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IGNOU BECC-104 Previous Year Solved Question Paper in Hindi

IGNOU BECC-104 Previous Year Solved Question Paper in English

Q1. (a) Find the solution of the following equation system using Cramer’s rule : 7x₁ − x₂ − x₃ = 0 10x₁ − 2x₂ + x₃ = 8 6x₁ + 3x₂ − 2x₃ = 7 (b) Define the concept of orthogonality of vectors.

Ans. (a) To solve the given system of equations using Cramer’s rule, we first need to find the determinant of the coefficient matrix. The given system of equations is: 7x₁ − x₂ − x₃ = 0 10x₁ − 2x₂ + x₃ = 8 6x₁ + 3x₂ − 2x₃ = 7 This can be written in matrix form AX = B, where: A = , X = , B = Step 1: Calculate the determinant of the coefficient matrix A, denoted as |A|. |A| = 7((-2)(-2) – (1)(3)) – (-1)( (10)(-2) – (1)(6) ) + (-1)( (10)(3) – (-2)(6) ) |A| = 7(4 – 3) + 1(-20 – 6) – 1(30 + 12) |A| = 7(1) + (-26) – 42 |A| = 7 – 26 – 42 = -61 Since |A| ≠ 0, a unique solution exists. Step 2: Calculate the determinants |A₁|, |A₂|, and |A₃|. To find |A₁|, replace the first column of A with B: |A₁| = = 0 – (-1)((8)(-2) – (1)(7)) + (-1)((8)(3) – (-2)(7)) |A₁| = 1(-16 – 7) – 1(24 + 14) = -23 – 38 = -61 To find |A₂|, replace the second column of A with B: |A₂| = = 7((8)(-2) – (1)(7)) – 0 + (-1)((10)(7) – (8)(6)) |A₂| = 7(-16 – 7) – (70 – 48) = 7(-23) – 22 = -161 – 22 = -183 To find |A₃|, replace the third column of A with B: |A₃| = = 7((-2)(7) – (8)(3)) – (-1)((10)(7) – (8)(6)) + 0 |A₃| = 7(-14 – 24) + 1(70 – 48) = 7(-38) + 22 = -266 + 22 = -244 Step 3: Find the values of the variables. According to Cramer’s rule: x₁ = |A₁| / |A| = -61 / -61 = 1 x₂ = |A₂| / |A| = -183 / -61 = 3 x₃ = |A₃| / |A| = -244 / -61 = 4 Thus, the solution to the system is x₁ = 1, x₂ = 3, and x₃ = 4 . (b) Concept of Orthogonality of Vectors The orthogonality of vectors is a fundamental concept in linear algebra. Two vectors are said to be orthogonal if they are perpendicular to each other, i.e., the angle between them is 90 degrees. Mathematically, the orthogonality of two non-zero vectors is defined through their dot product (or scalar product) . Definition: Let u and v be two vectors in an n-dimensional Euclidean space ℝⁿ. u = (u₁, u₂, …, uₙ) v = (v₁, v₂, …, vₙ) These two vectors u and v are said to be orthogonal if and only if their dot product is zero. u ⋅ v = 0 The dot product is calculated as: u ⋅ v = u₁v₁ + u₂v₂ + … + uₙvₙ = Thus, u and v are orthogonal if Geometric Interpretation: Geometrically, the dot product is related to the angle (θ) between the two vectors: u ⋅ v = ||u|| ||v|| cos(θ) Where ||u|| and ||v|| are the magnitudes (or lengths) of the vectors u and v, respectively. If u ⋅ v = 0, and both u and v are non-zero vectors, then: ||u|| ||v|| cos(θ) = 0 cos(θ) = 0 This implies that θ = 90° or π/2 radians. This means the vectors are perpendicular to each other. Example: Consider two vectors in ℝ³: u = (2, 3, -1) v = (4, -2, 2) Their dot product is: u ⋅ v = (2)(4) + (3)(-2) + (-1)(2) = 8 – 6 – 2 = 0 Since u ⋅ v = 0, the vectors u and v are orthogonal.

