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If -1<x<0<y<1, which of the following inequalities must be tr
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01 Sep 2025, 02:11
01 Sep 2025, 02:11
Expert Reply
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Question Stats:
33% (02:00) correct
66% (01:59) wrong based on 6 sessions
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If $\(-1<x<0<y<1\)$, which of the following inequalities must be true?
Indicate all such inequalities.
(A) $\(x+y>0\)$
(B) $\(x y<0\)$
(C) $\(x^2+y^2>x+y\)$
(D) $\(2 x+x^2<y\)$
Indicate all such inequalities.
(A) $\(x+y>0\)$
(B) $\(x y<0\)$
(C) $\(x^2+y^2>x+y\)$
(D) $\(2 x+x^2<y\)$
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Re: If -1<x<0<y<1, which of the following inequalities must be tr
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02 Oct 2025, 12:27
02 Oct 2025, 12:27
why is C not correct?
Re: If -1<x<0<y<1, which of the following inequalities must be tr
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02 Oct 2025, 14:53
02 Oct 2025, 14:53
Expert Reply
Since $y$ is greater than $O$ but less than 1 , it is certainly the case that $\(y>y^2\)$. On the other hand, since $x$ is negative, it's also the case that $\(x<x^2\)$. .
So we can't know whether the inequality in this answer choice is true or not: it depends on whether the difference between $\(y\)$ and $\(y^2\)$ is greater than the difference between $\(x\)$ and $\(x^2\)$.
For example: if $x\(=-\frac{1}{2}\)$ and $\(y=\frac{3}{4}\)$, then $\(x^2+y^2=\frac{13}{16}>x+y=\frac{1}{4}\)$.
True.
But if $\(x=-\frac{1}{8}\)$ and $\(y=\frac{1}{2}\)$, then
$\(x^2+y^2=\frac{17}{64}<x+y=\frac{3}{8}\)$. False.
So we can't know whether the inequality in this answer choice is true or not: it depends on whether the difference between $\(y\)$ and $\(y^2\)$ is greater than the difference between $\(x\)$ and $\(x^2\)$.
For example: if $x\(=-\frac{1}{2}\)$ and $\(y=\frac{3}{4}\)$, then $\(x^2+y^2=\frac{13}{16}>x+y=\frac{1}{4}\)$.
True.
But if $\(x=-\frac{1}{8}\)$ and $\(y=\frac{1}{2}\)$, then
$\(x^2+y^2=\frac{17}{64}<x+y=\frac{3}{8}\)$. False.
