Is a + a^(-1) > 2? (1) a > 0 (2) a < 1

Is \(a + \frac{1}{a} -2 > 0\)
\((\sqrt{a} - \frac{1}{\sqrt{a}}) ^ 2 > 0\)

The above equation fails if we have a = 1. So, B is necessary.
If a < 0 \(\sqrt{a}\) is a complex number. I don't know if GMAT allows that.

So,

Given expression: \(a + \frac{1}{a}\)

If a is negative, the expression becomes negative and hence less than 2.

Assuming a is positive,
Note that the product of the numbers is a*(1/a) = 1. Given a constant product, the sum of two positive numbers is minimum when they are equal. When a = 1/a, a^2 = 1 and hence a = 1
When a = 1, a + 1/a = 2
So minimum value of a + 1/a is 2. For all other positive values of a, the sum will be more than 2.
Both statements together tell us that a is positive and not 1. So in every case, the sum a + 1/a will be greater than 2. Together they are sufficient to answer the question with "Yes".

Answer (C)

Statement 2 is not sufficient, since the answer is 'no' for negative values of a, and 'yes' for all positive values except a=1, as we'll see below.

When, using Statement 1 alone, we know a > 0, we can rephrase the question by multiplying by a on both sides; since a > 0, we don't need to be concerned about whether to reverse the inequality:

Is a + a^(-1) > 2 ?
Is a^2 + 1 > 2a ?
Is a^2 - 2a + 1 > 0 ?
Is (a - 1)^2 > 0 ?

So we've rephrased our question, and it's now easier to see which values of a will make the answer to our question 'yes'. We have a square on the left side of the inequality, so the left side is always positive, except when it is exactly equal to 0. It is only exactly 0 when a = 1, so the answer to our question is 'yes' except if a=1. So we need Statement 2 as well, to rule out the possibility that a=1, and the answer is C.

1: Insufficient. If a = 1, a+1/a = 2. Otherwise, for all other values of a, a+1/a > 2.
2: Insufficient. If 0 < a < 1, a+1/a > 2. If a<0, a+1/a<0<2.
Together it is sufficient. Answer is C.

MANHATTAN GMAT OFFICIAL SOLUTION:

Because there are no constraints on a, we should consider the Standard Number Testing Set of numbers. The inequality can be rephrased as follows:
Is a + a^(-1) > 2? --> Is a + 1/a > 2?

Combining the two constraints, we have 0 < a < 1. Within the Standard Number Testing Set, only 1/2 is within this range. Also, because the answer to the question is “No” at a = 1 and “Undefined”at a = 0, we should apply the boundary principle—test numbers close to the boundaries of the range.



It seems that at the extreme edges, the values are greater than 2, so Statements (1) & (2) appear to be SUFFICIENT.

The correct answer is C.

Ahhh! Missed out on a = 1, presumed a = 1 satisfies the question and picked A.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is a + a^(-1) > 2?

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

We get a^3+a>2a^2?, or a(a^2-2a+1)>0?, or a(a-1)^2>0? and ultimately as a cannot equal 1, whether a>0, if we modify the question.
Hence the answer becomes (C) as
condition 1 answers the question 'no' if a=1, but 'yes' when a=2, making it insufficient.
It needs to work with condition 2, so the answer becomes (C).


Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.

HI Karishma
Please help me here
by solving we get
(a-1)^>0 which we need to prove.

now A is in sufficient as A can be 1.
but in B
put any value less than 1 square will make it >0 . Please let me know what m i missing.

i will go with C...
a+1/a>2?
1. a>0. if a=0.1, and 1/a=10 then yes, the result is greater than 2. but if a=1, then 1/a=1. 1+1=2 and the answer to the question is no. A/D out.

2. a<1.
a=0.1 1/a=10 => yes
a=-2. 1/a =-1/2 => -2.5<2 so no.
B is out.

1+2.
0<a<1
so a must be a non-integer between 0 and 1. so always will be >2.
for ex. a=0.1, 1/a=10. yes
a=0.25, 1/a=4, yes.

C must be the answer.

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