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. $$\frac{2}{{{{\text{s}}^2} + 1}}$$
Answer: Option C
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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