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28 Aug 2021, 05:12
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
55% (01:26) correct
45%
(01:33)
wrong
based on 92
sessions
History
Date
Time
Result
Not Attempted Yet
28 Aug 2021, 07:26
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sjuniv32
Statement 1:
Divide x^4 on both sides.
\(x^3 < x\)
We test four intervals created from (1) Magnitude >1 or <1, and (2) Positive vs negative. Plugging in x = -2 and x = 0.5 will satisfy this inequality which means x < -1 or 0 < x < 1 is the solution. Insufficient.
Statement 2:
Divide x^4 on both sides.
\(x < x^4\)
Again try testing the 4 intervals. x > 1 or x < 0 works. Insufficient.
Combined:
Combined, they only share the interval x < -1. Sufficient.
Ans: C
19 Sep 2021, 10:50
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sjuniv32
sjuniv32
Number line
If 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
____________________________________xx________________________________________________________
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