Q2. Given production function Q = 2L¹ᐟ²K¹ᐟ², where L stands for labour and K stands for capital. Find : (a) Marginal product of two factors (b) Nature of returns to scale (c) If total production is exhausted or not

Ans. The given production function is of the Cobb-Douglas type: Q = 2L¹ᐟ²K¹ᐟ². (a) Marginal product of two factors The marginal product of a factor is the change in total output resulting from employing one additional unit of that factor, holding other factors constant. It is calculated by taking the partial derivative of the production function. 1. Marginal Product of Labour (MPL): MPL is found by partially differentiating Q with respect to L, holding K constant. MPL = ∂Q/∂L = ∂(2L¹ᐟ²K¹ᐟ²)/∂L MPL = 2 (1/2) L⁽¹ᐟ²⁻¹⁾ * K¹ᐟ² MPL = L⁻¹ᐟ²K¹ᐟ² MPL = K¹ᐟ²/L¹ᐟ² = (K/L)¹ᐟ² 2. Marginal Product of Capital (MPK): MPK is found by partially differentiating Q with respect to K, holding L constant. MPK = ∂Q/∂K = ∂(2L¹ᐟ²K¹ᐟ²)/∂K MPK = 2 L¹ᐟ² (1/2) * K⁽¹ᐟ²⁻¹⁾ MPK = L¹ᐟ²K⁻¹ᐟ² MPK = L¹ᐟ²/K¹ᐟ² = (L/K)¹ᐟ² (b) Nature of returns to scale Returns to scale describe what happens to output when all inputs are increased by the same proportion. To check this, we multiply all inputs (L and K) by a constant factor λ (where λ > 1). New output, Q’ = f(λL, λK) Q’ = 2(λL)¹ᐟ²(λK)¹ᐟ² Q’ = 2 λ¹ᐟ² L¹ᐟ² λ¹ᐟ² K¹ᐟ² Q’ = λ⁽¹ᐟ²⁺¹ᐟ²⁾ * (2L¹ᐟ²K¹ᐟ²) Q’ = λ¹ * Q Since Q’ = λQ, the output increases by exactly the same proportion (λ) as the inputs were increased. This is a homogeneous function of degree 1. When a production function is homogeneous of degree 1, it exhibits Constant Returns to Scale (CRS) . This means if you double all your inputs, your output will also exactly double. (c) If total production is exhausted or not (Product Exhaustion Theorem) The product exhaustion theorem, also known as the economic application of Euler’s Theorem, addresses whether the total output is fully distributed when factors are paid their marginal products. According to Euler’s theorem, if a function f(x₁, x₂) is homogeneous of degree k, then: x₁ (∂f/∂x₁) + x₂ (∂f/∂x₂) = k * f(x₁, x₂) In our case, the production function Q(L, K) is homogeneous of degree 1 (k=1). Therefore, according to Euler’s theorem: L (∂Q/∂L) + K (∂Q/∂K) = 1 * Q L MPL + K MPK = Q Let’s verify this with our calculated marginal products: LHS = L (K¹ᐟ²/L¹ᐟ²) + K (L¹ᐟ²/K¹ᐟ²) LHS = L¹⁻¹ᐟ² K¹ᐟ² + K¹⁻¹ᐟ² L¹ᐟ² LHS = L¹ᐟ²K¹ᐟ² + K¹ᐟ²L¹ᐟ² LHS = 2L¹ᐟ²K¹ᐟ² And since the original production function is Q = 2L¹ᐟ²K¹ᐟ², we see that: LHS = Q = RHS This means that if labor is paid a wage equal to its marginal product (MPL) and capital is paid a return equal to its marginal product (MPK), the sum of these payments will be exactly equal to the total output (Q). Therefore, yes, the total product is exhausted . This is known as the product exhaustion theorem and it holds under constant returns to scale.

Q3. If a two-product firm faces the demand and cost functions given below : Q₁ = 40 – 2P₁ + P₂ Q₂ = 15 + P₁ – P₂ C = Q₁² + Q₁Q₂ + Q₂² (a) Find the output levels that satisfy first order condition for maximum profit. (b) Check the second order sufficient condition. (c) Find the maximum profit level.

