Examine the following command: ipadm create-addr -T static -a 192.168.1.112/26 net0/v6 Which two statements are true?

Question

Options

• AThe interface is plumbed.
• BThe interface is marked down.
• CThe netmask value is ffffffc0.
• DThe link local IPv6 address fe80::112 is created.
• EMulticast datagrams are not enabled on this interface.

• A
  85% (23)
• B
  7% (2)
• D
  4% (1)
• E
  4% (1)

Explanation

Running ipadm create-addr -T static -a 192.168.1.112/26 net0/v6 on Oracle Solaris automatically plumbs the underlying interface (net0) and brings it UP - making A correct. The /26 prefix means 26 consecutive binary 1s: 11111111.11111111.11111111.11000000, which converts to ff.ff.ff.c0 = ffffffc0 - making C correct.

Why the distractors are wrong:

• B is wrong because ipadm create-addr brings the interface UP, not down. Marking it down would require a separate ipadm disable-addr command.
• D is wrong because 192.168.1.112 is an IPv4 address - the address object name net0/v6 is just a label, not an instruction to create IPv6. Link-local IPv6 addresses are also derived from the MAC address via EUI-64, not from the IPv4 address.
• E is wrong because a plumbed, active interface has multicast enabled by default in Solaris; you would need to explicitly disable it.

Memory tip: For subnet math, count the prefix length as 1-bits from the left across 32 bits - /26 fills three full octets (ff.ff.ff) plus two more bits in the last octet (11000000 = c0). For ipadm, remember the mnemonic: "create-addr = create + plumb + up" - it does all three in one shot.

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