Let $x = 2^k$

$T(x) = 3T\left(\frac{x}{2}\right) + 1$

We can apply Master's theorem case 1 with $a = 3$ and $b = 2$ as $f(x) = 1 = O\left(x^{\log_2 3 - \epsilon} \right), \epsilon > 0 $

So, $T(x) = \Theta \left(x^{\log_2 3}\right) \\= \Theta \left( {2^k}^{\log_2 3} \right)\\= \Theta \left({2^{\log_2 3}}^k \right)\\ = \Theta \left(3^k \right)$

So, only option possible is **B**.

We can also directly solve as follows:

$T(x) = 3T\left(\frac{x}{2}\right) + 1\\= 9T \left (\frac{x}{4}\right) + 1 + 3 \\ \dots \\= 3^{\log_2 2^k} + \left( 1 + 3 + 9 + \dots + 3^{\log_2 {2^k-1}}\right)\\ \left(\text{recursion depth is }\log_2 x \text{ and } x = 2^k\right) \\= 3^{k} + \frac{3^{\log_2 {2^k}} - 1}{3-1} \\ \left(\text{Sum to n terms of GP with } a = 1 \text{ and } r = 3 \right)\\=3^k + \frac{3^k - 1}{2} \\=\frac{3. 3^k - 1}{2} \\=\frac{3^{k+1} -1}{2} $

OR

$T\left(2^k\right) = 3T\left(2^{k-1}\right) + 1 \\= 3^2T\left(2^{k-2}\right) + 1 +3 \\ \dots \\= 3^k T\left(2^{k-k}\right) + \left( 1 + 3 + 9 + \dots + 3^{k-1}\right) \\ \left(\text{recursion depth is }k\right)\\= 3^k + \frac{3^{k -1}} {3-1}\\\left(\text{Sum to n terms of GP with } a = 1 \text{ and } r = 3 \right) \\=3^k + \frac{3^k -1}{2}\\=\frac{3. 3^k - 1}{2} \\=\frac{3^{k+1} -1}{2} $