CUET UG Applied Mathematics Booster Test 2 - Integration by Partial Fractions -

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CUET UG Applied Mathematics Booster Test 2 - 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 2 - Integration by Partial Fractions

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QUESTION 1 OF 20

Which of the following mathematical expressions correctly strictly qualifies as a proper rational function ready for direct partial fraction decomposition?

Answer Review

1 Which of the following mathematical expressions correctly strictly qualifies as a proper rational function ready for direct partial fraction decomposition?

A \(\frac{x^{2}+3x+2}{x^{2}+7x+12}\)

B \(\frac{x^{3}+2}{x^{2}+1}\)

C \(\frac{e^{x}}{x+1}\)

D \(\frac{x}{x^{2}+1}\)

💡 Short Explanation

A proper rational function has the numerator degree less than the denominator degree. Both numerator and denominator must be polynomials. Such functions are directly suitable for partial fraction decomposition.

✅ Correct Answer Explanation

A rational function is of the form \(\frac{P(x)}{Q(x)},\) where \(P(x)\)and \(Q(x)\)are polynomials. A proper rational function satisfies \(deg⁡(P(x))<deg⁡(Q(x)).\) Among the given options, \(\frac{x}{x^{2}+1}\) has numerator degree \(1\) and denominator degree \(2\). Hence it is a proper rational function and is ready for partial fraction decomposition. Therefore, Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Numerator and denominator have equal degrees, making it an improper rational function.
Option B → Numerator degree is greater than the denominator degree, so it is improper.
Option C → \(e^{x}\)is not a polynomial; therefore, the expression is not a rational function.

🎯 Strategy

Used

Elimination

Application:

Compare the degrees of the numerator and denominator and verify that both are polynomials.

Final Logic:

Only Option D satisfies the definition of a proper rational function.

🧠 Memory Trick

"Proper → Numerator Power Lower."

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

2 (Match the Following)
Match the specific improper or complex rational expression in List I with its initial algebraic decomposition requirement in List II.

List I	List II
1. (x² + 3x + 2)/(x² + 7x + 12)	a. Use long division to obtain: 1 − (4x + 10)/(x² + 7x + 12)
2. (x³ + x)/(x² − 1)	b. Perform variable substitution first.
3. 1/[x(log x)² + 2]	c. Use long division to obtain: 1 − 1/(x² + 1)
4. x²/(x² + 1)	d. Use long division to obtain: x + 2x/(x² − 1)

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

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

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

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

💡 Short Explanation

Before integrating a rational function, identify whether long division, partial fractions, or substitution is required. Improper rational functions must first be simplified using long division. Expressions involving logarithmic functions are generally simplified through substitution before integration.

✅ Correct Answer Explanation

The correct matching is: 1. \(\frac{x^{2}+3x+2}{x^{2}+7x+12}\)→ a. Use long division to get \(1+\frac{-4x-10}{x^{2}+7x+12}\) 2. \(\frac{x^{3}+x}{x^{2}-1}\)→ d. Use long division to get a linear term \(x+\)remainder 3. \(\frac{1}{x(log⁡x)^{2}+2}\)→ b. Perform variable substitution first 4. \(\frac{x^{2}}{x^{2}+1}\)→ c. Use long division to get \(1-\frac{1}{x^{2}+1}\) Thus, the correct sequence is: 1-a, 2-d, 3-b, 4-c Hence, Option C is the correct answer.

❓ Why Other Options Are Incorrect

Option A (1-d, 2-a, 3-b, 4-c) → Incorrect because Question 1 does not simplify to a linear quotient. It simplifies to \(1+\frac{-4x-10}{x^{2}+7x+12}\), while Question 2 is the one that gives a linear quotient after long division.
Option B (1-b, 2-c, 3-a, 4-d) → Incorrect because the first expression does not require substitution, the logarithmic expression requires substitution, and the remaining decompositions are mismatched.
Option D (1-c, 2-b, 3-d, 4-a) → Incorrect because the decomposition methods are incorrectly assigned. The logarithmic expression is matched with long division instead of substitution, and the rational expressions are wrongly paired.

