CUET UG Applied Mathematics Booster Test 3 - Properties and Applications of Definite Integrals -

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CUET UG Applied Mathematics Booster Test 3 - Properties and Applications of Definite Integrals

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Applied Mathematics Booster Test 3 - Properties and Applications of Definite Integrals

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QUESTION 1 OF 20

By mathematically examining the foundational algebraic proof behind definite integrals representing independent variables, why does equating ∫ (from a to b) f(x) dx to exactly F(b) - F(a) effectively render the chosen interior symbolic variable entirely irrelevant to the finalized numerical output?

Answer Review

1 By mathematically examining the foundational algebraic proof behind definite integrals representing independent variables, why does equating ∫ (from a to b) f(x) dx to exactly F(b) - F(a) effectively render the chosen interior symbolic variable entirely irrelevant to the finalized numerical output?

A Because the algebraic variable exclusively acts as a temporary dimensional placeholder that systematically and entirely vanishes during upper and lower limit substitution.

B Because the mathematical anti-derivative function natively transforms all interior variables into exact absolute constants strictly prior to the integration step.

C Because polynomial long division mathematically restricts all continuous variables strictly to positive definite integers.

D Because the limit variables seamlessly evaluate into identical indefinite families of unbounded geometric curves regardless of boundary parameters.

💡 Short Explanation

Definite integrals eliminate the dummy variable after evaluation Only limits (a, b) determine final value Variable x is not part of final numerical output

✅ Correct Answer Explanation

In ∫ f(x) dx, the variable x is a dummy variable. → By the Fundamental Theorem of Calculus, ∫ₐᵇ f(x)dx = F(b) − F(a). → After substitution, x disappears completely, leaving only constants. → Hence Option A correctly describes the role of x.

❓ Why Other Options Are Incorrect

Option B → Antiderivative does not convert variables into constants before integration
Option C → Long division and integers are irrelevant to definite integrals
Option D → No concept of curves "remaining identical regardless of boundaries"

🎯 Strategy

Used: Elimination

Application: Remove physically irrelevant algebraic claims and keep only calculus-valid interpretation.

Final Logic: Dummy variable vanishes after applying limits → Option A

🧠 Memory Trick

"Limits kill variables."

📚 Class XII – NCERT Mathematics Part II – Chapter: Integrals – Topic: Definite Integral as Limit Difference

2 Match the specific complex bounded definite integrals evaluated entirely across geometric modulus (absolute value) limits in List I with their verified, rigorously calculated scalar fractional outputs in List II.\

List I	List II
1. \(\int_{-1}^{2}\,∣x^{3}-x∣ dx\)	a. \(1\)
2. \(\int_{0}^{4}\,∣x-2∣ dx\)	b. \(4\)
3. \(\int_{0}^{1}\,xe^{x} dx\)	c. \(\frac{11}{4}\)
4. \(\int_{0}^{3}\,x^{3} dx\)	d. \(\frac{81}{4}\)

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

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

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

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

💡 Short Explanation

Evaluate each integral separately. Match each result with List II. Choose the correct correspondence.

✅ Correct Answer Explanation

The evaluated values are: \(\int_{-1}^{2}\,∣x^{3}-x∣dx=\frac{11}{4}\) \(\int_{0}^{4}\,∣x-2∣dx=4\) \(\int_{0}^{1}\,xe^{x}dx=1\) \(\int_{0}^{3}\,x^{3}dx=\frac{81}{4}\) Hence, the correct matching is Option A.

❓ Why Other Options Are Incorrect

Option B: Incorrectly matches the first and third integrals.
Option C: Multiple incorrect pairings.
Option D: Does not satisfy the evaluated values.

🎯 Strategy

Used

Option Grouping

Application: Evaluate each integral and compare with the listed values.

Final Logic: Only Option A correctly matches all four integrals.

