IGNOU BECC-102 Solved Question Paper PDF Download

IGNOU BECC-102 Previous Year Solved Question Paper in Hindi

IGNOU BECC-102 Previous Year Solved Question Paper in English

Section—A

Q1. Given the following demand and supply functions : Q _d = 10 – 2P _t Q _s = -5 + 3P _t-1 where Q _d is quantity demanded; Q _s is quantity supplied; P is price; and t is time period. Find: (a) the time path of price. (b) determine if the time path is stable, unstable or regular.

Ans. Given the demand and supply functions: Demand function: Q _dt = 10 – 2P _t Supply function: Q _st = -5 + 3P _t-1 In market equilibrium, quantity demanded equals quantity supplied (Q _dt = Q _st ). 10 – 2P _t = -5 + 3P _t-1 Now, we rearrange the equation to solve for P _t : 2P _t = 10 + 5 – 3P _t-1 2P _t = 15 – 3P _t-1 P _t = -1.5 P _t-1 + 7.5 This is a first-order linear difference equation of the general form Y _t = bY _t-1 + a. (a) The time path of price. To find the time path, we first find the particular solution (the equilibrium price) and then the complementary function. 1. Particular Solution (Equilibrium Price): In equilibrium, P _t = P _t-1 = P _e . P _e = -1.5 P _e + 7.5 P _e + 1.5 P _e = 7.5 2.5 P _e = 7.5 P _e = 7.5 / 2.5 = 3 Thus, the equilibrium price (P _e ) is 3. 2. Complementary Function: This is the solution to the homogeneous part, P _t = -1.5 P _t-1 . The solution is of the form P _t = A(b) ^t , where b = -1.5. So, the complementary function is A(-1.5) ^t , where A is a constant that depends on the initial condition. 3. General Solution: The general solution is the sum of the particular solution and the complementary function. P _t = A(-1.5) ^t + P _e P _t = A(-1.5) ^t + 3 The constant A can be determined using the initial price P ₀ (at t=0): P ₀ = A(-1.5) ⁰ + 3 P ₀ = A + 3 A = P ₀ – 3 Therefore, the definitive time path of price is: P _t = (P ₀ – 3)(-1.5) ^t + 3 (b) Determine if the time path is stable, unstable or regular. The stability of the time path for a difference equation P _t = bP _t-1 + a depends on the absolute value of the coefficient ‘b’.

• If |b| < 1, the path is stable and converges to the equilibrium.
• If |b| > 1, the path is unstable and diverges from the equilibrium.
• If |b| = 1, the path has regular or uniform oscillations.

In our equation, b = -1.5.

|b| = |-1.5| = 1.5

Since

|b| = 1.5 > 1

, the time path of price is

unstable

.

Furthermore, since b is negative (b < 0), the path is oscillatory. Because it is both unstable and oscillatory, it is described as having

explosive oscillations

. The price will move further and further away from the equilibrium price of 3, changing direction in each period. It is not a regular path.

Q2. (a) The total cost function of a firm is : C(x) = x³/3 – 5x² + 28x + 10, where x is the output. A tax at the rate of ₹ 2 per unit of output is imposed and the producer adds it to his cost. If the market demand function is given by P = 2530 – 5x, where P is price per unit of output, find : (i) the profit maximizing level of output x (ii) the equilibrium price (iii) maximum profit. (b) The demand for furniture (x) is given as: x = 100 + 0.5y, where y represents consumer income. Find income elasticity of demand when y = 10,000.

