CUET UG Applied Mathematics Booster Test 3 - Integration by Partial Fractions -

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CUET UG Applied Mathematics Booster Test 3 - Integration by Partial Fractions

⏱️ 30 Minutes

❓ 20 Questions

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

CUET UG Applied Mathematics Booster Test 3 - Integration by Partial Fractions

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QUESTION 1 OF 20

In evaluating algebraic parameter constraints, if \(P(x)\)is an explicit polynomial of formal degree \(m\) and \(Q(x)\)is a polynomial of precise degree \(n\), what exact rigorous relationship exclusively forces \(\frac{P(x)}{Q(x)}\)into an improper rational classification?

Answer Review

1 In evaluating algebraic parameter constraints, if \(P(x)\)is an explicit polynomial of formal degree \(m\) and \(Q(x)\)is a polynomial of precise degree \(n\), what exact rigorous relationship exclusively forces \(\frac{P(x)}{Q(x)}\)into an improper rational classification?

A \(m=n+1\)

B \(m\geq n\)

C \(m<n\)

D \(m=0\)

💡 Short Explanation

Compare the degrees of numerator and denominator. Improper rational functions have a higher or equal numerator degree. Long division is then required.

✅ Correct Answer Explanation

A rational function is improper when the degree of the numerator is greater than or equal to the degree of the denominator. Therefore, \(m\geq n\) is the correct condition.

❓ Why Other Options Are Incorrect

Option A: Represents only one special case.
Option C: Defines a proper rational function.
Option D: Degree zero alone does not determine the classification.

🎯 Strategy

Used

Elimination

Application: Eliminate options that do not satisfy the general definition of an improper rational function.

Final Logic: Numerator degree must be greater than or equal to denominator degree.

🧠 Memory Trick

Improper ⇒ Numerator ≥ Denominator

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Improper Rational Functions

2 Match the specific complex integral algebraic manipulation action in List I needed to simplify the respective integrand expression in List II.

List I	List II
1. — Express as a polynomial quotient plus a proper fraction using long division.	a. — \(\int \frac{e^{x}}{e^{2x}+e^{x}} dx\)
2. — Decompose perfectly into distinct linear factors using \(\frac{A}{x-a}+\frac{B}{x-b}\).	b. — \(\int \frac{1}{x^{2}-3x-18} dx\)
3. — Complete the square specifically within the denominator.	c. — \(\int \frac{x^{2}+3x+2}{x^{2}+7x+12} dx\)
4. — Execute a preliminary change of variable before decomposing.	d. — \(\int \frac{4x-10}{\left(x-3)(x-4\right)} dx\)

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

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

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

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

💡 Short Explanation

Improper rational expressions require long division. Linear factors require partial fraction decomposition. Quadratic expressions are simplified by completing the square. Exponential expressions are simplified using substitution.

✅ Correct Answer Explanation

The correct matching is: 1. Express as a polynomial quotient plus a proper fraction using long division → c. \(\int \frac{x^{2}+3x+2}{x^{2}+7x+12} dx\) 2. Decompose perfectly into distinct linear factors → d. \(\int \frac{4x-10}{\left(x-3)(x-4\right)} dx\) 3. Complete the square specifically within the denominator → b. \(\int \frac{1}{x^{2}-3x-18} dx\) 4. Execute a preliminary change of variable before decomposing → a. \(\int \frac{e^{x}}{e^{2x}+e^{x}} dx\) Hence, Option C is the correct answer.

❓ Why Other Options Are Incorrect

Option A: Incorrect because substitution, partial fractions, and long division are matched with inappropriate integrals.
Option B: Incorrect because long division is incorrectly assigned to the quadratic expression, and substitution is mismatched.
Option D: Incorrect because the algebraic manipulation techniques are not paired with the appropriate integrands.

🎯 Strategy

Option Grouping

Application:

Identify the algebraic form of each integrand before choosing the integration technique:
Same degree numerator and denominator → Long Division
Distinct linear factors → Partial Fractions
Quadratic denominator → Complete the Square
Exponential expression → Substitution

Final Logic:

Recognizing the algebraic structure first leads directly to the correct preprocessing technique for each integral.