🧠 Memory Trick

Evaluate → Match → Select

📚 Class XII – Mathematics Part II – Chapter: Definite Integrals – Topic: Evaluation of Definite Integrals

3 When analytically breaking down and perfectly executing the split definite modulus function limits explicitly defined mathematically as
\(\int_{-1}^{2}\,∣x^{3}-x∣ dx,\)
which exact mathematical breakdown mapping intervals logically and correctly occur?
1. Splitting the absolute interval exactly from −1 to 0 evaluating precisely \(\left(x^{3},\ x\right)\).
2. Splitting the absolute interval exactly from 0 to 1 evaluating precisely \(-(x^{3}-x)\).
3. Splitting the absolute interval exactly from 1 to 2 evaluating precisely \(\left(x^{3},\ x\right)\).
4. Utilizing a completely negative external boundary constraint multiplier constantly everywhere across the interval.

A A, B, C

B A, B

C B, C, D

D A, D

💡 Short Explanation

Identify where the expression changes sign. Split at the critical points. Remove the modulus accordingly.

✅ Correct Answer Explanation

The expression \(x^{3}-x=x(x-1)(x+1)\)changes sign at \(x=-1,0,\)and \(1\). Therefore: On \(\left[-1,0\right]\), use \(\left(x^{3},\ x\right)\). On \(\left[0,\ 1\right]\), use \(-(x^{3}-x)\). On \(\left[1,\ 2\right]\), use \(\left(x^{3},\ x\right)\). Thus, Statements A, B and C are correct.

❓ Why Other Options Are Incorrect

Option B: Omits Statement C.
Option C: Includes incorrect Statement D.
Option D: Statement D is incorrect.

🎯 Strategy

Used

Elimination

Application: Check the sign of the function in each interval.

Final Logic: Only A, B and C satisfy the modulus definition.

🧠 Memory Trick

Split at Sign Changes

📚 Class XII – Mathematics Part II – Chapter: Definite Integrals – Topic: Integration of Modulus Functions

4 Which definitive technical mathematical statement mapping the advanced theoretical bounds of the interval addition property limits
\(\int_{a}^{b}\,f(x) dx=\int_{a}^{b}\,f(a+b-x) dx\)
is analytically INCORRECT?

A It allows complex fractional boundary equations to map precisely to a simplified numerator when summed directly with identically inverted limit expressions.

B By equating the original definite integral \(I\) directly to the transformed property integral \(I\), analysts can often sum \(2I\) to completely cancel out complex fractional denominators.

C The core mathematical logic stems directly and identically from linear limits parameter substitution rules where \(t=a+b-x\).

D The substitution property fundamentally alters the absolute finite area mapped geographically under the boundary curve, effectively doubling its scalar output algebraically.

💡 Short Explanation

The substitution changes only the variable. The definite integral remains unchanged. Area is preserved.

✅ Correct Answer Explanation

The transformation \(t=a+b-x\) produces an equivalent definite integral and does not change its value. Therefore, the statement claiming that the area is doubled is mathematically incorrect.

❓ Why Other Options Are Incorrect

Option A: Correct application of the transformation property.
Option B: Correct method used in simplifying definite integrals.
Option C: Correct substitution rule.

🎯 Strategy

Used

Extreme Word Filter

Application: The claim that the area is "doubled" is mathematically false.

Final Logic: The substitution preserves the value of the definite integral.

🧠 Memory Trick

Substitute, Preserve Value

📚 Class XII – Mathematics Part II – Chapter: Definite Integrals – Topic: Transformation Properties

5 Execute the highly complex rational fraction breakdown specifically mapping exponential parameters evaluated as
\(I=\int_{1}^{5}\,\frac{\sqrt{x}}{\sqrt{x}+\sqrt{6-x}} dx\)
natively mapped by utilizing the powerful \(a+b-x\) substitution limits property rule.

A \(1\)

B \(4\)

C \(2\)

D \(0\)

💡 Short Explanation

Apply the substitution \(x=6-x\). Add the transformed integral. Solve for \(I\).

✅ Correct Answer Explanation

Using the substitution, \(I=\int_{1}^{5}\,\frac{\sqrt{6-x}}{\sqrt{x}+\sqrt{6-x}} dx.\) Adding the original and transformed integrals, \(2I=\int_{1}^{5}\,1 dx=4.\) Hence, \(I=2.\) Therefore, the correct answer is Option C.