Ans. (a) Profit Maximization Analysis 1. Setting up the functions: Initial Total Cost function: C(x) = x³/3 – 5x² + 28x + 10 Tax per unit = ₹ 2 New Total Cost function after tax (C _T (x)): C _T (x) = C(x) + (Tax x) = (x³/3 – 5x² + 28x + 10) + 2x C _T (x) = x³/3 – 5x² + 30x + 10 Demand function: P = 2530 – 5x Total Revenue function (R(x)): R(x) = P x = (2530 – 5x)x R(x) = 2530x – 5x² Profit function (π(x)): π(x) = R(x) – C _T (x) π(x) = (2530x – 5x²) – (x³/3 – 5x² + 30x + 10) π(x) = 2530x – 5x² – x³/3 + 5x² – 30x – 10 π(x) = -x³/3 + 2500x – 10 2. Maximizing Profit: For profit maximization, we differentiate the profit function with respect to x and set it to zero (First-Order Condition). dπ/dx = d/dx (-x³/3 + 2500x – 10) = -x² + 2500 Set dπ/dx = 0 => -x² + 2500 = 0 x² = 2500 x = ±√2500 = ±50 Since output (x) cannot be negative, we take x = 50 . We check the Second-Order Condition to ensure this is a maximum. d²π/dx² = d/dx (-x² + 2500) = -2x At x = 50, d²π/dx² = -2(50) = -100, which is < 0. Since the second derivative is negative, profit is maximized at x = 50. (i) The profit-maximizing level of output: x = 50 units. (ii) The equilibrium price: We substitute x = 50 into the demand function: P = 2530 – 5(50) = 2530 – 250 P = ₹ 2280 (iii) The maximum profit: We substitute x = 50 into the profit function: π(50) = -(50)³/3 + 2500(50) – 10 π(50) = -125000/3 + 125000 – 10 π(50) = ( -125000 + 375000 ) / 3 – 10 π(50) = 250000 / 3 – 10 π(50) = 83333.33 – 10 π(50) = ₹ 83323.33 (b) Income Elasticity of Demand Note: The question paper has a typo `x = 100 + .5 4`. The Hindi version `x = 100 + 0.5y` is used for this solution. The demand function is given as: x = 100 + 0.5y The income level is given as: y = 10,000 The formula for income elasticity of demand (E _y ) is: E _y = (dx/dy) (y/x) 1. Find dx/dy: From x = 100 + 0.5y, we differentiate x with respect to y: dx/dy = 0.5 2. Find the value of x at y = 10,000: x = 100 + 0.5(10,000) x = 100 + 5000 x = 5100 3. Calculate the income elasticity: Now we substitute these values into the elasticity formula: E _y = (0.5) (10,000 / 5100) E _y = 5000 / 5100 E _y = 50 / 51 E _y ≈ 0.98 Therefore, when consumer income is ₹ 10,000, the income elasticity of demand for furniture is approximately 0.98. This indicates that the good is a normal good (since E _y > 0) and is income-inelastic (since E _y < 1).

Q3. (a) Differentiate between homogeneous and non-homogeneous difference equations. (b) Solve the following difference equation : y _t+1 – 8y _t = 4.(3) ^t

Ans. (a) Homogeneous vs. Non-homogeneous Difference Equations An n-th order linear difference equation can be written in its general form as: y _t+n + a ₁ y _t+n-1 + … + a _n y _t = c Here, y is the dependent variable, t is time (the independent variable), and a _i are coefficients. Homogeneous Difference Equation: A difference equation is called homogeneous if there is no term that is independent of the dependent variable (y) and its lags. In other words, if the term ‘c’ in the general form above is identically zero for all t (c ≡ 0), the equation is homogeneous. Example: y _t+1 + 5y _t = 0 The solution to a homogeneous equation is called the complementary function. It describes the intrinsic dynamics of the system, determining whether the system naturally converges to or diverges from an equilibrium (which is zero for homogeneous systems). Non-homogeneous Difference Equation: A difference equation is non-homogeneous if it contains a term that is independent of the dependent variable (y) and its lags. If the term ‘c’ in the general form is not identically zero (c ≠ 0), the equation is non-homogeneous. This term ‘c’ is sometimes called the ‘forcing term’ of the equation. Example: y _t+1 + 5y _t = 12 The general solution to a non-homogeneous equation is the sum of two parts: 1. The Complementary Function (y _c ): This is the general solution of the corresponding homogeneous equation (found by setting c=0). 2. The Particular Solution (y _p ): This is any specific solution to the full non-homogeneous equation. It represents the intertemporal equilibrium level of the system. Key Distinction:

• Structure: A homogeneous equation contains only terms involving the dependent variable and its lags. A non-homogeneous equation has an additional constant or a function of the independent variable.
• Solution: The solution to a homogeneous equation (the complementary function) describes the system’s inherent dynamics. The solution to a non-homogeneous equation (complementary function + particular solution) describes both the inherent dynamics and the movement towards an equilibrium determined by external forces.

