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NEW QUESTION: 1
人気のある商用仮想化プラットフォームでは、仮想ハードウェアを作成できます。仮想マシンにとって、この仮想ハードウェアは実際のハードウェアと区別がつきません。仮想化されたTPMを実装することにより、次の信頼できるシステムの概念のどれを実装できますか？
A. 信頼のルートがないソフトウェアベースのトラストアンカー
B. 継続的な信頼の連鎖
C. ソフトウェアベースの信頼のルート
D. ハードウェアの信頼のルートを持つ信頼の連鎖
Answer: D
Explanation:
A Trusted Platform Module (TPM) is a microchip designed to provide basic security-related functions, primarily involving encryption keys. The TPM is usually installed on the motherboard of a computer, and it communicates with the remainder of the system by using a hardware bus.
A vTPM is a virtual Trusted Platform Module; a virtual instance of the TPM. IBM extended the current TPM V1.2 command set with virtual TPM management commands that allow us to create and delete instances of TPMs. Each created instance of a TPM holds an association with a virtual machine (VM) throughout its lifetime on the platform.
The TPM is the hardware root of trust.
Chain of trust means to extend the trust boundary from the root(s) of trust, in order to extend the collection of trustworthy functions. Implies/entails transitive trust. Therefore a virtual TPM is a chain of trust from the hardware TPM (root of trust).
NEW QUESTION: 2
A社は、B社のビジネスパートナーを訪問して、A社の会議室で利用可能なイーサネットポートを利用することを許可しています。 このアクセスは、パートナーがB社のネットワークにVPNを確立できるようにするために提供されています。 A社のセキュリティアーキテクトは、B社のパートナーが利用可能なポートから直接インターネットにアクセスできるようにする一方、A社の従業員は、同じポートからA社の社内ネットワークにアクセスできるようにしたいと考えています。 これを可能にするために採用できるのはどれですか？
A. SIEM
B. NAC
C. ACL
D. SAML
E. MAC
Answer: B
NEW QUESTION: 3
What are two valid Redundancy options for the Citrix License Server? (Choose two.)
A. Use Citrix Application Delivery Management (ADM) Load Balancing.
B. Create a Microsoft cluster with multiple nodes.
C. Create a standby License Server.
D. Use Citrix Application Delivery Management (ADM) High Availability (HA).
E. Configure a Citrix License Server High Availability (HA) pair.
Answer: B,C
NEW QUESTION: 4
Given:
Which three values will appear in the output?
A. a2
B. b1
C. b2
D. a1
E. 0
F. 1
Answer: A,B,F
Explanation:
Staticmethod of base class is invoked >>
A myA = new B();
System.out.print(myA.doA() + myA.doA2() + myA.a);
class B String doA() { return "b1 "; }
class A protected static String doA2 () { return "a2 "; } class B int a = 7;