We have two divide$--$conquer algorithms, let's call them A and B$,$ where A calls B$.$ Their time complexities, TA$($n$)$ and TB$($n$),$ are expressed by the following recurrence relations TA$($n$)=27$ TA$($n$/9)+$ TB$($n$)$ TB$($n$)=8$ TB$($n$/2)+\backslash$theta $($n$3$log$3$n$)$ Then Question $1$Answer a$.$ TA$($n$)=\backslash$Theta $($n$1\mathrm{.}5$log$3$n$)$ b$.$ TA$($n$)=\backslash$Theta $($n$1\mathrm{.}5$log$5$n$)$ c$.$ TA$($n$)=\backslash$Theta $($n$3)$ d$.$ TA$($n$)=\backslash$Theta $($n$3$log$5$n$)$ e$.$ TA$($n$)=\backslash$Theta $($n$3$log$3$n$)$ f$.$ TA$($n$)=\backslash$Theta $($n$3$log$4$n$)$