Resolving into Partial fraction MCQs With Answer

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Partial fraction decomposition is an essential algebraic tool for B.Pharm students working with rational expressions in pharmaceutical mathematics. Mastering partial fractions helps simplify complex rational functions for integration, inverse Laplace transforms, and solving differential equations encountered in pharmacokinetics, drug release modelling, and formulation calculations. This topic covers recognizing proper vs improper fractions, handling distinct and repeated linear factors, treating irreducible quadratics with linear numerators, and techniques like the cover-up method and equating coefficients. Clear understanding improves accuracy in analytical and numerical tasks in pharmacy studies. The following focused, keyword-rich practice set emphasizes partial fraction decomposition, rational expressions, and applied integration. Now let’s test your knowledge with 50 MCQs on this topic.

Q1. What is the primary purpose of partial fraction decomposition?

  • To convert irrational functions into polynomials
  • To express a rational function as a sum of simpler fractions
  • To factor numerators into linear factors only
  • To find complex roots of a polynomial

Correct Answer: To express a rational function as a sum of simpler fractions

Q2. A rational function is called proper when which condition holds?

  • Degree of numerator is greater than degree of denominator
  • Degree of numerator equals degree of denominator
  • Degree of numerator is less than degree of denominator
  • Denominator factors into linear factors only

Correct Answer: Degree of numerator is less than degree of denominator

Q3. If a rational function is improper, what is the first step before partial fraction decomposition?

  • Use the cover-up method directly
  • Perform polynomial long division to obtain a proper fraction
  • Factor the numerator completely
  • Differentiate numerator and denominator

Correct Answer: Perform polynomial long division to obtain a proper fraction

Q4. For distinct linear factors (x-a)(x-b) in the denominator, the partial fraction form is:

  • A/(x^2 + 1) + B/(x^2 + 2)
  • Ax + B/(x-a) + Cx + D/(x-b)
  • A/(x-a) + B/(x-b)
  • (Ax+B)/(x-a) + (Cx+D)/(x-b)

Correct Answer: A/(x-a) + B/(x-b)

Q5. For a repeated linear factor (x-a)^3, the partial fraction terms include:

  • A/(x-a) only
  • A/(x-a) + B/(x-a)^2 + C/(x-a)^3
  • (Ax+B)/(x-a)^3 only
  • A/(x-a) + B/(x-a)^3 only

Correct Answer: A/(x-a) + B/(x-a)^2 + C/(x-a)^3

Q6. When the denominator has an irreducible quadratic factor (x^2+1), the numerator for that term should be:

  • A constant only
  • A linear expression Ax + B
  • A quadratic expression Ax^2 + Bx + C
  • Zero

Correct Answer: A linear expression Ax + B

Q7. The general rule for numerator degree over a factor is:

  • Numerator degree must be equal to factor degree
  • Numerator degree must be one more than factor degree
  • Numerator degree must be less than factor degree
  • Numerator degree must be zero

Correct Answer: Numerator degree must be less than factor degree

Q8. The cover-up method is best applied to find coefficients when:

  • Denominator has distinct linear factors only
  • Denominator contains repeated factors
  • Denominator contains irreducible quadratics
  • Numerator degree is higher than denominator degree

Correct Answer: Denominator has distinct linear factors only

Q9. In the equating coefficients method you:

  • Divide numerator by denominator directly
  • Plug random x-values without forming equations
  • Multiply through by the denominator and match coefficients of powers
  • Only use complex numbers

Correct Answer: Multiply through by the denominator and match coefficients of powers

Q10. A common application of partial fractions in pharmaceutical mathematics is:

  • Estimating molecular mass
  • Integrating rational functions arising in pharmacokinetic models
  • Determining pH from pKa values
  • Calculating osmotic pressure

Correct Answer: Integrating rational functions arising in pharmacokinetic models

Q11. Decompose 1/(x(x+1)). What are the terms?

