$($b$)$ The recursion T$($n$)=1$ for any n $<=10$ and T$($n$)=$ n

$3+$ n

$2\mathrm{.}93+8$T$($n$/2)$ solves to:

i$.\backslash$Theta $($n

$3$

$)$

ii$.\backslash$Theta $($n

$3$

log n$)$

iii. $\backslash$Theta $($n

$3$

$($log n$)$

$2\mathrm{.}93)$

iv$.\backslash$Theta $($n

$2\mathrm{.}93)$

v$.\backslash$Theta $($n

$2\mathrm{.}93+3)$