A delayed unit step function is defined as \[{\text{u}}\left( {{\text{t}} - {\text{a}}} \right) = \left\{ {\begin{array}{*{20}{c}} {0,}&{{\text{for t}} < {\text{a}}} \\ {1,}&{{\text{for t}} \geqslant {\text{a}}} \end{array}} \right..\]      Its Laplace transform is

The Laplace transform F(s) of the exponential function. f(t) = e^at when t ≥ 0, where a is a constant and (s - a) > 0, is

A. $$\frac{1}{{{\text{s}} + {\text{a}}}}$$

B. $$\frac{1}{{{\text{s}} - {\text{a}}}}$$

C. $$\frac{1}{{{\text{a}} - {\text{s}}}}$$

D. $$\infty $$

Evaluate $$\int\limits_0^\infty {\frac{{\sin {\text{t}}}}{{\text{t}}}{\text{dt}}} $$

A. $$\pi $$

B. $$\frac{\pi }{2}$$

C. $$\frac{\pi }{4}$$

D. $$\frac{\pi }{8}$$

If the Laplace transform of $${{\text{e}}^{\omega {\text{t}}}}$$  is $$\frac{1}{{{\text{s}} - \omega }},$$  the Laplace transform of tcosh t is

A. $$\frac{{1 + {{\text{s}}^2}}}{{{{\left( {{{\text{s}}^2} - 1} \right)}^2}}}$$

B. $$\frac{{{\text{st}}}}{{\left( {{{\text{s}}^2} - 1} \right)}}$$

C. $$\frac{{1 - {{\text{s}}^2}}}{{{{\left( {{{\text{s}}^2} - 1} \right)}^2}}}$$

D. $$\frac{{1 + {{\text{s}}^2}}}{{1 - {{\text{s}}^2}}}$$

The function f(t) satisfies the differential equation $$\frac{{{{\text{d}}^2}{\text{f}}}}{{{\text{d}}{{\text{t}}^2}}} + {\text{f}} = 0$$   and the auxiliary conditions, f(0) = 0, $$\frac{{{\text{df}}}}{{{\text{dt}}}}\left( 0 \right) = 4.$$  The Laplace transform of f(t) is given by

A. $$\frac{2}{{{\text{s}} + 1}}$$

B. $$\frac{4}{{{\text{s}} + 1}}$$

C. $$\frac{4}{{{{\text{s}}^2} + 1}}$$

D. $$\frac{2}{{{{\text{s}}^2} + 1}}$$