🧠 Memory Trick

S = Substitute

📚 Bottom of Form

3 When executing the algorithm
\(P(x)=Q(x)\times T(x)+R(x)\)
to resolve an improper rational evaluation, which of the following definitive mathematical principles correctly hold true?
A. The algebraic expression equates to \(\frac{P(x)}{Q(x)}=T(x)+\frac{R(x)}{Q(x)}\).
B. \(T(x)\)integrates as standard polynomial terms.
C. The degree of the polynomial remainder \(R(x)\)is less than the degree of divisor \(Q(x)\).
D. The expression \(\frac{R(x)}{Q(x)}\)forms a proper rational function.

A A, D

B A, B, C

C A, B, C, D

D B, C, D

💡 Short Explanation

Polynomial division gives quotient and remainder. The remainder has a smaller degree. The remainder fraction is proper.

✅ Correct Answer Explanation

All four statements are true properties of polynomial division and improper rational functions.

❓ Why Other Options Are Incorrect

Option A: Omits true statements B and C.
Option B: Omits true statement D.
Option D: Omits true statement A.

🎯 Strategy

Used

Elimination

Application: Verify each statement individually.

Final Logic: Since every statement is correct, all must be included.

🧠 Memory Trick

Divide → Quotient + Proper Fraction

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Polynomial Division

4 Which definitive theoretical statement regarding the partial fraction decomposition sequence of rational integrands is demonstrably INCORRECT?

A Decomposition can strictly and only be performed directly on rational fractions categorized mathematically as proper.

B Integration cannot proceed analytically on proper rational fractions unless all complex irreducible quadratic roots are artificially forced into linear polynomials.

C The polynomial dividend must always perfectly equal the product of the divisor and quotient, adding the valid remainder.

D Every improper fraction algebraically converts to a predictable polynomial summed with an integrable proper rational component.

💡 Short Explanation

Irreducible quadratics need not be factorised into linear terms. They are handled directly in partial fractions. Hence the statement is false.

✅ Correct Answer Explanation

Irreducible quadratic factors are decomposed using linear numerators. Converting them into linear factors is unnecessary.

❓ Why Other Options Are Incorrect

Option A: Correct after reducing improper fractions.
Option C: Standard polynomial division identity.
Option D: Correct property of improper rational functions.

🎯 Strategy

Used

Extreme Word Filter

Application: The word "unless" makes the statement excessively restrictive.

Final Logic: Absolute statements are often incorrect.

🧠 Memory Trick

Quadratics Stay Quadratic

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Partial Fractions

5 Extracting complex coefficients directly from mixed components, evaluate the precise sum of \(A+B+C\) when functionally decomposing
\(\frac{3x-2}{{\left((x+1)(x-2\right)}^{2})}=\frac{A}{x+1}+\frac{B}{x-2}+\frac{C}{{\left(x-2\right)}^{2}}.\)

A \(0\)

B \(\frac{4}{3}\)

C \(-\frac{5}{9}\)

D \(1\)

💡 Short Explanation

Determine the constants. Add \(A\), \(B\), and \(C\). Simplify the result.

✅ Correct Answer Explanation

After determining the constants using partial fractions, the sum \(A+B+C\) evaluates to \(\frac{4}{3}\).

❓ Why Other Options Are Incorrect

Option A: Incorrect sum.
Option C: Represents an incorrect value.
Option D: Does not equal the required sum.

🎯 Strategy

Used

Substitution

Application: Substitute suitable values of \(x\) to obtain the constants efficiently.

Final Logic: The computed constants give the required sum.

🧠 Memory Trick

Find → Add → Simplify

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Partial Fraction Constants

6 What specific geometric operation inherently validates the equating coefficients method utilized for extracting partial fraction constants \(A\) and \(B\)?

A Differentiating the constrained denominator explicitly to generate a logarithmic inverse.

B Expanding an identity algebraically so corresponding power terms strictly equal each other everywhere across the continuous real number line.

C Binding the rational function exclusively to a closed interval matrix.

D Converting real roots exclusively into dynamic imaginary parameters.

💡 Short Explanation

Equating coefficients compares like powers of \(x\). Polynomial identities hold for every value of \(x\). Corresponding coefficients must be equal.

✅ Correct Answer Explanation

The equating coefficients method is based on expanding both sides of a polynomial identity and comparing coefficients of corresponding powers of \(x\). Since the identity is true for all values of \(x\), the coefficients must be equal.

❓ Why Other Options Are Incorrect

Option A: Differentiation is not used to equate coefficients.
Option C: Closed intervals are unrelated to the method.
Option D: Imaginary roots are not required.

🎯 Strategy

Used

Elimination

Application: Remove options unrelated to polynomial identities.

Final Logic: Polynomial identities justify equating corresponding coefficients.

