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If k > 1, which of the following must be equal to 2/((k + 1)^(1/2)

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Bunuel
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If k > 1, which of the following must be equal to 2/((k + 1)^(1/2) [#permalink] New post  16 Jun 2015, 03:11
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If k > 1, which of the following must be equal to \(\frac{2}{\sqrt{k+1}+\sqrt{k-1}}\)?


A. 2

B. \(2\sqrt{2k}\)

C. \(2\sqrt{k+1}+\sqrt{k-1}\)

D. \(\frac{\sqrt{k+1}}{\sqrt{k-1}}\)

E. \(\sqrt{k+1}-\sqrt{k-1}\)
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Re: If k > 1, which of the following must be equal to 2/((k + 1)^(1/2) [#permalink] New post  16 Jun 2015, 05:32
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If k > 1, which of the following must be equal to \(\frac{2}{\sqrt{k+1}+\sqrt{k−1}}\)?

\(\frac{2}{\sqrt{k+1}+\sqrt{k−1}}\)
Rationalizing the denominator

We get
\(\sqrt{k+1}-\sqrt{k−1}\)

Ans : E
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Re: If k > 1, which of the following must be equal to 2/((k + 1)^(1/2) [#permalink] New post  16 Jun 2015, 09:13
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Bunuel wrote:
If k > 1, which of the following must be equal to \(\frac{2}{\sqrt{k+1}+\sqrt{k-1}}\)?

A. 2

B. \(2\sqrt{2k}\)

C. \(2\sqrt{k+1}+\sqrt{k-1}\)

D. \(\frac{\sqrt{k+1}}{\sqrt{k-1}}\)

E. \(\sqrt{k+1}-\sqrt{k-1}\)

Kudos for a correct solution.


Rationalization is one of the methods, Second is here

Since k>1 , \(\frac{2}{\sqrt{k+1}+\sqrt{k-1}}\)

Let's Substitute k = 2 in the expression, the expression now becomes

\(\frac{2}{\sqrt{k+1}+\sqrt{k-1}}\) = \(\frac{2}{\sqrt{2+1}+\sqrt{2-1}}\)
i.e. \(\frac{2}{\sqrt{k+1}+\sqrt{k-1}}\) = \(\frac{2}{\sqrt{3}+1}\) = 2/(1.732+1)
i.e. \(\frac{2}{\sqrt{k+1}+\sqrt{k-1}}\) = 2/(2.732) = approx 0.74

Check option with k=2

A. 2 > 0.74INCORRECT

B. \(2\sqrt{2k}\) = \(2\sqrt{4}\) = 4 > 0.74 INCORRECT

C. \(2\sqrt{k+1}+\sqrt{k-1}\) = \(2\sqrt{3}+\sqrt{2-1}\)>0.74 INCORRECT

D. \(\frac{\sqrt{k+1}}{\sqrt{k-1}}\) = \(\frac{\sqrt{2+1}}{\sqrt{2-1}}\)= 1.7 > 0.74 INCORRECT

E. \(\sqrt{k+1}-\sqrt{k-1}\) = \(\sqrt{2+1}-\sqrt{2-1}\)= 1.73-1 = 0.73 CORRECT

Answer: option
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E
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Re: If k > 1, which of the following must be equal to 2/((k + 1)^(1/2) [#permalink] New post  16 Jun 2015, 10:21
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If k > 1, which of the following must be equal to \(\frac{2}{\sqrt{k+1}+\sqrt{k-1}}\)?

Solution -

Divide the numerator and denominator with \(\sqrt{k+1}-\sqrt{k-1}\) in the above equation and solve the equation.

Results, \(\sqrt{k+1}-\sqrt{k-1}\).

ANS E.

Thanks,

Kudos please.
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Re: If k > 1, which of the following must be equal to 2/((k + 1)^(1/2) [#permalink] New post  16 Jun 2015, 10:57
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Bunuel wrote:
If k > 1, which of the following must be equal to \(\frac{2}{\sqrt{k+1}+\sqrt{k-1}}\)?

A. 2

B. \(2\sqrt{2k}\)

C. \(2\sqrt{k+1}+\sqrt{k-1}\)

D. \(\frac{\sqrt{k+1}}{\sqrt{k-1}}\)

E. \(\sqrt{k+1}-\sqrt{k-1}\)

Kudos for a correct solution.


This one can be easily solved by rationalizing the expression \(\frac{2}{\sqrt{k+1}+\sqrt{k-1}}\).

