IBM C1000-138 dumps - in .pdf

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  • Exam Code: C1000-138
  • Exam Name: IBM API Connect v10.0.3 Solution Implementation
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  • Exam Code: C1000-138
  • Exam Name: IBM API Connect v10.0.3 Solution Implementation
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NEW QUESTION: 1
A new high security printer is being installed at a small office, with the requirement that only connectivity over direct line of sight be allowed. Which of the following technologies would BEST achieve this goal?
A. Infrared
B. 802.11n
C. Bluetooth
D. FireWire
Answer: A

NEW QUESTION: 2
10.0.0.0/24サブネット内の奇数番号のホストから許可される標準のアクセス制御エントリはどれですか。
A. 許可10.0.0.0.0.0.0.1
B. 許可10.0.0.1.0.0.0.254
C. 許可10.0.0.0.255.255.255.254
D. 許可10.0.0.1.0.0.0.0
Answer: B
Explanation:
Explanation
Remember, for the wildcard mask, 1s are I DON'T CARE, and 0s are I CARE. So now let's analyze a simple ACL:
access-list 1 permit 172.23.16.0 0.0.15.255
Two first octets are all 0's meaning that we care about the network 172.23.x.x. The third octet of the wildcard mask, 15 (0000 1111 in binary), means that we care about first 4 bits but don't care about last 4 bits so we allow the third octet in the form of 0001xxxx (minimum:00010000 = 16; maximum: 0001111 = 31).

The fourth octet is 255 (all 1 bits) that means I don't care.
Therefore network 172.23.16.0 0.0.15.255 ranges from 172.23.16.0 to 172.23.31.255.
Now let's consider the wildcard mask of 0.0.0.254 (four octet: 254 = 1111 1110) which means we only care the last bit. Therefore if the last bit of the IP address is a "1" (0000 0001) then only odd numbers are allowed. If the last bit of the IP address is a "0" (0000 0000) then only even numbers are allowed.
Note: In binary, odd numbers are always end with a "1" while even numbers are always end with a "0".
Therefore in this question, only the statement "permit 10.0.0.1 0.0.0.254" will allow all oddnumbered hosts in the 10.0.0.0/24 subnet.

NEW QUESTION: 3
A systems administrator establishes a CIFS share on a UNIX device to share data to Windows systems. The security authentication on the Windows domain is set to the highest level. Windows users are stating that they cannot authenticate to the UNIX share. Which of the following settings on the UNIX server would correct this problem?
A. Accept only LM
B. Refuse LM and only accept NTLMv2
C. Accept only NTLM
D. Refuse NTLMv2 and accept LM
Answer: B
Explanation:
In a Windows network, NT LAN Manager (NTLM) is a suite of Microsoft security protocols that provides authentication, integrity, and confidentiality to users. NTLM is the successor to the authentication protocol in Microsoft LAN Manager (LANMAN or LM), an older Microsoft product, and attempts to provide backwards compatibility with LANMAN. NTLM version 2 (NTLMv2), which was introduced in Windows NT 4.0 SP4 (and natively supported in Windows 2000), enhances NTLM security by hardening the protocol against many spoofing attacks, and adding the ability for a server to authenticate to the client.
This question states that the security authentication on the Windows domain is set to the highest level. This will be NTLMv2. Therefore, the answer to the question is to allow NTLMv2 which will enable the Windows users to connect to the UNIX server. To improve security, we should disable the old and insecure LM protocol as it is not used by the Windows computers.
Incorrect Answers:
B: The question states that the security authentication on the Windows domain is set to the highest level. This will be NTLMv2, not LM.
C: The question states that the security authentication on the Windows domain is set to the highest level. This will be NTLMv2, not LM so we need to allow NTLMv2.
D: The question states that the security authentication on the Windows domain is set to the highest level. This will be NTLMv2, not NTLM (version1).
References:
https://en.wikipedia.org/wiki/NT_LAN_Manager

NEW QUESTION: 4
Which of the following BEST describes 'partners' in the phrase "people, processes, products and partners"?
A. internal departments
B. suppliers, manufacturers and vendors
C. the facilities manager
D. customers
Answer: B

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