Ans. To maximize profit, the firm needs to express the profit function (π) as a function of Q₁ and Q₂ and then find the values of Q₁ and Q₂ that maximize it. Profit (π) = Total Revenue (TR) – Total Cost (TC) π = (P₁Q₁ + P₂Q₂) – (Q₁² + Q₁Q₂ + Q₂²) Step 1: Express P₁ and P₂ in terms of Q₁ and Q₂ (Inverse Demand Functions). We have: 1) Q₁ = 40 – 2P₁ + P₂ 2) Q₂ = 15 + P₁ – P₂ From equation (2), express P₁: P₁ = Q₂ + P₂ – 15 Substitute this into equation (1): Q₁ = 40 – 2(Q₂ + P₂ – 15) + P₂ Q₁ = 40 – 2Q₂ – 2P₂ + 30 + P₂ Q₁ = 70 – 2Q₂ – P₂ => P₂ = 70 – Q₁ – 2Q₂ Now substitute this expression for P₂ back into the expression for P₁: P₁ = Q₂ + (70 – Q₁ – 2Q₂) – 15 => P₁ = 55 – Q₁ – Q₂ Step 2: Write the profit function in terms of Q₁ and Q₂. TR = P₁Q₁ + P₂Q₂ TR = (55 – Q₁ – Q₂)Q₁ + (70 – Q₁ – 2Q₂)Q₂ TR = 55Q₁ – Q₁² – Q₁Q₂ + 70Q₂ – Q₁Q₂ – 2Q₂² TR = 55Q₁ + 70Q₂ – Q₁² – 2Q₁Q₂ – 2Q₂² Now the profit function: π(Q₁, Q₂) = TR – C π = (55Q₁ + 70Q₂ – Q₁² – 2Q₁Q₂ – 2Q₂²) – (Q₁² + Q₁Q₂ + Q₂²) π = 55Q₁ + 70Q₂ – 2Q₁² – 3Q₁Q₂ – 3Q₂² (a) First-Order Condition (FOC) For maximum profit, the partial derivatives of the profit function with respect to Q₁ and Q₂ must be zero. ∂π/∂Q₁ = 55 – 4Q₁ – 3Q₂ = 0 — (i) ∂π/∂Q₂ = 70 – 3Q₁ – 6Q₂ = 0 — (ii) Now we solve this system of two linear equations for Q₁ and Q₂. Multiply equation (i) by 2: 110 – 8Q₁ – 6Q₂ = 0 — (iii) Subtract equation (ii) from equation (iii): (110 – 8Q₁ – 6Q₂) – (70 – 3Q₁ – 6Q₂) = 0 40 – 5Q₁ = 0 5Q₁ = 40 => Q₁ = 8 Substitute Q₁ = 8 into equation (i): 55 – 4(8) – 3Q₂ = 0 55 – 32 – 3Q₂ = 0 23 – 3Q₂ = 0 3Q₂ = 23 => Q₂ = 23/3 ≈ 7.67 Thus, the output levels satisfying the first-order condition are Q₁ = 8 and Q₂ = 23/3 . (b) Second-Order Sufficient Condition (SOC) For a maximum, we must check that the Hessian matrix is negative definite. π₁₁ = ∂²π/∂Q₁² = -4 π₂₂ = ∂²π/∂Q₂² = -6 π₁₂ = ∂²π/(∂Q₁∂Q₂) = -3 The conditions are: 1) π₁₁ < 0 2) |H| = π₁₁π₂₂ – (π₁₂)² > 0 Checking: 1) π₁₁ = -4 < 0 (Condition met) 2) |H| = (-4)(-6) – (-3)² = 24 – 9 = 15 > 0 (Condition met) Since both conditions are met, the second-order sufficient condition is satisfied, and the profit is maximized at Q₁ = 8, Q₂ = 23/3. (c) Find the maximum profit level Substitute Q₁ = 8 and Q₂ = 23/3 into the profit function to calculate the maximum profit. π = 55Q₁ + 70Q₂ – 2Q₁² – 3Q₁Q₂ – 3Q₂² π = 55(8) + 70(23/3) – 2(8)² – 3(8)(23/3) – 3(23/3)² π = 440 + 1610/3 – 2(64) – 8(23) – 3(529/9) π = 440 + 1610/3 – 128 – 184 – 529/3 π = (440 – 128 – 184) + (1610/3 – 529/3) π = 128 + 1081/3 π = 384/3 + 1081/3 π = 1465/3 ≈ 488.33 Thus, the maximum profit is 1465/3 or approximately 488.33 .

Q4. Explain the chain rule in the case of multivariate functions using a suitable example.

