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Is x + 1/x > 2?

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Is x + 1/x > 2?

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[Math Revolution GMAT math practice question]

Is x + 1/x > 2?

1) x > 0
2) x ≠ 1

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Max@Math Revolution wrote:
[Math Revolution GMAT math practice question]

Is x + 1/x > 2?

1) x > 0
2) x ≠ 1
Very nice problem, Max. Congrats!

$$x + {1 \over x}\,\,\mathop > \limits^? \,\,2$$

$$\left( 1 \right)\,\,x > 0\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,x = 1\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,x = 2\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.$$

$$\left( 2 \right)\,\,x \ne 1\,\,\left\{ \matrix{
\,\left( {{\mathop{\rm Re}\nolimits} } \right){\rm{Take}}\,\,\,x = 2\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,x = - 1\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.$$

$$\left( {1 + 2} \right)\,\,\,0\,\,\mathop < \limits^{x\, \ne \,1} \,\,{\left( {x - 1} \right)^2}\,\, = \,\,\,{x^2} - 2x + 1\,\,\,\,\mathop \Leftrightarrow \limits^{\,x\, > \,\,0} \,\,\,\,0 < {{{x^2} - 2x + 1} \over x} = x - 2 + {1 \over x}\,\,\,\,\, \Leftrightarrow \,\,\,\,x + {1 \over x} > 2\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\,\,$$


This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.

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Max@Math Revolution wrote:
[Math Revolution GMAT math practice question]

Is x + 1/x > 2?

1) x > 0
2) x ≠ 1
For x + 1/x to be greater than 2, x must be a positive number.

We have x + 1/x > 2

=> (x^2 + 1)/x > 2
=> x^2 + 1 > 2x
=> x^2 - 2x + 1 > 0
=> (x - 1)^2 > 0

We see that except for x = 1, (x - 1)^2 is a positive number. Thus, x ≠ 1.

So, two conditions are must: 1. x > 0 and 2. x ≠ 1.

The correct answer: C

Hope this helps!

-Jay
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=>

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question.

x + 1/x > 2
=> x^3 + x > 2x^2 after multiplying both sides by x^2
=> x^3 - 2x^2 + x > 0
=> x^3 - 2x^2 + x > 0
=> x(x^2 - 2x + 1) > 0
=> x(x-1)^2 > 0
=> x > 0 and x ≠ 1

Thus, we need both conditions together for sufficiency.

Therefore, C is the answer.
Answer: C

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The one-and-only World’s First Variable Approach for DS and IVY Approach for PS with ease, speed and accuracy.
Only $99 for 3 month Online Course
Free Resources-30 day online access & Diagnostic Test
Unlimited Access to over 120 free video lessons-try it yourself
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