(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
Inequality
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Frankenstein
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Hi,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
(2x^2+2x^2)/x^2 = 4
Hence, E
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MBA.Aspirant
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(x + 1/x)^2 - (x - 1/x)^2jayanti 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^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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Let x=1.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
= (1 + 1/1)² - (1 - 1/1)²
= 2² - 0²
= 4.
The correct answer is E.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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