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IGNOU MEC-203 Previous Year Solved Question Paper in Hindi

IGNOU MEC-203 Previous Year Solved Question Paper in English

Q1. (a) Given the following demand function: q=Ap ^ε , where ε < 0. Obtain the price elasticity of demand. (b) Given a linearly homogeneous production function: Q=AK ^α L ^β where A, α, β > 0 and α + β = 1. Obtain marginal productivities of L and K and also verify the Euler theorem.

Ans. (a) Price Elasticity of Demand The given demand function is: q = Ap ^ε where q is the quantity demanded, p is the price, A is a positive constant, and ε is the elasticity coefficient (ε < 0). The price elasticity of demand (E _d ) is defined as the responsiveness of the percentage change in quantity demanded to a percentage change in price. The formula is: E _d = (% Change in Quantity Demanded) / (% Change in Price) = (Δq/q) / (Δp/p) = (dq/dp) * (p/q) To calculate this, we first need to find the derivative of the demand function (q) with respect to price (p): dq/dp = d/dp (Ap ^ε ) Using the power rule, we get: dq/dp = A ε p ^(ε-1) Now, we substitute this derivative into the elasticity formula: E _d = (A ε p ^(ε-1) ) * (p/q) We know that q = Ap ^ε . Substituting this into the formula: E _d = (Aεp ^ε-1 ) * (p / (Ap ^ε )) Now we simplify the expression: E _d = (Aεp ^ε-1+1 ) / (Ap ^ε ) E _d = (Aεp ^ε ) / (Ap ^ε ) Cancelling Ap ^ε from the numerator and denominator, we get: E _d = ε Thus, for the given demand function, the price elasticity of demand is constant and equal to ε . This type of demand function is known as a constant-elasticity demand function . Since it is given that ε < 0, this is consistent with the law of demand, which states an inverse relationship between price and quantity demanded. (b) Marginal Productivities and Euler’s Theorem The given production function is a linearly homogeneous production function of the Cobb-Douglas type: Q = AK ^α L ^β where Q is output, K is capital, L is labour, and A, α, β are positive constants. It is also given that α + β = 1, which indicates constant returns to scale. 1. Marginal Productivities: Marginal Product of Labour (MP _L ): This is the change in output from using one additional unit of labour, holding capital constant. It is calculated by taking the partial derivative of Q with respect to L. MP _L = ∂Q/∂L = ∂/∂L (AK ^α L ^β ) MP _L = AK ^α * (βL ^β-1 ) MP _L = β * (AK ^α L ^β / L) Since Q = AK ^α L ^β , we can write this as: MP _L = β(Q/L) This is β times the average product of labour (Q/L). Marginal Product of Capital (MP _K ): This is the change in output from using one additional unit of capital, holding labour constant. It is calculated by taking the partial derivative of Q with respect to K. MP _K = ∂Q/∂K = ∂/∂K (AK ^α L ^β ) MP _K = A (αK ^α-1 ) L ^β MP _K = α * (AK ^α L ^β / K) Since Q = AK ^α L ^β , we can write this as: MP _K = α(Q/K) This is α times the average product of capital (Q/K). 2. Verification of Euler’s Theorem: Euler’s theorem states that if a function Q(K, L) is homogeneous of degree ‘n’, then: K (∂Q/∂K) + L (∂Q/∂L) = nQ In our case, the production function is homogeneous of degree (α + β). Since it is given that α + β = 1, the function is homogeneous of degree 1. Therefore, we need to verify that: K MP _K + L MP _L = 1 * Q Substituting the expressions we found for MP _K and MP _L : Left Hand Side = K [α(Q/K)] + L [β(Q/L)] Cancelling K and L: LHS = αQ + βQ Factoring out Q: LHS = (α + β)Q Since we are given that α + β = 1: LHS = (1)Q = Q Thus, Left Hand Side = Right Hand Side (Q = Q). Hence, Euler’s theorem is verified . This demonstrates that if the factors of production (labour and capital) are paid their marginal products, the total output will be exactly exhausted (the product exhaustion theorem).