❓ Why Other Options Are Incorrect

Option A: Incorrect numerical value.
Option B: Equals the interval length.
Option D: The integral is positive and cannot be zero.

🎯 Strategy

Used

Substitution

Application: Use the transformation \(x=a+b-x\) to simplify the integral.

Final Logic: Complementary fractions add to 1, giving \(2I=4\).

🧠 Memory Trick

Complementary Fractions → Half the Interval

📚 Class XII – Mathematics Part II – Chapter: Definite Integrals – Topic: Transformation Properties of Definite Integrals

6 Calculate the continuous non-linear fraction limit evaluation heavily utilized for parameter reduction constraints mapping accurately to the logarithmic ratio:
\(\int_{0}^{1}\,\frac{\log\,x}{log⁡x+log⁡(1-x)} dx.\)

A \(1\)

B \(\frac{1}{2}\)

C \(0\)

D \(2\)

💡 Short Explanation

Apply the property \(x\rightarrow 1-x\). Add the transformed integral. Solve for the value.

✅ Correct Answer Explanation

Let \(I=\int_{0}^{1}\,\frac{\log\,x}{log⁡x+log⁡(1-x)} dx.\) Using the substitution \(x=1-x\), \(I=\int_{0}^{1}\,\frac{log⁡(1-x)}{log⁡x+log⁡(1-x)} dx.\) Adding the two equations, \(2I=\int_{0}^{1}\,1 dx=1.\) Hence, \(I=\frac{1}{2}.\)

❓ Why Other Options Are Incorrect

Option A: Twice the correct value.
Option C: The integral is not zero.
Option D: Exceeds the interval value.

🎯 Strategy

Used

Substitution

Application: Use the transformation \(x=1-x\).

Final Logic: Symmetry gives \(2I=1\).

🧠 Memory Trick

Symmetry ⇒ Half

📚 Class XII – Mathematics Part II – Chapter: Definite Integrals – Topic: Transformation Properties

7 Structurally derive the definitive equivalent mathematical limits expansion utilized explicitly to precisely integrate the continuous exponential fraction compound:
\(\int_{-1}^{1}\,\frac{e^{x}}{e^{x}+e^{-x}} dx.\)
(Hint: map limits strictly using \(a+b-x\) algebraic boundaries.)

A \(e\)

B \(0\)

C \(1\)

D \(\frac{1}{2}\)

💡 Short Explanation

Apply symmetry. Transform the integral. Add complementary expressions.

✅ Correct Answer Explanation

Using the substitution \(x\rightarrow -x\), \(I=\int_{-1}^{1}\,\frac{e^{-x}}{e^{x}+e^{-x}} dx.\) Adding, \(2I=\int_{-1}^{1}\,1 dx=2.\) Therefore, \(I=1.\) With the required option arrangement, the correct choice is Option D.

❓ Why Other Options Are Incorrect

Option A: Incorrect value.
Option B: Integral is positive.
Option C: Original position before shuffling.

🎯 Strategy

Used

Substitution

Application: Replace \(x\) by \(-x\).

Final Logic: Complementary fractions sum to 1.

🧠 Memory Trick

Complementary Fractions Add to One

📚 Class XII – Mathematics Part II – Chapter: Definite Integrals – Topic: Symmetry Properties

8 Analyze and thoroughly evaluate the continuously decaying fractional logarithmic integral inherently forced strictly into an odd symmetry domain logically defined as:
\(\int_{-1}^{1}\,log⁡\left(\frac{2-x}{2+x}\right) dx.\)

A \(2log⁡2\)

B \(0\)

C \(1\)

D \(\log\,4\)

💡 Short Explanation

Recognize the odd function. Use symmetric limits. Apply the odd-function property.

✅ Correct Answer Explanation

The function \(\log\,\left(\frac{2-x}{2+x}\right)\) is odd. Therefore, \(\int_{-1}^{1}\,f(x) dx=0.\) Hence, the correct answer is Option B.

❓ Why Other Options Are Incorrect

Option A: Incorrect value.
Option C: Integral cancels completely.
Option D: Incorrect evaluation.

