f(x) denotes the number of prime numbers less than or equal

f(x) = number of prime number less than or equal to a positive integer x.
Hence, f(2) = no. of prime number < or = 2
so, f(2) = 1
f(3) = 2
f(5) = 3 [i.e between 0 and 5, there are only 3 prime numbers; 2, 3, and 5].
m(a,b) is a function that takes 2 argument 'a' and 'b'.
Compare both argument and return the argument that has the highest value.
$$I.e\ m\left(a,b\right)=\left(a\ if\ a\ \ge b\right)\ OR\ \left(b\ if\ a\le b\right)$$
For m[f(x), 5] = 5
$$This\ means\ that\ f\left(x\right)\le5.\ Since\ x\ is\ a\ positive\ integer\ 0\le f\left(x\right)\le5$$
This impies f(x) can either be 0, 1, 2, 3, 4 or 5. Definitely, f(x) have at most, 5 prime numbers which can be 2, 3, 5, 7, 11.
Maximum value of x = 12 because 13 is another prime number and that will make f(x) = 6.
Minimum value of x = 1
All possible values of x = 1, 2, 3, ..., 12
Therefore, the summation of all the values of x that satisifies ;
$$m\left[f\left(x\right),5\right]=5=>S_n=\frac{n\left(a+l\right)}{2}$$
nth term = 12
a = first term
l = last term
$$S_{12}=\frac{12\left(1+12\right)}{2}=\frac{12\cdot13}{2}=78$$

So, the answer to this is option E. Thank you.

=>

Since m(f(x),5)=5, we have f(x) ≤ 5 or f(x) = 0, 1, 2, 3, 4 or 5.
f(1) = 0, f(2) = 1, f(3) = 2, f(4) = 2, f(5) = 3, … , f(11) = 5, f(12) = 5, f(13) = 6. f(13) = 6 is out of the range of f(x) ≤ 5.
So, all possible values of x are 1, 2, 3, … , 12 and we have 1 + 2 + 3 + … + 12 = 78.

Therefore, E is the answer.
Answer: E