Laplace Transform of derivatives MCQs With Answer

The Laplace Transform of derivatives is a core mathematical tool for B. Pharm students studying pharmacokinetics, modeling drug concentration and solving linear differential equations. Understanding properties like L{f’} = sF(s) − f(0), L{f”} = s^2F(s) − s f(0) − f'(0), and transforms involving higher-order derivatives helps in converting time-domain rate equations into algebraic equations in the s-domain. Mastery of initial-condition handling, inverse Laplace techniques, and related properties (shifting, scaling, and differentiation in the s-domain) accelerates problem solving in dosage regimen design and compartmental models. This set of focused Laplace Transform of derivatives MCQs with answers targets common pitfalls and application-based questions for B. Pharm students. Now let’s test your knowledge with 50 MCQs on this topic.

Q1. What is the Laplace transform of f'(t)?

  • sF(s) – f(0)
  • F'(s)
  • 1/s * F(s)
  • F(s) – f(0)

Correct Answer: sF(s) – f(0)

Q2. What is the Laplace transform of f”(t)?

  • sF(s) – f(0)
  • s^2F(s) – s f(0) – f'(0)
  • F”(s)
  • s^2F(s) – f(0)

Correct Answer: s^2F(s) – s f(0) – f'(0)

Q3. Which property relates multiplication by t in time domain to differentiation in s-domain?

  • L{t f(t)} = -dF(s)/ds
  • L{t f(t)} = dF(s)/ds
  • L{t f(t)} = s dF(s)/ds
  • L{t f(t)} = -s dF(s)/ds

Correct Answer: L{t f(t)} = -dF(s)/ds

Q4. If L{f(t)} = F(s), what is L{f”'(t)}?

  • s^3 F(s) – s^2 f(0) – s f'(0) – f”(0)
  • s^3 F(s) – f(0) – f'(0) – f”(0)

Correct Answer: s^3 F(s) – s^2 f(0) – s f'(0) – f”(0)

Q5. When solving a first-order ODE for drug concentration C'(t) + kC(t) = R(t), using Laplace transforms, what term accounts for the initial concentration C(0)?

  • -C(0)
  • k C(0)
  • s C(0)
  • C(0) multiplied by s in transformed equation for C'(t)

Correct Answer: C(0) multiplied by s in transformed equation for C'(t)

Q6. For f(t)=t, what is F(s)=L{t}?

  • 1/s
  • 1/s^2
  • 1/s^3
  • 0

Correct Answer: 1/s^2

Q7. Which expression gives L{d^n f / dt^n} in terms of F(s) and initial conditions?

  • s^n F(s) – s^{n-1} f(0) – … – f^{(n-1)}(0)
  • F^{(n)}(s) + initial terms
  • (-1)^n d^n F(s)/ds^n
  • s F(s) – f(0)

Correct Answer: s^n F(s) – s^{n-1} f(0) – … – f^{(n-1)}(0)

Q8. How does the Laplace transform help in solving linear ODEs with constant coefficients in pharmacokinetic models?

  • Converts differential equations to algebraic equations in s-domain
  • Removes need for initial conditions
  • Only useful for non-homogeneous equations
  • Makes time domain calculations unnecessary but increases complexity

Correct Answer: Converts differential equations to algebraic equations in s-domain

Q9. Given f(0)=2 and L{f’} = sF(s)-2, if L{f’} = 6/(s+1), what is sF(s)-2?

  • 6/(s+1)
  • 6s/(s+1)
  • (6+2s)/(s+1)
  • 2/(s+1)

Correct Answer: 6/(s+1)

Q10. Which of the following is the inverse Laplace of 1/(s+a)?

  • e^{-at}
  • e^{at}
  • sin(at)
  • cos(at)

Correct Answer: e^{-at}

Q11. In L{f'(t)} = sF(s) – f(0), what role does f(0) play in solving initial value problems?

  • Represents initial condition that must be included to get correct solution
  • Can be ignored for linear ODEs
  • Only needed if f(0) ≠ 0 and time-shifted
  • It is absorbed into F(s) automatically

Correct Answer: Represents initial condition that must be included to get correct solution

Q12. If F(s)=1/s and f(0)=0, what is L{f’}?

  • 0
  • 1
  • s*(1/s) – 0 = 1
  • 1/s^2

Correct Answer: s*(1/s) – 0 = 1

Q13. Which transform property is useful for handling derivatives of a product t^n f(t)?

