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26 Mar 2016, 15:24
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Thank you for the replies!
Sorry, but something is still not 100% clear to me, so the meaning of "undefined" is not like "can't possibly be done, so don't even consider it" but more like a valid answer? Isn't undefined a sort of a dead end? Or is it just "correct" to write X#0 with no particular practical reason?
Sorry, but something is still not 100% clear to me, so the meaning of "undefined" is not like "can't possibly be done, so don't even consider it" but more like a valid answer? Isn't undefined a sort of a dead end? Or is it just "correct" to write X#0 with no particular practical reason?
26 Mar 2016, 15:39
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iliavko
The question asks: is x^2/|x| < 1 if -1 < x < 1. Now, if x is any number but 0 from -1 to 1, then the answer is YES. But if x is 0, then we cannot answer the question, because if x = 0 , then x^2/|x| is undefined.
17 Sep 2018, 10:28
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Hi,
here are my two cents for this question.
This looks long but actually i did solve in 45 secs, but took another 15 secs because of the range that combined statement 1 and 2 gave.
If you are not comfortable by reducing the inequality , though Bunuel has superbly explained it we can also think in terms of this approach.
we are given that x\(\neq{0}\)
IS \(\frac{x^2}{|x|} <1\) ?
lets try to understand what we are being asked by trying to rephrase the question
we can write this as
\(\frac{x^2}{|x|} -1 <0\)
now
\(\frac{x^2-|x|} {|x|}<0\)
Since |x| is positive we can multiply both sides by |x| to get rid of it in denominator
\({x^2-|x|}\) <0 ------(a)
now for |x|= x if x \(\geq{0}\)
|x|= - x if x <0
so above equation (a) for x\(\geq{0}\) becomes
\({x^2-x}\) <0
if we plot it on number line we will have
0<x<1
and for x<0 the equation (a) becomes
\({x^2 +x}\) <0
if we plot it on number line we will have
-1<x<0
combining we get that inequality hold if
-1<x<1 except for x\(\neq{0}\)
So question is actually asking this does x lie in this range for this inequality
Now if we see statement 1 and 2 we can say they are clearly insufficient . Combining 1 and 2 we get
-1<x<1
But since we are given that x\(\neq{0}\) so in this range ) is not considered as apart of the values of x for which the inequality holds.
PS : There can be several ways to solve the question, but if you follow one process for all questions there are chances you master that technique and solve quick enough to mark the answer confidently. No point in knowing multiple techniques and mastering none.
Probus
here are my two cents for this question.
This looks long but actually i did solve in 45 secs, but took another 15 secs because of the range that combined statement 1 and 2 gave.
If you are not comfortable by reducing the inequality , though Bunuel has superbly explained it we can also think in terms of this approach.
we are given that x\(\neq{0}\)
IS \(\frac{x^2}{|x|} <1\) ?
lets try to understand what we are being asked by trying to rephrase the question
we can write this as
\(\frac{x^2}{|x|} -1 <0\)
now
\(\frac{x^2-|x|} {|x|}<0\)
Since |x| is positive we can multiply both sides by |x| to get rid of it in denominator
\({x^2-|x|}\) <0 ------(a)
now for |x|= x if x \(\geq{0}\)
|x|= - x if x <0
so above equation (a) for x\(\geq{0}\) becomes
\({x^2-x}\) <0
if we plot it on number line we will have
0<x<1
and for x<0 the equation (a) becomes
\({x^2 +x}\) <0
if we plot it on number line we will have
-1<x<0
combining we get that inequality hold if
-1<x<1 except for x\(\neq{0}\)
So question is actually asking this does x lie in this range for this inequality
Now if we see statement 1 and 2 we can say they are clearly insufficient . Combining 1 and 2 we get
-1<x<1
But since we are given that x\(\neq{0}\) so in this range ) is not considered as apart of the values of x for which the inequality holds.
PS : There can be several ways to solve the question, but if you follow one process for all questions there are chances you master that technique and solve quick enough to mark the answer confidently. No point in knowing multiple techniques and mastering none.
Probus
