Hi swerve,

The restriction that K > 1 actually has no impact on this question. If you choose K=1, then you'll end up 2/√2 = √2.

There's only one answer that equals √2 when you TEST K=1...

Final Answer: E

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

swerve wrote:If k > 1, which of the following must be equal to
$$\frac{2}{\sqrt{k+1}+\sqrt{k-1}}?$$

$$A.\ 2$$
$$B.\ 2\sqrt{k}$$
$$C.\ 2\sqrt{k+1}+\sqrt{k-1}$$
$$D.\ \frac{\sqrt{k+1}}{\sqrt{k-1}}$$
$$E.\ \sqrt{k+1}-\sqrt{k-1}$$

Recall that the conjugate of an expression replaces the "plus" sign between the terms with a "minus" sign. Thus, the conjugate of √(k + 1) + √(k - 1) is √(k + 1) - √(k - 1). Multiplying both the numerator and the denominator of the given expression by the conjugate of the denominator allows us to rationalize the denominator (i.e., rid the denominator of any radical signs). Thus, we have:
2[√(k + 1) - √(k - 1)]/{[√(k + 1) + √(k - 1)][√(k + 1) - √(k - 1)]}

2[√(k + 1) - √(k - 1)]/[(k + 1) - (k - 1)]

2[√(k + 1) - √(k - 1)]/2

√(k + 1) - √(k - 1)

Answer: E