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GATE2020-ME-2: 4

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Arjun asked Feb 24, 2017
GATE2015-2-4
The Laplace transform of $e^{i5t}$ where $i=\sqrt{-1}$, is $\dfrac{s-5i}{s^2-25} \\$ $\dfrac{s+5i}{s^2+25} \\$ $\dfrac{s+5i}{s^2-25} \\$ $\dfrac{s-5i}{s^2+25} $
The Laplace transform of $e^{i5t}$ where $i=\sqrt{-1}$, is$\dfrac{s-5i}{s^2-25} \\$$\dfrac{s+5i}{s^2+25} \\$$\dfrac{s+5i}{s^2-25} \\$$\dfrac{s-5i}{s^2+25} $
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go_editor asked Mar 1, 2021
GATE Mechanical 2021 Set 2 | Question: 2
If the Laplace transform of a function $f(t)$ is given by $\frac{s+3}{\left ( s+1 \right )\left ( s+2 \right )}$, then $f(0)$ is $0$ $\frac{1}{2}$ $1$ $\frac{3}{2}$
If the Laplace transform of a function $f(t)$ is given by $\frac{s+3}{\left ( s+1 \right )\left ( s+2 \right )}$, then $f(0)$ is$0$$\frac{1}{2}$$1$$\frac{3}{2}$
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Arjun asked Feb 26, 2017
GATE2017 ME-2: 5
The Laplace transform of $te^{t}$ is $\dfrac{s}{(s+1)^{2}} \\$ $\dfrac{1}{(s-1)^{2}} \\$ $\dfrac{1}{(s+1)^{2}} \\$ $\dfrac{s}{(s-1)}$
The Laplace transform of $te^{t}$ is$\dfrac{s}{(s+1)^{2}} \\$$\dfrac{1}{(s-1)^{2}} \\$$\dfrac{1}{(s+1)^{2}} \\$$\dfrac{s}{(s-1)}$
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Arjun asked Feb 19, 2017
GATE Mechanical 2014 Set 4 | Question: 5
Laplace transform of $\cos(\omega t)$ is $\dfrac{s}{s^2+\omega ^2}$. The Laplace transform of $e^{-2t} \cos(4t)$ is $\dfrac{s-2}{(s-2)^2+16} \\$ $\dfrac{s+2}{(s-2)^2+16} \\$ $\dfrac{s-2}{(s+2)^2+16} \\$ $\dfrac{s+2}{(s+2)^2+16}$
Laplace transform of $\cos(\omega t)$ is $\dfrac{s}{s^2+\omega ^2}$. The Laplace transform of $e^{-2t} \cos(4t)$ is$\dfrac{s-2}{(s-2)^2+16} \\$$\dfrac{s+2}{(s-2)^2+16} \\...
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