🧠 Memory Trick

Same powers → Same coefficients

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Equating Coefficients Method

7 Structurally derive the explicit linear algebraic system evaluated to compute coefficients \(A\) and \(B\) for the distinct decomposition equation
\(4x-10=A(x-4)+B(x-3).\)

A \(A+B=0, 4A-3B=10\)

B \(A+B=-10, -4A-3B=4\)

C \(A+B=4, -4A-3B=-10\)

D \(A-B=4, 4A+3B=10\)

💡 Short Explanation

Expand both sides. Compare coefficients of \(x\). Compare constant terms.

✅ Correct Answer Explanation

Expanding, \(A(x-4)+B(x-3)=(A+B)x-(4A+3B).\) Comparing coefficients with \(4x-10\), \(A+B=4,-4A-3B=-10.\)

❓ Why Other Options Are Incorrect

Option A: Incorrect coefficient equations.
Option B: Coefficients are interchanged.
Option D: Does not match the expanded identity.

🎯 Strategy

Used

Substitution

Application: Expand first, then compare coefficients.

Final Logic: Matching coefficients gives the required equations.

🧠 Memory Trick

Expand → Compare → Solve

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Partial Fraction Decomposition

8 Calculate the exact continuous definite evaluation resulting strictly from the integration mapped as
\(\int \frac{4}{3(x-2)^{2}} dx.\)

A \(\frac{4}{3(x-2)}+C\)

B \(\frac{4}{3}log⁡∣x-2∣+C\)

C \(-\frac{4}{3(x-2)}+C\)

D \(-\frac{4}{3}log⁡∣(x-2)^{2}∣+C\)

💡 Short Explanation

Rewrite as \({\left(x-2\right)}^{-2}\). Apply the power rule. Add the constant of integration.

✅ Correct Answer Explanation

Using \(\int (x-a)^{-2}dx=-\frac{1}{x-a}+C,\) we obtain \(\int \frac{4}{3(x-2)^{2}}dx=-\frac{4}{3(x-2)}+C.\)

❓ Why Other Options Are Incorrect

Option A: Incorrect sign.
Option B: Logarithmic integration is not applicable.
Option D: The integrand is not of the form \(\frac{1}{x-a}\).

🎯 Strategy

Used

Substitution

Application: Recognise the standard integral of \({\left(x-a\right)}^{-2}\).

Final Logic: Apply the power rule correctly.

🧠 Memory Trick

Power \(-2\) gives negative reciprocal

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Standard Integrals

9 Determine the precise algebraic numerator arrangement mapped to an irreducible non-linear quadratic parameter precisely evaluated as
\(x^{2}+1\)
located in the denominator.

A \(A\)

B \(Ax^{2}+B\)

C \(\frac{A}{x}\)

D \(Bx+C\)

💡 Short Explanation

Irreducible quadratics require a linear numerator. Degree of numerator is one less than denominator. Standard partial fraction rule applies.

✅ Correct Answer Explanation

For an irreducible quadratic factor such as \(x^{2}+1\), the numerator must be linear. Therefore, the correct form is \(Bx+C.\)

❓ Why Other Options Are Incorrect

Option A: Constant numerator is insufficient.
Option B: Numerator degree is too high.
Option C: Not the required algebraic form.

🎯 Strategy

Used

Elimination

Application: Recall the standard numerator form for irreducible quadratics.

Final Logic: Quadratic denominator ⇒ Linear numerator.

🧠 Memory Trick

Quadratic Below → Linear Above

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Partial Fractions with Quadratic Factors

10 Calculate the algebraic parameter count mathematically necessitated to correctly deconstruct a rational function containing one explicit linear root and one irreducible quadratic polynomial:
\(\frac{px^{2}+qx+c}{\left(x+a)(x^{2}+b\right)}.\)

A Two parameters: \(A\) and \(B\)

B Four parameters: \(A,B,C,\)and \(D\)

C One parameter: \(A\)

D Three parameters: \(A,B,\)and \(C\)

💡 Short Explanation

Linear factor contributes one constant. Irreducible quadratic contributes two constants. Total parameters = three.

✅ Correct Answer Explanation

The decomposition is \(\frac{A}{x+a}+\frac{Bx+C}{x^{2}+b},\) which contains three unknown constants: \(A\), \(B\), and \(C\).

❓ Why Other Options Are Incorrect

Option A: One parameter is missing.
Option B: One unnecessary parameter is included.
Option C: Insufficient parameters.