Ans. Chain Rule for Multivariate Functions The chain rule is a fundamental rule in calculus used for finding the derivative of composite functions. In the multivariate case, the rule applies when we have a function that depends on several variables, and those variables are themselves functions of one or more other variables. Explanation of the Rule: Let z = f(x, y) be a function where x and y are both differentiable. Furthermore, let x and y themselves be functions of another variable t , i.e., x = g(t) and y = h(t) . Then z becomes a composite function of t. The chain rule states that the total derivative of z with respect to t is given by: This formula can be interpreted as follows:

• The total change in z resulting from a small change in t is the sum of two parts.
• First part : This is the rate of change of z transmitted through x due to the change in t. measures the sensitivity of z to x, and measures the sensitivity of x to t.
• Second part : This is the rate of change of z transmitted through y due to the change in t. measures the sensitivity of z to y, and measures the sensitivity of y to t.

Economic Example: Change in Utility over Time

A suitable economic example involves considering a consumer’s utility, which depends on the consumption of two goods, x and y. Suppose the consumer’s income changes over time, causing the consumption of the goods to also be a function of time.

1. Utility Function:

Let the consumer’s utility function be:

U = f(x, y) = x⁰.⁵y⁰.⁵

2. Consumption over Time:

Suppose the consumer’s income and preferences change over time (t), resulting in a consumption pattern as follows:

x(t) = 4t² (Consumption of good x increases with time)

y(t) = 100 – t² (Consumption of good y decreases with time)

Question:

What is the rate of change of utility (dU/dt) at time t = 2?

Solution:

We will use the chain rule:

Step A: Find the partial derivatives:

∂U/∂x = 0.5x⁻⁰.⁵y⁰.⁵

∂U/∂y = 0.5x⁰.⁵y⁻⁰.⁵

Step B: Find the derivatives with respect to time:

dx/dt = d(4t²)/dt = 8t

dy/dt = d(100 – t²)/dt = -2t

Step C: Substitute into the chain rule formula:

dU/dt = (0.5x⁻⁰.⁵y⁰.⁵)(8t) + (0.5x⁰.⁵y⁻⁰.⁵)(-2t)

Step D: Evaluate at t = 2:

First, find the values of x and y at t=2:

x(2) = 4(2)² = 16

y(2) = 100 – (2)² = 96

Now substitute t=2, x=16, y=96 into the expression for dU/dt:

dU/dt|ₜ=₂ = [0.5

(16)⁻⁰.⁵

(96)⁰.⁵]
(8

2) + [0.5

(16)⁰.⁵

(96)⁻⁰.⁵]
(-2

2)

dU/dt|ₜ=₂ = [0.5

(1/4)

√96]
(16) + [0.5

4

(1/√96)]
(-4)

dU/dt|ₜ=₂ = [0.125

4√6]
16 – [2

(1/4√6)]
(4)

dU/dt|ₜ=₂ = (0.5√6) * 16 – (2 / √6)

dU/dt|ₜ=₂ = 8√6 – 2/√6

dU/dt|ₜ=₂ = 8√6 – (2√6 / 6)

dU/dt|ₜ=₂ = 8√6 – √6/3

dU/dt|ₜ=₂ = (24√6 – √6)/3 = 23√6 / 3 ≈ 18.78

Conclusion:

In this example, the chain rule helps us understand how the consumer’s total satisfaction (utility) changes as underlying economic conditions (like income or preferences, represented here by t) evolve. At t=2, utility is increasing at a rate of approximately 18.78 units per unit of time.

