F(s) is the Laplace transform of the function $f (t) = 2 t^{2} e^{- t}$ . Then F(1) is (correct to two decimal places.)
0.5

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Solution

The correct option is A 0.5
$f (t) = 2 t^{2} e^{- t}$
$∵ L {t^{n}} = \frac{n!}{S^{n + 1}}$
$\Rightarrow L {t^{2}} = \frac{2!}{S^{3}} = \frac{2}{s^{3}}$

$T h e n L {2. (t^{2})} = \frac{4}{S^{3}}$
Hence
$L {e^{- t} . (2 t^{2})} = \frac{4}{(s + 1)^{3}} = F (s)$
$S o F (1) = \frac{4}{(1 + 1)^{3}} = \frac{1}{2} = 0.5$

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