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16 Sep 2014, 00:11
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Is \(x\) greater than 1?
(1) \(\frac{1}{x} \gt -1\)
(2) \(\frac{1}{x^5} \gt \frac{1}{x^3}\)
(1) \(\frac{1}{x} \gt -1\)
(2) \(\frac{1}{x^5} \gt \frac{1}{x^3}\)
16 Sep 2014, 00:11
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Official Solution:
Is \(x\) greater than 1?
(1) \(\frac{1}{x} \gt - 1\).
Multiply \(\frac{1}{x} \gt - 1\) by \(x^2\) (we can safely do that because a nonzero number in even power is positive) to get \(x > -x^2\):
\(x +x^2 > 0\);
\(x(1 + x) > 0\).
Two cases:
a. Both multiples are positive, (\(x > 0\) and \(x + 1 > 0\)), so \(x > 0\);
b. Both multiples are negative, (\(x < 0\) and \(x + 1 < 0\)), so \(x < -1\);
So, \(\frac{1}{x} \gt - 1\) is true when \(x < -1\) or when \(x > 0\). Thus, \(x\) may or may not be greater than 1. Not sufficient.
(2) \(\frac{1}{x^5} > \frac{1}{x^3}\).
Multiply \(\frac{1}{x^5} > \frac{1}{x^3}\) by \(x^6\) (we can safely do that because a nonzero number in even power is positive) to get \(x > x^3\)
\(x - x^3 > 0\)
\(x(1 - x^2) > 0\)
Two cases:
a. Both multiples are positive, (\(x > 0\) and \(1 - x^2 > 0\)), so \(0 < x < 1\). (Here is why: \(1 - x^2 > 0\) is the same as \(1 > x^2\), which gives \(-1 < x < 1\), which together with \(x > 0\) gives \(0 < x < 1\));
b. Both multiples are negative, (\(x < 0\) and \(1 - x^2 < 0\)), so \(x < -1\) (Here is why: \(1 - x^2 < 0\) is the same as \(1 < x^2\), which gives \(x < -1\) or \(x > 1\), which together with \(x < 0\) gives \(x < -1\)).
So, \(\frac{1}{x^5} > \frac{1}{x^3}\) is true when \(0 < x < 1\) or when \(x < -1\).
ANY \(x\) from these ranges will be less than 1. So, the answer to our original question (Is \(x\) greater than 1?), is NO. Sufficient.
Answer: B
Is \(x\) greater than 1?
(1) \(\frac{1}{x} \gt - 1\).
Multiply \(\frac{1}{x} \gt - 1\) by \(x^2\) (we can safely do that because a nonzero number in even power is positive) to get \(x > -x^2\):
\(x +x^2 > 0\);
\(x(1 + x) > 0\).
Two cases:
a. Both multiples are positive, (\(x > 0\) and \(x + 1 > 0\)), so \(x > 0\);
b. Both multiples are negative, (\(x < 0\) and \(x + 1 < 0\)), so \(x < -1\);
So, \(\frac{1}{x} \gt - 1\) is true when \(x < -1\) or when \(x > 0\). Thus, \(x\) may or may not be greater than 1. Not sufficient.
(2) \(\frac{1}{x^5} > \frac{1}{x^3}\).
Multiply \(\frac{1}{x^5} > \frac{1}{x^3}\) by \(x^6\) (we can safely do that because a nonzero number in even power is positive) to get \(x > x^3\)
\(x - x^3 > 0\)
\(x(1 - x^2) > 0\)
Two cases:
a. Both multiples are positive, (\(x > 0\) and \(1 - x^2 > 0\)), so \(0 < x < 1\). (Here is why: \(1 - x^2 > 0\) is the same as \(1 > x^2\), which gives \(-1 < x < 1\), which together with \(x > 0\) gives \(0 < x < 1\));
b. Both multiples are negative, (\(x < 0\) and \(1 - x^2 < 0\)), so \(x < -1\) (Here is why: \(1 - x^2 < 0\) is the same as \(1 < x^2\), which gives \(x < -1\) or \(x > 1\), which together with \(x < 0\) gives \(x < -1\)).
So, \(\frac{1}{x^5} > \frac{1}{x^3}\) is true when \(0 < x < 1\) or when \(x < -1\).
ANY \(x\) from these ranges will be less than 1. So, the answer to our original question (Is \(x\) greater than 1?), is NO. Sufficient.
Answer: B
05 Dec 2014, 05:01
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amitjathar
No, that's not correct.
You cannot multiply 1/x > -1 by x because you don't know the sign of x. If x is positive then yes, 1 > -x but if x is negative, then when multiplying by it you should flip the sign and you get 1 < -x.
Similarly you cannot reduce 1/x^5 > 1/x^3 by 1/x^3, because you don't know its sign.
Never multiply (or reduce) an inequality by a variable (or the expression with a variable) if you don't know its sign.
Other tips on Inequalities are here: inequalities-tips-and-hints-175001.html
