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# If x > 0, what is the least possible value for x + 1/x ?

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Joined: 02 Sep 2009
Posts: 56251
If x > 0, what is the least possible value for x + 1/x ?  [#permalink]

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20 Oct 2014, 07:09
1
11
00:00

Difficulty:

45% (medium)

Question Stats:

63% (01:24) correct 37% (01:35) wrong based on 416 sessions

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Tough and Tricky questions: Algebra.

If x > 0, what is the least possible value for x + 1/x ?

(A) 0.5
(B) 1
(C) 1.5
(D) 2
(E) 2.5

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Re: If x > 0, what is the least possible value for x + 1/x ?  [#permalink]

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20 Oct 2014, 22:30
7
4
coolredwine wrote:
I did this question using plugging in.

Any other short way to do this question?

Note that the product of the terms x and 1/x is a constant x*(1/x) = 1
So the sum will be minimum when the terms are equal.

x = 1/x
x^2 = 1
x = 1 (since x must be positive)

Least possible value of x +1/x = 1 + 1/1 = 2
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Re: If x > 0, what is the least possible value for x + 1/x ?  [#permalink]

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20 Oct 2014, 09:20
I did this question using plugging in.

Any other short way to do this question?
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Joined: 17 Oct 2010
Posts: 1
Re: If x > 0, what is the least possible value for x + 1/x ?  [#permalink]

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20 Oct 2014, 10:30
2
1
x+1/x =(X^2+1)/x =(X^2+1-2x+2x)/x =(x^2-2x+1)/x +2x/x =(x-1)^2/x +2

As (x-1)^2 is always greater than zero, hence the expression x+1/X will always be greater than or equal to 2. So the least possible value is 2 at x=1
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Re: If x > 0, what is the least possible value for x + 1/x ?  [#permalink]

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21 Oct 2014, 03:20
3
$$x + \frac{1}{x}$$

x is reciprocal of $$\frac{1}{x}$$ & vice versa. To have $$x + \frac{1}{x}$$ the lowest, we require to find the number whose reciprocal is same as that of it

$$x = \frac{1}{x}$$

$$x^2 = 1$$

x = 1

1 is the only number whose reciprocal = 1

Least addition = 1 + 1 = 2

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Re: If x > 0, what is the least possible value for x + 1/x ?  [#permalink]

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25 Nov 2015, 15:24
Hi All,

This question is built around the concept of reciprocals. While some Test Takers might be able to spot the logic involved, most would probably need to TEST VALUES so that they can 'see' what the rule/pattern is:

We're told that X > 0. We're asked for the LEAST possible value of X + 1/X.

To start, X and 1/X are reciprocals, meaning that when you plug in a value for X, you'll get the inverse of that value for 1/X.

For example...
X = 10
1/X = 1/10
Sum = 10.1

X = 1/2
1/X = 1/(1/2) = 2
Sum = 2.5

From these two example, we can see that the sum of the two terms WILL be greater than 1. It's just a matter of how 'close to 1' we can get the sum...

When...
X = 1
1/X = 1
So the sum is 1+1 = 2

There's no other example that will get us any lower than 2, so that must be the answer.

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Re: If x > 0, what is the least possible value for x + 1/x ?  [#permalink]

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12 Dec 2015, 18:25
The trick is to remember that reciprocals of 0<x<1 will be >1
Also we know that x is positive
the minimum value will be at x = 1 and 1/x = 1 i.e. 2
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Re: If x > 0, what is the least possible value for x + 1/x ?  [#permalink]

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08 Mar 2016, 10:06
Bunuel wrote:

Tough and Tricky questions: Algebra.

If x > 0, what is the least possible value for x + 1/x ?

(A) 0.5
(B) 1
(C) 1.5
(D) 2
(E) 2.5

x+1/x = (x^2 +1) /x = p (say)
= x^2-px+1=0
for the real roots B^2-4AC must be positive
hence p^2-4=9(for least value)
=> p=2 or -2
hene x=2
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Re: If x > 0, what is the least possible value for x + 1/x ?  [#permalink]

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04 Mar 2017, 05:11
Bunuel wrote:

Tough and Tricky questions: Algebra.

If x > 0, what is the least possible value for x + 1/x ?

(A) 0.5
(B) 1
(C) 1.5
(D) 2
(E) 2.5

Aah how I missed Derivatives!

I know it is out of syllabus, but if you take derivative wrt x -> 1 - 1/(x^2) should be 0 for minima (2nd derivative is +ve so minima) -> x = +-1 -> x = 1

Am I the only one who solved this problem using this approach?
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Re: If x > 0, what is the least possible value for x + 1/x ?  [#permalink]

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05 Mar 2017, 22:48
1
dhavalpgk wrote:
Bunuel wrote:

Tough and Tricky questions: Algebra.

If x > 0, what is the least possible value for x + 1/x ?

(A) 0.5
(B) 1
(C) 1.5
(D) 2
(E) 2.5

Aah how I missed Derivatives!

I know it is out of syllabus, but if you take derivative wrt x -> 1 - 1/(x^2) should be 0 for minima (2nd derivative is +ve so minima) -> x = +-1 -> x = 1

Am I the only one who solved this problem using this approach?

Derivatives are not tested on GMAT so every GMAT question can be solved without Derivatives.
If you remember your Derivatives concepts well, you are free to use them to solve appropriate questions since all you need to do is give the answer, but most people would be ignoring them.
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Re: If x > 0, what is the least possible value for x + 1/x ?  [#permalink]

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08 Apr 2018, 02:40
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Re: If x > 0, what is the least possible value for x + 1/x ?   [#permalink] 08 Apr 2018, 02:40
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