Q.$74$ = option **B**

Q.$75$ = option **C**

Time complexity of $f1$ is given by

$T(n) = T(n-1) + T(n-2)$, (multiplication by $2$ and $3$ won't affect complexity as it is a constant time operation)

$T(0) = T(1) = 1$

The solution to this (fibonacci series) is given by Golden ratio. https://en.wikipedia.org/wiki/Golden_ratio which is $O(2^n)$. (Using theta in question must be a mistake)

Time complexity of $f2$ is $\Theta(n)$ as here all recursive calls are avoided by saving the results in an array (dynamic programming).

So, answer to $74$ is (**B**).

$75$. Both $f1$ and $f2$ are calculating the same function. So,

$f1(2) = 2f1(1) + 3f1(0) = 2$

$f1(3) = 2f1(2) + 3f1(1) = 7$

$f1(4) = 20$

$f1(5) = 61$

$f1(6) = 182$

$f1(7) = 547$

$f1(8) = 1640 = f2(8)$