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Inequality

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Source: — Problem Solving |
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by knight247 » Fri Aug 05, 2011 4:57 am
(x+1/x)^2-(x-1/x)^2
=(x^2+1)^2/x^2-(x^2-1)^2/x^2
={(x^2+1)^2-(x^2-1)^2}/x^2

Using the formula (x+y)^2=x^2+2xy+y^2 and (x-y)=x^2-2xy+y^2 we get

(x^2+2x^2+1-x^2+2x^2-1)/x^2

2x^2/x^2=2

Hence D
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by Frankenstein » Fri Aug 05, 2011 5:00 am
knight247 wrote:(x+1/x)^2-(x-1/x)^2
=(x^2+1)^2/x^2-(x^2-1)^2/x^2
={(x^2+1)^2-(x^2-1)^2}/x^2

Using the formula (x+y)^2=x^2+2xy+y^2 and (x-y)=x^2-2xy+y^2 we get

(x^2+2x^2+1-x^2+2x^2-1)/x^2

2x^2/x^2=2

Hence D
Hi,
(2x^2+2x^2)/x^2 = 4

Hence, E
Cheers!

Things are not what they appear to be... nor are they otherwise
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by sushbis » Fri Aug 05, 2011 5:30 am
No need for any calculations..by looking at the prob itself you can eliminate all x's :)
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by MBA.Aspirant » Fri Aug 05, 2011 7:03 am
jayanti wrote:If x ≠ 0, then (x + 1/x)^2 - (x - 1/x)^2

A. 1/x
B. x
C. 1
D. 2
E. 4
(x + 1/x)^2 - (x - 1/x)^2

x^2 +2 + 1/x^2 - (x^2 - 2 + 1/x^2)


x^2 + 2 + 1/x^2 - x^2 + 2 - 1/x^2

= 4
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by GMATGuruNY » Fri Aug 05, 2011 8:10 am
jayanti wrote:If x ≠ 0, then (x + 1/x)^2 - (x - 1/x)^2

A. 1/x
B. x
C. 1
D. 2
E. 4
Let x=1.

(x + 1/x)² - (x - 1/x)²
= (1 + 1/1)² - (1 - 1/1)²
= 2² - 0²
= 4.

The correct answer is E.
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