d ϕ = 0

and

2 \cot θ d θ = (\frac{1}{r}) d r

, we can integrate to find the equation of the streamlines:

ϕ = C_{1} and r = C_{2} \sin^{2} θ

Here,

C_{1}

and

C_{2}

are arbitrary constants.

The equation

ϕ =

constant shows that the required streamlines lie in a plane passing through the axis of symmetry

θ = 0

Question:-06 (a) Solve the heat equation

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

subject to the conditions

\begin{aligned} u (0, t) = u (l, t) = 0 \\ u (x, 0) = x (l - x), 0 \leq x \leq l \end{aligned}

Answer:

Given the heat equation:

\frac{\partial u}{\partial t} = \frac{\partial^{2} u}{\partial x^{2}}, 0 < x < l, t > 0 (1)

with the following conditions:

\begin{aligned} u (0, t) = u (l, t) = 0 (2) \\ u (x, 0) = x (l - x), 0 \leq x \leq l (3) \end{aligned}

Step 1: Separation of Variables

Assume a solution of the form:

u (x, t) = X (x) T (t) (4)

Where

T (t)

and

X (x)

are functions of

t

and

x

respectively.

Step 2: Apply Separation of Variables to the Heat Equation

Substituting

(4)

into

(1)

, we get:

\frac{\partial u}{\partial t} = X T, \frac{\partial^{2} u}{\partial t^{2}} = X^{''} T

Therefore

X T = X^{''} T

\frac{T^{'}}{T} = \frac{X^{″}}{X} = μ (5)

This results in two separate ordinary differential equations:

X^{″} - μ X = 0 (6)

T - μ T = 0 (7)

Step 3: Boundary Conditions for $X (x)$

Using (2) and (4)

\equiv u (0, t) = X (0) T (t) = 0

And

u (l, t) = X (l) T (t) = 0

T (t) = 0

then

u (x, t) = 0

which is a contradictory to (3)

T (t) \neq 0

The boundary conditions

(2)

lead to

X (0) = 0

and

X (l) = 0

, denoted as

(8)

. These conditions will help determine the solutions for

X (x)

Step 4: Solve for $X (x)$ – Case 1: $μ = 0$

For

μ = 0

, we have:

X^{″} = 0 ⟹ X (x) = A x + B

Applying

(8)

, we find

A = 0

and

B = 0

, which leads to

X (x) = 0

, contradicting

(3)

. Thus,

μ = 0

is not a valid case.

Step 5: Solve for $X (x)$ – Case 2: $μ = λ^{2}$ ( $λ > 0$ )

For

μ = λ^{2}

, we have:

X^{″} - λ^{2} X = 0

This is a standard second-order linear differential equation, and its solutions are of the form:

X (x) = A e^{λ x} + B e^{- λ x}

Applying

(8)

, we find

A = 0

and

B = 0

, leading to

X (x) = 0

, which contradicts

(3)

. Thus,

μ = λ^{2}

is not a valid case.

Step 6: Solve for $X (x)$ – Case 3: $μ = - λ^{2}$ ( $λ \neq 0$ )

(6) \equiv X^{(x)} = A \cos λ x + B \sin λ x \to (10)

(8) \equiv X (0) = A = 0, X (l) = B \sin λ l = 0

There exist

B \neq 0

such that

\sin λ l = 0

\begin{aligned} \Rightarrow λ l = n π; n = 0, \pm 1, \pm 2 \\ \Rightarrow λ = \frac{n π}{l}; n = 1, 2, 3, \dots \\ (10) \equiv X (x) = B \sin (\frac{n π x}{l}); n = 1, 2, 3, \dots \end{aligned}

Now

T + λ^{2} T = 0

\Rightarrow T = C e^{- λ^{2} t} = e^{- \frac{λ^{2} x^{2} t}{l^{2}}} \to (12)

(11)

and (12)

\equiv X_{n} (x) = B_{n} \sin (\frac{n π x}{l}); n = 1, 2, 3, \dots

Step 7: Solve for $T (t)$

The time-dependent equation

(7)

can be solved as follows:

T_{n} (t) = C_{n} e^{- (\frac{n^{2} π^{2}}{l^{2}}) t} (12)

Step 8: General Solution for $u (x, t)$

The general solution for

u (x, t)

is a combination of the solutions obtained in steps 6 and 7:

u_{n} (x, t) = B_{n} \sin (\frac{n π x}{l}) e^{- (\frac{n^{2} π^{2}}{l^{2}}) t} for n = 1, 2, 3, \dots (13)

The more general form is:

u (x, t) = \sum_{n = 1}^{\infty} D_{n} \sin (\frac{n π x}{l}) e^{- (\frac{n^{2} π^{2}}{l^{2}}) t} (14)

\begin{aligned} Here D_{n} = \frac{2}{l} \int_{0} x (l - x) \sin (\frac{n π x}{l}) d x \\ = \frac{2}{l} \int_{0}^{l} (x l - x^{2}) \sin (\frac{n π x}{l}) d x \\ {= \frac{2}{l} [- \frac{(x l - x^{2}) l}{n π} \cos (\frac{n π x}{l}) + \int \frac{l - 2 x}{n π} l \cos (\frac{n π x}{l}) d x]]}_{0}^{l} \\ = \frac{2}{l} {[\frac{(x^{2} - x l) l}{n π} \cos \frac{n π x}{l} + \frac{l}{n π} [\frac{(l - 2 x) l}{n π} \sin (\frac{n π x}{l}) + \frac{l}{n π} \int 2 \sin \frac{n π x}{l} d x]]}_{0}^{l} \\ = \frac{2}{l} \times \frac{l}{n π} {[\frac{l (l - 2 x)}{n π} \sin (\frac{n π x}{l}) + \frac{l}{n π} \times \frac{l}{n π} (- 2) \cos (\frac{n π x}{l})]}_{0}^{l} l_{l}^{l} \\ = \frac{2}{n π} [\frac{l^{2}}{n π} \sin (\frac{n π x}{l}) - \frac{2 x l}{n π} \sin (\frac{n π x}{l}) - \frac{2 l^{2}}{n^{2} π^{2}} \cos (\frac{n π x}{l})]_{0} \\ D_{n} = \frac{2}{n π} [- \frac{2 l^{2}}{n^{2} π^{2}} (\cos n π - 1)] \end{aligned}

