Practice high-yield Mathematics & Quantitative Ability MCQs on Transform Theory — Part 2 with worked academic solutions for CSS, NTS, PPSC, FPSC, PMS.
Questions
9
Difficulty Level
Exam-Level
Target Exams
CSS, NTS, PPSC...
Estimated Time
11 mins
Exam-Level CSS NTS PPSC FPSC PMS
Verified for Commission Syllabus • Oct 14, 2026
9
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Question 1 of 9
★ CSS
★ NTS
★ CSS
★ NTS
The Laplace transform of $$i\left( t \right)$$ is given by $$I\left( s \right) = \frac{2}{{s\left( {1 + s} \right)}}.$$ As $$t \to \infty $$ the value of $$i\left( t \right)$$ tends to
Correct Answer: Option C — 2
Detailed Academic Explanation:
2 is the correct answer to “The Laplace transform of $$i\left( t \right)$$ is given by $$I\left( s \right) = \frac{2}{{s\left( {1 + s} \right)}}.$$ As $$t \to \infty $$ the value of $$i\left( t \right)$$ tends to”. This question is standard in competitive examination syllabi for Transform Theory.
💡Key Conceptual Anchor:Transform Theory
Question 2 of 9
★ CSS
★ NTS
★ CSS
★ NTS
The bilateral Laplace transform of $${e^{ - 1}}u\left( {t + 2} \right)$$ is
Correct Answer: Option A — $$\frac{{{e^{2\left( {s + 1} \right)}}}}{{s + 1}},\,\,\,\,\operatorname{Re} \left( s \right) > - 1$$
Detailed Academic Explanation:
$$\frac{{{e^{2\left( {s + 1} \right)}}}}{{s + 1}},\,\,\,\,\operatorname{Re} \left( s \right) > - 1$$ is the correct answer to “The bilateral Laplace transform of $${e^{ - 1}}u\left( {t + 2} \right)$$ is”. This question is standard in competitive examination syllabi for Transform Theory.
💡Key Conceptual Anchor:Transform Theory
Question 3 of 9
★ CSS
★ NTS
★ CSS
★ NTS
The bilateral Laplace transform of $${e^t}\cos 2tu\left( { - t} \right) + {e^{ - t}}u\left( t \right) + {e^{\frac{t}{2}}}u\left( t \right)$$ is
$$\frac{{1 - s}}{{{{\left( {s - 1} \right)}^2} + 4}} + \frac{1}{{s + 1}} + \frac{1}{{s - 0.5}},\,0.5 < \operatorname{Re} \left( s \right) < 1$$ is the correct answer to “The bilateral Laplace transform of $${e^t}\cos 2tu\left( { - t} \right) + {e^{ - t}}u\left( t \right) + {e^{\frac{t}{2}}}u\left( t \right)$$ is”. This question is standard in competitive examination syllabi for Transform Theory.
💡Key Conceptual Anchor:Transform Theory
Question 4 of 9
★ CSS
★ NTS
★ CSS
★ NTS
The bilateral Laplace transform of u(-t + 3) is
Correct Answer: Option B — $$\frac{{ - {e^{ - 3s}}}}{s},\,\,\,\,\,\operatorname{Re} \left( s \right) < 0$$
Detailed Academic Explanation:
$$\frac{{ - {e^{ - 3s}}}}{s},\,\,\,\,\,\operatorname{Re} \left( s \right) < 0$$ is the correct answer to “The bilateral Laplace transform of u(-t + 3) is”. This question is standard in competitive examination syllabi for Transform Theory.
💡Key Conceptual Anchor:Transform Theory
Question 5 of 9
★ CSS
★ NTS
★ CSS
★ NTS
The Laplace transform of $$x\left( t \right)$$ is $$X\left( s \right) = {e^{ - 2s}}\frac{{6{s^2} + s}}{{{s^2} + 2s - 2}}$$ The initial value of $$x\left( t \right)$$ is
Correct Answer: Option D — 0
Detailed Academic Explanation:
0 is the correct answer to “The Laplace transform of $$x\left( t \right)$$ is $$X\left( s \right) = {e^{ - 2s}}\frac{{6{s^2} + s}}{{{s^2} + 2s - 2}}$$ The initial value of $$x\left( t \right)$$ is”. This question is standard in competitive examination syllabi for Transform Theory.
