[Math Revolution GMAT math practice question]
Is x + 1/x > 2?
1) x > 0
2) x ≠1
Is x + 1/x > 2?
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- Max@Math Revolution
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Very nice problem, Max. Congrats!Max@Math Revolution wrote:[Math Revolution GMAT math practice question]
Is x + 1/x > 2?
1) x > 0
2) x ≠1
$$x + {1 \over x}\,\,\mathop > \limits^? \,\,2$$
$$\left( 1 \right)\,\,x > 0\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,x = 1\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,x = 2\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr} \right.$$
$$\left( 2 \right)\,\,x \ne 1\,\,\left\{ \matrix{
\,\left( {{\mathop{\rm Re}\nolimits} } \right){\rm{Take}}\,\,\,x = 2\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr
\,{\rm{Take}}\,\,x = - 1\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\, \hfill \cr} \right.$$
$$\left( {1 + 2} \right)\,\,\,0\,\,\mathop < \limits^{x\, \ne \,1} \,\,{\left( {x - 1} \right)^2}\,\, = \,\,\,{x^2} - 2x + 1\,\,\,\,\mathop \Leftrightarrow \limits^{\,x\, > \,\,0} \,\,\,\,0 < {{{x^2} - 2x + 1} \over x} = x - 2 + {1 \over x}\,\,\,\,\, \Leftrightarrow \,\,\,\,x + {1 \over x} > 2\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\,\,$$
This solution follows the notations and rationale taught in the GMATH method.
Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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For x + 1/x to be greater than 2, x must be a positive number.Max@Math Revolution wrote:[Math Revolution GMAT math practice question]
Is x + 1/x > 2?
1) x > 0
2) x ≠1
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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- Max@Math Revolution
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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
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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