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05 Feb 2019, 04:19
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Difficulty:
55%
(hard)
Question Stats:
20% (00:18) correct
80%
(01:37)
wrong
based on 5
sessions
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If \(\frac{1}{y}< x < 0\), then which one of the following must be true?
A. \(1 < x^2\)
B. \(x^2 < x\)
C. \(−1 < x^3 < 0\)
D. \(\frac{1}{x} > -1\)
E. \(x^3 < x\)
A. \(1 < x^2\)
B. \(x^2 < x\)
C. \(−1 < x^3 < 0\)
D. \(\frac{1}{x} > -1\)
E. \(x^3 < x\)
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Archit3110
Major Poster
05 Feb 2019, 06:14
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Bunuel
good question :
I am not sure of my soln.. given
\(\frac{1}{y}< x < 0\)
we can say
1/xy<1<0 --- ( a)
or 1<xy<0---( b)
or we can say that from (a) xy is -ve and from (b) xy lies between 1 & 0 and is +ve
only option which I think satisfies these two parameters is option B
IMO B
05 Feb 2019, 06:32
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Bunuel
A. \(1 < x^2......x^2-1>0.....(x-1)(x+1)>0\) Now, x-1<0 since x<0, therefore, x+1<0 or x<-1.. But x could be anything <0. say y is -2, so x>-1/2
B. \(x^2 < x......\) This means x>0.. Never true, as x<0
C. \(−1 < x^3 < 0......-1<x<0\).... Say y = -1/8, so x>1/(-1/8) or x>-8.. Thus x could be -7, -3 and so on.
D. \(\frac{1}{x} > -1.........\frac{x+1}{x} > 0\)...... x<0, so x+1<0 or x<-1.. We have seen in A, need not be true
E. \(x^3 < x\) this is same as \(x^2>1\), multiply it by negative x \(x^3<x\).. same as A.. need not be true.
or (x^3-x)<0......x(x^2-1)<0... x<0, so x^2-1>0, that is x^2>1 or x<-1...
None seems to be true. May be there is some restriction on variables x and y. Bunuel
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