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If 1 < x < 0, which of the following is greatest? x x3 1 x 1 x3 1
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29 Aug 2024, 02:06
29 Aug 2024, 02:06
Expert Reply
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Question Stats:
60% (01:04) correct
40% (00:53) wrong based on 10 sessions
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If \(–1 < x < 0\), which of the following is greatest?
A. \(x\)
B. \(x^3\)
C. \(1 – x
\)
D. \(1 – x^3\)
E. \(1 – x^4\)
A. \(x\)
B. \(x^3\)
C. \(1 – x
\)
D. \(1 – x^3\)
E. \(1 – x^4\)
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Re: If 1 < x < 0, which of the following is greatest? x x3 1 x 1 x3 1
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29 Aug 2024, 09:28
29 Aug 2024, 09:28
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C
Re: If 1 < x < 0, which of the following is greatest? x x3 1 x 1 x3 1
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10 Dec 2024, 15:15
10 Dec 2024, 15:15
Expert Reply
We know $\(-1<x<0\)$; we need to check that which of the given options is greatest.
As x is negative the value of \($\mathrm{x}^{\text {odd \)$ will also be negative and also as x lies between -1 and 0 , the higher odd power will result in higher negative value.
Out of the given options (A) \& (B) will be negative so cannot be the greatest, so the comparison remains among the remaining three options.
We know $\(\mathrm{x}^3>\mathrm{x}\)$ but $\(-\mathrm{x}^3<-\mathrm{x} \Rightarrow 1-\mathrm{x}^3<1-\mathrm{x}\)$, so option (C) is greater than option (D).
Finally we need to compare option (C) with option (E).
$\(\mathrm{x}^4>\mathrm{x}\)$ but $\(-\mathrm{x}^4<-\mathrm{x} \Rightarrow 1-\mathrm{x}^4<1-\mathrm{x}\)$ which means $\(1-\mathrm{x}\)$ is the greatest.
Hence the answer is (C).
As x is negative the value of \($\mathrm{x}^{\text {odd \)$ will also be negative and also as x lies between -1 and 0 , the higher odd power will result in higher negative value.
Out of the given options (A) \& (B) will be negative so cannot be the greatest, so the comparison remains among the remaining three options.
We know $\(\mathrm{x}^3>\mathrm{x}\)$ but $\(-\mathrm{x}^3<-\mathrm{x} \Rightarrow 1-\mathrm{x}^3<1-\mathrm{x}\)$, so option (C) is greater than option (D).
Finally we need to compare option (C) with option (E).
$\(\mathrm{x}^4>\mathrm{x}\)$ but $\(-\mathrm{x}^4<-\mathrm{x} \Rightarrow 1-\mathrm{x}^4<1-\mathrm{x}\)$ which means $\(1-\mathrm{x}\)$ is the greatest.
Hence the answer is (C).
