Biased binary search complexity We know Binary search on a set of n element array performs O(log(n)).
We have this recursive equation through which the search space is reduced by half in each iteration, after a single comparison.
T(n) = T(n/2) + 1 

Either by applying Master's Theorem or analytically we arrive at the complexity of it as log(n)
If I introduce bias, in the search by unequally partitioning the array instead of equal halves, How would the worst case time complexity change?
The unequal partitions are t * n and (1-t) * n where 0<=t<=1
This is a mathematics question as it involves asymptotic analysis and hence I don't wish my question to be downvoted.
 A: In the worst case scenarios, where $t = 0$ or $t = 1$, and assuming that at least one value will always be checked, then as I stated in my comment, the average number of checks would be $\frac{n}{2}$ and the maximum number would be $n$. However, for $0 \lt t \lt n$, where $tn \gg 1$, then as indicated in the comment by Joppy, the number of steps in the worst case scenario would be logarithmic.
To see this, let
$$u = \max(t, 1-t) \tag{1}\label{eq1}$$
The worst case would be that the value is always in the larger partition after each step, with $n_i$ being the size of this partition after $i$ steps, and with $n_0 = n$. Thus,
$$n_i = un_{i-1} \; \forall \; i \ge 1 \tag{2}\label{eq2}$$
As such, $n_1 = un$, $n_2 = un_1 = u^2 n$, etc., to get that
$$n_i = u^i n \; \forall \; i \ge 0 \tag{3}\label{eq3}$$
The search will eventually end after $m$ steps where
$$u^m n \approx 1 \tag{4}\label{eq4}$$
Note this is similar to the concept of the amount of time for exponential decay to cause a substance to eventually disappear, with the worst case number of unbiased binary searches being about the number of half-life periods. To use a value $\gt 1$ for logarithms, let
$$v = \frac{1}{u} \tag{5}\label{eq5}$$
so \eqref{eq4} becomes
$$n \approx v^m \; \implies \; m \approx \log_v(n) = \log_v(e)\ln(n) \tag{6}\label{eq6}$$
For $t = u = \frac{1}{2}$, then $v = 2$ and \eqref{eq6} gives $m \approx (1.44269)\ln(n)$. With a relatively extreme case of $t = \frac{1}{1001}$, then $u = \frac{1000}{1001}$ and $v = 1.001$, so \eqref{eq6} gives $m \approx (1000.49991)\ln(n)$, i.e., about $693.5$ times as large. Nonetheless, it's still generally considerably faster than just checking each value since, for $n = 10^{12}$, \eqref{eq6} gives $m \approx (1000.49991)(27.63102) \approx 27644.8$, so it's about $\frac{10^{12}}{27644.8} \approx 3.61731 \times 10^7$ times faster.
A: The cases $t=0$ and $t=1$ do not correspond to a terminating algorithm. Otherwise, the behavior remains logarithmic.
In the worst case, the array is every time reduced according to the smaller of $t$ and $1-t$; WLOG, let $t$.
After $k$ steps, we approximately reduce from $n$ to $nt^k\sim 1$, so that $k\sim-\log_tn$.
