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25 Jul 2024, 00:36
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
55% (01:23) correct
44% (01:15) wrong based on 29 sessions
History
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Not Attempted Yet
If \(x<y<0\), which of the following inequalities must be true?
A. \(y+1<x\)
B. \(y-1<x\)
C. \(xy^2<x\)
D. \(xy<y^2\)
E. \(xy<x^2\)
A. \(y+1<x\)
B. \(y-1<x\)
C. \(xy^2<x\)
D. \(xy<y^2\)
E. \(xy<x^2\)
07 Sep 2024, 22:22
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In all Must be true Qs - our objective has to disprove the option to eliminate them. Whenever we cant disprove - that is the answer
If x<y<0 which of the following inequalities must be true?
We understand from stem both x and y are negative. They can be integers/fractions - we cant say.
A. y+1<x = No, if y = -1 and x = -5 then -1+ 1 > -5 Hence not the option
B. y−1<x = No if y = -1 and x = -5 then -1 - 1 > -5 Hence not the option
C. xy^2<x = No, if y = -1 and x = -5 then - 5* 1 = -5 Hence not the option
D. xy<y^2 = No, if y = -1 and x = -5 then 5 > 1 Hence not the option
E. xy<x2 Correct
We can also prove it algebraic way
x<y - Given
Multiplying a negative number on both sides changes sing of inequality - so multiply by x on both sides
x^2>xy
If x<y<0 which of the following inequalities must be true?
We understand from stem both x and y are negative. They can be integers/fractions - we cant say.
A. y+1<x = No, if y = -1 and x = -5 then -1+ 1 > -5 Hence not the option
B. y−1<x = No if y = -1 and x = -5 then -1 - 1 > -5 Hence not the option
C. xy^2<x = No, if y = -1 and x = -5 then - 5* 1 = -5 Hence not the option
D. xy<y^2 = No, if y = -1 and x = -5 then 5 > 1 Hence not the option
E. xy<x2 Correct
We can also prove it algebraic way
x<y - Given
Multiplying a negative number on both sides changes sing of inequality - so multiply by x on both sides
x^2>xy
13 Nov 2024, 00:19
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OE
The conditions stated in the question, \(x<y<0\) , tell you that x and y are negative numbers and that y is greater than x. Keep this in mind as you evaluate each of the inequalities in the answer choices, to see whether the inequality must be true.
Choice A: \(x+1<x,\) According to the conditions given in the question, y is greater than x. Since y is greater than x and y + 1 is greater than y, it follows that y + 1 is greater than x. So it cannot be true that . Therefore, Choice A is not the correct answer.
Choice B: \(y-1<x\). While it is true that both x and y - 1 are less than y, it may not be true that \(y-1-<x\) x. Consider what happens if y = -2 and x = -7. In this case, the inequality \(y-1<x\) becomes \(-3<-7,\) which is false. Therefore, Choice B is not the correct answer.
Choice C: Note that \(y^2\) is positive and x is negative, so \(xy^2\) is negative. Is the negative number \(xy^2\) less than the negative number x? It depends on whether \(y^2>1\) or \(y^2<1\). Consider what happens if x = -4 and y = \(-\frac{1}{2}\), where \(y^2<1\). In this case, the inequality \(xy^2<x\) becomes \(-1<-4\), which is false. Therefore, Choice C is not the correct answer.
Choice D: Since y is a negative number, multiplying both sides of the inequality \(x<y\) by y reverses the inequality, resulting in the inequality \(xy>y^2\). So it cannot be true that \(xy>y^2\) . Therefore, Choice D is not the correct answer.
Since Choices A through D have been eliminated, the correct answer is Choice E.
Choice E: You can show that the inequality in Choice E must be true as follows: Multiply both sides of the given inequality \(x<y\) by x to obtain the inequality \(x^2<xy\), reversing the direction of the inequality because x is negative. Therefore, the inequality \(xy<y^2\) must be true, and the correct answer is Choice E.
The conditions stated in the question, \(x<y<0\) , tell you that x and y are negative numbers and that y is greater than x. Keep this in mind as you evaluate each of the inequalities in the answer choices, to see whether the inequality must be true.
Choice A: \(x+1<x,\) According to the conditions given in the question, y is greater than x. Since y is greater than x and y + 1 is greater than y, it follows that y + 1 is greater than x. So it cannot be true that . Therefore, Choice A is not the correct answer.
Choice B: \(y-1<x\). While it is true that both x and y - 1 are less than y, it may not be true that \(y-1-<x\) x. Consider what happens if y = -2 and x = -7. In this case, the inequality \(y-1<x\) becomes \(-3<-7,\) which is false. Therefore, Choice B is not the correct answer.
Choice C: Note that \(y^2\) is positive and x is negative, so \(xy^2\) is negative. Is the negative number \(xy^2\) less than the negative number x? It depends on whether \(y^2>1\) or \(y^2<1\). Consider what happens if x = -4 and y = \(-\frac{1}{2}\), where \(y^2<1\). In this case, the inequality \(xy^2<x\) becomes \(-1<-4\), which is false. Therefore, Choice C is not the correct answer.
Choice D: Since y is a negative number, multiplying both sides of the inequality \(x<y\) by y reverses the inequality, resulting in the inequality \(xy>y^2\). So it cannot be true that \(xy>y^2\) . Therefore, Choice D is not the correct answer.
Since Choices A through D have been eliminated, the correct answer is Choice E.
Choice E: You can show that the inequality in Choice E must be true as follows: Multiply both sides of the given inequality \(x<y\) by x to obtain the inequality \(x^2<xy\), reversing the direction of the inequality because x is negative. Therefore, the inequality \(xy<y^2\) must be true, and the correct answer is Choice E.
