CUET UG Applied Mathematics Booster Test 1

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CUET UG Applied Mathematics Booster Test 1 - Cost, Revenue and Rates

⏱️ 30 Minutes

❓ 20 Questions

📌 Answers are locked once submitted — results and explanations appear at the end.

CUET UG Applied Mathematics Booster Test 1 - Cost, Revenue and Rates

1 / 20

QUESTION 1 OF 20

A company produces 'x' units and the variable cost is given by V(x) = x² - 2x. What is the numerical variable cost when 10 units are produced?

Answer Review

1 A company produces 'x' units and the variable cost is given by V(x) = x² - 2x. What is the numerical variable cost when 10 units are produced?

A 120

B 100

C 80

D 20

💡 Short Explanation

Substitute \(x=10\) into the variable cost function. Calculate \({10}^{2}-2(10)\). The variable cost equals 80, making Option C correct.

✅ Correct Answer Explanation

The variable cost function is V(x) = x² - 2x. Substituting x = 10, we get V(10) = 10² − 2(10) = 100 − 20 = 80. Therefore, Option C is correct. Options A (120), B (100), and D (20) result from incorrect substitution or calculation and do not satisfy the given function.

❓ Why Other Options Are Incorrect

Option A) 120 → This value results from an incorrect calculation. Substituting x = 10 into the function does not produce 120.
Option B) 100 → This considers only the square term 10² and ignores the −2x term, giving an incomplete evaluation.
Option D) 20 → This represents only the value of 2x when x = 10, not the complete variable cost function.

🎯 Strategy

Used: Substitution

Application: Substitute the given production level directly into the cost function and simplify carefully to avoid arithmetic mistakes.

Final Logic: Since V(10) = 100 − 20 = 80, the correct answer is Option C.

🧠 Memory Trick

Mnemonic: "Substitute → Square → Subtract."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Cost Function and Marginal Cost (Economic Applications)

2 Match the cost models in List I with their expressions in List II, given variable cost \(V(x)=x^{2}-2x\) and a fixed cost of ₹15000:

List I	List II
1. (A) Variable Cost \(V(x)\)	a. \(x^{2}-2x\)
2. (B) Fixed Cost \(k\)	b. 15080
3. (C) Total Cost \(C(x)\)	c. 15000
4. (D) Total Cost for \(x=10, C(10)\)	d. \(x^{2}-2x+15000\)

A 1-a, 2-c, 3-d, 4-b

B 1-b, 2-a, 3-c, 4-d

C 1-a, 2-b, 3-c, 4-d

D 1-d, 2-c, 3-a, 4-b

💡 Short Explanation

Variable cost is \(x^{2}-2x\). Fixed cost remains ₹15000. Total cost equals variable cost plus fixed cost. At \(x=10\), total cost is ₹15080.

✅ Correct Answer Explanation

Variable cost is \(V(x)=x^{2}-2x\), so 1 → a. The fixed cost is constant at ₹15000, so 2 → c. Therefore, the total cost function is \(C(x)=x^{2}-2x+15000\), giving 3 → d. Evaluating at x=10, \(C(10)=80+15000=15080\), so 4 → b. Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option B) 1-b, 2-a, 3-c, 4-d→ Incorrect because it mismatches every expression. Variable cost is not 15080, fixed cost is not \(x^{2}-2x\), and total cost is not merely ₹15000.
Option C) 1-a, 2-b, 3-c, 4-d → Incorrect because 2-b treats the fixed cost as ₹15080, which is actually the total cost at x=10. The remaining matches are also incorrect.
Option D) 1-d, 2-c, 3-a, 4-b → Incorrect because 1-d assigns the total cost function as the variable cost, while C-I assigns the variable cost expression as the total cost.

🎯 Strategy

Used: Option Grouping

Application: First identify the known expressions for variable cost, fixed cost, total cost, and numerical total cost, then match each pair systematically.

Final Logic: Correct matching is 1-a, 2-c, 3-d, 4-b , which corresponds to Option A.

