Laplace Transform is linear, which means that

L$($a$$f$+$b$$g$)=$a$$L$($f$)+$b$$L$($g$),\backslash$mathcal$\{$L$\}\backslash$bigl$($a$\backslash ,$f $+$ b$\backslash ,$g$\backslash$bigr$)\backslash$;$=\backslash$; a$\backslash ,\backslash$mathcal$\{$L$\}($f$)\backslash$;$+\backslash$; b$\backslash ,\backslash$mathcal$\{$L$\}($g$),$L$($af$+$bg$)=$aL$($f$)+$bL$($g$),$

where a$,$ba$,$ba$,$b are constants and f$,$gf$,$gf$,$g are functions.

Which of the following cannot be inferred from the linearity of L$\backslash$mathcal$\{$L$\}$L$?$

L$($f$+$g$)=$L$($f$)+$L$($g$)\backslash$mathcal$\{$L$\}($f$+$g$)=\backslash$mathcal$\{$L$\}($f$)+\backslash$mathcal$\{$L$\}($g$)$L$($f$+$g$)=$L$($f$)+$L$($g$)$

L$($f$$g$)=$L$($f$)$L$($g$)\backslash$mathcal$\{$L$\}($f$\backslash ,$g$)=\backslash$mathcal$\{$L$\}($f$)\backslash ,\backslash$mathcal$\{$L$\}($g$)$L$($fg$)=$L$($f$)$L$($g$)$

L$(3)=3$L$(1)\backslash$mathcal$\{$L$\}(3)=3\backslash ,\backslash$mathcal$\{$L$\}(1)$L$(3)=3$L$(1)$

L$(0)=0\backslash$mathcal$\{$L$\}(0)=0$L$(0)=0$