🎯 Strategy

Used

Option Grouping

Application: Count unknown constants contributed by each denominator factor.

Final Logic: \(1+2=3\) parameters.

🧠 Memory Trick

Linear = 1, Quadratic = 2

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Partial Fraction Decomposition

11 By observing explicit equating coefficient formulas, derive the exact value of \(A\) for the complex identity setup:
\(3x-2=A(x^{2}-4x+4)+B(x^{2}-x-2)+C(x+1),\)
given the constraints
\(A+B=0,-4A-B+C=3,4A-2B+C=-2.\)

A \(\frac{5}{9}\)

B \(-\frac{5}{9}\)

C \(1\)

D \(\frac{4}{3}\)

💡 Short Explanation

Solve the simultaneous equations. Use substitution or elimination. Determine the value of \(A\).

✅ Correct Answer Explanation

Solving the given system of equations gives \(A=-\frac{5}{9}.\) Hence, Option B is correct.

❓ Why Other Options Are Incorrect

Option A: Incorrect sign.
Option C: Does not satisfy the equations.
Option D: Incorrect value of \(A\).

🎯 Strategy

Used

Substitution

Application: Express one variable in terms of another and solve.

Final Logic: The simultaneous equations uniquely determine \(A\).

🧠 Memory Trick

Equate → Solve → Substitute

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Equating Coefficients

12 Systematically process the equated constants for decomposing
\(\frac{1}{\left(x-1)(x+3\right)}\)
to identify the corresponding absolute numerical fractional values of \(A\) and \(B\).

A \(A=\frac{1}{2}, B=-\frac{1}{2}\)

B \(A=4, B=-4\)

C \(A=\frac{1}{4}, B=-\frac{1}{4}\)

D \(A=-\frac{1}{4}, B=\frac{1}{4}\)

💡 Short Explanation

Decompose into partial fractions. Compare coefficients. Solve for \(A\) and \(B\).

✅ Correct Answer Explanation

Using \(\frac{1}{\left(x-1)(x+3\right)}=\frac{A}{x-1}+\frac{B}{x+3},\) solving gives \(A=\frac{1}{4},B=-\frac{1}{4}.\)

❓ Why Other Options Are Incorrect

Option A: Values are doubled.
Option B: Incorrect constants.
Option D: Signs are interchanged.

🎯 Strategy

Used

Substitution

Application: Substitute suitable values of \(x\).

Final Logic: The constants satisfy the identity.

🧠 Memory Trick

Cover-up gives constants

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Partial Fractions

13 Integrate and structurally simplify the explicitly derived fraction mapping natively evaluated as
\(\int \left[-\frac{5}{9(x+1)}+\frac{5}{9(x-2)}\right]dx.\)

A \(\frac{5}{9}log⁡∣\frac{x+1}{x-2}∣+C\)

B \(log⁡∣\frac{x-2}{x+1}∣+C\)

C \(\frac{5}{9}log⁡∣\frac{x-2}{x+1}∣+C\)

D \(-\frac{5}{9}log⁡∣\frac{x-2}{x+1}∣+C\)

💡 Short Explanation

Integrate each fraction separately. Apply the logarithmic formula. Combine logarithms.

✅ Correct Answer Explanation

Using \(\int \frac{1}{x-a} dx=log⁡∣x-a∣+C,\) the integral becomes \(\frac{5}{9}log⁡∣x-2∣-\frac{5}{9}log⁡∣x+1∣=\frac{5}{9}log⁡∣\frac{x-2}{x+1}∣+C.\)

❓ Why Other Options Are Incorrect

Option A: Numerator and denominator are reversed.
Option B: Missing the coefficient \(\frac{5}{9}\).
Option D: Incorrect overall sign.

🎯 Strategy

Used

Substitution

Application: Apply the standard logarithmic integral.

Final Logic: Difference of logs becomes the log of a quotient.

🧠 Memory Trick

Minus logs → Quotient

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Logarithmic Integrals

14 Systematically calculate the complete combined integration specifically defining the evaluated function parameters:
\(\int \left[-\frac{2}{x-3}+\frac{6}{x-4}\right]dx.\)

A \(2log⁡∣x-3∣-6log⁡∣x-4∣+C\)

B \(-\frac{2}{{\left(x-3\right)}^{2}}+\frac{6}{{\left(x-4\right)}^{2}}+C\)

C \(log⁡∣\frac{{\left(x-3\right)}^{2}}{{\left(x-4\right)}^{6}}∣+C\)

D \(-2log⁡∣x-3∣+6log⁡∣x-4∣+C\)

💡 Short Explanation

Integrate each term. Apply \(\int \frac{1}{x-a}dx\). Add the constant.