Q2. Solve the following system of equations using the Cramer’s rule : 5x₁ – 2x₂ + 3x₃ = 6 2x₁ – 5x₃ = 2 4x₁ – 5x₂ + 6x₃ = 7

Ans. The given system of linear equations is: 5x₁ – 2x₂ + 3x₃ = 6 2x₁ + 0x₂ – 5x₃ = 2 4x₁ – 5x₂ + 6x₃ = 7 Cramer’s Rule is a method used to find the solution of a system of linear equations, provided the determinant of the coefficient matrix is non-zero. The solution is given by: x₁ = |A₁| / |A|, x₂ = |A₂| / |A|, x₃ = |A₃| / |A| Step 1: Find the determinant of the coefficient matrix, |A|. The coefficient matrix A is:

 | 5 -2 3 |A = | 2 0 -5 | | 4 -5 6 |

The determinant |A| can be calculated by expanding along the first row: |A| = 5 (0 6 – (-5) (-5)) – (-2) (2 6 – (-5) 4) + 3 (2 (-5) – 0 * 4) |A| = 5 (0 – 25) + 2 (12 + 20) + 3 * (-10 – 0) |A| = 5 (-25) + 2 (32) + 3 * (-10) |A| = -125 + 64 – 30 |A| = -91 Since |A| ≠ 0, a unique solution exists. Step 2: Calculate |A₁|. To find |A₁|, we replace the first column of A with the constant vector B = [6, 2, 7]ᵀ:

 | 6 -2 3 |A₁ =| 2 0 -5 | | 7 -5 6 |

|A₁| = 6 (0 6 – (-5) (-5)) – (-2) (2 6 – (-5) 7) + 3 (2 (-5) – 0 * 7) |A₁| = 6 (-25) + 2 (12 + 35) + 3 * (-10) |A₁| = -150 + 2 * (47) – 30 |A₁| = -150 + 94 – 30 |A₁| = -86 Step 3: Calculate |A₂|. To find |A₂|, we replace the second column of A with B:

 | 5 6 3 |A₂ =| 2 2 -5 | | 4 7 6 |

|A₂| = 5 (2 6 – (-5) 7) – 6 (2 6 – (-5) 4) + 3 (2 7 – 2 * 4) |A₂| = 5 (12 + 35) – 6 (12 + 20) + 3 * (14 – 8) |A₂| = 5 (47) – 6 (32) + 3 * (6) |A₂| = 235 – 192 + 18 |A₂| = 61 Step 4: Calculate |A₃|. To find |A₃|, we replace the third column of A with B:

 | 5 -2 6 |A₃ =| 2 0 2 | | 4 -5 7 |

|A₃| = 5 (0 7 – 2 (-5)) – (-2) (2 7 – 2 4) + 6 (2 (-5) – 0 * 4) |A₃| = 5 (0 + 10) + 2 (14 – 8) + 6 * (-10) |A₃| = 5 (10) + 2 (6) – 60 |A₃| = 50 + 12 – 60 |A₃| = 2 Step 5: Solve for the variables. Now we use Cramer’s rule to find the values of x₁, x₂, and x₃: x₁ = |A₁| / |A| = -86 / -91 = 86/91 x₂ = |A₂| / |A| = 61 / -91 = -61/91 x₃ = |A₃| / |A| = 2 / -91 = -2/91 Thus, the solution to the system of equations is: x₁ = 86/91, x₂ = -61/91, x₃ = -2/91

Q3. (a) Explain how the Harrod-Domar model can be solved with the help of a differential equation. (b) With the help of an example, find out the general solution of a first order differential equation.

Ans. (a) The Harrod-Domar Model and Differential Equations The Harrod-Domar model is a Keynesian model of economic growth. It explains how an economy’s growth rate depends on the savings ratio and the capital-output ratio. The core equation of the model can be derived and solved through a differential equation. Key Assumptions of the Model:

1. Savings Function: Aggregate savings (S) are a constant proportion (s) of national income (Y). S = sY, where ‘s’ is the marginal propensity to save (0 < s < 1).
2. Investment and Capital Stock: Net investment (I) is equal to the rate of change of the capital stock (K). I = ΔK = dK/dt
3. Production Function: The capital-output ratio (v) is assumed to be constant. This means ‘v’ units of capital are required to produce one unit of output. v = K/Y or K = vY Differentiating this gives another expression for investment: I = dK/dt = v(dY/dt)

Formation and Solution of the Differential Equation:

In equilibrium, planned savings must equal planned investment in an economy:

S = I

Substituting the expressions from the assumptions above:

sY = v(dY/dt)

Rearranging this equation gives us a first-order linear differential equation for income (Y):

dY/dt = (s/v)Y

or

(dY/dt) – (s/v)Y = 0

This differential equation can be solved by separation of variables:

(1/Y) dY = (s/v) dt

Integrating both sides:

∫(1/Y) dY = ∫(s/v) dt

ln(Y) = (s/v)t + C

where C is the constant of integration.

Solving for Y:

Y(t) = e

^{(s/v)t + C}

= e

^C

* e

^(s/v)t

If we consider the initial income at time t=0 to be Y₀, then Y(0) = Y₀ = e

^C

.

