Free BECC-102 Solved Assignment | July 2023-January 2024 | Mathematical Methods in Economics-I | IGNOU

BECC-102 Solved Assignment

A monopolist faces the demand curve $Q = 60 - P / 2$ . The cost function is $C = Q^{2}$ . Find the output that maximises this monopolist’s profits. What are the prices at profits and that output? Find the elasticity of demand at the profit maximising output.

A monopolist firm has the following total revenue and total cost functions:

\begin{array}{ll} R = - {mQ}^{2} + nQ & (m, n > 0) \\ C = a Q^{2} + bQ + c & (a, b, c > 0) \end{array}

Suppose that the government plans to levy an excise tax on the product of this firm and wishes to maximise the total tax revenue

T

from this source. What tax rate

t

(rupees per unit of output) should the government choose?
2. A firm in a perfectly competitive market has the following cost function:

C = 1 / 3 q^{3} - 5 q^{2} + 30 q + 10

If the market-clearing price is 6 , obtain the profit maximising level of output.
OR
Given the demand function

P_{D} = 27 - Q^{2}

and supply function

P_{S} = 2 Q + 3

. Assuming perfect competition, find (i) the consumers’ surplus, (ii) the producers’ surplus.

Assignment B
Answer the following Middle Category questions in about 250 words each. Each question carries 10 marks. Word limit will not apply in the case of numerical questions.

3 \times 10 = 30

Given the total cost function $T C = 9 q^{2} + 2 q + 8100$
(a) Find marginal cost (MC) and average coat (AC) as functions of $q$
(b) Show that when $MC < AC, AC$ is falling, and when $MC > AC, AC$ is rising

OR
Given the aggregate consumption function

C = 0.9 Y + 100

(where C is aggregate consumption and

Y

is aggregate income)
(a) Find the marginal propensity to consume (MPC) and average propensity to consume (APC)
(b) Find the elasticity of consumption with respect to income, and show that it equals MPC/APC
4 Let

X = {1, 3, 5}

and

Y = {2, 4, 6}

Find,

X U Y

and the Cartesian Product of

X

and

Y

Given

A = {1, 2},, B = {3, 4, 5}

and

C = {3, 5, 6, 7, 8}

, show that
(i)

A \cup B = B \cup A

(ii)

(A \cap B) \cap C = A \cap (B \cap C)

5. Create a truth table for
(a)

A \Leftrightarrow B

(b) the converse of ‘

A

implies

B

‘.

OR
Find the Euclidean distance between
(i)

(2, 3)

and

(4, 1)

(ii)

(2, 3, 4)

and

(4, 1, - 5)

Assignment

C

Answer the following Short Category questions in about 100 words each

5 \times 6 = 30

6. What is a point of inflexion ? Does

f (x) = x^{3}

have a point of inflexion at

x = 0

?
7. Find the integral of

(y^{2} - 1) d x - 2 d y = 0

Evaluate the Limits of

\frac{X^{2} - X - 2}{X (X - 2)} As X \to 2

If the demand function for a good is $Q = 140 - 5 P$ , what is the price elasticity of demand at $P =$ 15 rupees?
How long will it take a given sum of money (Say in Rupees) to increase 4 times its present value when compounded half yearly at $7 %$ rate of interest?

Expert Answer

Question:-1

A monopolist faces the demand curve $Q = 60 - P / 2$ . The cost function is $C = Q^{2}$ . Find the output that maximises this monopolist’s profits. What are the prices at profits and that output? Find the elasticity of demand at the profit-maximising output.

Answer:

To solve this problem, we need to find the profit-maximizing output for the monopolist, the price at that output, and the elasticity of demand at the profit-maximizing output. Let’s go through this step by step.

Step 1: Express the Revenue Function

The demand curve is given by:

Q = 60 - \frac{P}{2}

We can rearrange this to express the price

P

as a function of quantity

Q

P = 120 - 2 Q

Revenue

R

is given by:

R = P \times Q = (120 - 2 Q) \times Q = 120 Q - 2 Q^{2}

Step 2: Express the Cost Function

The cost function is given by:

C (Q) = Q^{2}

Step 3: Express the Profit Function

Profit

Π

is revenue minus cost:

Π (Q) = R (Q) - C (Q) = (120 Q - 2 Q^{2}) - Q^{2} = 120 Q - 3 Q^{2}

Step 4: Find the Profit-Maximizing Output

To maximize profit, we take the derivative of the profit function with respect to

Q

and set it equal to zero:

\frac{d Π}{d Q} = 120 - 6 Q = 0

Solving for

Q

Q = \frac{120}{6} = 20

So, the profit-maximizing output is

Q = 20

Step 5: Find the Price at Profit-Maximizing Output

Substitute

Q = 20

into the demand function to find the corresponding price

P

P = 120 - 2 (20) = 120 - 40 = 80

So, the profit-maximizing price is

P = 80

Step 6: Calculate the Profit at the Profit-Maximizing Output

Substitute

Q = 20

into the profit function:

Π (20) = 120 (20) - 3 (20)^{2} = 2400 - 1200 = 1200

So, the maximum profit is

Π = 1200

Step 7: Calculate the Elasticity of Demand

The price elasticity of demand

E

is given by:

E = \frac{d Q}{d P} \times \frac{P}{Q}

First, we find

\frac{d Q}{d P}

from the demand curve

Q = 60 - \frac{P}{2}

\frac{d Q}{d P} = - \frac{1}{2}

Now, substitute the values

P = 80

and

Q = 20

into the elasticity formula:

E = - \frac{1}{2} \times \frac{80}{20} = - \frac{1}{2} \times 4 = - 2

So, the elasticity of demand at the profit-maximizing output is

E = - 2

Summary

Profit-maximizing output: $Q = 20$
Profit-maximizing price: $P = 80$
Maximum profit: $Π = 1200$
Elasticity of demand at profit-maximizing output: $E = - 2$

Question:-1 (OR)

A monopolist firm has the following total revenue and total cost functions:

\begin{array}{ll} R = - {mQ}^{2} + nQ & (m, n > 0) \\ C = a Q^{2} + bQ + c & (a, b, c > 0) \end{array}

Suppose that the government plans to levy an excise tax on the product of this firm and wishes to maximise the total tax revenue

T

from this source. What tax rate

t

(rupees per unit of output) should the government choose?