🎯 Strategy

Identification of Integration Technique

Application:

Before integrating:
Compare the degree of numerator and denominator.
If degree of numerator ≥ degree of denominator, perform long division.
If the denominator factors, use partial fractions after simplification.
If the integrand contains a composite function such as \(\log\,x\), apply substitution.

Final Logic:

Only Option C correctly identifies the initial algebraic technique required for each expression before integration.

🧠 Memory Trick

Proper Rational Function → Partial Fractions

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Integration by Partial Fractions, Long Division Method, and Substitution Method (Rational Functions).

3 (Multiple Correct)
When reducing the improper fraction
\(\frac{x^{2}+3x+2}{x^{2}+7x+12}\)
using long division, which of the following algebraic components are accurate?
A. The quotient \(T(x)\)equals \(1\).
B. The remainder \(R(x)\)equals \(-4x-10\).
C. The divisor is \(x^{2}+7x+12\).
D. The remainder \(R(x)\)equals \(x+4\).

A A, C, D

B A, B

C B, C, D

D A, B, C

💡 Short Explanation

Apply polynomial long division. Identify quotient, divisor and remainder. Verify each statement independently.

✅ Correct Answer Explanation

Long division gives \(\frac{x^{2}+3x+2}{x^{2}+7x+12}=1+\frac{-4x-10}{x^{2}+7x+12}.\) Therefore, Quotient \(=1\). Remainder \(=-4x-10\). Divisor \(=x^{2}+7x+12\). Thus, Statements A, B and C are correct. Hence, Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Includes Statement D, which is incorrect.
Option B → Omits Statement C.
Option C → Includes the incorrect Statement D.

🎯 Strategy

Used

Option Grouping

Application:

Verify each statement separately before selecting the correct combination.

Final Logic:

Only Statements A, B and C are correct.

🧠 Memory Trick

"Divide → Quotient + Proper Fraction."

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

4 Which of the following statements about algebraic reduction to proper form is INCORRECT?

A The quotient \(T(x)\)is integrated using standard polynomial power rules.

B The remainder \(R(x)\)can be factored over \(Q(x)\)using partial fractions.

C \(T(x)+\frac{R(x)}{Q(x)}\)perfectly preserves the mathematical equivalence of the original expression.

D The fractional part \(\frac{R(x)}{Q(x)}\)has a numerator degree equal to its denominator degree.

💡 Short Explanation

The remainder must have a lower degree than the divisor. This makes the resulting fraction proper. Equal degrees would still make it improper.

✅ Correct Answer Explanation

After polynomial long division, \(\frac{P(x)}{Q(x)}=T(x)+\frac{R(x)}{Q(x)},\) where \(deg⁡(R)<deg⁡(Q).\) Therefore, the numerator degree of the remainder fraction is strictly less than the denominator degree. Hence, the statement in Option D is incorrect.

❓ Why Other Options Are Incorrect

Option A → Correct. The quotient is integrated using the power rule.
Option B → Correct. The resulting proper fraction can be decomposed using partial fractions.
Option C → Correct. Long division preserves the original expression exactly.

🎯 Strategy

Used

Elimination

Application:

Recall the defining property of the remainder after polynomial division.

Final Logic:

A proper fraction always satisfies
\(deg⁡(R)<deg⁡(Q).\)

🧠 Memory Trick

"Remainder Degree Must Remain Lower."

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Reduction to Proper Rational Form

5 For the mixed algebraic identity
\(\frac{3x-2}{{\left((x + 1)(x - 2\right)}^{2})}=\frac{A}{x+1}+\frac{B}{x-2}+\frac{C}{{\left(x-2\right)}^{2}},\)
what equation correctly eliminates the denominators for coefficient equating?