So multiplying and dividing the expression by \(\sqrt{k+1}-\sqrt{k-1}\).

We get, \(\frac{2}{[k+1]-[k-1]}\) * \(\sqrt{k+1}-\sqrt{k-1}\).

=\(\frac{2}{2}\) * \(\sqrt{k+1}-\sqrt{k-1}\)

that is \(\sqrt{k+1}-\sqrt{k-1}\).
Hence E.
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Re: If k > 1, which of the following must be equal to 2/((k + 1)^(1/2) [#permalink] New post  18 Jun 2015, 04:13
Bunuel wrote:
If k > 1, which of the following must be equal to \(\frac{2}{\sqrt{k+1}+\sqrt{k-1}}\)?

A. 2

B. \(2\sqrt{2k}\)

C. \(2\sqrt{k+1}+\sqrt{k-1}\)

D. \(\frac{\sqrt{k+1}}{\sqrt{k-1}}\)

E. \(\sqrt{k+1}-\sqrt{k-1}\)

Kudos for a correct solution.


Solution :
Rationalizing denominator by multiplying Numerator and denominator by \({\sqrt{k+1}-\sqrt{k-1}}\)

So we get \(\sqrt{k+1}-\sqrt{k-1}\) in numerator

Option E
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Re: If k > 1, which of the following must be equal to 2/((k + 1)^(1/2) [#permalink] New post  18 Jun 2015, 17:36
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Hi All,

This question has an interesting 'quirk' to it.... Even though the question tells us that K > 1, if K = 1 then you'll still end up with the correct answer....

TESTing K = 1 in the prompt gives us...

2/(√2 + √0) =
2/(√2) =

Now we have to multiply both the numerator and denominator by (√2), which simplifies to...

2(√2)/2 =
√2

So we're looking for an answer that equals √2 when K = 1....

Answer A: 2 NOT a match
Answer B: 2√2 NOT a match
Answer C: 2√2 + 0 NOT a match
Answer D: √2/0 = undefined NOT a match
Answer E: √2 - 0 This IS a MATCH

Final Answer:
Show Spoiler
E


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Re: If k > 1, which of the following must be equal to 2/((k + 1)^(1/2) [#permalink] New post  20 Jun 2015, 10:49
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Bunuel wrote:
If k > 1, which of the following must be equal to \(\frac{2}{\sqrt{k+1}+\sqrt{k-1}}\)?

A. 2

B. \(2\sqrt{2k}\)

C. \(2\sqrt{k+1}+\sqrt{k-1}\)

D. \(\frac{\sqrt{k+1}}{\sqrt{k-1}}\)

E. \(\sqrt{k+1}-\sqrt{k-1}\)

Kudos for a correct solution.


Let us try this by taking values
Let k=3
Equation becomes=2/2+\sqrt{2}
=\sqrt{2}/\sqrt{2}+1
=2-\sqrt{2}

Now put k=3 in the given options.
Only Option E gives the value as 2-\sqrt{2}
Answer E
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Re: If k > 1, which of the following must be equal to 2/((k + 1)^(1/2) [#permalink] New post  22 Jun 2015, 07:25
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Bunuel wrote:
If k > 1, which of the following must be equal to \(\frac{2}{\sqrt{k+1}+\sqrt{k-1}}\)?

A. 2

B. \(2\sqrt{2k}\)

C. \(2\sqrt{k+1}+\sqrt{k-1}\)

D. \(\frac{\sqrt{k+1}}{\sqrt{k-1}}\)

E. \(\sqrt{k+1}-\sqrt{k-1}\)

Kudos for a correct solution.


MANHATTAN GMAT OFFICIAL SOLUTION:

Since there are variables in the answer choices (VIC), we should pick a number and test the choices. If k = 2, then \(\frac{2}{\sqrt{k+1}+\sqrt{k-1}}=\frac{2}{\sqrt{2+1}+\sqrt{2-1}}=\frac{2}{\sqrt{3}+\sqrt{1}}\approx{\frac{1}{1.7+1}}=\frac{2}{2.7}\) which is less than 1. Now test the answer choices and try to match the target:


The correct answer is E.

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Re: If k > 1, which of the following must be equal to 2/((k + 1)^(1/2) [#permalink] New post  04 Oct 2015, 08:51
multiply both numerator and denominator by sqrt(K+1) - sqrt(k-1) ;
and u will get the answer
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Re: If k > 1, which of the following must be equal to 2/((k + 1)^(1/2) [#permalink] New post  21 Jun 2023, 12:27
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