Step 9: Apply Initial Condition

Using

(2)

and

(13)

, we find that

D_{n}

is given by:

D_{n} = \frac{8 l^{2}}{n^{3} π^{3}} for n odd, D_{n} = 0 for n even (15)

Step 10: Final Solution

The final solution for the given heat equation with the initial and boundary conditions is:

u (x, t) = \sum_{n = 1}^{\infty} (- \frac{8 l^{2}}{n^{3} π^{3}}) \sin (\frac{n π x}{l}) e^{- (\frac{n^{2} π^{2}}{l^{2}}) t} (16)

This is the solution that satisfies the heat equation along with the specified initial and boundary conditions.

Question:-06 (b) Find a combinatorial circuit corresponding to the Boolean function

f (x, y, z) = [x \cdot (\bar{y} + z)] + y

and write the input/output table for the circuit.

Answer:

Introduction: In this problem, we need to find a combinatorial circuit corresponding to the given Boolean function and create the input/output table for the circuit.
Assumptions: We assume that the Boolean function is well-defined and uses standard Boolean operations.

Definition: A combinatorial circuit is a network of interconnected logic gates that perform a specific digital logic operation based on the input signals. The output of the circuit depends only on the current input values.

Method/Approach: To design the combinatorial circuit, we will first simplify the given Boolean function if possible, and then implement it using basic logic gates. After that, we will construct an input/output table for the circuit.

Work/Calculations:

The given Boolean function is:

f (x, y, z) = [x \cdot (\bar{y} + z)] + y

We can simplify the expression using the distributive law:

f (x, y, z) = x \bar{y} + x z + y

Now, we can implement this simplified Boolean function using basic logic gates:

Use a NOT gate to find the complement of $y, \bar{y}$ .
Use an AND gate to find the product of $x$ and $\bar{y}, x \bar{y}$ .
Use an AND gate to find the product of $x$ and $z, x z$ .
Use an OR gate to find the sum of $x \bar{y}$ and $x z$ .
Finally, use an OR gate to find the sum of the previous result and $y$ .
Now, let’s create the input/output table for the circuit:

$x$	$y$	$z$	$\bar{y}$	$x \bar{y}$	$x z$	$x \bar{y} + x z$	$f (x, y, z)$
0	0	0	1	0	0	0	0
0	0	1	1	0	0	0	0
0	1	0	0	0	0	0	1
0	1	1	0	0	0	0	1
1	0	0	1	1	0	1	1
1	0	1	1	1	1	1	1
1	1	0	0	0	0	0	1
1	1	1	0	0	1	1	1

Conclusion: We have designed a combinatorial circuit corresponding to the Boolean function

f (x, y, z) = x \bar{y} + x z + y

, and constructed the input/output table for the circuit.

Question:-06 (c) Find the moment of inertia of a right circular solid cone about one of its slant sides (generator) in terms of its mass

M

, height

h

and the radius of base as

a

Answer:

Step 1: Establish Mass and Density Relationship
Let

M

be the mass of a right circular cone with height

h

and base radius

a

. If

α

is the semi-vertical angle and

ρ

is the density of the cone, then we have the relationship:

M = \frac{1}{3} ρ π h^{3} \tan^{2} α \to (1)

Step 2: Coordinate System Setup
Take the vertex of the cone as the origin,

x

-axis along the axis OD of the cone, and

y

-axis perpendicular to OD. The slant side OA makes an angle

α

with OX.

Step 3: Express Moment of Inertia Components
The moment of inertia of the cone about

O A

can be expressed as:

I_{O A} = A \cos^{2} α + B \sin^{2} α - F \sin 2 α \to (2)

Where:

A

is the moment of inertia of the cone about OX,

B

M \cdot I

. of the cone about

OY

F

P \cdot I

. of the cone about

OX

and

OY

Step 4: Calculate $A$ (Moment of Inertia about OX)
(i) Moment of Inertia (

I_{O X}

) of the elementary disc about OX:

\begin{aligned} I_{O X} & = \frac{1}{2} δ m \cdot c p^{2} = \frac{1}{2} ρ π x^{4} \tan^{4} α δ x \end{aligned}

(ii) Moment of Inertia (

A

) of the cone about OX:

\begin{aligned} A & = \int_{0}^{h} \frac{1}{2} ρ π x^{4} \tan^{4} α d x \\ = \frac{1}{10} ρ π h^{5} \tan^{4} α \\ = \frac{3 M}{10} h^{2} \tan^{2} α (from equation (1)) \\ = \frac{3}{10} M a^{2} (because \tan α = \frac{a}{h}) \end{aligned}

