Question

If x is very large compared to y, then the value of k if $\sqrt{\frac{x}{x+y}}\sqrt{\frac{x}{x-y}}$ = $1+\frac{y^2}{k{x}^2}$

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Answer 1

We have to re-arrange the expression so that we can apply binomial expansion. Since we are given x is large compared to y, we can expect the fraction $\frac{y}{x}$. So we will divide by x or powers of x

$\sqrt{\frac{x}{x+y}}\sqrt{\frac{x}{x-y}}$ = $\sqrt{\frac{x^2}{x^2-y^2}}$

= $\sqrt{\frac{1}{1-\frac{y^2}{x^2}}}$

= $(1-\frac{y^2}{x^2}{)}^{\frac{-1}{2}}$

= $1+\frac{1}{2}\frac{y^2}{x^2}$

(we neglect the higher power)

$\Rightarrow k$ = 2