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Given two equations: x2 9 0 and y2 16 > 0. Which of the followin
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19 Sep 2024, 01:27
19 Sep 2024, 01:27
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Given two equations: x2 – 9 ≤ 0 and y2 – 16 > 0. Which of the following options are true about xy, if x and y are integers?
A. The least positive integral value of xy is 4
B. The least positive integral value of xy is 5
C. The highest negative integral value of xy is –5
D. The least negative integral value of xy is –5
A. The least positive integral value of xy is 4
B. The least positive integral value of xy is 5
C. The highest negative integral value of xy is –5
D. The least negative integral value of xy is –5
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Re: Given two equations: x2 9 0 and y2 16 > 0. Which of the followin
[#permalink]
11 Dec 2024, 10:24
11 Dec 2024, 10:24
Expert Reply
We know $\(x^2-9 \leq 0 \& y^2-16>0\)$; we need to check from the options that which of them is true.
The equation $\(x^2-9 \leq 0\)$ is same as $\(x^2 \leq 9 \Rightarrow-3<x<3\)$ whereas the second equation $\(y^2-16>0 \Rightarrow y>4\)$ or $\(y<-4\)$
The minimum positive integral value of the product of $\(x\)$ and $\(y\)$ will come if we take either $x$ and $y$ both negative or both positive. The minimum possible negative values of $\(x\)$ and $\(y\)$ are -2 and -3 , so we get the product $\(\mathrm{xy}=-2 \times-3=6 \&\)$ the minimum possible positive values of x and y are 1 and 5 respectively, so we get the product $\(x y=1 \times 5=5\)$.
So, the minimum positive integral value of the product of x and y is 5 .
Similarly to find the maximum negative integral value of $\(x\)$ and $\(y\)$, we would consider exactly one of $\(x\)$ and $\(y\)$ negative and take their minimum possible values. If we take $\(x=-1\)$ and $\(y=5\)$, we get the product $\(\mathrm{xy}=-1 \times 5=-5 \&\)$ if we take $\(\mathrm{x}=1\)$ and $\(\mathrm{y}=-5\)$, we get $\(\mathrm{xy}=1 \times-5=-5\)$
So, the maximum possible negative integral value of the product of x and y is -5 .
Hence options (B) \& (C) are correct.
The equation $\(x^2-9 \leq 0\)$ is same as $\(x^2 \leq 9 \Rightarrow-3<x<3\)$ whereas the second equation $\(y^2-16>0 \Rightarrow y>4\)$ or $\(y<-4\)$
The minimum positive integral value of the product of $\(x\)$ and $\(y\)$ will come if we take either $x$ and $y$ both negative or both positive. The minimum possible negative values of $\(x\)$ and $\(y\)$ are -2 and -3 , so we get the product $\(\mathrm{xy}=-2 \times-3=6 \&\)$ the minimum possible positive values of x and y are 1 and 5 respectively, so we get the product $\(x y=1 \times 5=5\)$.
So, the minimum positive integral value of the product of x and y is 5 .
Similarly to find the maximum negative integral value of $\(x\)$ and $\(y\)$, we would consider exactly one of $\(x\)$ and $\(y\)$ negative and take their minimum possible values. If we take $\(x=-1\)$ and $\(y=5\)$, we get the product $\(\mathrm{xy}=-1 \times 5=-5 \&\)$ if we take $\(\mathrm{x}=1\)$ and $\(\mathrm{y}=-5\)$, we get $\(\mathrm{xy}=1 \times-5=-5\)$
So, the maximum possible negative integral value of the product of x and y is -5 .
Hence options (B) \& (C) are correct.
gmatclubot
Re: Given two equations: x2 9 0 and y2 16 > 0. Which of the followin [#permalink]
11 Dec 2024, 10:24
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