A \(3x-2=A(x+1)+B(x-2)+C\)

B \(3x-2=A(x-2)+B(x+1)+C(x-2)^{2}\)

C \(1=A(x-2)^{2}+B(x+1)(x-2)+C(x+1)\)

D \(3x-2=A(x-2)^{2}+B(x+1)(x-2)+C(x+1)\)

💡 Short Explanation

Multiply both sides by the common denominator. Cancel the denominators. Obtain an identity for equating coefficients.

✅ Correct Answer Explanation

Multiplying \(\frac{3x-2}{{\left(\right)}^{2}}=\frac{A}{x+1}+\frac{B}{x-2}+\frac{C}{{\left(x-2\right)}^{2}}\) by \({\left(\right)}^{2}\) gives \(3x-2=A(x-2)^{2}+B(x+1)(x-2)+C(x+1).\) Hence, Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Denominators are not completely eliminated.
Option B → Incorrect multiplication of the terms after clearing denominators.
Option C → The left-hand side should remain \(3x-2\), not 1.

🎯 Strategy

Used

Substitution / Algebraic Identity

Application:

Multiply both sides by the least common denominator before comparing coefficients.

Final Logic:

Clearing denominators correctly produces the identity in Option D.

🧠 Memory Trick

"Clear Denominators First, Compare Later."

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

6 When establishing the identical relationship
\(px+q=A(x+b)+B(x+a),\)
what geometric or algebraic condition ensures this works to find constants?

A It is an equation valid for only one value of \(x\).

B It represents a single specific root of a parabola.

C It bounds the region strictly to positive variables.

D It is an identity strictly true for all real values of \(x\), allowing coefficient comparison.

💡 Short Explanation

Partial fraction decomposition produces an identity. An identity is valid for every permissible value of \(x\). Therefore, coefficients of corresponding powers of \(x\) can be compared.

✅ Correct Answer Explanation

After clearing the denominators in partial fraction decomposition, the resulting equation is an identity, not an ordinary equation. For example, \(px+q=A(x+b)+B(x+a)\) holds for every value of \(x\) in the domain. Since the identity is true for all values of \(x\), the coefficients of corresponding powers of \(x\) must be equal. This allows us to determine the unknown constants \(A\) and \(B\). Therefore, Option D is correct.

❓ Why Other Options Are Incorrect

Option A → An identity is valid for all permissible values of \(x\), not just one.
Option B → The equation does not represent the root of a parabola.
Option C → There is no restriction to only positive values of \(x\).

🎯 Strategy

Used

Concept Identification

Application:

Recognize the difference between an algebraic identity and an ordinary equation.

Final Logic:

Coefficient comparison is possible only because the equation is an identity.

🧠 Memory Trick

"Identity → True for Every \(x\)."

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

7 Evaluate the integration coefficients to correctly break down the distinct linear fraction
\(\frac{4x-10}{\left(x-3)(x-4\right)}.\)

A \(A=2, B=-6\)

B \(A=1, B=2\)

C \(A=4, B=10\)

D \(A=-2, B=6\)

💡 Short Explanation

Write the expression in partial fractions. Substitute suitable values of \(x\). Solve for the constants.

✅ Correct Answer Explanation

Assume \(\frac{4x-10}{\left(x-3)(x-4\right)}=\frac{A}{x-3}+\frac{B}{x-4}.\) Multiplying both sides by \((x-3)(x-4),\) gives \(4x-10=A(x-4)+B(x-3).\) Putting \(x=3\), \(2=-AA=-2.\) Putting \(x=4\), \(6=B.\) Thus, \(A=-2,B=6.\) Therefore, Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Signs of both constants are incorrect.
Option B → These values do not satisfy the identity.
Option C → These constants do not satisfy the equation after substitution.

🎯 Strategy

Used

Substitution

Application:

Choose values of \(x\) that make one factor zero to determine each constant directly.