Step 5: Calculate $B$ (Moment of Inertia about $OY$ )
(i) Moment of Inertia (

I_{O Y}

) of the elementary disc about parallel diameter PQ and about OY:

\begin{aligned} I_{O Y} & = \frac{1}{4} δ m \cdot c p^{2} + δ m \cdot O C^{2} \\ = (\frac{1}{4} x^{2} \tan^{2} α + x^{2}) \cdot ρ π x^{2} \tan^{2} α δ x \\ = \frac{1}{4} (\tan^{2} α + 4) ρ π x^{4} \tan^{2} α δ x \end{aligned}

(ii) Moment of Inertia (

B

) of the cone about OY:

\begin{aligned} B & = \int_{0}^{h} \frac{1}{4} (\tan^{2} α + 4) ρ π x^{4} \tan^{2} α d x \\ = \frac{1}{20} ρ π h^{5} (\tan^{2} α + 4) \tan^{2} α \\ = \frac{3}{20} M h^{2} (\tan^{2} α + 4) (from equation (1)) \\ = \frac{3}{20} M (a^{2} + 4 h^{2}) (because \tan α = \frac{a}{h}) \end{aligned}

Step 6: Calculate $F$ (Moment of Inertia about $OX$ and $OY$ )
(i)

F = P \cdot I

. of the cone about

OX

and

OY = 0

, by symmetry about

OX

.
(ii)

\cos α = \frac{O D}{O A} = \frac{O D}{\sqrt{O D^{2} + A D^{2}}} = \frac{h}{\sqrt{h^{2} + a^{2}}}

(iii)

\sin α = \frac{A D}{O A} = \frac{a}{\sqrt{h^{2} + a^{2}}}

From (2), the Moment of Inertia of the cone about the slant side is given by:

= \frac{3 M a^{2}}{20} \times \frac{6 h^{2} + a^{2}}{h^{2} + a^{2}}

Conclusion:

In conclusion, we have determined the moment of inertia of a right circular solid cone about one of its slant sides (generator) in terms of its mass

M

, height

h

, and the radius of the base

a

. The moment of inertia

I_{O A}

about the generator

O A

is given by:

I_{O A} = \frac{3 M a^{2}}{20} \times \frac{6 h^{2} + a^{2}}{h^{2} + a^{2}}

This expression represents the moment of inertia of the cone about the generator

O A

, providing a mathematical description of its rotational behavior with respect to an axis along the generator.

Question:-07 (a) Find the general solution of the partial differential equation

(D^{2} + D D^{'} - 6 D^{' 2}) z = x^{2} \sin (x + y)

where

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

and

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

Answer:

Given the partial differential equation:

(D^{2} + D D^{'} - 6 D^{' 2}) z = x^{2} \sin (x + y) (1)

Where

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

and

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

Step 1: Factoring the Differential Operator

The given equation can be factored as:

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

Step 2: Auxiliary Equation and Solutions for $m$

The auxiliary equation for this factored operator is:

(m + 3) (m - 2) = 0

Solving for

m

, we find

m = - 3

and

m = 2

Step 3: Complementary Function (C.F.)

The complementary function is given by:

C . F = ϕ_{1} (y - 3 x) + ϕ_{2} (y + 2 x)

where

ϕ_{1}

and

ϕ_{2}

are arbitrary functions.

Step 4: Particular Integral (P.I.)

\begin{aligned} P. I. & = \frac{1}{D^{2} + D D^{'} - 6 D^{' 2}} x^{2} \sin (x + y) \\ = Im [\frac{1}{(D - 2 D^{'}) (D + 3 D^{'})} x^{2} e^{i (x + y)}] \\ = Im [e^{i (x + y)} \frac{1}{(D + i - 2 D^{'} - 2 i) (D + i + 3 D^{'} + 3 i)} x^{2}] \\ = Im [\frac{1}{4 i} e^{i (x + y)} \frac{1}{(D - 2 D^{'} - i)} {(1 + \frac{D + 3 D^{'}}{4 i})}^{- 1} x^{2}] \\ = Im [\frac{1}{4 i} e^{i (x + y)} \frac{1}{(D - 2 D^{'} - i)} (1 - \frac{D + 3 D^{'}}{4 i} - \frac{D^{2} + 6 D D^{'} + D^{' 2}}{16} + \dots) x^{2}] \\ = Im [\frac{1}{4 i} e^{i (x + y)} \frac{1}{(D - 2 D^{'} - i)} (x^{2} - \frac{2 x}{4 i} - \frac{2}{16})] \\ = Im [\frac{1}{4} e^{i (x + y)} {1 + i (D - 2 D^{'})}^{- 1} (x^{2} + \frac{i}{2} x - \frac{1}{8})] \\ = Im [\frac{1}{4} e^{i (x + y)} {1 - i (D - 2 D^{'}) - (D^{2} - 4 D D^{'} + 4 D^{' 2}) + \dots} (x^{2} + \frac{i}{2} x - \frac{1}{8})] \\ = Im [\frac{1}{4} e^{i (x + y)} {x^{2} + \frac{i}{2} x - \frac{1}{8} - i (2 x + \frac{i}{2}) - 2}] \\ = \frac{1}{4} Im [{\cos (x + y) + i \sin (x + y)} (x^{2} - \frac{13}{8} - i \frac{3}{2} x)] \\ = \frac{1}{4} [(x^{2} - \frac{13}{8}) \sin (x + y) - \frac{3}{2} x \cos (x + y)] . \end{aligned}

General Solution

The general solution for the given partial differential equation is:

z = ϕ_{1} (y - 3 x) + ϕ_{2} (y + 2 x) + (\frac{x^{2}}{4} - \frac{13}{32}) \sin (x + y) - \frac{3 x}{8} \cos (x + y)

This solution satisfies the given partial differential equation.