(b) Solve the following difference equation: y

_t+1

– 8y

_t

= 4.(3)

^t

This is a first-order linear non-homogeneous difference equation. Its general solution (y

_t

) is the sum of the complementary function (y

_c

) and the particular solution (y

_p

).

y

_t

= y

_c

+ y

_p

1. Finding the Complementary Function (y

_c

):

We solve the corresponding homogeneous equation:

y

_t+1

– 8y

_t

= 0

The solution is of the form y

_c

= A(b)

^t

.

Substituting y

_t

= b

^t

into the equation:

b

^t+1

– 8b

^t

= 0

b

^t

(b – 8) = 0

The characteristic equation is b – 8 = 0, which gives

b = 8

.

Thus, the complementary function is:

y

_c

= A(8)

^t

, where A is an arbitrary constant.

2. Finding the Particular Solution (y

_p

):

The non-homogeneous term is 4(3)

^t

. We assume a trial solution of a similar form:

Let y

_p

= C(3)

^t

, where C is a constant to be determined.

Now, substitute y

_p

into the original equation:

y

_t+1

will be C(3)

^t+1

.

C(3)

^t+1

– 8[C(3)

^t

] = 4(3)

^t

C(3)(3)

^t

– 8C(3)

^t

= 4(3)

^t

Dividing through by (3)

^t

:

3C – 8C = 4

-5C = 4

C = -4/5

Therefore, the particular solution is:

y

_p

= (-4/5)(3)

^t

3. Finding the General Solution (y

_t

):

The general solution is the sum of y

_c

and y

_p

:

y

_t

= A(8)

^t

– (4/5)(3)

^t

This is the complete solution to the given difference equation. The value of the constant A can be found if an initial condition (like y

₀

) is provided.

Q4. (a) Evaluate: ∫(from 0 to 1) x²eˣ dx (b) Given the demand function P = 20 – 5x and the supply function P = 4 + 3x, where P is price and x is quantity. Find: (i) Consumer surplus (ii) Producer surplus

Ans. (a) Evaluation of a Definite Integral We need to evaluate ∫(from 0 to 1) x²eˣ dx. We will use Integration by Parts , the formula for which is: ∫u dv = uv – ∫v du. First Pass: Let u = x² and dv = eˣ dx. Then du = 2x dx and v = ∫eˣ dx = eˣ. Applying the formula: ∫x²eˣ dx = x²eˣ – ∫eˣ(2x) dx ∫x²eˣ dx = x²eˣ – 2∫xeˣ dx Second Pass: Now we need to evaluate ∫xeˣ dx, again using integration by parts. Let u = x and dv = eˣ dx. Then du = dx and v = eˣ. ∫xeˣ dx = xeˣ – ∫eˣ dx = xeˣ – eˣ Third Step: Substitute this result back into the equation from the first step: ∫x²eˣ dx = x²eˣ – 2(xeˣ – eˣ) ∫x²eˣ dx = x²eˣ – 2xeˣ + 2eˣ Fourth Step: Now we evaluate the definite integral with the limits from 0 to 1: [x²eˣ – 2xeˣ + 2eˣ]¹₀ First, substitute x = 1: (1)²e¹ – 2(1)e¹ + 2e¹ = e – 2e + 2e = e Next, substitute x = 0: (0)²e⁰ – 2(0)e⁰ + 2e⁰ = 0 – 0 + 2(1) = 2 The final result is the value at the upper limit minus the value at the lower limit: Value = (e) – (2) = e – 2 Thus, the value of ∫(from 0 to 1) x²eˣ dx is e – 2. (b) Consumer and Producer Surplus 1. Find Equilibrium Price and Quantity: At equilibrium, the price for demand equals the price for supply. Demand function: P = 20 – 5x Supply function: P = 4 + 3x 20 – 5x = 4 + 3x 20 – 4 = 3x + 5x 16 = 8x x _e = 2 (Equilibrium quantity) Now we substitute x _e = 2 into either function to find the equilibrium price (P _e ): P _e = 4 + 3(2) = 4 + 6 = 10 or P _e = 20 – 5(2) = 20 – 10 = 10 P _e = 10 (Equilibrium price) (i) Consumer Surplus (CS): This is the area below the demand curve and above the equilibrium price line. It can be calculated with the following integral: CS = ∫(from 0 to x _e ) [Demand function – P _e ] dx CS = ∫(from 0 to 2) [(20 – 5x) – 10] dx CS = ∫(from 0 to 2) (10 – 5x) dx Integrating: CS = [10x – 5x²/2]²₀ CS = (10(2) – 5(2)²/2) – (10(0) – 5(0)²/2) CS = (20 – 5(4)/2) – 0 CS = 20 – 10 = 10 Alternatively (Area of a triangle): CS = ½ base height Base = x _e = 2 Height = (P-intercept of demand) – P _e = 20 – 10 = 10 CS = ½ 2 10 = 10 (ii) Producer Surplus (PS): This is the area above the supply curve and below the equilibrium price line. PS = ∫(from 0 to x _e ) [P _e – Supply function] dx PS = ∫(from 0 to 2) [10 – (4 + 3x)] dx PS = ∫(from 0 to 2) (6 – 3x) dx Integrating: PS = [6x – 3x²/2]²₀ PS = (6(2) – 3(2)²/2) – 0 PS = (12 – 3(4)/2) – 0 PS = 12 – 6 = 6 Alternatively (Area of a triangle): PS = ½ base height Base = x _e = 2 Height = P _e – (P-intercept of supply) = 10 – 4 = 6 PS = ½ 2 6 = 6 Therefore, the consumer surplus is 10 and the producer surplus is 6.