Final Logic:

Substituting \(x=3\) and \(x=4\) gives \(A=-2\) and \(B=6\).

🧠 Memory Trick

"Root of One Factor → Find the Other Constant."

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

8 Which partial fraction integration format is inherently generated by decomposing and integrating a repeated factor term
\(\frac{C}{{\left(x+a\right)}^{2}}?\)

A \(Clog⁡∣x+a∣+C_{1}\)

B \(\frac{Cx}{x+a}+C_{1}\)

C \(Clog⁡∣(x+a)^{2}∣+C_{1}\)

D \(-\frac{C}{x+a}+C_{1}\)

💡 Short Explanation

Rewrite the denominator using a negative power. Apply the power rule of integration. Substitute back to obtain the result.

✅ Correct Answer Explanation

The integral is \(\int \frac{C}{{\left(x+a\right)}^{2}} dx=C\int (x+a)^{-2} dx.\) Using the power rule, \(\int (x+a)^{-2} dx=-\frac{1}{x+a}+C.\) Therefore, \(-\frac{C}{x+a}+C_{1}.\) Hence, Option D is correct.

❓ Why Other Options Are Incorrect

Option A → This is the integral of \(\frac{C}{x+a}\).
Option B → Does not differentiate to the given integrand.
Option C → Logarithmic integration applies only when the denominator has power 1.

🎯 Strategy

Used

Pattern Recognition

Application:

Recognize the repeated linear factor as a power function and apply the power rule.

Final Logic:

Integrating \({\left(x+a\right)}^{-2}\)gives a reciprocal with a negative sign.

🧠 Memory Trick

"Power −2 → Negative Reciprocal."

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Integration of Repeated Linear Partial Fractions

9 For the proper fraction form
\(\frac{px^{2}+qx+c}{\left(x+a)(x^{2}+b\right)},\)
how many unique, unknown scalar constants must be solved using simultaneous equations?

A One

B Two

C Four

D Three

💡 Short Explanation

The linear factor contributes one constant. The quadratic factor contributes two constants. Total unknown constants = 3.

✅ Correct Answer Explanation

The standard decomposition is \(\frac{px^{2}+qx+c}{\left(x+a)(x^{2}+b\right)}=\frac{A}{x+a}+\frac{Bx+C}{x^{2}+b}.\) The unknown constants are \(A, B, C.\) Thus, there are \(3\) unknown constants to determine. Therefore, Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Only one constant is insufficient.
Option B → Two constants cannot represent the complete decomposition.
Option C → Four constants are unnecessary for this denominator.

🎯 Strategy

Used

Pattern Recognition

Application:

Recall the standard decomposition form for mixed linear and irreducible quadratic factors.

Final Logic:

One constant + two constants = three unknowns.

🧠 Memory Trick

"Linear = 1, Quadratic = 2, Total = 3."

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

10 Identify the fully decomposed structure of a mixed rational fraction containing three distinct linear roots:
\(\frac{px^{2}+qx+c}{\left(x+a)(x+b)(x+c\right)}.\)

A \(\frac{A}{{\left(x+a\right)}^{2}}+\frac{B}{x+b}+\frac{C}{x+c}\)

B \(\frac{Ax+B}{x+a}+\frac{C}{x+c}\)

C \(\frac{A}{x+a}+\frac{Bx+C}{\left(x+b)(x+c\right)}\)

D \(\frac{A}{x+a}+\frac{B}{x+b}+\frac{C}{x+c}\)

💡 Short Explanation

Each distinct linear factor contributes one constant numerator. There are three distinct linear factors. Hence, three simple fractions are formed.