Question:-07 (b) The velocity of a train which starts from rest is given by the following table, the time being reckoned in minutes from the start and the velocity in

km /

hour :

$t$ (minutes)	2	4	6	8	10	12	14	16	18	20
$v (km /$ hour $)$	16	$28 \cdot 8$	40	$46 \cdot 4$	$51 \cdot 2$	32	$17 \cdot 6$	8	$3 \cdot 2$	0

Using Simpson’s

\frac{1}{3} rd

rule, estimate approximately in

km

the total distance run in 20 minutes.

Answer:

Estimation of Total Distance Using Simpson’s 1/3 Rule:

Given the velocity of a train as a function of time:

$t$ (minutes)	2	4	6	8	10	12	14	16	18	20
$v (km /$ hour $)$	16	$28 \cdot 8$	40	$46 \cdot 4$	$51 \cdot 2$	32	$17 \cdot 6$	8	$3 \cdot 2$	0

We have the corresponding values of

x

and

y

$x$	2	4	6	8	10	12	14	16	18	20
$y$	16	28.8	40	46.4	51.2	32	17.6	8	3.2	0

Step 1: Application of Simpson’s 1/3 Rule

Using Simpson’s

\frac{1}{3}

Rule for numerical integration, we estimate the total distance covered in 20 minutes as follows:

\begin{aligned} \int y d x & = \frac{h}{3} [(y_{0} + y_{9}) + 4 (y_{1} + y_{3} + y_{5} + y_{7}) + 2 (y_{2} + y_{4} + y_{6} + y_{8})] \\ \int y d x & = \frac{2}{3} [(16 + 0) + 4 \times (28.8 + 46.4 + 32 + 8) + 2 \times (40 + 51.2 + 17.6 + 3.2)] \\ \int y d x & = \frac{2}{3} [(16 + 0) + 4 \times (115.2) + 2 \times (112)] \\ \int y d x & = 467.2 \end{aligned}

Step 2: Conclusion

Using Simpson’s

\frac{1}{3}

Rule, the estimated total distance covered by the train in 20 minutes is approximately 467.2 kilometers.

Question:-07 (c) Two point vortices each of strength

k

are situated at

(\pm a, 0)

and a point vortex of strength

- \frac{k}{2}

is situated at the origin. Show that the fluid motion is stationary and also find the equations of streamlines. If the streamlines, which pass through the stagnation points, meet the

x

-axis at

(\pm b, 0)

, then show that

3 \sqrt{3} {(b^{2} - a^{2})}^{2} = 16 a^{3} b

Answer:

Analysis of Fluid Motion with Vortices:

Given the configuration of vortices in the fluid:

Two point vortices with strength $k$ are located at $(\pm a, 0)$ .
A point vortex with strength $- \frac{k}{2}$ is situated at the origin.

Step 1: Complex Potential

The complex potential of the fluid motion is given by:

W = \frac{i k}{2 π} \log (z - a) + \frac{i K}{2 π} \log (z + a) - \frac{i K}{4 π} \log z

Simplified as:

W = \frac{i K}{2 π} [\log (z^{2} - a^{2}) - \frac{1}{2} \log z] \to (1)

Step 2: Vortex at Point A

For the motion of the vortex at point A, the complex potential is:

\begin{aligned} W^{'} = W - \frac{i K}{2 π} \log (z - a) \\ = \frac{i K}{2 π} [\log (z + a) - \frac{1}{2} \log z] \end{aligned}

Calculating

- \frac{d W^{'}}{d z}

z = a

- \frac{i K}{2 π} {[\frac{1}{z + a} - \frac{1}{2 z}]}_{z = a} = 0

This implies

u_{A} - i v_{A} = 0

, indicating that the vortex at A is at rest. The same holds for vortices at O and B.

Step 3: Stationary Fluid Motion

Hence, the fluid motion is stationary, which establishes the first part of the problem.

Step 4: Equations of Streamlines

To determine the equations of streamlines, we start with:

\begin{aligned} ϕ + i Ψ = \frac{i K}{2 π} [\log (z^{2} - a^{2}) - \frac{1}{2} \log z], where b y (1) \\ = \frac{i K}{2 π} [\log (x^{2} - y^{2} - a^{2} + 2 i x y) - \frac{1}{2} \log (x + i y)] \end{aligned}

Simplifying, we get:

Ψ = \frac{K}{4 π} [\log {{(x^{2} - y^{2} - a^{2})}^{2} + 4 x^{2} y^{2}} - \frac{1}{2} \log (x^{2} + y^{2})]

Streamlines are given by

Ψ =

constant.

Step 5: Equation of Streamlines

That is

\log [\frac{{(x^{2} - y^{2} - a^{2})}^{2} + 4 x^{2} y^{2}}{\sqrt{x^{2} + y^{2}}}] = \log c

{(x^{2} - y^{2} - a^{2})}^{2} + 4 x^{2} y^{2} = c {(x^{2} + y^{2})}^{\frac{1}{2}}

{(x^{2} - y^{2})}^{2} + a^{4} - 2 a^{2} (x^{2} - y^{2}) + 4 x^{2} y^{2} = c {(x^{2} + y^{2})}^{\frac{1}{2}}

{(x^{2} + y^{2})}^{2} - 2 a^{2} (x^{2} - y^{2}) + a^{4} = c {(x^{2} + y^{2})}^{\frac{1}{2}} \to (2)

Step 6: Stagnation Points

To find stagnation points, we set:

\frac{d W}{d z} = 0, i.e. \frac{2 z}{z^{2} - a^{2}} - \frac{1}{2 z} = 0, at b y (1)