Thus, the solution is:

Y(t) = Y₀ * e

^(s/v)t

This solution demonstrates that in the Harrod-Domar model, for equilibrium to be maintained, income must grow at a constant rate. This growth rate, known as the

‘Warranted Rate of Growth’ (G

_W

)

, is equal to G

_W

= s/v. The differential equation thus helps to show that for steady growth, income must grow exponentially.

(b) General Solution of a First-Order Differential Equation (with Example)

A standard form for a first-order linear differential equation is:

dy/dx + P(x)y = Q(x)

A standard method for solving this type of equation is by using an

‘Integrating Factor’ (IF)

.

Method of Solution:

1. Calculate the Integrating Factor (IF): The IF is calculated by the formula: IF = e ^∫P(x)dx
2. Multiply the equation by the IF: Multiply both sides of the original equation by the IF. IF (dy/dx) + IF P(x)y = IF * Q(x) After this step, the left-hand side of the equation is always the derivative of (y IF) with respect to x, i.e., d/dx(y IF).
3. Integrate both sides: ∫d/dx(y IF) dx = ∫IF Q(x) dx y IF = ∫IF Q(x) dx + C
4. Solve for y: Isolate y to obtain the general solution. y = (1/IF) [∫IF Q(x) dx + C] The ‘C’ here is the constant of integration, and its presence is why the solution is called a ‘general solution’.

Example:

Find the general solution of the following differential equation:

dy/dx + 3y = 9

Step 1: Identify P(x) and Q(x).

Comparing with the standard form dy/dx + P(x)y = Q(x), we find:

P(x) = 3

Q(x) = 9

Step 2: Calculate the Integrating Factor (IF).

IF = e

^∫P(x)dx

= e

^∫3dx

= e

^3x

Step 3: Multiply the equation by the IF.

e

^3x

(dy/dx) + e

^3x

(3y) = 9e

^3x

The left side is equal to d/dx(y * e

^3x

).

d/dx(y * e

^3x

) = 9e

^3x

Step 4: Integrate both sides with respect to x.

∫d/dx(y * e

^3x

) dx = ∫9e

^3x

dx

y

e

^3x

= 9

(e

^3x

/ 3) + C

y * e

^3x

= 3e

^3x

+ C

Step 5: Solve for y.

Divide both sides of the equation by e

^3x

:

y = (3e

^3x

/ e

^3x

) + (C / e

^3x

)

y = 3 + Ce

^-3x

This is the

general solution

of the given first-order differential equation.

Q4. Distinguish between a parameter and a statistic. Explain the concept of the sampling distribution of a statistic using an example.

Ans. Parameter vs. Statistic In statistical analysis, parameter and statistic are two fundamental concepts, but they relate to the distinction between a population and a sample.

• Parameter:
  • A parameter is a numerical value that describes a characteristic of an entire population .
  • It is a fixed, constant value , but in practice, it is usually unknown because it is often impossible or impractical to collect data from the whole population.
  • Parameters are typically denoted using Greek letters.
  • Examples:
    • Population mean (μ)
    • Population standard deviation (σ)
    • Population proportion (P)
• Statistic:
  • A statistic is a numerical value that describes a characteristic of a sample , which is a subset of the population.
  • It is a variable value ; its value depends on the particular sample drawn from the population. If we take another sample from the same population, the value of the statistic will likely be different.
  • A statistic is used to estimate the corresponding population parameter.
  • Statistics are typically denoted using Roman letters.
  • Examples:
    • Sample mean (x̄)
    • Sample standard deviation (s)
    • Sample proportion (p̂)

In essence,

parameters describe populations, while statistics describe samples.

We use known statistics to make inferences about unknown parameters.

The Sampling Distribution of a Statistic

The sampling distribution of a statistic is the theoretical probability distribution of the values of that statistic, calculated from all possible samples of the same size, drawn from the same population. It is a crucial concept that describes the sample-to-sample variability.

Example: The Sampling Distribution of the Sample Mean (x̄)

Let’s imagine a scenario where we want to know the average age (a population parameter, μ) of all undergraduate students at a large university. Let’s assume the true (unknown) average age is μ = 21 years.

1. Taking a Sample: We can’t survey the entire population, so we take a random sample of, say, n=50 students and calculate their average age. Let’s say for this first sample, we get a sample mean x̄₁ = 21.5 years.
2. Repeating the Process: Now, imagine we repeat this process over and over. We take another random sample of 50 students and find its mean, x̄₂, which might be 20.9 years. We take a third sample and find x̄₃ = 21.2 years, and so on, thousands of times.
3. Building the Distribution: We would now have a very large collection of sample means: {21.5, 20.9, 21.2, 21.8, 20.7, …}. If we create a histogram of all these different sample means, that histogram would approximate the sampling distribution of the sample mean .