✅ Correct Answer Explanation

When the denominator consists of three distinct linear factors, \((x+a)(x+b)(x+c),\) the standard decomposition is \(\frac{A}{x+a}+\frac{B}{x+b}+\frac{C}{x+c}.\) Each linear factor receives a constant numerator. Therefore, Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Includes a repeated linear factor, which is not present.
Option B → Uses a linear numerator over a linear factor, which is incorrect.
Option C → Incorrectly combines two linear factors into one denominator.

🎯 Strategy

Used

Pattern Recognition

Application:

Match the denominator with the standard partial fraction decomposition rule.

Final Logic:

Each distinct linear factor receives one constant numerator.

🧠 Memory Trick

"Three Linear Factors → Three Constant Fractions."

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

11 During the decomposition
\(3x-2=A(x^{2}-4x+4)+B(x^{2}-x-2)+C(x+1),\)
which set correctly equates the coefficients of \(x^{2}\)?

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

B \(4A-2B+C=-2\)

C \(A+B=3\)

D \(A+B=0\)

💡 Short Explanation

Expand the right-hand side. Collect like terms. Compare the coefficients of \(x^{2}\).

✅ Correct Answer Explanation

Expanding, \(A(x^{2}-4x+4)+B(x^{2}-x-2)+C(x+1)\) gives \((A+B)x^{2}+(-4A-B+C)x+(4A-2B+C).\) Since the left-hand side is \(3x-2,\) its coefficient of \(x^{2}\)is 0. Therefore, \(A+B=0.\) Hence, \(A+B=0\) and Option D is correct.

❓ Why Other Options Are Incorrect

Option A → This equation is obtained by comparing the coefficients of \(x\), not \(x^{2}\).
Option B → This equation comes from comparing the constant terms.
Option C → The coefficient of \(x^{2}\)on the left-hand side is 0, not 3.

🎯 Strategy

Used

Equating Coefficients

Application:

Expand the polynomial identity and compare the coefficients of corresponding powers of \(x\).

Final Logic:

Since the coefficient of \(x^{2}\)is zero, \(A+B=0\).

🧠 Memory Trick

"Same Powers → Same Coefficients."

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

12 Given
\(A+B=0,-4A-B+C=3,4A-2B+C=-2,\)
determine the explicit value of \(C\).

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

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

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

D \(1\)

💡 Short Explanation

Solve the simultaneous equations. First determine \(A\) and \(B\). Substitute them to obtain \(C\).

✅ Correct Answer Explanation

From \(A+B=0,\) we get \(B=-A.\) Substituting into \(-4A-B+C=3\) and \(4A-2B+C=-2,\) gives \(-3A+C=3,6A+C=-2.\) Subtracting, \(9A=-5A=-\frac{5}{9}.\) Hence, \(B=\frac{5}{9}.\) Substituting into \(-3A+C=3,\) gives \(C=\frac{4}{3}.\) Therefore, the mathematically correct value is \(\frac{4}{3}.\) Note: The provided answer key is incorrect. The correct answer is Option A, not Option C.

❓ Why Other Options Are Incorrect

Option B → Does not satisfy the given simultaneous equations.
Option C → This is actually the value of \(A\), not \(C\).
Option D → Does not satisfy the equations.

🎯 Strategy

Used

Substitution

Application:

Reduce the system to two equations in one variable and solve systematically.

Final Logic:

The simultaneous equations give
\(C=\frac{4}{3}.\)

🧠 Memory Trick

"Solve \(A\) First → Then Find \(C\)."

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Determination of Constants in Partial Fractions

13 Integrate the decomposed fractions:
\(\int \left[-\frac{2}{x-3}+\frac{6}{x-4}\right]dx.\)

A \(-2log⁡∣x-4∣+6log⁡∣x-3∣+C\)

B \(2log⁡∣x-3∣-6log⁡∣x-4∣+C\)

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

D \(-2log⁡∣x-3∣+6log⁡∣x-4∣+C\)

💡 Short Explanation

Integrate each partial fraction separately. Apply the standard logarithmic integral. Combine the results.