Q5. Find the inverse of : B =

Ans. The inverse of a matrix B, denoted B⁻¹, is given by the formula: B⁻¹ = (1/|B|) * adj(B) where |B| is the determinant of matrix B and adj(B) is the adjugate matrix of B. Given matrix: B = Step 1: Calculate the determinant |B|. We expand along the first row: |B| = 4 – 1 + (-1) * |B| = 4 ((3)(7) – (2)(0)) – 1 ((0)(7) – (2)(3)) – 1 * ((0)(0) – (3)(3)) |B| = 4 (21 – 0) – 1 (0 – 6) – 1 * (0 – 9) |B| = 4(21) – 1(-6) – 1(-9) |B| = 84 + 6 + 9 |B| = 99 Since |B| ≠ 0, the inverse exists. Step 2: Find the matrix of cofactors. The cofactor Cᵢⱼ = (-1)ⁱ⁺ʲ * Mᵢⱼ, where Mᵢⱼ is the minor. C₁₁ = (-1)¹⁺¹ = 1 * (21 – 0) = 21 C₁₂ = (-1)¹⁺² = -1 * (0 – 6) = 6 C₁₃ = (-1)¹⁺³ = 1 * (0 – 9) = -9 C₂₁ = (-1)²⁺¹ = -1 * (7 – 0) = -7 C₂₂ = (-1)²⁺² = 1 * (28 – (-3)) = 31 C₂₃ = (-1)²⁺³ = -1 * (0 – 3) = 3 C₃₁ = (-1)³⁺¹ = 1 * (2 – (-3)) = 5 C₃₂ = (-1)³⁺² = -1 * (8 – 0) = -8 C₃₃ = (-1)³⁺³ = 1 * (12 – 0) = 12 The matrix of cofactors is: C = Step 3: Find the adjugate matrix adj(B). The adj(B) is the transpose of the cofactor matrix C. adj(B) = Cᵀ = Step 4: Calculate the inverse matrix B⁻¹. B⁻¹ = (1/|B|) * adj(B) B⁻¹ = (1/99) * Thus, the inverse of B is: B⁻¹ = =

Q6. Find total differential of the following functions : (a) y = x₁² / (x₂x₃) (b) y = x₁ / (x₁ + x₂)

Ans. The total differential, dy, of a function y = f(x₁, x₂, …, xₙ) approximates the total change in the function’s value for small changes in the variables. Its formula is: dy = (∂y/∂x₁)dx₁ + (∂y/∂x₂)dx₂ + … + (∂y/∂xₙ)dxₙ (a) y = x₁² / (x₂x₃) We need to find the partial derivatives of y with respect to x₁, x₂, and x₃. 1. Partial derivative with respect to x₁: ∂y/∂x₁ = (2x₁)/(x₂x₃) 2. Partial derivative with respect to x₂: y = x₁²x₂⁻¹x₃⁻¹ ∂y/∂x₂ = x₁²(-1)x₂⁻²x₃⁻¹ = -x₁² / (x₂²x₃) 3. Partial derivative with respect to x₃: ∂y/∂x₃ = x₁²x₂⁻¹(-1)x₃⁻² = -x₁² / (x₂x₃²) Now, substitute these partial derivatives into the total differential formula: dy = (∂y/∂x₁)dx₁ + (∂y/∂x₂)dx₂ + (∂y/∂x₃)dx₃ dy = [2x₁ / (x₂x₃)] dx₁ – [x₁² / (x₂²x₃)] dx₂ – [x₁² / (x₂x₃²)] dx₃ (b) y = x₁ / (x₁ + x₂) We need to find the partial derivatives of y with respect to x₁ and x₂. We will use the quotient rule: d(u/v) = (v(du) – u(dv)) / v² 1. Partial derivative with respect to x₁: u = x₁, v = x₁ + x₂ ∂y/∂x₁ = [(x₁ + x₂)(1) – x₁(1)] / (x₁ + x₂)² ∂y/∂x₁ = [x₁ + x₂ – x₁] / (x₁ + x₂)² ∂y/∂x₁ = x₂ / (x₁ + x₂)² 2. Partial derivative with respect to x₂: u = x₁, v = x₁ + x₂ ∂y/∂x₂ = [(x₁ + x₂)(0) – x₁(1)] / (x₁ + x₂)² ∂y/∂x₂ = -x₁ / (x₁ + x₂)² Now, substitute these partial derivatives into the total differential formula: dy = (∂y/∂x₁)dx₁ + (∂y/∂x₂)dx₂ dy = [x₂ / (x₁ + x₂)²] dx₁ – [x₁ / (x₁ + x₂)²] dx₂ This can also be written as: dy = (1 / (x₁ + x₂)²) * (x₂dx₁ – x₁dx₂)

Q7. Explain the following concepts : (a) Compensated demand function (b) Shephard’s lemma

Ans. (a) Compensated Demand Function The compensated demand function, also known as the Hicksian Demand Function , is a crucial concept in microeconomics. It shows how much of a good a consumer would demand if, after a change in the price of the good, they were “compensated” in such a way that they remain on their original level of utility. Key Characteristics:

• Holds Utility Constant: Unlike the ordinary (Marshallian) demand function, which holds income constant, the compensated demand function holds utility constant.
• Derived from Expenditure Minimization: It is the solution to the consumer’s expenditure minimization problem. The consumer seeks to achieve a certain utility level (U) with the minimum possible expenditure (E). Minimize E = PₓX + PᵧY Subject to U(X, Y) = U* (a constant utility level)
• Substitution Effect Only: Since the consumer’s real income (in the form of utility) is held constant after a price change, the compensated demand curve illustrates only the substitution effect . It does not include the income effect.
• Slope: As a result, the compensated demand curve is always downward sloping (a negative relationship between price and quantity), even for Giffen goods, because the substitution effect is always negative.
• Mathematical Form: It is represented as hᵢ(P, U), meaning the compensated demand for good i is a function of all prices (P) and a fixed utility level (U).

In essence, the compensated demand function is a theoretical construct that helps isolate the pure substitution effect of a price change.

(b) Shephard’s Lemma

Shephard’s Lemma establishes a powerful and direct relationship between the expenditure function and the compensated demand function. It is a key result of duality theory, which links the utility maximization and expenditure minimization problems.

Statement of the Lemma:

Shephard’s Lemma states that if a consumer’s expenditure function e(P, U) is differentiable with respect to prices (P), then the compensated (Hicksian) demand function for good

i

is the partial derivative of that expenditure function with respect to the price of good

i

(Pᵢ).

Mathematically:

hᵢ(P, U) = ∂e(P, U) / ∂Pᵢ

Where:

• hᵢ(P, U) is the compensated demand function for good i.
• e(P, U) is the minimum expenditure necessary to achieve a utility level U, given a set of prices P.
• ∂e(P, U) / ∂Pᵢ is the partial derivative of the expenditure function with respect to the i-th component of the price vector P.

Significance and Use:

• Derivation of Demand Functions: If we know a consumer’s expenditure function, we can directly derive their compensated demand functions using Shephard’s Lemma. This is a very convenient shortcut.
• Duality Relationship: The lemma highlights the duality relationship between utility and expenditure. Just as Roy’s Identity derives Marshallian demand from the indirect utility function, Shephard’s Lemma derives Hicksian demand from the expenditure function.
• Application in Welfare Economics: Since compensated demand curves provide the basis for measuring welfare changes (like equivalent variation and compensating variation), Shephard’s Lemma is a crucial tool for calculating these measures.

Q8. If the utility function is u = ax + by + cxy, find the ratio of marginal utilities of the two goods x and y.

Ans. The given utility function is: u(x, y) = ax + by + cxy where a, b, and c are constants. The Marginal Utility (MU) of a good is the change in total utility from consuming one additional unit of that good, holding the quantity of the other good constant. It is calculated by taking the partial derivative of the utility function. 1. Find the Marginal Utility of good x (MUₓ): MUₓ is calculated by partially differentiating u with respect to x, treating y as a constant. MUₓ = ∂u/∂x = ∂(ax + by + cxy)/∂x MUₓ = ∂(ax)/∂x + ∂(by)/∂x + ∂(cxy)/∂x MUₓ = a 1 + 0 + cy 1 MUₓ = a + cy 2. Find the Marginal Utility of good y (MUᵧ): MUᵧ is calculated by partially differentiating u with respect to y, treating x as a constant. MUᵧ = ∂u/∂y = ∂(ax + by + cxy)/∂y MUᵧ = ∂(ax)/∂y + ∂(by)/∂y + ∂(cxy)/∂y MUᵧ = 0 + b 1 + cx 1 MUᵧ = b + cx 3. Find the ratio of marginal utilities: The ratio of the marginal utilities of the two goods x and y is MUₓ / MUᵧ. This is also equal to the Marginal Rate of Substitution (MRS). Ratio = MUₓ / MUᵧ Substituting the expressions derived for MUₓ and MUᵧ: Ratio (MRSₓᵧ) = (a + cy) / (b + cx) Thus, the ratio of the marginal utilities of the two goods x and y is (a + cy) / (b + cx) . This ratio depends on the quantities of x and y consumed, which means the marginal rate of substitution is not constant and varies along an indifference curve.