Key Characteristics of this Sampling Distribution (according to the Central Limit Theorem):

• Center: The mean of this distribution of sample means (μ _x̄ ) would be equal to the population mean (μ). In our example, the average of all the possible x̄’s would be very close to 21 years.
• Spread: The standard deviation of the distribution of sample means, called the Standard Error of the Mean , is σ _x̄ = σ/√n, where σ is the population standard deviation and n is the sample size. This shows that larger samples (bigger n) lead to less variability in the mean (smaller standard error).
• Shape: According to the Central Limit Theorem (CLT) , if the sample size is sufficiently large (typically n ≥ 30), the sampling distribution of the sample mean will be approximately normal, regardless of the shape of the original population distribution.

It is this concept of the sampling distribution that allows us to make inferences and conduct hypothesis tests about a population parameter based on information from a single sample. It helps us quantify how reliable our sample statistic is as an estimate of the population parameter.

Q5. Given A= | 6 6 | | 5 -3 | Find out characteristic roots and characteristic vectors.

Ans. The given matrix is:

 | 6 6 |A = | 5 -3 |

To find the characteristic roots (eigenvalues) and characteristic vectors (eigenvectors) , we follow these steps: Step 1: Form the characteristic equation and find the eigenvalues The eigenvalues (λ) are found by solving the characteristic equation |A – λI| = 0 , where I is the identity matrix. A – λI = | 6 6 | – λ | 1 0 | = | 6 6 | – | λ 0 | = | 6-λ 6 | | 5 -3 | | 0 1 | | 5 -3 | | 0 λ | | 5 -3-λ | Now we set the determinant of this matrix to zero: |A – λI| = (6 – λ)(-3 – λ) – (6)(5) = 0 -18 – 6λ + 3λ + λ² – 30 = 0 λ² – 3λ – 48 = 0 This is a quadratic equation. We can find the values of λ using the quadratic formula: λ = [-b ± √(b² – 4ac)] / 2a Here, a=1, b=-3, c=-48. λ = [ -(-3) ± √((-3)² – 4 1 (-48)) ] / 2 * 1 λ = [ 3 ± √(9 + 192) ] / 2 λ = [ 3 ± √(201) ] / 2 Thus, the eigenvalues are: λ₁ = (3 + √201) / 2 and λ₂ = (3 – √201) / 2 Step 2: Find the eigenvectors for each eigenvalue The eigenvector V = [x, y]ᵀ is found by solving the equation (A – λI)V = 0 . For eigenvalue λ₁ = (3 + √201) / 2: (A – λ₁I)V = 0

| 6-λ₁ 6 | |x| = |0|| 5 -3-λ₁ | |y| |0|

Using the first row: (6 – λ₁)x + 6y = 0 (6 – (3 + √201)/2)x + 6y = 0 ((12 – 3 – √201)/2)x + 6y = 0 ((9 – √201)/2)x = -6y (9 – √201)x = -12y If we choose x = 12, then y = -(9 – √201) = √201 – 9. So, an eigenvector corresponding to λ₁ is: V₁ = k * [ 12, (√201 – 9) ]ᵀ , where k is any non-zero scalar. For eigenvalue λ₂ = (3 – √201) / 2: (A – λ₂I)V = 0

| 6-λ₂ 6 | |x| = |0|| 5 -3-λ₂ | |y| |0|

Using the first row: (6 – λ₂)x + 6y = 0 (6 – (3 – √201)/2)x + 6y = 0 ((12 – 3 + √201)/2)x + 6y = 0 ((9 + √201)/2)x = -6y (9 + √201)x = -12y If we choose x = 12, then y = -(9 + √201). So, an eigenvector corresponding to λ₂ is: V₂ = k * [ 12, -(9 + √201) ]ᵀ , where k is any non-zero scalar.

Q6. Bring out the salient features of the Poisson’s distribution.