Section—B

Q5. (a) Explain the following : (i) Theorem (ii) Proof (b) Briefly discuss various types of proofs.

Ans. (a) Explanation of Theorem and Proof (i) Theorem: In mathematics and logic, a theorem is a statement that has been proven to be true on the basis of previously established statements, such as other theorems, and generally accepted statements, such as axioms. A theorem is essentially a non-obvious statement whose truth has been demonstrated through logical deduction. The structure of a theorem typically consists of two parts:

• Hypotheses: The conditions or premises that are assumed to be true.
• Conclusion: The statement that logically follows if the hypotheses are true.

For example, the Pythagorean theorem states that “In any right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides.” Here, the hypothesis is that we have a right-angled triangle, and the conclusion is the relationship between the sides.

(ii) Proof:

A

proof

is a rigorous and formal logical argument that demonstrates a theorem or statement is true beyond any doubt, provided the axioms are true. A proof begins with the hypotheses and proceeds through a step-by-step logical deduction to arrive at the conclusion. Each step must be justified by a definition, an axiom, or a previously proven theorem. The purpose of a mathematical proof is not only to show that a statement is true but also to explain why it is true. It is the foundation of mathematical knowledge, turning conjecture and intuition into certainty.

(b) Various Types of Proofs

Several strategies or methods are used in mathematics to prove a theorem. Some of the main types are:

1.

Direct Proof:

This is the most straightforward method of proof. In it, one starts with the hypotheses (the givens) and proceeds directly to the conclusion through a series of logical steps. Each step follows logically from the preceding one or from known facts.

Example:

Proving that the sum of two even numbers is always even.

2.

Proof by Contradiction (Reductio ad absurdum):

In this method, we start by assuming that the statement we want to prove is false. We then use logical deduction to show that this assumption leads to a logical contradiction (e.g., 1=0). Since a logical contradiction is impossible, our initial assumption (that the statement was false) must be wrong, and therefore the statement must be true.

Example:

Proving that the square root of 2 is an irrational number.

3.

Proof by Contrapositive:

To prove a statement of the form “If P, then Q,” we instead prove its logically equivalent contrapositive statement “If not Q, then not P.” Sometimes, the contrapositive statement is easier to prove directly.

Example:

To prove “If n² is even, then n is even,” we prove its contrapositive, “If n is odd, then n² is odd.”

4.

Proof by Induction:

This technique is used to prove statements that are true for all natural numbers (or an infinite sequence of integers). It consists of two main steps:

• Base Case: Showing that the statement holds for the first number (e.g., n=1).
• Inductive Step: Assuming the statement is true for some arbitrary number k (the inductive hypothesis), and then proving that it must also be true for the next number, k+1.

Example:

Proving the formula for the sum of the first n natural numbers is n(n+1)/2.

Q6. (a) Find the equation of the circle with centre (2, -3) and radius 5. (b) Show that y² + 4x = -8 is a parabola.

