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 hereSince 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 0.74Check option with k=2

A. 2 > 0.74

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

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

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

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

CORRECTAnswer: option

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