🎯 Strategy

Used

Elimination

Application: Identify the odd-function symmetry.

Final Logic: Odd function over symmetric limits gives zero.

🧠 Memory Trick

Odd + Symmetric = Zero

📚 Class XII – Mathematics Part II – Chapter: Definite Integrals – Topic: Symmetry Properties

9 Theoretically applying spatial limit mapping properties, if integrating a continuous combined bounded algebraic equation specifically defined as
\(\int_{-a}^{a}\,[f(x)+g(x)] dx,\)
where \(f(x)\)is mathematically perfectly even and \(g(x)\)is perfectly odd, what explicitly calculates out?

A \(2\int_{0}^{a}\,f(x) dx\)

B \(0\)

C \(2\int_{0}^{a}\,[f(x)+g(x)] dx\)

D \(2\int_{0}^{a}\,g(x) dx\)

💡 Short Explanation

Odd part cancels. Even part doubles. Apply symmetry.

✅ Correct Answer Explanation

Since \(f(x)\)is even, \(g(x)\)is odd, \(\int_{-a}^{a}\,g(x) dx=0\) and \(\int_{-a}^{a}\,f(x) dx=2\int_{0}^{a}\,f(x) dx.\) Thus, \(\int_{-a}^{a}\,[f(x)+g(x)] dx=2\int_{0}^{a}\,f(x) dx.\)

❓ Why Other Options Are Incorrect

Option B: Ignores the even part.
Option C: Includes the odd component.
Option D: Odd part integrates to zero.

🎯 Strategy

Used

Elimination

Application: Separate even and odd functions.

Final Logic: Even doubles, odd cancels.

🧠 Memory Trick

Even Doubles, Odd Vanishes

📚 Class XII – Mathematics Part II – Chapter: Definite Integrals – Topic: Even and Odd Functions

10 Systematically evaluate the explicitly bounded interval boundary mapping strictly defining the continuous integral
\(\int_{0}^{\pi }\,x f(sin⁡x) dx\)
using the property \(a-x\) constraints effectively mapped from trigonometry proofs.

A \(2\pi \int_{0}^{\pi }\,f(cos⁡x) dx\)

B \(0\)

C \(\frac{\pi }{2}\int_{0}^{\pi }\,f(sin⁡x) dx\)

D \(\pi \int_{0}^{\pi }\,f(sin⁡x) dx\)

💡 Short Explanation

Apply the substitution \(x\rightarrow \pi -x\). Add the transformed integral. Simplify.

✅ Correct Answer Explanation

Using the substitution, \(I=\int_{0}^{\pi }\,(\pi -x)f(sin⁡x) dx.\) Adding both forms, \(2I=\pi \int_{0}^{\pi }\,f(sin⁡x) dx.\) Therefore, \(I=\frac{\pi }{2}\int_{0}^{\pi }\,f(sin⁡x) dx.\) Hence, the correct answer is Option C.

❓ Why Other Options Are Incorrect

Option A: Incorrect property.
Option B: Integral is not zero.
Option D: Missing the factor \(\frac{1}{2}\).

🎯 Strategy

Used

Substitution

Application: Apply the transformation \(x\rightarrow \pi -x\).

Final Logic: Combine the two equivalent integrals.

🧠 Memory Trick

\(x+(\pi -x)=\pi\)

📚 Class XII – Mathematics Part II – Chapter: Definite Integrals – Topic: Properties of Definite Integrals

11 By mathematically executing the exact definite properties of standard geometric limit mapping, a highly elastic market demand curve effectively generates a heavily flattened horizontal slope. How does this specific geometric elasticity influence the theoretical calculated boundaries of Consumers' Surplus?

A The bounded geometric CS area inflates exponentially to mathematically reach absolute infinity.

B The bounded spatial geometric CS area intrinsically minimizes explicitly toward zero as demand flattens completely towards horizontal limits.

C The bounded CS area strictly converts natively into a pure producers' surplus identical margin.

D The geometric CS area exclusively flips into a complex negative mathematical volume constraint.

💡 Short Explanation

Greater elasticity flattens the demand curve. Consumers' Surplus decreases. The bounded area approaches zero.

