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If x < 0 and 0 < y < 1, which of the following has the greatest value?
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30 Aug 2024, 03:14
30 Aug 2024, 03:14
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50% (01:39) correct
50% (01:52) wrong based on 18 sessions
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If x < 0 and 0 < y < 1, which of the following has the greatest value?
A. \(x^2\)
B. \((xy)^2\)
C. \((\frac{x}{y})^2\)
D. \(\dfrac{x^2}{y^2}\)
E. \(x^2y\)
A. \(x^2\)
B. \((xy)^2\)
C. \((\frac{x}{y})^2\)
D. \(\dfrac{x^2}{y^2}\)
E. \(x^2y\)
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Re: If x < 0 and 0 < y < 1, which of the following has the greatest value?
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30 Aug 2024, 10:16
30 Aug 2024, 10:16
It can be seen that x^2 is a multiplicative term in all options, so we can just ignore it. So, basically it boils down to comparing 1,y^2, y^-2 and y.
Since, 0<y<1, y^-2 > 1 and hence it is the greatest (all others are less than 1). Hence, (x/y)^2 is the greatest. Hence, C is correct.
C and D are the same expression, so D should also be correct. Hence, imo, C and D are correct options.
Since, 0<y<1, y^-2 > 1 and hence it is the greatest (all others are less than 1). Hence, (x/y)^2 is the greatest. Hence, C is correct.
C and D are the same expression, so D should also be correct. Hence, imo, C and D are correct options.
Re: If x < 0 and 0 < y < 1, which of the following has the greatest value?
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10 Dec 2024, 12:52
10 Dec 2024, 12:52
Expert Reply
We know $\(-1<\mathrm{x}<0\)$; 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).
