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# If k > 1, which of the following must be equal to

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Math Expert
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If k > 1, which of the following must be equal to  [#permalink]

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16 Jun 2015, 02:11
00:00

Difficulty:

35% (medium)

Question Stats:

73% (01:16) correct 27% (01:58) wrong based on 244 sessions

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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. $$\sqrt{k+1}-\sqrt{k-1}$$

Kudos for a correct solution.

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Re: If k > 1, which of the following must be equal to  [#permalink]

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16 Jun 2015, 04:32
1
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  [#permalink]

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16 Jun 2015, 08:13
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.

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

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Re: If k > 1, which of the following must be equal to  [#permalink]

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16 Jun 2015, 09:21
1
1
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.

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Re: If k > 1, which of the following must be equal to  [#permalink]

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16 Jun 2015, 09:57
1
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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If k > 1, which of the following must be equal to  [#permalink]

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18 Jun 2015, 03:13
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  [#permalink]

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18 Jun 2015, 16:36
2
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

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Re: If k > 1, which of the following must be equal to  [#permalink]

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20 Jun 2015, 09:49
1
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}
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Posts: 51121
Re: If k > 1, which of the following must be equal to  [#permalink]

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22 Jun 2015, 06:25
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:

Attachment:

2015-06-22_1824.png [ 60.23 KiB | Viewed 2541 times ]

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Re: If k > 1, which of the following must be equal to  [#permalink]

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04 Oct 2015, 07:51
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  [#permalink]

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05 Oct 2018, 23:44
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