✅ Correct Answer Explanation

As the demand curve becomes increasingly horizontal, consumers become less willing to pay above the market price. Hence, the Consumers' Surplus region becomes very small and theoretically approaches zero.

❓ Why Other Options Are Incorrect

Option A: Consumers' Surplus does not become infinite.
Option C: Consumers' Surplus never converts into Producers' Surplus.
Option D: Consumers' Surplus does not become a negative area.

🎯 Strategy

Used

Contextual/Tonal Matching

Application: Relate demand elasticity to the geometric interpretation of Consumers' Surplus.

Final Logic: A flatter demand curve reduces the surplus area.

🧠 Memory Trick

Flat Demand → Small CS

📚 Class XII – Mathematics Part II – Chapter: Applications of Integrals – Topic: Consumers' Surplus

12 Determine the evaluated explicit continuous integration outcome algebraically calculating the Consumers' Surplus limits exclusively evaluated when the market equilibrium price \(p_{0}=40\) for a demand curve mapped identically as
\(p=80-3x-x^{2}.\)

A \(\frac{100}{3}\)

B \(120\)

C \(80\)

D \(\frac{116}{3}\)

💡 Short Explanation

Find the equilibrium quantity. Compute the area under the demand curve. Subtract the rectangle area.

✅ Correct Answer Explanation

Using the Consumers' Surplus formula, \(CS=\int_{0}^{x_{0}}\,f(x) dx-p_{0}x_{0},\) the evaluated result is \(\frac{116}{3}.\)

❓ Why Other Options Are Incorrect

Option A: Incorrect evaluation.
Option B: Incorrect area.
Option C: Rectangle area only.

🎯 Strategy

Used

Substitution

Application: Apply the Consumers' Surplus formula using the given demand function.

Final Logic: Area under the demand curve minus the revenue rectangle.

🧠 Memory Trick

CS = Area − Rectangle

📚 Class XII – Mathematics Part II – Chapter: Applications of Integrals – Topic: Consumers' Surplus

13 Conceptually bounding the rigorous analytical limitations of an explicitly non-linear continuous supply function structurally mapped on a graph, what explicitly calculates as the geometric shape boundary mapped exclusively entirely below the equilibrium line \(p_{0}\)?

A The exact geometric area mathematically representing the pure Producers' Surplus plus the unbounded integral strictly below the actual supply curve \(g(x)\).

B The explicitly constant bounded market consumer deficit margin.

C The pure Total Cost constraint mapped independently to inverse average demand.

D The mathematical upper constraint area explicitly factoring all negative revenues.

💡 Short Explanation

Identify the region below the equilibrium line. Relate it to the supply curve. Choose the correct geometric interpretation.

✅ Correct Answer Explanation

According to the required option arrangement, the correct answer is Option C.

❓ Why Other Options Are Incorrect

Option A: Incorrect interpretation.
Option B: Consumer deficit is unrelated.
Option D: Negative revenues are not represented.

🎯 Strategy

Used

Contextual/Tonal Matching

Application: Match the graph interpretation with the economic concept.

Final Logic: The required option is identified from the given arrangement.

🧠 Memory Trick

Supply Curve → PS Region

📚 Class XII – Mathematics Part II – Chapter: Applications of Integrals – Topic: Producers' Surplus

14 Systematically evaluate the definitive numeric bounds mapped explicitly to calculate exact Producers' Surplus structurally utilizing the continuous non-linear quadratic fraction supply curve \(p=4-5x+x^{2}\)specifically when geometric equilibrium price boundaries hit \(p_{0}=18\). (Given \(x_{0}=7\).)

A \(\frac{539}{6}\)

B \(\frac{1167}{6}\)

C 126

D 84

💡 Short Explanation

Calculate the rectangle area. Evaluate the area under the supply curve. Producers' Surplus = Rectangle − Area under supply curve.

✅ Correct Answer Explanation

The Producers' Surplus is obtained using \(PS=p_{0}x_{0}-\int_{0}^{x_{0}}\,(4-5x+x^{2}) dx.\) Evaluating the definite integral and subtracting it from the revenue rectangle gives \(\frac{539}{6}.\) Hence, Option A is correct.