🧠 Memory Trick

Mnemonic: "Total Cost = Variable Cost + Fixed Cost."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Cost, Revenue and Profit Functions (Economic Applications)

3 If the price per unit is given by \(p=5-x\), which of the following statements about the revenue formula are true?
1) The revenue function \(R(x)\)evaluates to \(5x-x^{2}\).
2) The revenue function \(R(x)\)evaluates to \(5-x^{2}\).
3) The marginal revenue \(MR\) evaluates to \(5-2x\).
4) \(R(x)\)is found using the formula \(p+x\).

A 1, 2, 4

B 1, 2

C 1, 3

D 1, 2, 3, 4

💡 Short Explanation

Revenue equals price × quantity. Thus, \(R(x)=x(5-x)=5x-x^{2}\). Differentiating gives \(MR=5-2x\). Therefore, only statements 1 and 3 are correct.

✅ Correct Answer Explanation

Revenue is calculated as Price × Quantity. Since \(p=5-x\), the revenue function is \(R(x)=x(5-x)=5x-x^{2}\), making Statement A correct. Differentiating gives \(MR=\frac{dR}{dx}=5-2x\), so Statement 3 is also correct. Statement 2 omits the multiplication by quantity, while Statement 4 incorrectly uses addition instead of multiplication. Hence, Option C is correct.

❓ Why Other Options Are Incorrect

Option A) 1, 2, 4 → Incorrect because Statements 2 and 4 are false. Revenue is not \(5-x^{2}\), and it is never calculated as price + quantity.
Option B) 1, 2 → Incorrect because Statement 2 is false, while Statement 3 is true since marginal revenue is the derivative of the revenue function.
Option D) 1, 2, 3, 4 → Incorrect because it includes Statements 2 and 4, both of which contradict the correct revenue formula and its definition.

🎯 Strategy

Used: Substitution

Application: Substitute the given price function into the revenue formula \(R=x\times p\), then differentiate to identify the correct statements.

Final Logic: Since \(R(x)=5x-x^{2}\)and \(MR=5-2x\), only A and C are correct.

🧠 Memory Trick

Mnemonic: "Revenue = Price × Quantity; MR = Derivative of Revenue."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Revenue Function and Marginal Revenue (Economic Applications)

4 Identify the incorrect statement regarding the price and output relation when price \(p=30-2x\).

A The Revenue function is \(R(x)=30x-2x^{2}\).

B The Marginal Revenue is evaluated by finding \(dR/dx\).

C The Marginal Revenue evaluates to \(MR=30-4x\).

D The Marginal Revenue is always a constant value regardless of \(x\).

💡 Short Explanation

Revenue equals price multiplied by quantity. Differentiate revenue to obtain marginal revenue. Since \(MR=30-4x\), it depends on output and is not constant.

✅ Correct Answer Explanation

Given \(p=30-2x\), the revenue function is \(R(x)=x(30-2x)=30x-2x^{2}\), so Option A is correct. Marginal Revenue is obtained by differentiation, giving \(MR=30-4x\), making Options B and C correct. Option D is incorrect because marginal revenue varies with the value of \(x\).

❓ Why Other Options Are Incorrect

Option A) The Revenue function is \(R(x)=30x-2x^{2}\). → Correct because revenue is calculated as Price × Quantity, giving the stated expression.
Option B) The Marginal Revenue is evaluated by finding \(dR/dx\). → Correct because marginal revenue is defined as the derivative of the revenue function with respect to output.
Option C) The Marginal Revenue evaluates to \(MR=30-4x\). → Correct since differentiating \(30x-2x^{2}\)gives \(30-4x\).

🎯 Strategy

Used: Elimination

Application: Verify each statement using the revenue formula and differentiation. Eliminate statements that match the mathematical definitions and identify the one contradicting the concept.

Final Logic: Since \(MR=30-4x\) changes with output, it cannot be constant. Therefore, Option D is the incorrect statement.

