CORRECTED (Previously B) Your network contains an Active Directory domain named contoso.com. The
domain contains two member servers named Server1 and Server2. All servers run Windows Server 2012 R2.
Server1 and Server2 have the Failover Clustering feature installed. The servers are configured as nodes in a
failover cluster named Cluster1. Cluster1 contains a cluster disk resource.
A developer creates an application named App1. App1 is NOT a cluster-aware application. App1 runs as a
service. App1 stores date on the cluster disk resource.
You need to ensure that App1 runs in Cluster1. The solution must minimize development effort.
Which cmdlet should you run?
A.
Add-ClusterGenericServiceRole
B.
Add-ClusterGenericApplicationRole
C.
Add-ClusterScaleOutFileServerRole
D.
Add-ClusterServerRole
Explanation:
Explanation/Reference:Configure high availability for an application that was not originally designed to run in a failover cluster.
If you run an application as a Generic Application, the cluster software will start the application, then
periodically query the operating system to see whether the application appears to be running. If so, it is
presumed to be online, and will not be restarted or failed over
http://technet.microsoft.com/en-us/library/ee460976.aspx
Answer should be A. App1 runs as a service, therefore it would be add-clustergenericservicerole. Nothing else matters. It runs as a service therefore it is a generic service role because it is not cluster aware.
5
0
A.
Add-ClusterGenericServiceRole
2
0
I also believe the answer should be A
https://docs.microsoft.com/en-us/powershell/module/failoverclusters/add-clustergenericservicerole?view=win10-ps
3
0
I also think the answer should be “Add-ClusterGenericServiceRole”. Both Add-ClusterGenericServiceRole and Add-ClusterGenericApplicationRole seem to do exactly the same, with the difference that ClusterGenericApplicationRole checks for services while ClusterGenericApplicationRole checks for processes. If the application runs as a service, then the Failover cluster needs to monitor a service, not a process.
3
1