Ans. The Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events occurring in a fixed interval of time or space, if these events occur with a known constant mean rate and independently of the time since the last event. It is named after the French mathematician Siméon Denis Poisson. The salient features of the Poisson distribution are as follows:

1. Discrete Distribution: It is a discrete distribution, meaning the variable can only take integer values (0, 1, 2, 3, …). It models the number of occurrences of an event, which are countable.
2. Single Parameter (λ): The distribution is completely defined by a single parameter, lambda (λ) . λ represents the average number or expected value of events occurring in the given fixed interval.
3. Probability Mass Function (PMF): The probability of exactly k events occurring in an interval is given by the PMF: P(X = k) = (e ^-λ * λ ^k ) / k! Where:
   • k is the number of successes (events) (k = 0, 1, 2, …).
   • e is Euler’s number (approximately 2.71828).
   • λ is the average number of events.
4. Equality of Mean and Variance: A unique and important feature of the Poisson distribution is that both its mean and its variance are equal to its parameter λ. Mean (E[X]) = λ Variance (Var[X]) = λ
5. Shape and Skewness: The distribution is typically positively skewed or skewed to the right. As the value of λ increases, the distribution becomes more symmetric and starts to resemble a Normal distribution.
6. A Limiting Case of Binomial: The Poisson distribution can be used as an approximation to the binomial distribution when the number of trials (n) is very large and the probability of success (p) is very small, such that their product, np, is a finite value λ (λ = np). This is known as the law of “rare events”.
7. Additive Property: If X and Y are two independent Poisson random variables with parameters λ₁ and λ₂ respectively, then their sum (X+Y) will also follow a Poisson distribution with parameter (λ₁ + λ₂).

Example:

A call center receives an average of 5 calls per hour. The probability of receiving exactly 7 calls in a given hour can be calculated using the Poisson distribution (with λ=5, k=7).

Q7. Define rank of a matrix. Determine the rank of the matrix: A = | -3 6 2 | | 1 5 4 | | 4 -8 2 |

Ans. Rank of a Matrix The rank of a matrix is a fundamental concept that signifies the amount of “information” contained within the matrix. It has several equivalent definitions:

• Linear Independence Definition: The rank of a matrix is the maximum number of linearly independent vectors in its set of row vectors or column vectors. The row rank and column rank of a matrix are always equal.
• Determinant Definition: The rank of a matrix is the order (size) of the largest square sub-matrix that has a non-zero determinant. This sub-matrix is called a ‘minor’.
• Echelon Form Definition: When a matrix is converted to its row echelon form using row operations, the number of non-zero rows is the rank of the matrix.

In short, the rank tells us how many of the matrix’s rows/columns provide truly unique information.

Determining the Rank of the Given Matrix

The given matrix is:

 | -3 6 2 |A = | 1 5 4 | | 4 -8 2 |

The most direct method to find the rank of matrix A is to calculate its determinant. A is a 3×3 matrix.

• If |A| ≠ 0, then the three rows/columns are linearly independent, and the rank will be 3.
• If |A| = 0, then the rows/columns are linearly dependent, and the rank will be less than 3 (either 2 or 1).

Calculation of the determinant |A|: We expand along the first row: |A| = -3 (5 2 – 4 (-8)) – 6 (1 2 – 4 4) + 2 (1 (-8) – 5 * 4) |A| = -3 (10 – (-32)) – 6 (2 – 16) + 2 * (-8 – 20) |A| = -3 (10 + 32) – 6 (-14) + 2 * (-28) |A| = -3 (42) – 6 (-14) – 56 |A| = -126 + 84 – 56 |A| = -42 – 56 |A| = -98 Since the determinant of matrix A is non-zero (-98), it means that the largest non-zero minor of matrix A is the 3×3 matrix itself. By definition, the rank of a matrix is the order of the largest non-zero minor. In this case, that is 3. Therefore, the rank of the given matrix A is 3 .

Q8. Using Taylor’s approach, find Taylor’s series expansion for the function f(x, y, z) = xyz around the point (1, 1, 1).