🧠 Memory Trick

Mnemonic: "MR = dR/dx → If \(x\) is present, MR changes."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Revenue Function and Marginal Revenue (Economic Applications)

5 For a manufacturing firm, the total cost model combines variable costs dependent on production and absolute fixed costs. If a firm's \(V(x)=x(x+30)\)and storage cost is ₹1500, what is the correct construction of the cost function \(C(x)\)?

A \(x^{2}+30x\)

B \(x^{2}+30x+1500\)

C \(x+1530\)

D \(x^{2}+1500x+30\)

💡 Short Explanation

Total cost equals variable cost plus fixed cost. Expand \(V(x)=x(x+30)\)to obtain \(x^{2}+30x\). Add ₹1500 fixed cost, giving \(C(x)=x^{2}+30x+1500\).

✅ Correct Answer Explanation

The variable cost is \(V(x)=x(x+30)=x^{2}+30x\). Total cost is obtained by adding the fixed (storage) cost of ₹1500 to the variable cost. Hence, \(C(x)=x^{2}+30x+1500\). Option B is correct. Option A omits the fixed cost, Option C is not derived from the given function, and Option D incorrectly multiplies the fixed cost by \(x\).

❓ Why Other Options Are Incorrect

Option A) \(x^{2}+30x\)→ Incorrect because it represents only the variable cost. The fixed storage cost of ₹1500 must also be included to obtain the total cost.
Option C) \(x+1530\)→ Incorrect because it is a linear expression and does not result from expanding the given quadratic variable cost function.
Option D) \(x^{2}+1500x+30\)→ Incorrect because the fixed cost is a constant amount, not a coefficient of \(x\). The terms are incorrectly arranged.

🎯 Strategy

Used: Substitution

Application: First expand the given variable cost function, then apply the formula Total Cost = Variable Cost + Fixed Cost.

Final Logic: Expanding \(x(x+30)\)gives \(x^{2}+30x\); adding ₹1500 results in \(x^{2}+30x+1500\). Therefore, Option B is correct.

🧠 Memory Trick

Mnemonic: "TC = VC + FC" (Total Cost = Variable Cost + Fixed Cost).

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Cost Function and Economic Applications

6 Using the cost function example \(C(x)=x^{2}+30x+1500\), what is the Marginal Cost (MC) when 20 toys are produced?

A 70

B 50

C 1500

D 2500

💡 Short Explanation

Marginal Cost is the derivative of the total cost function. Differentiate \(C(x)=x^{2}+30x+1500\). Substitute \(x=20\) to obtain \(MC=70\).

✅ Correct Answer Explanation

Marginal Cost (MC) is the derivative of the total cost function with respect to output. Differentiating \(C(x)=x^{2}+30x+1500\) gives \(MC=\frac{dC}{dx}=2x+30\). Substituting \(x=20\), \(MC=2(20)+30=70\). Therefore, Option A is correct. Options B, C, and D do not represent the derivative evaluated at \(x=20\).

❓ Why Other Options Are Incorrect

Option B) 50 → Incorrect because substituting \(x=20\) into \(2x+30\) gives 70, not 50. This value results from an incorrect calculation.
Option C) 1500 → Incorrect because 1500 is the fixed cost in the total cost function. Fixed cost disappears after differentiation and does not affect marginal cost.
Option D) 2500 → Incorrect because this is unrelated to the marginal cost calculation. It is neither the derivative nor the total cost at \(x=20\).

🎯 Strategy

Used: Substitution

Application: Differentiate the cost function first to obtain the marginal cost expression, then substitute the given production level into the derivative.

Final Logic: Since \(MC=2x+30\), substituting \(x=20\) gives \(70\). Hence, Option A is correct.

🧠 Memory Trick

Mnemonic: "Differentiate first, substitute later."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Marginal Cost (Economic Applications)

7 In revenue function formation, if an operator charges ₹136 per passenger with a discount factor influencing price, the revenue \(R(x)\)will systematically be a product of the number of passengers \(x\) and:

A The total fixed costs

B The variable cost \(V(x)\)

C The adjusted price per unit \(p\)

D The marginal cost \(MC\)

💡 Short Explanation

Revenue is calculated by multiplying quantity by price per unit. Here, quantity is the number of passengers \(x\). Therefore, revenue equals \(x\times p\), where \(p\) is the adjusted price.