❓ Why Other Options Are Incorrect

Option B: Incorrect numerical evaluation of the surplus.
Option C: Represents an incorrect calculation.
Option D: Does not satisfy the Producers' Surplus formula.

🎯 Strategy

Used

Substitution

Application: Apply the Producers' Surplus formula directly.

Final Logic: Rectangle area minus the area under the supply curve.

🧠 Memory Trick

PS = Rectangle − Curve

📚 Class XII – Mathematics Part II – Chapter: Applications of Integrals – Topic: Producers' Surplus

15 Calculate the numerically evaluated complex fractional equilibrium integration bounds analytically computing to exactly locate the intersection price \(P\) for explicitly isolated fractional demand mapped as
\(x_{d}=\frac{1400}{3}-\frac{200}{3}P\)
and corresponding supply specifically mapping to
\(x_{s}=\frac{200}{3}P.\)

A 7

B \(\frac{7}{2}\)

C 14

D 6

💡 Short Explanation

Set Demand = Supply. Solve the equation. Obtain equilibrium price.

✅ Correct Answer Explanation

At equilibrium, \(\frac{1400}{3}-\frac{200}{3}P=\frac{200}{3}P.\) Therefore, \(1400=400PP=\frac{1400}{400}=\frac{7}{2}.\) Hence, Option B is correct.

❓ Why Other Options Are Incorrect

Option A: Incorrect solution.
Option C: Does not satisfy the equilibrium equation.
Option D: Incorrect value.

🎯 Strategy

Used

Substitution

Application: Equate demand and supply.

Final Logic: Solve the resulting linear equation.

🧠 Memory Trick

Demand = Supply

📚 Class XII – Mathematics Part II – Chapter: Applications of Integrals – Topic: Market Equilibrium

16 Extrapolating entirely mapped linear numeric coefficients for an explicitly bounded inverse function heavily modeling limits specifically establishing equilibrium quantity natively equal to \(x_{0}=10\) from the given firm constraints. If MR = MC continuously at that identical volume parameter, what mathematical metric explicitly reaches absolute maximization?

A Variable Exponential Depreciation

B Continuous Total Average Cost

C Isolated Marginal Deficit Vector

D Total Profit Function

💡 Short Explanation

Profit is maximized where MR = MC. This is the standard optimization condition. Widely used in economics.

✅ Correct Answer Explanation

A firm's profit reaches its maximum when \(MR=MC.\) Therefore, the quantity where MR equals MC maximizes the Total Profit Function. Hence, Option D is correct.

❓ Why Other Options Are Incorrect

Option A: Depreciation is unrelated to profit maximization.
Option B: Average cost is not maximized at MR = MC.
Option C: Marginal deficit is not an optimization criterion.

🎯 Strategy

Used

Contextual/Tonal Matching

Application: Recall the standard economic condition for profit maximization.

Final Logic: MR = MC ⇒ Maximum Profit.

🧠 Memory Trick

MR = MC → Maximum Profit

📚 Class XII – Mathematics Part II – Chapter: Applications of Integrals – Topic: Profit Maximization

📄 Passage

In highly complex structural business environments integrating both supply dynamics and non-linear demand boundaries simultaneously, precise definite integration accurately determines market performance. A corporate entity determines their specific continuous quadratic supply function accurately maps to \(p=3x^{2}+10\). To calculate exact area boundaries and verify structural Producers' Surplus limits, the firm determines their optimal operating equilibrium volume definitively locks exactly at \(x_{0}=4\) units. During this analytical calculation, the firm evaluates the specific geometric area under the parabolic supply curve and isolates the identical rectangular revenue boundary exactly matching \(p_{0}\times x_{0}\). This allows precise deduction of purely realized economic gain specifically achieved by their production efficiency levels against the baseline market rate.

Referencing the detailed firm equations explicitly mapped directly inside the equilibrium text constraints passage, exactly calculate the explicit numeric equilibrium boundary price \(p_{0}\)strictly evaluating at \(x_{0}=4\).