Ans. Taylor Series Expansion The Taylor series expansion of a multi-variable function f(x, y, z) around a point (a, b, c) is a way to approximate the function in terms of the values of the function and its partial derivatives at that point. The series is given by: f(x, y, z) = f(a,b,c) + [ (x-a)f _x + (y-b)f _y + (z-c)f _z ] + (1/2!) [ (x-a)²f _xx + … + 2(x-a)(y-b)f _xy + … ] + … where all partial derivatives are evaluated at the point (a, b, c). Here, the given function is f(x, y, z) = xyz and the point of expansion is (a, b, c) = (1, 1, 1) . Step 1: Evaluate the function and its partial derivatives at the point (1, 1, 1). Zeroth-order: f(x, y, z) = xyz => f(1, 1, 1) = 1 1 1 = 1 First-order partial derivatives: f _x = yz => f _x (1, 1, 1) = 1 * 1 = 1 f _y = xz => f _y (1, 1, 1) = 1 * 1 = 1 f _z = xy => f _z (1, 1, 1) = 1 * 1 = 1 Second-order partial derivatives: f _xx = 0 f _yy = 0 f _zz = 0 f _xy = z => f _xy (1, 1, 1) = 1 f _xz = y => f _xz (1, 1, 1) = 1 f _yz = x => f _yz (1, 1, 1) = 1 Third-order partial derivatives: f _xyz = 1 All other third-order derivatives are zero (e.g., f _xxy = 0). Fourth-order and higher derivatives: All fourth-order and higher derivatives are zero. This means the Taylor series will be a finite series. Step 2: Substitute the values into the Taylor series formula. x-a = x-1, y-b = y-1, z-c = z-1 f(x, y, z) = f(1,1,1) + [ (x-1)f _x + (y-1)f _y + (z-1)f _z ] + (1/2!) [ 2(x-1)(y-1)f _xy + 2(x-1)(z-1)f _xz + 2(y-1)(z-1)f _yz ] + (1/3!) [ 3! * (x-1)(y-1)(z-1)f _xyz ] + … (higher order terms which are zero) Plugging in the values: f(x, y, z) = 1 + [ (x-1)(1) + (y-1)(1) + (z-1)(1) ] + (1/2) [ 2(x-1)(y-1)(1) + 2(x-1)(z-1)(1) + 2(y-1)(z-1)(1) ] + (1/6) [ 6(x-1)(y-1)(z-1)(1) ] Simplifying: f(x, y, z) = 1 + (x-1) + (y-1) + (z-1) + (x-1)(y-1) + (x-1)(z-1) + (y-1)(z-1) + (x-1)(y-1)(z-1) This is the exact Taylor series expansion for f(x, y, z) = xyz around the point (1, 1, 1). It is not an approximation but an exact equality because all higher-order terms become zero. If we were to expand the expression on the right-hand side, it would simplify to exactly xyz.

Q9. What is meant by continuity of a function ? Discuss the properties of a continuous function.

Ans. Continuity of a Function In mathematics, continuity is a property of the behavior of a function. Intuitively, a function is said to be continuous if its graph has no interruptions, gaps, or holes, i.e., it can be drawn without lifting the pen from the paper. Formal Definition: A function f(x) is said to be continuous at a point x = c if it satisfies the following three conditions:

1. f(c) is defined: The value of the function must exist at point c.
2. The limit exists: The limit of f(x) as x approaches c must exist. This means the left-hand limit must equal the right-hand limit: lim _x→c⁻ f(x) = lim _x→c⁺ f(x)
3. The limit equals the function value: The value of the limit at the point must be equal to the value of the function at that point: lim _x→c f(x) = f(c)

If any of these conditions are not met, the function is said to be

discontinuous

at the point

c

. A function is said to be continuous on an interval if it is continuous at every point in that interval.

Properties of Continuous Functions

Continuous functions possess several important properties that make them very useful in analysis. If

f(x)

and

g(x)

are two functions that are continuous at

x = c

, then:

• Sum and Difference: The function f(x) ± g(x) is also continuous at x = c .
• Product: The function f(x) * g(x) is also continuous at x = c .
• Scalar Multiplication: For any constant k , the function k * f(x) is also continuous at x = c .
• Quotient: The function f(x) / g(x) is also continuous at x = c , provided that the denominator is not zero, i.e., g(c) ≠ 0 .
• Composition of Functions: If g is continuous at c and f is continuous at g(c) , then the composite function (f o g)(x) = f(g(x)) is also continuous at c .

Important Theorems for Functions Continuous on a Closed Interval:

When a function is continuous on a closed interval [a, b], it exhibits some very powerful properties:

• Intermediate Value Theorem (IVT): If f is continuous on [a, b], then for any value L between f(a) and f(b) , there exists at least one number c in the interval (a, b) such that f(c) = L . In simple terms, the function takes on all values between its starting and ending values.
• Extreme Value Theorem (EVT): If f is continuous on a closed interval [a, b], then f attains both an absolute maximum and an absolute minimum value on that interval. This guarantees the existence of optimal solutions in optimization problems.

In economics, continuity is a critical assumption in many models, such as production functions, utility functions, and cost functions.