✅ Correct Answer Explanation

Revenue is defined as the product of quantity sold and price per unit, i.e., \(R(x)=x\times p\). If the ticket price changes because of a discount factor, the adjusted price \(p\) is used in the formula. Therefore, Option C is correct. Fixed cost, variable cost, and marginal cost are cost concepts and are not used directly to calculate revenue.

❓ Why Other Options Are Incorrect

Option A) The total fixed costs → Incorrect because fixed costs are expenses that remain constant regardless of passenger numbers. They do not determine revenue.
Option B) The variable cost \(V(x)\)→ Incorrect because variable cost represents production or operating expenses. Revenue depends on selling price and quantity, not production cost.
Option D) The marginal cost \(MC\)→ Incorrect because marginal cost measures the additional cost of producing one extra unit. It is unrelated to the formula for total revenue.

🎯 Strategy

Used: Conceptual/Tonal Matching

Application: Recognize the standard economic definition of revenue and match it with the option describing price per unit multiplied by quantity.

Final Logic: Revenue is always Quantity × Price per Unit. Therefore, the correct answer is Option C.

🧠 Memory Trick

Mnemonic: "R = P × Q" (Revenue = Price × Quantity).

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Revenue Function and Economic Applications

8 In a revenue function example, if \(p=30-2x\), what is the exact Marginal Revenue when 5 commodities are produced?

A 30

B 20

C 10

D 5

💡 Short Explanation

Revenue equals price multiplied by quantity. Differentiate the revenue function to obtain Marginal Revenue. Substitute \(x=5\); the result is 10, so Option C is correct.

✅ Correct Answer Explanation

Given the price function \(p=30-2x\), the revenue function is \(R(x)=x(30-2x)=30x-2x^{2}\). Differentiating gives the Marginal Revenue, \(MR=\frac{dR}{dx}=30-4x\). Substituting \(x=5\), \(MR=30-20=10\). Therefore, Option C is correct. Options A, B, and D do not satisfy the derivative calculation.

❓ Why Other Options Are Incorrect

Option A) 30 → Incorrect because it is the constant term of the Marginal Revenue expression. The value of \(x=5\) must also be substituted.
Option B) 20 → Incorrect because substituting \(x=5\) into \(30-4x\) gives 10, not 20. It results from an incorrect evaluation.
Option D) 5 → Incorrect because 5 is the number of commodities produced, not the Marginal Revenue. The derivative must be evaluated at \(x=5\).

🎯 Strategy

Used: Substitution

Application: First derive the Marginal Revenue expression by differentiating the revenue function, then substitute the given production level.

Final Logic: Since \(MR=30-4x\), substituting \(x=5\) gives \(10\). Hence, Option C is correct.

🧠 Memory Trick

Mnemonic: "Revenue → Differentiate → Substitute."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Marginal Revenue (Economic Applications)

9 The instantaneous rate concept dictates that if \(y=f(x)\)is a real function, the limit of change in \(y\) over a very small change in \(x\) represents:

A The total area under the curve

B The average moving slope

C The derivative \(dy/dx\)

D A fixed numerical constant

💡 Short Explanation

Instantaneous rate of change is defined using a limit. It measures how rapidly a function changes at a point. This limit is called the derivative, \(dy/dx\).

✅ Correct Answer Explanation

The derivative is defined as the limit of the ratio of the change in a function to the change in its variable as the change approaches zero. It represents the instantaneous rate of change or the slope of the tangent at a point. Therefore, Option C is correct. Options A, B, and D do not describe the mathematical definition of a derivative.

❓ Why Other Options Are Incorrect

Option A) The total area under the curve → Incorrect because the area under a curve is determined using integration, not differentiation. It does not represent an instantaneous rate of change.
Option B) The average moving slope → Incorrect because an average slope is calculated over an interval using a secant line. The question asks for the instantaneous rate at a single point.
Option D) A fixed numerical constant → Incorrect because a derivative generally varies with the value of \(x\). It is constant only for specific linear functions.

