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by anshulseth » Mon May 04, 2009 5:11 am
If x&#8800;0, is |x| <1?
(1) x2<1
(2) |x| < 1/x

OA: D

Please explain the working.
Last edited by anshulseth on Mon May 04, 2009 7:51 am, edited 1 time in total.
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by btg760 » Mon May 04, 2009 6:22 am
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.
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by anshulseth » Mon May 04, 2009 7:51 am
My bad. Have edited it.
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by iriijei.idwimd » Mon May 04, 2009 10:50 am
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.
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by maihuna » Mon May 04, 2009 11:11 am
|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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by dumb.doofus » Mon May 04, 2009 12:14 pm
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
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.
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