Ans. (a) Equation of the Circle The standard equation of a circle with center at (h, k) and radius r is given by: (x – h)² + (y – k)² = r² In the given question, we have: Center (h, k) = (2, -3) Radius (r) = 5 Substituting these values into the standard equation: (x – 2)² + (y – (-3))² = 5² (x – 2)² + (y + 3)² = 25 This is the equation of the circle in standard form. If we wish to expand this to get the general form of the equation, we square the binomials: (x² – 2 x 2 + 2²) + (y² + 2 y 3 + 3²) = 25 (x² – 4x + 4) + (y² + 6y + 9) = 25 Rearranging the terms: x² + y² – 4x + 6y + 4 + 9 – 25 = 0 x² + y² – 4x + 6y – 12 = 0 This is the equation of the circle in general form. Both answers are correct, but the standard form (x – 2)² + (y + 3)² = 25 is more informative as it directly shows the center and radius. (b) Showing the given equation is a parabola The given equation is: y² + 4x = -8 A parabola is defined as the locus of a point that is equidistant from a fixed point (the focus) and a fixed line (the directrix). The standard equations of a parabola have two main forms: 1. (y – k)² = 4a(x – h) (A horizontal parabola, opening left or right) 2. (x – h)² = 4a(y – k) (A vertical parabola, opening up or down) Here, (h, k) is the vertex of the parabola. We will attempt to rearrange the given equation into one of these standard forms. y² + 4x = -8 Isolate the y² term on one side and the x term on the other: y² = -4x – 8 Now, factor out a common multiplier from the x-terms to make the coefficient of x equal to 1: y² = -4(x + 2) Comparing this equation with the standard horizontal parabola equation, (y – k)² = 4a(x – h) , we can see that it matches perfectly. Here:

• k = 0
• h = -2
• 4a = -4, which implies a = -1

Since the given equation y² + 4x = -8 can be written in the standard form of a parabola, it represents a parabola.

Specifically, it is a

horizontal parabola

with its

vertex at (-2, 0)

. Since the value of ‘a’ is negative, it

opens to the left

.

Q7. (a) Briefly explain the concept of present value. How is it related to discounting principle ? (b) A firm has purchased an item on a fixed payment plan of ₹ 20,000 per year for 8 years. Payments are to be made at the beginning of each year. What is the present value of the total cash flow of payments for an interest rate of 20% per year?

Ans. (a) Present Value and the Discounting Principle Present Value (PV): Present value is a core financial concept stating that an amount of money today is worth more than the same amount in the future. This is due to the “time value of money,” which holds that money available at the present time can be invested and earn a return, growing to a larger amount in the future. Present value, therefore, is the current worth of a future sum of money or stream of cash flows given a specified rate of return (the discount rate). It answers the question: “How much money would I need to invest today to have a certain amount in the future?” Relation to Discounting Principle: Discounting is the process used to calculate the present value. It is the exact opposite of compounding. While compounding calculates the future value (FV) of a present sum, discounting calculates the present value (PV) of a future sum. The “discounting principle” is the application of this process. The basic formula for discounting is: PV = FV / (1 + r)ⁿ Where:

• PV = Present Value
• FV = Future Value (the amount to be received after n periods)
• r = The discount rate (interest rate) per period
• n = The number of periods

This formula shows that to convert a future sum into its present value, it is “discounted” or divided by (1 + r)ⁿ. The higher the discount rate and the longer the time period, the lower the present value of a future sum will be.

(b) Present Value of an Annuity

This is a problem of an annuity, as it involves regular payments for a fixed period. Since the payments are made at the

beginning

of each year, it is an

Annuity Due

.

Given values:

Yearly Payment (Pmt) = ₹ 20,000

Number of periods (n) = 8 years

Interest rate (r) = 20% per annum = 0.20

The formula for the Present Value of an Annuity Due (PV

_due

) is:

PV

_due

= Pmt

[ (1 – (1 + r)⁻ⁿ) / r ]
(1 + r)

We will calculate this step-by-step:

1. Calculate (1 + r)⁻ⁿ:

(1 + 0.20)⁻⁸ = (1.2)⁻⁸

1.2⁻⁸ ≈ 0.2325679

2. Calculate the bracketed part (the ordinary annuity factor):

[ (1 – 0.2325679) / 0.20 ]

= [ 0.7674321 / 0.20 ]

≈ 3.83716

3. Adjust for an Annuity Due:

Now, multiply this factor by (1 + r):

3.83716

(1 + 0.20)

= 3.83716

1.2

≈ 4.604592

4. Calculate the Present Value:

PV

_due

= Pmt

(Annuity due factor)

PV

_due

= 20,000

4.604592

PV

_due

= 92,091.84

Therefore, the present value of the total cash flow of payments is approximately

₹ 92,091.84

.

