If x ≠ 0, is x^2/|x| < 1?
(1) x < 1
(2) x > −1
Can some one explain how |x| works
set 4 Q 23
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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.
# 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.
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- Master | Next Rank: 500 Posts
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- Joined: Fri Apr 13, 2007 2:25 am
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- Master | Next Rank: 500 Posts
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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
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
Regards
Samir
Samir