🎯 Strategy

Used: Conceptual/Tonal Matching

Application: Identify the key phrase "instantaneous rate of change", which is the standard definition of the derivative in differential calculus.

Final Logic: Instantaneous rate of change is represented by the derivative \(dy/dx\). Therefore, Option C is correct.

🧠 Memory Trick

Mnemonic: "Instantaneous = Derivative = Tangent Slope."

📚 Class XI – NCERT Mathematics – Chapter: Limits and Derivatives – Topic: Derivative as an Instantaneous Rate of Change

10 When utilizing the derivative as a rate of change, if the cost \(C\) changes with production \(x\), then \(\frac{dC}{dx}\)specifically measures the:

A Average cost

B Marginal cost

C Total variable cost

D Fixed property tax

💡 Short Explanation

The derivative measures the instantaneous rate of change. Differentiating the cost function gives the additional cost of producing one more unit. This quantity is called Marginal Cost (MC).

✅ Correct Answer Explanation

The derivative \(\frac{dC}{dx}\)represents the instantaneous rate of change of the cost function with respect to production. In economics, this derivative is known as the Marginal Cost (MC), which measures the additional cost incurred in producing one extra unit of output. Therefore, Option B is correct. Options A, C, and D do not describe the derivative of the cost function.

❓ Why Other Options Are Incorrect

Option A) Average cost → Incorrect because average cost is calculated as Total Cost ÷ Number of Units (\(C(x)/x\)). It is not obtained by differentiating the cost function.
Option C) Total variable cost → Incorrect because total variable cost is a component of the total cost function. Its derivative gives marginal cost, not the variable cost itself.
Option D) Fixed property tax → Incorrect because a fixed property tax does not vary with production. Its derivative with respect to output is zero and does not represent marginal cost.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Recognize the economic meaning of the derivative of a cost function. In economics, \(\frac{dC}{dx}\)is universally defined as Marginal Cost.

Final Logic: Since \(\frac{dC}{dx}\)measures the additional cost of producing one more unit, the correct answer is Option B.

🧠 Memory Trick

Mnemonic: "Derivative of Cost = Marginal Cost (MC)."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Marginal Cost (Economic Applications)

11 For a spherical balloon undergoing surface area change, A = 4πr². What is the rate of change of the surface area with respect to r when r = 2 cm?

A 8π cm²/cm

B 16π cm²/cm

C 4π cm²/cm

D 2π cm²/cm

💡 Short Explanation

The derivative of surface area is obtained using differentiation. Since \(A=4\pi r^{2}\), \(dA/dr=8\pi r\). Substituting \(r=2\) gives \(16\pi\). Differentiate the formula. Substitute the given radius. Evaluate the derivative.

✅ Correct Answer Explanation

Surface area of a sphere is \(A=4\pi r^{2}\). Differentiating with respect to \(r\) gives \(dA/dr=8\pi r\). At \(r=2\), \(dA/dr=16\pi\). Therefore, Option B is correct. Options A, C, and D represent incorrect derivative values or incomplete substitution.

❓ Why Other Options Are Incorrect

Option A → 8π cm²/cm is the derivative only when \(r=1\), not at \(r=2\).
Option C → 4π cm²/cm results from incorrect differentiation.
Option D → 2π cm²/cm is unrelated to the derivative of \(4\pi r^{2}\).

🎯 Strategy

Used: Substitution

Application: Differentiate first, then substitute the given value of the radius into the derivative.

Final Logic: \(8\pi \times 2=16\pi\), so Option B is correct.

🧠 Memory Trick

"Sphere Area → 8πr after differentiation."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Rate of Change

12 For a sphere undergoing volume change, V = (4/3)πr³ and S = 4πr². Utilizing chain differentiation, the rate of change of volume with respect to its surface area (dV/dS) simplifies to:

A r

B 2r

C r/2

D 4r

💡 Short Explanation

Use the chain rule. Compute \(dV/dr=4\pi r^{2}\)and \(dS/dr=8\pi r\). Their ratio gives \(dV/dS=r/2\). Differentiate both expressions. Apply chain rule. Simplify the ratio.