Answer:

To determine the optimal excise tax rate

t

that maximizes the total tax revenue

T

for the government, we need to follow these steps:

Understand the firm’s profit function and how the tax affects it.
Find the firm’s profit-maximizing output under the tax.
Determine the total tax revenue as a function of the tax rate $t$ .
Maximize the total tax revenue with respect to $t$ .

Step 1: Understand the Firm’s Profit Function

Given the total revenue (

R

) and total cost (

C

) functions:

R = - m Q^{2} + n Q, C = a Q^{2} + b Q + c

where

m, n, a, b,

and

c

are positive constants. If an excise tax

t

per unit of output is imposed, the cost function becomes:

C_{t} = a Q^{2} + b Q + c + t Q

Step 2: Firm’s Profit Function with Tax

The firm’s profit

Π

under the tax is:

Π (Q) = R - C_{t} = (- m Q^{2} + n Q) - (a Q^{2} + b Q + c + t Q)

Simplifying this, we get:

Π (Q) = - m Q^{2} + n Q - a Q^{2} - b Q - c - t Q

Π (Q) = - (m + a) Q^{2} + (n - b - t) Q - c

Step 3: Find the Profit-Maximizing Output

To find the profit-maximizing output

Q^{*}

, we take the derivative of

Π (Q)

with respect to

Q

and set it equal to zero:

\frac{d Π}{d Q} = - 2 (m + a) Q + (n - b - t) = 0

Solving for

Q

Q^{*} = \frac{n - b - t}{2 (m + a)}

Step 4: Determine Total Tax Revenue

Total tax revenue

T

is given by the tax rate

t

multiplied by the quantity produced

Q^{*}

T (t) = t \cdot Q^{*} = t (\frac{n - b - t}{2 (m + a)})

T (t) = \frac{t (n - b - t)}{2 (m + a)}

Step 5: Maximize Total Tax Revenue with Respect to $t$

To find the tax rate

t

that maximizes

T (t)

, take the derivative of

T (t)

with respect to

t

and set it equal to zero:

\frac{d T}{d t} = \frac{n - b - 2 t}{2 (m + a)} = 0

Solving for

t

n - b - 2 t = 0

2 t = n - b

t = \frac{n - b}{2}

Conclusion

The tax rate

t

that the government should choose to maximize total tax revenue from this monopolist firm is:

t = \frac{n - b}{2}

This result shows that the optimal excise tax rate is half of the difference between the coefficient

n

of the linear term in the revenue function and the coefficient

b

of the linear term in the cost function.

Question:-2

A firm in a perfectly competitive market has the following cost function:

C = 1 / 3 q^{3} - 5 q^{2} + 30 q + 10

If the market-clearing price is 6, obtain the profit-maximising level of output.

Answer:

To find the profit-maximizing level of output for a firm in a perfectly competitive market, we need to determine the output level

q

where marginal cost (MC) equals the market-clearing price

P

Step 1: Calculate Marginal Cost (MC)

Given the total cost function:

C (q) = \frac{1}{3} q^{3} - 5 q^{2} + 30 q + 10

The marginal cost (MC) is the derivative of the total cost function with respect to

q

M C = \frac{d C}{d q}

Calculating the derivative:

M C = \frac{d}{d q} (\frac{1}{3} q^{3} - 5 q^{2} + 30 q + 10)

M C = q^{2} - 10 q + 30

Step 2: Set Marginal Cost Equal to Market Price

In a perfectly competitive market, firms maximize profit by producing the output level where marginal cost equals the market-clearing price:

M C = P

Given that the market-clearing price

P

is 6, we set MC equal to 6:

q^{2} - 10 q + 30 = 6

Simplify this equation:

q^{2} - 10 q + 30 - 6 = 0

q^{2} - 10 q + 24 = 0

Step 3: Solve for the Profit-Maximizing Output Level $q$

To find the values of

q

, we solve the quadratic equation:

q^{2} - 10 q + 24 = 0

The quadratic formula is:

q = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

Here,

a = 1

b = - 10

, and

c = 24

. Plugging in these values:

q = \frac{- (- 10) \pm \sqrt{(- 10)^{2} - 4 (1) (24)}}{2 (1)}

q = \frac{10 \pm \sqrt{100 - 96}}{2}

q = \frac{10 \pm \sqrt{4}}{2}

q = \frac{10 \pm 2}{2}

So, the two possible solutions for

q

are:

q = \frac{10 + 2}{2} = 6 and q = \frac{10 - 2}{2} = 4

Step 4: Determine the Profit-Maximizing Output

To determine which of these values is the profit-maximizing output, we need to check the profit at these outputs:

The profit

Π (q)

is given by:

Π (q) = R (q) - C (q) = P \cdot q - C (q)

Let’s calculate the profit for

q = 4

and

q = 6

For

q = 4

C (4) = \frac{1}{3} (4)^{3} - 5 (4)^{2} + 30 (4) + 10 = \frac{64}{3} - 80 + 120 + 10 = \frac{64}{3} + 50

C (4) = \frac{64 + 150}{3} = \frac{214}{3} \approx 71.33

R (4) = 6 \times 4 = 24

Π (4) = 24 - 71.33 = - 47.33

For

q = 6

C (6) = \frac{1}{3} (6)^{3} - 5 (6)^{2} + 30 (6) + 10 = \frac{216}{3} - 180 + 180 + 10 = 72 + 10 = 82

R (6) = 6 \times 6 = 36

Π (6) = 36 - 82 = - 46

Since both profits are negative, the firm incurs a loss in both cases. However, the loss is smaller when

q = 6

Conclusion

The profit-maximizing level of output for the firm, given the market-clearing price of 6, is $q = 6$ . While the firm incurs a loss at this output level, it is the smallest possible loss given the current market conditions and cost structure.