3 z^{2} + a^{2} = 0 or z = \pm \frac{i a}{\sqrt{3}}

Solving for

z

, we find stagnation points at

(0, \frac{a}{\sqrt{3}})

and

(0, - \frac{a}{\sqrt{3}})

Step 7: Meeting the $x$ -axis

The streamlines given by (2) will pass through the stagnation points only if:

{(\frac{a^{2}}{3})}^{2} - 2 a^{2} (0 - \frac{a^{2}}{3}) + a^{4} = c {(0 + \frac{a^{2}}{3})}^{\frac{1}{2}}

This leads to the value of

c

c = 16 \frac{a^{3}}{3 \sqrt{3}}

Step 8: Relationship between $a$ and $b$

The streamlines (2) will pass through

(\pm b, 0)

if:

b^{4} - 2 a^{2} (b^{2} - 0) + a^{4} = c {(b^{2} + 0)}^{\frac{1}{2}} = b \frac{16 a^{3}}{3 \sqrt{3}}

b^{4} - 2 a^{2} b^{2} + a^{4} = (\frac{16 a^{3} b}{3 \sqrt{3}})

Which simplifies to:

3 \sqrt{3} {(b^{2} - a^{2})}^{2} = 16 a^{3} b

This concludes the solution.

Question:-08 (a) Reduce the following partial differential equation to a canonical form and hence solve it :

y u_{x x} + (x + y) u_{x y} + x u_{y y} = 0

Answer:

Partial Differential Equation and Canonical Form:

We are given the partial differential equation:

y u_{x x} + (x + y) u_{x y} + x u_{y y} = 0

Step 1: Canonical Form Analysis

We start by defining

y r + (x + y) s + x t = 0

as equation (1). We’ll compare this with the general form

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

Here, we have

R = y

S = x + y

, and

T = x

To determine the nature of (1), we evaluate

S^{2} - 4 R T

S^{2} - 4 R T = (x + y)^{2} - 4 x y = (x - y)^{2} > 0 for x \neq y

This shows that (1) is hyperbolic.

Step 2: Characteristic Equations

Next, we consider the characteristic equation for (1):

R λ^{2} + S λ + T = 0

Substituting the values of

R

S

, and

T

, we get:

y λ^{2} + (x + y) λ + x = 0

This quadratic equation has solutions

λ = - 1

and

λ = - \frac{x}{y}

The corresponding characteristic equations are:

\frac{d y}{d x} - 1 = 0

and

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

Integrating these, we obtain:

y - x = c_{1} (for the first equation)

and

\frac{y^{2}}{2} - \frac{x^{2}}{2} = c_{2} (for the second equation)

Step 3: Reducing to Canonical Form

To reduce equation (1) to its canonical form, we introduce new variables:

u = y - x and V = \frac{y^{2}}{2} - \frac{x^{2}}{2} (equation 2)

