If x≠0, is |x| <1?
(1) x2<1
(2) |x| < 1/x
OA: D
Please explain the working.
|x|
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Can you please complete the question? Second statement is incomplete.
For 1st statement, proceed like this:-
Question asked is |x| < 1, that means, x should lies between -1 and 1
Since x^2 < 1 therefore, x will be always in the range -1 to 1. Hence statement (1) is sufficient.
For 1st statement, proceed like this:-
Question asked is |x| < 1, that means, x should lies between -1 and 1
Since x^2 < 1 therefore, x will be always in the range -1 to 1. Hence statement (1) is sufficient.
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I think the 1st option is already explained in prev post.
lets take 2nd option
given |x| < 1/x , need to determine |x| <1
|x| < 1/x is never true for any -ve numbers.
|-100| < -1/100 ==> 100 < -1/100 (false)
try positive numbers ( 0 to 1, > 1)
for x>1 , say 5 == 5 < 1/5 false.
try 0 to 1, this holds true
when x belongs to (0,1) we can say |x| < 1.
lets take 2nd option
given |x| < 1/x , need to determine |x| <1
|x| < 1/x is never true for any -ve numbers.
|-100| < -1/100 ==> 100 < -1/100 (false)
try positive numbers ( 0 to 1, > 1)
for x>1 , say 5 == 5 < 1/5 false.
try 0 to 1, this holds true
when x belongs to (0,1) we can say |x| < 1.
- dumb.doofus
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You cannot multiply like what you have done in the part that I have highlighted above.. in inequalities, sign does matter. and without knowing the sign of x, you cannot multiply coz the inequality may change.maihuna wrote:|x| < 1/x
revers:
1/|x| >x
1>x|x|
if x<0, 1>-x^2 or x^2>-1 alwas true
x>0=> 1>x^2 x^2<1 so |x|<1
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