For option a)
f(n)=$n^{2}$
g(n)=$n^{3}$
then f(n)=O(g(n))
but g(n) $\neq$ O(f(n)) because on using any constant K we can’t prove that
$n^{3}$<=K$n^{2}$ (because for that K = n should be there but it is not allowed as K is constant.)
so option a) is false.
For option b)
f(n)=$n^{2}$ and
let, S(n)= O(f(n)) hence
S(n)=O($n^{2}$) that means
S(n)<= K. $n^{2}$ where K is any constant ,
since S(n) may be equal or less than f(n)
hence in,
f(n)+(S(n))=$\Theta$(f(n))
we can neglect S(n) so it can be written as f(n)=$\Theta$(f(n)) .
so option b) is True.
For option c)
f(n)=$n^{2}$ and
let, S(n)= o(f(n)) hence,
S(n)=o($n^{2}$) that means
S(n)< K. $n^{2}$ where K is any constant ,
since S(n) is strictly less than f(n)
hence in,
f(n)+(S(n))=$\Theta$(f(n))
we can neglect S(n) so it can be written as f(n)=$\Theta$(f(n)) .
so option c) is True.
For option d)
log n)! and (log log n)! are not polynomially bounded functions.
The factorial function grows at an exponential rate, meaning the value of (log n)! and (log log n)! increases rapidly as n or log n increases. In contrast, a polynomially bounded function is one that can be upper-bounded by a polynomial function of the input size.
Therefore, (log n)! and (log log n)! are not polynomially bounded functions.