Q8. Discuss the basic setup of the Cobweb model.

Ans. The Cobweb Model is an economic model that explains why prices might be subject to periodic fluctuations in certain types of markets. It is a classic dynamic model of cyclical supply and demand, particularly applicable to markets where there is a time lag between the decision to produce and the actual supply, such as in agriculture. Basic Setup and Assumptions: The Cobweb model is built on the following key assumptions: 1. Time Lag in Production: The most crucial assumption is that the quantity supplied cannot be adjusted instantaneously. Producers must make decisions about production in advance, and the actual crop or product comes to the market later. This means that the quantity supplied in the current period (Qˢ _t ) is based on the price from the previous period (P _t-1 ). Mathematically: Qˢ _t = f(P _t-1 ) 2. Naive Producers: The model assumes that producers are “naive” or have static expectations. They base their current production plans solely on the market price of the previous period. They do not attempt to forecast future prices and do not learn that their own production actions will influence the future price. 3. Instantaneous Demand Adjustment: In contrast, consumers react immediately. The quantity demanded in the current period (Qᵈ _t ) depends on the current price in that same period (P _t ). Mathematically: Qᵈ _t = g(P _t ) 4. Market Clearing: In each period, the market price adjusts to a level where the quantity demanded equals the available quantity supplied. Since supply is predetermined, the price adjusts along the demand curve to clear the market. Mathematically: Qᵈ _t = Qˢ _t Dynamics of the Model: The combination of these assumptions creates a dynamic process in prices and quantities.

• Suppose the price in period t-1 was high.
• Naive producers, encouraged by this high price, increase their production for period t (Qˢ _t increases).
• When this increased supply hits the market in period t, the price must fall to sell it all (P _t decreases).
• Now, faced with a low price in period t, producers cut back their production for period t+1.
• When this reduced supply comes to market in period t+1, the price rises again.

This process creates oscillations in prices and quantities over time. When plotted on a supply and demand diagram, these points trace a pattern that can resemble a spider’s web, hence the name “Cobweb Model.”

The outcome of the model (convergence, divergence, or regular oscillation) depends on the relative slopes of the demand and supply curves.

Q9. (a) If the 7th term of an A.P. is equal to its 11th term, show that the 18th term of the A.P. is zero. (b) Find the 28th term from the end of the A.P.: 6, 9, 12, 15, …, 102.

Ans. (a) Proving a Property of an A.P. Note: There is a likely typo in the question. “7th term… is equal to its 11th term” (a₇ = a₁₁) would imply a common difference of zero, which would not make the 18th term zero (unless the first term was zero). A standard problem is “7 times the 7th term is equal to 11 times the 11th term”. We will solve this corrected, logical version of the problem. Assumed Question: If 7 times the 7th term of an A.P. is equal to 11 times its 11th term, show that its 18th term is zero. Let the first term of the A.P. be ‘a’ and the common difference be ‘d’. The n-th term (aₙ) of an A.P. is given by: aₙ = a + (n – 1)d According to the question: 7 × (7th term) = 11 × (11th term) 7 × a₇ = 11 × a₁₁ Substituting the formulas for a₇ and a₁₁: a₇ = a + (7 – 1)d = a + 6d a₁₁ = a + (11 – 1)d = a + 10d Now, put these into the condition: 7(a + 6d) = 11(a + 10d) Solve the equation: 7a + 42d = 11a + 110d Rearrange the terms: 7a – 11a = 110d – 42d -4a = 68d a = – (68/4)d a = -17d Now, we need to find the 18th term (a₁₈). a₁₈ = a + (18 – 1)d a₁₈ = a + 17d Substitute the value of ‘a’ we found (-17d): a₁₈ = (-17d) + 17d a₁₈ = 0 Hence, it is proved that the 18th term of the A.P. is zero. (b) Finding the term from the end Note: The A.P. in the English paper (6, 9, 2, 5, 8) is incorrect. We will use the correct A.P. from the Hindi version: 6, 9, 12, 15, …, 102. The given Arithmetic Progression (A.P.) is: 6, 9, 12, 15, …, 102. Here: First term (a) = 6 Common difference (d) = 9 – 6 = 3 Last term (l) = 102 We need to find the 28th term from the end. There are two methods: Method 1: Reverse the A.P. If we reverse the A.P., the 28th term from the end becomes the 28th term from the beginning of the new series. The reversed series is: 102, 99, 96, …, 6. For this new A.P.: First term (a’) = 102 Common difference (d’) = 99 – 102 = -3 Now find the 28th term (T₂₈) of this new series: T₂₈ = a’ + (n – 1)d’ T₂₈ = 102 + (28 – 1)(-3) T₂₈ = 102 + 27(-3) T₂₈ = 102 – 81 T₂₈ = 21 Method 2: Using the formula A direct formula for the m-th term from the end is: l – (m – 1)d Here, l = 102, m = 28, and d = 3. 28th term from the end = 102 – (28 – 1) 3 = 102 – (27) 3 = 102 – 81 = 21 Therefore, the 28th term from the end of the given A.P. is 21.

