What would be the time complexity for following recurrence?
You asked for a solution with the Master Theorem, so let's have a look;

Define your Function as $T(n) = a \cdot{} T \left( \frac{n}{b} \right) + f(n)$. Here you have $a = 1$, $b = 2$ and $f(n) = 2^n$.

Master Theorem has 3 main cases (e.g. shown here):

Case 1 - $f(n) \in \mathcal{O} \left( n^{\log_b(a)-\epsilon{}} \right)$ with $\epsilon &gt; 0$

Case 2 - $f(n) \in \Theta \left( n^{\log_b(a)} \log^k n \right)$ with $k \geq 0 $

Case 3 - $f(n) \in \Omega \left( n^{\log_b(a) + \epsilon}  \right)$ and $a \cdot f \left( \frac{a}{b} \right) \leq c \cdot f(n) $ for large $n$ and some constant $c &lt; 1$

Let's check, which one applies: Inserting $a$ and $b$ into $\log_b(a)$ gives $\log_2(1) = 0$ we get $n^{\log_b(a)} = n^0 = 1$ resulting into $f(n) \in \Omega \left( n^{\log_b(a) + \epsilon}  \right)$.

The other precondition also holds obviously (just put in the numbers for $a$ and $b$ and find a $c$ for yourself).

This results in the 3rd case of the Master Theorem, giving $T(n) \in \Theta (f(n)) = \Theta (2^n) $.