✅ Correct Answer Explanation

Using the chain rule, \(\frac{dV}{dS}=\frac{dV/dr}{dS/dr}=\frac{4\pi r^{2}}{8\pi r}=\frac{r}{2}.\) Thus Option C is correct. Options A, B, and D ignore the denominator obtained while differentiating the surface area.

❓ Why Other Options Are Incorrect

Option A → r ignores division by \(8\pi r\).
Option B → 2r results from incorrect simplification.
Option D → 4r has no basis in chain differentiation.

🎯 Strategy

Used: Elimination

Application: Differentiate both quantities separately before forming their ratio.

Final Logic: \(\frac{4\pi r^{2}}{8\pi r}=\frac{r}{2}\), making Option C correct.

🧠 Memory Trick

"Volume over Surface → Half the radius."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Related Rates

📄 Passage

A cylindrical vessel of radius 0.5 m is filled with oil at the rate of 0.25π m³/min. Let r be the radius, h be the height, and V be the volume of the oil at time t. The volume of a cylinder is given by V = πr²h. Because the radius of the rigid cylinder remains constant, only the height h and volume V change with time. We apply the related rates concept using time-based differentiation to determine the speed at which the oil level rises.

Applying related rates using the chain rule on the volume formula V = πr²h (where r is constant), the derivative dV/dt equates to:

A πr²

B πr²(dh/dt)

C 2πr(dr/dt)h

D dh/dt

💡 Short Explanation

Since the cylinder radius is constant, only height changes with time. Differentiating \(V=\pi r^{2}h\) gives \(dV/dt=\pi r^{2}(dh/dt)\). Radius is constant. Differentiate with respect to time. Apply related rates.

✅ Correct Answer Explanation

The cylinder has a fixed radius, so \(dr/dt=0\). Differentiating \(V=\pi r^{2}h\) with respect to time gives \(\frac{dV}{dt}=\pi r^{2}\frac{dh}{dt}.\) Hence, Option B is correct. Options A, C, and D omit required differentiation or include unnecessary terms.

❓ Why Other Options Are Incorrect

Option A → πr² lacks the time derivative.
Option C → 2πr(dr/dt)h applies when radius changes, which is not true.
Option D → dh/dt omits the multiplying factor \(\pi r^{2}\).

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Notice that the passage states the radius remains constant.

Final Logic: Constant radius implies \(dr/dt=0\), so only Option B remains.

🧠 Memory Trick

"Rigid cylinder → Only height changes."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Related Rates

📄 Passage

In this time-based rate problem, substituting r = 0.5 and dV/dt = 0.25π yields the rate at which the surface of the oil is rising (dh/dt) as:

A 0.5 m/min

B 1 m/min

C 0.25 m/min

D 2 m/min

💡 Short Explanation

Using \(dV/dt=\pi r^{2}(dh/dt)\), substitute \(r=0.5\) and \(dV/dt=0.25\pi\). Solving gives \(dh/dt=1\) m/min. Use the related-rate formula. Substitute values. Solve for height rate.

✅ Correct Answer Explanation

Given \(0.25\pi =\pi (0.5)^{2}\frac{dh}{dt}=\pi (0.25)\frac{dh}{dt}.\) Cancelling common factors gives \(\frac{dh}{dt}=1 m/min.\) Thus Option B is correct. Options A, C, and D result from incorrect substitution or arithmetic.

❓ Why Other Options Are Incorrect

Option A → 0.5 m/min comes from incorrect division.
Option C → 0.25 m/min ignores the radius squared.
Option D → 2 m/min overestimates the calculated rate.

🎯 Strategy

Used: Substitution

Application: Insert the numerical values into the differentiated equation before solving.

Final Logic: Direct substitution gives \(dh/dt=1\), making Option B correct.