Now, we express partial derivatives in terms of

u

and

V

\begin{aligned} \begin{aligned} p & = \frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial z}{\partial v} \frac{\partial v}{\partial x} = - (\frac{\partial z}{\partial u} + x \frac{\partial z}{\partial v}), using (2) \to (3) \\ q & = \frac{\partial z}{\partial y} = \frac{\partial z}{\partial u} \frac{\partial u}{\partial y} + \frac{\partial z}{\partial v} \frac{\partial v}{\partial y} = \frac{\partial z}{\partial u} + y \frac{\partial z}{\partial v}, using (2) \to (4) \\ r & = \frac{\partial^{2} z}{\partial x^{2}} = \frac{\partial}{\partial x} (\frac{\partial z}{\partial x}) = - \frac{\partial}{\partial x} (\frac{\partial z}{\partial u}) - \frac{\partial}{\partial x} (x \frac{\partial z}{\partial v}) using (3) \\ = - \frac{\partial}{\partial x} (\frac{\partial z}{\partial u}) - [x \frac{\partial}{\partial x} (\frac{\partial z}{\partial v}) + \frac{\partial z}{\partial v}] = - \frac{\partial}{\partial x} (\frac{\partial z}{\partial u}) - x \frac{\partial}{\partial x} (\frac{\partial z}{\partial v}) - \frac{\partial z}{\partial v} \\ \Rightarrow - [\frac{\partial}{\partial u} (\frac{\partial z}{\partial u}) \frac{\partial u}{\partial x} + \frac{\partial}{\partial v} (\frac{\partial z}{\partial u}) \frac{\partial v}{\partial x}] - x [\frac{\partial}{\partial u} (\frac{\partial z}{\partial v}) \frac{\partial u}{\partial x} + \frac{\partial}{\partial v} (\frac{\partial z}{\partial v}) \frac{\partial v}{\partial x}] - \frac{\partial z}{\partial v} \\ = - (- \frac{\partial^{2} z}{\partial u^{2}} - x \frac{\partial^{2} z}{\partial v \partial u}) - x (- \frac{\partial^{2} z}{\partial u \partial v} - x \frac{\partial^{2} z}{\partial v^{2}}) - \frac{\partial z}{\partial v} using (2) \\ v & = \frac{\partial^{2} z}{\partial u^{2}} + 2 z \frac{\partial^{2} z}{\partial u \partial v} + x^{2} \frac{\partial^{2} z}{\partial v^{2}} - \frac{\partial z}{\partial v} \end{aligned} \\ \begin{aligned} Now, t = \frac{\partial^{2} z}{\partial y^{2}} = \frac{\partial}{\partial y} (\frac{\partial z}{\partial y}) = \frac{\partial}{\partial y} (\frac{\partial z}{\partial u} + y \frac{\partial z}{\partial v}) \\ = \frac{\partial}{\partial y} (\frac{\partial z}{\partial u}) + \frac{\partial}{\partial y} (y \frac{\partial z}{\partial v}) b y (4) \\ = \frac{\partial}{\partial y} (\frac{\partial z}{\partial u}) + y \frac{\partial}{\partial y} (\frac{\partial z}{\partial v}) + \frac{\partial z}{\partial v} \\ = \frac{\partial}{\partial u} (\frac{\partial z}{\partial u}) (\frac{\partial u}{\partial y}) + \frac{\partial}{\partial v} (\frac{\partial z}{\partial u}) \frac{\partial v}{\partial y} + y {\frac{\partial}{\partial u} (\frac{\partial z}{\partial v}) \frac{\partial u}{\partial y} + \frac{\partial}{\partial v} (\frac{\partial z}{\partial v}) \frac{\partial v}{\partial y}} + \frac{\partial z}{\partial v} \\ \Rightarrow \frac{\partial^{2} z}{\partial u^{2}} + y \frac{\partial^{2} z}{\partial u \partial v} + y (\frac{\partial^{2} z}{\partial u \partial v} + y \frac{\partial^{2} z}{\partial v^{2}}) + \frac{\partial z}{\partial v} \\ t = \frac{\partial^{2} z}{\partial u^{2}} + 2 y \frac{\partial^{2} z}{\partial u \partial v} + y^{2} \frac{\partial^{2} z}{\partial v^{2}} + \frac{\partial z}{\partial v} \end{aligned} \\ \begin{aligned} Also s = \frac{\partial^{2} z}{\partial x \partial y} = \frac{\partial}{\partial x} (\frac{\partial z}{\partial y}) = \frac{\partial}{\partial x} (\frac{\partial z}{\partial u} + y \frac{\partial z}{\partial v}) using \\ = \frac{\partial}{\partial x} (\frac{\partial z}{\partial u}) + \frac{\partial}{\partial x} (y \frac{\partial z}{\partial v}) = \frac{\partial}{\partial x} (\frac{\partial z}{\partial u}) + y \frac{\partial}{\partial x} (\frac{\partial z}{\partial v}) \\ = \frac{\partial}{\partial u} (\frac{\partial z}{\partial u}) \frac{\partial u}{\partial x} + \frac{\partial}{\partial v} (\frac{\partial z}{\partial v}) \frac{\partial v}{\partial x} + y {\frac{\partial}{\partial u} (\frac{\partial z}{\partial v}) \frac{\partial u}{\partial x} + \frac{\partial}{\partial v} (\frac{\partial z}{\partial v}) \frac{\partial v}{\partial x}} \\ \Rightarrow - \frac{\partial^{2} z}{\partial u^{2}} - (x + y) \frac{\partial^{2} z}{\partial u \partial v} - x y \frac{\partial^{2} z}{\partial v^{2}} \to (7) \end{aligned} \end{aligned}

Using (5), (6) and (7) in (1) we get

\begin{aligned} y (\frac{\partial^{2} z}{\partial u^{2}} + (2 x) \frac{\partial^{2} z}{\partial u \partial v} + (x^{2}) \frac{\partial^{2} z}{\partial v^{2}} - \frac{\partial z}{\partial v}) + (x + y) {- \frac{\partial^{2} z}{\partial u^{2}} - (x + y) \frac{\partial^{2} z}{\partial u \partial v} - (x y) \frac{\partial^{2} z}{\partial v^{2}}} + {\frac{\partial^{2} z}{\partial u^{2}} + 2 y \frac{\partial^{2} z}{\partial u \partial v} + y^{2} \frac{\partial^{2} z}{\partial v^{2}} - \frac{\partial z}{\partial v}} = 0 \\ Or {u x y - (x + y)} \frac{\partial^{2} z}{\partial u \partial v} - y \frac{\partial z}{\partial v} + x \frac{\partial z}{\partial v} = 0 \end{aligned}

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

Step 4: Canonical Form Equation

u^{2} \frac{\partial^{2} z}{\partial u \partial v} + u \frac{\partial z}{\partial v} = 0

u \frac{\partial^{2} z}{\partial u \partial v} + \frac{\partial z}{\partial v} = 0 \to

(8)
[ since

u \neq 0

and

y - x = u, b y (2)

]
(8) is the required canonical form of (1)
Solution of (8) multiplying both sides of ( 8 ) by v we get

u v (\frac{\partial^{2} z}{\partial u \partial v}) + v (\frac{\partial z}{\partial v}) = 0 or (u v D D^{'} + v D^{'}) z = 0 \to (9)

Where

D \equiv \frac{\partial}{\partial u}

and

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

. to reduce (9) in to linear equation with constant coefficients, we take new variable

x

and

y

as follows:

Step 5: Linear Equation with Constant Coefficients

To simplify further, we introduce new variables

x

and

y

as follows:

Let

u = e^{x}

and

v = e^{y}

so that

x = \log u, y = \log v \to (10)

Let

D_{1} \equiv \frac{\partial}{\partial x}

and

D_{1}^{'} \equiv \frac{\partial}{\partial y}

then (9) reduces to

(\begin{array}{ll} D_{1} D_{1}^{'} + D_{1}^{'} \end{array}) z = 0 or D_{1}^{'} (D + 1) z = 0

Step 6: General Solution

The general solution of the linear equation with constant coefficients is:

z = e^{- x} ϕ_{1} (y) + ϕ_{2} (x) = u^{- 1} ϕ_{1} (\log v) + ϕ_{2} (\log u)

z = u^{- 1} Ψ_{1} (v) + Ψ_{2} (u) = (y - x)^{- 1} Ψ_{1} (y^{2} - x^{2}) + Ψ_{2} (y - x)

Here,

Ψ_{1}

and

Ψ_{2}

are arbitrary functions.