Q9. If: u’ = [5, 1, 3] v’ = [3, 1, -1] w’ = [7, 5, 8] x = [x₁, x₂, x₃] Compute : (a) u’v (b) w’x

Ans. In the given question, u’, v’, and w’ are presented as row vectors. The vector x is also shown as a list of components. The notation ‘u’v’ typically means the dot product (or inner product) of the two vectors u and v. We will assume that ‘v’ and ‘x’ here represent column vectors, as the product of a row vector and a column vector yields a scalar, which is equivalent to the dot product. If u’=[u₁, u₂, u₃] is a row vector and v= is a column vector, their product u’v = u₁v₁ + u₂v₂ + u₃v₃. Given vectors: u’ = [5, 1, 3] v = (as the transpose of v’=[3,1,-1]) w’ = [7, 5, 8] x = (as the transpose of [x₁, x₂, x₃]) (a) Compute u’v u’v is equivalent to the dot product of vectors u and v. u’v = [5, 1, 3] * u’v = (5 3) + (1 1) + (3 * -1) u’v = 15 + 1 – 3 u’v = 13 (b) Compute w’x w’x is equivalent to the dot product of vectors w and x. w’x = [7, 5, 8] * w’x = (7 x₁) + (5 x₂) + (8 * x₃) w’x = 7x₁ + 5x₂ + 8x₃ This result is a linear expression because the vector x contains variables. In economics, such a computation is often used to represent a budget constraint (e.g., p’x = M) or to evaluate a linear function.

Q10. Solve the following differential equation : (d²y/dt²) + (dy/dt) + y = 0

Ans. The given equation is a second-order, linear, homogeneous differential equation with constant coefficients. y” + y’ + y = 0 To solve this type of equation, we use the characteristic equation method. We assume a solution of the form y = eʳᵗ, which gives y’ = reʳᵗ and y” = r²eʳᵗ. Substituting these into the original equation: r²eʳᵗ + reʳᵗ + eʳᵗ = 0 eʳᵗ(r² + r + 1) = 0 Since eʳᵗ can never be zero, we must solve the characteristic equation: r² + r + 1 = 0 This is a quadratic equation which we can solve using the quadratic formula: r = [-b ± √(b² – 4ac)] / 2a Where a = 1, b = 1, and c = 1. r = [-1 ± √(1² – 4 1 1)] / (2*1) r = [-1 ± √(1 – 4)] / 2 r = [-1 ± √(-3)] / 2 r = [-1 ± i√3] / 2 We get two complex conjugate roots: r₁ = -1/2 + i(√3/2) r₂ = -1/2 – i(√3/2) When the roots of the characteristic equation are complex and of the form α ± iβ, the general solution to the differential equation is given by: y(t) = e^(αt) * (A cos(βt) + B sin(βt)) In our case, α = -1/2 and β = √3/2. Substituting these values into the general solution formula, we get: y(t) = e^(-t/2) [A cos(√3/2 t) + B sin(√3/2 * t)] where A and B are two arbitrary constants. The values of these constants could be determined if initial conditions were provided. This solution represents a damped oscillation, where the amplitude of the oscillations decreases over time due to the e^(-t/2) term.

Q11. Determine if the following functions are homogeneous and of what degree ? (a) f(x, y) = x³ – xy⁵/² (b) f(x, y, w) = x⁴ – 5yw³

Ans. A function f(x₁, x₂, …) is said to be homogeneous of degree k if for any scalar λ > 0, the following condition holds: f(λx₁, λx₂, …) = λᵏ * f(x₁, x₂, …) (a) f(x, y) = x³ – xy⁵/² To check for homogeneity, we replace x with λx and y with λy: f(λx, λy) = (λx)³ – (λx)(λy)⁵/² f(λx, λy) = λ³x³ – λ x λ⁵/² * y⁵/² f(λx, λy) = λ³x³ – λ¹⁺⁵/² * xy⁵/² f(λx, λy) = λ³x³ – λ⁷/² * xy⁵/² It is not possible to factor out a single power of λ from this expression to make it of the form λᵏ * f(x, y). λ³x³ – λ⁷/²xy⁵/² ≠ λᵏ (x³ – xy⁵/²) for any k. This is because the degree of the first term (x³) is 3, while the degree of the second term (xy⁵/²) is 1 + 5/2 = 7/2. Since the terms have different degrees, the function is not homogeneous. Conclusion: The function f(x, y) = x³ – xy⁵/² is not homogeneous. (b) f(x, y, w) = x⁴ – 5yw³ To check for homogeneity, we replace x with λx, y with λy, and w with λw: f(λx, λy, λw) = (λx)⁴ – 5(λy)(λw)³ f(λx, λy, λw) = λ⁴x⁴ – 5(λy)(λ³w³) f(λx, λy, λw) = λ⁴x⁴ – 5λ¹⁺³yw³ f(λx, λy, λw) = λ⁴x⁴ – 5λ⁴yw³ Now we can factor out λ⁴ from the expression: f(λx, λy, λw) = λ⁴(x⁴ – 5yw³) Since f(x, y, w) = x⁴ – 5yw³, we can write: f(λx, λy, λw) = λ⁴ * f(x, y, w) This satisfies the definition of a homogeneous function with k = 4. This is possible because each term in the function has the same degree:

• The degree of the first term (x⁴) is 4.
• The degree of the second term (-5yw³) is the sum of the power of y (1) and the power of w (3), which is 1 + 3 = 4.

Conclusion: The function f(x, y, w) = x⁴ – 5yw³ is homogeneous of degree 4.

Q12. Write short notes on the following : (a) Vector space (b) Homothetic functions

Ans. (a) Vector Space A vector space (or linear space) is a mathematical structure composed of a set of objects called vectors (V), a set of scalars from a field (F), and two operations—vector addition and scalar multiplication. This structure is the foundation of linear algebra and allows for the generalization of many mathematical and physical concepts. A set V is called a vector space over a field F if it satisfies the following ten axioms: Axioms for Vector Addition (for u, v, w ∈ V):

1. Closure under addition: If u and v are in V, then u + v is also in V.
2. Associativity of addition: (u + v) + w = u + (v + w).
3. Commutativity of addition: u + v = v + u.
4. Existence of additive identity: There exists a zero vector 0 in V such that for all u ∈ V, u + 0 = u.
5. Existence of additive inverse: For each u ∈ V, there exists a vector -u in V such that u + (-u) = 0 .

Axioms for Scalar Multiplication (for a, b ∈ F and u, v ∈ V):

6. Closure under scalar multiplication: If a is a scalar and u is a vector in V, then au is also in V.
7. Distributivity over vector addition: a(u + v) = au + av.
8. Distributivity over scalar addition: (a + b)u = au + bu.
9. Associativity of scalar multiplication: a(bu) = (ab)u.
10. Scalar multiplication identity: For the multiplicative identity 1 of the field F, 1u = u.

The most common example is ℝⁿ, the set of all real vectors of n-tuples, which forms a vector space over the field of real numbers. In economics, the set of commodity bundles, the set of price vectors, etc., are often treated as vector spaces.

(b) Homothetic Functions

A homothetic function is a generalization of a homogeneous function. A function is called homothetic if it is a positive monotonic transformation of a homogeneous function.

Definition:

A function H(x) is homothetic if it can be written in the form H(x) = g(f(x)), where:

• f(x) is a homogeneous function.
• g(z) is a positive, continuously differentiable, and strictly increasing function, i.e., g'(z) > 0.

Key Property and Economic Importance:

The most important property of homothetic functions, particularly in utility and production theory, is that their

Marginal Rate of Substitution (MRS)

or

Marginal Rate of Technical Substitution (MRTS)

depends only on the ratio of the inputs, not on their absolute levels.

If the utility function U(x₁, x₂) is homothetic, then:

MRS = MU₁/MU₂ = h(x₁/x₂)

This means that along any ray from the origin (where the ratio x₁/x₂ is constant), the slope of the indifference curves is the same.

Economic Implications:

1. Linear Expansion Paths: For a consumer with homothetic preferences or a firm with a homothetic production function, the expansion path (income-consumption curve or output expansion path) is a straight line from the origin. This means that as income or budget increases, the consumer increases consumption of all goods in the same proportion.
2. Shape of Indifference Curves: All indifference curves of a homothetic function are “radial blow-ups” of one another. They change in size as one moves away from the origin, but their shape (pattern of slopes) remains the same.
3. Effect on Demand: For a homothetic utility function, the fraction of income spent on each good depends only on prices, not on income.

Example: If f(L, K) = L⁰.⁵K⁰.⁵ is a homogeneous production function, then H(L, K) = [ln(L⁰.⁵K⁰.⁵) + 5]³ is a homothetic function, as it is a monotonic transformation of f.