🧠 Memory Trick

"Divide volume rate by πr²."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Related Rates

15 A boy is walking towards a lamp post. Because the distance 'x' between the boy and the lamp post is decreasing over time, the rate of change dx/dt is assigned a value of:

A +0.5 m/sec

B -0.5 m/sec

C 0 m/sec

D 1 m/sec

💡 Short Explanation

When distance decreases with time, its derivative is negative. Therefore, approaching the lamp post means \(dx/dt\) has a negative value. Distance decreases. Rate becomes negative. Motion is toward the lamp.

✅ Correct Answer Explanation

The derivative \(dx/dt\) indicates how distance changes over time. Since the boy walks towards the lamp post, the distance continuously decreases. Therefore, the rate must be negative. Hence Option B is correct. Options A, C, and D do not represent decreasing distance correctly.

❓ Why Other Options Are Incorrect

Option A → +0.5 m/sec indicates increasing distance.
Option C → 0 m/sec means no movement.
Option D → 1 m/sec is positive, representing motion away from the lamp.

🎯 Strategy

Used: Contextual/Tonal Matching

Application: Interpret the phrase "walking towards" to determine the sign of the derivative.

Final Logic: Towards means decreasing distance, so \(dx/dt<0\). Therefore, Option B is correct.

🧠 Memory Trick

"Towards → Negative rate."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Related Rates and Motion

16 In shadow problems, if the shadow length y relates to distance x by y = x/4, and the boy walks towards the lamp at dx/dt = -0.5 m/sec, the rate at which the shadow length decreases is:

A 1/8 m/sec

B 1/4 m/sec

C 1/2 m/sec

D 1 m/sec

💡 Short Explanation

Differentiate \(y=\frac{x}{4}\)with respect to time. Since \(dy/dt=\frac{1}{4}(dx/dt)\)and \(dx/dt=-0.5\), the shadow decreases at \(1/8\) m/sec. Differentiate the relation. Substitute the given rate. Take the magnitude of decrease.

✅ Correct Answer Explanation

Given \(y=\frac{x}{4}\), \(\frac{dy}{dt}=\frac{1}{4}\frac{dx}{dt}=\frac{1}{4}(-0.5)=-\frac{1}{8} m/sec.\) The negative sign indicates decreasing length, so the rate of decrease is 1/8 m/sec. Hence Option A is correct. Options B, C, and D overestimate the rate.

❓ Why Other Options Are Incorrect

Option B → 1/4 m/sec ignores multiplying by the given speed.
Option C → 1/2 m/sec equals the walking speed, not the shadow's rate.
Option D → 1 m/sec has no relation to the given equation.

🎯 Strategy

Used: Substitution

Application: Differentiate first and substitute the given value of \(dx/dt\).

Final Logic: \(\frac{1}{4}\times 0.5=\frac{1}{8}\), so Option A is correct.

🧠 Memory Trick

"Shadow = One-fourth of distance."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Related Rates

17 A toy firm calculates its total cost as C(x) = x² + 30x + 1500. The Marginal Cost (MC) equation representing the derivative is:

A 2x

B x + 30

C 2x + 30

D 30x

💡 Short Explanation

Marginal Cost is the derivative of the total cost. Differentiate each term: \(d(x^{2})/dx=2x\), \(d(30x)/dx=30\), and the constant becomes zero. Differentiate term-wise. Constant becomes zero. Add derivative terms.

✅ Correct Answer Explanation

Marginal Cost is \(MC=\frac{dC}{dx}.\) Differentiating, \(\frac{d}{dx}(x^{2}+30x+1500)=2x+30.\) Thus Option C is correct. Option A ignores the linear term, Option B differentiates incorrectly, and Option D is not the derivative.

❓ Why Other Options Are Incorrect

Option A → 2x omits the derivative of \(30x\).
Option B → x + 30 incorrectly differentiates \(x^{2}\).
Option D → 30x is not obtained by differentiation.

🎯 Strategy

Used: Substitution

Application: Apply basic differentiation rules to each term separately.

Final Logic: \(MC=dC/dx=2x+30\), so Option C is correct.

