Answer
14 comparisons
Work Step by Step
Let $f (n)$ be the quantity of examinations required in a binary search of a rundown of $n $components.
We know that recurrance relation for binary search with $n$ elements is: $f(n) = f(n/2) +2$.
In this manner $f(64) = f(32) + 2$
$= (f(16) + 2) +2 $
$= (f(8) + 4) +2$
$=( f(4) + 6) +2 $
$ = (f(2) + 8) +2$
$= (f(1) + 10) +2 $
$f(1)=2$ since two comparisons are required for one number
so $= 2 + 12 $
$= 14.$