  • Time multiplication: L{t^n f(t)} = (-1)^n d^n F(s)/ds^n
  • Frequency shift property
  • Convolution theorem
  • Initial value theorem

Correct Answer: Time multiplication: L{t^n f(t)} = (-1)^n d^n F(s)/ds^n

Q14. What is L{δ'(t)} where δ is the Dirac delta? (Assume δ’ denotes derivative)

  • s
  • 1
  • s times L{δ(t)} minus δ(0)
  • s * 1 = s

Correct Answer: s

Q15. Which method is commonly combined with Laplace transforms to invert rational F(s) back to f(t)?

  • Partial fraction decomposition
  • Taylor series expansion
  • Numerical integration only
  • Fourier transform

Correct Answer: Partial fraction decomposition

Q16. For a second-order linear ODE describing drug kinetics, initial values required for Laplace solution are:

  • f(0) and f'(0)
  • f(0) only
  • f”(0) only
  • No initial values

Correct Answer: f(0) and f'(0)

Q17. Which of the following is the Laplace of e^{-at} f(t) (frequency shift)?

  • F(s + a)
  • F(s – a)
  • e^{-a s} F(s)
  • F(s) / (s + a)

Correct Answer: F(s + a)

Q18. Given f(t)=cos(ωt), what is L{f'(t)}?

  • s * s/(s^2 + ω^2) – 1
  • -ω sin(ωt) transformed to -ω^2/(s^2 + ω^2)
  • L{-ω sin(ωt)} = -ω * (ω/(s^2+ω^2)) = -ω^2/(s^2+ω^2)
  • 0

Correct Answer: L{-ω sin(ωt)} = -ω * (ω/(s^2+ω^2)) = -ω^2/(s^2+ω^2)

Q19. If F(s)=1/(s^2 + 4), what is L{f”(t)} using initial conditions f(0)=0, f'(0)=2?

  • s^2 F(s) – s*0 – 2 = s^2/(s^2+4) – 2
  • s^2/(s^2+4)
  • sF(s)-0
  • 2/(s^2+4)

Correct Answer: s^2/(s^2+4) – 2

Q20. Which theorem relates the value of f(0+) to the limit of sF(s) as s → ∞?

  • Initial value theorem
  • Final value theorem
  • Convolution theorem
  • Shifting theorem

Correct Answer: Initial value theorem

Q21. What is the Laplace transform of a derivative multiplied by a step u(t-a): L{d/dt [u(t-a) g(t-a)]}?

  • e^{-a s} [s G(s) – g(0+)]
  • [s G(s) – g(0+)]
  • e^{-a s} G(s)
  • G'(s)

Correct Answer: e^{-a s} [s G(s) – g(0+)]

Q22. Which expression is true for the derivative of the Laplace transform F(s) with respect to s?

  • dF/ds = -L{t f(t)}
  • dF/ds = L{t f(t)}
  • dF/ds = -L{f'(t)}
  • dF/ds = L{f'(t)}

Correct Answer: dF/ds = -L{t f(t)}

Q23. In pharmacokinetics, using Laplace transforms simplifies solving compartment models by:

  • Converting time derivatives into algebraic s-terms, incorporating initial concentrations
  • Eliminating need to consider elimination rate constants
  • Making solutions time-invariant always
  • Replacing concentrations with steady-state values only

Correct Answer: Converting time derivatives into algebraic s-terms, incorporating initial concentrations

Q24. What is the Laplace transform of t f'(t) in terms of F(s)?

  • -d/ds[sF(s) – f(0)]
  • dF/ds
  • -dF/ds
  • s dF/ds

Correct Answer: -d/ds[sF(s) – f(0)]

Q25. Which approach is correct to invert F(s) that is non-rational (e.g., involves sqrt(s))?

  • Use known transform pairs, Bromwich integral or tables, and possibly numerical inversion
  • Always use partial fractions
  • There is no inverse Laplace
  • Differentiate F(s) until it becomes rational

Correct Answer: Use known transform pairs, Bromwich integral or tables, and possibly numerical inversion

Q26. Given L{f(t)} = 3/(s+2), what is L{f'(t)} if f(0)=1?