Question:-2 (OR)

Given the demand function $P_{D} = 27 - Q^{2}$ and supply function $P_{S} = 2 Q + 3$ . Assuming perfect competition, find (i) the consumers’ surplus, (ii) the producers’ surplus.

Answer:

To find the consumer surplus and producer surplus in a perfectly competitive market, we first need to determine the equilibrium price and quantity where the demand function equals the supply function.

Step 1: Find the Equilibrium Price and Quantity

Given the demand function

P_{D} = 27 - Q^{2}

and the supply function

P_{S} = 2 Q + 3

, the equilibrium occurs where

P_{D} = P_{S}

27 - Q^{2} = 2 Q + 3

Rearrange this equation to find

Q

27 - 3 = Q^{2} + 2 Q

24 = Q^{2} + 2 Q

Q^{2} + 2 Q - 24 = 0

To solve for

Q

, we can use the quadratic formula:

Q = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}

where

a = 1

b = 2

, and

c = - 24

. Plugging in these values:

Q = \frac{- 2 \pm \sqrt{2^{2} - 4 (1) (- 24)}}{2 (1)}

Q = \frac{- 2 \pm \sqrt{4 + 96}}{2}

Q = \frac{- 2 \pm \sqrt{100}}{2}

Q = \frac{- 2 \pm 10}{2}

So, the solutions for

Q

are:

Q = \frac{- 2 + 10}{2} = 4 and Q = \frac{- 2 - 10}{2} = - 6

Since quantity

Q

cannot be negative, we take

Q = 4

as the equilibrium quantity.

Now, substitute

Q = 4

back into either the demand or supply function to find the equilibrium price

P

Using the supply function

P_{S} = 2 Q + 3

P = 2 (4) + 3 = 8 + 3 = 11

So, the equilibrium price

P = 11

and equilibrium quantity

Q = 4

Step 2: Calculate Consumer Surplus (CS)

Consumer surplus is the area between the demand curve and the equilibrium price, up to the equilibrium quantity.

The demand curve is

P_{D} = 27 - Q^{2}

. To find the maximum price consumers are willing to pay (where

Q = 0

P_{D} = 27 - (0)^{2} = 27

The consumer surplus is the area of the triangle with height

27 - 11 = 16

and base

4

Consumer Surplus (CS) = \frac{1}{2} \times Base \times Height = \frac{1}{2} \times 4 \times 16 = 32

Step 3: Calculate Producer Surplus (PS)

Producer surplus is the area between the supply curve and the equilibrium price, up to the equilibrium quantity.

The supply curve is

P_{S} = 2 Q + 3

. To find the minimum price producers are willing to accept (where

Q = 0

P_{S} = 2 (0) + 3 = 3

The producer surplus is the area of the triangle with height

11 - 3 = 8

and base

4

Producer Surplus (PS) = \frac{1}{2} \times Base \times Height = \frac{1}{2} \times 4 \times 8 = 16

Summary

Consumers’ Surplus (CS): $32$
Producers’ Surplus (PS): $16$

These values represent the benefits to consumers and producers, respectively, from participating in the market at the equilibrium price and quantity.

Question:-3

Given the total cost function $T C = 9 q^{2} + 2 q + 8100$

(a) Find marginal cost (MC) and average cost (AC) as functions of

q

(b) Show that when $MC < AC, AC$ is falling, and when $MC > AC, AC$ is rising

Answer:

Let’s go step-by-step to solve each part of the question.

(a) Finding Marginal Cost (MC) and Average Cost (AC)

Given the total cost function:

T C = 9 q^{2} + 2 q + 8100

Marginal Cost (MC)

Marginal Cost (MC) is the derivative of the Total Cost (TC) with respect to quantity

q

M C = \frac{d (T C)}{d q}

Calculating the derivative:

M C = \frac{d}{d q} (9 q^{2} + 2 q + 8100)

M C = 18 q + 2

Average Cost (AC)

Average Cost (AC) is the total cost divided by the quantity

q

A C = \frac{T C}{q}

Calculating the average cost:

A C = \frac{9 q^{2} + 2 q + 8100}{q}

A C = 9 q + \frac{2}{q} + 8100

(b) Relationship Between Marginal Cost (MC) and Average Cost (AC)

We need to show that:

When $MC < AC$ , $AC$ is falling.
When $MC > AC$ , $AC$ is rising.

Relationship Between MC and AC

To understand the relationship between MC and AC, let’s analyze the change in AC when MC is different from AC. The Average Cost

A C

will change based on how the marginal cost

M C

compares to it.

We can determine this by analyzing the derivative of the AC function.

\frac{d (A C)}{d q}

First, let’s write down the AC function again for clarity:

A C = 9 q + \frac{2}{q} + 8100

Now, let’s compute the derivative of AC with respect to

q

\frac{d (A C)}{d q} = \frac{d}{d q} (9 q + \frac{2}{q} + 8100)

\frac{d (A C)}{d q} = 9 - \frac{2}{q^{2}}

Analysis of $\frac{d (A C)}{d q}$

When $M C < A C$ :

$M C = 18 q + 2, A C = 9 q + \frac{2}{q} + 8100$

To show when $A C$ is falling, $\frac{d (A C)}{d q}$ should be negative. The expression $\frac{d (A C)}{d q} = 9 - \frac{2}{q^{2}}$ tells us that $A C$ will be falling when:

$9 - \frac{2}{q^{2}} < 0 ⟹ 9 < \frac{2}{q^{2}} ⟹ q^{2} < \frac{2}{9}$

In this case, when $q^{2} < \frac{2}{9}$ , the change in average cost $\frac{d (A C)}{d q}$ is negative, and hence $A C$ is falling.
When $M C > A C$ :

Now, let’s look at when $M C > A C$ . This happens when:

$18 q + 2 > 9 q + \frac{2}{q} + 8100$

Simplifying the inequality:

$9 q + 2 > \frac{2}{q} + 8100$

Rearranging terms, we see that $M C > A C$ as $q$ increases past the point where $9 q = \frac{2}{q} + 8100$ . When $q^{2} > \frac{2}{9}$ , $\frac{d (A C)}{d q} > 0$ , meaning $A C$ is rising.

