ISACA CRISC Q&A - in .pdf

  • CRISC pdf
  • Exam Code: CRISC
  • Exam Name: Certified in Risk and Information Systems Control
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NEW QUESTION: 1
DRAG DROP
You are developing an ASP.NET MVC application.
You must handle any first chance exceptions that the application throws. The exception handler has the following requirements:
Catch any first chance exceptions thrown by the default app domain.

Display the name of the app domain that caused the exception.

Display the message for the exception.

You need to implement the exception handler.
How should you complete the relevant code? To answer, drag the appropriate code segment to the correct location or locations. Each code segments may be used once, more than once, or not at all. You may need to drag the split bar between panes or scroll to view content.
Select and Place:

Answer:
Explanation:

Explanation/Reference:
Explanation:
Note: Example code with inline lambda expression for the handler:
class Example
{
static void Main()
{
AppDomain.CurrentDomain.FirstChanceException +
(object source, FirstChanceExceptionEventArgs e) =>
{
Console.WriteLine("FirstChanceException event raised in {0}: {1}",
AppDomain.CurrentDomain.FriendlyName, e.Exception.Message);
};
Example with handler defined separately:
static void FirstChanceHandler(object source, FirstChanceExceptionEventArgs e)
{
Console.WriteLine("FirstChanceException event raised in {0}: {1}",
AppDomain.CurrentDomain.FriendlyName, e.Exception.Message);
}
Reference: https://docs.microsoft.com/en-us/dotnet/framework/app-domains/how-to-receive-first-chance- exception-notifications

NEW QUESTION: 2



A. Option A
B. Option C
C. Option D
D. Option B
Answer: B
Explanation:
The clone domain controller uses the security context of the source domain controller (the domain controller whose copy it represents) to contact the Windows Server 2012 R2 Primary Domain Controller (PDC) emulator operations master role holder (also known as flexible single master operations, or FSMO). The PDC emulator must be running Windows Server 2012 R2, but it does not have to be running on a hypervisor. http://technet.microsoft.com/en-us/library/hh831734.aspx

NEW QUESTION: 3
Given the network diagram.

Which address would successfully summarize only the networks seen?
A. 192.168.12.0/20
B. 192.168.0.0/24
C. 192.168.16.0/21
D. 192.168.8.0/21
E. 192.168.8.0/20
F. These networks cannot be summarized.
Answer: D
Explanation:
let's suppose it is a /20 then we would have addresses from 192.168.0.1 to 192.168.15.255
Now let's suppose it is a /21 then we would have addresses from 192.168.8.1 to
192.168.15.255
So both summaries encompass the networks we want to summarize but the second one is the most restrictive one as it only encompasses the networks we were asked to summarize and not others so it is the correct summary.
In fact just count the number of subnets which is 8 and find the exponent of 2 which is 8, that gives you 3 and to find out the summary mask just do /24 - 3 = /21.
if you make it in binary you'll find out the answer too:
the interesting octet is 3rd one so let's convert in binary.
8 0000 1000
9 0000 1001
1 0 0000 1010
1 1 0000 1011
1 2 0000 1100
1 3 0000 1101
1 4 0000 1110
1 5 0000 1111
I've bolded the bits in common there are 5 so 16+5=21 which is mask and address is
1 92.168.8.0

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