Samacheer Kalvi Class 11 Business Maths Solutions Chapter 6 Applications of Differentiation Exercise 6.6

Choose The Correct Answer.

Question 1. Average fixed cost of the cost function \( C(x) = 2x^3 + 5x^2 - 14x + 21 \) is:
(a) \( \frac{2}{3} \)
(b) \( \frac{5}{x} \)
(c) \( -\frac{14}{x} \)
(d) \( \frac{21}{x} \)
Answer: (d) \( \frac{21}{x} \)
In simple words: The average fixed cost is the fixed part of the cost divided by x. Here, 21 is the fixed cost, so it's \( \frac{21}{x} \).

🎯 Exam Tip: Remember that fixed costs are the part of the total cost that does not change with the quantity produced (x), represented by the constant term in the cost function.

Question 2. Marginal revenue of the demand function \( p = 20 - 3x \) is:
(a) \( 20 - 6x \)
(b) \( 20 - 3x \)
(c) \( 20 + 6x \)
(d) \( 20 + 3x \)
Answer: (a) \( 20 - 6x \)
In simple words: First, find the total revenue by multiplying price (p) by quantity (x). Then, take the derivative of this total revenue with respect to x to find the marginal revenue, which shows how revenue changes with one more unit sold.

🎯 Exam Tip: To find marginal revenue, always start by calculating the total revenue function \( R = px \) and then differentiate it with respect to \( x \).

Question 3. If demand and the cost function of a firm are \( p = 2 - x \) and \( C = -2x^2 + 2x + 7 \) then its profit function is:
(b) \( x^2 - 7 \)
(c) \( -x^2 + 7 \)
(d) \( -x^2 - 7 \)
Answer: (b) \( x^2 - 7 \)
In simple words: To get the profit, you take the money you earn (revenue) and subtract the money you spend (cost). In this case, the profit function is found by subtracting the given cost function from the revenue function.

🎯 Exam Tip: Profit is always calculated as Total Revenue minus Total Cost. Be careful with signs when subtracting polynomial functions.

Question 4. If the demand function is said to be inelastic, then:
(a) \( | \eta_d | > 1 \)
(b) \( | \eta_d | = 1 \)
(c) \( | \eta_d | < 1 \)
(d) \( | \eta_d | = 0 \)
Answer: (c) \( | \eta_d | < 1 \)
In simple words: Inelastic demand means that when the price changes, the quantity people want to buy does not change much. This happens when the elasticity of demand is less than 1.

🎯 Exam Tip: Understand the different ranges of elasticity: \( | \eta_d | < 1 \) for inelastic, \( | \eta_d | > 1 \) for elastic, and \( | \eta_d | = 1 \) for unit elastic demand.

Question 5. The elasticity of demand for the demand function \( x = \frac{1}{p} \) is:
(a) 0
(b) 1
(c) \( -\frac{1}{p} \)
(d) \( \infty \)
Answer: (b) 1
In simple words: The elasticity of demand tells us how much the quantity demanded changes when the price changes. For this specific demand function, the elasticity always comes out to be 1, which means it is unit elastic.

🎯 Exam Tip: The formula for point elasticity of demand is \( \eta_d = \frac{-p}{x} \frac{dx}{dp} \). Remember to first find \( \frac{dx}{dp} \) from the given demand function.

Question 6. Relationship among MR, AR and \( \eta_d \) is:
(a) \( \eta_d = \frac{AR}{AR - MR} \)
(b) \( \eta_d = AR - MR \)
(c) \( MR = AR = \eta_d \)
(d) \( AR = \frac{MR}{\eta_d} \)
Answer: (a) \( \eta_d = \frac{AR}{AR - MR} \)
In simple words: This equation shows how marginal revenue (MR), average revenue (AR), and the elasticity of demand \( (\eta_d) \) are all connected in economics. It helps to understand how a company's revenue changes based on price adjustments and consumer response.

🎯 Exam Tip: This formula, derived from the relationship \( MR = AR(1 - \frac{1}{\eta_d}) \), is crucial for understanding revenue maximization and market behavior. Make sure to memorize it.

