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08 May 2021, 15:55
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akhil911
(1) a neg. This means we have two choices range =a^2-a^5 (if a is a negative integer), range =a^2-a (if a is a negative fraction)
NS
(2) a^5 <a. this means either a is a negative integer or a positive fraction NS
(1) and (2) A is negative and (either neg integer or pos fract) so a is neg integer so range = a^2-a^5 sufficient OA is C
23 Sep 2021, 23:16
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2) a^5<a for which range would be different for the cases
(i) [b]if 0 < a <1
(ii) if a < -1 ; So,not sufficient[/b]
How do you get this range?
Can someone elaborate. Bunuel GMATBusters chetan2u VeritasKarishma
I am getting a different range.
a<0 or a<1 or a<1 or a^2<-1
Not sure what am I doing wrong.
(i) [b]if 0 < a <1
(ii) if a < -1 ; So,not sufficient[/b]
How do you get this range?
Can someone elaborate. Bunuel GMATBusters chetan2u VeritasKarishma
I am getting a different range.
a<0 or a<1 or a<1 or a^2<-1
Not sure what am I doing wrong.
24 Sep 2021, 07:47
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homersimpsons
Hi
\(a^5<a……….. a^5-a<0………..a(a^4-1)<0……a(a-1)(a+1)(a^2+1)<0\)
Now a^2+1>0, so we are left with \(a(a-1)(a+1)<0\)
Two cases
1) -*-*- is -.
So all three a, a-1 and a+1 are negative. Thus, let’s check the largest => a+1<0……a<-1
2) -*-*- is -.
So only one of a, a-1 and a+1 is negative. That will happen when only the smallest is negative => a-1<0……a<1.
But the second largest is positive. => a>0
Range 0<a<1