Conclusion

If $M C < A C$ : $A C$ is falling because $\frac{d (A C)}{d q} < 0$ .
If $M C > A C$ : $A C$ is rising because $\frac{d (A C)}{d q} > 0$ .

This demonstrates the relationship between marginal cost and average cost in terms of how they affect the direction of average cost changes.

Question:-3 (OR)

Given the aggregate consumption function $C = 0.9 Y + 100$ (where C is aggregate consumption and $Y$ is aggregate income)

(a) Find the marginal propensity to consume (MPC) and average propensity to consume (APC)

(b) Find the elasticity of consumption with respect to income, and show that it equals MPC/APC

Answer:

Let’s go through each part of the problem step-by-step.

(a) Finding the Marginal Propensity to Consume (MPC) and Average Propensity to Consume (APC)

Given the aggregate consumption function:

C = 0.9 Y + 100

where

C

is aggregate consumption and

Y

is aggregate income.

Marginal Propensity to Consume (MPC)

The Marginal Propensity to Consume (MPC) is the change in consumption resulting from a change in income. It is the coefficient of

Y

in the consumption function.

From the consumption function

C = 0.9 Y + 100

, the MPC is:

MPC = 0.9

Average Propensity to Consume (APC)

The Average Propensity to Consume (APC) is the ratio of total consumption to total income:

APC = \frac{C}{Y}

Using the consumption function:

APC = \frac{0.9 Y + 100}{Y}

Simplifying this:

APC = 0.9 + \frac{100}{Y}

(b) Finding the Elasticity of Consumption with Respect to Income

The elasticity of consumption with respect to income measures the percentage change in consumption resulting from a percentage change in income.

The formula for elasticity of consumption

(E_{C})

with respect to income is given by:

E_{C} = \frac{d C / d Y}{C / Y}

where

d C / d Y

is the marginal propensity to consume (MPC) and

C / Y

is the average propensity to consume (APC).

Finding the Derivative $d C / d Y$

The derivative of the consumption function

C = 0.9 Y + 100

with respect to

Y

is:

\frac{d C}{d Y} = 0.9

Calculating Elasticity

Now, plug the values into the elasticity formula:

E_{C} = \frac{\frac{d C}{d Y}}{C / Y} = \frac{0.9}{\frac{C}{Y}}

Substitute the APC expression for

C / Y

E_{C} = \frac{0.9}{0.9 + \frac{100}{Y}}

To demonstrate that

E_{C}

equals

MPC / APC

, let’s compute

MPC / APC

\frac{MPC}{APC} = \frac{0.9}{0.9 + \frac{100}{Y}}

This matches the expression for

E_{C}

. Therefore, we have shown that:

E_{C} = \frac{MPC}{APC}

Summary

Marginal Propensity to Consume (MPC): $0.9$
Average Propensity to Consume (APC): $0.9 + \frac{100}{Y}$
Elasticity of consumption with respect to income ( $E_{C}$ ): $\frac{0.9}{0.9 + \frac{100}{Y}}$
Equality of Elasticity and MPC/APC: $E_{C} = \frac{MPC}{APC}$

Question:-4

Let $X = {1, 3, 5}$ and $Y = {2, 4, 6}$

Find,

X \cup Y

and the Cartesian Product of

X

and

Y

Answer:

Let’s find the union and the Cartesian product of the sets

X

and

Y

1. Union of $X$ and $Y$ ( $X \cup Y$ )

The union of two sets

X

and

Y

, denoted by

X \cup Y

, is the set of all elements that are in

X

, in

Y

, or in both.

Given:

X = {1, 3, 5}, Y = {2, 4, 6}

The union

X \cup Y

includes all elements from both sets:

X \cup Y = {1, 2, 3, 4, 5, 6}

2. Cartesian Product of $X$ and $Y$ ( $X \times Y$ )

The Cartesian product of two sets

X

and

Y

, denoted by

X \times Y

, is the set of all ordered pairs

(x, y)

where

x

is an element of

X

and

y

is an element of

Y

Given:

X = {1, 3, 5}, Y = {2, 4, 6}

The Cartesian product

X \times Y

is:

X \times Y = {(1, 2), (1, 4), (1, 6), (3, 2), (3, 4), (3, 6), (5, 2), (5, 4), (5, 6)}

Summary

Union ( $X \cup Y$ ): ${1, 2, 3, 4, 5, 6}$
Cartesian Product ( $X \times Y$ ): ${(1, 2), (1, 4), (1, 6), (3, 2), (3, 4), (3, 6), (5, 2), (5, 4), (5, 6)}$

Question:-4 (OR)

Given $A = {1, 2}, B = {3, 4, 5}$ and $C = {3, 5, 6, 7, 8}$ , show that

(i)

A \cup B = B \cup A

(ii)

(A \cap B) \cap C = A \cap (B \cap C)

Answer:

To show these set operations, we’ll go through each part step-by-step using the definitions of set union (

\cup

) and intersection (

\cap

(i) Proving $A \cup B = B \cup A$

The union of two sets

A

and

B

, denoted

A \cup B

, is the set of elements that are in either

A

B

, or both.

