Option D should be the correct answer.
ASSUMPTION: $0$ indexing of array $L$.
Pseudo code for finding the correct position of $x$ in array $L$ will be as follows:
if (x <= 0)
return (m + 1);
else
return MODIFIED_BIN_SEARCH(0, n - 1, x);
Code for Modified Binary Search:
MODIFIED_BIN_SEARCH (low, high, x){
if (low > high)
return mid; //Exit point for unsuccessful searches.
mid = ( low + high )/2;
if ( x == a[mid])
return mid; //Exit point for successful searches.
else if (x > a [mid])
return MODIFIED_BIN_SEARCH (low, mid - 1, x);
else
return MODIFIED_BIN_SEARCH (mid + 1, high, x);
}
In the above pseudo code for finding correct position of $x$ in array $L$
If $x <= 0$, only one comparison will be required,
If $x > 0$,
$1$ comparison for checking $x > 0$ and,
$O(\log n)$ comparisons in MODIFIED_BIN_SEARCH(1, n, x) will be required irrespective of whether the search is successful or unsuccessful.
Worst Case : $x > 0$ & search for $x$ in array $L$ is unsuccessful.
Clearly worst cases will require $1 + O(\log n)$ which in turn, is same as $O(\log n)$ comparisons.
Hence option D should be correct.
Example of Unsuccessful Search:
Let $n = 5$,
First $n$ elements of $L$ be: $8, 6, 5, 3, 1$, and
$x = 2$.
Then clearly $x > 0$,
also MODIFIED_BIN_SEARCH(0, 4, 2) will return $4$,
so pseudo code for finding correct position of $x$ in array $L$ will return $4$, which is correct.
Array $L$ after inserting $2$ will be: $8, 6, 5, 3, 2, 1$.
Example of Successful Search :
Here we are inserting an element that is already present in $L$,
Let $n = 5$,
First $n$ elements of $L$ be: $8, 6, 5, 3, 1$, and
$x = 6$.
Here also clearly $x > 0$,
MODIFIED_BIN_SEARCH(0, 4, 2) will return position of $6$ in $L$ that is $1$,
so pseudo code for finding correct position of $x$ in array $L$ will return $1$, which is correct.
Array $L$ after inserting $x$ will be: $8, 6, 6, 5, 3, 1$.