Q10. Find the stationary values and check whether they are maximum or minimum for the following functions : (i) z = 8x² + 6xy + 7y² (ii) z = 4x² – 2y² + 7xy

Ans. Stationary values of a function occur at points where the slope of the function is zero. At these points, the function’s value can be a maximum, a minimum, or a saddle point. Method:

1. First-Order Conditions (FOC): Set the partial derivatives of the function with respect to x and y equal to zero and solve for the values of x and y. This will give us the stationary points. ∂z/∂x = 0 and ∂z/∂y = 0
2. Second-Order Conditions (SOC): To classify the stationary point, we need the second-order partial derivatives: z _xx , z _yy , and z _xy . We then calculate the Hessian determinant, D: D = z _xx * z _yy – (z _xy )²
3. Classification:
   • If D > 0 and z _xx < 0, the point is a local maximum .
   • If D > 0 and z _xx > 0, the point is a local minimum .
   • If D < 0, the point is a saddle point .
   • If D = 0, the test is inconclusive.

(i) z = 8x² + 6xy + 7y²

FOC:

z

_x

= ∂z/∂x = 16x + 6y = 0 …(1)

z

_y

= ∂z/∂y = 6x + 14y = 0 …(2)

From equation (1), y = -16x/6 = -8x/3.

Substituting this into equation (2):

6x + 14(-8x/3) = 0

6x – 112x/3 = 0

(18x – 112x) / 3 = 0

-94x = 0 => x = 0

When x = 0, then y = 0.

Thus, the only stationary point is

(0, 0)

.

SOC:

z

_xx

= ∂²z/∂x² = 16

z

_yy

= ∂²z/∂y² = 14

z

_xy

= ∂²z/∂x∂y = 6

D = z

_xx

* z

_yy

– (z

_xy

)²

D = (16)(14) – (6)²

D = 224 – 36 = 188

Classification:

Since D = 188 > 0 and z

_xx

= 16 > 0, the stationary point (0, 0) is a

local minimum

.

The stationary value is z(0,0) = 8(0)² + 6(0)(0) + 7(0)² =

0

.

(ii) z = 4x² – 2y² + 7xy

FOC:

z

_x

= ∂z/∂x = 8x + 7y = 0 …(1)

z

_y

= ∂z/∂y = -4y + 7x = 0 …(2)

From equation (2), 7x = 4y => x = (4/7)y.

Substituting this into equation (1):

8((4/7)y) + 7y = 0

(32/7)y + 7y = 0

(32y + 49y) / 7 = 0

81y = 0 => y = 0

When y = 0, then x = 0.

Thus, the only stationary point is

(0, 0)

.

SOC:

z

_xx

= ∂²z/∂x² = 8

z

_yy

= ∂²z/∂y² = -4

z

_xy

= ∂²z/∂x∂y = 7

D = z

_xx

* z

_yy

– (z

_xy

)²

D = (8)(-4) – (7)²

D = -32 – 49 = -81

Classification:

Since D = -81 < 0, the stationary point (0, 0) is a

saddle point

. It is neither a maximum nor a minimum.

The stationary value is z(0,0) = 4(0)² – 2(0)² + 7(0)(0) =

0

.

Q11. What are the types of bias in a sample survey ?

Ans. In a sample survey, bias refers to a systematic error that leads to incorrect conclusions about a population. This error is not random but stems from a particular aspect of the survey process, resulting in a consistent overestimation or underestimation of the population parameter. Biases in sample surveys can be broadly classified into two main types: 1. Sampling Bias or Selection Bias This occurs when the method used to collect the sample results in a sample that is not representative of the population. Some common forms are:

• Undercoverage: This happens when some groups in the population are left out of or are inadequately represented in the sampling frame. For example, a survey conducted using only landline phones will miss people who only have mobile phones, potentially under-representing younger demographics.
• Voluntary Response Bias: This occurs when the sample consists of people who self-select to participate in a survey. These individuals often hold stronger or more extreme opinions than the general public. Online polls and TV call-in shows are classic examples.
• Convenience Sampling: This involves choosing individuals who are the easiest to reach. For example, surveying students in just one class to represent all students at a university. This rarely yields a representative sample.

2. Non-sampling Bias or Response Bias

This type of bias is independent of the sampling process and can arise during data collection, processing, or analysis. It can occur even in a perfect random sample.

• Non-response Bias: This occurs when individuals selected for the sample are unwilling to participate or cannot be contacted. The people who do not respond may differ in significant ways from those who do. For instance, in a survey about income, high-income individuals might be less willing to participate, leading to an underestimation of average income.
• Response Bias: This occurs when respondents provide inaccurate, misleading, or untruthful answers. It can be caused by several factors:
  • Social Desirability Bias: Respondents tend to give answers that they believe are more socially acceptable rather than their true opinions. For example, people might over-report voting or making charitable donations.
  • Wording of Questions: Leading, ambiguous, or biased questions can influence the answers. “Do you support more police to stop rising crime?” might get a different answer than “Do you support spending more tax dollars on police?”
  • Interviewer Bias: The presence, behavior, or even the tone of voice of the interviewer can subconsciously influence a respondent’s answers.
• Measurement Error: This arises from flaws in the instruments used for data collection. For example, a poorly calibrated weighing scale will cause a systematic error in measurements.