Given:

A = {1, 2}, B = {3, 4, 5}

First, let’s compute

A \cup B

and

B \cup A

$A \cup B$ :

$A \cup B = {1, 2} \cup {3, 4, 5} = {1, 2, 3, 4, 5}$
$B \cup A$ :

$B \cup A = {3, 4, 5} \cup {1, 2} = {3, 4, 5, 1, 2}$

Since union operation is commutative (order does not matter), the sets

A \cup B

and

B \cup A

contain the same elements:

A \cup B = {1, 2, 3, 4, 5} = B \cup A

Thus, we have shown that:

A \cup B = B \cup A

(ii) Proving $(A \cap B) \cap C = A \cap (B \cap C)$

The intersection of two sets

A

and

B

, denoted

A \cap B

, is the set of elements that are in both

A

and

B

Given:

A = {1, 2}, B = {3, 4, 5}, C = {3, 5, 6, 7, 8}

Let’s compute

(A \cap B) \cap C

and

A \cap (B \cap C)

Find $A \cap B$ :

$A \cap B = {1, 2} \cap {3, 4, 5} = {}$

(There are no common elements between $A$ and $B$ , so the intersection is an empty set.)
Compute $(A \cap B) \cap C$ :

$(A \cap B) \cap C = {} \cap {3, 5, 6, 7, 8} = {}$

(An intersection with an empty set results in an empty set.)
Find $B \cap C$ :

$B \cap C = {3, 4, 5} \cap {3, 5, 6, 7, 8} = {3, 5}$

(The common elements between $B$ and $C$ are 3 and 5.)
Compute $A \cap (B \cap C)$ :

$A \cap (B \cap C) = {1, 2} \cap {3, 5} = {}$

(Again, there are no common elements between $A$ and $B \cap C$ , resulting in an empty set.)

Since both

(A \cap B) \cap C

and

A \cap (B \cap C)

result in the empty set:

(A \cap B) \cap C = A \cap (B \cap C) = {}

Thus, we have shown that:

(A \cap B) \cap C = A \cap (B \cap C)

This illustrates the associativity of the intersection operation.

Question:-5(a)

Create a truth table for

(a)

A \Leftrightarrow B

(b) the converse of ‘

A

implies

B

‘.

Answer:

Let’s create the truth tables for each of the statements.

(a) Truth Table for $A \Leftrightarrow B$

The biconditional statement

A \Leftrightarrow B

(also known as "if and only if") is true when both

A

and

B

have the same truth value (both true or both false). It is false when

A

and

B

have different truth values.

Here is the truth table for

A \Leftrightarrow B

$A$	$B$	$A \Leftrightarrow B$
T	T	T
T	F	F
F	T	F
F	F	T

(b) Truth Table for the Converse of ‘ $A$ implies $B$ ‘

The converse of "

A

implies

B

" is "

B

implies

A

". In logical notation, the implication "

A

implies

B

" is written as

A \Rightarrow B

, and the converse is

B \Rightarrow A

The implication

B \Rightarrow A

is true except when

B

is true and

A

is false.

Here is the truth table for the converse of "

A

implies

B

" (

B \Rightarrow A

$A$	$B$	$B \Rightarrow A$
T	T	T
T	F	T
F	T	F
F	F	T

Summary

(a) $A \Leftrightarrow B$ : The biconditional is true when $A$ and $B$ have the same truth values.
(b) Converse of ‘ $A$ implies $B$ ‘ ( $B \Rightarrow A$ ): The converse is true except when $B$ is true and $A$ is false.

Question:-5 (OR)

Find the Euclidean distance between

(i)

(2, 3)

and

(4, 1)

(ii)

(2, 3, 4)

and

(4, 1, - 5)

Answer:

To find the Euclidean distance between two points, we use the Euclidean distance formula.

Euclidean Distance Formula

In 2D (two-dimensional space), the Euclidean distance between points $(x_{1}, y_{1})$ and $(x_{2}, y_{2})$ is given by:

$d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}}$
In 3D (three-dimensional space), the Euclidean distance between points $(x_{1}, y_{1}, z_{1})$ and $(x_{2}, y_{2}, z_{2})$ is given by:

$d = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2} + (z_{2} - z_{1})^{2}}$

Let’s use these formulas to calculate the distances for the given points.

(i) Distance between $(2, 3)$ and $(4, 1)$

Given points:

(x_{1}, y_{1}) = (2, 3)

and

(x_{2}, y_{2}) = (4, 1)

Applying the 2D distance formula:

d = \sqrt{(4 - 2)^{2} + (1 - 3)^{2}}

Calculate the differences:

d = \sqrt{(2)^{2} + (- 2)^{2}}

d = \sqrt{4 + 4} = \sqrt{8}

Simplify the square root:

d = 2 \sqrt{2}

So, the Euclidean distance between

(2, 3)

and

(4, 1)

is $2 \sqrt{2}$ .

(ii) Distance between $(2, 3, 4)$ and $(4, 1, - 5)$

Given points:

(x_{1}, y_{1}, z_{1}) = (2, 3, 4)

and

(x_{2}, y_{2}, z_{2}) = (4, 1, - 5)

Applying the 3D distance formula:

d = \sqrt{(4 - 2)^{2} + (1 - 3)^{2} + (- 5 - 4)^{2}}

Calculate the differences:

d = \sqrt{(2)^{2} + (- 2)^{2} + (- 9)^{2}}

d = \sqrt{4 + 4 + 81}

d = \sqrt{89}

So, the Euclidean distance between

(2, 3, 4)

and

(4, 1, - 5)

is $\sqrt{89}$ .

Question:-6

What is a point of inflexion? Does $f (x) = x^{3}$ have a point of inflexion at $x = 0$ ?

Answer:

What is a Point of Inflexion?