✅ Correct Answer Explanation

Applying the logarithmic integral to each term gives \(-2log⁡∣x-3∣+6log⁡∣x-4∣+C.\)

❓ Why Other Options Are Incorrect

Option A: Incorrect signs.
Option B: Incorrect integration formula.
Option C: Not equivalent to the required expression.

🎯 Strategy

Used

Substitution

Application: Integrate each fraction independently.

Final Logic: Sum of standard logarithmic integrals.

🧠 Memory Trick

Coefficient stays outside log

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Standard Logarithmic Integrals

15 Structurally simplify the definitive integral equation mapping exactly to the complex operational function:
\(\int \left[1-\frac{4}{x-3}+\frac{6}{x-4}\right]dx.\)

A \(x-4log⁡∣x-3∣+6log⁡∣x-4∣+C\)

B \(x+4log⁡∣x-3∣-6log⁡∣x-4∣+C\)

C \(x-2log⁡∣x-3∣+6log⁡∣x-4∣+C\)

D \(1-4log⁡∣x-3∣+6log⁡∣x-4∣+C\)

💡 Short Explanation

Integrate the constant. Integrate each rational term. Combine all results.

✅ Correct Answer Explanation

Since \(\int 1 dx=x,\) the complete integral is \(x-4log⁡∣x-3∣+6log⁡∣x-4∣+C.\)

❓ Why Other Options Are Incorrect

Option B: Incorrect signs.
Option C: Wrong coefficient.
Option D: Constant is not integrated correctly.

🎯 Strategy

Used

Substitution

Application: Integrate each component separately.

Final Logic: Constant gives \(x\); fractions give logarithms.

🧠 Memory Trick

Constant → \(x\), Fraction → Log

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Integration Using Partial Fractions

16 Before analytically initiating the complex partial fraction evaluation mapping, mathematically divide and exact the continuous quotient function strictly derived from processing
\(\frac{x^{3}+2x}{x^{2}-1}.\)

A Quotient \(T(x)=x^{2}\)

B Quotient \(T(x)=2x\)

C Quotient \(T(x)=x\)

D Quotient \(T(x)=1\)

💡 Short Explanation

Apply polynomial long division. Divide the leading terms first. The quotient obtained is \(x\).

✅ Correct Answer Explanation

Divide \(\frac{x^{3}+2x}{x^{2}-1}.\) The leading term is \(\frac{x^{3}}{x^{2}}=x.\) Thus, \(x(x^{2}-1)=x^{3}-x.\) Subtracting, \((x^{3}+2x)-(x^{3}-x)=3x.\) Hence, \(\frac{x^{3}+2x}{x^{2}-1}=x+\frac{3x}{x^{2}-1},\) so the quotient is \(T(x)=x.\) Therefore, Option C is correct.

❓ Why Other Options Are Incorrect

Option A: The quotient degree is too high.
Option B: Does not satisfy the long division process.
Option D: Ignores the leading-term division.

🎯 Strategy

Used

Substitution

Application: Divide the highest-degree terms first to obtain the quotient.

Final Logic: The quotient begins with \(\frac{x^{3}}{x^{2}}=x\).

🧠 Memory Trick

Leading terms decide the quotient.

📚 Class XII – Mathematics Part II – Chapter: Integral

📄 Passage

Advanced equilibrium business models rely heavily on precise mathematical total revenue frameworks derived structurally from non-linear Marginal Revenue (MR) inputs. If a corporate enterprise identifies that their Marginal Revenue evaluates natively as the quadratic function

Following the detailed passage parameters, strictly evaluate the precise algebraic equation resulting sequentially from fully substituting \(x=2\) into the raw integrated revenue formula
\(R(x)=80x+10x^{2}+x^{3}+C\)
to equate identically to \(240\).

A \(80(2)+20(2)^{2}+3(2)^{3}=240\)

B \(80x+10x^{2}+x^{3}=240\)

C \(160+40+8+C=240\)

D \(80+20+3+C=240\)

💡 Short Explanation

Substitute \(x=2\). Simplify each term. Equate to 240.

✅ Correct Answer Explanation

Substituting \(x=2\) into \(R(x)=80x+10x^{2}+x^{3}+C\) gives \(160+40+8+C=240.\) Hence, Option C is correct.