  • 1/x + 1/(x+1)
  • 1/x – 1/(x+1)
  • -1/x + 1/(x+1)
  • -1/x – 1/(x+1)

Correct Answer: 1/x – 1/(x+1)

Q12. For (x+2)/(x(x+1)) the decomposition A/x + B/(x+1) gives A and B as:

  • A=1, B=1
  • A=2, B=-1
  • A=-2, B=3
  • A=0, B=1

Correct Answer: A=2, B=-1

Q13. Which statement is true about an improper fraction like (x^2+1)/x?

  • It is already proper and ready for decomposition
  • It must be simplified by polynomial long division first
  • It cannot be integrated
  • Denominator must be factored into quadratics

Correct Answer: It must be simplified by polynomial long division first

Q14. Decompose (2x+3)/(x^2-1) into partial fractions; the constants A and B for A/(x-1)+B/(x+1) are:

  • A=1, B=2
  • A=5/2, B=-1/2
  • A=-1/2, B=5/2
  • A=2, B=3

Correct Answer: A=5/2, B=-1/2

Q15. Decompose 1/(x^2(x+1)) = A/x + B/x^2 + C/(x+1). What are A, B, C?

  • A=1, B=1, C=-1
  • A=-1, B=1, C=1
  • A=0, B=1, C=0
  • A=2, B=-1, C=0

Correct Answer: A=-1, B=1, C=1

Q16. Which condition requires polynomial division before decomposition?

  • Numerator degree is less than denominator degree
  • Numerator degree equals denominator degree
  • Numerator degree is greater than or equal to denominator degree
  • Denominator is prime

Correct Answer: Numerator degree is greater than or equal to denominator degree

Q17. For (x^2+1) repeated twice as (x^2+1)^2, the proper numerator forms are:

  • A/(x^2+1) + B/(x^2+1)^2
  • (Ax+B)/(x^2+1) + (Cx+D)/(x^2+1)^2
  • A/(x^2+1)^2 only
  • A/(x^2+1) + B/(x^2+1)^3

Correct Answer: (Ax+B)/(x^2+1) + (Cx+D)/(x^2+1)^2

Q18. Partial fractions are often used to invert Laplace transforms in pharmacokinetics because:

  • They convert rational Laplace-domain functions into simpler time-domain terms
  • They eliminate the need for initial conditions
  • They provide exact pKa values
  • They avoid integration altogether

Correct Answer: They convert rational Laplace-domain functions into simpler time-domain terms

Q19. Decompose 3/(x(x+2)). The values of A and B in A/x + B/(x+2) are:

  • A=1, B=2
  • A=3/2, B=3/2
  • A=3/2, B=-3/2
  • A=-3/2, B=3/2

Correct Answer: A=3/2, B=-3/2

Q20. Which method is typically the fastest for finding coefficients for distinct linear factors?

  • Polynomial long division
  • Equating coefficients for every power of x
  • Cover-up method (Heaviside method)
  • Partial differentiation

Correct Answer: Cover-up method (Heaviside method)

Q21. If denominator factors include irreducible quadratics, the decomposition must include:

  • Only constant numerators for each quadratic
  • Linear numerators Ax + B for each quadratic factor
  • Only repeated linear terms
  • Exponentials in the numerator

Correct Answer: Linear numerators Ax + B for each quadratic factor

Q22. For the rational function (x^2+1)/(x^3-x) with denominator x(x-1)(x+1), how many unknown constants are needed in partial fractions?

  • 1
  • 2
  • 3
  • 4

Correct Answer: 3

Q23. When equating coefficients after clearing denominators, the number of independent linear equations equals:

  • The highest power of x in the expanded numerator plus one
  • The number of distinct denominator factors only
  • Always two
  • The number of denominator factors squared

Correct Answer: The highest power of x in the expanded numerator plus one

Q24. Decompose (2x+1)/(x^2+3x+2) where denominator = (x+1)(x+2). The constants A and B in A/(x+1) + B/(x+2) are:

  • A=1, B=1
  • A=-1, B=3
  • A=2, B=-1
  • A=0, B=2

Correct Answer: A=-1, B=3

Q25. Simplify before decomposing: x^2/(x(x^2+1)) reduces to which simpler expression?