  • s*(3/(s+2)) – 1 = 3s/(s+2) – 1
  • 3/(s+2) – 1
  • s*(3/(s+2))
  • 0

Correct Answer: s*(3/(s+2)) – 1 = 3s/(s+2) – 1

Q27. Which statement about differentiation under the Laplace integral sign is true?

  • Differentiating F(s) with respect to s corresponds to multiplication by -t in time domain
  • Differentiation with respect to s has no time-domain interpretation
  • It corresponds to time derivative f'(t)
  • It always simplifies inversion

Correct Answer: Differentiating F(s) with respect to s corresponds to multiplication by -t in time domain

Q28. What is the Laplace transform of the time derivative of a convolution: L{d/dt (f * g)}?

  • s F(s) G(s) – (f * g)(0)
  • F(s) G(s)
  • d/ds [F(s) G(s)]
  • 0

Correct Answer: s F(s) G(s) – (f * g)(0)

Q29. Which of these is the correct Laplace transform pair for f'(t) when f(t)=e^{-2t}? (f(0)=1)

  • L{f’} = s*(1/(s+2)) – 1 = s/(s+2) – 1
  • L{f’} = 1/(s+2)
  • L{f’} = -2/(s+2)
  • L{f’} = s/(s+2)

Correct Answer: L{f’} = s*(1/(s+2)) – 1 = s/(s+2) – 1

Q30. Which property helps when dealing with derivatives of shifted functions u(t-a)g(t-a)?

  • Time-shifting property with multiplication by e^{-as} in s-domain
  • Frequency differentiation property
  • Convolution with delta function
  • Scaling property

Correct Answer: Time-shifting property with multiplication by e^{-as} in s-domain

Q31. Compute L{t^2 f(t)} in terms of F(s).

  • L{t^2 f(t)} = d^2 F/ds^2
  • L{t^2 f(t)} = (-1)^2 d^2 F/ds^2 = d^2 F/ds^2
  • L{t^2 f(t)} = -dF/ds
  • L{t^2 f(t)} = s^2 F(s)

Correct Answer: L{t^2 f(t)} = (-1)^2 d^2 F/ds^2 = d^2 F/ds^2

Q32. For a linear system with input R(t) and state variable C(t), Laplace transforms convert C'(t) terms to:

  • s C(s) – C(0)
  • s C(s)
  • C(0)
  • C(s)/s

Correct Answer: s C(s) – C(0)

Q33. If L{f(t)} = F(s) and f(0)=0, which of the following is L{∫_0^t f(τ) dτ}?

  • F(s)/s
  • s F(s)
  • 1/F(s)
  • F(s) * s

Correct Answer: F(s)/s

Q34. How does one include initial derivative f'(0) when transforming a second-order differential equation?

  • Appears as a subtractive term -f'(0) in L{f”}
  • Does not appear
  • Appears multiplied by s only
  • Appears as +f'(0)

Correct Answer: Appears as a subtractive term -f'(0) in L{f”}

Q35. Which Laplace pair is correct for f(t)=t e^{at}?

  • L{t e^{at}} = 1/(s-a)^2
  • L{t e^{at}} = 1/(s-a)
  • L{t e^{at}} = (s-a)^{-3}
  • L{t e^{at}} = s/(s-a)^2

Correct Answer: L{t e^{at}} = 1/(s-a)^2

Q36. When using Laplace to solve for concentration, the algebraic equation often requires solving for C(s) as:

  • C(s) = (terms from input and initial conditions) / (polynomial in s)
  • C(s) = integral of input only
  • C(s) = constant only
  • C(s) cannot be found using Laplace

Correct Answer: C(s) = (terms from input and initial conditions) / (polynomial in s)

Q37. What is the Laplace transform of cosh(bt)?

  • s/(s^2 – b^2)
  • s/(s^2 + b^2)
  • b/(s^2 – b^2)
  • 1/(s – b)

Correct Answer: s/(s^2 – b^2)

Q38. Which of the following is true for stability analysis using Laplace transforms in compartment models?

  • Roots of denominator (poles) determine transient behavior and stability
  • Only numerator matters
  • Laplace cannot analyze stability
  • Stability is determined by F(0) only

Correct Answer: Roots of denominator (poles) determine transient behavior and stability

Q39. What is L{f”(t) + 3 f'(t) + 2 f(t)} in terms of F(s) and initial conditions f(0)=a, f'(0)=b?

