The Laplace transform of ei5t where $${\text{i}} = \sqrt { - 1} ,$$ is
A. $$\frac{{{\text{s}} - 5{\text{i}}}}{{{{\text{s}}^2} - 25}}$$
B. $$\frac{{{\text{s}} + 5{\text{i}}}}{{{{\text{s}}^2} + 25}}$$
C. $$\frac{{{\text{s}} + 5{\text{i}}}}{{{{\text{s}}^2} - 25}}$$
D. $$\frac{{{\text{s}} - 5{\text{i}}}}{{{{\text{s}}^2} + 25}}$$
Answer: Option B
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. $$\frac{\pi }{8}$$
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. $$\frac{{1 + {{\text{s}}^2}}}{{1 - {{\text{s}}^2}}}$$
A. $$\frac{2}{{{\text{s}} + 1}}$$
B. $$\frac{4}{{{\text{s}} + 1}}$$
C. $$\frac{4}{{{{\text{s}}^2} + 1}}$$
D. $$\frac{2}{{{{\text{s}}^2} + 1}}$$
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