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11 Feb 2023, 03:06
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If \(|x| ≠ 1\), is \(\frac{x}{x + 1} > \frac{-x}{1 - x}\) ?
(1) \(x > 1\)
(2) \(\frac{x}{x + 1} < 1\)
(1) \(x > 1\)
(2) \(\frac{x}{x + 1} < 1\)
11 Feb 2023, 12:33
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Is \(\frac{x}{(x+1)}>−\frac{x}{(1−x) }\)?
|x|≠1 implies x ≠ 1 or -1
(1) x > 1
We can check the above inequality for different values of x(say) = 1.5 or 2 or 10.
x = 1.5 . Inequality is \(\frac{1.5}{2.5} > \frac{-1.5}{(-0.5)}\) i.e., 0.6 > 3 which is not true. Similarly we can prove that for any value of x >1, the above inequality is false.
Hence statement 1 is sufficient
(2) \(\frac{x}{(x+1)}\)<1
this inequality is true for all values of x>-1
For ex - \(\frac{-0.5}{0.5 }< 1\) is true; 0/1 < 1 is true ; \(\frac{5}{6 }< 1\) is true.
we have already seen that for x >1 our main inequality is not satisfied. Now lets check for values of x between -1 and 1
x = -0.5
\(\frac{-0.5}{(0.5) }> \frac{-(-0.5)}{[1- (- 0.5)]}\)
= -1 > \(\frac{1}{3 }-\) Not true
x = 0.4
\(\frac{0.4}{1.4} > \frac{-0.4 }{ 0.6 }\)
\(\frac{2}{7}> - 2/3\) - True
therefore statement 2 is Not sufficient
Answer A
|x|≠1 implies x ≠ 1 or -1
(1) x > 1
We can check the above inequality for different values of x(say) = 1.5 or 2 or 10.
x = 1.5 . Inequality is \(\frac{1.5}{2.5} > \frac{-1.5}{(-0.5)}\) i.e., 0.6 > 3 which is not true. Similarly we can prove that for any value of x >1, the above inequality is false.
Hence statement 1 is sufficient
(2) \(\frac{x}{(x+1)}\)<1
this inequality is true for all values of x>-1
For ex - \(\frac{-0.5}{0.5 }< 1\) is true; 0/1 < 1 is true ; \(\frac{5}{6 }< 1\) is true.
we have already seen that for x >1 our main inequality is not satisfied. Now lets check for values of x between -1 and 1
x = -0.5
\(\frac{-0.5}{(0.5) }> \frac{-(-0.5)}{[1- (- 0.5)]}\)
= -1 > \(\frac{1}{3 }-\) Not true
x = 0.4
\(\frac{0.4}{1.4} > \frac{-0.4 }{ 0.6 }\)
\(\frac{2}{7}> - 2/3\) - True
therefore statement 2 is Not sufficient
Answer A
12 Feb 2023, 05:02
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Bunuel
\(\frac{x}{x + 1} > \frac{-x}{1 - x}\)
\(\frac{x}{x + 1} > \frac{-x}{-(x+1)}\)
\(\frac{x}{x + 1} > \frac{x}{(x+1)}\)
This is true for all values of x except for x = 0.
Hence, we can translate the question to "Is x = 0?"
Statement 1
(1) \(x > 1\)
\(x \neq 0,\) hence the statement is sufficient to answer the question.
Sufficient.
Statement 2
(2) \(\frac{x}{x + 1} < 1\)
\(\frac{x}{x + 1} < 1\)
\(\frac{x}{x + 1} - 1 < 0\)
\(\frac{x - x - 1}{x + 1} < 0\)
\(\frac{- 1}{x + 1} < 0\)
Therefore x + 1 > 0
x > -1.
x can be 0 or it can be any other number. Hence, the statement is not sufficient.
Option A