✅ Correct Answer Explanation

Using \(\int \frac{1}{x-a} dx=log⁡∣x-a∣+C,\) we obtain \(\int \left[-\frac{2}{x-3}+\frac{6}{x-4}\right]dx=-2log⁡∣x-3∣+6log⁡∣x-4∣+C.\) Hence, \(-2log⁡∣x-3∣+6log⁡∣x-4∣+C.\) Therefore, Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Logarithmic terms are interchanged.
Option B → Both coefficients have incorrect signs.
Option C → Represents differentiation-related expressions, not the integral.

🎯 Strategy

Used

Pattern Recognition

Application: Recognize each term as the standard form \(\int \frac{1}{x-a} dx\).

Final Logic: Integrate each fraction independently.

🧠 Memory Trick

"Linear denominator → Natural Log."

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

14 Evaluate the mixed decomposed integral mathematically evaluated as:
\(\int \left[-\frac{5}{9(x+1)}+\frac{5}{9(x-2)}+\frac{4}{3(x-2)^{2}}\right]dx.\)

A \(-\frac{5}{9}log⁡∣x+1∣+\frac{5}{9}log⁡∣x-2∣+\frac{4}{3(x-2)}+C\)

B \(\frac{5}{9}log⁡∣x+1∣-\frac{5}{9}log⁡∣x-2∣-\frac{4}{3(x-2)^{2}}+C\)

C \(-\frac{5}{9}log⁡∣x+1∣+\frac{5}{9}log⁡∣x-2∣+4log⁡∣x-2∣+C\)

D \(-\frac{5}{9}log⁡∣x+1∣+\frac{5}{9}log⁡∣x-2∣-\frac{4}{3(x-2)}+C\)

💡 Short Explanation

Integrate logarithmic terms directly. Rewrite the repeated factor using the power rule. Combine all terms.

✅ Correct Answer Explanation

Using \(\int \frac{1}{x+a} dx=log⁡∣x+a∣+C\) and \(\int \frac{1}{{\left(x-2\right)}^{2}} dx=-\frac{1}{x-2}+C,\) we get \(-\frac{5}{9}log⁡∣x+1∣+\frac{5}{9}log⁡∣x-2∣-\frac{4}{3(x-2)}+C.\) Hence, Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Incorrect sign for the repeated-factor integral.
Option B → Incorrect logarithmic signs and repeated-factor result.
Option C → Incorrectly integrates the repeated-factor term as a logarithm.

🎯 Strategy

Used

Substitution / Pattern Recognition

Application: Apply standard formulas separately to each decomposed term.

Final Logic: The repeated factor contributes a negative reciprocal.

🧠 Memory Trick

"Log for Power 1, Reciprocal for Power 2."

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Integration of Partial Fractions with Repeated Factors

15 If the source evaluation simplifies an integral to
\(\int \left[1-\frac{2}{x-3}+\frac{6}{x-4}\right]dx,\)
what is the final integrated mathematical expression?

A \(1-2log⁡∣x-3∣+6log⁡∣x-4∣+C\)

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

C \(x+2log⁡∣x-3∣-6log⁡∣x-4∣+C\)

D \(x-2log⁡∣x-3∣+6log⁡∣x-4∣+C\)

💡 Short Explanation

Integrate each term individually. Constant 1 integrates to \(x\). Remaining terms become logarithms.

✅ Correct Answer Explanation

Term-by-term integration gives \(\int 1 dx=x,\int -\frac{2}{x-3} dx=-2log⁡∣x-3∣,\int \frac{6}{x-4} dx=6log⁡∣x-4∣.\) Therefore, \(x-2log⁡∣x-3∣+6log⁡∣x-4∣+C.\) Hence, Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Integral of 1 is \(x\), not 1.
Option B → Integral of 1 is not \(\frac{x^{2}}{2}\).
Option C → Logarithmic signs are incorrect.