  • x/(x^2+1)
  • 1/(x^2+1)
  • x^2/(x^2+1)
  • 1/x

Correct Answer: x/(x^2+1)

Q26. Is the partial fraction decomposition of a given proper rational function over the reals unique (with real coefficients)?

  • Yes, it is unique
  • No, there are infinitely many decompositions
  • Only unique for linear denominators
  • Unique only if numerator is zero

Correct Answer: Yes, it is unique

Q27. Which of the following integrals typically requires partial fractions to evaluate?

  • Integral of e^x dx
  • Integral of 1/(x^2+1) dx
  • Integral of (3x+1)/(x^2-1) dx
  • Integral of sin x dx

Correct Answer: Integral of (3x+1)/(x^2-1) dx

Q28. For integrating rational functions with irreducible quadratic denominators, partial fractions often lead to which types of primitives?

  • Only logarithmic terms
  • Logarithmic and arctangent terms
  • Exponential terms only
  • Trigonometric integrals only

Correct Answer: Logarithmic and arctangent terms

Q29. Decompose (5x+3)/(x^2-4x+3) where denominator factors as (x-1)(x-3). The constants A and B in A/(x-1)+B/(x-3) are:

  • A=2, B=1
  • A=-4, B=9
  • A=-1, B=4
  • A=4, B=-9

Correct Answer: A=-4, B=9

Q30. For (x^2+1)(x+1) in the denominator, the partial fraction structure is:

  • A/(x+1) + B/(x^2+1)
  • (Ax+B)/(x^2+1) + C/(x+1)
  • A/(x+1) + B/(x^2+1)^2
  • A/(x+1)^2 + (Bx+C)/(x^2+1)

Correct Answer: (Ax+B)/(x^2+1) + C/(x+1)

Q31. For denominator factor x^3, the partial fraction terms should include:

  • A/x + B/x^2 only
  • A/x + B/x^2 + C/x^3
  • A/(x^3) only
  • A + Bx + Cx^2

Correct Answer: A/x + B/x^2 + C/x^3

Q32. The cover-up method cannot directly determine coefficients for which of the following?

  • Distinct linear factors
  • Repeated linear factors
  • Simple poles
  • Rational functions with all simple roots

Correct Answer: Repeated linear factors

Q33. Which substitution strategy helps quickly find coefficients when using equating coefficients?

  • Plugging the denominator roots into the cleared equation
  • Plugging random irrational numbers only
  • Only using complex conjugates
  • Not substituting any values

Correct Answer: Plugging the denominator roots into the cleared equation

Q34. In inversion of Laplace transforms, the residues at poles correspond to:

  • Roots of the numerator only
  • Coefficients in the partial fraction expansion that give time-domain amplitudes
  • The degree of the denominator only
  • Irrelevant constants

Correct Answer: Coefficients in the partial fraction expansion that give time-domain amplitudes

Q35. Decompose (4x+1)/(x^2-x) into A/x + B/(x-1). Values of A and B are:

  • A=1, B=3
  • A=-1, B=5
  • A=-1, B=5
  • A=2, B=2

Correct Answer: A=-1, B=5

Q36. Which of the following requires polynomial division before decomposition?

