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22 May 2003, 00:57
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
64% (00:54) correct
36%
(01:00)
wrong
based on 508
sessions
History
Date
Time
Result
Not Attempted Yet
06 Jun 2014, 01:41
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S1
After looking at the equations that this equations con only hold true for 1, -1.
S1 holds true for both.
S2
This equation is true only when F=1.
So, the integer that we are looking for is 1.
Hence, B.
After looking at the equations that this equations con only hold true for 1, -1.
S1 holds true for both.
S2
This equation is true only when F=1.
So, the integer that we are looking for is 1.
Hence, B.
06 Jun 2014, 02:01
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stolyar
Find an integer F.
(1) F = 1/F
(2) |F| = 1/F
(1) F = 1/F
(2) |F| = 1/F
St:1 F=1/F or F- 1/F =0
F^2- 1/F=0, Now F cannot be equal to 0
So F^2-1= 0 or F =+1 or -1
Not sufficient
St 2 |F|=1/F
Now LHS is mod value and hence always non-negative. Therefore \(|F|\geq{0}\) or \(1/F\geq{0}\)
Note that F cannot be equal to zero as otherwise the expression becomes undefined.
Therefore |F| >0 or 1/F >0 -----Thus F>0
Since F>0 So we have F= 1/F or F^2=1 and F =1 as F >0
Ans is B
C is Trap answer here