🧠 Memory Trick

"Cost → Differentiate once."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Marginal Cost

18 The price per unit of a commodity is p = 30 - 2x. Finding the Marginal Revenue (MR) requires evaluating dR/dx. What is the MR expression?

A 30 - 2x

B 30 - 4x

C 30

D -4x

💡 Short Explanation

Revenue equals price multiplied by quantity. Thus \(R=x(30-2x)=30x-2x^{2}\). Differentiating gives \(MR=30-4x\). Form revenue. Differentiate. Simplify.

✅ Correct Answer Explanation

Revenue is \(R=x(30-2x)=30x-2x^{2}.\) Differentiating, \(MR=\frac{dR}{dx}=30-4x.\) Therefore Option B is correct. Option A is only the price equation, while Options C and D are incomplete derivatives.

❓ Why Other Options Are Incorrect

Option A → 30 − 2x represents price, not marginal revenue.
Option C → 30 ignores differentiation of \(-2x^{2}\).
Option D → -4x omits the constant derivative.

🎯 Strategy

Used: Elimination

Application: Construct the revenue function before differentiating.

Final Logic: Differentiate \(30x-2x^{2}\)to obtain 30−4x.

🧠 Memory Trick

"Revenue first, derivative next."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Marginal Revenue

19 Exploring cost optimization concepts, if C(x) = 300x - 10x² + (1/3)x³, finding the output 'x' at which Marginal Cost is minimum requires setting the second derivative of Cost equal to zero. What is this 'x'?

A 10

B 20

C 30

D 5

💡 Short Explanation

Marginal Cost is \(C^{'}(x)\). Its minimum occurs where \(C^{''}(x)=0\). Here \(C^{''}(x)=2x-20\), giving \(x=10\). Differentiate twice. Set second derivative to zero. Solve for \(x\).

✅ Correct Answer Explanation

\(C^{'}(x)=300-20x+x^{2}.\) Differentiate again: \(C^{''}(x)=2x-20.\) Setting \(2x-20=0\) gives \(x=10.\) Hence Option A is correct. Options B, C, and D do not satisfy the required condition.

❓ Why Other Options Are Incorrect

Option B → 20 gives \(C^{''}(20)=20\neq 0\).
Option C → 30 also fails the condition.
Option D → 5 gives a negative value of the second derivative.

🎯 Strategy

Used: Substitution

Application: Differentiate twice and substitute the condition \(C^{''}(x)=0\).

Final Logic: Only x = 10 satisfies the equation.

🧠 Memory Trick

"Minimum MC → Second derivative equals zero."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Optimization Problems

20 In revenue change analysis, if a company charges ₹6000 per unit minus ₹75 for each unit in excess, R(x) = x(9750 - 75x). The largest order size for maximum revenue occurs when R'(x) = 0. What is 'x'?

A 50

B 60

C 65

D 75

💡 Short Explanation

Expand the revenue function and differentiate. Setting \(R^{'}(x)=0\) gives the production level where revenue is maximum. Solving the equation gives \(x=65\). Differentiate revenue. Set derivative to zero. Solve for output.

✅ Correct Answer Explanation

Given \(R(x)=x(9750-75x)=9750x-75x^{2}.\) Then, \(R^{'}(x)=9750-150x.\) Setting \(R^{'}(x)=0\), \(9750=150x,x=65.\) Thus, Option C is correct. Options A, B, and D do not satisfy the maximum revenue condition.

❓ Why Other Options Are Incorrect

Option A → 50 gives a positive derivative, so revenue is still increasing.
Option B → 60 is below the optimum.
Option D → 75 is beyond the maximum point where revenue decreases.

🎯 Strategy

Used: Substitution

Application: Differentiate the revenue function and solve the resulting linear equation.

Final Logic: \(9750-150x=0\Rightarrow x=65\), making Option C correct.

🧠 Memory Trick

"Maximum Revenue → Set MR = 0."

📚 Class XII – NCERT Mathematics – Chapter: Application of Derivatives – Topic: Revenue Maximization

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