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03 Feb 2021, 01:15
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B
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Difficulty:
65%
(hard)
Question Stats:
53% (01:45) correct
47%
(01:32)
wrong
based on 181
sessions
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Is \(g< g^2 < g^3\), for all real values of \(g\)?
(1) \(g < g^3\)
(2) \(g^3 > g^2\)
Are You Up For the Challenge: 700 Level Questions
(1) \(g < g^3\)
(2) \(g^3 > g^2\)
Are You Up For the Challenge: 700 Level Questions
03 Feb 2021, 05:36
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Bunuel
What can we make out of \(g<g^2<g^3\)?
a) g cannot be 0 as all values will be equal or
b) g cannot be negative because \(g^3>g^2\) or
c) g cannot be less than or equal to 1 as the number becomes lesser as the power increases.
Thus we are looking at whether g>1.
(1) \(g < g^3\)
If g is positive, then surely g>1.
If g is negative and >-1, then surely g<1.
So not sufficient
OR
\(g^3-g>0.......g(g^2-1)>0\)
Case I g>0, then \(g^2-1>0 \) or g>1.....Yes, g is greater than 1.
Case II g<0, then \(g^2-1<0......g^2<1......-1<g<0\)......No g is not greater than 1.
(2) \(g^3 > g^2\)
\(g^3-g^2>0.......g^2(g-1)>0\)
Only one case possible as g^2 >0.
Thus, \(g-1>0 \) or g>1.....Yes, g is greater than 1.
When a number greater than 1 is raised to higher powers, the value becomes greater, that is => \(g^3>g^2>g\)
Sufficient
B
03 Feb 2021, 11:48
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Is g< g^2 < g^3, for all real values of g?
It will be true if g>1.
Stat1: g < g^3
g*(g^2 - 1) >0 or, g<-1 or g>1, so, g^2>g >g^3 or, g< g^2 < g^3. Not sufficient.
Stat2: g^3 > g^2
g^2 *(g-1)>0 or, g>1, so, g< g^2 < g^3. Sufficient.
So, I think B.
It will be true if g>1.
Stat1: g < g^3
g*(g^2 - 1) >0 or, g<-1 or g>1, so, g^2>g >g^3 or, g< g^2 < g^3. Not sufficient.
Stat2: g^3 > g^2
g^2 *(g-1)>0 or, g>1, so, g< g^2 < g^3. Sufficient.
So, I think B.
