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

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Re: If x > 0, what is the least possible value for x + 1/x ? [#permalink]
I did this question using plugging in.

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

GMAT assassins aren't born, they're made,
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Re: If x > 0, what is the least possible value for x + 1/x ? [#permalink]
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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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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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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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]
stonecold wrote:
Bunuel wrote:

Tough and Tricky questions: Algebra.

@stonecold

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

Could you explain $$p^2-4=9$$
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Re: If x > 0, what is the least possible value for x + 1/x ? [#permalink]
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

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

Since AM > GM; AM-> Arithmetic mean; GM-> Geometric Mean
AM of x & 1/x > GM of x & 1/x
$$(x + 1/x)/2 >= \sqrt{x*1/x}$$
x + 1/x >= 2

IMO D
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Re: If x > 0, what is the least possible value for x + 1/x ? [#permalink]
For any positive number , the sum of the number and its reciprocal is always greater than or equal to 2.
2<=1+1/x.
the equality of this relationship will occur only when 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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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

Solution:

The minimum value of the sum of the terms x and 1/x (that have positive values) occurs when the two terms are equal to each other. Therefore, we have:

x = 1/x

x^2 = 1

x = 1

So the minimum value of x + 1/x is 1 + 1/1 = 2.

Alternate Solution:

Let x + 1/x = k. Since x and 1/x are positive, so is k. Let’s multiply each side of the equality by x (which is possible since x > 0):

x^2 + 1 = kx

x^2 - kx + 1 = 0

In order for this quadratic to have roots, we must have b^2 - 4ac = (-k)^2 - 4(1)(1) = k^2 - 4 ≥ 0. Thus, k^2 ≥ 4 which implies k ≥ 2 or k ≤ -2. The latter is not possible since k is positive, thus k ≥ 2. The smallest value of k satisfying this inequality is k = 2. Therefore, the smallest value of x + 1/x is 2.

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Re: If x > 0, what is the least possible value for x + 1/x ? [#permalink]
Using differentials ,
x + 1/x = 0
differentiating wrt x
1 - 1/x^2=0
x^2=1
x=1 or x=-1
acceptable solution is x=1 as x>0
Putting x=1 in the original equation , we get the value as 2
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Re: If x > 0, what is the least possible value for x + 1/x ? [#permalink]
KarishmaB wrote:
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

Dear karishma please explain relation between product of x and 1/x and there sum

Posted from my mobile device
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Re: If x > 0, what is the least possible value for x + 1/x ? [#permalink]
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puneetfitness wrote:
KarishmaB wrote:
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

Dear karishma please explain relation between product of x and 1/x and there sum

Posted from my mobile device

When two positive numbers have a constant sum, their product is maximum when the numbers are equal.
When two positive numbers have a constant product, their sum is minimum when the numbers are equal.

Given, a + b = 10, the product ab will be maximum when a = b i.e. when a = b = 5. So maximum value of ab = 25.
Given ab = 16, the sum will be minimum when a = b i.e. when a = b = 4. So minimum value of (a + b) = 8.

P.S. - Please tag me using "KarishmaB" (without quotes) to ensure that I see your question. Else I may not know to open the thread.
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Re: If x > 0, what is the least possible value for x + 1/x ? [#permalink]
puneetfitness wrote:
KarishmaB wrote:
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

Dear karishma please explain relation between product of x and 1/x and there sum

Posted from my mobile device

You can use AM GM inequality

with more than one variable for two non-negative numbers x and y, is the statement that

(x+y)/2≥ root(xy)

with equality if and only if x = y.
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Re: If x > 0, what is the least possible value for x + 1/x ? [#permalink]
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Re: If x > 0, what is the least possible value for x + 1/x ? [#permalink]
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