A point of inflexion (or inflection point) is a point on the graph of a function where the curvature changes sign. In other words, it is where the function changes from being concave (curved upwards) to convex (curved downwards) or vice versa.

Mathematically, a point

x = c

is a point of inflexion if the second derivative of the function

f (x)

changes sign as

x

passes through

c

. This implies that around this point, the function changes concavity:

Concave up if $f^{″} (x) > 0$
Concave down if $f^{″} (x) < 0$

For a point

c

to be a point of inflexion,

f^{″} (c)

is typically zero or undefined, and the sign of

f^{″} (x)

must change on either side of

c

Does $f (x) = x^{3}$ Have a Point of Inflexion at $x = 0$ ?

To determine if

f (x) = x^{3}

has a point of inflexion at

x = 0

, we will examine the second derivative of

f (x)

First derivative $f^{'} (x)$ :

f (x) = x^{3}

f^{'} (x) = 3 x^{2}

Second derivative $f^{″} (x)$ :

f^{″} (x) = \frac{d}{d x} (3 x^{2}) = 6 x

Now, let’s analyze the second derivative

f^{″} (x) = 6 x

to determine the concavity:

At $x = 0$ , $f^{″} (0) = 6 \times 0 = 0$ .

Check the sign change around $x = 0$ :

For $x > 0$ , $f^{″} (x) = 6 x > 0$ , indicating the function is concave up.
For $x < 0$ , $f^{″} (x) = 6 x < 0$ , indicating the function is concave down.

Since the second derivative

f^{″} (x)

changes sign from negative to positive as

x

passes through 0, there is indeed a change in concavity at

x = 0

Conclusion

Yes,

f (x) = x^{3}

has a point of inflexion at

x = 0

because the second derivative

f^{″} (x) = 6 x

changes sign at that point.

Question:-7

Find the integral of

(y^{2} - 1) d x - 2 d y = 0

Answer:

To solve the integral of the differential equation

(y^{2} - 1) d x - 2 d y = 0

, we can rewrite this equation in a more standard differential form and solve for a potential function.

Step 1: Rewrite the Equation

The given differential equation is:

(y^{2} - 1) d x - 2 d y = 0

Rearrange it to isolate terms involving

d x

and

d y

(y^{2} - 1) d x = 2 d y

Divide both sides by

2

to simplify:

\frac{y^{2} - 1}{2} d x = d y

Step 2: Separate Variables

Now, separate the variables

x

and

y

d x = \frac{2}{y^{2} - 1} d y

Integrate both sides with respect to their respective variables:

\int d x = \int \frac{2}{y^{2} - 1} d y

Step 3: Integrate Both Sides

The integral on the left side is straightforward:

\int d x = x + C_{1}

Now, let’s focus on the integral on the right side:

\int \frac{2}{y^{2} - 1} d y

This integral can be solved using partial fractions. The expression

\frac{2}{y^{2} - 1}

can be factored as:

y^{2} - 1 = (y - 1) (y + 1)

Thus, we write:

\frac{2}{y^{2} - 1} = \frac{A}{y - 1} + \frac{B}{y + 1}

To find

A

and

B

, multiply both sides by

(y - 1) (y + 1)

2 = A (y + 1) + B (y - 1)

Set up equations by substituting convenient values for

y

Let $y = 1$ :

2 = A (1 + 1) + B (1 - 1) ⟹ 2 = 2 A ⟹ A = 1

Let $y = - 1$ :

2 = A (- 1 + 1) + B (- 1 - 1) ⟹ 2 = - 2 B ⟹ B = - 1

Thus, we have:

\frac{2}{y^{2} - 1} = \frac{1}{y - 1} - \frac{1}{y + 1}

Now, integrate each term separately:

\int \frac{1}{y - 1} d y - \int \frac{1}{y + 1} d y = \ln | y - 1 | - \ln | y + 1 | + C_{2}

Combine the logarithms:

\ln | \frac{y - 1}{y + 1} | + C_{2}

Step 4: Combine the Results

Putting it all together:

x + C_{1} = \ln | \frac{y - 1}{y + 1} | + C_{2}

Combine the constants

C_{1}

and

C_{2}

into a single constant

C = C_{2} - C_{1}

x = \ln | \frac{y - 1}{y + 1} | + C

Therefore, the integral of the differential equation

(y^{2} - 1) d x - 2 d y = 0

is:

x = \ln | \frac{y - 1}{y + 1} | + C

or equivalently:

\ln | \frac{y - 1}{y + 1} | = x - C

Question:-8

Evaluate the Limits of

\frac{X^{2} - X - 2}{X (X - 2)} as X \to 2

Answer:

To evaluate the limit

lim_{X \to 2} \frac{X^{2} - X - 2}{X (X - 2)},

we first need to simplify the expression. Direct substitution of

X = 2

into the expression leads to a

\frac{0}{0}

indeterminate form. Therefore, we should try to simplify the expression by factoring.

Step 1: Factor the Numerator

The numerator

X^{2} - X - 2

can be factored as:

X^{2} - X - 2 = (X - 2) (X + 1) .

Step 2: Simplify the Expression

Substitute the factorization into the original expression:

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

We can cancel the common factor

(X - 2)

in the numerator and the denominator (provided

X \neq 2

\frac{(X - 2) (X + 1)}{X (X - 2)} = \frac{X + 1}{X}, X \neq 2.

Step 3: Evaluate the Limit

Now, we evaluate the simplified expression as

X \to 2

lim_{X \to 2} \frac{X + 1}{X} = \frac{2 + 1}{2} = \frac{3}{2} .

Conclusion

The limit of

\frac{X^{2} - X - 2}{X (X - 2)}

X \to 2

is:

lim_{X \to 2} \frac{X^{2} - X - 2}{X (X - 2)} = \frac{3}{2} .

Question:-9

If the demand function for a good is $Q = 140 - 5 P$ , what is the price elasticity of demand at $P = 15$ rupees?

