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31 May 2024, 01:30
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D
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Difficulty:
45%
(medium)
Question Stats:
66% (01:42) correct
34%
(01:58)
wrong
based on 297
sessions
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If \(\frac{x}{(x – 2)(x + 1)} ≤ 0\), which of the following specifies all the possible values of x ?
A. x < –1
B. 0 ≤ x < 2
C. x ≤ –1 and 0 ≤ x < 2
D. x < –1 and 0 ≤ x < 2
E. x ≤ –1 and x ≥ 2
A. x < –1
B. 0 ≤ x < 2
C. x ≤ –1 and 0 ≤ x < 2
D. x < –1 and 0 ≤ x < 2
E. x ≤ –1 and x ≥ 2
31 May 2024, 02:42
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\(\frac{x}{(x-2)(x+1)}≤0\)
\(x = 2\) and \(x = -1\) will result in a denominator of zero, which will yield an answer that is undefined. One can plot these values, along with 0, on a number line as points of change.
-1 0 2
------|----------|----------|------
-1 0 2
A value that is higher than 2: using \(x = 3\) will yield \(\frac{3}{4}\). Therefore \(x>2\) will always be positive.
A value that is between 0 and 2: using \(x = 1\) will yield \(-\frac{1}{2}\). Therefore \(0<x<2\) will be negative.
x = 0: Will yield \(0\), as zero divided by any non-zero number will result in zero.
A value that is between 0 and -1: using \(x = -\frac{1}{2}\) will yield \(\frac{4}{10}\). Therefore \(-1<x<0\) will be positive.
A value that is lower than -1: using \(-2\) will yield \(-\frac{1}{2}\). Therefore \(x<-1\) will always be negative.
From the above one knows that when \(-1<x\) and when \(0≤x<2\) the inequality \(\frac{x}{(x-2)(x+1)}≤0\) will hold.
ANSWER D
\(x = 2\) and \(x = -1\) will result in a denominator of zero, which will yield an answer that is undefined. One can plot these values, along with 0, on a number line as points of change.
-1 0 2
------|----------|----------|------
-1 0 2
A value that is higher than 2: using \(x = 3\) will yield \(\frac{3}{4}\). Therefore \(x>2\) will always be positive.
A value that is between 0 and 2: using \(x = 1\) will yield \(-\frac{1}{2}\). Therefore \(0<x<2\) will be negative.
x = 0: Will yield \(0\), as zero divided by any non-zero number will result in zero.
A value that is between 0 and -1: using \(x = -\frac{1}{2}\) will yield \(\frac{4}{10}\). Therefore \(-1<x<0\) will be positive.
A value that is lower than -1: using \(-2\) will yield \(-\frac{1}{2}\). Therefore \(x<-1\) will always be negative.
From the above one knows that when \(-1<x\) and when \(0≤x<2\) the inequality \(\frac{x}{(x-2)(x+1)}≤0\) will hold.
ANSWER D
31 May 2024, 03:22
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I think better way is going with options elimination here,
if we put value of x = -1 in given inequality we will get ratio in the form of -1/0 which is not defined. hence eliminate option C & E.
After this, lower the value of x to -2, Put x=-2 and solve you will see inequality is valid hence option A & D are possible for x<-1 Eliminate option B.
Now lets check for option D. other inequality 0 ≤ x < 2 you will see for x=0, x=1 and x=3/2 inequality is valid, hence we can safely mark option D as answer.
if we put value of x = -1 in given inequality we will get ratio in the form of -1/0 which is not defined. hence eliminate option C & E.
After this, lower the value of x to -2, Put x=-2 and solve you will see inequality is valid hence option A & D are possible for x<-1 Eliminate option B.
Now lets check for option D. other inequality 0 ≤ x < 2 you will see for x=0, x=1 and x=3/2 inequality is valid, hence we can safely mark option D as answer.