Question 7. For the cost function \( C = \frac{1}{25} e^{5x} \), the marginal cost is:
(a) \( \frac{1}{25} \)
(b) \( \frac{1}{5} e^{5x} \)
(c) \( \frac{1}{125} e^{5x} \)
(d) \( 25e^{5x} \)
Answer: (b) \( \frac{1}{5} e^{5x} \)
In simple words: To find the marginal cost, you need to take the derivative of the total cost function with respect to x. This tells you how much extra it costs to make one more item.

🎯 Exam Tip: When differentiating exponential functions like \( e^{ax} \), remember to multiply by the derivative of the exponent (a in this case) using the chain rule.

Question 8. Instantaneous rate of change of \( y = 2x^2 + 5x \) with respect to x at \( x = 2 \) is:
(a) 4
(b) 5
(c) 13
(d) 9
Answer: (c) 13
In simple words: The instantaneous rate of change is like finding the speed of something at an exact moment. You find the derivative of the function first, then put the given value of x into the derivative to get the exact rate.

🎯 Exam Tip: First find the derivative \( \frac{dy}{dx} \) of the function, and then substitute the given value of x into the derivative to calculate the instantaneous rate of change.

Question 9. If the average revenue of a certain firm is Rs. 50 and its elasticity of demand is 2, then their marginal revenue is:
(a) Rs. 50
(b) Rs. 25
(c) Rs. 100
(d) Rs. 75
Answer: (b) Rs. 25
In simple words: Marginal revenue is the extra money you get from selling one more item. We can find it using a special formula that connects average revenue and how sensitive demand is to price changes.

🎯 Exam Tip: Use the formula \( MR = AR \left[ 1 - \frac{1}{\eta_d} \right] \) for calculating marginal revenue. Make sure to substitute the correct values for average revenue (AR) and elasticity of demand \( (\eta_d) \).

Question 10. Profit P(x) is maximum when:
(a) MR = MC
(b) MR = 0
(c) MC = AC
(d) TR = AC
Answer: (a) MR = MC
In simple words: A company makes the most profit when the extra money gained from selling one more unit (marginal revenue) is equal to the extra cost of making that unit (marginal cost). This is a key principle in economics.

🎯 Exam Tip: The condition for profit maximization is always where marginal revenue (MR) equals marginal cost (MC).

Question 11. The maximum value of \( f(x) = \sin x \) is:
(a) 1
(b) \( \frac{\sqrt{3}}{2} \)
(c) \( \frac{1}{\sqrt{2}} \)
(d) \( -\frac{1}{\sqrt{2}} \)
Answer: (a) 1
In simple words: The sine function goes up and down, but it never goes higher than 1 and never lower than -1. So, its biggest possible value is 1.

🎯 Exam Tip: Remember the range of trigonometric functions. The sine function \( \sin x \) has a maximum value of 1 and a minimum value of -1.

Question 12. If \( f(x, y) \) is a homogeneous function of degree n, then \( x \frac{\partial f}{\partial x}+y \frac{\partial f}{\partial y} \) is equal to:
(a) \( (n - 1)f \)
(b) \( n(n - 1)f \)
(c) \( nf \)
Answer: (c) \( nf \)
In simple words: This is Euler's Theorem for homogeneous functions, which states that if you multiply each variable by its partial derivative and sum them up, you get the function itself multiplied by its degree. It's a useful rule in advanced math.

🎯 Exam Tip: Euler's Theorem for homogeneous functions is a fundamental result in multivariable calculus. Remember that for a homogeneous function \( f(x,y) \) of degree \( n \), the relationship is always \( x \frac{\partial f}{\partial x} + y \frac{\partial f}{\partial y} = nf \).

Question 13. If \( u = 4x^2 + 4xy + y^2 + 4x + 32y + 16 \), then \( \frac{\partial^2 u}{\partial y \partial x} \) is equal to:
(a) \( 8x + 4y + 4 \)
(b) 4
(c) \( 2y + 32 \)
(d) 0
Answer: (b) 4
In simple words: To find this, you first take the derivative of the function with respect to x, treating y as a constant. Then, take the derivative of that new result with respect to y, treating x as a constant. This finds the "mixed" partial derivative.

🎯 Exam Tip: For mixed partial derivatives like \( \frac{\partial^2 u}{\partial y \partial x} \), calculate \( \frac{\partial u}{\partial x} \) first, and then differentiate that result with respect to \( y \). Be careful to treat other variables as constants during each differentiation step.