Answer:

To find the price elasticity of demand at a specific price, we need to calculate the elasticity using the demand function. The price elasticity of demand measures the responsiveness of the quantity demanded to a change in price.

Price Elasticity of Demand Formula

The price elasticity of demand

E_{d}

is given by the formula:

E_{d} = \frac{d Q}{d P} \times \frac{P}{Q}

where:

$\frac{d Q}{d P}$ is the derivative of the quantity demanded $Q$ with respect to the price $P$ ,
$P$ is the price,
$Q$ is the quantity demanded at that price.

Step 1: Find the Derivative of the Demand Function

Given the demand function:

Q = 140 - 5 P

First, find the derivative of

Q

with respect to

P

\frac{d Q}{d P} = - 5

Step 2: Find $Q$ at $P = 15$

Now, calculate the quantity demanded

Q

when the price

P = 15

Q = 140 - 5 (15) = 140 - 75 = 65

Step 3: Calculate the Price Elasticity of Demand

Substitute

\frac{d Q}{d P} = - 5

P = 15

, and

Q = 65

into the elasticity formula:

E_{d} = \frac{d Q}{d P} \times \frac{P}{Q} = (- 5) \times \frac{15}{65}

Simplify this expression:

E_{d} = - 5 \times \frac{15}{65} = - 5 \times \frac{3}{13} = - \frac{15}{13}

Conclusion

The price elasticity of demand at

P = 15

rupees is:

E_{d} = - \frac{15}{13} \approx - 1.15

This means that at

P = 15

rupees, the demand is elastic, as the absolute value of the elasticity is greater than 1 (

| E_{d} | > 1

). This indicates that the quantity demanded is relatively responsive to changes in price.

Question:-10

How long will it take a given sum of money (Say in Rupees) to increase 4 times its present value when compounded half-yearly at $7 %$ rate of interest?

Answer:

To determine how long it will take for a given sum of money to increase four times its present value when compounded half-yearly at a

7 %

interest rate, we can use the formula for compound interest.

Compound Interest Formula

The compound interest formula is:

A = P {(1 + \frac{r}{n})}^{n t}

where:

$A$ is the amount of money accumulated after $n$ years, including interest.
$P$ is the principal amount (the initial sum of money).
$r$ is the annual nominal interest rate (as a decimal).
$n$ is the number of times the interest is compounded per year.
$t$ is the time the money is invested for in years.

Problem Details

In this problem:

The amount $A$ is 4 times the principal $P$ , so $A = 4 P$ .
The annual interest rate $r$ is $7 % = 0.07$ .
Interest is compounded half-yearly, so $n = 2$ .

Step-by-Step Solution

Set Up the Equation:

Plug the values into the compound interest formula:

$4 P = P {(1 + \frac{0.07}{2})}^{2 t}$
Simplify the Equation:

Divide both sides by $P$ (assuming $P \neq 0$ ):

$4 = {(1 + \frac{0.07}{2})}^{2 t}$
Calculate the Rate per Period:

Compute the rate per half-year period:

$1 + \frac{0.07}{2} = 1 + 0.035 = 1.035$

So the equation becomes:

$4 = (1.035)^{2 t}$
Take the Logarithm of Both Sides:

To solve for $t$ , take the natural logarithm (or common logarithm) of both sides:

$\ln (4) = \ln ((1.035)^{2 t})$

Using the logarithmic identity $\ln (a^{b}) = b \ln (a)$ :

$\ln (4) = 2 t \ln (1.035)$
Solve for $t$ :

Divide both sides by $2 \ln (1.035)$ :

$t = \frac{\ln (4)}{2 \ln (1.035)}$
Calculate $t$ :

Now compute the value using a calculator:

$\ln (4) \approx 1.3863, \ln (1.035) \approx 0.0344$

Plugging these values in:

$t = \frac{1.3863}{2 \times 0.0344} = \frac{1.3863}{0.0688} \approx 20.15$

Conclusion

It will take approximately 20.15 years for a given sum of money to increase to four times its present value when compounded half-yearly at a

7 %

interest rate.

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BECC-102 Solved Assignment

Expert Answer

Question:-1

Answer:

Step 1: Express the Revenue Function

Step 2: Express the Cost Function

Step 3: Express the Profit Function

Step 4: Find the Profit-Maximizing Output

Step 5: Find the Price at Profit-Maximizing Output

Step 6: Calculate the Profit at the Profit-Maximizing Output

Step 7: Calculate the Elasticity of Demand

Summary

Question:-1 (OR)

A monopolist firm has the following total revenue and total cost functions:

Answer:

Step 1: Understand the Firm’s Profit Function

Step 2: Firm’s Profit Function with Tax

Step 3: Find the Profit-Maximizing Output

Step 4: Determine Total Tax Revenue

Step 5: Maximize Total Tax Revenue with Respect to t t ttt

Conclusion

Question:-2

A firm in a perfectly competitive market has the following cost function:

Answer:

Step 1: Calculate Marginal Cost (MC)

Step 2: Set Marginal Cost Equal to Market Price

Step 3: Solve for the Profit-Maximizing Output Level q q qqq

Step 4: Determine the Profit-Maximizing Output

Conclusion

Question:-2 (OR)

Given the demand function P D = 27 − Q 2 P D = 27 − Q 2 P_(D)=27-Q^(2)P_D=27-Q^2PD=27−Q2 and supply function P S = 2 Q + 3 P S = 2 Q + 3 P_(S)=2Q+3P_S=2 Q+3PS=2Q+3. Assuming perfect competition, find (i) the consumers’ surplus, (ii) the producers’ surplus.

