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.

My bad. Have edited it.

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.

|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

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.