  • (s^2 + 3s + 2) F(s) – s a – b – 3 a
  • (s^2 + 3s + 2) F(s)
  • F(s) – a – b
  • (s^2 + 3s + 2) F(s) – a – b

Correct Answer: (s^2 + 3s + 2) F(s) – s a – b – 3 a

Q40. Inverse Laplace of (s+1)/(s^2 + 4s + 5) corresponds to which time function?

  • e^{-2t} cos(t) + e^{-2t} sin(t)
  • e^{-t}
  • cos(t)
  • e^{-2t}

Correct Answer: e^{-2t} cos(t) + e^{-2t} sin(t)

Q41. Which initial/final value theorem statement is correct?

  • lim_{t→0+} f(t) = lim_{s→∞} s F(s); lim_{t→∞} f(t) = lim_{s→0} s F(s) if poles permit
  • lim_{t→0+} f(t) = lim_{s→0} s F(s)
  • Both limits always exist
  • These theorems don’t involve sF(s)

Correct Answer: lim_{t→0+} f(t) = lim_{s→∞} s F(s); lim_{t→∞} f(t) = lim_{s→0} s F(s) if poles permit

Q42. If F(s)= (s+3)/(s^2+4s+5), what is the time-domain derivative f'(t) at t=0+ using initial value theorem?

  • lim_{s→∞} s[sF(s) – f(0)] gives f'(0+) using transforms of derivative
  • Directly equals F(0)
  • Cannot be found
  • Equal to f(0)

Correct Answer: lim_{s→∞} s[sF(s) – f(0)] gives f'(0+) using transforms of derivative

Q43. Which factor complicates inversion when initial conditions are non-zero?

  • Presence of additional polynomial terms from initial conditions in numerator
  • They always simplify inversion
  • Initial conditions remove poles
  • They convert problem into convolution only

Correct Answer: Presence of additional polynomial terms from initial conditions in numerator

Q44. When applying Laplace to systems of ODEs (e.g., two compartments), the transformed algebraic system is solved for:

  • Each compartment’s Laplace variable C_i(s) incorporating initial amounts
  • Only total mass
  • Only steady state values
  • Time derivatives directly

Correct Answer: Each compartment’s Laplace variable C_i(s) incorporating initial amounts

Q45. Which of the following is L{sin(ωt)} and useful when differentiating in s-domain?

  • ω/(s^2 + ω^2)
  • s/(s^2 + ω^2)
  • 1/(s + ω)
  • 0

Correct Answer: ω/(s^2 + ω^2)

Q46. What is the effect on the Laplace transform of adding a known polynomial initial term (e.g., f(0) and f'(0)) to the numerator?

  • Shifts the inverse solution by adding impulses or polynomial terms in time domain
  • Removes poles
  • No effect on inversion
  • Transforms denominator only

Correct Answer: Shifts the inverse solution by adding impulses or polynomial terms in time domain

Q47. How to handle non-zero initial derivatives when using Laplace for pharmacokinetic equations with dosing at t=0?

  • Include derivative initial values explicitly in transformed equations L{f’} and L{f”}
  • Set them to zero always
  • Ignore them if doses are large
  • Use Fourier transform instead

Correct Answer: Include derivative initial values explicitly in transformed equations L{f’} and L{f”}

Q48. Which is true for L{d/dt [e^{at} f(t)]} in terms of F(s)?

  • Transformed as (s-a) F(s-a) – f(0)
  • Equal to s F(s)
  • Equal to F(s-a)
  • No simple relation exists

Correct Answer: Transformed as (s-a) F(s-a) – f(0)

Q49. Which technique is useful when inverse Laplace yields convolution integrals in time domain?

  • Recognize convolution and use convolution theorem to compute time-domain convolution
  • Avoid convolution by numerical inversion only
  • Convolution never appears for derivatives
  • Use only partial fractions

Correct Answer: Recognize convolution and use convolution theorem to compute time-domain convolution

Q50. Which statement best summarizes the utility of Laplace transforms for derivatives in B. Pharm curriculum?

  • They convert time-domain differential rate equations with initial conditions into solvable algebraic equations in s-domain, aiding pharmacokinetic modeling
  • They only apply to pure mathematics, not pharmacology
  • They are obsolete compared to numerical methods
  • They remove the role of initial conditions entirely

Correct Answer: They convert time-domain differential rate equations with initial conditions into solvable algebraic equations in s-domain, aiding pharmacokinetic modeling

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