❓ Why Other Options Are Incorrect

Option A: Uses the MR equation instead of the integrated revenue function.
Option B: Does not substitute \(x=2\).
Option D: Uses incorrect numerical values.

🎯 Strategy

Used

Substitution

Application: Replace \(x\) with the given value.

Final Logic: Direct substitution gives the required equation.

🧠 Memory Trick

Revenue = Integrate, then Substitute

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Application of Integration

📄 Passage

Deriving accurately from the evaluated text passage equations, what represents the finalized strict analytical Total Revenue formula \(R(x)\)immediately after calculating explicitly that \(C=32\)?

A \(R(x)=80+20x+3x^{2}+32\)

B \(R(x)=80x+20x^{2}+3x^{3}+32\)

C \(R(x)=40x-\frac{10x^{3}}{3}+32\)

D \(R(x)=80x+10x^{2}+x^{3}+32\)

💡 Short Explanation

Integrate MR. Find \(C\). Write the final revenue function.

✅ Correct Answer Explanation

Integrating \(MR=80+20x+3x^{2}\) gives \(R(x)=80x+10x^{2}+x^{3}+C.\) Using \(C=32\), \(R(x)=80x+10x^{2}+x^{3}+32.\)

❓ Why Other Options Are Incorrect

Option A: Represents the MR function, not total revenue.
Option B: Incorrect integration.
Option C: Unrelated expression.

🎯 Strategy

Used

Substitution

Application: Insert the calculated constant into the integrated function.

Final Logic: Revenue function is obtained after integration and applying the boundary condition.

🧠 Memory Trick

Integrate MR → Get TR

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

19 Extrapolate and evaluate the deeply complex mixed fractional numerical integration specifically defined natively as
\(\int \frac{1}{e^{2x}-1} dx.\)

A \(log⁡∣\frac{e^{x}-1}{e^{x}}∣+C\)

B \(\frac{1}{2}log⁡∣e^{2x}-1∣+C\)

C \(\frac{1}{2}log⁡∣\frac{e^{x}-1}{e^{x}+1}∣+C\)

D \(log⁡∣e^{x}-1∣-x+C\)

💡 Short Explanation

Substitute \(t=e^{x}\). Convert to a rational integral. Apply partial fractions.

✅ Correct Answer Explanation

Using \(t=e^{x},dt=e^{x} dx,\) the integral reduces to a rational function. Partial fraction decomposition gives \(\int \frac{1}{e^{2x}-1} dx=\frac{1}{2}log⁡∣\frac{e^{x}-1}{e^{x}+1}∣+C.\)

❓ Why Other Options Are Incorrect

Option A: Incomplete simplification.
Option B: Incorrect logarithmic form.
Option D: Does not match the integral.

🎯 Strategy

Used

Substitution

Application: Convert the exponential expression into a rational function.

Final Logic: Substitute first, then apply partial fractions.

🧠 Memory Trick

\(e^{x}=t\) first

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Integration by Substitution and Partial Fractions

20 In an equilibrium economic function calculating exact fractional constants, if the quantity demanded and quantity supplied functionally intersect exactly at 30 units when the standard market price structurally evaluates to ₹8 per unit, what specific mathematical condition formally locates the equilibrium point?

A Setting Marginal Revenue explicitly equal to zero.

B Integrating the constant price variable exclusively.

C Setting quantity demanded mathematically equal to quantity supplied \(\left(x_{d},\ x_{s}\right)\).

D Converting linear constraints directly into an improper rational exponent.

💡 Short Explanation

Equilibrium occurs where demand equals supply. Both quantities become equal. The intersection gives the equilibrium point.

✅ Correct Answer Explanation

The market equilibrium is obtained when the quantity demanded equals the quantity supplied. The given price and quantity satisfy this equilibrium condition.

❓ Why Other Options Are Incorrect

Option A: Relates to optimisation, not equilibrium.
Option B: Integration is not required.
Option D: Has no role in equilibrium analysis.

🎯 Strategy

Used

Elimination

Application: Eliminate options unrelated to market equilibrium.

Final Logic: Equilibrium is defined by equality of demand and supply.

🧠 Memory Trick

Demand = Supply ⇒ Equilibrium

📚 Class XII – Mathematics Part II – Chapter: Applications of Integrals – Topic: Mathematical Applications in Economics

CUET UG Applied Mathematics Booster Test 3 - Integration by Partial Fractions

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