Which of the following commands changes the number of days before the ext3 filesystem on /dev/sda1 has to run through a full filesystem check while booting?
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A. B. C. D. E.C
The correct answer is A. tune2fs -d 200 /dev/sda1
Explanation:
The ext3 filesystem has a feature called "filesystem check," which is performed automatically at boot time or after a certain number of mounts or time period, depending on the settings.
The tune2fs command is used to modify various filesystem parameters on ext2, ext3, and ext4 filesystems. In this question, we need to change the number of days before the ext3 filesystem on /dev/sda1 has to run through a full filesystem check while booting.
The available options in the question are:
A. tune2fs -d 200 /dev/sda1 B. tune2fs -c 200 /dev/sda1 C. tune2fs -i 200 /dev/sda1 D. tune2fs -n 200 /dev/sda1 E. tune2fs --days 200 /dev/sda1
Option A is the correct answer because the -d option is used to set the maximum mount count between two filesystem checks, and it is expressed in days rather than mounts. Therefore, the correct command to change the number of days before the ext3 filesystem on /dev/sda1 has to run through a full filesystem check while booting is:
tune2fs -d 200 /dev/sda1
Option B is incorrect because the -c option is used to set the number of mounts after which a filesystem check is performed, not the number of days.
Option C is incorrect because the -i option is used to set the maximum interval between two filesystem checks, expressed in days, weeks, or months, but not the number of days before the next check.
Option D is incorrect because the -n option is used to set the number of blocks reserved for the filesystem's use, not the number of days before the next filesystem check.
Option E is incorrect because there is no --days option in the tune2fs command.