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If a ≠ 0, is a + a−1 > 2? (1) a > 0 (2) a < 1 13 Feb 2020, 09:15
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

65% (hard)

Question Stats:

51% (01:41) correct 49% (01:43) wrong based on 199 sessions

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If \(a ≠ 0\), is \(a + a^{−1} > 2\)?

(1) \(a > 0\)
(2) \(a < 1\)
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C
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If a ≠ 0, is a + a−1 > 2? (1) a > 0 (2) a < 1 13 Feb 2020, 09:29
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If \(a ≠ 0\), is \(a + a^{−1} > 2\)?

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


Question : is \(a + a^{−1} > 2\)?

CONCEPT: For any positive value of a, \(a + \frac{1}{a} ≥ 2\)

Statement1: \(a > 0\)

\(a + a^{−1}\) may be Greater than or Equal to 2 (equal for a=1) hence

NOT SUFFICIENT

Statement 2: \(a < 1\)

If 0 < a < 1 then \(a + a^{−1} > 2\)

If a < 0 then \(a + a^{−1} < -2\) hence

NOT SUFFICIENT

Combining the two statements

0 < a < 1, therefore \(a + \frac{1}{a} > 2\)

SUFFICIENT

Answer: Option C
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If a ≠ 0, is a + a−1 > 2? (1) a > 0 (2) a < 1 13 Feb 2020, 12:20
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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.

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C
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If a ≠ 0, is a + a−1 > 2? (1) a > 0 (2) a < 1 13 Feb 2020, 13:49
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Asad
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.

Show Spoiler
C
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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If a ≠ 0, is a + a−1 > 2? (1) a > 0 (2) a < 1 13 Feb 2020, 15:24
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GMATGuruNY
Asad
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.

Show Spoiler
C
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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If a ≠ 0, is a + a−1 > 2? (1) a > 0 (2) a < 1 23 Oct 2020, 02:42
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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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If a ≠ 0, is a + a−1 > 2? (1) a > 0 (2) a < 1 28 Aug 2022, 13:45
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