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

Question

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}

sanguine2016 · Accepted Answer

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}{ - }  *  \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.

ManojReddy · Answer

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

GMATinsight · Answer

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 > 0.74  INCORRECT

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  E

balamoon · Answer

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.

vishwaprakash · Answer

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

EMPOWERgmatRichC · Answer

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:  E

GMAT assassins aren't born, they're made,
Rich

KS15 · Answer

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

Bunuel · Answer

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: https://gmatclub.com/forum/download/file.php?id=27044 The correct answer is E. 2015-06-22_1824.png

rohitkumar77 · Answer

multiply both numerator and denominator by sqrt(K+1) - sqrt(k-1) ;
and u will get the answer

bumpbot · Answer

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

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