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Let's compare \(x^2\) and \(x^{-2}\) as both are going to be positive... But |X|<1, so x^2<1.... .. As x^2 is positive, divide both sides by x^2, so \(\frac{x^2}{x^2}\)<\(\frac{1}{x^2}\)........\(\frac{1}{x^2>1}\)........\(x^{-2}>1\) So \(x^2,1,x^{-2}\) , x is negative and reciprocal of x.......\(x^{-1}<x\)
Re: If -1<x<0, Which of the following is listed in ascending order?
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21 Oct 2018, 17:57
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We can find the correct order by plugging in a value for \(x\). Suppose \(x =\frac{-1}{2.}\) Then \(-1 < x < 0,\) and \(x^2 = \frac{1}{4}\), \(x^{-1} = -2\) and \(x^{-2} = 4.\) Thus, \(x^{-1} < x < x^2 < 1 < x^{-2}\)
Therefore, the answer is A. Answer: A
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53% (01:30) correct 
