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Updated on: 15 Apr 2025, 01:55
Originally posted by NextstopISB on 15 Apr 2025, 00:55.
Last edited by Bunuel on 15 Apr 2025, 01:55, edited 1 time in total.
Last edited by Bunuel on 15 Apr 2025, 01:55, edited 1 time in total.
Edited the question.
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
51% (01:08) correct
49%
(01:23)
wrong
based on 225
sessions
History
Date
Time
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I. \(f(x) = \sqrt{(x/100)^2}\)
II. \(f(x) = 25x^{\frac{1}{2}}\)
III. \(f(x) = \frac{x^2-32}{(x-5) (x + 5)}\)
Which of the above functions has as its domain the set of all real numbers?
A. I only
B. III only
C. I and II only
D. I and III only
E. I II and III
1000003053.jpg [ 304.95 KiB | Viewed 1397 times ]
II. \(f(x) = 25x^{\frac{1}{2}}\)
III. \(f(x) = \frac{x^2-32}{(x-5) (x + 5)}\)
Which of the above functions has as its domain the set of all real numbers?
A. I only
B. III only
C. I and II only
D. I and III only
E. I II and III
Attachment:
1000003053.jpg [ 304.95 KiB | Viewed 1397 times ]
Updated on: 15 Apr 2025, 01:54
Originally posted by GMATinsight on 15 Apr 2025, 01:50.
Last edited by GMATinsight on 15 Apr 2025, 01:54, edited 1 time in total.
Last edited by GMATinsight on 15 Apr 2025, 01:54, edited 1 time in total.
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NextstopISB
CONCEPT:
1) A value becomes non-real if the value is NEGATIVE INSIDE SQUARE ROOT, such as √(-3) is non-real
2) Also, a function is NOT DEFINED when the denominator becomes zero such as 1/(x-2) is not defined for x = 2
1) A value becomes non-real if the value is NEGATIVE INSIDE SQUARE ROOT, such as √(-3) is non-real
2) Also, a function is NOT DEFINED when the denominator becomes zero such as 1/(x-2) is not defined for x = 2
I. \(f(x) = √(\frac{x}{100})^2\)
This function is valid for positive as well as Negative values of x as squaring the negative value results in a positive value under the square root.
hence, it is DEFINED for all Real values of x
II. \(f(x) = √[25x]\)
this function will result in a non-real value for any negative value of x hence it is NOT DEFINED for all real values of x
III. \(f(x) = \frac{(x^2-32)}{((x-5) (x + 5))}\)
This is not defined for x = 5 and x=-5 for which the denominator becomes zero
Hence, Answer: Option A
Related Concept Video:
15 Apr 2025, 01:54
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Thank you for the youtube video explanation
GMATinsight
