Detailed Video Solution: https://youtu.be/6cP7omPN4qQ
We can easily solve it by considering cases:
Case 1: When $A= 0, B = 0$
In this case, $f(A,B,C,D) = C$
So, $f = 1$ for the following input combinations: $(A,B,C,D) = (0,0,1,0), (0,0,1,1)$ i.e. $f(A,B,C,D)=1$ for minterms $m_2, m_3.$
Hence, Options $A,C$ are eliminated.
Now, in the remaining two options, we consider whether $f(A,B,C,D)$ is $1$ or not, for minterm $m_9.$
So, consider case 2: When $A= 1, B = 0$
In this case, $f(A,B,C,D) = \overline{C}$
So, $f = 1$ for the following input combinations: $(A,B,C,D) = (1,0,0,0), (1,0,0,1)$ i.e. $f(A,B,C,D)=1$ for minterms $m_8, m_9.$
Hence, Option $B$ is also eliminated.
Final answer is Option D.
Detailed Video Solution: https://youtu.be/6cP7omPN4qQ