SECTION-A
Answer any ten of the following questions: 2 x 10 = 20
1. Define set. Give an example of null set.
A set is a well-defined collection of distinct objects
. A null set (or empty set) is a set containing no elements, denoted by { } or ∅. Example: A set of humans living on the Sun.2. Mention two conditions of continuity of a function.
- The limit of the function f(x) as x approaches 'a' must exist.
- The limit must be equal to the function's value at that point, i.e., Lt (x→a) f(x) = f(a).
- Singular Matrix: A square matrix whose determinant is zero (|A| = 0).
- Non-singular Matrix: A square matrix whose determinant is not zero (|A| ≠ 0).
- First-order condition: The first derivative must be zero (dy/dx = 0).
- Second-order condition: For a maximum, d²y/dx² < 0; for a minimum, d²y/dx² > 0.
- ∂z/∂x = 4x + 5y + 8
- ∂z/∂y = 5x + 20
- (a) x + C
- (b) log x + C
- Domain: {1, 2, 3, 4, 5, 6, 7}.
- Range: {2, 3, 4, 5, 6, 7, 8}.
- Diagonal: Non-zero elements only on the main diagonal.
- Identity: A diagonal matrix with all 1s on the main diagonal.
- Scalar: A diagonal matrix where all diagonal elements are the same constant.
- Profit Maximization: MR = MC.
- Elasticity of Demand: Ed = -(p/x) * (dx/dp).
- Consumer Surplus: CS = ∫ f(x)dx - p₀x₀.
- For matrix inverse, always check if the determinant is non-zero first.
- In optimization problems, never forget to verify the second-order condition (max/min).
- Show step-by-step substitution for IS-LM models to avoid calculation errors.
3. If A = {1, 2, 3, 4} and B = {6, 7, 8}, find A ∩ B.
Intersection (A ∩ B) contains elements common to both sets
. Since there are no common elements, A ∩ B = ∅ (Null set).4. Define symmetric matrix.
A square matrix A is called symmetric if it is equal to its transpose, i.e., A = A'
.5. Distinguish between singular matrix and non-singular matrix.
6. Find rank of the matrix A = [[8, 7, 0], [0, 7, 3], [2, 5, 2]].
The rank is the dimension of the largest non-zero minor
. Calculating the determinant: |A| = 8(14-15) - 7(0-6) + 0 = 8(-1) + 42 = 34. Since |A| ≠ 0, the Rank = 3.7. What is convex function?
A function is convex if a line segment between any two points on its graph lies above or on the graph
. Mathematically, its second derivative is non-negative (f''(x) ≥ 0).8. Differentiate y = a^x.
The derivative is dy/dx = a^x log a
.9. Mention the order conditions for maximum-minimum values.
10. Define total derivative.
A total derivative measures the rate of change of a multi-variable function with respect to a variable when all other variables also depend on that same variable
.11. Find partial derivatives of z = (x + 4)(2x + 5y).
z = 2x² + 5xy + 8x + 20y
.12. Find total differential of z = √(x + y).
Using dz = (∂z/∂x)dx + (∂z/∂y)dy
.dz = [1 / (2√(x+y))] dx + [1 / (2√(x+y))] dy
.13. Define integration.
Integration is the inverse process of differentiation, used to find the total accumulation or area under a curve
.14. Marginal cost is (1 + x + 6x²). Find total cost function if fixed cost is 100.
TC = ∫(1 + x + 6x²)dx = x + x²/2 + 2x³ + C
. Since FC = 100, C = 100.TC = 2x³ + 0.5x² + x + 100
.15. Integrate: (a) ∫1dx (b) ∫(1/x)dx.
SECTION-B
Answer any five of the following questions: 10 x 5 = 50
16. (a) Prove (A ∪ B) × C = (A × C) ∪ (B × C). (b) Cartesian product. (c) R = {(x,y): y=x+1} on A={1..8}. [3+3+4=10]
(a) Proof: Let (x, y) ∈ (A ∪ B) × C. Then x ∈ (A ∪ B) and y ∈ C
. This implies (x ∈ A or x ∈ B) and y ∈ C. Thus (x ∈ A and y ∈ C) or (x ∈ B and y ∈ C), meaning (x, y) ∈ (A × C) ∪ (B × C).(b) Cartesian Product: The set of all ordered pairs (a, b) where a ∈ A and b ∈ B
. Example: If A={1,2} and B={3}, A×B = {(1,3), (2,3)}.(c) Relation: R = {(1,2), (2,3), (3,4), (4,5), (5,6), (6,7), (7,8)}
.18. (a) Find inverse of matrix A = [[1, 2, 3], [1, 3, 5], [1, 5, 12]]. [5]
First, find |A|: 1(36-25) - 2(12-5) + 3(5-3) = 11 - 14 + 6 = 3
.Next, find Adjoint A by transposing the matrix of cofactors
.A⁻¹ = (1/|A|) Adjoint A = (1/3) [[11, -9, 1], [-7, 9, -2], [2, -3, 1]]
.19. (a) IS-LM: 0.4Y + 150i = 209; 0.1Y - 250i = 35. (b) Matrix types. [5+5=10]
(a) Equilibrium: Multiply second equation by 4: 0.4Y - 1000i = 140
. Subtracting from first: 1150i = 69, so i = 0.06. Substitute i: 0.1Y - 250(0.06) = 35 → 0.1Y - 15 = 35 → 0.1Y = 50 → Y = 500.(b) Matrix Illustrations:
20. (c) Demand law x = 20/(p+1). Find Elasticity of Demand (Ed) at p=3. [3]
Ed = -(p/x) * (dx/dp)
.dx/dp = -20 / (p+1)²
. At p=3, x = 20/4 = 5 and dx/dp = -20/16 = -1.25.Ed = -(3/5) * (-1.25) = 0.75
.21. (b) Cost C(x)=250+0.005x², Revenue R=4x. Maximize profit. [5+3=8]
Profit (π) = R - C = 4x - 0.005x² - 250
.dπ/dx = 4 - 0.01x = 0 → x = 400
.Second derivative: -0.01 < 0 (Maximum confirmed)
.Max Profit: 4(400) - 0.005(160000) - 250 = 1600 - 800 - 250 = 550
.25. (a) MC = 4 + 0.08x, MR = 12. Fixed cost = 0. Total profit. [5]
Profit is maximized where MR = MC
. 12 = 4 + 0.08x → 8 = 0.08x → x = 100.Total Profit = ∫(MR - MC)dx from 0 to 100 = ∫(8 - 0.08x)dx
.= [8x - 0.04x²] from 0 to 100 = 800 - 400 = 400
.25. (b) Demand p = 85 - 4x - x². Consumer's Surplus (CS) if x₀ = 5. [5]
Price at x₀ = 5: p₀ = 85 - 20 - 25 = 40
.CS = ∫(85 - 4x - x²)dx from 0 to 5 - (p₀ * x₀)
.= [85x - 2x² - x³/3] from 0 to 5 - (40 * 5)
.= (425 - 50 - 41.67) - 200 = 333.33 - 200 = 133.33
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