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To prepare for the F5 303 exam, candidates must have a solid understanding of application security and network security concepts. They should have a good knowledge of BIG-IP ASM and its features and functionalities. Candidates can also take advantage of various training programs and resources offered by F5 Networks to prepare for the exam.
F5 303 exam is designed for those individuals looking to become a BIG-IP ASM specialist. 303 Exam covers a range of topics including security, application delivery, and network administration. BIG-IP ASM Specialist certification is designed for network professionals who want to validate their skills and knowledge in deploying and managing the F5 BIG-IP ASM solution. 303 exam is intended for individuals who have experience working with BIG-IP ASM and are looking to take their skills to the next level.
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F5 303 certification exam is a computer-based test that consists of multiple-choice questions. It is a proctored exam that is administered at Pearson VUE testing centers around the world. 303 Exam is timed, with a total of 90 minutes given to complete the exam. Candidates must achieve a passing score of at least 69% to earn the certification.
NEW QUESTION # 242
A virtual server is configured to offload SSL from a pool of backend servers. When users connect to the virtual server, they successfully establish an SSL connection but no content is displayed. A packet trace performed on the server shows that the server receives and responds to the request.
What should a BIG-IP Administrator do to resolve the problem?
Answer: B
NEW QUESTION # 243
New Syslog servers have been deployed in an organization. The BIG-IP Administrator must reconfigure the BIG-IP system to send log messages to these servers.
In which location in the Configuration Utility can the BIG-IP Administrator make the needed configuration changes to accomplish this?
Answer: D
NEW QUESTION # 244
An application owner claims an LTM device is delaying delivery of an HTTP application. The LTM device has two VLANs, an internal and an external. The application servers reside on the internal VLAN. The virtual server and clients reside on the external VLAN. With appropriate filters applied, which solution is most efficient for obtaining packet captures in order to investigate the claim of delayed delivery?
Answer: C
NEW QUESTION # 245
Refer to the exhibit.
An LTMSpecialist configures the two syslog destination Syslog destination #1 can receive messages but the syslog destination #2 can NOT receive messages.
Which command sill correct the issue?
Answer: B
NEW QUESTION # 246
The LTM device is configured to provide load balancing to a set of web servers that implement access control lists (ACL) based on the source IP address of the client. The ACL is at the network level and the web server is configured to send a TCP reset back to the client if it is NOT permitted to connect.
The virtual server is configured with the default OneConnect profile.
The ACL is defined on the web server as:
Permit: 192.168.136.0/24
Deny: 192.168.116.0/24
The packet capture is taken of two individual client flows to a virtual server with IP address
192.168.136.100.
Client A - Src IP 192.168.136.1 - Virtual Server 192.168.136.100:
Clientside:
09:35:11.073623 IP 192.168.136.1.55684 > 192.168.136.100.80: S
869998901:869998901(0) win 8192 <mss 1460,nop,wscale 2,nop,nop,sackOK>
09:35:11.073931 IP 192.168.136.100.80 > 192.168.136.1.55684: S
2273668949:2273668949(0) ack 869998902 win 4380 <mss 1460,nop,wscale
0,sackOK,eol>
09:35:11.074928 IP 192.168.136.1.55684 > 192.168.136.100.80: . ack 1
win 16425
09:35:11.080936 IP 192.168.136.1.55684 > 192.168.136.100.80: P
1:299(298) ack 1 win 16425
09:35:11.081029 IP 192.168.136.100.80 > 192.168.136.1.55684: . ack 299
win 4678
Serverside:
09:35:11.081022 IP 192.168.136.1.55684 > 192.168.116.128.80: S
685865802:685865802(0) win 4380 <mss 1460,nop,wscale 0,sackOK,eol>
09:35:11.081928 IP 192.168.116.128.80 > 192.168.136.1.55684: S
4193259095:4193259095(0) ack 685865803 win 5840 <mss
1460,nop,nop,sackOK,nop,wscale 6>
09:35:11.081943 IP 192.168.136.1.55684 > 192.168.116.128.80: . ack 1
win 4380
09:35:11.081955 IP 192.168.136.1.55684 > 192.168.116.128.80: P
1:299(298) ack 1 win 4380
09:35:11.083765 IP 192.168.116.128.80 > 192.168.136.1.55684: . ack 299
win 108
Client B - Src IP 192.168.116.1 - Virtual Server 192.168.136.100:
Clientside:
09:36:11.244040 IP 192.168.116.1.55769 > 192.168.136.100.80: S
3320618938:3320618938(0) win 8192 <mss 1460,nop,wscale
2,nop,nop,sackOK>
09:36:11.244152 IP 192.168.136.100.80 > 192.168.116.1.55769: S
3878120666:3878120666(0) ack 3320618939 win 4380 <mss 1460,nop,wscale
0,sackOK,eol>
09:36:11.244839 IP 192.168.116.1.55769 > 192.168.136.100.80: . ack 1
win 16425
09:36:11.245830 IP 192.168.116.1.55769 > 192.168.136.100.80: P
1:299(298) ack 1 win 16425
09:36:11.245922 IP 192.168.136.100.80 > 192.168.116.1.55769: . ack 299
win 4678
Serverside:
09:36:11.245940 IP 192.168.136.1.55684 > 192.168.116.128.80: P
599:897(298) ack 4525 win 8904
09:36:11.247847 IP 192.168.116.128.80 > 192.168.136.1.55684: P
4525:5001(476) ack 897 win 142
Why was the second client flow permitted by the web server?
Answer: C
NEW QUESTION # 247
......
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