sjuniv32 wrote:
Is x < 0 ?
(1) \(x^7 < x^5\)
(2) \(x^5 < x^8\)
sjuniv32Number lineIf you want to make use of number properties
A) x>1….As power increases, value of x increases. => \(x^3>x^2>x\)
B) 0<x<1….As power increases, value of x decreases. => \(x^3<x^2<x\)
C) -1<x<0….As odd power increases, value of x increases, but as even power increases, the value of x decreases but remains more than odd power => \(x^2>x^4>x^3>x\)
D) x<-1….As odd power increases, value of x decreases, but as even power increases, the value of x increases and remains more than odd power always => \(x^4>x^2>x>x^3>x^5\)
(1) \(x^7 < x^5\)
Cases B and D
Insufficient
(2) \(x^5 < x^8\)
Cases B and D
Insufficient
Combined
Case D. x<-1
Sufficient
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Algebraic(1) \(x^7 < x^5……..x^7-x^5<0……..x^5(x^2-1)<0\)
Two cases
a) \(x^5<0 \ \ or \ \ x<0\), then \(x^2-1>0…..|x|>1, \ \ that \ \ is \ \ x>1 \ \ or \ \ x<-1\). As x<0, x<-1.
b) \(x^5>0 \ \ or \ \ x>0\), then \(x^2-1<0…..|x|<1, \ \ that \ \ is \ \ -1<x<1\). As x>0, x>1.
Insufficient
(2) \(x^5 < x^8……….x^8-x^5>0……x^5(x^3-1)>0.\)
Two cases
a) \(x^5<0 \ \ or \ \ x<0\), then \(x^3-1<0…….x^3<1………x<1\). As x<0, the range is x<0.
b) \(x^5>0 \ \ or \ \ x>0\), then \(x^3-1>0…..x^3>1…. x>1\). As x>0, x>1.
Insufficient
Combined
1) x<0…..statement I gives x<-1 and statement II gives x<0. Common range x<-1.
2) x>0…..statement I gives 0<x<1 and statement II gives x>1. NO Common range , so not possible .
Thus, x<-1<0.
Sufficient
C