  • (x+1)/(x^2+1)
  • (x^3+1)/(x^2+1)
  • 1/(x^2+1)
  • (2x+3)/(x^2-1)

Correct Answer: (x^3+1)/(x^2+1)

Q37. For a single irreducible quadratic factor in the denominator, the numerator must be of degree:

  • Zero (constant)
  • One (linear)
  • Two (quadratic)
  • Three

Correct Answer: One (linear)

Q38. Partial fraction decomposition in pharmaceutical modeling can help to:

  • Directly compute particle size
  • Simplify expressions for drug concentration vs time for analytical solutions
  • Measure solubility experimentally
  • Replace numerical methods always

Correct Answer: Simplify expressions for drug concentration vs time for analytical solutions

Q39. For a simple pole r of P(x)/Q(x) where Q(r)=0 and Q'(r)≠0, the partial fraction coefficient at (x-r) equals:

  • P(r)/Q(r)
  • P(r)/Q'(r)
  • Q'(r)/P(r)
  • Zero

Correct Answer: P(r)/Q'(r)

Q40. Using the cover-up method, the coefficient corresponding to (x-2) for 6/(x(x-2)(x+3)) is:

  • 6/(2*5) = 3/5
  • 6/(0*5) = undefined
  • 6/(2*1) = 3
  • 6/(5) = 6/5

Correct Answer: 6/(2*5) = 3/5

Q41. Why is polynomial long division used before partial fractions when numerator degree ≥ denominator degree?

  • To convert the integrand into a polynomial plus a proper rational function
  • To factor the denominator further
  • To find complex roots of the numerator
  • It is never used

Correct Answer: To convert the integrand into a polynomial plus a proper rational function

Q42. Over the real numbers, any polynomial denominator can be factored into:

  • Only linear factors
  • Only irreducible quartic factors
  • Linear and irreducible quadratic factors
  • Exponential factors

Correct Answer: Linear and irreducible quadratic factors

Q43. For factor x^2+4 in the denominator, the correct general numerator form is:

  • A (constant)
  • Ax + B (linear)
  • Ax^2 + Bx + C (quadratic)
  • Zero

Correct Answer: Ax + B (linear)

Q44. Partial fractions assist in solving linear ODEs in pharmacokinetics because they:

  • Avoid the need to consider initial conditions
  • Allow inverse Laplace transforms to be taken term-by-term
  • Make the ODE nonlinear
  • Only give numeric approximations

Correct Answer: Allow inverse Laplace transforms to be taken term-by-term

Q45. Integrate (2x+3)/(x^2-1) dx via partial fractions; the antiderivative is:

  • 2.5 ln|x-1| – 0.5 ln|x+1| + C
  • ln|x^2-1| + C
  • arctan(x) + C
  • 0

Correct Answer: 2.5 ln|x-1| – 0.5 ln|x+1| + C

Q46. Partial fraction decomposition is unique provided you restrict coefficients to:

  • Complex numbers only
  • Real numbers only (when working over reals)
  • Integers only
  • Rational functions again

Correct Answer: Real numbers only (when working over reals)

Q47. Decompose (x+5)/(x^2+4x+3) where denominator factors (x+1)(x+3); A and B in A/(x+1)+B/(x+3) are:

  • A=2, B=-1
  • A=1, B=4
  • A=-2, B=3
  • A=0, B=5

Correct Answer: A=2, B=-1

Q48. When using equating coefficients for (Ax+B)/(x^2+1) + C/(x+1), you typically obtain a linear system in A, B, C by:

  • Comparing coefficients of x^2, x, and constant after clearing denominators
  • Taking derivatives and equating
  • Setting x to infinity only
  • Integrating both sides

Correct Answer: Comparing coefficients of x^2, x, and constant after clearing denominators

Q49. The number of unknown constants in a partial fraction expansion equals:

  • The sum of the degrees of the distinct irreducible factors in the denominator
  • The number of numerator terms
  • Always three
  • The product of the degrees of numerator and denominator

Correct Answer: The sum of the degrees of the distinct irreducible factors in the denominator

Q50. Decompose (3x^2+2x+1)/(x(x^2+1)) into A/x + (Bx+C)/(x^2+1). Values of A, B, C are:

  • A=1, B=2, C=2
  • A=2, B=1, C=0
  • A=0, B=3, C=1
  • A=1, B=1, C=0

Correct Answer: A=1, B=2, C=2

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