Kaplan Professional Engineer (PE) Licensure Examination Comprehensive Practice Question Bank Multiple-Choice Questions — Advanced/Hard Difficulty

Kaplan Professional Engineer (PE) Licensure
Examination Comprehensive Practice Question Bank
Multiple-Choice Questions — Advanced/Hard Difficulty


EXAM TITLE: Principles and Practice of Engineering (PE) Licensure Examination —
Comprehensive Review for Professional Engineer Certification




SECTION 1: MATHEMATICS AND ENGINEERING FUNDAMENTALS (Questions 1–20)


Question 1


A second-order linear differential equation has the characteristic equation r² + 4r + 13 = 0. The


general solution to the homogeneous equation is:


A) y = C₁e^(−2x)cos(3x) + C₂e^(−2x)sin(3x)


B) y = C₁e^(2x)cos(3x) + C₂e^(2x)sin(3x)


C) y = C₁e^(−4x) + C₂e^(−9x)


D) y = C₁e^(−2x) + C₂xe^(−2x)


CorreCt Answer: A

,Rationale: The characteristic equation r² + 4r + 13 = 0 has roots r = [−4 ± √(16−52)]/2 = [−4 ±


√(−36)]/2 = −2 ± 3i. For complex conjugate roots a ± bi, the general solution is y =


e^(ax)[C₁cos(bx) + C₂sin(bx)]. Here a = −2 and b = 3, giving y = e^(−2x)[C₁cos(3x) + C₂sin(3x)].




Question 2


What is the Laplace transform of f(t) = t²e^(−3t)?


A) 2/(s+3)³


B) 2/(s−3)³


C) 6/(s+3)⁴


D) 2s/(s²+9)²


CorreCt Answer: A


Rationale: The Laplace transform of tⁿe^(at) is n!/(s−a)^(n+1). For t²e^(−3t), n=2 and a=−3,


giving L{t²e^(−3t)} = 2!/(s+3)³ = 2/(s+3)³. This is a fundamental transform used in solving


differential equations for engineering systems.




Question 3

,The matrix A = [[3, 1], [1, 3]] has eigenvalues:


A) λ₁ = 2, λ₂ = 4


B) λ₁ = 3, λ₂ = 3


C) λ₁ = 1, λ₂ = 9


D) λ₁ = −2, λ₂ = 6


CorreCt Answer: A


Rationale: The characteristic equation is det(A − λI) = (3−λ)² − 1 = 0 → (3−λ)² = 1 → 3−λ = ±1 → λ


= 2 or λ = 4. Eigenvalues are fundamental in structural dynamics, control systems, and stress


analysis.




Question 4


For a normally distributed dataset with mean μ = 50 and standard deviation σ = 8, what is the


probability that a randomly selected value is between 42 and 58?


A) 0.3413


B) 0.6826

, C) 0.9544


D) 0.9974


CorreCt Answer: B


Rationale: Values between 42 and 58 represent μ ± 1σ (50 ± 8). For a normal distribution,


approximately 68.26% of data falls within ±1 standard deviation of the mean. This is the


empirical rule (68-95-99.7 rule) applicable to engineering quality control and reliability analysis.




Question 5


The partial derivative ∂z/∂x for z = x²y + sin(xy) is:


A) 2xy + ycos(xy)


B) 2xy + xcos(xy)


C) x² + ycos(xy)


D) 2xy − ycos(xy)


CorreCt Answer: A