💡Key Conceptual Anchor:Transform Theory
Question 6 of 9
★ CSS
★ NTS
★ CSS
★ NTS
The bilateral Laplace transform of $${e^{\left( {3t + 6} \right)}}u\left( {t + 3} \right)$$ is
Correct Answer: Option C — $$\frac{{{e^{3\left( {s - 1} \right)}}}}{{s - 3}},\,\,\,\,\,\operatorname{Re} \left( s \right) > 3$$
Detailed Academic Explanation:
$$\frac{{{e^{3\left( {s - 1} \right)}}}}{{s - 3}},\,\,\,\,\,\operatorname{Re} \left( s \right) > 3$$ is the correct answer to “The bilateral Laplace transform of $${e^{\left( {3t + 6} \right)}}u\left( {t + 3} \right)$$ is”. This question is standard in competitive examination syllabi for Transform Theory.
💡Key Conceptual Anchor:Transform Theory
Question 7 of 9
★ CSS
★ NTS
★ CSS
★ NTS
The Laplace transform of $$x\left( t \right)$$ is $$X\left( s \right) = \frac{{{e^{ - 3s}}\left( {2{s^2} + 1} \right)}}{{s\left( {s + 1} \right)\left( {s + 4} \right)}}$$ The final value of $$x\left( t \right)$$ is
Correct Answer: Option B — $$\frac{1}{4}$$
Detailed Academic Explanation:
$$\frac{1}{4}$$ is the correct answer to “The Laplace transform of $$x\left( t \right)$$ is $$X\left( s \right) = \frac{{{e^{ - 3s}}\left( {2{s^2} + 1} \right)}}{{s\left( {s + 1} \right)\left( {s + 4} \right)}}$$ The final value of $$x\left( t \right)$$ is”. This question is standard in competitive examination syllabi for Transform Theory.
💡Key Conceptual Anchor:Transform Theory
Question 8 of 9
★ CSS
★ NTS
★ CSS
★ NTS
The Laplace transform of $$x\left( t \right)$$ is $$X\left( s \right) = \frac{{2{s^2} + 3}}{{{s^2} + 5s + 1}}$$ The initial value of $$x\left( t \right)$$ is
Correct Answer: Option A — 0
Detailed Academic Explanation:
0 is the correct answer to “The Laplace transform of $$x\left( t \right)$$ is $$X\left( s \right) = \frac{{2{s^2} + 3}}{{{s^2} + 5s + 1}}$$ The initial value of $$x\left( t \right)$$ is”. This question is standard in competitive examination syllabi for Transform Theory.
💡Key Conceptual Anchor:Transform Theory
Question 9 of 9
★ CSS
★ NTS
★ CSS
★ NTS
The inverse Laplace transform of the function $$\frac{{s + 5}}{{\left( {s + 1} \right)\left( {s + 3} \right)}}$$ is
$$2{e^{ - t}} - {e^{ - 3t}}$$ is the correct answer to “The inverse Laplace transform of the function $$\frac{{s + 5}}{{\left( {s + 1} \right)\left( {s + 3} \right)}}$$ is”. This question is standard in competitive examination syllabi for Transform Theory.
💡Key Conceptual Anchor:Transform Theory
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Algebra — Part 1
Master the next curriculum module with 40 exam-standard MCQs.
💡Examiner's Advisory for Mathematics & Quantitative Ability
In competitive commission exams (FPSC, PPSC, KPPSC, SPSC, CSS, PMS), questions are frequently phrased with distractor options designed to catch rote learners. Whenever you answer an MCQ, read the Detailed Academic Explanation and Conceptual Anchor to understand why the remaining options are incorrect.