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# If a ≠ 0, is a + a−1 > 2? (1) a > 0 (2) a < 1

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Re: If a ≠ 0, is a + a−1 > 2? (1) a > 0 (2) a < 1 [#permalink]
GMATGuruNY wrote:
If $$a ≠ 0$$, is $$a + a^{−1} > 2$$?

(1) $$a > 0$$
(2) $$a < 1$$

CRITICAL POINTS occur when the two sides of the inequality are EQUAL or when the inequality is UNDEFINED.
Given $$a + a^{−1} > 2$$:
The two sides are equal when a=1, since $$1 + 1^{−1} = 1 + \frac{1}{1^1} = 2$$.
The inequality is undefined when a=0, since $$0^{-1} = \frac{1}{0^1} =$$ undefined.

To determine which ranges satisfy the inequality, test one value to the left and one value to the right of each critical point.
Here, we must test a<0, 0<a<1 and a>1.
If we test a=-1, a=1/2 and a=2, only a=1/2 and a=2 satisfy $$a + a^{−1} > 2$$, implying that the valid ranges are 0<a<1 and a>1.

Question stem, rephrased:
Is $$a$$ a positive value other than 1?

Statement 1:
If a=2, the answer to the rephrased question stem is YES.
If a=1, the answer to the rephrased question stem is NO.
INSUFFICIENT.

Statement 2:
If a=1/2, the answer to the rephrased question stem is YES.
If a=-1, the answer to the rephrased question stem is NO.
INSUFFICIENT.

Statements combined:
Since 0<a<1, the answer to the rephrased question stem is YES.
SUFFICIENT.

GMATGuruNY
Thank you so much for your response. Sir, could explain a bit the highlighted part? i mean: how do someone convinced that we must test those values?
Also, why it is other than 1
Thanks__
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Re: If a ≠ 0, is a + a−1 > 2? (1) a > 0 (2) a < 1 [#permalink]
GMATGuruNY wrote:
If $$a ≠ 0$$, is $$a + a^{−1} > 2$$?

(1) $$a > 0$$
(2) $$a < 1$$

CRITICAL POINTS occur when the two sides of the inequality are EQUAL or when the inequality is UNDEFINED.
Given $$a + a^{−1} > 2$$:
The two sides are equal when a=1, since $$1 + 1^{−1} = 1 + \frac{1}{1^1} = 2$$.
The inequality is undefined when a=0, since $$0^{-1} = \frac{1}{0^1} =$$ undefined.

To determine which ranges satisfy the inequality, test one value to the left and one value to the right of each critical point.
Here, we must test a<0, 0<a<1 and a>1.
If we test a=-1, a=1/2 and a=2, only a=1/2 and a=2 satisfy $$a + a^{−1} > 2$$, implying that the valid ranges are 0<a<1 and a>1.

Question stem, rephrased:
Is $$a$$ a positive value other than 1?

Statement 1:
If a=2, the answer to the rephrased question stem is YES.
If a=1, the answer to the rephrased question stem is NO.
INSUFFICIENT.

Statement 2:
If a=1/2, the answer to the rephrased question stem is YES.
If a=-1, the answer to the rephrased question stem is NO.
INSUFFICIENT.

Statements combined:
Since 0<a<1, the answer to the rephrased question stem is YES.
SUFFICIENT.

GMATGuruNY
Thank you so much for your response. Sir, could explain a bit the highlighted part? i mean: how do someone convinced that we must test those values?
Also, why it is other than 1
Thanks__

The critical points, plotted on a number line:
<------0-------1------>
0 and 1 are the only values where the $$a + a^{−1}$$ is undefined or where $$a + a^{−1} = 2$$.
Implication:
In each of the colored ranges, $$a + a^{−1} < 2$$ or $$a + a^{−1} > 2$$.
To determine in which ranges $$a + a^{−1} > 2$$, we must test only one value in each of the colored ranges.

a=-1 is in the red range.
When a=-1, $$a + a^{−1} < 2$$.
Thus, the red range (a<0) is not valid.

a=1/2 is in the green range.
When a=1/2, $$a + a^{−1} > 2$$.
Thus, the green range (0<a<1) is valid.

a=2 is in the blue range.
When a=2, $$a + a^{−1} > 2$$.
Thus, the blue range (a>1) is valid.

Result:
$$a + a^{−1} > 2$$ when $$a$$ is in the green range (0<a<1) or in the blue range (a>1).
Together, the green and blue ranges include every positive value except for 1.
Thus, the answer to the question stem is YES if $$a$$ is ANY POSITIVE VALUE OTHER THAN 1.

Question stem. rephrased:
Is $$a$$ a positive value other than 1?
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Re: If a ≠ 0, is a + a−1 > 2? (1) a > 0 (2) a < 1 [#permalink]
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Bunuel
I tried to solve it using algebra. Here's my process. Is it right?
a+a^-1>2
or a+1/a-2>0
or (a-1)^2. a^-1>0
So, the question becomes is 1/a>0 when a≠1?
1. a>0, It works in all the cases except when a=1. so, Not sufficient
2. a<1, a can be + or -. So, Not sufficient.
(1) + (2) = 0<a<1 , So, Clearly a is + & a≠1. So, Sufficent . Answer. (C)
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Re: If a 0, is a + a1 > 2? (1) a > 0 (2) a < 1 [#permalink]
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Re: If a 0, is a + a1 > 2? (1) a > 0 (2) a < 1 [#permalink]
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