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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:
65%
(hard)
Question Stats:
36% (01:18) correct
64%
(01:30)
wrong
based on 66
sessions
History
Date
Time
Result
Not Attempted Yet
Is \(uv^5 < 0 \)
(1) \(uv < 0\)
(2) \(u + v^5 = 0\)
(1) \(uv < 0\)
(2) \(u + v^5 = 0\)
Updated on: 06 Feb 2024, 11:52
Originally posted by Shubhradeep on 05 Feb 2024, 13:59.
Last edited by Shubhradeep on 06 Feb 2024, 11:52, edited 1 time in total.
Last edited by Shubhradeep on 06 Feb 2024, 11:52, edited 1 time in total.
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Correct option should be A
we have to find out whether u\(v^{5}\) < 0
Statement 1:
uv<0
now, u\(v^{5}\) = uv*\(v^{4}\) ; now \(v^{4}\) is always >0
so, negative*positive = negative
Hence, u\(v^{5}\) < 0 Sufficient
Statement 2:
U+\(v^{5}\)=0
=> \(v^{5}\) = -u
=> u\(v^{5}\) = -u*u = -\(u^{2}\) which is always negative (negative of a square)
But as u=v=0 can also be a case for this statement we can't conclude that u\(v^{5}\) < 0 (in this case u\(v^{5}\)=0)
Hence, insufficient
So, only statement 1 is sufficient but not statement 2.
Hence, correct option is A
we have to find out whether u\(v^{5}\) < 0
Statement 1:
uv<0
now, u\(v^{5}\) = uv*\(v^{4}\) ; now \(v^{4}\) is always >0
so, negative*positive = negative
Hence, u\(v^{5}\) < 0 Sufficient
Statement 2:
U+\(v^{5}\)=0
=> \(v^{5}\) = -u
=> u\(v^{5}\) = -u*u = -\(u^{2}\) which is always negative (negative of a square)
But as u=v=0 can also be a case for this statement we can't conclude that u\(v^{5}\) < 0 (in this case u\(v^{5}\)=0)
Hence, insufficient
So, only statement 1 is sufficient but not statement 2.
Hence, correct option is A
05 Feb 2024, 19:12
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IMO -D
Each statement is sufficient to answer the question.
Each statement is sufficient to answer the question.
