NONE of the options are fully correct — but if option A had used 10.2.1.1 or 10.2.1.2 instead of .3, it would be right and if Option D had used a correct subnet mask (255.255.255.252), it would be correct too.
A. interface e0/0 description to XX-AXXX:XXXXX ip address 10.2.1.3 255.255.255.252
Given address belongs to subnet 10.2.1.0/30, and it’s the broadcast address inside that subnet.
Wrong answer.
B. interface e0/0 description to XX-AXXX:XXXXX ip address 192.168.1.1 255.255.255.248
Given address belongs to subnet 192.168.1.0/29.
Wrong answer.
C. interface e0/0 description to XX-AXXX:XXXXX ip address 172.16.1.4 255.255.255.248
Given address belongs to subnet 172.16.1.0/29.
Wrong answer.
D. interface e0/0 description to XX-AXXX:XXXXX ip address 209.165.201.2 225.255.255.252
Given address belongs to subnet 209.165.201.0/30, which ranges from 209.165.201.1 to 209.165.201.2.
Correct answer.
The answer is A. Yes it it a board cast address but it’s also a point to point link, which doesn’t need to utilize the broadcast IP. So we can assign it to a device instead. Plus the scenario doesn’t mention using public IP addresses go a routers to communicate. It’s could be a vpn connection for all we know.
The Answers B and C are out because their submask have the prefix /29, in A the 10.2.1.3 is the broadcast of 10.2.1.0 network. The answer correct is D because 209.165.201.2 is the 2nd available IP for 209.165.201.0 network.
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