Answer:

Step 1: Find the Equilibrium Price and Quantity

Step 2: Calculate Consumer Surplus (CS)

Step 3: Calculate Producer Surplus (PS)

Summary

Question:-3

Given the total cost function T C = 9 q 2 + 2 q + 8100 T C = 9 q 2 + 2 q + 8100 TC=9q^(2)+2q+8100T C=9 q^2+2 q+8100TC=9q2+2q+8100

(b) Show that when MC < AC , AC MC < AC , AC MC < AC,AC\mathrm{MC}<\mathrm{AC}, \mathrm{AC}MC<AC,AC is falling, and when MC > AC , AC MC > AC , AC MC > AC,AC\mathrm{MC}>\mathrm{AC}, \mathrm{AC}MC>AC,AC is rising

Answer:

(a) Finding Marginal Cost (MC) and Average Cost (AC)

Marginal Cost (MC)

Average Cost (AC)

(b) Relationship Between Marginal Cost (MC) and Average Cost (AC)

Relationship Between MC and AC

Analysis of d ( A C ) d q d ( A C ) d q (d(AC))/(dq)\frac{d(AC)}{dq}d(AC)dq

Conclusion

Question:-3 (OR)

Given the aggregate consumption function C = 0.9 Y + 100 C = 0.9 Y + 100 C=0.9Y+100\mathrm{C}=0.9 \mathrm{Y}+100C=0.9Y+100 (where C is aggregate consumption and Y Y YYY is aggregate income)

Answer:

(a) Finding the Marginal Propensity to Consume (MPC) and Average Propensity to Consume (APC)

Marginal Propensity to Consume (MPC)

Average Propensity to Consume (APC)

(b) Finding the Elasticity of Consumption with Respect to Income

Finding the Derivative d C / d Y d C / d Y dC//dYdC/dYdC/dY

Calculating Elasticity

Summary

Question:-4

Let X = { 1 , 3 , 5 } X = { 1 , 3 , 5 } X={1,3,5}X=\{1,3,5\}X={1,3,5} and Y = { 2 , 4 , 6 } Y = { 2 , 4 , 6 } Y={2,4,6}Y=\{2,4,6\}Y={2,4,6}

Answer:

1. Union of X X XXX and Y Y YYY ( X ∪ Y X ∪ Y X uu YX \cup YX∪Y)

2. Cartesian Product of X X XXX and Y Y YYY ( X × Y X × Y X xx YX \times YX×Y)

Summary

Question:-4 (OR)

Given A = { 1 , 2 } , B = { 3 , 4 , 5 } A = { 1 , 2 } , B = { 3 , 4 , 5 } A={1,2},B={3,4,5}A=\{1,2\}, B=\{3,4,5\}A={1,2},B={3,4,5} and C = { 3 , 5 , 6 , 7 , 8 } C = { 3 , 5 , 6 , 7 , 8 } C={3,5,6,7,8}C=\{3,5,6,7,8\}C={3,5,6,7,8}, show that

Answer:

(i) Proving A ∪ B = B ∪ A A ∪ B = B ∪ A A uu B=B uu AA \cup B = B \cup AA∪B=B∪A

(ii) Proving ( A ∩ B ) ∩ C = A ∩ ( B ∩ C ) ( A ∩ B ) ∩ C = A ∩ ( B ∩ C ) (A nn B)nn C=A nn(B nn C)(A \cap B) \cap C = A \cap (B \cap C)(A∩B)∩C=A∩(B∩C)

Question:-5(a)

Create a truth table for

Answer:

(a) Truth Table for A ⇔ B A ⇔ B A<=>BA \Leftrightarrow BA⇔B

(b) Truth Table for the Converse of ‘ A A AAA implies B B BBB ‘

Summary

Question:-5 (OR)

Find the Euclidean distance between

Answer:

Euclidean Distance Formula

(i) Distance between ( 2 , 3 ) ( 2 , 3 ) (2,3)(2, 3)(2,3) and ( 4 , 1 ) ( 4 , 1 ) (4,1)(4, 1)(4,1)

Step 5: Maximize Total Tax Revenue with Respect to $t$

Step 3: Solve for the Profit-Maximizing Output Level $q$

Given the demand function $P_{D} = 27 - Q^{2}$ and supply function $P_{S} = 2 Q + 3$ . Assuming perfect competition, find (i) the consumers’ surplus, (ii) the producers’ surplus.

Given the total cost function $T C = 9 q^{2} + 2 q + 8100$

(b) Show that when $MC < AC, AC$ is falling, and when $MC > AC, AC$ is rising

Analysis of $\frac{d (A C)}{d q}$

Given the aggregate consumption function $C = 0.9 Y + 100$ (where C is aggregate consumption and $Y$ is aggregate income)

Finding the Derivative $d C / d Y$

Let $X = {1, 3, 5}$ and $Y = {2, 4, 6}$

1. Union of $X$ and $Y$ ( $X \cup Y$ )

2. Cartesian Product of $X$ and $Y$ ( $X \times Y$ )

Given $A = {1, 2}, B = {3, 4, 5}$ and $C = {3, 5, 6, 7, 8}$ , show that

(i) Proving $A \cup B = B \cup A$

(ii) Proving $(A \cap B) \cap C = A \cap (B \cap C)$

(a) Truth Table for $A \Leftrightarrow B$

(b) Truth Table for the Converse of ‘ $A$ implies $B$ ‘

(i) Distance between $(2, 3)$ and $(4, 1)$

(ii) Distance between $(2, 3, 4)$ and $(4, 1, - 5)$

What is a point of inflexion? Does $f (x) = x^{3}$ have a point of inflexion at $x = 0$ ?

Does $f (x) = x^{3}$ Have a Point of Inflexion at $x = 0$ ?

If the demand function for a good is $Q = 140 - 5 P$ , what is the price elasticity of demand at $P = 15$ rupees?

Step 2: Find $Q$ at $P = 15$

How long will it take a given sum of money (Say in Rupees) to increase 4 times its present value when compounded half-yearly at $7 %$ rate of interest?