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here u go...

by sportypk » Sat May 31, 2008 12:01 pm
I m fairly new here posting thanks for the headsup.

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

OA is D

Could someone explain this
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by netigen » Sat May 31, 2008 12:57 pm
Question rephrase: is -1<x<1 ?

(a) this is straightforward as x will be between -1 and 1 so suff

(b) |x| < 1/x
when x = +ve
x < 1/x or x^2 < 1 so sufficient
when x = -ve
-x < 1/x or -x^2 > 1 or x^2 < -1 this is a not possible case hence reject
so B is also sufficient.

I think plugging numbers would have been an easier approach in this case
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