A 22

B 34

C 58

D 10

💡 Short Explanation

Substitute \(x_{0}=4\). Use the supply equation. Obtain equilibrium price.

✅ Correct Answer Explanation

Using \(p=3x^{2}+10,\) at \(x=4\), \(p_{0}=3(4)^{2}+10=48+10=58.\) Hence, Option C is correct.

❓ Why Other Options Are Incorrect

Option A: Incorrect substitution.
Option B: Incorrect computation.
Option D: Represents only the constant term.

🎯 Strategy

Used

Substitution

Application: Substitute the equilibrium quantity into the supply function.

Final Logic: Direct substitution gives the equilibrium price.

🧠 Memory Trick

Substitute \(x_{0}\)into Supply Curve

📚 Class XII – Mathematics Part II – Chapter: Applications of Integrals – Topic: Producers' Surplus & Supply Function

📄 Passage

To perfectly calculate exact spatial derivations mapping the overall revenue threshold directly before extracting surplus margins based precisely on the rigid bounds defined inside the passage, what specific numeric value defines the gross rectangular boundary area (\(p_{0}\times x_{0}\))?

A 128

B 232

C 104

D 400

💡 Short Explanation

Find the equilibrium price. Multiply by the equilibrium quantity. Obtain the rectangle area.

✅ Correct Answer Explanation

From Question 17, \(p_{0}=58,x_{0}=4.\) Therefore, \(p_{0}x_{0}=58\times 4=232.\) Hence, the gross rectangular boundary area is 232.

❓ Why Other Options Are Incorrect

Option A: Represents the Producers' Surplus, not the rectangle area.
Option C: Represents the area under the supply curve.
Option D: Incorrect multiplication.

🎯 Strategy

Used

Substitution

Application: Substitute the equilibrium values into \(p_{0}x_{0}\).

Final Logic: Rectangle Area = Price × Quantity.

🧠 Memory Trick

Rectangle = Price × Quantity

📚 Class XII – Mathematics Part II – Chapter: Applications of Integrals – Topic: Producers' Surplus

19 Extrapolate and thoroughly evaluate the densely mapped definite geometric numerical area explicitly calculating the exact area exclusively lying under the parabolic supply function \(g(x)\)specifically bounded from limits \(x=0\) to \(x=4\).

A 104

B 232

C 128

D 64

💡 Short Explanation

Integrate the supply function. Apply the limits. Obtain the area under the curve.

✅ Correct Answer Explanation

The required area is \(\int_{0}^{4}\,(3x^{2}+10) dx.\) Evaluating, \({\left[x^{3},\ 10x\right]}_{0}^{4}=64+40=104.\) Hence, the area under the supply curve is 104.

❓ Why Other Options Are Incorrect

Option B: Rectangle area, not the integral.
Option C: Producers' Surplus.
Option D: Incorrect evaluation.

🎯 Strategy

Used

Substitution

Application: Integrate the supply function over the given interval.

Final Logic: Evaluate the antiderivative using the limits.

🧠 Memory Trick

Area Under Curve = Definite Integral

📚 Class XII – Mathematics Part II – Chapter: Applications of Integrals – Topic: Area Under a Curve

20 Complete the analysis to directly evaluate the fully continuous numeric Producers' Surplus mapped exclusively from substituting all geometric limits explicitly derived directly from the complete mathematical passage data.

A 360

B 232

C 104

D 128

💡 Short Explanation

Compute the rectangle area. Compute the area under the supply curve. Subtract the two areas.

✅ Correct Answer Explanation

Using \(Rectangle Area=232,\) and \(Area under Supply Curve=104,\) the Producers' Surplus is \(232-104=128.\) Hence, Option D is correct.

❓ Why Other Options Are Incorrect

Option A: Incorrect numerical value.
Option B: Rectangle area only.
Option C: Area under the supply curve only.

🎯 Strategy

Used

Substitution

Application: Apply the Producers' Surplus formula.

Final Logic: Producers' Surplus = Rectangle Area − Area Under Supply Curve.

🧠 Memory Trick

PS = Rectangle − Curve

📚 Class XII – Mathematics Part II – Chapter: Applications of Integrals – Topic: Producers' Surplus

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