set 4 Q 23

[radhika1306]

If x &#8800; 0, is x^2/|x| < 1?
(1) x < 1
(2) x > &#8722;1

Can some one explain how |x| works

[Bharat]

Two thumb rules for modulus function:
# 1: x^2 = |x|^2
# 2: x^1/2 = |x|
# 3: |x| < y is equivalent to -y < x < y

Solution:
x^2/|x| = |x^2/x| = |x| -------------(since x is non zero)

hence, the given condition is: |x| < 1
or - 1 < x < 1

so both 1 & 2 together are sufficient.

[radhika1306]

correct thanks

[samirpandeyit62]

IMO C

both alone are insufficient

combine then x can between -1 & 0 or 0 & 1

in x^2/|x| since both x^2 & |x| are positive

so any value between 0 & -1 e.g -.9 or 0 & 1 eg 0.4

will give a result as the same value for this expression

hence the answer should be C