To design a reliable survey, researchers must carefully consider and minimize all these potential sources of bias.

Q12. Write short notes on any two of the following : (i) Mean Value Theorem (ii) Bolzano-Weierstrass Theorem (iii) Properties of Indefinite Integrals (iv) Cramer-Rao Inequality

Ans. (i) Mean Value Theorem (MVT) The Mean Value Theorem is a fundamental theorem of calculus. It connects the average rate of change of a function over an interval to the instantaneous rate of change at a point within that interval. Statement of the Theorem: If a function f is:

1. Continuous on the closed interval [a, b], and
2. Differentiable on the open interval (a, b),

then there exists at least one point

c

in the interval (a, b) for which:

f'(c) = [f(b) – f(a)] / (b – a)

Geometric Interpretation:

Geometrically, the theorem states that there is at least one tangent line to the curve y = f(x) that is parallel to the secant line connecting the points (a, f(a)) and (b, f(b)). The slope of the secant line is [f(b) – f(a)] / (b – a), and the slope of the tangent at point

c

is f'(c). The MVT guarantees that there is at least one point on the curve where the instantaneous velocity equals the average velocity. This theorem serves as a foundation for proving many other important results in calculus.

(ii) Bolzano-Weierstrass Theorem

The Bolzano-Weierstrass theorem is a foundational result in real analysis, closely related to the notion of compactness.

Statement of the Theorem:

Every

bounded sequence

in Euclidean space ℝⁿ has a

convergent subsequence

.

Explanation:

This means that if you have an infinite sequence of points that are all contained within a finite “space” (e.g., all numbers in a sequence are between -10 and 10), you can always find a smaller, infinite subsequence that gets closer and closer to some specific point (a limit point).

Importance:

The theorem is crucial in mathematical analysis and economics. It is essential for proving the Extreme Value Theorem, which states that a continuous function on a compact set attains its maximum and minimum values. It is a key tool in problems proving the existence of an optimum or an equilibrium.

(iii) Properties of Indefinite Integrals

The indefinite integral of a function f(x), denoted ∫f(x)dx, is the family of all antiderivatives of f(x). If F'(x) = f(x), then ∫f(x)dx = F(x) + C, where C is the “constant of integration”.

The main properties of indefinite integrals, which simplify the process of integration, are as follows:

1. Constant Multiple Rule: ∫k f(x) dx = k ∫f(x) dx That is, a constant can be factored out of the integral.
2. Sum/Difference Rule: ∫[f(x) ± g(x)] dx = ∫f(x) dx ± ∫g(x) dx The integral of a sum (or difference) of two functions is the sum (or difference) of their individual integrals.
3. Power Rule: ∫xⁿ dx = (xⁿ⁺¹ / (n+1)) + C, for n ≠ -1. This is one of the most frequently used integration rules.
4. Inverse relationship with differentiation: d/dx [∫f(x)dx] = f(x). This is part of the Fundamental Theorem of Calculus, showing that differentiation and integration are inverse processes.

These properties are essential for solving various problems in economics, such as calculating consumer and producer surplus, determining capital stock from investment, and more.

(iv) Cramer-Rao Inequality

The Cramer-Rao inequality (or Cramer-Rao Lower Bound) is a fundamental result in statistical estimation theory. It provides a lower bound on the variance of any

unbiased estimator

of a deterministic parameter.

Statement of the Inequality:

Let θ be an unknown parameter to be estimated from data X, and let T(X) be any unbiased estimator of θ (i.e., E[T(X)] = θ). Then, the variance of T(X) must satisfy:

Var(T(X)) ≥ 1 / I(θ)

Here, I(θ) is called the

Fisher Information

. It measures the amount of information that the sample data provides about the parameter θ.

Importance and Implication:

The main importance of the Cramer-Rao inequality is that it sets a benchmark for the “efficiency” of estimators.

• It tells us the best we can possibly do, i.e., how small the variance of an unbiased estimator can be.
• An unbiased estimator that achieves this lower bound (i.e., whose variance equals 1/I(θ)) is called an efficient estimator or a Minimum Variance Unbiased Estimator (MVUE) .
• It provides a standard for comparing different unbiased estimators. The estimator whose variance is closest to the Cramer-Rao lower bound is considered more efficient.