Question 14. If \( u = x^3 + 3xy^2 + y^3 \) then \( \frac{\partial^2 u}{\partial y \partial x} \) is:
(a) 3
(b) \( 6y \)
(c) \( 6x \)
(d) 2
Answer: (b) \( 6y \)
In simple words: We need to find the change in the rate of change. First, we find how 'u' changes when 'x' changes. Then, we see how that change itself changes when 'y' changes. This involves two steps of differentiation.

🎯 Exam Tip: When computing mixed partial derivatives, remember that for most well-behaved functions (where the second partial derivatives are continuous), the order of differentiation does not matter, i.e., \( \frac{\partial^2 u}{\partial y \partial x} = \frac{\partial^2 u}{\partial x \partial y} \).

Question 15. If \( u = e^{x^{2}} \) then \( \frac{\partial u}{\partial x} \) is equal to:
(a) \( 2x e^{x^{2}} \)
(b) \( e^{x^{2}} \)
(c) \( 2 e^{x^{2}} \)
(d) 0
Answer: (a) \( 2x e^{x^{2}} \)
In simple words: To differentiate this, we use the chain rule. We differentiate the outer function (the exponential) and then multiply by the derivative of the inner function (the exponent \( x^2 \)).

🎯 Exam Tip: Remember the chain rule for differentiation: if \( y = f(g(x)) \), then \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \). For \( e^{f(x)} \), the derivative is \( e^{f(x)} \cdot f'(x) \).

Question 16. Average cost is minimum when:
(a) Marginal cost = marginal revenue
(b) Average cost = marginal cost
(c) Average cost = Marginal revenue
(d) Average Revenue = Marginal cost
Answer: (b) Average cost = marginal cost
In simple words: A firm produces at its most efficient level, where the average cost per unit is the lowest, when the cost to produce one more unit (marginal cost) is exactly equal to the average cost of all units produced so far. This point is often called the efficient scale of production.

🎯 Exam Tip: The average cost curve is U-shaped, and its minimum point occurs where the marginal cost curve intersects it from below. So, \( AC_{min} \) occurs when \( AC = MC \).

Question 17. A company begins to earn profit at:
(a) Maximum point
(b) Breakeven point
(c) Stationary point
(d) Even point
Answer: (b) Breakeven point
In simple words: The breakeven point is when a company's total income equals its total costs. This means it's making neither a profit nor a loss. Any sales after this point will start generating profit.

🎯 Exam Tip: The breakeven point is a critical concept in business, marking the threshold where total revenue covers total costs, and beyond which profit generation begins.

Question 18. The demand function is always:
(a) Increasing function
(b) Decreasing function
(c) Non-decreasing function
(d) Undefined function
Answer: (b) Decreasing function
In simple words: Usually, if the price of something goes up, people want to buy less of it, and if the price goes down, people want to buy more. This inverse relationship means the demand function typically slopes downwards, making it a decreasing function.

🎯 Exam Tip: In most economic models, the demand function is decreasing, indicating an inverse relationship between price and quantity demanded (the law of demand).

Question 19. If \( q = 1000 + 8p_1 - p_2 \) then, \( \frac{\partial q}{\partial p_1} \) is:
(a) -1
(b) 8
(c) 1000
(d) \( 1000 - p_2 \)
Answer: (b) 8
In simple words: This is a partial derivative, which means we find how 'q' changes only when 'p1' changes, while we hold 'p2' and any other numbers constant. In this case, the rate of change is simply the coefficient of \( p_1 \).

🎯 Exam Tip: When taking a partial derivative with respect to one variable (e.g., \( p_1 \)), treat all other variables (e.g., \( p_2 \)) and constants as if they are fixed numbers.

Question 20. If R = 5000 units/year, \( C_1 = 20 \) paise, \( C_3 = \) Rs. 20 then EOQ is:
(a) 5000
(b) 100
(c) 1000
(d) 200
Answer: (c) 1000
In simple words: EOQ stands for Economic Order Quantity. It is the best amount of goods to order at one time to keep costs low. We use a formula that looks at how many items are needed, the cost to order, and the cost to hold them. Remember to convert all costs to the same currency unit.

🎯 Exam Tip: Always ensure that all cost components \( C_1 \) and \( C_3 \) are in the same currency unit (e.g., Rupees). Convert paise to Rupees before applying the EOQ formula \( EOQ = \sqrt{\frac{2RC_3}{C_1}} \).