Searching in sorted array with $O(\log n)$ Recently been practicing some recent exams, there was a problem I could not comprehend the given answer, the question is as follows:

Suppose array $A[1..n]$ consist of $n$ distinct integers that is
sorted in ascending order, then How many of below problems could be
solved in $O(\log n)$?

and there were three statements;


Find an $i$ index so that $A[i]= i $
Find an $i$ index so that $A \left \lfloor{i}\right \rfloor = 3i +2$
Find an $i$ index so that $A[i]= 4i^{2} + 3i + 5$


The answer sheet stated as one statement is true, Also there was no clue on which statement is true.

What is my issue?
So I think the first and second statements could be true, since the array is sorted then we can use Binary search with $O(\log n)$ to find the element with index $i$, also the first two statements are linear, thus I think that's what the question wants, If it's not then I don't know what the question exactly wants, any explanation would be handful.
You can answer the first question with binary search. Lets consider you're now discussing the interval $[l, r]$, and $m = \frac{(l + r)}{2}$. If $A[m] = m$, you have found an answer and you shall stop the search. If $A[m] &lt; m$, then the answer must be in the interval $[m + 1, r]$, because if $i \le m$, then $A[i] \le A[m] - (m - i) &lt; m - (m - i) = i$ (since the array is sorted in strictly ascending order and the elements are integers), thus no $i \in [l, m]$ can be the answer. Otherwise, if $A[m] &gt; m$, then the answer must be in the interval $[l, m - 1]$, because if $i \ge m$, then $A[i] \ge A[m] + (i - m) &gt; m + (i - m) = i$.
And I don't think you can solve the other tasks in $O(lg(n))$ using binary search. Consider these two examples :
$A = [5, 6, 12, 14, 15]$
$B = [4, 9, 12, 14, 17]$
As we know, binary search initially examines the middle element, in these cases, $12$. $12 \ne 3 \times 3 + 2$, thus the search has to continue. In both cases, decision must be the same, since the middle element is the same, therefore, if the algorithm decides to check the interval $[4, 5]$, the algorithm will respond NO to the array $A$, although $5 = 3 \times 1 + 2$, and if the algorithm decides to search the interval $[1, 2]$, it yields NO to the array $B$, while $17 = 3 \times 5 + 2$.
It is also true about the third task, with examples :
$A = [13, 49, 51, 52, 180]$
$B = [12, 48, 51, 52, 193]$
I hope my answer was helpful.