🎯 Strategy

Used

Pattern Recognition

Application: Integrate each decomposed term separately.

Final Logic: Constant becomes \(x\); fractions become logarithms.

🧠 Memory Trick

"Integrate One First, Then the Logs."

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

16 Before performing partial fractions on the improper term
\(\frac{x^{3}+x}{x^{2}-1},\)
polynomial division yields a quotient and a remainder fraction. What is the quotient term \(T(x)\)resulting from this specific polynomial division?

A \(1\)

B \(x^{2}\)

C \(2x\)

D \(x\)

💡 Short Explanation

Compare the leading terms of the numerator and denominator. Perform polynomial long division. The first term of the quotient is obtained by dividing the highest-degree terms.

✅ Correct Answer Explanation

Divide \(\frac{x^{3}+x}{x^{2}-1}.\) The leading terms give \(\frac{x^{3}}{x^{2}}=x.\) Thus, the quotient begins with \(T(x)=x.\) Indeed, \(x(x^{2}-1)=x^{3}-x,\) and subtracting from the numerator gives the remainder \((x^{3}+x)-(x^{3}-x)=2x.\) Hence, \(\frac{x^{3}+x}{x^{2}-1}=x+\frac{2x}{x^{2}-1}.\) Therefore, Option D is correct.

❓ Why Other Options Are Incorrect

Option A → The quotient is linear, not constant.
Option B → The quotient degree cannot exceed the degree difference.
Option C → \(2x\) is the remainder numerator, not the quotient.

🎯 Strategy

Used

Polynomial Division

Application: Divide the leading terms first to determine the quotient.

Final Logic: The quotient is obtained from \(x^{3}\div x^{2}=x\).

🧠 Memory Trick

"Leading Term ÷ Leading Term = First Quotient Term."

📚 Class XII – Mathematics Part II – Chapter: Integrals – Topic: Polynomial Long Division Before Partial Fractions

📄 Passage

In analyzing applied economics through integral calculus, Marginal Cost (MC) acts as a derivative indicating the specific cost incurred to produce one additional unit. By integrating this MC function, we recover the Total Cost function \(C(x)\), inherently containing a mathematical constant \(C\). This constant accurately represents the firm's Fixed Cost, which remains firmly present even when the production volume \(x\) is strictly zero. For complex fractional marginal cost models such as

If a firm's marginal cost is explicitly given by
\(MC=17+\frac{200}{x+1},\)
and fixed initial costs are natively ₹1000, what determines the explicit final Cost function \(C(x)\)?

A \(C(x)=17x+200log⁡∣x+1∣+2000\)

B \(C(x)=17x^{2}+200log⁡∣x+1∣+1000\)

C \(C(x)=17+200log⁡∣x+1∣+1000\)

D \(C(x)=17x+200log⁡∣x+1∣+1000\)

💡 Short Explanation

Integrate the marginal cost. Add the constant of integration. Use the fixed cost to determine the constant.

✅ Correct Answer Explanation

Since \(MC=\frac{dC}{dx},\) we integrate: \(C(x)=\int \left(17,\ \frac{200}{x+1}\right)dx.\) Therefore, \(C(x)=17x+200log⁡∣x+1∣+C.\) Given the fixed cost \(C(0)=1000,\) we get \(1000=0+200log⁡1+C=C.\) Hence, \(C(x)=17x+200log⁡∣x+1∣+1000.\) Therefore, Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Uses an incorrect fixed cost.
Option B → Incorrectly integrates the constant term.
Option C → Omits the term \(17x\).

🎯 Strategy

Used

Substitution

Application: Integrate first, then use the initial condition.

Final Logic: The fixed cost determines the integration constant.

🧠 Memory Trick

"Integrate MC, Then Apply Fixed Cost."

📚 Class XII – Mathematics Part II – Chapter: Applications of Integrals – Topic: Marginal Cost and Total Cost

📄 Passage

According to the text constraints, how is the Average Cost function mathematically defined using the total cost model?