Question:-08 (b) Using Runge-Kutta method of fourth order, solve the differential equation

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

with

y (0) = 1

, at

x = 0 \cdot 2

. Use four decimal places for calculation and step length

0 \cdot 1

Answer:

Given Problem:

We are given the initial value problem:

y^{'} = x + y^{2}, y (0) = 1, h = 0.1, Find y (0.2)

Fourth-Order Runge-Kutta Method:

We will solve this problem using the fourth-order Runge-Kutta method.

Step 1: Initial Evaluation

Calculate $k_{1}$ using the formula:

$k_{1} = h \cdot f (x_{0}, y_{0}) = 0.1 \cdot f (0, 1) = 0.1 \cdot 1 = 0.1$
Calculate $k_{2}$ using the formula:

$k_{2} = h \cdot f (x_{0} + \frac{h}{2}, y_{0} + \frac{k_{1}}{2}) = 0.1 \cdot f (0.05, 1.05) = 0.1 \cdot 1.1525 = 0.1153$
Calculate $k_{3}$ using the formula:

$k_{3} = h \cdot f (x_{0} + \frac{h}{2}, y_{0} + \frac{k_{2}}{2}) = 0.1 \cdot f (0.05, 1.0576) = 0.1 \cdot 1.1686 = 0.1169$
Calculate $k_{4}$ using the formula:

$k_{4} = h \cdot f (x_{0} + h, y_{0} + k_{3}) = 0.1 \cdot f (0.1, 1.1169) = 0.1 \cdot 1.3474 = 0.1347$
Now, calculate $y_{1}$ using the formula for the fourth-order Runge-Kutta method:

$y_{1} = y_{0} + \frac{1}{6} (k_{1} + 2 k_{2} + 2 k_{3} + k_{4})$

Substituting the values:

$y_{1} = 1 + \frac{1}{6} (0.1 + 2 \cdot 0.1153 + 2 \cdot 0.1169 + 0.1347) = 1.1165$

Step 2: Iteration for $y (0.2)$

Now, we will use the newly calculated

(x_{1}, y_{1})

in place of

(x_{0}, y_{0})

and repeat the process to find

y (0.2)

Calculate $k_{1}$ using the formula:

$k_{1} = h \cdot f (x_{1}, y_{1}) = 0.1 \cdot f (0.1, 1.1165) = 0.1 \cdot 1.3466 = 0.1347$
Calculate $k_{2}$ using the formula:

$k_{2} = h \cdot f (x_{1} + \frac{h}{2}, y_{1} + \frac{k_{1}}{2}) = 0.1 \cdot f (0.15, 1.1838) = 0.1 \cdot 1.5514 = 0.1551$
Calculate $k_{3}$ using the formula:

$k_{3} = h \cdot f (x_{1} + \frac{h}{2}, y_{1} + \frac{k_{2}}{2}) = 0.1 \cdot f (0.15, 1.1941) = 0.1 \cdot 1.5758 = 0.1576$
Calculate $k_{4}$ using the formula:

$k_{4} = h \cdot f (x_{1} + h, y_{1} + k_{3}) = 0.1 \cdot f (0.2, 1.2741) = 0.1 \cdot 1.8233 = 0.1823$
Now, calculate $y_{2}$ using the formula for the fourth-order Runge-Kutta method:

$y_{2} = y_{1} + \frac{1}{6} (k_{1} + 2 k_{2} + 2 k_{3} + k_{4})$

Substituting the values:

$y_{2} = 1.1165 + \frac{1}{6} (0.1347 + 2 \cdot 0.1551 + 2 \cdot 0.1576 + 0.1823) = 1.2736$

Final Result:

Therefore, the solution to the initial value problem

y^{'} = x + y^{2}, y (0) = 1

with

h = 0.1

y (0.2) = 1.2736

Question:-08 (c) Verify that

w = i k \log {(z - i a) / (z + i a)}

is the complex potential of a steady flow of fluid about a circular cylinder, where the plane

y = 0

is a rigid boundary. Find also the force exerted by the fluid on unit length of the cylinder.

Answer:

Complex Potential and Streamlines:

We have the complex potential

w = ϕ + i Ψ

given by:

w = i k \log | \frac{z - i a}{z + i a} | (1)

Hence, we can extract the stream function

Ψ

from

w

as follows:

Ψ = k \log | \frac{z - i a}{z + i a} |

This leads to streamlines defined by:

Ψ = constant = K λ, say

That is,

| \frac{z - i a}{z + i a} | = λ

, which can be expressed as:

| z - i a | = λ | z + i a | (2)

Equation (2) describes non-intersecting co-axial circles with

z = \pm i a

as limiting points. In particular, for

λ = 1

, (2) represents the straight line:

| z - i a | = | z + i a |

i.e. | x + i (y - a) | = | x + i (y + a) |,

Simplifying this, we get:

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

This demonstrates that

y = 0

is a rigid boundary.

