set 4 Q 23

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set 4 Q 23

by radhika1306 » Tue Oct 16, 2007 2:10 pm
If x &#8800; 0, is x^2/|x| < 1?
(1) x < 1
(2) x > &#8722;1

Can some one explain how |x| works

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by Bharat » Tue Oct 16, 2007 10:29 pm
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.

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by radhika1306 » Wed Oct 17, 2007 4:33 am
correct thanks :)

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by samirpandeyit62 » Wed Oct 17, 2007 4:39 am
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
Regards
Samir