Q10. What do you understand by a convex combination? Use the concept of convex combination to explain the concept of a convex set.

Ans. Convex Combination A convex combination is a specific type of linear combination of points (or vectors) in a vector space. If we have a set of points {x₁, x₂, …, xₙ}, a convex combination of these points is a new point ‘x’ that can be expressed as: x = λ₁x₁ + λ₂x₂ + … + λₙxₙ This combination must satisfy two conditions: 1. All the coefficients (λᵢ) must be non-negative, i.e., λᵢ ≥ 0 for all i = 1, …, n. 2. The sum of all coefficients must be equal to 1, i.e., Σ λᵢ = 1 . Geometrically, the convex combination of two points, x₁ and x₂, given by x = λx₁ + (1-λ)x₂ (where 0 ≤ λ ≤ 1), represents all the points that lie on the line segment connecting x₁ and x₂. When λ=0, x=x₂; when λ=1, x=x₁; and when λ is between 0 and 1, x is a point between them. Explanation of a Convex Set using Convex Combination The concept of a convex combination provides a precise and powerful way to define a convex set . Definition: A set ‘S’ is called a convex set if, for any two points chosen from within the set, the entire line segment connecting those two points also lies completely within the set ‘S’. Using the concept of convex combination, this definition can be expressed as follows: A set S is convex if, for any two points x₁ and x₂ belonging to the set S (i.e., x₁, x₂ ∈ S), any convex combination of them is also in S. Mathematically: S is a convex set if for all x₁, x₂ ∈ S, and for all λ ∈ [0, 1], the point λx₁ + (1-λ)x₂ is also in S. Explanation: This definition essentially says that you cannot “leave” the set by drawing a straight line between any two of its points. If you can, the set is not convex (it is called concave or non-convex). Examples:

• A circle (and its interior), a triangle, a rectangle are all convex sets . You can draw a line between any two interior points, and that line will remain entirely inside the shape.
• A star shape or a crescent moon shape are non-convex sets . You can find two points (e.g., on two different tips of the star) such that the line segment connecting them passes through the area outside the shape.

In economics, this concept is crucial in defining production possibility sets and budget sets, which are typically assumed to be convex.

Section—C

Q11. Evaluate the following limits : (a) lim (x→0) (3x² – x) / (x³ + 1) (b) lim (x→1) (x² – 1) / (2x² – 7x + 5)