Derivative of $w$ with Respect to $z$ :

From equation (1), we can calculate the derivative of

w

with respect to

z

w = i k {\log (z - i a) - \log (z + i a)}

\frac{d w}{d z} = i k (\frac{1}{z - i a} - \frac{1}{z + i a}) = \frac{2 k a}{z^{2} + a^{2}} (3)

Force Exerted by the Fluid:

Consider a circular section

C

of the cylinder and the rigid plane. If the pressure thrusts on the circular disc are represented by

(x, y)

, we can use Blasius’ theorem to calculate

X

and

Y

X - i Y = \frac{1}{2} i ρ \int_{c} {(\frac{d w}{d z})}^{2} d z = 2 k^{2} a^{2} ρ i \int_{c} \frac{d z}{{(z^{2} + i^{2})}^{2}} (4)

By Cauchy’s residue theorem, we have:

\int_{c} \frac{d z}{{(z^{2} + a^{2})}^{2}} = 2 π i \times (sum of the residues)

Therefore, equation (4) becomes:

X - i Y = - 4 k^{2} a^{2} ρ π \times (sum of the residues) (5)

We proceed to find the residues of

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

. The only pole of

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

is at

z = \pm i a

, but only

z = i a

lies within the boundary

C

. Hence, we calculate the residue at

z = i a

\begin{aligned} \frac{1}{z^{2} + a^{2}} = \frac{1}{(z - i a) (z + i a)} = \frac{1}{2 i a} (\frac{1}{z - i a} - \frac{1}{z + i a}) \\ \frac{1}{{(z^{2} + a^{2})}^{2}} = - \frac{1}{4 a^{2}} {\frac{1}{(z - i a)^{2}} + \frac{1}{(z + i a)^{2}} - \frac{2}{(z - i a) (z + i a)}} \\ = - \frac{1}{4 a^{2}} {\frac{1}{(z - i a)^{2}} + \frac{1}{(z + i a)^{2}} - \frac{1}{2 i a} (\frac{1}{z - i a} - \frac{1}{z + i a})} \end{aligned}

Hence, the residue of

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

z = i a

\frac{1}{8 i a^{3}}

Therefore, equation (5) becomes:

X - i Y = - (4 k^{2} a^{2} ρ π) \times (\frac{1}{8 i a^{3}}) = (\frac{π ρ k^{2}}{2 a}) i

This implies that

X = 0

and

Y = - (\frac{π ρ k^{2}}{2 a})

. Thus, the fluid exerts a downward force on the cylinder, and the numerical value of this force per unit length is

(\frac{π ρ k^{2}}{2 a})

Free UPSC Mathematics Optional Paper-2 2022 Solutions: View Online | UPSC Maths Solution | IAS Maths Solution

Introduction

Work/Calculations

Step 1: Define a Bijective Function f : G → G ′ f : G → G ′ f:G rarrG^(‘)f: G \to G’f:G→G′

Step 2: Show that f f fff Preserves the Group Operation

Conclusion

Complex Contour Integral

Finding the Poles

Calculating the Residues

Residue at z = i a z = i a z=iaz = iaz=ia

Residue at z = i b z = i b z=ibz = ibz=ib

Sum of Residues

Applying Cauchy’s Residue Theorem

Evaluating the Integral on the Circular Arc

Final Answer

Optimal Solution of the Given Problem

Dual Problem

Optimal Solution of the Dual Problem from the Optimal Table of the Given Problem

Initial Definitions and Equation Transformation

Calculating Partial Derivatives

Final Differential Equation

Conclusion

Conversion of Numbers to Different Number Systems

(i) Conversion of ( 1093.21875 ) 10 ( 1093.21875 ) 10 (1093.21875)_(10)(1093.21875)_{10}(1093.21875)10 to Octal

Step 1: Convert the Integer Part to Octal

Step 2: Convert the Fractional Part to Octal

Step 3: Combine the Integer and Fractional Parts

(ii) Conversion of ( 1693 ⋅ 0628 ) 10 ( 1693 ⋅ 0628 ) 10 (1693*0628)_(10)(1693 \cdot 0628)_{10}(1693⋅0628)10 to Hexadecimal

Step 1: Convert the Integer Part to Hexadecimal

Step 2: Convert the Fractional Part to Hexadecimal

Step 3: Combine the Integer and Fractional Parts

Expressing the Boolean Function F ( x , y , z ) = x y + x ′ z F ( x , y , z ) = x y + x ′ z F(x,y,z)=xy+x^(‘)zF(x, y, z) = xy + x’zF(x,y,z)=xy+x′z in a Product of Maxterms Form

Given Function

Step 1: Apply Distributive Law

Step 2: Apply the Identity Law

Step 3: Apply the Null Law

Step 4: Apply the Idempotent Law

Final Expression

Step 1: Define Kinetic Energy T T TTT

Step 2: Define Central Force F F FFF

Step 3: Define Potential Energy V V VVV

Step 4: Define Lagrangian L L LLL

Step 5: Partial Derivatives of L L LLL

Step 6: Equation of Motion in r r rrr

Step 7: Equation of Motion in θ θ theta\thetaθ

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Step 1: Define a Bijective Function $f : G \to G^{'}$

Step 2: Show that $f$ Preserves the Group Operation

Residue at $z = i a$

Residue at $z = i b$

(i) Conversion of $(1093.21875)_{10}$ to Octal

(ii) Conversion of $(1693 \cdot 0628)_{10}$ to Hexadecimal

Expressing the Boolean Function $F (x, y, z) = x y + x^{'} z$ in a Product of Maxterms Form

Step 1: Define Kinetic Energy $T$

Step 2: Define Central Force $F$

Step 3: Define Potential Energy $V$

Step 4: Define Lagrangian $L$

Step 5: Partial Derivatives of $L$

Step 6: Equation of Motion in $r$

Step 7: Equation of Motion in $θ$