A Average Cost \(=\frac{d}{dx}[C(x)]\)

B Average Cost \(=C(x)\times x\)

C Average Cost \(=\int C(x) dx\)

D Average Cost \(=\frac{C(x)}{x}\)

💡 Short Explanation

Average Cost means cost per unit. Divide total cost by output. This follows directly from the passage.

✅ Correct Answer Explanation

Average Cost is defined as \(AC=\frac{Total Cost}{Number of Units}.\) Hence, \(AC=\frac{C(x)}{x}.\) Therefore, Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Gives Marginal Cost.
Option B → Produces an incorrect quantity.
Option C → Integration is not used to compute Average Cost.

🎯 Strategy

Used

Contextual/Tonal Matching

Application: Identify the mathematical definition stated in the passage.

Final Logic: Average Cost equals Total Cost divided by Quantity.

🧠 Memory Trick

"Average = Total ÷ Quantity."

📚 Class XII – Mathematics Part II – Chapter: Applications of Integrals – Topic: Average Cost Function

19 Evaluate the direct partial fraction integral
\(\int \frac{1}{x^{2}-a^{2}} dx\)
utilizing standard linear factor decomposition mapped to logarithms.

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

B \(log⁡∣x+\sqrt{x^{2}-a^{2}}∣+C\)

C \(\frac{1}{a}{tan⁡}^{-1} \left(\frac{x}{a}\right)+C\)

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

💡 Short Explanation

Factor the denominator. Apply partial fractions. Integrate each linear fraction.

✅ Correct Answer Explanation

Since \(x^{2}-a^{2}=(x-a)(x+a),\) partial fractions give \(\frac{1}{x^{2}-a^{2}}=\frac{1}{2a}\left(\frac{1}{x-a},\ \frac{1}{x+a}\right).\) Integrating, \(\int \frac{1}{x^{2}-a^{2}} dx=\frac{1}{2a}log⁡∣\frac{x-a}{x+a}∣+C.\) Therefore, Option D is correct.

❓ Why Other Options Are Incorrect

Option A → Equivalent only up to a negative sign convention; not the standard form listed here.
Option B → Formula for a different integral.
Option C → Corresponds to \(\int \frac{1}{x^{2}+a^{2}} dx\).

🎯 Strategy

Used

Partial Fraction Decomposition

Application: Factor the denominator and integrate each simple fraction.

Final Logic: Difference of logarithms gives the required answer.

🧠 Memory Trick

"Difference of Squares → Difference of Logs."

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

20 A marginal cost is defined simply as
\(MC=x,\)
and the total operational cost of producing \(3\) units equals ₹7800. What evaluates the integration constant \(C\)?

A 7800

B 7791

C 0

D 7795.5

💡 Short Explanation

Integrate the marginal cost. Use the given total cost at \(x=3\). Solve for the constant.

✅ Correct Answer Explanation

Since \(MC=\frac{dC}{dx}=x,\) integrating, \(C(x)=\frac{x^{2}}{2}+C.\) Given \(C(3)=7800,\) we obtain \(7800=\frac{9}{2}+C=4.5+C.\) Hence, \(C=7800-4.5=7795.5.\) Therefore, \(7795.5\) and Option D is correct.

❓ Why Other Options Are Incorrect

Option A → This is the total cost, not the integration constant.
Option B → Incorrect arithmetic.
Option C → Ignores the given initial condition.

🎯 Strategy

Used

Substitution

Application: Integrate first, then substitute the given point.

Final Logic: The known cost at \(x=3\) uniquely determines the constant.

🧠 Memory Trick

"Integrate First, Substitute Later."

📚 Class XII – Mathematics Part II – Chapter: Applications of Integrals – Topic: Determining Constants from Initial Conditions

CUET UG Applied Mathematics Booster Test 2 - Integration by Partial Fractions

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