Ans. (a) lim (x→0) (3x² – x) / (x³ + 1) This is a limit of a rational function. To evaluate the limit, we first try to substitute the value of x directly. Substituting x = 0 into the function: Numerator: 3(0)² – 0 = 0 Denominator: (0)³ + 1 = 1 Since the denominator is not zero, we can evaluate the limit by direct substitution: Limit = 0 / 1 = 0 Therefore, lim (x→0) (3x² – x) / (x³ + 1) = 0 (b) lim (x→1) (x² – 1) / (2x² – 7x + 5) First, we try direct substitution of x = 1: Numerator: (1)² – 1 = 1 – 1 = 0 Denominator: 2(1)² – 7(1) + 5 = 2 – 7 + 5 = 0 We get the form 0/0 , which is an indeterminate form . This means we need further analysis to evaluate the limit. We can do this in two ways: factorization or L’Hôpital’s Rule. Method 1: Factorization Since both the numerator and denominator are zero at x=1, we know that (x-1) is a factor of both. Factorizing the numerator: x² – 1 = (x – 1)(x + 1) (using a² – b² = (a-b)(a+b)) Factorizing the denominator (2x² – 7x + 5): We can split the middle term: 2x² – 2x – 5x + 5 = 2x(x – 1) – 5(x – 1) = (2x – 5)(x – 1) Now substitute the factored forms back into the limit: lim (x→1) [ (x – 1)(x + 1) ] / [ (2x – 5)(x – 1) ] Since x is approaching 1 but is not equal to 1, we can cancel the (x-1) term from the numerator and denominator: lim (x→1) (x + 1) / (2x – 5) Now, substitute x = 1 into the remaining expression: = (1 + 1) / (2(1) – 5) = 2 / (2 – 5) = 2 / -3 = -2/3 Method 2: L’Hôpital’s Rule Since we have the indeterminate form 0/0, we can differentiate the numerator and the denominator separately and then take the limit. d/dx (x² – 1) = 2x d/dx (2x² – 7x + 5) = 4x – 7 Therefore, the limit is equal to: lim (x→1) (2x) / (4x – 7) Now, substitute x = 1: = 2(1) / (4(1) – 7) = 2 / (4 – 7) = 2 / -3 = -2/3 Both methods yield the same result.

Q12. Explain the following : (a) Phase diagrams (b) Order of a difference equation

Ans. (a) Phase Diagrams A phase diagram is a powerful graphical tool used for the qualitative analysis of autonomous differential or difference equations. It helps us understand the long-term behavior of a variable (such as its stability, convergence, or divergence) without needing to solve the equation explicitly. For a first-order difference equation, such as y _t+1 = f(y _t ) , a phase diagram is constructed as follows: 1. On a 2-dimensional graph, y _t is plotted on the horizontal axis and y _t+1 is plotted on the vertical axis. 2. Two key lines are drawn on this graph: a. The Phase Line: This is the graph of the equation y _t+1 = f(y _t ). It shows how the current value of y (y _t ) determines its next value (y _t+1 ). b. The 45-Degree Line: This is the graph of the equation y _t+1 = y _t . Along this line, the value of the variable does not change from one period to the next. Analysis:

• Steady States: The intersection points of the phase line and the 45-degree line represent the steady states or equilibrium points . At these points, y _t+1 = y _t , meaning the system ceases to change.
• Dynamics: Starting from any initial value y₀, we can trace the path of y over time. From y₀, we move vertically to the phase line (to find the value of y₁), then horizontally to the 45-degree line (to transfer this value to the y-axis for the next step), and repeat the process. This “staircase” or “cobweb” pattern shows whether the value of y is converging towards an equilibrium (stable equilibrium) or moving away from it (unstable equilibrium).

Phase diagrams provide an immediate visual representation of a system’s stability.

(b) Order of a Difference Equation

The

order

of a difference equation is the difference between the highest and lowest time subscripts of the dependent variable appearing in the equation.

In simpler terms, it tells us how many previous periods’ values the current value of the variable depends on.

Definition:

If a difference equation has the form:

F(y

_t+n

, y

_t+n-1

, …, y

_t

, t) = 0

then the order of this equation is ‘n’, which is the difference between the highest subscript (t+n) and the lowest subscript (t).

Examples:

• y _t+1 = a y _t + c This is a first-order equation because the difference between the highest subscript (t+1) and the lowest (t) is 1. It indicates that the value of y depends on its value in only one preceding period.
• y _t+2 + a₁ y _t+1 + a₂ y _t = c This is a second-order equation because the difference between the highest subscript (t+2) and the lowest (t) is 2. It indicates that the value of y depends on its values in the two preceding periods.
• y _t = 5 y _t-3 This is a third-order equation because the difference between the subscripts is t – (t-3) = 3.

Significance:

The order of a difference equation is important because it determines the number of

initial conditions

required to find a unique solution. An n-th order equation requires ‘n’ initial conditions (e.g., y₀, y₁, …, y

_n-1

) to be solved for a specific path.